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G/ 


THE  LIBRARY 

OF 

THE  UNIVERSITY 
OF  CALIFORNIA 

GIFT  OF 

Henry  Strickland 


STRENGTH   OF   MATERIALS 


McGraw-Hill  BookCompany 


Electrical  World         The  Engineering  andMining  Journal 
En5ineering  Record  Engineering  News 

Railway  Age  Gazette  American  Machinist 

Signal  E,ngin<?<?r  American  Engineer 

Electric  Railway  Journal  Coal  Age 

Metallurgical  and  Chemical  Engineering  Power 


STRENGTH  OF  MATERIALS 


BY 

JAMES   E.    BOYD,    M.S. 

Professor  of  Mechanics,  The  Ohio  State  University. 


FIRST  EDITION 

FOURTH  IMPRESSION,  CORRECTED 

TOTAL  ISSUE,  7,000 


McGRAW-HILL   BOOK   COMPANY,  INC. 

239   WEST   39TH   STREET,    NEW   YORK 

6  BOUVERIE  STREET,  LONDON,  E.  C. 

1911 


COPYRIGHT,  1911, 

BY 
McGRAW-HILL  BOOK  COMPANY 


First  Printing,  December,  1911 
Second  Printing,  August,  1912 
Third  Printing,  February,  1914 
Fourth  Printing,  February,  1915 


REP.GEN.ua 

ACCESS.  NO. 


,   F, 

GIFT 


Stanbopc 

H.GILSON  COMPANY 
BOSTON,  U.S.A. 


B-r 


PREFACE. 


THIS  book  is  intended  to  give  the  student  a  grasp  of  the  physical 
and  mathematical  ideas  underlying  the  Mechanics  of  Materials, 
together  with  enough  of  the  experimental  facts  and  simple  appli: 
cations  to  sustain  his  interest,  fix  his  theory,  and  prepare  him 
for  the  technical  subjects  as  given  in  works  on  Machine  Design, 
Reinforced  Concrete,  or  Stresses  in  Structures. 

It  is  assumed  that  the  reader  has  completed  the  Integral 
Calculus,  and  has  taken  a  course  in  Theoretical  Mechanics  which 
includes  statics  and  the  moment  of  inertia  of  plane  areas.  Chap- 
ters XVI  and  XVII  give  a  brief  discussion  of  center  of  gravity 
and  moment  of  inertia.  Students  who  have  not  mastered  these 
subjects  should  study  these  chapters  before  taking  up  Chapter  V 
(preferably  before  beginning  Chapter  I). 

The  problems,  which  are  given  with  nearly  every  article,  form 
an  essential  part  of  the  development  of  the  subject.  They  were 
prepared  with  the  twofold  object  of  fixing  the  theory  and  enabling 
the  student  to  discover  for  himself  important  facts  and  applica- 
tions. The  first  problems  of  each  set  usually  require  the  use  of 
but  one  new  principle,  —  the  one  given  in  the  text  which  immedi- 
ately precedes;  the  later  problems  aim  to  combine  this  principle 
with  others  previously  studied  and  with  the  fundamental  opera- 
tions of  Mathematics  and  Mechanics.  The  constants  given  in 
the  data  or  derived  from  the  results  of  the  problems  fall  within 
the  range  of  the  figures  obtained  from  actual  tests  of  materials. 
Many  of  the  problems  are  taken  directly  from  such  measurements. 
Some  of  them  are  from  tests  made  by  the  author  or  his  colleagues 
at  the  Ohio  State  University;  others  are  from  bulletins  of  the 
University  of  Illinois  Engineering  Experiment  Station,  from 
" Tests  of  Metals"  at  the  Watertown  Arsenal,  and  from  the  Trans- 
actions of  the  American  Society  of  Civil  Engineers. 

This  book  is  designed  for  use  with  "  Cambria  Steel,"  to  which 
references  are  made  by  title  instead  of  by  page,  so  that  they  are 
adapted  to  any  edition  of  the  handbook. 


M828229 


vi  PREFACE 

The  author  acknowledges  his  indebtedness  for  suggestions  and 
criticisms  to  Professors  C.  T.  Morris,  E.  F.  Coddington,  Robert 
Meiklejohn,  K.  D.  Swartzel,  and  many  others  of  the  Faculty  of 
the  College  of  Engineering;  and  to  Professor  Horace  Judd  of  the 
Department  of  Mechanical  Engineering  for  the  material  for  several 
of  the  half-tones.  He  also  expresses  his  obligations  to  the  books 
which  have  helped  to  mold  his  ideas  of  the  subject,  —  Johnson's 
" Materials  of  Construction,"  Swing's  "Strength  of  Materials,"  and 
especially  the  textbooks  which  he  has  used  with  his  classes,  —  Mer- 
riman's  "  Mechanics  of  Materials,"  Heller's  "  Stresses  in  Struc- 
tures," and  Goodman's  "Mechanics  Applied  to  Engineering." 

The  symbols  used  in  the  mathematical  expressions  are  much 
the  same  as  in  Heller's  "  Stresses  in  Structures." 

COLUMBUS,  OHIO, 

November  6,  1911. 


CONTENTS. 


PAGE 


PREFACE v 

NOTATION jd 

CHAPTER  I.  —  STRESSES 1 

Strength  of  Materials;  Tension;  Compression;  Stress  —  Total 
Stress;  Unit  Stress;  Working  Stress  —  Allowable  Unit  Stress; 
Deformations  —  Unit  Deformation;  Elastic  Limit;  Modulus  of 
Elasticity;  Physical  Meaning  of  E;  Resilience;  Modulus  of  Re- 
silience; Poisson's  Ratio;  Change  in  Volume  inside  the  Elastic 
Limit. 

CHAPTER  II.  —  STRESS  BEYOND  THE  ELASTIC  LIMIT 15 

Stress-strain  Diagrams;  Elastic  Limit  and  Yield  Point;  Johnson's 
•  Apparent  Elastic  Limit;  Calculation  of  the  Modulus  of  Elasticity; 
Ultimate  Strength  and  Breaking  Strength;  Per  cent  of  Elongation 
and  Reduction  of  Area;  Apparent  and  Actual  Unit  Stress;  Curves 
of  Various  Structural  Materials;  Factor  of  Safety;  Effect  of  Form 
on  the  Ultimate  Strength ;  Effect]of  Stresses  beyond  the  Yield  Point. 

CHAPTER  III.  —  SHEAR 37 

Shear  and  Shearing  Stress;  Shearing  Deformation;  Modulus  of 
Elasticity  in  Shear;  Shear  Caused  by  Tension  or  Compression; 
Shearing  Forces  in] Pairs;  Compressive  and  Tensile  Stress  Caused 
by  Shear;  Methods  of  Failure;  Bearing  Strength  —  Failure  by 
Cutting. 

CHAPTER  IV.  —  RIVETED  JOINTS 54 

Kinds  of  Stress;  Bearing  Stress;  Lap  Joint  with  Single  Row  of 
Rivets;  Butt  Joint;  Rivets  in  More  than  One  Row;  Efficiency  of 
a  Riveted  Joint. 

CHAPTER  V.  —  BEAMS 66 

Definition  of  a  Beam;  Kinds  of  Beams;  Internal  Stresses  in  Beams; 
External  Moment  and  Resisting  Moment;  Experiments  Illustrating 
Moment  and  Shear;  Shear  Diagrams ;  Moment  Diagrams;  Relation 
of  Moment  and  Shear;  Area  of  Shear  Diagram  Equals  Moment. 

vii 


viii  CONTENTS 

PAGE 

CHAPTER  VI.  —  STRESSES  IN  BEAMS 85 

Nature  of  Stresses;  Relation  of  Moment  to  Stress;  Unit  Stress  in 
the  Outer  Fibers;  Location  of  Neutral  Axis;  Section  Modulus; 
Graphic  Representation  of  Stress  Distribution;  Stress  beyond  the 
Elastic  Limit;  Modulus  of  Rupture. 

CHAPTER  VII.  —  DEFLECTION  OF  BEAMS 99 

Deflection  and  Moment;  Deflection  in  Rectangular  Coordinates; 
Beam  Supported  at  Two  Points;  Cantilever  with  Load  on  the  Free 
End;  Cantilever  with  Uniformly  Distributed  Load;  Beam  Sup- 
ported at  the  Ends,  Uniformly  Loaded;  Beam  Supported  at  the 
Ends  with  Concentrated  Load  at  the  Middle;  Beam  Supported  at 
the  Ends,  Load  at  any  Point  between  Supports;  Beam  Supported 
at  the  Ends,  Two  Equal  Loads  Symmetrically  Placed;  Any  Beam 
with  Two  Supports;  Stiffness  of  Beams. 

CHAPTER  VIII.  —  BEAMS  WITH  MORE  THAN  Two  SUPPORTS 119 

Relation  of  Deflection  to  Stress;  Cantilever  Supported  at  One  End; 
Cantilever  Supported  at  One  End,  Concentrated  Load;  General 
Equations  of  Moment  and  Shear;  Point  of  Inflection;  Cantilever 
Fixed  at  Left  End,  Supported  at  Right  End,  Uniformly  Loaded; 
Beam  Fixed  at  Both  Ends,  Uniformly  Loaded;  Beam  Fixed  at 
Both  Ends,  Concentrated  Load  kl  from  the  Left  End;  Theorem  of 
Three  Moments;  Theorem  of  Three  Moments  for  Distributed 
Loads;  Calculation  of  Moments;  Calculation  of  Shear;  Theorem 
of  Three  Moments  for  Single  Concentrated  Load  on  Each  Span; 
Superimposed  Loads;  Moments  in  Different  Planes. 

CHAPTER  IX.  —  SHEAR  IN  BEAMS 139 

Direction  of  Shear;  Intensity  of  Shearing  Stress;  Resultant  of 
Shearing  and  Tensile  Stress;  Maximum  Resultant  Stress;  Resultant 
Stress  in  Beams;  Failure  of  Beams. 

CHAPTER  X.  —  BEAMS  OF  SPECIAL  FORM 153 

Beams  of  Constant  Strength;  Cantilever  with  Load  on  the  End; 
Shearing  and  Bearing  Stresses  at  the  End;  Cantilever  .with  Uni- 
formly Distributed  Load;  Beams  of  Constant  Strength,  Sup- 
ported at  the  Ends;  Deflection  of  Beams  of  Constant  Strength; 
Cantilever  of  Constant  Depth  with  Load  on  End;  Cantilever  of 
Constant  Breadth  with  Load  on  End;  Cast-iron  Beams;  Beams 
of  Two  or  More  Materials;  Reinforced  Concrete  Beams;  Resist- 
ing Moment  in  Reinforced  Concrete;  Resultant  Tensile  Stress. 

CHAPTER    XI.  —  BENDING    COMBINED    WITH    TENSION    OR    COM- 
PRESSION       169 

Transverse  and  Longitudinal  Loading;  Eccentric  Loading;  Maxi- 
mum Eccentricity  without  Reversing  Stress;  Resultant  Load  not 
on  Principal  Axis. 


CONTENTS  ix 

PAGE 

CHAPTER  XII.  —  COLUMNS 180 

Definition;  Column  Theory;  Application  of  Column  Formulas; 
Euler's  Formula;  Effect  of  Ultimate  Strength  of  Material  on 
Strength  of  Columns;  Classification  of  Columns;  End  Conditions 
in  Actual  Columns;  Some  Experiments  Showing  the  Effect  of  End 
Conditions.  * 

CHAPTER  XIII.  —  COLUMN  FORMULAS  USED  BY  ENGINEERS 202 

Straight-line  Formulas;]  Algebraic  Derivation  of  the  Straight-line 
Formulas;  Ultimate  Strength;  Straight-line  Formulas  for  Square 
or  Fixed  Ends;  Rankine's  or  Gordon's  Formulas;  Ritter's  Rational 
Constant;  General  Conclusions;  Failure  of  Beams  Due  to  Flexure 
on  the  Compression  Side;  Failure  Due  to  Buckling  of  the  Web. 

CHAPTER  XIV.  —  TORSION 223 

Torque;  Shearing  Stress  in  a  Shaft;  Relation  of  Torque  to  Angle 
of  Twist;  Relation  of  Torque  to  Work;  Torsion  Combined  with 
Bending  or  Tension;  Helical  Springs. 

CHAPTER  XV.  —  RESILIENCE  IN  BENDING  AND  TORSION 231 

Beams;  Expression  for  Internal  Work;  Cantilever  with  Uniformly 
Distributed  Load;  Beams  with  Variable  Section;  Leaf  Springs; 
Resilience  in  Torsion;  Sections  of  Maximum  Resilience. 

CHAPTER  XVI.  —  CENTER  OF  GRAVITY 239 

Center  of  Gravity;  Determination  of  Center  of  Gravity  by  Balanc- 
ing; Center  of  Gravity  by  Moments;  Center  of  Gravity  of  Con- 
tinuous Bodies;  Center  of  Gravity  of  Plane  Areas. 

CHAPTER  XVII.  —  MOMENT  OF  INERTIA 251 

Definition;  Radius  of  Gyration;  Transfer  of  Axis;  Moment  of 
Inertia  of  Thin  Plate;  Moment  of  Inertia  of  Plane  Area;  Change  of 
Direction  of  Axis;  Product  of  Inertia;  Transformation  of  Axes  for 
Product  of  Inertia;  Direction  of  Axis  for  Maximum  Moment  of 
Inertia;  Moment  of  Inertia  of  a  Prism  or  Pyramid. 

CHAPTER  XVIII.  —  COMPUTATION  WITHOUT  INTEGRALS 268 

Areas  Which  Cannot  be  Integrated;  Approximate  Computation; 
Center  of  Gravity  by  Experiment;  Moment  of  Inertia  by  Experi- 
ment; Moment  of  Inertia  of  a  Plane  Section. 

CHAPTER  XIX.  —  REPEATED  STRESSES 275 

Lag  of  Deformation  behind  Stress;  Watertown  Arsenal  Tests 
of  Eyebars;  Failure  under  Repeated  Stress;  Design  for  Varying 
Stress;  Impact  Stresses. 


NOTATION. 


The  symbols  which  are  frequently  used  in  this  book  are: 

a  =  radius  of  circle;  pitch  of  rivets. 
A  =  area  of  cross  section. 

6  =  breadth;  breadth  of  rectangular  section;  base  of  triangle. 
B  =  some  special  value  of  b. 

C,  Ci,  Cz  =  integration  constants. 

d  =  depth;  height  of  rectangular  section ;  diameter;  distance  between  parallel 

axes. 
D  =  some  special  depth. 

e  =  eccentricity  of  load  on  column. 
E  =  modulus  of  elasticity. 

h  =  height. 
H  =  product  of  inertia. 

7  =  moment  of  inertia. 

Ix  =  moment  of  inertia  with  respect  to  the  X  axis. 

70  =  moment  of  inertia  with  respect  to  an  axis  through  the  center  of  gravity. 

J  =  polar  moment  of  inertia  of  a  plane  area. 

k  =  a  constant  coefficient  (in  Chapters  VII,  VIII). 

k  =  radius  of  gyration  (in  Chapter  XVII) . 
K,  Ki,  K2  =  constants  of  integration. 

I  =  length;  length  of  beam  between  supports;  length  of  column  between 
points  of  counterflexure. 

L  =  length;  total  length  of  column. 
M,  MO  =  moment,  moment  at  origin  of  coordinates. 
M^  Mb,  Mc  =  moment  over  three  consecutive  supports. 
Mi,  M2,  Mz,  etc.  =  moment  over  first,  second,  and  third  supports,  etc. 
Mt  =  torque. 
N  =  normal  force  at  surface. 

P,  PI,  etc.  =  concentrated  loads  or  forces. 

r  =  radius  of  gyration  (in  Chapters  XII,  XIII) . 

r  =  distance  from  origin  (in  Chapters  XVI,  XVII). 
R  =  reaction  at  support;  resultant  force. 
Ri  =  reaction  at  left  support. 
#2  =  reaction  at  right  or  second  support. 

xi 


xii  NOTATION 

s  =  unit  stress. 

st,  sc,  ss  =  unit  tensile,  compressive,  and  shearing  stress. 
su  =  ultimate  unit  stress. 

s'  =  unit  stress  resulting  from  shear  and  compression  or  tension. 
S,  S0  =  total  vertical  shear,  total  shear  at  origin. 

i  =  thickness. 
T  =  tension. 

v  =  distance  from  neutral  axis. 

Vi  =  distance  of  extreme  lower  fibers  from  neutral  axis. 
v2  =  distance  of  extreme  upper  fibers  from  neutral  axis. 
Vz  =  some  particular  value  of  v. 
w  =  distributed  load  per  unit  length. 
W  =  total  load  uniformly  distributed. 
x,  y,  2  =  coordinates  of  center  of  gravity. 

y  =  deflection  of  points  on  beam  or  column. 
#max  =  maximum  deflection  in  a  beam  or  column. 

d  =  unit  deformation. 

p  =  Poisson's  ratio;  density;  radius  of  curvature. 
8,  <f>  =  angles  in  figure. 


STRENGTH  OF  MATERIALS. 


CHAPTER  I. 
STRESSES. 

1.  Strength  of  Materials.  —  That  branch  of  Mechanics  which 
treats  of  the  changes  in  form  and  dimensions  of  elastic  solids  and 
the  forces  which  cause  these  changes  is  called  The  Mechanics  of 
Materials.     When  the  physical  constants  and  the  results  of  ex- 
perimental tests  upon  the  materials  of  construction  are  included 
with  the  theoretical  discussion  of  the  ideal  elastic  solid,  the  entire 
subject  is  called  The  Strength  of  Materials  or  The  Resistance  of 
Materials. 

2.  Tension.  —  Support  one  end  of  a  band  of  soft  rubber,  and 
attach  a  small  hook  to  the  other  end,  as  shown  in  Fig.  1.     Now 
apply  a  small  weight  to  the  hook. 

The  rubber  band  is  stretched;  its 
length  is  increased  by  an  amount 
a,  while  its  cross  section  is  dimin- 
ished. Add  a  second  weight.  If 
the  second  weight  is  equal  to  the 
first  one,  the  elongation  b,  which 
it  causes,  is  equal  to  that  caused 
by  the  first  weight.  Remove  the 
weights,  and  the  rubber  band  re- 
turns to  its  original  length  and 
cross  section. 

If  steel,  iron,  wood,  concrete, 
stone,  or  other  solid  material  is 
used  instead  of  rubber,  the  re- 
sults are  similar.  There  is  this 
apparent  difference:  while  the  rub- 
ber may  be  stretched  to  twice  or 

three  times  its  original  length  and  still  return  to  its  original  size 
and  shape  after  the  load  is  removed,  one  of  the  other  materials 

1 


Fig.  1 .  —  Rubber  Bands  in 
Tension. 


2  STRENGTH  OF  MATERIALS 

may  be  stretched  only  a  very  small  amount  (usually  less  than 
0.002  of  its  length),  without  receiving  a  permanent  change  in  its 
dimensions.  Again,  the  force  required  to  produce  a  relatively 
small  increase  in  the  length  of  a  rod  of  wood  or  steel,  for  instance, 
is  many  times  greater  than  that  necessary  to  double  the  length 
of  a  soft  rubber  band  of  equal  cross  section.  These  differences 
between  the  behavior  of  soft  rubber  and  other  solid  materials 
are  differences  of  degree  and  not  of  kind.  Essentially  they  are 
alike. 

The  rubber  bands  shown  in  Fig.  1  are  subjected  to  the  action 
of  two  forces:  the  force  of  the  weights  pulling  downward,  and  the 
reaction  of  the  support  pulling  upward.  The  bands  are  in  tension. 
A  body  is  said  to  be  in  tension  when  it  is  subjected  to  two  sets  of 
forces  whose  resultants  are  in  the  same  straight  line,  opposite  in 
direction,  and  directed  away  from  each  other. 

3.  Compression.  —  When  a  body  is  subjected  to  two  sets  of 

forces  whose  resultants  are  in  the  same  straight 
line,  opposite  in  direction,  and  directed  toward 
each  other,  it  is  said  to  be  in  compression.  In 
Fig.  2,  the  block  B  is  in  compression  under  the 
action  of  the  50  pounds  pushing  down  and  the 
reaction  of  the  support  pushing  up.  The  effect 
of  compression  upon  a  body  is  to  shorten  it  in 
the  line  of  the  forces  and  increase  its  dimen- 
sions in  the  plane  perpendicular  to  this  line. 

Tension  and  compression  may  be  repre- 
sented as  in  Fig.  3,  where  the  arrows  repre- 
sent the  forces,  and  the  small  rectangles  represent  the  bodies, 
or  portions  of  a  body,  upon  which 
the  forces  act.  The  rectangles  are 
often  omitted;  a  pair  of  arrows  with  Fi  3 

their  heads  together  indicate  compres- 
sion, and  a  pair  with  their  heads  in  the  opposite  sense  indicate 
tension. 

4.  Stress;  Total  Stress.  —  The  force   exerted  by  one   body 
upon  another  at  their  surface  of  contact   is  called  the  stress 
between  the  bodies  or  the  total  stress  between  the  bodies.     If  a 
single  body  be  regarded  as  cut  by  an  imaginary  surface,-  the  force 
exerted  across  this  surface  by  either  portion  of  the  body  upon  the 
other  portion  is  the  total  internal  stress  in  the  body  at  the  section. 


COMPRESSION 


Fig.  4.  — Tensile 
Stress. 


STRESSES  3 

In  the  case  of  an  internal  stress,  if  the  forces  are  such  that  the 

portions  of  the  body  are  pushed  together  at  the  imaginary  surface, 

the  stress  is  compressive.      If  the  forces  tend 

to  pull  the  portions  apart,  the  stress  is  tensile. 

Compressive  stress  at  the  surface  of  contact  of 

two  separate  bodies  is  called  bearing  stress. 
All  parts  of   the  bar  AB,  Fig.  4,  are  under 

tensile  stress.     The  total  tensile  stress  at  any 

section  CD  is  the  load  L  and  the  weight  of  the 

hook  and  of  that  portion  of  the  bar  below  the 

section. 

All  parts  of  the  block  in  Fig.  5  are  in  com- 
pression.     The  total  compressive  stress  at  any 

section  JK  is  10  pounds  plus  the  weight  of  the 

portion  of  the  block  above  the  section;  or,  since 

action  and  reaction  are  equal,  it  is  the  upward  reaction  at  the 

base  minus  the  weight  of  the  portion  below  JK. 

5.   Unit  Stress;  Intensity  of  Stress.  —  The  unit  stress  at  any 

surface  is  the  total  stress  at  the  surface  divided  by  its  area. 
Unit  stress  is  frequently  called  intensity  of 
stress.  In  American  engineering  practice,  unit 
stresses  are  usually  expressed  in  pounds  per 
square  inch.  Compressive  stresses  in  masonry 
are  sometimes  given  in  tons  per  square  foot; 
the  bearing  pressure  of  masonry  upon  soils  is 
always  so  expressed.  English  engineers  fre- 
quently use  long  tons  per  square  inch  to  express 
the  intensity  of  stress  in  steel  and  similar  solids. 
Continental  engineers,  of  course,  use  kilograms 

per  square  centimeter.     Physicists,  the  world  over,  prefer  dynes 

per   square  centimeter.      In  the  case  of  tensile  or  compressive 

stresses,  the  surface  considered  is  a  plane  section  perpendicular 

to  the  direction  of  the  forces,  unless  otherwise  stated. 


V  10  Lbs 

K 

\ 

\ 

J.r™ 


Fi     5  _  Com  res 
sive  Stress. 


PROBLEMS. 

1.  The  rod  AB,  Fig.  4,  is  circular  and  2  inches  in  diameter.     If  the  load  L 
is  5000  pounds,  and  the  weight  of  the  hook  and  the  lower  part  of  the  rod  are 
neglected,  what  is  the  unit  stress  at  any  section? 

Ans.  1592  pounds  per  square  inch. 

2.  If  in  Fig.  4  the  diameter  of  the  rod  is  3  inches,  what  must  be  the  load  L 
to  produce  an  intensity  of  stress  of  3500  pounds  per  square  inch? 


4  STRENGTH  OF  MATERIALS 

3.  If  a  section  of  the  block  in  Fig.  5  is  %  inch  by  f  inch,  what  is  the  unit 
stress  near  the  top?  Ans.  32  pounds  per  square  inch. 

4.  A  6-inch  by  8-inch  oak  post  12  feet  high  supports  a  load  of  6  tons  at  the 
top.    The  oak  weighs  48  pounds  per  cubic  foot.     Find  the  unit  compressive 
stress  near  the  top  and  near  the  bottom. 

Ans.  254  pounds  per  square  inch  near  the  bottom. 

5.  A  concrete  wall  18  inches  thick  and  12  feet  high  carries  a  load  of  3300 
pounds  per  running  foot.     The  concrete  weighs  150  pounds  per  cubic  foot. 
Find  the  bearing  pressure  on  the  soil  immediately  under  the  foot  of  the  wall. 

Ans.  2  tons  per  square  foot. 

6.  A  2-inch  round  steel  rod  is  subjected  to  a  pull  of  100,000  pounds.    Find 
the  unit  stress  in  pounds  per  square  inch,  long  tons  per  square  inch,  kilograms 
per  square  centimeter,  and  dynes  per  square  centimeter. 

The  total  load  upon  a  structure  is  often  given  in  terms  of  the 
volume  of  some  material.  It  is  therefore  advisable  to  learn  the 
approximate  density  of  some  of  the  common  substances.  The 
figures  given  below  will  be  used  in  problems  throughout  this  book 
unless  otherwise  stated,  and  should  be  memorized. 

Material.  Weight  per  cubic 

foot  in  pounds. 

Water 62.5 

Wrought  Iron 480 

Steel 490 

Cast  Iron 450 

Concrete 150 

Brick 125 

Structural  timber 48 

A  steel  rod  one  inch  square  and  one  foot  long  is  calculated  as 
weighing  3.4  pounds.  A  wrought-iron  rod  of  one  square  inch 
cross  section  weighs  10  pounds  per  yard. 

PROBLEMS. 

7.  A  1-inch  round  steel  rod  100  feet  long  hangs  vertical  and  supports  a 
tank  weighing  125  pounds,  which  contains  14  cubic  feet  of  water.     Find  the 
unit  stress  in  the  rod  near  the  top.  Ans.  1613  pounds  per  square  inch. 

8.  A  brick  wall  12  inches  thick,  30  feet  high,  and  15  feet  long  is  supported 
by  a  steel  beam  15  feet  long  and  19  square  inches  cross  section.     The  ends  of 
the  beajn  are  supported  on  two  concrete  columns  each  12  inches  square  and 
10  feet  high.    Find  the  unit  stress  near  the  bottom  of  the  columns. 

9.  A  floor  is  supported  by  long-leaf  yellow-pine  joists  2  inches  by  12  inches 
by  16  feet,  spaced  16  inches  center  to  center.     The  floor  is  inch  oak  over  inch 
hemlock.     The  lath  and  plaster  weighs  14  pounds  per  square  foot.     Find  the 
load  on  each  joist  including  its  own  weight.     (Take  weights  of  timber  from 
tables  of  the  United  States  Department  of  Agriculture,  given  in  Cambria.) 

6.  Working  Stress;  Allowable  Unit  Stress.  —  Working  stresses 
are  the  unit  stresses  to  which  the  materials  of  a  structure  or 


STRESSES  5 

machine  are  subjected.  The  allowable  unit  stress  for  a  given 
material  is  the  maximum  working  stress  which,  in  the  judgment 
of  some  engineer  or  other  authority,  should  be  applied  to  that 
material.  As  examples  of  allowable  unit  stresses,  the  building 
laws  of  New  York  city  (see  Cambria)  and  the  American  Railway 
Engineering  and  Maintenance  of  Way  Association  recommend 
16,000  pounds  per  square  inch  as  the  allowable  unit  tensile  stress 
in  structural  steel.  The  United  States  Department  *)f  Agricul- 
ture gives  1000  pounds  per  square  inch  as  the  allowable  com- 
pressive  stress,  parallel  to  the  grain,  in  long-leaf  yellow  pine,  and 
215  pounds  per  square  inch  across  the  grain. 

PROBLEMS. 

For  allowable  unit  stresses  in  the  following  problems  use  the  values  recom- 
mended by  the  United  States  Department  of  Agriculture  for  timber,  and  those 
of  the  New  York  building  laws  for  other  materials,  unless  otherwise  specified. 

1.  Find  the  total  load  in  compression  parallel  to  the  grain  which  may  be 
applied  to  a  4-inch  by  4-inch  short  block  of  white  oak. 

Ans.  12,800  pounds. 

2.  What  should  be  the  length  of  a  6-inch  by  6-inch  block  of  short-leaf 
yellow  pine  to  support  a  compressive  load  of  12,000  pounds  across  the  grain? 

Ans.  9.3  inches. 

3.  What  should  be  the  thickness  of  an  eye-bar  of  rolled  steel,  5  inches  wide, 
to  support  80,000  pounds  in  tension?  Ans.  1  inch. 

4.  What  is  the  total  tension  which  may  be  safely  applied  to  a  1-inch  wrought- 
iron  bolt?     (See  table  of  Franklin  Institute  standard  dimensions  of  bolts  and 
nuts  in  Cambria.)  Ans.  6600  pounds. 

5.  An  8-inch  by  8-inch  short  oak  post  stands  on  a  cast-iron  base  plate 
which  is  supported  by  a  pier  of  1:2:4  Portland  cement  concrete.     If  the 
post  is  subjected  to  its  allowable  safe  load,  what  should  be  the  minimum  area 
of  the  pier?  Ans.  223  square  inches,  15  inches  square. 

6.  If  the  pier  in  Problem  5  is  16  inches  square  at  the  top  and  enlarges  to 
form  a  frustum  of  a  pyramid  3  feet  in  height,  what  must  be  the  dimensions 
of  the  base,  if  the  bearing  pressure  on  the  soil  shall  not  exceed  2  tons  per 
square  foot? 

7.  Deformations ;  Unit  Deformation.  —  The  changes  in  di- 
mensions which  occur  when  forces  are  applied  to  a  body  are  called 
deformations.     In  Fig.  1,  the  increase  of  length,  a,  which  takes 
place  when  the  first  load  is  applied  is  the  deformation  due  to  that 
load,  the  increase  b  is  the  deformation  due  to  the  second  load, 
and  a  +  b  is  the  deformation  due  to  the  two  loads.     The  def- 
ormation produced  by  a  tensile  force  or  pull  is  an  elongation; 
that  caused  by  a  compressive  force  or  push  is  a  compression.     A 


6  STRENGTH  OF  MATERIALS 

deformation  which  remains  after  the  force  is  removed  is  called 
a  set. 

Unit  deformation  in  a  body  is  the  deformation  per  unit  length. 
It  is  calculated  by  dividing  the  total  deformation  in  a  given 
length  by  the  original  length.  The  length  a  in  Fig.  1  divided  by 
the  original  length  of  the  band  is  the  unit  elongation  due  to  the 
first  load.  It  is  frequently  convenient  to  consider  unit  deforma- 
tion as  the*  ratio  of  the  deformation  to  the  original  length.  It  is 
then  called  the  relative  deformation. 

In  algebraic  equations  many  authors  represent  unit  deforma- 
tion by  the  letter  5  (pronounced  delta). 


PROBLEMS. 

1.  A  rod  is  subjected  to  a  tensile  stress  and  a  portion  of  the  rod,  originally 
8  inches  long,  is  stretched  0.0040  inch.     Find  the  unit  elongation. 

Ans.  0.00050  inch. 

2.  A  block  4  inches  long  is  compressed  0.0028  inch.     What  is  the  unit  com- 
pression? 

3.  A  change  of  temperature  of  20°  F.  causes  a  relative  elongation  of  0.00013 
in  an  iron  rod.     What  is  the  total  elongation  in  a  length  of  30  feet? 

Ans.  0.0468  inch. 

4.  In  the  tension  test  of  a  bar  of  cast  iron,  it  was  found  that  a  pull  of 
7000  pounds  per  square  inch  produced  an  elongation  of  0.0044  inch  in  a  length 
of  8  inches.     What  was  the  relative  elongation?  Ans.  0.00055. 


8.  Elastic  Limit.  —  When  a  stress  is  applied  to  a  solid  body 
and  then  removed,  the  body  returns  to  its  original  size  and  shape, 
provided  the  stress  has  not  exceeded  a  certain  limit.  If  the 
stress  has  gone  beyond  this  limit,  the  body  does  not  return 
entirely  to  its  original  dimensions,  but  retains  some  permanent 
deformation  or  set.  This  limit  is  called  the  elastic  limit  of  the 
material.  A  wrought-iron  rod  in  tension  is  stretched  about 
0.006  inch  in  a  length  of  8  inches  by  a  load  of  20,000  pounds  per 
square  inch.  When  the  load  is  removed,  it  returns  to  its  original 
length.  A  stress  of  20,000  pounds  per  square  inch,  or  the  corre- 
sponding unit  elongation  of  0.00075  inch,  is  below  the  elastic 
limit  of  wrought  iron.  If  the  load  is  increased  to  30,000  pounds 
per  square  inch,  the  elongation  in  8  inches  becomes,  perhaps, 
0.075  inch.  Upon  the  removal  of  the  load  the  rod  shortens  only 
0.009  inch  and  the  residual  0.066  inch  remains  as  a  permanent 
set.  The  elastic  limit  is  below  30,000  pounds  per  square  inch. 


STRESSES  7 

9.  Modulus  of  Elasticity.  —  For  all  stresses  below  the  elastic 
limit  the  unit  stress  bears  a  constant  ratio  to  the  unit  defor- 
mation. The  quotient  obtained  by  dividing  unit  stress  by  the 
accompanying  unit  deformation  is  called  the  modulus  of  elasticity 
of  the  material,  or  Young's  modulus.  In  algebraic  formulas, 
modulus  of  elasticity  is  represented  by  the  letter  E.  Writing 
the  above  definition  algebraically 

#  =  f,  *  Formula  I. 

o 

where  E  is  the  modulus  of  elasticity, 

s  is  the  unit  stress, 
d  is  the  unit  deformation. 

PROBLEMS. 

1.  A  steel  rod  of  1  square  inch  cross  section  is  tested  in  tension.    It  is 
found  that  a  pull  of  23,600  pounds  stretches  8  inches  of  the  rod  0.0064  inch. 
Find  the  unit  deformation  and  the  modulus  of  elasticity. 

Ans.  Modulus  of  elasticity,  29,500,000  pounds  per  square  inch. 

2.  A  wooden  block  2  inches  square  and  12  inches  long  is  tested  in  com- 
pression.    It  is  found  that  a  total  load  of  2800  pounds  shortens  10  inches  of 
the  block  0.0050  inch.     What  is  the  modulus  of  elasticity  of  this  wood? 

Ans.  1,400,000  pounds  per  square  inch. 

3.  A  steel  bar  1  inch  by  5  inches  is  elongated  0.024  inch  in  a  length  of  5  feet 
by  a  certain  load.     If  the  modulus  of  elasticity  of  this  steel  is  30,000,000 
pounds  per  square  inch,  what  is  the  unit  stress  and  the  total  load? 

Ans.  Total  load,  60,000  pounds. 

4.  In  a  tension  test  of  cast  iron  at  the  Watertown  Arsenal,  an  increase  of 
unit  stress  from  1000  pounds  per  square  inch  to  6000  pounds  per  square  inch 
produced  an  increase  in  length  of  0.0034  inch  in  a  gauged  length  of  10  inches. 
Find  E  for  this  cast  iron.  Ans.  14,700,000  pounds  per  square  inch. 

5.  A  2-inch  round  steel  rod  is  cooled  from  80°  F.  to  30°  F.  without  being 
allowed  to  contract.     If  the  coefficient  of  expansion  of  the  steel  is  0.0000067  per 
degree  Fahrenheit,  and  the  modulus  of  elasticity  is  29,000,000,  what  is  the 
total  tension  developed?  Ans.  30,520  pounds. 

6.  A  cast-iron  bar  2  inches  wide  and  \  inch  thick  is  placed  between  two 
steel  bars  each  2  inches  wide  by  \  inch  thick.     If  the  modulus  of  elasticity  of 
the  steel  is  30,000,000  and  that  of  the  cast  iron  15,000,000,  what  total  pull 
will  stretch  the  combined  bar  0.0044  inch  in  a  length  of  8  inches,  and  what  will 
be  the  unit  stress  in  each  material?  Ans.  Total  pull,  24,750  pounds. 

7.  A  steel  bar  4  inches  wide  and  1  inch  thick  is  placed  between  two  wrought- 
iron  bars  each  4  inches  wide  and  f  inch  thick,  and  a  total  pull  of  94,000  pounds 
is  applied  to  the  combination.     If  E  for  the  steel  is  30,000,000,  and  for  the 
iron,  27,000,000,  what  is  the  unit  tensile  stress  in  each? 

Ans.  10,000  pounds  per  square  inch  in  the  steel; 
9,000  pounds  per  square  inch  in  the  iron. 

*  Important  formulas,  which  should  be  memorized,  are  designated  by  the 
Roman  numerals  in  this  book. 


8  STRENGTH  OF  MATERIALS 

'  8.  A  1-inch  round  steel  rod  passes  through  a  wrought-iron  pipe  1  inch 
inside  diameter,  1.4  inches  outside  diameter,  and  12  inches  long.  A  nut  on  the 
rod  is  turned  so  as  to  produce  tension  in  the  rod  and  compression  in  the  pipe. 
Neglecting  the  elasticity  of  the  nut  and  of  the  projecting  part  of  the  rod,  how 
much  is  the  unit  stress  in  each  increased  when  the  nut  is  turned  30  degrees, 
there  being  eight  threads  to  the  inch,  and  E  the  same  as  in  Problem  7? 

10.   Physical  Meaning  of  E.  —  Formula  I  of  Article  9  may  be 
written 


If  s  be  made  equal  to  unity,  5  becomes  equal  to-^.     With  the 

£j 

common  engineering  units,  the  reciprocal  of  E  is  the  unit  defor- 
mation produced  by  a  unit  load  of  one  pound  per  square  inch. 
For  steel  having  a  modulus  of  30,000,000  pounds  per  square  inch, 
a  unit  stress  of  one  pound  per  square  inch  is  developed  when  the 
deformation  is  one  thirty-millionth  of  the  original  length. 

EXAMPLES. 

Solve  without  writing. 

1.  If  wood  having  a  modulus  of  1,200,000  pounds  per  square  inch  is  sub- 
jected to  a  tensile  stress  of  600  pounds  per  square  inch,  what  is  its  elongation 
per  inch  of  length?     What  is  the  total  elongation  in  a  length  of  5  feet? 

2.  A  4-inch  by  4-inch  wooden  block  is  subjected  to  a  compressive  force  of 
6400  pounds.     If  the  modulus  of  elasticity  parallel  to  the  fibers  is  1,600,000 
pounds  per  square  inch,-  what  is  the  unit  compression  and  the  total  com- 
pression in  a  length  of  20  inches? 

3.  If  steel  has  a  modulus  of  30,000,000,  what  is  the  unit  elongation  due  to 
a  load  of  15,000  pounds  per  square  inch? 

Formula  I  may  also  be  written 

s  =  Ed, 

which  defines  E  as  the  coefficient  which  multiplied  into  the  unit 
deformation  gives  the  unit  stress.  It  helps  to  fix  our  ideas  if  we 
consider  the  case  where  the  unit  deformation  is  0.001.  We  may 
then  define  the  modulus  as  1000  times  the  unit  stress  which  pro- 
duces a  unit  deformation  of  0.001  of  the  original  length. 

EXAMPLES. 

Solve  without  writing. 

4.  If  the  modulus  of  steel  is  30,000,000,  what  is  the  unit  stress  when  the 
unit  deformation  is  0.001?     If  the  unit  deformation  is  0.0005,  what  is  the  unit 
stress?    If  a  steel  rod  40  inches  long  is  stretched  0.008  inch,  what  is  the  unit 


STRESSES  9 

stress?  What  total  pull  will  stretch  a  bar  2  inches  square  0.024  inch  in  a  length 
of  5  feet? 

5.  If  the  modulus  of  white  oak  is  1,500,000,  what  is  the  unit  stress  which 
produces  a  unit  elongation  of  0.001?  Is  this  more  or  less  than  the  allowable 
unit  stress?  . 

If,  in  Formula  I,  5  be  made  unity,  s  becomes  equal  to  E.  From  this,  the 
modulus  of  elasticity  may  be  defined  as  the  unit  stress  which  would  produce  a 
deformation  equal  to  the  original  length,  if  such  deformation  were  possible 
without  breaking  the  material  or  exceeding  the  elastic  limit.  This  means  the 
unit  stress  which  would  double  the  length  of  a  bar  in  tension  or  reduce  to 
nothing  the  length  of  a  block  in  compression. 

ii.  Resilience.  —  When  force  acts  on  a  body  and  motion 
takes  place  in  the  direction  of  the  force,  the  force  does  work  on 
the  body.  The  amount  of  work  is  measured  by  the  product  of 
the  force  multiplied  by  the  distance  its  point  of  application  moves 
along  its  line  of  action.  When  a  body  is  deformed  by  a  force,  the 
amount  of  work  is  the  product  of  the  average  force  multiplied 
by  the  deformation.  When  the  stress  is  kept  below  the  elastic 
limit,  the  average  force  is  the  mean  of  the  initial  and  final  forces. 

PROBLEMS. 

1.  A  load  of  6000  pounds  is  applied  to  a  steel  rod  having  no  initial  load 
and  causes  an  elongation  of  0.024  inch.     What  is  the  work  in  foot  pounds? 

Ans.  6  foot  pounds. 

2.  An  additional  load  of  9000  pounds  is  applied  to  the  rod  of  Problem  1, 
producing  an  additional  elongation  of  0.036  inch.     What  is  the  additional 
work  done  on  the  rod  (a)  by  the  9000  pounds  alone?  (6)  by  the  9000  pounds 
and  the  6000  pounds  together?     (c)  Compare  the  result  of  (fo)  with  the  work 
done  if  the  entire  15,000  pounds  is  applied  at  once  to  the  rod  with  no  initial 
load. 

Ans.  (a)  13.5  foot  pounds;  (6)  31.5  foot  pounds;  (c)  37.5  foot  pounds. 

3.  A  load  of  60,000  pounds  in  tension  is  applied  to  a  steel  bar  of  4  square 
inches  cross  section.     If  the  modulus  of  the  steel  is  30,000,000,  how  much 
work  is  done  on  a  length  of  10  feet  of  the  bar?  Ans.  150  foot  pounds. 

4.  A  steel  rod  of  6  square  inches  cross  section  is  stretched  0.072  inch  in  a 
length  of  9  feet.     If  the  modulus  is  30,000,000,  what  is  the  total  load,  the 
average  force,  and  the  total  work  on  the  9-foot  length? 

Ans.  360  foot  pounds  of  work. 

The  work  done  in  deforming  an  elastic  body  is  stored  up  in 
the  body  as  elastic  energy,  which  may  be  given  up  in  restoring 
the  body  to  its  original  form  when  the  load  is  removed.  If  the 
stress  does  not  exceed  the  elastic  limit,  all  this  work  is  returned. 
If  the  stress  exceeds  the  elastic  limit,  some  of  the  work  is  con- 
verted into  heat  as  the  deformation  takes  place  and  cannot  be 
recovered  as  mechanical  energy. 


10  STRENGTH  OF  MATERIALS 

12.  Modulus  of  Resilience.  —  The  work  expended  in  deform- 
ing unit  volume  of  any  material  to  the  elastic  limit  is  called  the 
modulus  of  resilience  of  the  material.  It  is  the  elastic  potential 
energy  of  unit  volume  when  stressed  to  the  elastic  limit.  The 
modulus  of  resilience  is  a  measure  of  the  amount  of  energy  which 
may  be  stored  in  a  given  material  and  recovered  as  mechanical 
work  without  loss. 

If  we  consider  a  cubic  inch  of  material  subjected  to  unit  stress 

o  o 

s,  the  deformation  is  -=  and  the  average  force  is  ~  ;  the  total  work 

£j  - 

is  the  product 

I  X  1  =  A"  Formula  II. 

Cs2  \ 
energy  in  unit  volume  =  ~-^  J  gives  the  energy 

for  any  value  of  s  below  the  elastic  limit.  When  s  is  the  unit 
stress  at  the  elastic  limit,  the  expression  is  the  modulus  of  resil- 
ience. When  s  and  E  are  given  in  pounds  per  square  inch, 
Formula  II  gives  the  energy  in  inch  pounds  per  cubic  inch. 

The  total  elastic  energy  in  a  body,  all  parts  of  which  are  sub- 
jected to  a  unit  stress  s,  is  obtained  by  multiplying  the  total 
volume  of  the  body  by  the  energy  per  unit  volume,  and  is  inde- 
pendent of  the  form  of  body. 


PROBLEMS. 

1.  Find  the  modulus  of  resilience  of  structural  steel  having  a  modulus 
of  elasticity  of  29,000,000  and  an  elastic  limit  of  30,000  pounds  per  square 
inch.  Ans.  15.5  inch  pounds  per  cubic  inch. 

2.  Find  the  modulus  of  resilience  of  spring  steel   for  which  E  equals 
30,000,000  and  the  elastic  limit  is  100,000  pounds  per  square  inch. 

Ans.  166.7  inch  pounds. 

3.  A  2-inch  round  steel  rod  is  subjected  to  a  pull  of  100,000  pounds,  which 
produces  an  elongation  of  0.012  inch  in  a  12-inch  length.     What  is  the  total 
work  expended  on  the  12-inch  length?     What  is  the  work  per  cubic  inch? 

4.  What  is  the  modulus  of  resilience  of  wood  having  a  modulus  of  elas- 
ticity of  1,200,000  and  an  elastic  limit  of  3000  pounds  per  square  inch? 

5.  How  many  cubic  inches  of  steel  having  an  elastic  limit  of  80,000  pounds 
and  a  modulus  of  elasticity  of  30,000,000  are  required  to  store  100  foot  pounds 
of  energy?  Ans.  11.25  cubic  inches. 

6.  How  high  can  the  energy  which  may  be  stored  in  steel  as  used  in 
Problem  5  lift  its  own  weight?  Ans.  31.37  feet. 


STRESSES  11 

In  calculating  the  work  of  resilience,  we  used  the  average  force 
multiplied  by  the  deformation.  We  may  obtain  the  same  results 
by  means  of  the  Calculus. 

Let  x  represent  the  total  elongation  of  a  rod  of  length  I  and 
unit  cross  section;  and  let  dx  represent  an  infinitesimal  increment 
of  this  elongation.  When  the  elongation  is  x  the  unit  elongation 

.    x       ,  ,,          .,  .    Ex 

is  j  and  the  unit  stress  is  -y-  • 

The  work  done  in  causing  an  elongation  dx  in  the  rod  of  unit 
cross  section  is  the  product  of  this  unit  stress  multiplied  by  dx. 

7TT 

Increment  of  work  =  -y-  dx.  (1) 


Total  work  =  dx=  fl*»       =       (z*-**),  (2) 

where  x\  and  x2  are  the  initial  and  final  elongations  respectively. 
Substituting  for  x\  and  x2  their  values  in  terms  of  the  stress,  we 
get: 


Total  work  =  l-l   =      -          X  volume.  (3) 

\    2  tL  i     \    2  hj  / 

If  the  initial  stress  is  zero,  equation  (3)  becomes  Formula  II. 


PROBLEMS. 

7.  Derive  equation  (2)  by  means  of  average  force  without  integrating. 

8.  Derive  the  expression  for  total  work  and  work  per  unit  volume  in  a 
bar  of  length  I,  cross  section  A,  and  modulus  E,  when  the  total  load  changes 
from  PI  to  P2,  and  show  that  the  final  expression  for  total  work  is  the  same 
as  equation  (3)  above. 

13.  Poisson's  Ratio.  —  When  a  body  is  subjected  to  a  tensile 
stress  it  is  elongated,  the  amount  of  elongation,  provided  the 
unit  stress  does  not  exceed  the  elastic  limit,  being  proportional 
to  the  stress.  At  the  same  time  its  diameter  is  diminished.  The 
ratio  of  this  relative  decrease  in  diameter  to  the  unit  increase  in 
length  is  called  Poisson's  ratio.  The  value  of  this  ratio  varies 
with  the  material,  but  it  is  usually  in  the  neighborhood  of  j.  It 
is  about  0.27  for  steel.  If  a  steel  rod  is  elongated  0.001  of  its 
length,  its  diameter  is  diminished  about  0.00027  of  its  original 
value.  The  same  relation  holds  in  compression. 


12  STRENGTH  OF  MATERIALS 


PROBLEMS. 

1.  Taking  Poisson's  ratio  as  0.27  and  the  modulus  of  elasticity  as  30,000,000, 
find  the  decrease  in  diameter  of  a  2-inch  round  steel  rod  under  a  pull  of  100,000 
pounds.  ' 

2.  In  Problem  1,  if  the  unit  stress  is  proportional  to  the  unit  deformation, 
what  is  the  transverse  unit  compressive  stress? 

Ans.  8594  pounds  per  square  inch. 

3.  A  block  of  metal  2  inches  square  and  10  inches  long  is  subjected  to  a 
compressive  stress  parallel  to  its  length  which  makes  the  unit  deformation 
0.001.     If  Poisson's  ratio  is  |,  how  much  is  its  volume  diminished? 

Ans.  0.02  cubic  inch  nearly. 

4.  A  block  of  metal,  in  the  form  of  a  rectangular  parallelepiped  whose 
edges  correspond  with  the  axes  of  Cartesian  coordinates,  is  subjected  to  a 
compressive  stress  of  4800  pounds  per  square  inch  along  the  X  axis.     If  the 
modulus  of  elasticity  in  both  compression  and  tension  is  the  same  in  all  direc- 
tions, what  is  the  unit  stress  and  unit  deformation  in  each  of  the  principal 
directions  when  E  equals  30,000,000  and  Poisson's  ratio  is  j? 

Axis.  Unit  Deformation.  Unit  Stress. 

Ans.     X  0.00016  compression  4800  Ibs.  per  sq.  in. 

Y  0.00004  tension  1200  Ibs.  per  sq.  in. 

Z  0.00004  tension  1200  Ibs.  per  sq.  in. 

5.  Solve  Problem  4  if  the  applied  stress  is  6000  pounds  per  square  inch 
compression  along  the  X  axis  and  4800  pounds  per  square  inch  tension  along 
the  Y  axis. 

Axis.  Unit  Deformation. 

Ans.    X  0.00024  compression. 

F  0.00021  tension. 

Z  0.00001  tension. 

14.  Change  in  Volume  inside  the  Elastic  Limit.  —  If  a  body 
of  unit  dimensions  is  elongated  an  amount  d  due  to  an  external 
pull,  its  length  becomes  1+5,  and  its  transverse  dimensions 
become  1  —  p5,  where  p*  is  Poisson's  ratio.  Its  area  of  cross 
section  becomes  (1  —  p5)2  =  1  —  2  p5  +(p6)2.  Since  (pS)  is  small, 
being  never  greater  than  0.001,  (p5)2,  being  more  than  a  thou- 
sand times  smaller,  may  be  neglected  without  appreciable  error. 

The  volume  is  (1  +  6)  (1  -  2  P5)  =  1  +  (1  -  2  p)  8  -  2  pS2. 

The  last  term,  2  p52,  may  also  be  neglected. 

Final  volume  =  1  +  (1  -  2p)  5; 

Original  volume  =  1; 

Increment  of  volume  =  (1  —  2p)  5. 

PROBLEMS. 

1.  If  Poisson's  ratio  is  J,  show  that  the  relative  increase  in  volume  when 
a  tension  is  applied  is  one-half  as  great  as  the  relative  increase  in  length. 

*  Greek  letter  p,  pronounced  rho. 


STRESSES  .  13 

2.  In  the  case  of  compression,  find  the  ratio  of  the  increment  of  volume 
to  the  increment  of  length. 

3.  A  steel  bar  of  6  square  inches  cross  section  is  subjected  to  a  pull  of 
90,000  pounds.     If  Poisson's  ratio  is  0.27  and  E  is  30,000,000,  what  is  the 
increase  in  volume  of  a  10-inch  length? 

Solve  by  means  of  the  formula  above,  and  also  by  multiplying  together 
the  dimensions  without  omitting  any  figures,  and  compare  the  results. 

4.  Show  that  a  body  for  which  Poisson's  ratio  is  \  has  its  volume  un- 
changed by  a  direct  stress. 

Note  that  the  discussion  of  Articles  13  and  14  applies  to  stresses 
and  deformations  below  the  elastic  limit,  for  which  all  deforma- 
tion is  temporary.  For  stresses  beyond  the  elastic  limit,  produc- 
ing permanent  deformation  and  rearrangement  of  the  molecules 
of  the  body,  the  conditions  are  somewhat  different. 

MISCELLANEOUS  PROBLEMS. 

1.  A  stick  of   Douglas  fir  tested  in   tension  at   the  Watertown  Arsenal 
("  Tests  of  Metals,"  1896,  page  405)  showed  an  elongation  of  0.0427  inch  in  a 
gauged  length  of  200  inches  when  the  load  per  square  inch  changed  from  100 
pounds  to  500  pounds.     Find  E.  Ans.  1,874,000  pounds  per  square  inch. 

2.  A  second  stick  of   Douglas  fir  tested  in  tension  (1896,  pages   407-9) 
showed  an  elongation  of  0.1015  inch  in  a  gauged  length  of  200  inches,  and  a 
decrease  of  width  of  0.0020  inch  in  a  width  of  12  inches  when  the  load  changed 
from  100  pounds  to  1000  pounds  per  square  inch."     Find  the  modulus  of  elas- 
ticity in  tension  parallel  to  the  grain  and  Poisson's  ratio. 

Ans.  Poisson's  ratio,  0.33. 

3.  In  a  compressive  piece  cut  from  the  stick  of  Problem  2,  when  the  com- 
pressive  stress  changed  from  100  pounds  to  1000  pounds  per  square  inch  there 
was  a  compression  of  0.0230  inch  in  a  gauged  length  of  50  inches.     Find  Ec. 

4.  A  white-oak  stick  11.98  inches  by  9.95  inches  tested  in  compression 
(1896,  page  425)  was  shortened  0.0140  inch  in  a  gauged  length  of  50  inches 
when  the  load  was  increased  from  11,920  pounds  to  71,520  pounds.     Find  E. 

5.  A  block  of  the  same  oak  used  in  Problem  4  was  tested  in  compression 
across  the  grain.     When  the  unit  stress  changed  from  20  pounds  per  square 
inch  to  320  pounds  per  square  inch  the  compression  in  a  gauged  length  of 
6  inches  was  0.0091  inch.     Find  the  modulus  of  elasticity  of  oak  across  the 
grain.  Ans.  198,000  pounds  per  square  inch. 

6.  Two  blocks  of  Douglas  fir  were  tested  in  compression  across  the  grain. 
In  the  first  block  the  compression  was  normal  to  the  growth  rings,  and  the 
compression  in  a  gauged  length  of  6  inches  when  the  unit  load  changed  from 
20  pounds  to  300  pounds  was  0.0081  inch.     In  the  second  block  the  compres- 
sion was  tangent  to  the  growth  rings,  and  the  compression  in  6  inches  with 
the  same  change  of  load  was  0.0195  inch.     Find  E  for  each  case.     ("Tests  of 
Metals,"  1896,  pages  396-7.) 

7.  A  steel  column  100  feet  long  and  of  uniform  cross  section  stands  in  a 
vertical  position.     If  the  modulus  of  elasticity  is  30,000,000,  how  much  is  it 
shortened  by  its  own  weight?     Solve  by  Calculus  and  check  by  average  force. 

Ans.  0.0068  inch. 


14  .  STRENGTH  OF  MATERIALS 

8.  A  round  steel  rod  tapers  gradually  from  2  inches  diameter  to  1  inch 
diameter  in  a  length  of  10  inches.     If  E  is  30,000,000,  and  if  we  assume  that 
the  unit  stress  in  any  transverse  section  is  uniform  throughout  the  section, 
calculate  by  means  of  Integral  Calculus  the  elongation  in  this  10-inch  length 
due  to  a  pull  of  20,000  pounds.  Ans.  0.00424  inch. 

9.  A  plate  of  uniform  thickness  t  has  a  breadth  6  at  one  end  of  a  given 
length  I  and  a  breadth  c  at  the  other  end.     Find  the  expression  for  the  elonga- 

Pl  c 

tion  of  this  length  I  due  to  a  pull  P.  Ans.     ^—  —  —  r-  log  r  • 

10.  In  Problem  9,  E  is  30,000,000;  P,  40,000  pounds;  t  is  1  inch;  6  is  2  inches; 
c  is  3  inches;  Ms  12  inches.     Find  elongation  and  check  approximately  by 
comparing  with  a  bar  of  uniform  breadth  2.5  inches. 

The  ratio  of  unit  stress  which  is  equal  in  all  directions  to  the  unit  volume 
deformation  which  it  causes  is  called  the  modulus  of  volume  elasticity.  A  solid 
submerged  in  a  liquid  is  under  stress  of  this  kind. 

If  Poisson's  ratio  is  known,  the  modulus  of  volume  elasticity  may  be  com- 
puted from  the  modulus  in  tension  or  compression.  When  a  solid  is  subjected 
to  a  stress  in  all  directions,  the  unit  deformation  in  any  direction  is  the  sum 
of  the  deformation  in  that  direction  due  to  the  force  in  the  same  direction  plus 
the  deformation  due  to  the  force  in  each  of  the  other  principal  directions.  If 
5  is  the  deformation  along  the  X  axis  due  to  a  force  in  the  direction  of  that 
axis,  an  equal  force  along  the  Y  axis  produces  a  deformation  p5,  and  a  simi- 
lar force  along  the  Z  axis  has  the  same  effect.  When  the  force  is  compressive 
and  equal  in  all  directions,  the  unit  deformation  along  any  axis  —  5  -J-  pd  -\-  pS  = 
—  5  (1  —  2  p).  A  unit  cube  has  each  dimension  changed  from  unity  to 
1  —  5  (1  —  2  p).  The  volume  of  this  cube  becomes 

V  =  [1  -  5  (1  -  2p)]3  =  1  -  3  5  (1  -  2p),  etc. 

The  increment  of  volume  is  —  36(1  —  2  p). 

If  E  is  the  modulus  of  elasticity  in  tension  or  compression  when  the  force 
is  applied  along  only  one  direction, 


where  s  is  the  unit  pressure. 

The  increment  of  volume,  or  unit  volume  deformation  (since  the  original 
volume  was  unity),  becomes 


Dividing  the  unit  stress  by  the  unit  volume  deformation, 

E    -  E 

~3(1-2P)' 

where  Ev  is  the  modulus  of  volume  elasticity. 

1  27^ 

If  Poisson's  ratio  is  ^  ,  Ev  =  -5-  • 

PROBLEMS. 

1  1  .  Find  the  modulus  of  volume  elasticity  for  steel  for  which  E  =  29,000,000 
and  Poisson's  ratio  is  0.27.  .4ns.  21,000,000  nearly. 

12.  If  the  modulus  of  volume  elasticity  is  18,000,000  and  the  modulus  in 
compression  is  25,000,000,  find  Poisson's  ratio. 


CHAPTER  II. 
STRESS  BEYOND   THE   ELASTIC   LIMIT. 

15.  Stress-strain  Diagrams.  —  In  the  preceding  chapter  we 
have  considered  only  stresses  below  the  elastic  limit.  Within 
this  limit  the  unit  stress  is  proportional  to  the  unit  deformation 
and  Formula  I  holds  good.  Stresses  below  the  elastic  limit  are 
the  most  important  from  the  standpoint  of  the  engineer,  for  in 
well-designed  structures  the  unit  stress  seldom  exceeds  one-half 
this  limit.  It  is  desirable,  however,  to  know  what  takes  place 
above  the  elastic  limit  and  the  character  of  the  final  failure  of 
the  material.  To  secure  this  information,  tests  are  made  in 
which  a  series  of  loads  are  applied  to  a  piece  of  the  material  in 
question,  and  the  corresponding  deformations  are  observed  with 
suitable  measuring  apparatus.  Table  I  gives  a  part  *  of  the  re- 
sults of  a  tension  test  of  a  rod  of  f  machine  steel.  The  rod  was 
originally  20  inches  long  and  turned  to  a  diameter  of  1.31  inches. 
About  9  inches  of  the  rod  at  the  middle  was  turned  down  further 
to  a  diameter  of  1.115  inches.  A  length  of  8  inches  in  this  middle 
portion  was  taken  as  the  gauged  length  from  which  to  measure 
elongations.  The  rod  I  on  the  right  in  Fig.  6  (photographed 
from  a  rod  exactly  like  the  one  tested)  shows  the  original  form 
of  this  test  piece.  The  elongations  in  this  gauged  length  were 
measured  by  an  extensometer  reading  to  0.0001  inch  (see  John- 
son's "  Materials  of  Construction,"  Fig.  271).  As  there  are  two 
micrometers  in  this  extensometer,  we  are  warranted  in  giving 
the  gauge  readings  to  0.5  of  a  division.  When  the  load  reached 
78,000  pounds  per  square  inch,  the  extensometer  was  removed 
and  the  elongations  taken  with  an  ordinary  steel  scale  reading 
in  hundredths  of  an  inch.  After  fracture  the  rod  was  taken  from 

*  Readings  were  taken  at  2000-pound  intervals  from  56,000  to  76,000 
pounds  per  square  inch,  and  were  used  in  locating  the  curve  of  Fig.  6.  Read- 
ings were  also  taken  at  2000-pound  intervals  between  30,000  and  40,000 
pounds  per  square  inch,  as  it  was  suspected  that  the  yield  point  might  fall 
between  these  limits. 

t  An  analysis  of  this  steel,  made  by  Prof.  D.  J.  Demorest,  gave:  carbon, 
0.42  of  1  per  cent;  manganese,  0.71  of  1  per  cent.  The  rod  was  turned  from 
a  bar  of  hot-rolled  steel. 

15 


16 


STRENGTH  OF  MATERIALS 


the  machine,  the  two  portions  placed  together  as  shown  in  Fig.  6, 
II,  and  the  final  elongation  of  1.99  inches  measured.  Loads  were 
applied  and  measured  by  means  of  a  100,000-pound  Olsen  testing 
machine  (see  Johnson's  " Materials  of  Construction,"  Fig.  256). 


N 


II  I 

Fig.  6.  —  Steel  Rod  Tested  in  Tension. 

In  order  to  present  the  results  of  such  a  test  visually,  it  is  con- 
venient to  use  the  unit  stress  and  the  unit  elongation  as  the  co- 
ordinates in  a  curve  called  the  stress-strain  *  diagram,  or  simply 
stress  diagram. 

*  This  is  using  the  word  "strain"  in  its  correct  sense  as  a  synonym  for 
deformation.  A  permanent  deformation  or  set  is  frequently  designated  as  a 
strain.  The  term  "strain"  is  frequently  heard  where  stress  is  meant.  This 
incorrect  use  of  the  word  should  be  avoided, 


STRESS  BEYOND   THE  ELASTIC  LIMIT 


17 


TABLE   I. 
TENSION  TEST  OF  MACHINE   STEEL. 

Diameter,  1.115  inches;  area  of  section,  0.976  square  inch;  gauged  length, 

8  inches. 


Applied  load. 

Elongation. 

Total. 

Per  square  inch. 

In  gauged  length. 

Per  inch  length. 

Pounds. 

Pounds. 

Inch. 

Inch. 

0 

0 

0 

0 

2,928 

3,000 

.00085 

.00011 

4,880 

5,000 

.00145 

.00018 

9,760 

10,000 

.00260 

.00033 

14,640 

15,000 

.00410 

.00051 

19,520 

20,000 

.00535 

.00067 

24,400 

25,000 

.00665 

.00083 

29,280 

30,000 

.00795 

.00099 

34,160 

35,000 

.00920 

.00115 

39,040 

40,000 

.01075 

.00134 

40,992 

42,000 

.0114 

.00142 

42,944 

44,000 

.0144 

.00180 

44,896 

46,000 

.0356 

.00445 

44,000 

45,080 

.0734 

.00917 

44,500 

45,504 

.0965 

.01206 

45,000 

46,100 

.0973 

.01216 

45,872 

47,000 

.0981 

.01226 

46,848 

48,000 

.0991 

.01239 

47,824 

49,000 

.1013 

.01266 

48,800 

50,000 

.1163 

.01454 

50,752 

52,000 

.1273 

.01589 

52,704 

54,000 

.1381 

.01726 

54,656 

56,000 

.1552 

.01940 

64,416 

66,000 

.2601  (1)* 

.03250 

74,176 

76,000 

.4244  (2) 

.05530 

76,128 

78,000 

.50  (by 

.0625 

78,080 

80,000 

.59  scale) 

.074 

79,056 

81,000 

.70 

.0875 

80,032 

82,000 

.76 

.095 

81,008 

83,000 

.85 

.106 

81,984 

84,000 

.99 

.124 

83,000 

85,040 

1.24 

.155 

83,200 

85,240 

1.50 

.187 

82,000 

84,100 

1.64  (3) 

.205 

80,000 

82,000 

1.85  (4) 

.231 

72,000 

73,800  (broke) 

1.99  (5) 

.247 

(1)  Diameter,  1.097  inches. 

(3)  Begins  to  "  neck." 

(5)  Elongation  measured  after  fracture. 

Steel,  hot  rolled;  carbon,  0.42  per  cent. 


(2)  Diameter,  1.083  inches. 
(4)  Diameter  of  neck,  0.904  inch. 
Diameter  of  neck,  0.821  inch. 


18 


STRENGTH  OF  MATERIALS 


In  America,  the  unit  stress  in  pounds  per  square  inch  is  'used 
as  ordinate,  and  the  unit  deformation  is  taken  as  abscissa.  In 
England,  some  writers  use  unit  stress  as  abscissa  and  unit  defor- 
mation as  ordinate. 

Fig.  7  is  the  stress-strain  diagram  plotted  from  Table  I.  One 
division  on  the  horizontal  scale  represents  a  unit  elongation  of 
0.01,  and  one  division  on  the  vertical  scale  represents  a  unit  stress 
of  5000  pounds  per  square  inch. 

Fig.  8  is  a  part  of  the  stress-strain  diagram  from  the  same  table 
plotted  on  an  enlarged  scale;  one  division  on  the  horizontal  repre- 


95 
90 

I85 
2  80 

^*~ 

—  —  - 

—  • 

-^^ 

^^ 

-^ 

F_ 

—  •* 

-~- 

^ 

^, 

^ 

s 

\ 

/• 

G 

*« 

/ 

I 

"cn 

/ 

Err 

§55 
§-„ 

/ 

r 

C 

J 

o  40 
z  35 

B- 

D 

lo'5 

i» 

10 

.02   .04   .00   .03   .10   .12   .14   .16   .18   .20   .22   .24   . 

RELATIVE  ELONGATION 

Fig.  7.  —  Stress-strain  Diagram  of  Machine  Steel. 

sents  a  unit  elongation  of  0.0002  inch  per  inch  of  length  (one-fif- 
tieth as  much  as  in  Fig.  7) ;  one  division  on  the  vertical  represents 
a  unit  stress  of  2500  pounds  per  square  inch  (one-half  as  much 
as  in  Fig.  7). 

16.  Elastic  Limit,  and  Yield  Point.  —  The  point  B  in  Figs.  7 
and  8  represents  the  true  elastic  limit.  Up  to  that  point  the  curve 
is  a  straight  line.  In  other  words,  the  unit  stress  is  proportional 
to  the  unit  deformation  below  this  point.  Above  B,  the  curve 
deviates  from  the  straight  line.  Up  to  a  unit  stress  of  42,000 
pounds  per  square  inch  this  deviation  is  slight,  and  it  is  difficult 
to  locate  the  point  B  exactly.  From  42,000  to  44,000  pounds 
the  change  is  more  rapid.  This  may  be  seen  from  Table  I. 
Below  35,000  pounds  per  square  inch  the  increase  in  the  total 


STRESS   BEYOND    THE   ELASTIC   LIMIT 


19 


elongation  for  an  increment  of  5000  pounds  in  the  unit  stress 
is  about  13  divisions.  Between  35,000  and  40,000  pounds  unit 
stress  the  stretch  is  15.5  divisions.  Between  40,000  and  42,000 
the  rate  is  slightly  greater.  Between  42,000  and  44,000  pounds 
the  increase  amounts  to  30  divisions,  the  rate  being  over  five 
times  that  below  the  true  elastic  limit. 

At  C,  at  a  unit  stress  of  46,000  pounds  per  square  inch,  the 
curve  becomes  horizontal.     This  is  the  yield  point.     Beyond  the 


|*6 

ai 
cc 
<  40 

£ 


7 


RELATIVE  ELONGATIONS 

Fig.  8.  —  Part  of  Diagram  for  Machine  Steel. 

yield  point  the  curve  drops  to  a  unit  stress  of  about  45,000  pounds 
per  square  inch.  Not  only  is  there  an  increase  of  length  with  no 
increase  of  stress,  but  there  is  a  considerable  elongation  with  a 
diminished  stress.  In  changing  down  to  45,000  pounds  and  back 
again  to  45,500  pounds,  the  increase  in  length  is  nearly  twice 
as  great  as  the  entire  elongation  up  to  the  yield  point,  and  five 
times  as  great  as  the  elongation  from  zero  load  to  a  stress  of 
42,000  pounds  per  square  inch.  At  the  yield  point,  steel  and 
wrought  iron  stretch  like  taffy,  though  the  force  required  is 
probably  one  hundred  thousand  times  as  great. 


20  STRENGTH  OF  MATERIALS 

As  the  stress-strain  diagram  deviates  slowly  from  a  straight 
line  at  first,  it  is  difficult  to  locate  the  true  elastic  limit.  On  the 
other  hand,  the  stress  at  the  yield  point  may  be  easily  determined 
in  rapid  commercial  tests  and  without  delicate  apparatus  for 
measuring  the  elongations.  If  we  consider  Table  I,  we  find  that 
the  total  elongation  in  the  gauged  length  of  8  inches  is  about 
one-seventieth  of  an  inch  at  a  unit  stress  of  44,000  pounds,  and 
rises  to  more  than  one-thirtieth  of  an  inch  at  a  unit  stress  of 
46,000  pounds.  This  increase  in  length  may  easily  be  measured 
with  an  ordinary  scale,  so  that  the  yield  point  may  be  determined 
within  one  or  two  thousand  pounds  without  the  use  of  any 
extensometer  whatever.  Again,  just  beyond  the  yield  point 
the  elongation  is  increased  with  a  diminished  load.  This  may 
easily  be  determined  in  rapid  commercial  tests  in  which  the  test- 
ing machine  is  kept  running  continuously.  Before  reaching  the 
yield  point,  the  poise  on  the  beam  of  the  weighing  apparatus 
must  be  continually  moved  out  to  preserve  a  balance,  showing 
that  the  stress  is  increasing  with  the  elongation.  At  the  yield 
point  the  "  beam  drops  "  while  the  elongation  increases,  and  the 
poise  must  be  moved  backward  to  secure  a  balance.  In  iron  or 
steel  which  has  not  been  turned  or  polished,  and  is  therefore 
covered  with  a  coat  of  oxide,  the  yield  point  may  be  determined 
by  this  oxide  breaking  loose  and  falling.  We  sometimes  see  a 
portion  of  a  rod  reach  the  yield  point  before  the  remainder;  the 
oxide  falls  from  this  portion,  while  the  other  parts  of  the  bar  are 
unchanged  till  the  stress  becomes  a  little  greater.  The  curve 
in  such  a  rod  will  show  several  steps  or  bends  beyond  the  first 
yield  point,  corresponding  to  the  yield  points  of  the  various 
portions. 

Owing  to  the  fact  that  the  yield  point  may  be  determined 
so  easily,  by  methods  which  were  in  use  before  delicate  exten- 
someters  were  available,  the  term  "  elastic  limit  "  is  commonly 
applied  to  what  is  really  the  yield  point.  When  the  term  "  elas- 
tic limit  "  is  used  in  specifications,  yield  point  is  frequently  meant. 
The  point  B,  where  the  curve  deviates  from  the  straight  line, 
is  distinguished  by  some  writers  as  the  true  elastic  limit.  Since 
the  deformation  is  proportional  to  the  unit  stress  for  values 
below  this  point,  it  is  also  called  the  proportional  elastic  limit* 

*  See  Chapter  XIX  for  effect  of  time  on  the  form  of  the  stress-strain  dia- 
gram of  steel. 


STRESS   BEYOND    THE   ELASTIC   LIMIT         21 

17.  Johnson's  Apparent  Elastic  Limit.  —  Since  it  is  somewhat 
difficult  to  determine  the  true  elastic  limit  accurately,  especially 
in  hard  steel,  where  there  is  a  wide  range  between  this  point  and 
the  yield  point,  and  in  materials  (such  as  cast  iron)  which  have  no 
yield  point,  the  late  Prof.  J.  B.  Johnson  proposed  another  point 
which  he  called  the  "apparent  elastic  limit."*     He  denned  the 
apparent  elastic   limit  as  "the   point   on  the  stress  diagram  at 
which  the  rate  of  deformation  is  50  per  cent  greater  than  at  the 
origin."    It  is  that  point  on  the  curve  at  which  the  slope  of  the 
tangent  from  the  vertical  is  50  per  cent  greater  than  that  of 
the  straight-line  part  of  the  curve. 

This  term  has  not  yet  come  into  general  use  among  engineers. 
In  some  investigations  of  the  strength  of  materials,  it  has  been 
found  useful  in  comparing  the  results  of  different  tests. f 

1 8.  Calculation  of  the  Modulus  of  Elasticity.  —  The  stress- 
strain  diagram,  when  plotted  to  a  sufficiently  large  scale,  en- 
ables us  to  calculate  quickly  the  average  value  of  the  modulus 
of  elasticity.     If  the  straight  line  passes  through  the  origin,  we 
merely  find  the  value  of  stress  which  corresponds  to  some  con- 
venient unit  elongation  such  as  0.001  or  0.0005.     If  the  straight 
line  does  not  pass  through  the  origin,  we  get  the  difference  of 
stress  for  some  convenient  difference  of  elongation.     In  either 
case,  to  get  the  modulus,  we  divide  the  difference  in  stress  by 
the  corresponding  difference  in  elongation. 

PROBLEMS. 

1.  From  the  curve  of  Fig.  8  find  the  unit  stress  which  corresponds  to  the 
unit  elongation  0.001,  and  calculate  E  to  three  significant  figures. 

2.  From  Fig.  8  find  the  unit  elongation  which  corresponds  to  the  unit 
stress  of  25,000  pounds,  and  compute  E  to  three  significant  figures. 

3.  From  the  data  of  Table  I  plot  the  stress-strain  diagram  up  to  the  unit 
stress  of  42,000  pounds  per  square  inch  to  the  scales  1  inch  equals  a  unit  stress 
of  5000  pounds  per  square  inch  and  a  unit  elongation  of  0.0002  inch  per  inch 
of  length.    Use  paper  ruled  in  0.1  inch  units.     Draw  the  curve  as  a  light  line 
and  solve  Problems  1  and  2. 

It  is,  of  course,  not  necessary  to  plot  the  stress  diagram  in 
order  to  solve  for  E.  The  diagram  enables  us  to  compute  a 
good  average  value  with  a  single  calculation.  It  also  enables 

*  See  Johnson's  "  Materials  of  Construction,"  pages  18-20. 
t  See  work  of  H.  F.  Moore  in  Bulletin  No.  42  of  the  University  of  Illinois 
Engineering  Experiment  Station,  page  14. 


22  STRENGTH  OF  MATERIALS 

us  to  judge  of  the  accuracy  of  the  test  by  observing  how  closely 
the  points  fall  on  the  straight  line.  In  an  autographic  testing 
machine,  where  the  curve  is  automatically  drawn  as  the  piece 
is  tested,  it  is  important  to  be  able  to  determine  the  modulus 
from  the  curve. 

Where  the  readings  are  available,  as  in  Table  I,  these  may  be 
used  in  determining  E.  If  all  the  readings  were  exactly  correct, 
so  that  all  would  fall  exactly  on  the  straight  line  when  plotted 
to  any  scale,  any  one  such  reading  would  give  a  correct  result. 
Since  there  is  some  probable  error  in  balancing  the  scales  and 
setting  the  extensometers,  there  will  be  some  variation  in  the 
values  of  E  taken  from  the  different  intervals.  The  relative 
accuracy  is  practically  proportional  to  the  interval.  The  value 
of  E  from  a  20,000-pound  interval  should  be  given  twice  the 
weight  which  is  given  to  the  result  of  a  10,000-pound  interval. 

Of  course,  all  readings  for  calculating  E  must  be  taken  below 
the  true  elastic  limit.  If  the  curve  is  not  plotted,  this  may 
be  determined  by  observing  the  point  where  the  increment  of 
elongation  corresponding  to  a  given  increment  of  stress  per- 
manently increases.  As  shown  in  Article  16,  this  is  between 
35,000  and  40,000  pounds  per  square  inch  for  the  steel  of 
Table  I. 

PROBLEMS. 

4.  From  Table  I  calculate  E,  using  the  intervals  0-25,000  and  0-30,000 
pounds  and  the  exact  unit  elongation  calculated  from  the  total  elongation. 
What  relative  weights  may  be  given  to  the  two  results  in  averaging  them? 

Ans.  Relative  weight,  5  :  6.     Mean  E,  30,140,000. 

5.  Calculate  E  for  three  intervals  of  25,000  pounds  per  square  inch;  from 
0  to  25,000,  from  5000  to  30,000,  and  from  10,000  to  35,000.     Average  the 
results.  Ans.  Mean  E,  30,350,000. 

6.  Calculate  the  weighted  mean  for  all  the  readings  without  computing 
the  separate  values  of  E,  by  adding  together  all  the  micrometer  readings  and 
all  the  stresses  and  using  the  totals  as  the  unit  stress  and  the  total  elongations 
respectively.  Ans.  E  equals  29,990,000. 

7.  If  you  are  using  the  interval  of  30,000  pounds,  what  error  in  E  would 
an  error  of  one  division  in  the  extensometer  reading  produce?     What  would 
be  the  error  in  E  due  to  an  error  of  10  pounds  in  the  total  load? 

19.  Ultimate  Strength  and  Breaking  Strength.  —  The  point  F 
at  the  top  of  the  curve  of  Fig.  7,  representing  a  unit  stress  of  a 
little  more  than  85,000  pounds  per  square  inch,  gives  the  ulti- 
mate strength  of  the  steel  under  test.  The  rod  at  this  stress  was 
elongated  1.5  inches  in  the  gauged  length  of  8  inches,  and  the  diam- 


STRESS   BEYOND    THE   ELASTIC   LIMIT          23 

eter  was  practically  uniform  throughout  this  length.  Beyond  this 
elongation,  the  rod  began  to  "neck";  its  diameter  decreasing 
rapidly  at  one  section,  while  the  remainder  was  not  changed. 
When  the  load  had  dropped  to  about  82,000  pounds  per  square 
inch,  the  minimum  diameter  at  the  neck  was  0.904  inch,  while 
that  of  most  of  the  gauged  length  was  a  little  over  1  inch.  It 
finally  broke  at  a  total  load  of  72,000  pounds,  which,  in  terms 
of  the  original  area,  corresponds  to  a  unit  stress  of  73,800  pounds 
per  square  inch.  This  is  the  breaking  strength,  the  point  G  of 
Fig.  7. 

Most  materials,  such  as  wood,  cast  iron,  concrete,  and  hard 
steel,  do  not  neck;  the  ultimate  strength  corresponds  with  the 
breaking  strength. 

20.  Per  cent  of  Elongation  and  of  Reduction  of  Area.  —  The 
per  cent  of  elongation  in  ductile  materials,  such  as  wrought  iron 
and  steel,  is  an  important  factor.  (See  Cambria  for  the  minimum 
values  in  the  Manufacturers'  Standard  Specifications.  See  also 
latest  Proceedings  of  the  American  Society  for  Testing  Materials.) 
In  tension  tests  the  gauged  length  of  8  inches  is  subdivided  into 
inch  intervals  by  punch  marks,  Fig.  6,  I.  If  we  observe  the 
tested  rod,  Fig.  6,  II,  we  notice  a  considerable  variation  in  the 
distance  between  points  which  were  originally  one  inch  apart. 
Measurements  of  the  rod  are  as  follows : 

Interval.  Elongation. 

0-1 0.17  inch 

1-2 ' 19 

2-3 31 

3-4 54          included  neck. 

4-5 25 

5-6 19 

6-7 17 

7-8 17 

If  we  use  only  the  interval  3-4,  which  included  the  neck,  we  get 
an  elongation  of  54  per  cent.  If  we  take  the  4-inch  interval  0-4, 
we  get  30.2  per  cent.  'In  order  to  make  the  results  of  differ- 
ent tests  comparable  with  one  another,  the  Society  for  Testing 
Materials  has  adopted  8  inches  as  the  standard  gauged  length 
for  rods  f  inch  and  upwards.  (See  Manufacturers'  Specifications, 
Article  12,  in  Cambria.) 

The  per  cent  of  reduction  of  area  at  the  neck  is  also  of  inter- 
est. In  the  rod  of  Table  I,  the  original  diameter  was  1.115  inches 
and  the  final  diameter  was  0.821  inch.  The  final  area  of  the  neck 


24  STRENGTH  OF  MATERIALS 

was  54.2  per  cent  of  the  original  area.     The  reduction  was  45.8 
per  cent. 

Table  I  is  for  steel  containing  0.42  per  cent  carbon.  For 
structural  steel,  compare  the  annealed  rod  of  Table  V. 

PROBLEMS. 

1.  From  the  above  measurements  find  the  per  cent  of  elongation  for  the 
four  intervals  4-8,  which  do  not  include  the  neck.     Find  also  the  per  cent  of 
elongation  for  the  entire  gauged  length. 

2.  A  rod  of  soft  steel,  originally  0.874  inch  in  diameter,  was  tested  in  ten- 
sion.    After  fracture  under  a  load  of  28,000  pounds,  the  diameter  of  the  neck 
was  found  to  be  0.570  inch.     What  was  the  per  cent  of  reduction  of  area? 
What  was  the  breaking  strength? 

Ans.  57.5  per  cent;  46,700  pounds  per  square  inch. 

3.  In  Problem  2,  the  maximum  load  was  36,000  pounds.     Find  the  ultimate 
strength. 

21.  Apparent  and  Actual  Unit  Stress.  —  The  unit  stresses  in 
Table  I  were  calculated  by  dividing  the  total  load  by  the  origi- 
nal area  of  cross  section.  This  is  the  usual  custom,  and,  unless 
otherwise  stated,  all  tables  and  curves  are  given  in  this  way. 
Owing  to  the  fact  that  the  area  of  cross  section  is  permanently 
reduced  when  the  yield  point  is  reached,  the  actual  unit  stress 
is  different.  At  the  time  of  rupture  of  Table  I,  the  total  load 
was  72,000  pounds  and  the  apparent  unit  stress  was  73,800  pounds 
per  square  inch.  The  actual  diameter  of  the  neck  at  fracture 
was  0.821  inch,  which  gives  an  actual  unit  stress  of  136,000  pounds 
per  square  inch.  The  curve  of  Fig.  7  falls  from  F  to  G.  The 
actual  stress  increases  for  all  loads  except  at  the  yield  point. 

PRO'BLEMS. 

1.  Calculate  from  Table  I  the  actual  unit  stress  when  the  total  elongation 
was  1.85  inches. 

2.  Calculate  the  actual  unit  stress  when  the  apparent  unit  stress  was  76,000 
pounds  per  square  inch. 

To  get  the  actual  unit  stress  after  the  rod  begins  to  neck,  we 
must  measure  the  diameter  of  the  smallest  section  at  each  load. 
Before  necking  begins,  the  mean  area  of  the  cross  section  may 
be  computed  from  the  fact  that  the  volume  and  density  remain 
nearly  constant.  Accordingly,  the  area  of  cross  section  is  in- 
versely as  the  length,  and  the  ratio  of  actual  to  apparent  unit 
stress  is  directly  as  the  length.  If  A  is  the  original  cross  section 


STRESS   BEYOND    THE   ELASTIC   LIMIT          25 

and  A'  the  actual  area  corresponding  to  a  given  unit  elongation,  6, 
A  =  A'(\  +  5)  and 

Actual  stress  =  apparent  stress  multiplied  by  (1  +5). 

PROBLEMS. 

3.  Calculate  the  diameters  corresponding  to  unit  stresses  of  66,000  and 
76,000  and  compare  results  with  Table  I. 

4.  Compute  the  actual  unit  stress  for  Table  I  for  all  apparent  stresses  above 
45,000  pounds  per  square  inch,  and  plot  curve  with  coordinates  as  follows: 
One  inch  equals  a  unit  deformation  of  5  per  cent  and  a  unit  stress  of  20,000 
pounds  per  square  inch. 

The  student  may  notice  an  apparent  discrepancy  between  the  statements 
above  and  those  of  Article  14.  Poisson's  ratio  and  the  discussion  of  Article  14 
apply  only  to  stresses  and  deformations  inside  the  elastic  limits,  where  all 
deformations  are  temporary  and  relatively  very  small.  The  discussion  above 
refers  to  the  permanent  changes  beyond  the  elastic  limit.  The  temporary 
deformations,  to  which  Poisson's  ratio  applies,  are,  of  course,  superimposed 
on  these  permanent  deformations;  but  they  are  relatively  so  small  that  they 
can  only  be  measured  with  delicate  instruments.  In  the  test  of  Table  I  the 
unit  deformation  at  the  elastic  limit  was  a  little  over  0.001;  and  the  corre- 
sponding decrease  in  diameter  calculated  by  Poisson's  ratio  was  about  0.0003 
inch. 

The  density  does  not  remain  exactly  constant,  but  the  differences  are  beyond 
the  limits  of  accuracy  of  the  elongation  measurements. 

22.   Curves  of  Various  Structural  Materials.  —  The  curves  of 

Figs.  7  and  8  give  a  fair  average  idea  of  the  behavior  of  machine 
steel  in  tension.  Structural  steel,  with  a  smaller  per  cent  of  car- 
bon, has  a  yield  point  of  a  little  over  30,000  pounds  per  square 
inch  and  an  ultimate  strength  of  about  60,000  pounds  per  square 
inch.  Its  modulus  of  elasticity  is  about  29,000,000.  Tool  steel 
has  a  yield  point  above  50,000  pounds  and  an  ultimate  strength  of 
over  100,000  pounds.  The  heat  and  mechanical  treatment  have 
a  great  effect  upon  these  factors  but  very  little  effect  upon  the 
modulus  of  elasticity. 

Table  II  and  curve  II  of  Fig.  9  represent  the  behavior  of  cast 
iron  in  tension.  The  table  is  the  mean  of  the  tests  of  six  bars 
from  the  same  heat.  The  figures  represent  what  may  be  expected 
in  good  cast  iron. 

Table  II  is  from  the  average  of  six  tests,  specimens  8014, 
8041,  8050,  8051,  8053,  and  8063,  at  the  Watertown  Arsenal 
("  Tests  of  Metals,"  1905). 

The  average  ultimate  load  was  26,450  pounds  per  square  inch. 


26 


STRENGTH  OF  MATERIALS 


The  actual  initial  load  was  1000  pounds.  The  table  is  calculated 
on  the  assumption  that  the  elongation  from  0  to  1000  is  the 
same  as  from  1000  to  2000, 

The  curve  for  this  cast  iron  is  plotted  to  the  same  scale  as 
Fig.  8,  and  a  part  of  the  curve  of  steel  from  Fig.  8  is  drawn  for 
comparison.  The  dotted  line  shows  approximately  the  initial 


25 


II 


OJ          •«          to 

I  i  § 


UNIT  DEFORMATIONS 

Fig.  9. 

slope  of  the  cast-iron  stress  diagram.  The  curve  begins  to  bend 
almost  at  the  start,  and  it  is  difficult  to  locate  the  true  elastic 
limit.  There  is  no  yield  point,  and  the  material  breaks  with- 
out necking. 

PROBLEMS. 

1.  From  the  dotted  line  of  curve  II,  Fig.  9,  calculate  the  modulus  of  elas- 
ticity of  cast  iron.     Check  result  by  means  of  the  readings  of  Table  II. 

2.  From  the  curve  of  Fig.  9  find  Johnson's  apparent  elastic  limit  and  com- 
pare the  result  with  the  table.    Also  construct  a  curve  for  which  the  abscissas 


STRESS  BEYOND   THE  ELASTIC  LIMIT 


27 


are  unit  stress,  and  ordinates  are  the  differences  between  successive  unit  defor- 
mations and  56,  and  determine  the  apparent  elastic  limit  from  the  abscissa 
of  the  point  for  which  the  ordinate  is  28. 

3.  From  Tables  I  and  II  determine  the  ratio  of  the  elongation  of  steel  at 
rupture  to  that  of  cast  iron. 

4.  If  we  take  2000  pounds  as  the  true  elastic  limit  of  cast  iron  in  tension, 
what  is  its  modulus  of  resilience? 


TABLE  II. 

TENSION  TEST  OF  CAST  IRON. 
Diameter,  1.129  inches;  area,  1  square  inch;  gauged  length,  10  inches. 


Load  per  square  inch. 

Elongation. 

In  gauged  length. 

Per  inch  length. 

Pounds. 

Inch. 

Inch. 

1,000 

0.00056 

0.000056 

2,000 

.00112 

.000112 

3,000 

.00171 

.000171 

4,000 

.00236 

.000236 

5,000 

.00303 

.000303 

6,000 

.00374 

.000374 

7,000 

.00446 

.000446 

8,000 

.00526 

.000526 

9,000 

.00606 

.000606 

10,000 

.00691 

.000691 

11,000 

.00779 

.000779 

12,000 

.00871 

.000871 

13,000 

.00968 

.000968 

14,000 

.01061 

.    .001061 

15,000 

.01174 

.001174 

16,000 

.01283 

.001283 

17,000 

.01404 

.001404 

18,000 

.01544 

.001544 

19,000 

.01689 

.001689 

20,000 

.01851 

.001851 

21,000 

.02003 

.002003 

22,000 

.02182 

.002182 

23,000 

.02420 

.002420 

24,000 

.02626 

.002626 

Table  III,  and  curve  III  of  Fig.  9,  represent  the  behavior  of 
long-leaf  yellow  pine  in  compression.  Like  steel,  the  curve  for 
timber  is  a  straight  line  for  a  considerable  portion  of  its  length. 
In  other  respects  it  resembles  the  curve  for  cast  iron.  The  post 


28 


STRENGTH  OF  MATERIALS 


represented  by  Table  III  failed  outside  of  the  gauged  portion; 
the  ultimate  elongation  is,  therefore,  less  than  it  would  be  if  the 
failure  had  occurred  inside  of  this  length. 

TABLE  III. 

COMPRESSION  TEST  OF  LONG-LEAF  YELLOW  PINE. 

From  Watertown  Arsenal  Report,  1897,  page  420. 

Length  of  post,  10  feet.     Dimensions,  9.75  inches  by  9.77  inches. 

Area,  95.26  square  inches.     Gauged  length,  50  inches. 


Applied  load. 

Deformation. 

Total. 

Unit  stress  per 
square  inch. 

In  gauged  length. 

Unit  per  inch  length. 

Pounds. 

Pounds. 

Inch. 

Inch. 

9,526 

100 

0.0021 

0.000042 

19,052 

200 

.0044 

.000088 

28,578 

300 

.0067 

.000134 

38,104 

400 

.0091 

.000182 

47,630 

500 

.0116 

.000232 

57,156 

600      r 

.0141 

.000282 

66,682 

700 

.0165 

.000330 

76,208 

800 

.0191 

.000382 

85,734 

900 

.0215 

.000430 

95,260 

1000 

.0240 

.000480 

114,312 

1200 

.0290 

.000580 

133,364 

1400 

.0340 

.000680 

152,416 

1600 

.0389 

.000778 

171,468 

1800 

.0443 

.000886 

190,520 

2000 

.0495 

.000990 

209,572 

2200 

.0546 

.001092 

228,624 

2400 

.0601 

.001202 

247,676 

2600 

.0652 

.001304 

266,728 

2800 

.0705 

.001410 

285,780 

3000 

.0758 

.001516 

304,832 

3200 

.0811 

.001622 

323,884 

3400 

.0869 

.001738 

342,936 

3600 

.0932 

.001864 

361,988 

3800 

.1005 

.002010 

381,040 

4000 

.1077 

.002154 

400,092 

4200 

.1084 

.002168 

416,000 

4367 

Ultimate  strength 

Failed  by  crushing  at  end. 

PROBLEMS. 

5.  Find  E  of  yellow  pine  from  the  dotted  prolongation  of  curve  III,  and 
also  from  Table  III. 

6.  In  a  post  similar  to  that  of  Table  III  an  increment  of  load  amounting  to 


STRESS   BEYOND    THE   ELASTIC   LIMIT 


29 


500  pounds  per  square  inch  produced  a  deformation  of  0.0146  inch  in  a  gauged 
length  of  50  inches.     Find  E.  Ans.  1,710,000. 

7.  An  old  post  of  long-leaf  yellow  pine,  tested  at  the  Watertown  Arsenal, 
was  9.43  inches  by  9.35  inches,  and  was  chamfered  f  inch  at  each  corner. 
Using  the  area  to  the  nearest  square  inch,  find  the  ultimate  strength,  the  ulti- 
mate load  being  528,400  pounds. 


TABLE   IV. 

COMPRESSION  TEST  OF  1  :  2J  :  6  CONCRETE;  AGE, 


DAYS. 


Diameter  of  test  cylinder,  8  inches;  area,  50  square  inches. 
Total  length,  16  inches;  gauged  length,  10  inches. 


Applied  load. 

Deformation. 

Total. 

Per  square  inch. 

In  gauged  length. 

Per  inch  length. 

Pounds. 

Pounds. 

Inch. 

Inch. 

2,000 

40 

0.00013 

0.000013 

4,000 

80 

.00026 

.000026 

6,000 

120 

.00038 

.000038 

8,000 

160 

.00052 

.000052 

10,000 

200 

.00068 

.000068 

12,000 

240 

.00081 

.000081 

14,000 

280 

.00099 

.000099 

16,000 

320 

.00113 

.000113 

18,000 

360 

.00136 

.000136 

20,000 

400 

.00158 

.000158 

22,000 

440 

.00180 

.000180 

24,000 

480 

.00206 

.000206 

26,000 

520 

.00232 

.000232 

28,000 

560 

.00260 

.000260 

30,000 

600 

.00295 

.000295 

32,000 

640 

.00327 

.000327 

34,000 

680 

.00377 

.000377 

36,000 

720 

.00421 

.000421 

38,000 

760 

.00473 

.000473 

40,000 

800 

•   .00535 

.000535 

42,000 

840 

.00609 

.000609 

44,000 

880 

.00692 

.000692 

46,000 

920 

.00796 

.000796 

48,000 

960 

.00922 

.000922 

50,000 

1000 

.01058 

.001058 

52,000 

1080 

.01177 

.001177 

54,000 

1080 

.01323 

.001323 

56,000 

1120 

.01575 

.001575 

58,000 

1160 

.01847 

.001847 

60,000 

1200 

Failed 

Fig.  10  gives  some  comparative   curves  for  timber  and  con- 
crete.    Curve  I  is  the  long-leaf  yellow  pine  of  Table  III.     The 


30 


STRENGTH  OF  MATERIALS 


unit  deformations  are  represented  on  a  scale  twice  as  great  as  in 
Fig.  9,  and  the  unit  stresses,  by  a  scale  ten  times  as  great.  Curve 
II  of  Fig.  10  is  the  stress  diagram  for  a  sample  of  1 : 2.5 :  6  concrete 
in  compression,  the  readings  for  which  are  given  in  Table  IV. 


4500 
•rinnn 

;-S 

v_ 

-,"1 

^ 

)S  PER  SQUARE  INC 

X 

ft 

V<> 

X 

2^ 

x 

ri 

x^ 

^ 

'^ 

1 

g 

^ 

o 

^ 

IT  STRESS  IN  POON 

III! 

2 

g 

<( 

g 

t 

/ 

x^ 

& 

/ 

/< 

'9* 

I 

t 

/ 

<N 

I] 

/" 

fi 

*~~ 

~c< 

lit 

r<; 

_— 

lc 

in 

=C 

nT 

In 

e« 

sit 

/<.• 

^( 

E& 

('-, 

v 

3 

/ 

—  .—  • 

/ 

^ 

4*- 

'l: 

&J 

:(> 

4 

j? 

g 

•0002  .0004   .0006  .0008  .0010    .0012    .0014  .0016   .0013  .0020  .0022   .0024. 
UNIT  DEFORMATIONS 

Fig.  10. 

PROBLEMS. 

8.  Calculate  the  modulus  of  elasticity  of  concrete  from  the  dotted  pro- 
longation of  the  curve  in  Fig.  10. 

9.  From  the  area  included  between  the  curve  for  timber  in  Fig.  10,  the 
X  axis,  and  the  ordinate  corresponding  to  a  unit  elongation  0.001,  compute 
the  work  done  per  unit  volume  in  producing  this  elongation. 

23.  Factor  of  Safety.  —  In  Article  6,  the  allowable  unit  stress 
was  defined  as  depending  upon  the  judgment  of  some  authority. 
These  judgments  are  based  on  tests  of  materials  such  as  those 
of  Tables  I  to  IV. 

Working  stresses  should  never  exceed  the  true  elastic  limit. 
They  are  generally  based  on  the  ultimate  strength  of  the  material. 
The  ratio  of  the  ultimate  strength  of  a  given  material  to  the 
allowable  working  stress  is  called  the  factor  of  safety. 

PROBLEMS. 

1.  If  the  steel  of  Table  I  is  used  with  a  factor  of  safety  of  5,  what  is  the 
allowable  unit  stress? 

2.  If  cast  iron  such  as  that  of  Table  II  is  used  with  a  factor  of  safety  of 
10  in  tension,  what  is  the  allowable  unit  stress? 

Ans.  2500  pounds  per  square  inch. 


STRESS   BEYOND    THE   ELASTIC   LIMIT          31 

3.  A  concrete  pier  16  inches  square  carries  a  load  of  75,000  pounds.     Using 
Table  IV,  find  the  factor  of  safety.  Ans.  4. 

4.  A  steel  structure  is  designed  in  accordance  with  the  New  York  building 
laws.     The  steel  used  satisfies  the  minimum  requirements  of  the  Manufac- 
turers' Standard  Specifications  for  steel  for  bridges.     What  is  the  factor  of 
safety?  Ans.  3.5. 

The  value  of  the  factor  of  safety  which  should  be  used  depends 
upon  a  great  number  of  conditions.  Some  of  these  are: 

Repeated  stresses  slightly  beyond  the  true  elastic  limit  will 
finally  cause  failure,  so  that  a  body  subjected  to  varying  load 
should  have  its  allowable  stresses  well  below  this  limit.  The 
greater  the  variation  of  stress,  the  smaller  should  be  the  allow- 
able unit  stress. 

The  factor  of  safety  must  be  large  enough  to  allow  for  any 
deterioration  of  the  material  from  any  cause  during  the  time 
which  it  is  to  be  used.  This  includes  the  decay  of  timber,  the 
rusting  of  metal,  the  effect  of  frost  and  electrolysis. 

In  deciding  what  factor  of  safety  to  use,  the  uniformity  of  the 
material  must  be  taken  into  account.  Structural  steel  which 
has  an  ultimate  strength  of  65,000  pounds  per  square  inch  on 
an  average  will  seldom  vary  5000  pounds  on  either  side  of  this 
figure;  while  the  variation  of  timber  sufficiently  good  to  pass  a 
reasonable  inspection  may  be  50  per  cent  of  the  average  ulti- 
mate strength.  An  engineer,  in  designing  a  concrete  structure 
which  he  knows  will  be  built  under  competent  supervision,  will 
use  much  higher  unit  stresses  than  he  will  risk  where  such  inspec- 
tion is  wanting. 

The  factor  of  safety  must  also  depend  upon  the  amount  of 
injury  which  would  occur  if  the  material  failed.  We  would 
use  a  plank  in  a  scaffold  3  feet  high  with  a  much  lower  factor 
of  safety  than  we  would  consider  if  failure  meant  a  fall  of  100 
feet. 

The  factor  of  safety  must  allow  some  margin  for  unexpected 
loads.  Cases  have  occurred  where  a  wagon  bridge  has  failed 
when  used  as  a  grand  stand  to  watch  a  boat  race  or  fireworks. 
That  part  of  the  factor  of  safety  which  makes  allowance  for  lack 
of  ordinary  judgment  in  persons  using  the  machine  or  struc- 
ture is  called  the  "  fool  factor." 


32  STRENGTH  OF  MATERIALS 


PROBLEM. 

5.  Taking  the  figures  of  the  United  States  Department  of  Agriculture  as 
correct,  find  the  factor  of  safety  of  white  oak  and  long-leaf  yellow  pine  in  com- 
pression, with  the  grain  and  across  the  grain,  when  used  in  accordance  with  the 
New  York  building  laws. 

24.  Effect  of  Form  on  the  Ultimate  Strength.  —  We  have 
assumed  in  our  discussions  that  the  stress  across  any  section  is 
uniform.  This  is  true  in  a  rod  of  uniform  section  at  some  dis- 
tance from  the  surface  of  application  of  the  load,  provided  that 
the  line  of  resultant  force  coincides  with  the  axis  of  the  rod. 

Test  bars  are  made  of  uniform  section  throughout,  or  of 
uniform  section  for  some  distance  beyond  the  extremities  of  the 


[t-GA' 


GED  LENGTH 


BCD 


Fig.  11.  —  Stress  Distribution  in  Test  Fig.  12.  —  Abrupt  Change  of 

Bar.  Section. 

gauged  length.  (See  Fig.  6,  and  also  the  form  of  test  bar  adopted 
by  the  Society  for  Testing  Materials,  as  given  in  Cambria,  under 
Manufacturers'  Standard  Specifications,  or  in  the  Proceedings 
of  the  Society.)  Fig.  11  represents  one  end  of  such  a  bar.  The 
stress  which  may  be  uniformly  distributed  across  a  section  at  A 
is  unequally  distributed  at  sections  B  and  C,  and  becomes  uniform 
and  parallel  to  the  axis  at  D.  If  the  gauged  length  began  at  C 
at  the  beginning  of  the  parallel  portion,  the  measured  elongation 
would  be  too  high,  owing  to  the  fact  that  the  stress  is  greater 
than  the  average  near  the  surface.  This  effect  would  be  increased 
if  the  change  in  section  were  abrupt  as  in  Fig.  12. 

The  ultimate  strength  of  a  rod  at  such  a  change  of  section 
depends  upon  its  ductility.  If  rods  as  in  Fig.  12  are  made 
of  cast  iron  or  other  nonductile  material,  they  will  fail  at  section 
C  owing  to  the  concentration  of  stress  near  the  surface.  The 
more  abrupt  the  change  the  greater  the  concentration  and  the 
easier  the  failure.  If  the  rod  is  of  ductile  material,  such  as  struc- 
tural steel,  the  strength  at  C  will  be  increased  by  the  material  of 
the  larger  section  to  the  left.  A  ductile  substance  necks  before 
it  fails.  The  material  of  the  larger  section  tends  to  prevent 


STRESS  BEYOND   THE   ELASTIC  LIMIT'        33 


necking  in  the  smaller  sections  at  a  considerable  distance  to  the 
right  of  C. 

A  rod  of  ductile  material  with  a  short  reduced  area,  such  as  I 
and  II,  Fig.  13,  will  show  a  considerably  higher  ultimate  strength 
than  a  rod  in  which  the  minimum  section  is  longer,  as  in  III, 
Fig.  13. 

It  is  not  necessary  to  make  test  bars  of  the  form  shown  in 
Fig.  6;  any  bar  of  uniform  section  will  do,  and  many  tests  are 
made  of  such  bars  as  they  come  from  the  rolls.  There  is  this 
advantage  in  the  standard  form  shown  in  Fig.  6,  —  that  it  will 
fail  inside  the  gauged  length  on  account  of  the  resistance  to  neck- 
ing for  some  distance  from  the  larger  section.  A  bar  of  uni- 
form section  may  fail  outside  of  the  gauged  length. 


o 


ii 


i  ii  &  in 


in 


Fig.  13.  —  Reduced  Sections. 

It  is  hardly  necessary  to  state  that  all  changes  in  section  should 
be  gradual.  The  standard  form  of  bar,  as  adopted  by  the  Society 
for  Testing  Materials,  changes  from  large  to  small  section  on  the 
arc  of  a  circle  tangent  to  the  surface  of  the  smaller  section.  It  is 
easier  to  make  a  taper  from  one  size  to  the  other,  and  the  results 
are  practically  as  good. 

25.  Effect  of  Stresses  beyond  the  Yield  Point.  —  In  materials 
which  are  not  ductile,  any  stress  beyond  the  elastic  limit  produces 
a  permanent  injury.  In  ductile  materials,  especially  ssoft  iron 
and  steel,  this  is  not  the  case.  If  a  rod  of  steel  or  iron,  originally 
hot-rolled,  is  stressed  beyond  the  yield  point,  the  result  is  a  rais- 
ing of  the  yield  point.  If  a  rod  having  a  yield  point  of  35,000 
pounds  is  carried  up  to  50,000  pounds,  producing  a  large  per- 
manent set,  upon  testing  the  second  time  the  yield  point  will  be 
found  to  be  about  50,000  pounds.  When  a  high  elastic  limit 


34  STRENGTH  OF  MATERIALS 

and  yield  point  are  desired,  soft  steel  is  subjected  to  cold-rolling. 
The  effect  of  this  is  to  raise  the  yield  point  to  nearly  the  ultimate 
strength  of  the  steel.  The  ultimate  strength  is  also  raised  con- 
siderably in  terms  of  the  original  cold-rolled  section.  Cold-draw- 
ing as  employed  in  the  manufacture  of  wire  has  a  similar  effect. 


Fig.  14.  — Soft  Steel  in  Tension;  Left,  Cold-Rolled;  Right,  Annealed. 

Fig.  14  and  Table  V  show  the  effect  of  cold-rolling.  In  Fig.  14 
the  middle  rod  is  a  piece  of  nominal  J-inch  cold-rolled  shafting. 
The  left  rod  is  an  exactly  similar  one  after  testing  in  tension.  Its 
ultimate  strength  was  over  86,000  pounds  per  square  inch  and 
its  elongation  practically  10  per  cent.  On  the  right  is  a  third 
rod,  originally  like  the  others,  which  was  annealed  by  heating  to 
redness  and  cooling  slowly  to  destroy  the  effect  of  the  previous 
cold-rolling.  When  tested  in  tension  its  ultimate  strength  was 
found  to  be  60,000  pounds  per  square  inch  and  its  elongation 
22  per  cent. 


STRESS   BEYOND    THE   ELASTIC   LIMIT         35 


TABLE  V. 

TENSILE  TEST  OF  SOFT  STEEL;  COLD-ROLLED,  AND  ANNEALED. 
Diameter  of  rods,  0.874  inch;  area,  0.600  square  inch;  gauged  length,  8  inches. 


Total  load  in  pounds. 

Unit  elongation. 

Cold-rolled  rod. 

Annealed  rod. 

inch 

inch 

3,000 

0.000163 

0.000175 

6,000 

.000356 

.000356 

9,000 

.000512 

.000525 

12,000 

.000681 

.000712 

15,000 

.000861 

.000894 

18,000 

.00102 

.00106 

21,000 

.00118 

.00122 

22,200 

.00177 

22,800 

.00217 

22,200 

.00231 

22,800 

.00245 

23,400 

.00480 

24,000 

.00132 

.00750 

27,000 

.00150 

.034 

30,000 

.00166 

.050 

33,000 

.00184 

.081 

36,000 

.00202 

.164 

35,200 

.189 

33,000 

.201 

28,000 

.219 

39,000 

.00223 

42,000 

.00264 

45,000 

.00326 

48,000 

.00570 

49,500 

.00739 

47,000 

.00775 

49,000 

.01045 

49,500 

.01502 

50,000 

.01895 

51,000 

.02970 

51,900 

.054 

51,100 

.067 

48,000 

.079 

45,000 

.085 

39,000 

.099 

Diameter  of  neck  at  fracture  was  0.640  inch  in  cold-rolled  rod  and  0.570 
inch  in  annealed  rod.  Analysis:  carbon,  0.113  per  cent;  manganese,  0.495 
per  cent. 


36  STRENGTH    OF  MATERIALS 


PROBLEMS. 

1.  From  Table  V  find  E  for  each  rod  from  the  15,000-,  18,000-,  and  21,000- 
pound  loads. 

2.  Plot  curve  for  each  on  same  sheet  to  a  suitable  scale.     Plot  part  of  each 
-curve  to  a  scale  with  abscissas  enlarged  twenty-fold  and  determine  E.      Cal- 
culate the  apparent  and  actual  unit  stress  in  the  neck  at  rupture  for  each  case. 

The  fact  that  soft  steel  may  be  stressed  beyond  the  yield  point 
without  injury,  and  with  no  change  except  a  slight  reduction  of 
section  and  elevation  of  the  yield  point,  is  of  great  advantage  in 
its  use  in  structures.  In  a  heavy  structure  made  of  many  parts, 
there  is  always  some  adjustment  when  the  loads  are  first  applied. 
This  may  cause  an  overstraining  of  some  parts.  If  these  parts 
are  made  of  soft  steel,  they  can  yield  slightly,  permitting  other 
members  to  take  part  of  the  excess  load. 

Chapter  XIX  gives  additional  information  in  regard  to  the  effect  of  stress 
beyond  the  yield  point  and  the  nature  of  the  stress-strain  diagram  of  steel. 


CHAPTER  III. 


SHEAR. 

26.  Shear  and  Shearing  Stress.  —  We  have  learned  that  when 
a  body  is  subjected  to  a  pair  of  forces  in  the  same  line,  tensile 
stress  is  produced,  if  the  forces  are  directed  away  from  each  other, 
and  compressive  stress,  if  they  are  directed  towards  each  other. 

If  the  forces  are  in  parallel  lines  or  planes,  shearing  and  bending 
stresses  are  produced  in  the  portion  of  the  body  between  the 
planes  of  the  forces.  In  Fig.  15,  the  block  A  is  securely  held  by 
the  body  B  and  a  horizontal  force  P  is  applied  by  a  second  body 
C.  This  force  P  is  parallel  to  the  upper  surface  of  B.  The  body 
B  exerts  a  horizontal  force  on  the  block  which  is  equal  and  oppo- 
site to  the  force  in  C.  If  we  consider  that  portion  of  the  block  A 
between  the  plane  of  the  upper  surface  of  B  and  the  plane  EFG 
of  the  lower  surface  of  (7,  we  find  that 
it  is  subjected  to  a  pair  of  equal  and 
opposite  forces.  The  material  of  this 
portion  of  the  block  is  subjected  to 
shearing  and  bending  stresses.  The 
shearing  stresses  depend  upon  the  mag- 
nitude of  the  forces  and  the  area  of 
the  section  of  A .  The  bending  stresses 
depend  upon  these  and  also  upon  the 
distance  of  the  forces  apart.  If  the  body 
C  is  brought  very  close  to  B,  so  that  the  distance  between  the 
two  forces  P  and  P'  becomes  negligible,  the  unit  bending  stress 
becomes  small,  while  the  unit  shearing  stress  is  unchanged.  The 
average  unit  shearing  stress  is  calculated  by  dividing  the  force  P 
by  the  area  of  the  cross  section  EFG  or  the  area  of  any  section 
parallel  to  it.  We  notice  that  in  tension  or  compression  we 
divide  the  total  force  by  the  area  of  the  cross  section  perpendicular 
to  its  direction  to  get  the  unit  stress;  while  in  shear  we  divide  the 
total  force  by  the  area  of  the  cross  section  parallel  to  the  forces 
and  perpendicular  to  the  plane  which  includes  the  two  sets  of 
forces. 

37 


Fig.  15.  — Shear  and  Bend- 
ing. 


38 


STRENGTH  OF  MATERIALS 


In  this,  as  in  all  other  cases,  the  line  P  in  the  drawing  represents 
the  resultant  of  a  set  of  forces  distributed  over  an  area.  The 
resultant  P'  must  fall  some  distance  below  the  upper  surface  of 
B,  and  the  resultant  P  must  be  above  the  lower  surface  of  C.  It 
is,  therefore,  not  convenient  by  this  method  to  get  shearing  stress 
entirely  free  from  bending  or  compressive  stress.  We  will  find 
later  that  the  distribution  of  shearing  stress,  when  combined  with 
bending,  is  not  uniform  over  the  section;  but  for  the  present  we 
shall  take  no  account  of  this  variation,  and  shall  calculate  the 
average  shearing  stress  by  dividing  force  by  area. 


PROBLEMS. 

1.  A  1-inch  round  rod  projects  horizontally  from  a  wall.     A  ring  hung  on 
the  rod  supports  a  load  of  5000  pounds.     Find  the  average  unit  shearing  stress. 

Ans.  6366  pounds  per  square  inch. 

2.  A  bar  1  inch  wide  and  f  inch  thick  rests  upon  two  supports  and  carries 
a  load  of  400  pounds  midway  between  them.     Find  the  mean  unit  shearing 
stress. 

3.  A  4-inch  by  4-inch  wooden  block  has  a  notch  2  inches  deep  cut  in  one 
side,  the  edge  of  the  notch  being  6  inches  from  one  end  of  the  block.     A  pull 

of  1800  pounds  parallel  to  its  length  is  applied 
by  means  of  a  second  block  set  in  the  no'tch. 
Find  the  unit  shearing  stress. 

Ans.  75  pounds  per  square  inch. 

4.  If,  in  Problem  3,  the  grain  is  parallel  to 
the  length,  what  is  the  greatest  allowable  pull  for 
an  oak  block?  Ans.  4800  pounds. 

5.  Solve  Problem  4  for  long-leaf  yellow  pine 
and  for  hemlock.       State  your  authority  for 
allowable  unit  stresses. 

6.  A  2-inch  by  4-inch  long-leaf  yellow-pine 
block,  hung  vertical  and  supported  at  the  upper 
end,  has  a  hole  1  inch  square  perpendicular  to 
the  4-inch  face.     The  lower  edge  of  this  hole  is 
4 1  inches  from  the  lower  end  of  the  block.     If  a 
load  of  1800  pounds  is  hung  on  a  rod  passing 
through   this   hole,  what  is  the  unit  shearing 
stress  in  the  timber?     What  is  the  mean  unit 
shearing  stress  in  the  rod  if  the  load  is  symmetri- 
cal (Fig.  16)  ?( 

'Ans.  100  pounds  per  square  inch; 
900  pounds  per  square  inch. 

7.  The  head  of  a  1-inch  bolt  is  £  inch  thick.     Find  the  mean  unit  shear- 
ing stress  tending  to  strip  the  head  from  the  bolt  when  subjected  to  a  pull  of 
10,000  pounds.  Ans.  3640  pounds  per  square  inch. 

8.  In  Problem  7,  what  is  the  unit  tensile  stress  in  the  weakest  part  of  the 
bolt  if  the  pull  is  applied  by  means  of  a  nut? 


Fig.  16.  —  Shear  in  Timber. 


SHEAR 


39 


27.  Shearing  Deformations.  —  Consider  a  portion  D  of  block 
A  of  Fig.  15.  The  portion  extends  through  the  block  with  its 
long  dimension  perpendicular  to  the  plane  which  contains  the 
resultants  P  and  P'.  It  is  represented  on  an  enlarged  scale  by 
the  rectangle  HIJK,  Fig.  17.  When  the  shearing  forces  are 
applied  as  shown  in  Fig.  15,  it  is  distorted  to  the  form  HI'J'K. 
If  we  regard  HK  as  fixed,  the  total  displacement  of  any  point 
in  the  upper  line  is  equal  to  77'  or  JJ'.  The  unit  deformation 
is  the  ratio  of  this  horizontal  displacement,  77',  to  the  vertical 
distance,  HI.  The  unit  shear  is  the  tangent  of  the  angle  7777' 
or  JKJ'.  The  effect  of  the  shearing  forces  is  to  lengthen  the 
diagonal  HJ,  and  shorten  the  diagonal  IK. 


J  j 


H  K 

Fig.  17.  —  Shearing  Deformations. 


o) 

_JO 

| 

t 

1 

6" 

1 

I 

1 

1 

-k 

0) 

_(o 

A 

c 

Fig.  18. 


PROBLEMS. 


1.  Two  equal  bars,  AB  and  CD,  Fig.  18,  are  hinged  to  a  second  pair  of 
equal  bars,  AC  and  BD,  to  form  a  parallelogram.     A  sheet  of  rubber  6  inches 
wide  has  one  edge  securely  clamped  to  AB  and  the  other  edge  to  CD.     The 
length  of  AB,  center  to  center  of  hinges,  is  10  inches.     What  is  the  unit  shear 
when  B  is  displaced  0.2  inch  to  the  right  of  the  vertical  from  its  original  vertical 
position?  Ans.  Unit  shear,  0.02. 

2.  A  hollow  circular  shaft  5  inches  in  diameter  is  subjected  to  a  twisting 
moment,  and  it  is  found  that  two  sections  10  feet  apart  suffer  a  relative  dis- 
placement of  2  degrees.     What  is  the  total  shearing  displacement  of  the  fibers? 
What  is  the  unit  displacement? 

A       (  Total  displacement,  0.0873  inch; 
'  1  Unit  displacement,  0.00073. 

28.  Modulus  of  Elasticity  in  Shear.  —  The  modulus  of  elas- 
ticity in  shear  is  obtained  by  dividing  unit  shearing  stress  by 
unit  shearing  deformation,  just  as  the  modulus  of  elasticity  in 
tension  or  compression  is  computed  by  dividing  unit  stress  by 
unit  deformation.  The  modulus  of  elasticity  in  shear  is  also 
called  the  modulus  of  rigidity.  In  formulas  it  is  represented 
by  Ea,  when  it  is  desirable  to  distinguish  it  from  the  modulus  in 


40 


STRENGTH   OF  MATERIALS 


tension  or  compression.  The  latter  may  be  represented  by  Et 
and  Ec,  respectively.  Forces  applied  as  in  Fig.  15  do  not  give 
pure  shearing  stress.  It  is  only  in  the  case  of  torsion,  as  in 
Problem  2  of  Article  27,  that  we  get  pure  shear. 

PROBLEMS. 

1.  In  Problem  2  of  Article  27,  if  Es  equals  11,000,000,  what  is  the  unit 
shearing  stress?  Ans.  8030  pounds  per  square  inch. 

2.  If  the  modulus  of  elasticity  in  shear  in  a  given  4-inch  circular  shaft  is 
10,500,000,  what  is  the  maximum  allowable  unit  shear,  if  the  allowable  unit 
shearing  stress  is  9000  pounds  per  square  inch?  Ans.  0.000857. 

3.  In  Problem  2,  what  is  the  maximum  angle  of  twist  in  a  length  of  5  feet? 

MISCELLANEOUS  PROBLEMS. 

1.   In  Fig.  19,  A  and  B  are  short  compression  members  or  struts  of  yellow 
pine,  joined  together  at  the  top  by  a  bolt  or  pin  and  held  from  spreading  at 


Fig.  19. 


the  bottom  by  being  set  into  the  notches  in  the  bottom  chord  C.  If  the  load 
P  is  6000  pounds,  what  is  the  unit  compressive  stress  in  A  and  B1  What  is 
the  maximum  unit  tensile  stress  in  C?  What  must  be  the  length  of  the  section 
d  to  avoid  shearing,  if  C  is  made  of  oak?  Ans.  Length  of  d,  6.5  inches. 

2.  In  Problem  1,  what  must  be  the  thickness  of  the  supports  upon  which 
C  rests;  (a)  if  made  of  limestone;  (6)  if  made  of  Douglas  fir  parallel  to  the 
grain?  (c)  if  made  of  Douglas  fir  with  load  perpendicular  to  the  grain? 

3.  What  is  the  force  required  to  punch  a  f-inch  hole  in  a  £-inch  steel  plate 
if  the  ultimate  shearing  strength  of  the  plate  is  40,000  pounds  per  square  inch? 

4.  In  Problem  3,  what  is  the  unit  compressive  stress  in  the  punch? 

5.  A  set  of  punches  are  made  of  steel  having  a  compressive  strength  of 
150,000  pounds  per  square  inch  and  are  used  to  punch  steel  with  a  unit  shear- 
ing strength  of  40,000  pounds  per  square  inch.     What  is  the  smallest  hole 
which  can  be  punched  in  a  f-inch  plate?  Ans.  0.8  inch  diameter. 

6.  If  sc  is  the  compressive  strength  of  the  punch,  and  ss  the  shearing 
strength  of  the  plate,  derive  the  formula  which  will  express  the  relation  between 
the  thickness  of  the  plate  and  the  diameter  of  the  smallest  hole  which  can  be 
safely  punched. 


SHEAR   AND   COMPRESSION 


41 


29.  Shear  Caused  by  Compression  or  Tension.  —  Fig.  20  rep- 
resents a  block  subjected  to  a  compress!  ve  force  P  in  the  direc- 
tion of  its  length  and  an  equal  reaction  at  the  bottom.  Imagine 
the  block  cut  by  a  plane  normal  to  its  length  and  glued  together 
again.  If  we  consider  the  portion  of  the  block  above  the  section 
BCDE  as  a  free  body  and  resolve  vertically,  we  have  the  force 
P  acting  downwards  equal  to  the  upward  reaction  of  the  glued 
surface.  (Neglect  the  weight  of  the  portion  above  BCDE.)  If 
A  is  the  area  of  the  glued  surface,  the  unit  compressive  stress  is 

given  by 

P 


If  we  resolve  horizontally,  that  is,  parallel  to  any  line  in  BCDE, 
all  the  components  of  the  external  force  are  zero  and  the  unit 
shearing  stress  is  zero.  If  the  body  was  actually  made  of  two 
portions,  the  upper  portion  would  not  slide  on  the  lower  portion, 
no  matter  how  smooth  the  surfaces  of  contact  might  be. 


Fig.  20.  —  Section  Normal  to  Force.         Fig.  21.  —  Section  Inclined  to  Force. 

Now  consider  a  similar  body,  Fig.  21,  cut  by  a  plane  BC'D'E' 
which  makes  an  angle  4>  with  the  normal  plane.  Taking  the 
portion  above  the  plane  as  a  free  body,  as  before,  we  will  resolve 
the  external  force  P  perpendicular  and  parallel  to  the  plane. 
The  total  perpendicular  component  is  P  cos  <j>,  and  the  unit 
compressive  stress  is  this  component  divided  by  the  area  of  the 
section.  If  A  is  the  area  of  the  normal  section  BCDE,  the  area 
of  the  inclined  section  is  A  sec  <f>. 

Pcos 


Unit  compressive  stress  = 


P 

=  -    cos 


(1) 


A  sec<£ 

Resolving  parallel  to  the  line  BG,  which  makes  the  maximum 
angle  with  the  normal  plane,  the  component  of  P  is  equal  to 


42 


STRENGTH  OF  MATERIALS 


P  sin  4>.     The  unit  shearing  stress  is  this  component  divided  by 
the  area  of  the  inclined  section. 


Psin0       P  .  P     .    0 

S8   —  -:  -  =  —r  Sin  <£  COS  0  =  TT— - r  Sin  2  </>. 

.A  sec  </>      A  2  A 

The  same  relations  hold  for  tension  and  compression. 


(2) 


PROBLEMS. 

1.  Show  from  equations  (1)  and  (2)  that  shearing  stress  is  zero  and  com- 
pressive  stress  a  maximum  when  <f>  is  zero. 

2.  A  2-inch  by  2-inch  block  is  subjected  to  a  compressive  force  of  1800 
pounds  in  the  direction  of  its  length.     Find  the  unit  compressive  and  unit 
shearing  stresses  with  respect  to  a  plane  making  an  angle  of  20  degrees  with 
the  normal  cross  section.  ^ng  (  sc,  397  pounds  per  square  inch; 

'  (  ss,  145  pounds  per  square  inch. 

3.  A  4-inch  by  4-inch  long-leaf  yellow-pine  post  has  the  grain  15  degrees 
with  its  axis.     What  is  the  total  safe  load  considering  the  shear  parallel  to 
the  grain,  using  the  values  recommended  by  the  Association  of  Railway 
Superintendents  of  Buildings  and  Bridges?  Ans.  9600  pounds. 

4.  Prove  that  the  unit  shearing  stress  produced  by  a  single  tensile  or  com- 
pressive load  is  a  maximum  at  45  degrees  with  the  direction  of  the  load,  and 
that  the  maximum  unit  shearing  stress  is  one-half  of  the  compressive  or  tensile 
stress. 

30.  Shearing  Forces  in  Pairs.  —  Shearing  forces  applied  as  in 
Fig.  15  do  not  give  pure  shearing  stress.  If  we  consider  the 
block  shown  in  vertical  section  in  Fig.  22,  the  force  P  at  the  left 


\*-b 


Fig.  22. 


Fig.  23.  —  Equilibrium  in  Shear. 


will  turn  it  over,  unless  it  is  held  down  at  C  or  at  some  other 
point  to  the  left  of  F.  If  it  is  held  down  at  C,  there  is  an  equal 
and  opposite  force  at  F,  the  two  forces  at  C  and  F  forming  a 
couple  of  moment  equal  to  that  of  the  horizontal  forces  P  and 
P'.  Take  a  small  block  of  rectangular  section,  of  height  h  and 
breadth  6,  running  through  the  large  block  A  perpendicular  to 
the  plane  of  the  paper  (Figs.  22,  23).  Let  I  be  the  length  of  this 
small  block  perpendicular  to  the  plane  of  the  paper  in  Fig.  22, 
and  perpendicular  to  the  planes  of  all  the  applied  forces.  The 
top  and  bottom  surfaces  have  areas  bl  each.  The  area  of  the  left 
vertical  surface  is  hi.  Let  there  be  a  shearing  force  of  s  pounds 


SHEAR  43 

per  square  inch  directed  towards  the  right  in  the  upper  surface, 
and  a  stress  of  equal  intensity  in  the  lower  surface  directed  towards 
the  left.  The  total  shearing  force  in  each  of  these  surfaces  is 
sbl,  and  the  moment  of  the  couple  tending  to  turn  the  block  in  a 
clockwise  direction  is  the  product  of  one  of  these  total  forces  mul- 
tiplied by  the  distance  between  their  planes,  or  sblh.  If  the  block 
is  subjected  to  shearing  forces  only,  there  must  be  a  shear  down- 
ward at  the  left  surface  and  a  shear  upward  at  the  right  surface, 
to  prevent  rotation.  If  s'  is  the  intensity  of  one  of  these  verti- 
cal shearing  stresses,  the  total  force  on  each  of  these  vertical 
surfaces  is  the  product  of  s'  multiplied  by  the  area  hi',  and  the 
moment  is  s'hlb.  The  moments  of  the  horizontal  and  vertical 
forces  are  equal,  if  the  block  is  in  equilibrium. 

sbhl  =  s'hlb,     s  =  sf.  Formula  III. 

When  a  body  or  portion  of  a  body  is  subjected  to  pure  shear- 
ing stresses,  these  stresses  occur  in  pairs  at  right  angles  to  each 
other,  and  the  unit  shearing  stress  is  the  same  in  both  pairs. 

If  forces  are  applied  to  a  block  as  in  Fig.  22,  with  a  downward 
pull  at  C,  and  an  upward  push  at  Ft  a  small  block  of  section  bh 
will  be  subjected  to  pure  shear  if  it  is  in  the  vertical  plane  mid- 
way betwee'n  C  and  F.  If  the  small  block  is  located  to  the  left 
of  the  middle,  there  will  be  shear  combined  with  tension.  If  it 
is  on  the  right  side,  there  will  be  shear  and  compression. 

31.  Compressive  and  Tensile  Stress  Caused  by  Shear.  — 
Fig,  24  I,  represents  a  rectangular  parallelepiped  of  breadth  b, 
height  h,  and  length  I,  sub- 
jected to  pure  shearing 
stress.  The  shearing  stress 
acts  toward  the  right  paral- 
lel to  the  breadth  at  the  top 
and  toward  the  left  at  the 
bottom.  As  shown  in  Arti- 
cle 30,  there  is  also  a  shear-  c^  6._=y 
ing  stress  of  the  same  inten-  i 

sity  at  the  left  surface  acting      Fig.  24.  —  Shear  Causing  Compression, 
downward    and    an    equal 

shearing  stress  at  the  right  surface  acting  upward.  (If  the  di- 
rection of  one  of  these  shears  is  reversed,  they  must  all  be  re- 
versed to  produce  equilibrium.)  Now  consider  the  parallelepiped 


44 


STRENGTH  OF  MATERIALS 


divided  by  the  inclined  plane  containing  the  edges  CD  and  GF, 
and  treat  the  triangular  prism  to  the  left  of  this  plane  as  a  free 
body  in  equilibrium  under  the  action  of  the  forces  at  its  surface. 
These  forces  are  four  in  number:  the  shearing  force  H  in  the 
upper  surface  acting  toward  the  right,  the  shearing  force  V  in  the 
left  vertical  surface  acting  downward,  the  compressive  force  N 
acting  normal  to  the  inclined  surface  (Fig.  24,  II,  which  repre- 
sents all  the  forces  in  the  plane  of  the  paper),  and  a  shearing 
force  T  along  this  surface  parallel  to  the  diagonal  line  CG.  If 
s8  is  the  intensity  of  the  horizontal  and  vertical  shear, 

H  =  ssbl,     V  =  sshl 
Resolving  normal  to  the  inclined  plane, 

N  =H*mO+VcosO,  (1) 

N  =  sabl  sin  0  +  sshl  cos  6',  (2) 

where  6  is  the  angle  which  the  inclined  plane  makes  with  the 

horizontal  surface.     Since 

tan  6=7, 
o 

N  =  2  sabl  sin  0.  (3) 

To  get  the  unit  compressive  stress,  sc,  across  the  inclined  sur- 
face, divide  the  total  compression  N  by  the  area  of  the  surface 
U  sec  0: 

sc  =  2  s8  sin  0  cos  6  =  sa  sin  2  0.  (4) 

When  0  is  45  degrees,  we  get  the  maximum  value  of  the  compres- 
sive stress, 

sc  =  ss.  Formula  IV. 

In  like  manner,  if  we  consider  a  second  inclined  plane  perpendic- 
ular to  CG  and  parallel  to  CD,  we  get  a  tensile  stress  of  the  same 
value. 

ICOMP. 


f 

Fig.  26. 

When  a  body  is  subjected  to  pure  shear,  there  is  a  compressive 
stress  of  equal  intensity  across  planes  at  45  degrees  to  the  direc- 


SHEAR 


45 


tion  of  the  shearing  stresses,  and  tensile  stresses  of  the  same  inten- 
sity across  planes  at  45  degrees  to  the  shearing  planes  in  the 
opposite  directions.  This  is  shown  in  Fig.  25. 


PROBLEMS. 

1.  Prove  that  a  block,  subjected  to  a  compressive  stress  of  intensity  s  and  a 
tensile  stress  of  the  same  intensity  at  right  angles,  has  a  shearing  stress  of  the 
same  intensity  at  45  degrees  and  135  degrees  (Fig.  26). 

2.  A  2-inch  by  2-inch  block  is  cut  at  an  angle  of  14  degrees  and  glued  to- 
gether again.     If  a  pull  of  400  pounds  is  applied  parallel  to  its  length  (at  14 
degrees  with  the  glued  section),  what  is  the  unit  shearing  stress  in  the  glue? 
what  is  the  unit  tensile  stress  in  the  glue? 

Ans  J  Ss'  2^  P°uncls  Per  square  inch; 
;,  5.85  pounds  per  square  inch. 


Fig.  27. 

3.  If  a  compressive  force  of  800  pounds  is  applied  transversely  to  a  length 
of  8  inches  of  the  block  of  Problem  2  (Fig.  27),  while  the  longitudinal  stress 
is  acting,  find  the  unit  shearing  and  normal  stress  in  the  glue. 

Ans  \  Ss'  ^'2  Poun(^s  Per  sQuare  inch; 
'  1  sc,  41.1  pounds  per  square  inch. 

4.  Solve  Problem  3,  if  the  400  pounds  is  changed  to  compression. 


Fig.  28.  —  Shearing  Deformation. 

The  modulus  of  shearing  elasticity  may  be  calculated  from  the  modulus  in 
tension  or  compression  if  Poisson's  ratio  is  known. 

Fig.  28  is  the  front  elevation  of  a  block  of  square  section  subjected  to  shear- 
ing forces.  The  unit  shearing  displacement  is  the  tangent  of  the  angle  0 
between  the  lines  IH  and  I'H  of  Fig.  28,  II.  In  Fig.  28,  III,  we  have  kept  the 


46  STRENGTH   OF  MATERIALS 

directions  of  the  diagonals  constant  instead  of  the  base  line,  as  in  II,  so  that 

f\ 

the  line  HK'  makes  an  angle  ^  with  the  horizontal,  and  the  line  HI'  makes 
an  equal  angle  with  the  vertical.     We  will  determine  the  magnitude  of  this 

t\ 

angle  ^  • 

When  the  shearing  force  acts  on  the  body,  the  diagonal  HJ  is  lengthened 
to  HJ'  and  the  diagonal  IK  is  shortened  to  I'K'.  The  half-diagonals  HM  and 
MK,  originally  equal,  suffer  a  proportional  deformation. 

If  5  is  the  unit  linear  deformation  due  to  a  unit  direct  stress  s,  the  unit 
deformation  along  the  diagonal  HJ'  is  5  (1  +p).  This  is  made  up  of  the 
elongation  5  due  to  the  tension  along  this  diagonal  and  the  elongation  p5  due 
to  the  equal  compression  along  the  other  diagonal.  In  a  similar  way  the  unit 
deformation  along  the  diagonal  I'K  is  —  5  (1  +  p). 

The  tangent  of  the  angle  <£  which  the  line  HK'  (Fig.  28,  III)  makes  with 
the  diagonal  is  given  by: 

M'K'  =  MK  [1  -  5  (1  +  P)]  =  1  -  8  (1  +  P) 
:  HM' 


To  get  the  angle  -  we  will  subtract  the  angle  0  from  45  degrees. 

tan  45°  - 


For  small  angles,  tan  6  =  2  tan  =  =  2  5  (1  +  p). 
Since 


Es         s  E 


tan  0      2  (1  +  p) 
When  P  =  \,    Es=^. 

PROBLEMS. 

5.  If  E  for  steel  is  29,000,000  and  Poisson's  ratio  is  0.27,  find  the  modulus 
of  shearing  elasticity.  Ans.  E8  =  11,400,000. 

6.  If  E  is  30,000,000  and  Es  is  11,600,000,  find  Poisson's  ratio. 

32.  Methods  of  Failure.  —  We  have  shown  the  method  of  fail- 
ure of  soft  steel  in  tension.  It  necks  and  finally  breaks  normal 
to  the  length.  One  portion  is  slightly  concave,  especially  at  the 
outside,  showing  that  the  final  failure  here  is  by  shearing.  A  rod 
of  hard  steel  will  sometimes  shear  off  at  nearly  45  degrees  with 
the  direction  of  its  length.  In  Article  29  we  found  that  a  tensile 
or  compressive  stress  produces  a  shearing  stress  which  has  its 
maximum  value  at  45  degrees,  at  which  direction  it  is  one-half 


FAILURE 


47 


the  direct  stress.  If  any  material  is  less  than  half  as  strong  at  45 
degrees  in  shear  as  it  is  in  tension  or  compression  in  the  direction 
of  the  stress,  it  will  fail  by  shear.  As  the  shearing  stress  varies 
slowly  near  45  degrees,  the  direction  of  failure  may  differ  consider- 
ably from  that  angle.  If  the  shearing  strength  at  30  degrees  or 
at  60  degrees  is  less  than  86  per  cent  of  the  strength  at  45  degrees, 
failure  may  occur  along  one  of  these  direc- 
tions. Timber  has  a  small  shearing  strength 
parallel  to  the  growth  rings  and  a  larger 
strength  at  right  angles  to  them.  Fig.  29 
shows  a  small  piece  of  timber  which  was 
tested  in  tension.  The  lines  of  fracture  are 
oblique,  the  shear  taking  place  first  in  one 
direction  and  then  in  the  other. 

In  compression,  materials  which  are  not 
ductile  may  fail  by  shearing  at  about  45  de- 
grees with  the  direction  of  the  stress  or  by 
splitting  longitudinally.  The  shearing  takes 
place  as  in  tension  when  the  unit  compres- 
sive  stress  exceeds  twice  the  unit  shearing 
strength  at  45  degrees;  the  splitting  longi- 
tudinally depends  upon  the  relative  values  of 
the  compressive  strength  in  the  direction  of 
the  length,  the  tensile  strength  normal  to  the 
length,  and  Poisson's  ratio.  For  instance,  if 
Poisson's  ratio  for  concrete  is  0.2,  a  unit  com- 
pressive stress  of  2000  pounds  per  square  inch 
will  cause  a  unit  tensile  stress  of  400  pounds 
per  square  inch,  provided  the  moduli  of  elas- 
ticity in  tension  and  compression  are  the  same. 
Under  such  conditions  concrete  may  fail  by 
tension.  If  the  concrete  is  tested  by  com-  Fig<  29*  Timber  in 
pression  between  steel  plates  in  the  testing 
machine,  the  friction  of  the  plates  tends  to  prevent  expansion  at 
the  ends.  Fig.  30  shows  the  behavior  of  two  4-inch  by  4-inch' 
blocks  of  1 : 1  cement  mortar  under  a  compressive  load  parallel 
to  the  long  dimension.  Both  failed  by  splitting  lengthwise, 
and  both  sheared  to  form  a  pyramid-shaped  block  at  one  end. 
Cement  and  stone  cubes  usually  fail  by  shear.  Fig.  31  shows 
the  failure  of  two  paving  bricks  in  compression. 


48 


STRENGTH  OF  MATERIALS 


In  Fig.  32  the  short  block  illustrates  shear.  Only  short  blocks 
of  very  uniform  structure  will  fail  in  this  way  (by  shear  entirely 
across  the  section  in  one  plane).  Most  blocks  fail  by  a  combi- 


Fig.  30.  —  Cement  in  Compression. 

nation  of  shear  and  splitting,  as  shown  in  the  other  cases.  In 
fact,  all  of  these  are  .selected  samples.  Generally  the  shear  planes 
will  run  for  only  a  short  distance  and  then  split  or  run  the  other 
way. 

PROBLEMS. 

1.  The  white-oak  block  on  the  right  in  Fig.  32  was  originally  1.83  inches 
by  1.82  inches.  The  ultimate  load  was  19,314  pounds.  What  was  the  unit 
shearing  stress  at  45  degrees  to  the  normal  section?  what  was  it  at  35  degrees 
and  at  55  degrees  to  that  section? 


FAILURE 


49 


2.  A  white-oak  block  1.8  inches  square  was  cut  so  that  the  growth  rings 
made  an  angle  of  10  degrees  with  the  plane  of  the  ends.  When  tested  in  com- 
pression it  failed  by  shearing  along  the  growth  rings  under  a  load  of  4700 
pounds.  What  was  the  unit  shearing  strength  parallel  to  the  grain? 

Plastic  material  in  compression  may  expand  almost  indefinitely 
under  compression.  If  less  plastic,  there  is  considerable  expan- 
sion and  longitudinal  splitting. 

Fig.  33  shows  two  pieces  of  wrought-iron  tubing  which  failed 
in  this  way.  A  solid  rod,  unless  very  short,  with  the  resultant 
force  exactly  central,  will  fail  by  bending,  as  in  the  left  of  Fig.  33. 
If  too  short  to  bend  it  will  expand  indefinitely. 


Fig.  31.  —  Hard  Brick  in  Compression. 

33.  Bearing  Strength;  Failure  by  Cutting.  —  The  bearing 
strength  of  a  solid  depends  upon  the  relative  size  of  the  surface 
of  contact  and  the  entire  climensions  of  the  body.  In  the  treat- 
ment of  bearing  stress,  there  are  two  limiting  cases.  The  first 
is  that  shown  in  Fig.  34,  in  which  the  surface  of  contact  is  equal 
to  the  entire  cross  section  of  the  body  B,  and  the  length  in  the 
direction  of  the  applied  force  is  at  least  equal  to  the  thickness  of 
the  body.  In  this  case  the  bearing  strength  is  equal  to  the  com- 
pressive  strength.  Used  in  this  way,  a  soft  material  like  babbitt 
metal  would  have  little  bearing  strength.  Fig.  35  shows  the 
second  case.  Here  the  load  is  applied  to  a  small  portion  of  a 
body  which  is  of  unlimited  extent  or  is  confined  laterally  by 


50 


STRENGTH  OF  MATERIALS 


YELLOW  PINE 
IN    COMPRESSION 

/.66"x  i.ea" 

LOAD   /7,OOO* 


Fig.  32.  —  Timber  in  Compression. 


Fig  33.  —  Metal  in  Compression. 


BEARING   STRENGTH 


51 


another  body.  The  portion  outside  of  the  loaded  area  acts  as 
a  hoop  to  prevent  the  lateral  expansion.  In  this  form,  a  body 
composed  of  separate  particles  may  have  considerable  bearing 
strength,  depending  upon  the  friction.  Dry  sand  is  an  example. 


Fig.  34. 


Fig.  35. 


In  a  mass  of  wheat  or  flaxseed,  where  the  friction  is  smaller,  the 
bearing  strength  is  less. 

Fig.  36  shows  two  cases  intermediate  between  Figs.  34  and  35. 


t 


Fig.  36. 

Cutting  with  a  knife  or  chisel  depends  upon  the  bearing 
strength  of  the  tool  and  of  the  material  cut.  The  bearing  strength 
of  the  tool  under  the  conditions  of  Fig.  34  must  be  greater  than 


ii 


in 


Fig.  37.  —  Cutting. 


that  of  the  material  under  the  conditions  of  Fig.  35.  At  first 
there  is  a  depression  in  the  material  under  the  edge  of  the  tool, 
as  shown  in  Fig.  37,  I.  When  the  stress  in  the  material  passes 
the  bearing  strength,  it  is  permanently  pushed  back.  In  a  plas- 


52 


STRENGTH  OF  MATERIALS 


tic  nonporous  material,  some  of  the  substance  is  forced  up  by  the 
pressure,  as  shown  in  Fig.  37,  II.  In  a  porous  body  like  wood 
there  is  an  increase  in  density  adjacent  to  the  cutting  surface. 
The  wheel  of  a  wagon  cutting  in  soft  earth  illustrates  both  cases. 
If  the  earth  is  wet  clay,  we  have  an  illustration  of  the  plastic  non- 
porous  substance;  if  it  is  dry  loam,  it  approaches  the  other  case. 

When  a  cutting  tool  has  penetrated  a  little  distance,  it  acts 
as  a  wedge  and  exerts  a  tensile  stress  upon  the  material  in  front 
of  its  edge.  This  is  shown  in  Fig.  37,  III. 


Fig.  38.  —  Cutting  with  Shears. 


Fig.  39. 


Fig.  38  shows  the  behavior  of  a  pair  of  scissors  or  shears.  At 
the  beginning,  the  cutting  is  due  to  the  bearing  stress  on  the 
cutting  edges,  as  shown  in  Fig.  38,  I.  As  the  edges  penetrate 


a  m 


Fig.  40.  —  Slugs  Punched  from  Steel  Plates. 

into  the  material  the  bearing  force  is  increased  at  each  blade. 
These  forces  produce  shearing  stresses  in  all  portions  of  the  body 
in  the  plane  of  the  cutting  edges.  The  corresponding  shearing 
deformations  are  shown  by  the  dotted  lines  in  Fig.  38,  II.  Fig.  39 
represents  the  punching  of  a  metal  plate.  The  plate  is  bent 
a  little  at  first,  which  makes  the  surface  of  contact  a  narrow 
ring  at  the  edge  of  the  punch  and  die.  When  the  compressive 


PUNCHING  53 

stress  on  these  rings  exceeds  the  bearing  strength  of  the  plate, 
cutting  begins.  This  is  followed  by  shear,  as  in  the  case  of  cut- 
ting with  scissors. 

Fig.  40  shows  some  of  the  slugs  punched  from  steel  plates. 
Notice  the  curvature  at  the  ends.  In  the  case  of  the  small  diam- 
eter compared  with  the  length,  the  punch  failed  after  making 
about  a  dozen  holes. 


CHAPTER  IV. 
RIVETED   JOINTS. 

34.  Kinds  of  Stress.  —  Riveted  joints  afford  an  excellent  illus- 
tration of  tension,  compression,  and  shear,  and  of  the  manner 
of  transmission  of  stress.  Fig.  41  represents  a  pair  of  plates, 
each  of  breadth  b  and  thickness  t,  transmitting  a  pull  P  in  the 


y 


Fig.  41.  —  Stress  at  a  Bolted  Joint. 

direction  of  their  length.  The  plates  are  united  by  means  of  a 
pin  C,  which  fits  tightly  in  a  hole  in  the  lower  plate  and  passes 
through  a  hole  in  the  upper  plate.  If  we  consider  the  upper 
plate,  we  find  that  the  portion  to  the  left  of  the  pin  is  in  tension. 
The  intensity  of  this  tensile  stress  is  found  by  dividing  the 
pull  P  by  the  area  bt.  At  the  section  EH  in  the  plane  of  the 
center  of  the  hole,  the  stress  is  still  tension.  The  unit  stress  is 
greater  here,  for  the  area  of  cross  section  is  diminished  by  the 
material  cut  away  to  make  room  for  the  pin.  If  the  hole  is  in 
the  middle  of  the  section  and  in  the  line  of  the  pull,  half  of 
the  total  stress  is  transmitted  by  the  lower  section  EF  and  half 
by  the  upper  section  GH.  The  total  stress  which  passes  EF  as 

tension  passes  FK  as  shear.      The  intensity  of   this  shearing 

p 
stress  in  the  plate  may  be  calculated  by  dividing  the  pull,  —  > 

by  the  section  of  length  FK  and  thickness  t.     At  M,  the  surface 

54 


RIVETED   JOINTS 


55 


of  contact  of  the  pin  and  plate,  the  stress  is  compression.  The 
force  is  transmitted  as  a  shearing  stress  from  the  part  of  the 
pin  in  the  upper  plate  to  the  portion  in  the  lower  plate,  and 
finally  as  compression  to  the  lower  plate. 

It  helps  to  fix  our  ideas  if  we  regard  stress  as  flowing  like  an 
electric  current.     This  is  illustrated  in  Fig.  42.     We  may  regard 


Fig.  42.  —  Flow  of  Stress. 

the  circuit  as  closed  through  the  bodies  which  exert  the  pull 
on  the  plates. 

PROBLEMS. 

1.  The  plates  in  Fig.  41  are  each  2  inches  wide  and  £  inch  thick.     The  hole 
in  the  upper  plate  is  f  inch  in  diameter  and  that  in  the  lower  plate  is  f  inch 
in  diameter.     The  bolt  is  |  inch  in  diameter.     The  pull  is  4000  pounds.     Find 
the  unit  shearing  stress  in  the  bolt  and  the  maximum  tensile  stress  in  each  plate. 

M       (  ss,  9054  pounds  per  square  inch; 

(  st,  7111,  6400  pounds  per  square  inch. 

2.  In  Fig.  41  the  bolt  is  f  inch  in  diameter  and  exactly  fits  in  the  lower 
plate.     The  lower  plate  is  2  inches  wide.     What  must  be  its  thickness  in  order 
that  the  maximum  unit  tensile  stress  shall  be  equal  to  the  unit  shearing  stress 
in  the  bolt  joining  the  plates? 

In  calculating  the  unit  shearing  stress  in  the  plates  behind  the 
bolt  or  pin,  since  there  is  some  uncertainty  as  to  the  width  of 
the  bearing  surface  at  M ,  it  is  customary  to  take  the  distance 
MN  instead  of  FK  in  getting  the  shear  area. 


PROBLEM. 

3.   In  Problem  1  find  the  unit  stress  in  shear  in  the  upper  plate  to  the  right 
of  the  pin,  if  the  center  of  the  hole  is  1.5  inches  from  the  edge  of  the  plate. 

Ans.  3765  pounds  per  square  inch. 


56 


STRENGTH  OF  MATERIALS 


35.  Bearing  Stress.  —  In  calculating  the  unit  bearing  or  com- 
pressive  stress  at  the  surface  of  contact  of  the  pin  and  plate, 
it  is  customary  to  regard  the  bearing  area  as  the  product  of  the 
thickness  of  the  plate  multiplied  by  the  diameter  of  the  pin.  If 
d  is  the  diameter  of  the  pin  and  t  is  the  thickness  of  the  plate, 
the  bearing  area  is  td.  In  other  words,  it  is  the  projection  upon 
a  plane  parallel  to  the  axis  of  the  pin  of  that  portion  of  the  pin 
which  is  inside  of  the  plate.  Consider  Fig.  43,  in  which  a  rec- 


Fig.  43.  —  Bearing. 


Fig.  44.  —  Bearing. 


tangular  bar  of  thickness  d  is  placed  across  the  edge  of  a  plate 
of  thickness  t.  If  the  bar  crosses  the  plate  at  right  angles,  it 
is  plain  that  the  area  of  contact  is  td.  If,  as  in  Fig.  44,  the 
bar  passes  through  a  hole  in  the  plate,  the  bearing  area  is  the 
same;  and  if  the  forces  PI,  P2  are  balanced  with  respect  to  the 
center  of  the  plate,  the  bearing  stress  is  uniform  over  the  entire 


Fig.  45.  —  Bearing.  Fig.  46.  —  Bearing. 

area.  If  the  forces  are  not  balanced,  the  area  remains  the  same 
and  the  average  bearing  stress  is  the  same,  but  the  maximum 
stress  is  greater.  If  there  is  force  on  only  one  side  of  the 
plate,  the  maximum  bearing  stress  will  be  less  the  smaller  the 
distance  between  the  force  and  plate.  In  Fig.  45  we  have  a 
round  pin  or  bolt  passing  through  a  plate.  The  actual  area  is 
the  lower  half  of  the  surface  of  the  cylinder,  of  length  t  and 


RIVETED   JOINTS  57 

diameter  d.  The  reactions  R\,  R2,  etc.,  are  not  all  vertical,  but 
are  nearly  normal  to  the  surface  of  contact.  If,  as  in  the  case 
of  liquid  pressure,  these  reactions  were  exactly  normal  and  of 
equal  intensity,  the  resultant  of  their  vertical  components  would 
be  the  same  as  if  that  unit  pressure  were  exerted  on  the  hori- 
zontal projection  of  this  cylindrical  surface. 

Consider  a  rivet  in  the  form  of  an  isosceles  triangle  of  base  b 
and  equal  sides  c  (Fig.  46),  subjected  to  a  load  P  perpendicular 
to  the  side  b.  If  B  is  one-half  of  the  angle  opposite  b  and  N 
is  the  resultant  normal  force  on  one  side  c,  we  have  by  vertical 
resolutions : 

2  N  sin  0  =  P. 

The  unit  normal  pressure  on  each  side  is  found  by  dividing  N  by 
the  area  of  contact  tc.  Substituting  for  c  its  value  in  terms  of 

b  and  0,  we  get: 

p 

Unit  stress  =  •*-.  • 
ot 

This  is  on  the  ^assumption  that  there  is  no  friction  and  that  the 
bearing  stress  is  uniform. 

From  these  considerations  we  are  warranted  in  assuming  that 
the  unit  bearing  stress  on  a  pin  which  accurately  fits  the  hole 
is  obtained  by  regarding  the  projection  of  the  curved  surface  on 
a  plane  as  the  surface  of  contact.  This  is  the  common  practice 
of  engineers  in  computing  rivets. 

PROBLEMS. 

1.  In  Problem  1  of  Article  34,  what  is  the  unit  bearing  stress  on  the  pin? 

Ans.  10,667  pounds  per  square  inch. 

2.  In  Fig.  41  the  diameter  of  the  bolt  and  of  the  hole  in  the  lower  plate  is 
|  inch.     What  must  be  the  thickness  and  width  of  this  plate  in  order  that  the 
unit  shearing  stress  in  bolt  shall  be  one-half  of  the  unit  bearing  stress  and 
two-thirds  of  the  unit  tensile  stress  in  the  net  section  of  the  plate? 

36*  Lap  Joint  with  a  Single  Row  of  Rivets.  —  Fig.  47  shows 
a  lap  joint  with  a  single  row  of  rivets.  In  any  riveted  joint  the 
distance  a  from  center  to  center  of  adjacent  rivets  in  a  row  is 
called  the  pitch.  In  solving  problems,  it  is  often  convenient  to 
consider  a  single  strip  of  width  a  alone.  In  this  case  the  prob- 
lem of  a  lap  joint  with  a  single  row  of  rivets  becomes  the  same  as 
that  of  Article  34.  We  may  take  this  strip  as  extending  from 
center  to  center  of  adjacent  rivets,  as  in  the  lower  part  of  Fig.  47. 
In  this  case,  the  total  tension  is  transmitted  in  the  plate  between 


58 


STRENGTH  OF  MATERIALS 


the  two  rivets,  and  the  shear  is.  equally  divided  between  the 
upper  half  of  the  lower  rivet  and  the  lower  half  of  the  upper 
rivet.  Or  we  may  take  the  strip  as  in- 
cluding a  single  rivet,  as  in  the  upper  portion 
of  Fig.  47,  in  which  case  the  shear  is  trans- 
mitted by  a  single  rivet  and  the  tension  is 
divided. 

In  problems  in  riveting,  unless  otherwise  stated,  we 
shall  consider  the  rivet  as  exactly  filling  the  rivet  hole, 
and  that  the  holes  are  drilled  or  reamed  so  that  there 
is  no  injured  material  around  them  due  to  overstrain 
while    punching.      In   practice,   where  the   holes   are 
punched   and  not  reamed  it  is  customary   to  make 
Fig.  47. — Lap  Joint     some  allowance  for  this  injured  material, 
with  Single  Row          In  problems  where  the  width  of  the  plate  is  given, 
of  Rivets.  it  is  generally  better  to  consider  the  entire  plate  as 

a    unit.      In    Problem    1    below,    to    get    th^e    tensile 

stress  we  may  take  a  strip  4.25  inches  wide  transmitting  a  pull  of  15,000 
pounds. 


PROBLEMS. 

1.  Two  ^-inch  plates,  each  8  inches  wide,  are  united  by  five  f-inch  rivets 
in  a  single  row  to  form  a  lap  joint.     The  joint  transmits  a  pull  of  15,000  pounds. 
Find  the  unit  stress  in  the  gross  section  of  the  plates,  the  unit  tensile  stress  in 
the  net  section  between  the  rivets,  the  unit  shearing  stress  in  the  rivets,  and 
the  unit  bearing  stress. 

/  st,  7059  pounds  in  net  section; 
Ans.  <  ss,  6790  pounds; 
'  sc,  8000  pounds. 

2.  Two  £-mch  plates  are  united  to  form  a  lap  joint  by  a  single  row  of 
rivets.     The  pitch  is  3  inches  and  the  diameter  of  the  rivets  is  |  inch.     Find 
the  unit  tensile  stress  in  the  net  section  when  the  unit  shearing  stress  in  the 
rivets  is  7068  pounds  per  square  inch. 

3.  Two  f-inch  plates  are  united  by  a  single  row  of  1-inch  rivets  to  form  a 
lap  joint.     What  must  be  the  pitch  if  the  unit  shearing  stress  in  the  rivets  shall 
be  6000  pounds  per  square  inch  when  the  unit  tensile  stress  in  the  net  section 
of  the  plates  is  8000  pounds  per  square  inch,  and  what  is  the  unit  bearing 
stress?  Ans.  1.94  inches;  7540  pounds  per  square  inch. 


37.  Butt  Joint.  —  Fig.  48  represents  a  butt  joint,  with  double 
cover  plates,  with  a  single  row  of  rivets  on  each  side.  In  a  butt 
joint  with  double  cover  plates  the  rivets  are  in  double  shear. 
In  all  other  respects  the  problem  is  the  same  as  that  of  the  lap 
joint.  A  butt  joint  with  a  single  cover  plate  is  the  same  as  a 
pair  of  lap  joints  placed  tandem. 


RIVETED   JOINTS 


59 


PROBLEM. 

1.  Two  f-inch  plates  are  united  to  form  a  butt  joint  by  means  of  two  fV 
inch  cover  plates.  There  is  one  row  of  f-inch  rivets  on  each  side.  What  must 
be  the  pitch  if  the  tensile  stress  in  the  net  section  of  the  £-mch  main  plates 
between  the  rivets  shall  be  8000  pounds  per  square  inch  when  the  shearing 
stress  in  the  rivets  is  6000  pounds  per  square  inch?  What  is  the  bearing  stress 
between  rivets  and  ^-inch  plates?  also  between  rivets  and  cover  plates? 

Ans.  Pitch,  2.07  inches;  sc,  14,137  and  11,310  pounds  per  square  inch. 

Fig.  49  represents  a  set  of  tests  at  the  Watertown  Arsenal  in 
1885,  to  study  the  behavior  of  riveted  joints.  A  plate  of  width  6 
and  thickness  t  was  planed  down  for  a  portion  of  its  length  to 
some  convenient  width  and  united  to  a  pair  of  cover  plates,  thus 
forming  one-half  of  a  butt  joint.  Wrought-iron  rivets  were  used 
of  nominal  diameter  one-sixteenth  of  an  inch  less  than  the  diam- 
eter of  the  holes.  In  calculating  it  was  assumed  that  the  fin- 
ished rivets  entirely  filled  the  rivet  holes. 


S3- 


/t\ 


ATE]  !          I 


Fig.  48.  —  Butt  Joint  with  Single  Row 
of  Rivets  on  Each  Side. 


Fig.  49.  —  Half  of  Butt 
Joint. 


PROBLEMS. 

2.  In  test  piece  No.  1353  (Watertown  Arsenal,  1885,  page  867),  the  breadth 
b  was  14.90  inches;  the  tested  width,  14.39  inches;  the  actual  thickness  of  the 
plate,  0.248  inch.  There  were  five  rivets  in  inch  drilled  holes.  The  joint 
failed  by  tension  along  the  line  of  the  rivet  holes  under  a  pull  of  156,440  pounds. 
The  calculated  results  as  published  are: 

AREAS.  Square  inches. 

Gross  sectional  area  of  plate 3 . 569 

Net  sectional  area  of  plate 2 . 329 

Bearing  surface  of  rivets 1 . 240 

Shearing  area  of  rivets 7 . 854 

MAXIMUM  STRESS  ON  JOINT.  Pounds  per  sq.  in. 

Tension  in  gross  section  of  plate 43,830 

Tension  on  net  section  of  plate 67,170 

Compression  on  bearing  surface  of  rivets 126,160 

Shearing  on  rivets 19,920 

Verify  these  results. 


60  STRENGTH  OF  MATERIALS 

3.   In  test  piece  No.  1355  the  results  were: 

Tested  width  of  plate 15  inches. 

Actual  thickness 0.251  inch. 

Ultimate  load 167,200  pounds. 

There  were  five  rivets  in  1-inch  holes.  "Fractured  two  outside  sections  of 
plate  at  edge  along  line  of  riveting;  the  two  middle  sections  sheared  in  front 
of  the  rivets." 

Compute  all  unit  stresses  as  in  Problem  2. 

Fig.  50  is  a  copy  of  a  photograph  of  this  plate  after  failure. 
It  shows  failure  by  tension  across  the  net  section  and  shear  in 
front  of  the  rivets.  It  also  shows  elongation  of  the  rivet  holes 
due  to  bearing  pressure  on  the  plate,  combined  with  shear. 


Fig.  50.  —  Failure  of  Riveted  Plate.  f, 

In  order  to  compare  the  strength  of  the  material  in  the  net 
section  of  a  riveted  joint  with  the  ordinary  tension  tests,  two 
strips  were  sheared  from  each  sheet  of  steel,  one  lengthwise,  the 
other  crosswise  the  sheet.  These  were  planed  to  a  width  of  1.5 
inches  and  tested  in  the  usual  way. 

From  the  sheet  used  in  No.  1353  two  test  pieces  were  taken. 
These  gave  as  ultimate  tensile  strengths: 

Pounds  per  sq.  in. 

No.  1213,  lengthwise 59,180 

1224,  crosswise 60,840 

Four  test  strips  were  taken  from  the  sheet  used  for  No.  1355: 

Pounds  per  sq.  in. 

No.  1214,  lengthwise 58,680 

1220,  "        62,300 

1225,  crosswise 61,230 

1226,  "          60,890 


RIVETED   JOINTS  61 

Comparing  these  results  with  the  unit  stresses  in  the  net  sec- 
tion of  the  riveted  plates,  we  find  that  the  stress  in  the  test  pieces 
is  considerably  lower.  This  is  an  illustration  of  Article  24.  The 
net  section  in  the  riveted  plate  is  relatively  short  and  consequently 
is  kept  from  necking  as  it  would  in  a  longer  piece.  Notice  that 
there  is  no  certain  difference  between  the  pieces  lengthwise  and 
those  crosswise  the  plate.  This  is  explained  by  the  fact  that  in 
rolling  the  metal  it  was  worked  both  ways,  so  that  there  was  no 
definite  grain  in  one  direction. 

In  Problems  2  and  3,  the  design  was  such  that  there  was  rela- 
tively small  shearing  stress.  The  rivets  used  were  large  compared 
with  the  thickness  of  the  plate.  Problem  4,  below,  represents  a 
different  case  with  a  different  mode  of  failure. 

PROBLEMS. 

4.  In  a  test  piece  similar  to  Fig.  49  (Watertown  Arsenal,  1886,  page  1401), 
the  following  data  are  given;  tested  width,  13.11  inches;  thickness,  0.630  inch; 
five  rivets  in  1-inch  drilled  holes;  failed  by  shearing  the  rivets  under  a  pull 
of  295,500  pounds;  rivet  holes  elongated  0.31  inch,  0.32  inch,  0.26  inch,  0.25 
inch,  0.24  inch. 

Calculate  the  unit  stresses. 

Pounds  per  sq.  in. 

Ans.  Tensile  stress  in  net  section 57,840 

Bearing  stress 93,810 

Shearing  stress  on  rivets 37,620 

5.  .In  Problem  4,  the  cover  plates  were  0.384  inch  -thick.     Find  the  unit 
tensile  stress  in  the  net  section. 

Fig.  51  is  a  copy  of  a  photograph  of  a  rivet  which  failed  by 
shear  as  in  Problem  4  (Watertown  Arsenal,  "  Tests  of  Metals," 
1886,  page  1567). 

38.  Rivets  in  More  than  One  Row.  —  Rivets  are  frequently 
arranged  in  more  than  one  row.  The  rivets  in  the  second  row 
may  be  placed  directly  behind  those  in  the  first  row,  or  they  may 
be  arranged  zigzag  as  shown  in  Fig.  52.  The  figure  shows  the 
plates  and  rivet  holes  for  a  lap  joint  with  a  double  row  of  rivets. 
In  this  zigzag  arrangement  the  second  row  must  be  set  back 
sufficiently  far  that  the  sum  of  the  net  diagonal  intervals  c  and  e 
shall  considerably  exceed  the  net  interval  /  between  rivet  holes 
in  the  same  row.  If  the  rows  are  placed  too  close  together  the 
joint  is  likely  to  fail  along  the  zigzag  line. 

In  computing  problems  with  two  or  more  rows  of  rivets,  we 
assume  that  the  shearing  stress  is  the  same  in  the  rivets  of  all 


62 


STRENGTH  OF  MATERIALS 


TOWS.  If  we  consider  a  strip  of  width  a  equal  to  the  pitch,  ex- 
tending from  the  center  of  rivet  hole  1  to  the  center  of  rivet  hole 

2,  the  total  stress  transmitted  by 
this  strip  is  carried  by  the  lower 
half  of  rivet  1,  the  upper  half  of 
rivet  2,  and  the  whole  of  rivet  3. 
Or  we  may  take  a  strip  of  width 
a,  which  entirely  includes  one 
rivet  in  each  row.  In  any  case 
the  entire  stress  transmitted  by 
a  strip  of  gross  width  a  passes 
through  between  the  rivets  of 
the  first  row.  Half  of  this  stress 
is  carried  as  shear  to  the  lower 
plate  by  the  rivet  of  the  first  row 
and  the  other  half  passes  as  ten- 
sion through  the  second  row. 

Fig.  53  shows  an  arrangement 
of  three  rows,  with  twice  as  many 
rivets  in  the  second  row  as  in 
either  of  the  others.  In  solving 
a  problem  of  this  kind,  we  take 
as  the  unit  a  strip  of  width  equal 

to  the  pitch  in  the  outer  rows.  This  strip  may  extend  from  the 
middle  of  rivet  1  to  the  middle  of  rivet  5,  or  it  may  include  the 
whole  of  rivet  1  and  none  of  rivet  5.  In  either  case  the  strip 
embraces  two  rivets  in  the  middle  row  and  one  in  each  of  the 
others. 

PROBLEMS. 

1.  In  a  joint  similar  to  Fig.  52,  the  plates  are  f  inch  thick  and  are  united 
by  two  rows  of  |-inch  rivets  to  form  a  lap  joint.     The  pitch  is  2|  inches.     If  the 
unit  tensile  stress  in  the  gross  section  is  4800  pounds  per  square  inch,  find  the 
unit  tensile  stress  in  the  net  section  at  the  left  row  in  the  upper  plate,  the  unit 
tensile  stress  at  the  right  row  in  the  upper  plate,  the  unit  shearing  stress  in 
the  rivets,  and  the  unit  bearing  stress. 

Pounds  per  sq.  in. 

Ans.  Tensile  stress,  left  upper  and  right  lower 7200 

Tensile  stress,  right  upper  and  left  lower 3600 

Shearing  stress  in  rivets 6548 

Bearing  stress . .  : 7200 

2.  In  a  lap  joint  similar  to  Fig.  53,  the  pitch  in  the  outer  rows  is  5  inches, 
and  in  the  middle  row  2.5  inches.     The  plates  are  \  inch  thick  and  are  joined 


Fig.  51.  —  Failure  of  Rivet. 


RIVETED   JOINTS 


63 


by  f-inch  rivets.  When  the  unit  tensile  stress  in  the  gross  section  is  4000 
pounds  per  square  inch,  find  the  unit  shearing  stress  in  the  rivets  and  the  unit 
tensile  stress  in  the  net  section  at  the  right  row  in  the  upper  plate  arid  the  left 
row  in  the  lower  plate. 

ss,  5659  pounds  per  square  inch; 

st,  4706  pounds  per  square  inch. 
3.  In  Problem  2  find  the  unit  tensile  stress  in  the  net  section  at  the  middle 
row  in  each  plate.  Ans.  4286  pounds  per  square  inch. 


Ans. 


Fig.  52.  —  Lap  Joint  with  Double 
Row  of  Rivets. 


Fig.  53.  —  Rivets  in  Three 
Rows. 


4.  In  Problem  2  show  that  the  unit  tensile  stress  in  the  net  section  at  the 
left  row  in  the  upper  plate  is  one-fourth  that  in  the  right  row  in  the  same  plate. 

5.  A  butt  joint  is  formed  of  two  Hnch  plates  united  by  two  /2-inch  cover 
plates.     There  are  two  rows  of  f -inch  rivets  on  each  side,  the  inner  rows  of  3-inch 
pitch  and  the  outer  rows  of  6-inch  pitch.     The  unit  stress  in  the  gross  section 
of  the  Hnch  plates  is  6000  pounds  per  square  inch.     Find  the  unit  tensile 
stress  in  the  net  section  at  each  row  of  rivets  in  the  Hnch  plates  and  at  one 
of  the  inner  rows  in  the  cover  plates.     Find  the  unit  •  shearing  stress  in  the 
rivets  and  the  unit  bearing  stress  between  the  rivets  and  the  Hnch  plates. 

Pounds  per  sq.  in. 

in  Hnch  plates  in  outer  rows 7,024 

in  Hnch  plates  in  inner  rows 5,647 

Ans.    <tt  in  cover  plates  at  inner  rows 7,529 

in  all  rivets .4,989 

between  rivets  and  main  plates 13,714 

6.  In  Problem  5  what  should  be  the  thickness  of  the  cover  plates  if  the 
unit  tensile  stress  does  not  exceed  the  maximum  in  the  main  plates? 

7.  Taking  ultimate  strengths  from  Problems  2  and  4  of  Article  37,  what  is 
the  factor  of  safety  in  Problem  5  for  each  case? 

8.  Fig.  54  shows  a  joint  tested  at  the  Watertown  Arsenal  (1896,  pages 
258,  259).     The  average  thickness  of  the  plates  was  0.534  inch;  the  rivet  holes, 
f|  inch.     The  joint  failed  by  tension  along  line  shown  when  the  total  pull  was 
502,200  pounds.     Find  the  unit  tensile  stress  at  each  row  of  rivets  and  the 
unit  shearing  and  bearing  stress. 

39.  Efficiency  of  a  Riveted  Joint.  —  The  ratio  of  the  strength 
of  a  riveted  joint  to  the  strength  of  one  of  the  plates  which 
it  unites  is  called  the  efficiency  of  the  joint.  The  efficiency  may 


64  STRENGTH  OF  MATERIALS 

also  be  defined  as  the  ratio  of  the  unit  stress  in  the  gross  section, 
when  the  joint  is  stressed  to  its  allowed  limit,  to  the  allowable 
unit  stress  in  the  plates.  If  the  joint  is  so  designed  as  to  make 
it  at  least  as  strong  in  shear  and  compression  as  it  is  in  tension  at 
the  net  section,  the  efficiency  becomes  the  ratio  of  the  net  to  the 
gross  sections.  For  instance,  if  two  plates  are  united  by  f-inch 
rivets  with  a  pitch  of  3  inches,  the  efficiency  is  75  per  cent,  pro- 
vided the  thickness  of  the  plates  is  such  that  the  joint  is  no 
stronger  in  tension  than  it  is  in  shear  and  compression. 

PROBLEMS. 

In  the  following  problems,  unless  otherwise  stated,  we  shall  use  as  allowable 
unit  stresses:  st,  9000;  ss,  6000;  and  sc,  12,000  pounds  per  square  inch. 

1.  Two  f-inch  plates  are  united  by  a  single  row  of  1-inch  rivets  to  form  a 
butt  joint.     The  pitch  is  2£  inches.     What  is  the  efficiency  based  on  tensile 
stress  only?    What  pull  on  strip  of  pitch  width  will  produce  the  maximum  al- 
lowable tensile  stress?    What  are  the  shearing  and  bearing  stresses  at  this  pull? 

Ans.  57}  per  cent  efficiency. 

2.  Two  Hnch  plates  are  united  to  form  a  lap  joint  by  two  rows  of  f-inch 
rivets  spaced  3  inches  apart.     What  stress  determines  the  strength  of  the 
joint?    What  is  its  efficiency? 

Ans.  Weakest  in  shear.     Efficiency,  39.3  per  cent. 

3.  In  Problem  2  what  should  be  the  pitch  in  order  to  make  the  joint  equally 
strong  in  shear  and  in  tension?     What  is  the  efficiency? 

Ans.  1.93  inches;  61.1  per  cent. 

4.  In  Problem  3  is.  the  unit  bearing  stress  within  the  allowed  value? 

It  is  possible  to  design  a  riveted  joint  so  as  to  bring  tensile, 
compressive,  and  shearing  stresses  to  their  allowable  values  at 
the  same  load.  This  is  not  usually  done,  however,  as  it  would 
involve  inconvenient  sizes  of  rivets,  and  frequently  require  rivet 
holes  too  small  relatively  to  the  size  of  the  plate  to  be  punched. 
Usually  the  joint  is  designed  for  shear  and  tension  in  the  net 
section  and  then  examined  to  see  if  the  bearing  stress  is  within 
the  allowable  limit. 

PROBLEMS. 

5.  If  the  allowable  bearing  stress  is  twice  the  shearing  stress,  what  is  the 
minimum  thickness  of  plate  which  can  be  used  with  f-inch  rivets  in  a  lap  joint 
with  a  single  row  of  rivets?  Ans.  0.295  inch. 

6.  Derive  an  expression  for  thickness  of  plate  in  terms  of  diameter  of  rivet 
which  will  make  the  bearing  stress  twice  the  shearing  stress. 

=  ^-   for  single  shear; 
Ans. 

t  =  -r  for  double  shear. 


RIVETED   JOINTS 


65 


7.  Solve  Problem  6  if  the  allowable  shearing  stress  is  two-fifths  of  the 
bearing  stress. 

8.  Using  shearing  stress  two-fifths  the  bearing  stress  and  two-thirds  the 
tensile  stress,  and  using  rivets  to  the  nearest  sixteenth,  design  a  butt  joint  with 


LINE  OF  FRACTURE 


ZKPitch 


4 


holes 


x-k 


.534 


T  H&flr  T   ^T 

Fig.  54.  —  Butt  Joint  Tested  at  Watertown  Arsenal. 


double  cover  plates  with  one  row  of  rivets  equally  spaced  on  each  side  to  unite 
two  f-inch  plates.     Find  the  efficiency  of  the  joint. 

Ans.  1-inch  rivets;  pitch,  2.67  inches;  efficiency,  62.5  per  cent. 

9.   In  Problem  8  use  three  rows  of  rivets  having  a  pitch  ratio  1  :  2  :  4  on 

each  side.       Ans.  Efficiency,  92  per  cent;  cover  plates  at  least  T7g  inch  thick. 


CHAPTER  V. 
BEAMS. 

40.  Definition  of  a  Beam.  —  Fig.  55  is  a  front  elevation  of  a 
beam  supported  near  the  ends  and  carrying  a  single  concentrated 
load  P  in  addition  to  its  own  weight.  If  the  beam  is  uniform, 
its  own  weight  is  a  uniformly  distributed  load.  There  is  an 
upward  reaction  at  each  support.  A  beam  may  be  defined  as  a 
rigid  body  subject  to  transverse  loads  and  reactions. 


Bl  # 

Fig.  55.  —  Beam  Supported 
at  Ends. 


Fig.  56.  —  Cantilever. 


41.  Kinds  of  Beams.  —  Beams  may  be  classified  according 
to  the  character  of  the  support  and  the  method  of  loading. 
Fig.  56  represents  a  beam  fixed  at  one  end  and  free  at  the  other. 
This  kind  of  beam  is  called  a  cantilever.  Fig.  57  is  a  beam 


Fig.  57.  —  Beam  Fixed  at  Both 
Ends. 


1 


Fig.  58.  —  Beam  Fixed  at  One  End 
and  Supported  at  the  Other. 


fixed  at  both  ends.  Fig.  58  is  fixed  at  the  right  end  and  supported 
at  the  left.  Fig.  59  is  a  beam  which  overhangs  its  supports. 
A  beam  with  three  or  more  supports,  as  in  Fig.  60,  is  a  continuous 
beam. 


t 


Fig.  59.  —  Beam  Overhanging  its 
Supports. 


Fig.  60.  —  Continuous  Beam. 


66 


BEAMS  67 

The  figures  show  different  methods  of  loading  and  some  of  the 
,  ways  of  representing  the  loads  and  reactions  in  diagrams  and 
drawings. 

In  Figs.  55  and  58,  we  have  a  single  concentrated  load.  In 
Fig.  57,  there  are  two  concentrated  loads.  In  Fig.  56,  there  is 
a  uniformly  distributed  load  over  the  entire  length.  In  Fig.  59, 
there  is  a  uniformly  distributed  load  over  half  of  the  length,  and 
a  concentrated  load  at  the  right  end.  In  Fig.  60,  there  is  a 
distributed  load  over  part  of  the  left  portion  and  another  load 
of  greater  weight  per  unit  length  over  the  right  half. 

A  beam  is  not  necessarily  horizontal.  A  vertical  fence  post 
subjected  to  the  horizontal  force  of  the  wind  or  the  weight  of 
a  gate  is  a  cantilever  beam.  A  post  at  the  end  of  a  line  of  wire 
fence  is  a  vertical  beam  supported  at  one  end  and  partially 
fixed  at  the  other,  with  several  concentrated  loads  due  to  the 
tension  in  the  wire. 


1 

0 

c  3 

(D 

(DO 

Fig.  61. 

O 

Fig.  62. 

42.  Internal  Stresses  in  Beams.  —  If  a  log  of  wood,  supported 
near  the  ends,  is  cut  in  two  by  a  saw  running  horizontal,  the  saw 
cut  gradually  narrows  at  the  top.  In  a  large  log  it  becomes 
necessary  to  drive  a  wedge  into  the  cut  above  the  saw,  as  shown 
in  Fig.  61  to  prevent  it  from  "  pinching."  The  material  which 
has  been  removed  by  the  saw  was  under  compressive  stress.  If 
the  log  overhangs  the  support  as  in  Fig.  62,  the  cut  widens  and 
finally  splits  off  along  the  line  BC.  The  material  cut  in  this 
case  was  in  tension  and  the  final  failure  is  due  to  shear.  If  the 
log  overhangs  each  support  one-fourth  of  its  length  and  the  cut 
is  at  the  middle,  there  is  no  tendency  to  open  or  close.  There 
is  zero  stress  at  this  point  under  these  conditions. 

We  learned  in  Mechanics,  that  when  a  body  is  subjected  to 
a  set  of  coplanar  nonconcurrent  forces,  these  forces  may  be 
replaced  by  a  single  force  at  any  desired  point  in  the  plane  and  a 
single  couple.  We  learned  also  that  when  a  body  in  equilibrium 
is  cut  into  two  portions  at  any  section,  the  resultant  of  the  external 


68 


STRENGTH   OF   MATERIALS 


forces  upon  either  portion  is  equal  to  the  forces  across  this  sec- 
tion from  this  portion  to  the  other. 

Fig.  63  represents  a  cantilever  of  uniform  section,  weighing 
w  pounds  per  foot.  Let  us  consider  the  forces  at  a  section  EF 
at  a  distance  x  from  the  left  end.  If  we  consider  the  portion  to 
the  left  of  the  section  EF  as  a  free  body  in  equilibrium,  we  find 
that  the  only  external  force  is  the  weight  of  the  portion  vertically 

T 

downward  applied  at  its  center  of  gravity,  a  distance  ~  from  the 

Jt 

section.  This  weight  is  wx  pounds.  To  produce  equilibrium, 
there  must  be  an  equal  force  wx  vertically  upward  acting  from 
the  right  portion  to  the  left  across  the  section  EF.  There  is 
an  equal  force  acting  vertically  downward  from  the  left  portion 


w  Pounds  Per  Foot 


>lji 


Fig.  63.  —  Shear  at  Beam  Section. 

to  the  right.  This  last  force  is  the  single  force  above  mentioned, 
which  replaces  the  external  forces  acting  on  the  left  portion. 
It  is  called  the  vertical  shear  at  the  section.  The  vertical  shear 
at  any  section  of  a  beam  is  the  sum  of  the  vertical  components 
of  all  the  external  forces  to  the  left  of  the  section.  It  is  regarded 
as  positive  upward  and  negative  downward.  It  is,  of  course, 
equal  and  opposite  to  the  sum  of  the  vertical  components  of  all 
the  external  forces  acting  upon  the  portion  to  the  right  of  the 
section.  Either  portion  may  be  used  in  determining  the  magni- 
tude of  the  vertical  shear  The  left  portion  alone  determines 
the  sign.  These  are  merely  arbitrary  conventions,  but  they  are 
commonly  used  in  the  discussion  of  Strength  of  Materials.  In 
this  book,  total  vertical  shear  is  represented  in  algebraic  formulas 
by  the  letter  S  (many  authors  use  7).  The  author  finds  it 


BEAMS  69 

convenient  to  represent  shear  by  an  arrow  with  a  single  barb. 
The  one  on  the  left  of  EF,  in  Fig.  63,  indicates  that  the  shear 
from  the  left  portion  to  the  right  is  downward.  The  arrow 
on  the  right  of  the  section  indicates  that  there  is  a  shearing 
stress  upward  from  the  right  to  the  left  portion. 

PROBLEMS. 

1.  A  cantilever  beam  weighing  40  pounds  per  foot  is  fixed  at  the  right  end 
and  carries  a  load  of  100  pounds  2  feet  from  the  left  end.     Find  the  vertical 
shear  at  1  foot  and  at  3  feet  from  the  left  end. 

Ans.  At  1  foot,  S  =  —  40  pounds; 
At  3  feet,  S  =  -220  pounds. 

2.  A  cantilever  beam  weighing  30  pounds  per  foot  is  fixed  at  the  left  end 
and  carries  a  load  of  120  pounds  3  feet  from  the  right  end.     Find  the  total 
shear  2  feet  and  4  feet  from  the  right  end. 

Ans.  60  pounds  and  240  pounds. 

3.  A  beam  8  feet  long  weighing  30  pounds  per  foot  is  supported  at  the 
left  end  and  2  feet  from  the  right  end.     Find  the  total  shear  2  feet  and  4  feet 
from  the  left  end.  Ans.  20  pounds,  —40  pounds. 

When  the  external  forces  are  not  all  given,  it  is  necessary  to  determine  them 
for  at  least  one  side  of  the  section.  In  this  case  it  is  sufficient  if  one  reaction 
is  determined.  It  is  better  to  determine  both  reactions  by  moments  and  then 
check  by  vertical  resolutions.  Taking  moments  about  the  right  support,  we 
get  80  pounds  as  the  reaction  at  the  left  support.  Taking  moments  about 
the  left  support,  we  find  the  right  reaction  to  be  160  pounds.  In  taking 
moments  about  the  right  support,  consider  the  entire  beam  at  its  center  of 
gravity  2  feet  from  this  support;  and  do  not  divide  it  into  two  portions  of  6  feet 
and  2  feet. 

4.  A  beam  10  feet  long  weighing  40  pounds  per  foot  is  supported  at  the 
ends  and  carries  a  load  of  120  pounds  3  feet  from  the  left  end.     Find  the  re- 
actions at  the  supports  and  the  vertical  shear  2  feet  and  4  feet  from  the  left 
end.     Check  the  magnitude  of  the  shear  by  using  the  right  portion  as  a  free 
body.  Ans.  Rlf  284;  R2,  236;  S,  204  and  4  pounds. 

43.  External  Moment  and  Resisting  Moment.  —  In  Fig.  63 
the  external  force  of  gravity  at  the  center  of  the  left  portion  is 
resisted  by  the  vertical  shear  upward  at  the  section  EF.  This 
vertical  shear  from  the  right  portion  to  the  left  is  called  the 
resisting  shear.  This  resisting  shear  and  the  weight  of  the  por- 
tion together  form  a  couple  of  which  the  forces  are  each  wx  and 

/Y*  77)^7*^ 

the  moment  arm  —     The  moment  is  -_-•     This  is  the  external 
z  z 

moment.  We  may  get  this  moment  in  another  way.  Compute 
the  moment  of  all  the  external  forces  acting  on  the  portion  about 
a  horizontal  axis  perpendicular  to  the  beam  through  some  point 


70  STRENGTH  OF  MATERIALS 

in  the  section  EF.     The  moment  of  the  resisting  shear  is  zero, 
since  its  moment  arm  is  zero;  the  moment  of  the  weight  is 

x 

wx  multiplied  by  -•     The  external  moment  at  any  section  is 
2i 

the  moment  with  respect  to  a  horizontal  axis  in  the  section  of 
all  the  forces  acting  on  the  portion  of  the  body  to  the  left  of  the 
section.  The  external  moment  is  balanced  by  a  second  couple, 
the  forces  of  which  act  across  the  section  from  the  other  portion. 
This  is  the  resisting  moment.  In  Fig.  63  the  forces  of  the  resist- 
ing moment  are  tensile  at  the  top  and  compressive  at  the  bottom. 
In  Fig.  63  we  have  the  weight  wx  applied  at  the  center  of 
gravity  of  the  left  portion,  replaced  by  an  equal  force  applied  at 
EF  (the  external  shear),  and  a  couple  which  we  call  the  external 
moment.  The  effect  of  this  external  moment  is  to  tend  to  turn 
the  left  portion  counterclockwise.  It  is  convenient  to  give  this 
moment  the  negative  sign.  When  the  moment  of  the  external 
forces  tends  to  turn  the  left  portion  clockwise,  the  moment  is 
positive.  This  is  an  arbitrary  convention,  which  is  convenient 
in  calculating  deflections. 

If  the  student  prefers  to  think  of  counterclockwise  rotation  as  positive,  he 
may  fix  his  attention  on  the  resisting  moment.  When  the  resisting  moment 
acting  on  the  left  portion  of  a  beam  across  any  section  is  counterclockwise, 
the  moment  at  that  section  is  positive. 


PROBLEMS. 

1.  A  cantilever  beam  weighing  30  pounds  per  foot  is  fixed  at  the  right  end 
and  carries  a  concentrated  load  of  80  pounds  at  the  left  end.     Find  the  external 
moment  and  shear  3  feet  from  the  left  end. 

Ans.  Moment,  —375  foot  pounds;  shear,  —170  pounds. 

2.  A  cantilever  beam  weighing  20  pounds  per  foot  is  fixed  at  the  left  end 
and  carries  a  concentrated  load  of  100  pounds  at  the  right  end.     Find  the 
external  moment  and  shear  4  feet  from  the  right  end.     (Use  portion  to  the 
right  of  section  as  free  body.) 

Ans.  Moment,  —560  foot  pounds;  shear,  180  pounds. 

3.  A  beam  12  feet  long  weighing  20  pounds  per  foot  is  supported  at  the 
ends.     Find  the  external  moment  and  shear  2  feet  from  the  left  end,  2  feet 
from  the  right  end,  and  at  the  middle. 

Ans.  From  left  end.  Foot-pounds  moment.  Pounds  shear. 

2  feet  200  80 

6  feet  360  ,  0 

10  feet  200  -80 

4.  A  beam  20  feet  long  weighing  6  pounds  per  foot  is  supported  at  the  ends 
and  carries  a  load  of  80  pounds  6  feet  from  the  left  end.     Find  the  moment 


BEAMS  71 

and  shear  at  5  feet  from  the  left  end,  at  the  middle,  and  at  5  feet  from  the 
right  end.     Check  results,  using  the  portion  to  the  right  of  the  section. 

Ans.  From  left  end.  Foot-pounds  moment.  Pounds  shear. 

5  feet  505  86 

10  feet  540  -24 

15  feet  345  -54 

5.  A  beam  of  length  I  supported  at  the  ends  carries  a  load  P  at  the 
middle.     Find  the  moment  at  the  middle. 

PI 

Ans.  Moment  at  the  middle,  -j- 

6.  A  beam  of  length  I  supported  at  the  ends  carries  a  uniformly  distributed 
load  of  w  pounds  per  unit  length."    Find  the  moment  at  the  middle,  and  the 

shear  at  the  middle.  wl2 

Ans.  -TT->  0. 

o 

7.  A  beam  supported  at  the  ends  carries  a  uniformly  distributed  load  W. 

Find  the  moment  at  the  middle.  Wl 

Ans.  -g  • 

8.  Find  the  moment  at  the  fixed  end  of  a  cantilever  of  length  I  due  to  a 

load  P  at  the  free  end.  also  that  due  to  a  uniform  load  W.  nj        Wl 

Ans.  —PI, 2~' 

44.   Experiments  Illustrating  Moment  and  Shear.  —  In  Fig. 
64  we  have  a  front  elevation  of  a  cantilever  beam  which  has  been 


& 

/ 

•3L 

A 

oT^ 

z__ 

Fl  SHEAR  TENSION 
la^RESULTANT  TENSION  1 

hf^-RESULTANT  COMPRESSION 

m 

/ 

*  s         \ 

Fig.  64.  —  Shear  and  Moment  at  Beam  Section. 

cut  in  two  along  a  vertical  plane.  A  chain  A  is  attached  at  the 
top  connecting  the  two  portions  and  a  cylinder  B  is  placed  be- 
tween them  near  the  bottom.  The  chain  exerts  a  pull  in  the 
direction  of  its  length  and  the  cylinder  exerts  a  horizontal  push. 
When  the  chain  is  horizontal  the  left  portion  of  the  beam  is  not 
in  equilibrium,  as  there  is  no  resisting  shear  to  oppose  the  external 
shear.  A  second  cylinder  C  may  take  the  place  of  this  resisting 
shear.  If  this  cylinder  C  rests  on  a  scale  platform,  .we  may  de- 
termine the  resisting  shear  by  weighing.  If  the  chain  is  horizon- 
tal and  the  friction  of  the  cylinder  B  is  small,  we  find  the  resisting 
shear  equal  to  the  weight  of  the  portion  of  the  beam  to  the  left 
of  the  section. 


72  STRENGTH  OF  MATERIALS 

If  a  spring  balance  is  attached  to  the  chain  A,  we  may  measure 
the  tension  and  compute  the  resisting  moment.  We  find  that 
the  moment  of  the  tension  in  the  chain  about  the  axis  of  the 
cylinder  B  is  equal  to  the  moment  of  the  weight  of  the  left  por- 
tion about  the  axis  of  the  cylinder  C.  That  is,  the  resisting 
moment  and  the  external  moment  are  equal.  If  we  move  the 
chain  in  a  vertical  plane  about  its  left  end,  the  load  on  the  cylinder 
C  gradually  decreases.  In  the  position  shown  by  the  dotted  line 
the  weight  becomes  zero.  The  entire  vertical  shear  is  now 
carried  by  the  vertical  component  of  the  pull  on  the  chain.  This 
is  the  position  in  which  the  line  of  pull  of  the  chain  intersects  the 
horizontal  line  through  B  directly  below  the  center  of  gravity 
of  the  left  portion  of  the  beam.  This  is  the  condition  of  equi- 
librium of  three  forces.  In  an  actual  beam  there  is  tension  in  the 
upper  fibers.  This  tension  has  a  component  upward.  The  com- 
pression in  the  lower  fibers  has  also  a  component  upward.  We 
learned  in  Article  31  that  a  shearing  stress  is  equivalent  to  a 
tensile  and  a  compressive  stress  at  right  angles  to  each  other  and 
45  degrees  with  the  plane  of  shear.  The  real  stress  in  the  upper 
fibers  of  any  section  of  Fig.  64  is  the  resultant  of  the  horizontal 
tension  and  an  inclined  tension  due  to  shear.  In  the  same  way 
the  compression  below  the  middle  may  be  regarded  as  made  up 
of  a  horizontal  compression  and  an  inclined  compression  due 
to  shear.  This  is  shown  diagrammatically  at  the  section  EF, 
Fig.  64. 

Fig.  65  represents  a  beam  supported  near  the  ends.  The 
external  moment  is  positive;  hence  the  resisting  moment  is 


CT7 


COCNODOOOCu 


Fig.  65. 

counterclockwise.  The  horizontal  force  which  acts  across  any 
section  from  the  right  portion  to  the  left  must  push  at  the  top 
and  pull  at  the  bottom;  consequently  we  put  the  chain  at  the 
bottom  and  the  cylinder  near  the  top. 

The  shear  at  a  section  near  the  left  end  is  positive.     The  verti- 
cal forces  tend  to  move  the  left  portion  upward  relatively  to  the 


BEAMS  73 

right  portion.  To  resist  this  shear  without  changing  the  end 
reaction,  we  place  the  cylinder  C  on  the  top  of  the  left  portion 
and  support  the  right  portion  by  means  of  a  small  extension  H 
resting  on  (7. 

Fig.  66  represents  another  beam  supported  at  the  ends.     In- 
stead of  a  single  chain  under  the  middle  of  the  section,  we  have  a 


.INK 


Fig.  66. 

pair  of  links  pulling  against  pins  in  the  front  and  back  vertical 
surfaces.  Instead  of  the  cylinder  B,  we  have  a  prism  of  square 
section,  with  edges  in  contact  with  the  portions  of  the  beam.  If 
we  place  this  prism  with  its  diagonal  somewhat  inclined  to  the 
horizontal,  we  find  one  position  of  unstable  equilibrium  where  it 
will  support  both  shear  and  compression. 


PROBLEMS. 

1.  In  a  beam  similar  to  Fig.  64,  the  portion  to  the  left  of  the  section  is 
3  feet  long  and  weighs  16  pounds.     The  distance  from  the  center  of  the  cylinder 
B  to  the  chain  A  is  8  inches.     What  is  the  tension  in  the  chain  in  pounds  if  the 
cylinder  C  is  exactly  under  the  end?  Ans.  36  pounds. 

2.  In  Problem  1,  if  the  chain  A  is  attached  to  the  upper  corner  of  the  left 
portion,  at  what  angle  should  it  be  placed  in  order  that  there  may  be  no  load 
on  cylinder  C?  Ans.  The  angle  with  the  horizontal  whose  tangent  is  f . 

3.  In  Problem  1,  if  the  chain  is  attached  to  a  point  in  the  upper  surface  at 
a  distance  of  4  inches  to  the  left  of  the  right  end  of  the  portion,  at  what  angle 
should  it  be  placed  to  the  horizontal  in  order  that  there  may  be  no  load  on  C? 

4.  In  Problem  1,  replace  cylinder  B  by  a,  small  square  block  with  faces 
parallel  to  the  section,  and  remove  cylinder  C.     What  must  be  the  coefficient 
of  friction  between  the  block  and  the  portions  of  the  beam  in  order  that  the 
friction  will  support  the  shear?  Ans.  0.44. 

5.  In  Problem  4,  if  the  coefficient  of  friction  is  0.4,  what  is  the  maximum 
distance  between  the  block  and  chain  if  the  friction  supports  the  shear? 

6.  A  beam  6  feet  long  weighing  72  pounds  is  supported  at  the  ends.     The 
beam  is  cut  in  two  at  the  middle  and  supported  by  a  pair  of  links  and  a.square 
prism  as  shown  in  Fig.  66.     If  the  vertical  distance  between  the  plane  of  the 
links  and  the  axis,  of  the  prism  is  8  inches,  find  the  total  tension  on  the  links. 

Ans.  81  pounds. 

7.  The  beam  in  Problem  6  is  cut  2  feet  from  the  left  end.     The  tension 
links  are  horizontal.     What  is  the  angle  of  inclination  of  the  diagonal  of  the 


74 


STRENGTH  OF  MATERIALS 


compression  prism?  (Solve  by  resolutions  and  moments,  using  the  left  portion 
as  a  free  body.  Also  solve  by  finding  the  resultant  of  the  left  reaction  and  the 
weight  of  the  2  feet  of  beam,  and  the  condition  that  the  lines  of  the  remaining 
two  forces  intersect  on  this  resultant.) 

8.  A  uniform  beam  10  feet  long  and  1  foot  deep  rests  on  supports  each  2  feet 
from  the  ends.  The  beam  is  cut  in  two  at  the  middle  and  drops  slightly  till 
the  edges  touch  at  the  top.  What  must  be  the  coefficient  of  friction  at  the 
supports  to  hold  it  in  this  position? 

45.  Shear  Diagrams.  —  It  is  convenient  to  represent  the  total 
shear  at  all  sections  of  a  beam  by  means  of  curves  called  shear 
diagrams.  Fig.  67  is  the  shear  diagram  for  Problem  3  of  Article 


Fig.  67.  —  Shear  Diagram  for  Distributed  Load. 

43.  The  end  reactions  are  each  120  pounds.  We  begin  at  the 
left  support  and  take  a  section  infinitely  near  the  end.  The 
vertical  reaction,  Ri,  is  120  pounds  and  the  weight  of  the  portion 
is  negligible.  The  vertical  shear  is  120  pounds.  Accordingly, 
we  lay  off  an  ordinate  120  units  long,  using  some  convenient  scale. 
We  may  now  take  points  at  1-foot  intervals  on  the  beam  and 
compute  the  shear.  At  1  foot  from  the  end  it  is  100  pounds; 
at  2  feet  it  is  80  pounds,  etc.  We  notice  that  the  curve  which 
joins  these  points  is  a  straight  line,  since  the  ordinates  have  a 
constant  rate  of  change.  All  that  is  really  necessary  is  to  com- 
pute the  shear  at  the  ends  and  connect  the  points  by  a  straight 
line.  This  straight  line,  together  with  the  ordinates  at  the  ends, 
makes  the  shear  diagram. 

Fig.  68  is  the  shear  diagram  for  a  beam  10  feet  long,  supported 
at  the  ends,  with  a  uniform  load  due  to  its  own  weight  of  60 
pounds  per  foot  and  a  concentrated  load  of  200  pounds,  3  feet 
from  the  left  end.  By  moments  about  the  right  support,  we 


BEAMS 


75 


find  the  left  reaction,  R1}  to  be  440  pounds.  By  moments  about 
the  left  support,  we  find  R2  to  be  360  pounds.  The  sum  of  these 
reactions  is  equal  to  the  total  load,  affording  a  check. 


CO* Per  Foot 


260* 


Fig.  68.  —  Concentrated  and  Distributed  Loads. 

Infinitely  close  to  the  left  support  the  shear  is  440  pounds. 
It  drops  180  pounds  in  the  first  3  feet  and  is  260  pounds  infinitely 
close  to  the  load  of  200  pounds.  In  a  negligible  distance  in  pass- 
ing from  the  left  side  to  the  right  side  of  the  concentrated  load  it 
drops  an  additional  200  pounds,  so  that  it  becomes  60  pounds 
infinitely  close  to  the  load  on  the  right  side.  Here  the  shear 
diagram  is  a  vertical  line.  Beyond  the  concentrated  load  the 
shear  drops  at  the  rate  of  60  pounds  per  foot  for  the  remaining  7 
feet,  which  brings  it  to  —360  pounds  infinitely  close  to  the  right 
support.  The  reaction  of  360  pounds  raises  it  to  the  initial  line. 
The  shear  diagram  crosses  the  zero  or  initial  line  1  foot  to  the 
right  of  the  concentrated  load,  or  4  feet  from  the  left  support. 

Notice  that  the  shear  diagram  in  Fig.  68  drops  vertically  downward  under 
the  load  of  200  pounds,  and  we  speak  of  points  as  infinitely  near  the  load  on 
either  side.  This  would  mean  that  the  load  is  applied  along  a  mathematical 
line  running  across  the  beam.  The  actual  surface  of  contact  is  a  band  of  some 
width  running  across  the  beam,  and  the  actual  shear  diagram  is  something  like 
that  represented  by  the  dotted  lines. 


76 


STRENGTH  OF  MATERIALS 


PROBLEMS. 

1.  A  uniform  beam  15  feet  long  weighing  40  pounds  per  foot  is  supported 
at  the  ends  and  carries  a  concentrated  load  of  360  pounds  3  feet  from  the  left 
end.     Construct  the  shear  diagram,  using  as  abscissas  1  inch  equals  2  feet  of 
length,  and  as  ordinates  1  inch  equals  a  shear  of  100  pounds.     Find  the  equa- 
tion of  the  shear  diagram  on  each  side  of  the  load. 

Ans.  S  =  588  -  40  x]  S  =  228  -  40  x. 

2.  In  Problem  1  find  the  position  of  zero  shear  from  the  equations. 

3.  A  cantilever  beam  of  uniform  section,  weighing  20  pounds  per  foot,  is 
fixed  at  the  right  end  and  projects  10  feet  to  the  left.     It  carries  a  concen- 
trated load  of  80  pounds  3  feet  from  the  free  end.     Construct  the  shear  dia- 
gram, using  the  scale  of  Problem  1.     Find  the  equation  of  the  two  parts  of 
the  line,  taking  the  origin  at  the  left  end. 

4.  A  beam  24  feet  long,  supported  at  the  ends,  carries  a  load  of  4000  pounds 
8  feet  from  the  left  end  and  a  load  of  5000  pounds  5  feet  from  the  right  end. 
Construct  the  shear  diagram  to  the  scale  1  inch  equals  4  feet  of  length  and 
1000  pounds  shear.     Neglect  the  weight  of  the  beam. 


II 


III 


IV 


Fig.  69.  —  Shear  Diagrams. 


Shear  diagrams  are  usually  made  up  of  straight  lines.  These 
lines  are  horizontal  from  one  load  to  the  other  when  the  loads 
are  concentrated  and  the  weight  of  the  beam  is  neglected.  With 
uniformly  distributed  loads,  the  lines  slope  downward  from  left 
to  right.  (With  distributed  loads  pushing  up,  as  in  the  bottom 
of  a  boat  due  to  the  water  pressure,  the  lines  slope  upward.) 
Where  loads  are  distributed  not  uniformly,  as  in  the  case  of  the 
water  pressure  on  a  vertical  dam,  the  shear  diagram  is  curved. 

The  student  should  become  sufficiently  familiar  with  the 
simpler  shear  diagrams  to  be  able  to  recognize  the  character  of 
the  loading  at  a  glance. 


BEAMS 


77 


PROBLEM. 

5.  Describe  the  loading  and  the  character  of  support  which  gives  each  of 
the  shear  diagrams  of  Fig.  69. 

46.  Moment  Diagrams.  —  Moment  diagrams  are  constructed 
in  the  same  way  as  shear  diagrams,  using  external  moment  as 
ordinates.  In  this  book  we  shall  draw  positive  moment  upward, 
though  many  engineers  prefer  the  opposite. 

Shear  diagrams  are  easily  constructed,  as  they  usually  consist 
of  straight  lines.  Moment  diagrams  are  curved,  except  when 
the  loading  is  made  of  concentrated  loads  only. 


Distributed  Load  40*  Per  Foot 


MOMENT  DIAGRAM 


MOMENTS 


Fig.  70.  —  Single  Concentrated 
Load. 


Fig.  71.  —  Uniformly  Distributed 
Load. 


Fig.  70  shows  the  shear  and  moment  diagram  for  a  beam  sup- 

ported at  the  ends  and  carrying  a  load  P  at  the  middle.     The 

p 

weight  of  the  beam  is  neglected.     The  end  reactions  are  -^  .     The 

2i 

Px 

moment  at  any  section  at  a  distance  x  from  the  left  end  is  -^-t 

provided  x  is  not  greater  than  one-half  of  the  length.     Under  the 

PI 
load  the  moment  is  -j-.     The  moment  diagram  for  the  left  half 

of  the  beam  is  a  straight  line  through  the  points  (0, 


o'.  TJ* 


Beyond  the  load,  the  moment  is  due  to  the  reaction  at  the  left 
support   turning  clockwise  minus  that  due  to  the  load  at  the 


78  STRENGTH  OF  MATERIALS 

middle.     At  a  distance  x  from  the  left  end  when  x  is  greater  than 

J 

2' 


ivr 

Moment 


Px        T>t          l\       Pl       Px       P  n         ^ 

=  -y  -  P\x  -  ^  =  -^  -•  -^  ==  -^  (I  -  x). 


This  is  also  a  straight  line.     Notice  that  the  last  of  the  three 
expressions  for  moment  is  the  one  which  we  get  directly,  if  we 

use  the  portion  to  the  right  of  the  section  as  a  free  body.      The 

p 
right  reaction  is  -^  and  the  moment  arm  is  I  —  x.     The  sign  is 

A 

opposite,  as  it  should  be. 

Fig.  71  gives  the  shear  and  moment  diagrams  for  a  beam 
supported  at  the  ends,  with  a  uniformly  distributed  load.  The 
moment  diagram  is  a  parabola  with  the  vertex  at  the  top. 


PROBLEMS. 

1.  With  the  data  of  Fig.  71,  find  the  equation  of  the  moment  curve. 

2.  Find  the  equation  of  the  moment  curve  of  a  beam  supported  at  the 
ends,  with  a  uniformly  distributed  load  of  w  pounds  per  unit  length. 

, ,      wlx      wx2      wx ,,        N 
Ans.  M  =  —  -  —  =  —  (I  -  x). 

3.  A  beam  10  feet  long  is  supported  at  the  ends  and  carries  a  concentrated 
load  of  200  pounds  6  feet  from  the  left  end.     Neglecting  the  weight  of  the 
beam,  construct  the  shear  and  moment  diagrams  to  the  scale  1  inch  hori- 
zontally equals  2  feet  of  length,  1  inch  vertically  equals  50  pounds  shear  and 
100  foot  pounds  of  moment.     Derive  the  equation  of  each  curve. 

Ans.  On  the  right  of  the  load,  S  =  -120  pounds;  M  =  1200  -120z. 

4.  A  beam  of  uniform  section,  10  feet  long,  is  supported  at  the  ends.     It 
carries  a  uniform  load,  including  its  own  weight,  of  16  pounds  per  foot.     Cal- 
culate the  moment  for  each  foot  and  construct  the  shear  and  moment  diagrams, 
using  the  same  scale  as  in  Problem  3. 

5.  A  beam  10  feet  long,  supported  at  the  ends,  carries  a  distributed  load 
equal  to  that  of  Problem  4,  and  a  concentrated  load  equal  to  that  of  Problem  3. 
Using  the  same  scale  as  in  those  problems,  construct  the  shear  and  moment 
diagrams  for  the  combined  loadings. 

It  is  best  to  construct  the  moment  diagrams  for  the  concentrated  and  dis- 
tributed loads  separately.  Then  combine  the  two  by  adding  the  ordinates 
graphically.  Fig.  72  shows  the  shear  and  moment  diagrams  for  another  beam 
supported  at  the  ends,  with  a  uniform  load  and  two  concentrated  loads.  The 
moment  diagram  for  the  concentrated  loads  consists  of  three  straight  lines. 
To  construct  the  moment  diagram  for  the  distributed  loads,  it  is  sufficient  to 
compute  the  moments  for  foot  intervals  for  one  side  of  the  middle.  The 
resultant  moment  AD  at  the  2-foot  position  is  the  sum  of  the  ordinates 
AB  and  AC.  With  a  scale  or  compass  lay  off  from  B  the  distance  BD  equal 
to  AC.  This  may  be  done  for  every  half-foot  interval  cr  less. 


BEAMS 


79 


6.  From  your  curve  of  Problem  5  find  the  slope  of  the  moment  diagram 
at  5  feet  and  at  6  feet.  Find  also  the  position  of  maximum  moment  and  the 
corresponding  value  of  the  shear. 


MOMENT  OF 
CONCENTRATED  LOADS 


0'       i'        2'       3'       4'        5'       6'       7'        8'        9'       10' 

MOMENT 

Fig.  72.  —  Concentrated  and  Distributed  Loads. 


47.  Relation  of  Moment  and  Shear.  — -  Fig.  73  represents  a 
beam  supported  at  the  ends  and  carrying  a  concentrated  load 
P  at  a  distance  a  from  the  left  end  and  distributed  load  of  w 
per  unit  length.  The  reaction  at  the  left  support  is  -Ri.  Cal- 
culating the  moment  at  a  section  at  a  distance  x  from  the  left 
end  (where  x  is  greater  than  a),  we  get: 


M 


-  P  (x  -  a)  -      - 


Differentiating  with  respect  to  x,  we  get  the  derivative: 

dM 
dx 


i-  P  -  wx. 


(1) 


(2) 


80 


STRENGTH  OF  MATERIALS 


We  recognize  the  second  member  of  equation  (2)  as  expressing 
literally  the  definition  of  total  vertical  shear.     Hence 


dM 
dx 


=  S. 


Formula  V. 


The  derivative  of  the  moment  at  any  section  gives  the  shear 
at  the  section,  except  at  a  concentrated  load  or  reaction  where 
the  shear  diagram  is  vertical. 


Fig.  73.  —  Shear  Diagram. 


The  above  proof  of  Formula  V  is  not  general.  We  will  now  consider  Fig.  74 
for  a  proof  for  a  more  general  case.  This  figure  represents  a  beam  of  indefinite 
extent.  The  origin  of  coordinates  is  taken  in  the  vertical  line  0.  The  beam 
carries  loads  Pi,  P2,  etc.,  at  positions  ai,  a2,  and  a  distributed  load  of  w  per 


-a. 


Pounds  Per  Unit  Length       \ 


'A' 


Fig.  74. 


unit  length.  There  are  also  loads  and  reactions  to  the  left  of  the  origin.  The 
resultant  of  all  the  loads  to  the  left  of  the  origin,  whether  distributed  or  con- 
centrated, may  be  replaced  by  a  single  load  Q  at  some  distance  b  to  the  left 
of  the  origin.  The  resultant  of  all  the  reactions  on  the  left  of  the  origin  is  a 
single  force  R  acting  upward  at  some  point  at  a  distance  c  to  the  left  of  the 


BEAMS  81 

origin.  Computing  the  moment  with  respect  to  a  section  at  a  distance  x  to 
the  right  of  the  origin: 

M  =  R  (c  +  x)  -  Q  (b  +  x)  -  P!  (x  -  01)  -  P2  (x  -  a2)  -  ~  (3) 

Differentiating: 

^  =  R  -  Q  -  P1  -  P2  -  wx.  (4) 

The  second  member  of  (4)  is  the  shear  at  the  section,  hence 

^  =  S.  Formula  V. 

Where  there  are  several  concentrated  loads,  such  as  PI,  the  moment  equation 
for  points  to  the  right  of  the  origin  for  all  points  to  the  right  of  these  loads  may 
be  written: 


M  =  R  (c  +  x)  -  Q  (b  +  x)  -        P  (x  -  a)  -        ;  (5) 

and  the  shear  equation: 

S  =  R  -  Q  -  S  P  -  wx.  (6) 

There  might  be  an  infinite  number  of  loads  P,  so  that  the  equations  apply 
to  any  distribution  whatever.    Some  of  the  concentrated  loads  may  be  negative. 

Considering  Fig.  73, 
On  the  left  of  the  load,  On  the  right  of  the  load, 

M  =  R,x  -  ~,  M  =  RlX  -  ^  -  P  (x  -  a), 

dM  dM 

——     =    fa     —    WX,  -y—     =    Rl    —   WX    —    P. 

ax  ax 

If  we  let  re  =  a  in  each  shear  equation, 

dM  dM 

-j—  =  Ri  —  wa,  3—  =  Ri  —  wa  —  P. 

dx  dx 

the  shear  just  to  the  left  of  the  load;  the  shear  just  to  the  right  of  the  load. 


PROBLEMS. 

1.  In  Fig.  72  calculate  -r-  for  both  equations  at  the  point  x  equals  3  feet. 

Compare  results  with  the  slope  of  the  tangent  to  the  curve  and  with  the 
ordinates  of  the  shear  diagram. 

2.  Solve  Problem  1  for  the  8-foot  position. 

48.  Area  of  Shear  Diagram  Equals  Moment.  —  Since 


$  =  -T-  j  Sdx  =  dM,  and    I  Sdx  =   I  dM  =  M .      Formula  VI. 
dx  J  J 

In  Fig.  75,  I,  Sdx  is  the  area  of  a  vertical  strip  of  the  shear 
diagram  of  height  S  and  width  dx.  The  integral  of  Sdx 
between  the  limits  0  and  Xi  is  the  area  bounded  by  the  shear 
diagram,  the  X  axis,  and  the  ordinate  at  x\.  The  moment  at 
the  point  Xi  is  the  area  of  the  shear  diagram  to  the  left  of  that 
point.  Similarly,  the  moment  at  x2  is  the  area  of  the  shaded 


82 


STRENGTH  OF  MATERIALS 


portion  of  the  shear  diagram  in  Fig.  75,  II;  and  the  moment  at 
x3  is  the  area  of  the  shaded  portion  above  the  X  axis  minus  the 
shaded  portion  below  the  X  axis  in  Fig.  75,  III. 


Pounds  Per  Foot 


Fig.  75.  —  Shear  Areas. 

Let  us  check  this  for  Fig.  75,  II.     Writing  the  moment  equa- 
tion for  the  section  at  a  distance  xz  from  the  left  support : 

M  =  R,x2  -^-P  (z2  -  a)  (1) 

RiXz  is  the  area  of  the  rectangle  xz  long  by  Ri  high; 
-y-  is  the  area  of  the  triangle  CDF', 

P  (x2  —  a)  is  the  area  of  the  parallelogram  FGHK; 
hence  the  moment  equals  the  area  of  the  shear  diagram  in  this 
case. 


BEAMS  83 

PROBLEMS. 

1.  In  Fig.  70,  calculate  the  moment  at  the  middle  by  means  of  the  area  of 
the  shear  diagram.     Find  also  the  moment  at  one-fourth  the  length  and  at 
three-fourths  the  length  from  the  left  end. 

2.  In  Fig.  71,  find  the  moment  at   1   foot  from  the  left  end  by  means 
of  the  area  bounded  by  the  shear  line,  the  X  axis,  and  the  ordinates  x  =  0 
and  x  =  1  foot. 

3.  In  Fig.  72,  find  the  ordinates  of  the  moment  diagram  at  3  feet  and  at 
4  feet  from  the  left  end  by  means  of  the  corresponding  areas  of  the  shear 
diagram. 

4.  In  Fig.  72,  find  the  moment  at  3  feet  and  the  moment  at  6  feet. 

5.  Without  computing  the  moments  at  either  position,  show  that  the 
moment  at  2  feet,  in  Fig.  72,  is  60  foot  pounds  less  than  that  at  7  feet. 

Since  the  area  of  the  shear  diagram  to  the  left  of  any  point 
measures  the  moment  at  that  point,  and  areas  below  the  X  axis 
are  regarded  as  negative,  the  moment  has  its  greatest  numerical 
values  at  points  where  the  shear  is  zero.  This  is  true  whether 
the  shear  crosses  the  axis  obliquely,  giving  a  true  mathematical 
maximum  or  minimum,  or  crosses  vertically  under  a  concentrated 
load  or  over  a  support.  These  points  of  zero  shear  are  called  the 
dangerous  sections  in  a  beam. 

PROBLEMS. 

In  the  problems  below,  compute  the  moments  by  means  of  the  shear  diagram 
and  check  by  the  moment  equations. 

6.  A  beam  12  feet  long  weighing  100  pounds  per  foot  is  supported  at  the 
left  end  and  2  feet  from  the  right  end,  and  carries  a  load  of  600  pounds  2  feet 
from  the  left  end.     Calculate  the  moment  under  the  concentrated  load  and 
at  the  dangerous  sections. 

7.  A  beam  20  feet  long,  supported  at  the  ends,  carries  a  distributed  load, 
including  its  own  weight,  of  120  pounds  per  foot,  and  two  concentrated  loads, 
600  pounds  4  feet  from  the  left  end  and  720  pounds  5  feet  from  the  right  end. 
Find  the  moment  at  the  dangerous  section  and  under  each  concentrated  load. 

Ans.  At  dangerous  section,  M  =  9015  foot  pounds. 

8.  A  cantilever  fixed  at  the  right  end  projects  8  feet  and  carries  a  dis- 
tributed load,  including  its  own  weight,  of  60  pounds  per  foot  and  a  load  of 
300  pounds  2  feet  from  the  left  end.     Construct  the  moment  and  shear  dia- 
grams: 1  inch  equals  1  foot;  1  inch  equals  200  pounds  shear;  1  inch  equals 
1000  foot  pounds  of  moment.     Where  is  the  dangerous  section? 

9.  A  beam  14  feet  long  weighing  30  pounds  per  foot  is  supported  4  feet 
from  the  left  end  and  held  in  a  horizontal  position  by  a  downward  force  at  the 
left  end.     Draw  the  shear  and  moment  diagrams. 

10.  A  cantilever  beam  weighing  20  pounds  per  foot  is  fixed  at  the  left  end 
and  projects  10  feet.     Construct  the  moment  and  shear  diagrams. 

We  must  construct  the  shear  diagram  from  the  free  end.  We  may  also 
begin  at  the  right  end  to  get  the  moment  diagram.  Or  we  may  begin  at  the 


84 


STRENGTH   OF  MATERIALS 


wall  to  get  the  moment,  if  we  remember  that  there  must  be  a  negative  shear 
area  in  the  wall  equal  to  the  triangle  outside.  If  the  beam  touches  the  wall 
at  two  points  A  and  B  (Fig.  76),  the  problem  is  the  same  as  that  of  Problem  9. 


MOMENT 


-\j-looo  Foot  Pounds 
Fig.  76.  —  Cantilever  Fixed  at  Left  End. 


The  shear  diagram  is  CDEF.  If  the  contact  is  distributed,  the  shear  diagram 
is  that  shown  by  the  dotted  lines,  and  the  dangerous  section  is  at  C",  a  little 
to  the  left  of  the  face  of  the  wall. 

11.  A  beam  is  supported  at  the  ends  and  carries  two  equal  concentrated 
loads  at  one- third  the  length  from  the  left  end  and  one- third  the  length  from 
the  right  end.  Draw  moment  and  shear  diagrams,  neglecting  the  weight  of 
the  beam. 

12.  A  beam  12  feet  long  is  supported  3  feet  from  the  ends.  The  total 
load  is  60  pounds  per  foot.  Construct  the  shear  and  moment  diagrams: 
1  inch  equals  2  feet  length,  100  pounds  shear,  and  100  foot-pounds  moment. 


CHAPTER  VI. 
STRESSES  IN  BEAMS. 

49.  Nature  of  Stresses.  —  In  the  experiments  described  in 
Article  44,  we  found  tension  at  the  top  of  the  cantilever  beam 
and  compression  at  the  bottom.  With  the  beam  supported 
at  the  ends,  we  were  obliged  to  put  the  chain  below  and  the 
cylinder  above.  In  each  case,  the  compression  was  at  the  con- 
cave side  and  the  tension  at  the  convex  side.  At  any  section 
in  a  bent  beam  the  fibers  in  the  convex  side  are  elongated  and 
those  in  the  concave  side  are  compressed.  Between  these, 
there  is  a  surface  which  remains  unchanged  in  length.  This  is 
called  the  neutral  surface. 

Fig.  77  represents  a  beam  supported  at  the  ends  and  bent  by 
its  own  weight  or  other  loads  between  the  supports.  (The 
amount  of  bending  is  exaggerated.)  The  points  B,  B',  C,  C', 
lie  on  the  neutral  surface.  The  transverse  lines  joining  B  to 
E'  and  C  to  C'  are  neutral  axes  for  their  respective  sections. 
Experiments  show  that  a  plane  section  such  as  EFG  remains 
plane  when  the  beam  is  bent.  In  Fig.  77,  the  dotted  lines 
K'Mr,  M'N'  indicate  the  position,  before  the  beam  was  bent, 
of  a  plane  section  parallel  to  EFG.  After  bending,  its  position 
with  reference  to  EFG  is  shown  by  the  plane  KMN.  The 
change  consists  of  a  rotation  of  the  plane  KMN  through  an 
angle  A0  about  the  neutral  axis  CCf  from  the  position  K'M'N' 
to  the  position  KMN.  (There  is  also  a  slight  vertical  shift  in 
position,  but  this  does  not  affect  the  problem.)  If  we  consider 
a  portion  of  length  Al  extending  from  the  plane  EFG  to  the  plane 
KMN,  we  find  that  the  fibers  at  the  bottom  are  elongated  an 
amount  equal  to  the  distance  from  N'  to  N,  and  those  at  the 
top  are  shortened  an  amount  equal  to  the  distance  from  M'  to 
M.  In  the  figure  a  filament  of  infinitesimal  cross  section  dA  is 
shown  at  a  distance  v  above  the  neutral  surface.  This  filament 
is  compressed  an  amount  vA6  when  the  beam  is  bent.  If  v  is 
negative,  the  deformation  becomes  an  elongation.  (We  use  v 
to  represent  the  distance  of  any  element  of  area  above  the 

85 


86 


STRENGTH  OF  MATERIALS 


neutral  surface,  and  reserve  y  to  represent  the  deflection  of  the 
beam  from  its  original  position.) 


NEUTRAL  AXIS 


N 

SIDE  ELEVATION  FRONT  ELEVATION 

II  III 

Fig.  77.  —  Deformation  of  a  Bent  Beam. 

To  get  some  idea  of  the  magnitude  of  the  quantities  involved, 
let  us  consider  Fig.  78.  This  represents  a  beam  6  inches  wide  and 
8  inches  deep,  and  about  7  feet  long,  supported  at  two  points 
about  80  inches  apart.  An  extensometer  (not  shown)  is  attached 
at  two  points,  F  and  M,  40  inches  apart  and  1  inch  below  the 
top  of  the  beam.  A  second  extensometer  is  attached  at  G  and 
TV,  1  inch  from  the  bottom.  Two  loads  of  4000  pounds  each  are 
applied  16  inches  from  the  supports.  If  the  beam  is  made  of 


4000*                                                          4000S- 

L              in"       J,                                               ,Q>/                                            JL            -if."            I 

F- 
G- 

"                      in" 

tt 

-N 

Fig.  78, 

timber,  the  deflection  at  the  middle  is  about  0.08  inch.  (This 
deflection  is  too  small  to  show  in  the  drawing  unless  the  scale 
is  exaggerated.)  The  upper  extensometer  shows  a  shortening 
of  about  0.0180  inch  in  the  original  length  of  40  inches,  and  the 


STRESSES   IN   BEAMS  87 

lower  extensometer  shows  an  equal  elongation.  If  the  tension 
and  compression  are  exactly  equal,  the  neutral  surface  is  mid- 
way between  the  extensometers.  If  the  readings  are  unequal,  the 
location  of  the  neutral  surface  may  be  found  from  the  similar 
triangles,  such  as  MCM',  NCN'  (Fig.  77),  with  MM'  and  NN' 
known  from  the  extensometer  readings,  and  the  distance  between 
the  instruments  equal  to  MN.  In  case  the  readings  are  each 
0.0180  inch,  showing  that  the  neutral  axis  is  at  the  middle  of 
the  section,  4  inches  from  the  top,  the  compression  in  the  top 
fibers  is  four-thirds  as  great  as  at  M .  The  compression  at  1  inch 
from  the  neutral  surface  is  0.0060  inch;  and  at  a  distance  v  it  is 
0.0060  v.  The  unit  deformation  at  a  distance  v  from  the  neutral 
axis  is  0.00015  v. 

PROBLEMS. 

1.  A  beam  is  tested  as  shown  in  Fig.  78.     The  points  F  and  M  are  40  inches 
apart  and  6  inches  above  the  similar  points  G  and  N.     The  compression  read- 
ing on  the  upper  instrument  is  0.0198  inch,  and  the  extension  on  the  lower 
instrument  is  the  same.     What  is  the  unit  stress  4  inches  above  and  4  inches 
below  the  neutral  surface,  if  E  equals  1,500,000  pounds  per  square  inch? 

Ans.  99.0  pounds  per  square  inch. 

2.  In  Problem  1,  what  is  the  unit  stress  at  a  distance  unity  and  at  a  distance 
v  from  the  neutral  surface? 

3.  In  a  case  similar  to  Problem  1,  the  upper  instrument  shows  a  com- 
pression of  0.0140  inch  and  the  lower  instrument  an  extension  of  0.0160  inch. 
If  the  beam  is  8  inches  deep  how  far  is  the  neutral  surface  from  the  top  surface? 
What  is  the  unit  stress  at  the  top  and  at  the  bottom  of  the  beam? 

Ans.  3.8  inches  from  the  top;  sc  equals  0.000475  Ec. 

50.  Relation  of  Moment  to  Stress.  —  In  Fig.  77,  let  AZ 
represent  the  original  distance  between  the  planes  EFG  and 
K'M'N'.  This  is  the  distance  BC  in  the  figure.  Consider  a 
filament  of  cross  section  dA  connecting  these  planes  at  a  distance 
v  above  the  neutral  surface.  The  cross  section  of  this  filament 
may  be  of  infinitesimal  dimensions  in  both  directions,  as  in  Fig.  77, 
or  it  may  be  of  infinitesimal  height  dv  and  extend  the  entire 
width  of  the  beam.  If  this  filament  is  above  the  neutral  axis 
(in  a  beam  concave  upwards),  it  is  shortened  an  amount  vAd, 
and  if  it  is  below  the  neutral  axis  it  is  lengthened  vA0-when  the 
beam  is  bent.  The  unit  deformation,  being  the  ratio  of  the  total 
deformation  to  the  original  length,  is: 


88  STRENGTH  OF  MATERIALS 

The  unit  stress  in  a  filament  is  the  product  of  the  unit  deforma- 
tion by  the  modulus  of  elasticity  : 

«-^g-.  (2) 

The  total  stress  on  a  filament  of  area  dA  is  the  product  of  the 
unit  stress  by  the  area: 

total  stress  on  dA  =  Ev—.dA.  (3) 

The  moment  of  this  stress  with  respect  to  the  neutral  axis  BB' 
is  the  product  of  the  total  stress  on  dA  by  the  moment  arm  v; 

moment  of  stress  about  axis  =  Ev2  —  j  dA  =  E  -—r  v2dA  =  dM.     (4) 

The  total  moment  of  all  the  filaments  which  make  up  the 
beam  is  the  integral  of  dM  over  the  section  EFG.     Integrating 

over  this  area,  -T-  remains  constant  and 


where  v\,  v2  are  the  distances  of  the  lower  and  upper  surfaces  of 
the  beam  from  the  neutral  surface,  I  is  the  moment  of  inertia  of 
the  cross  section  EFG  or  KMN  with  respect  to  its  neutral  axis, 
and  A0  is  the  change  in  slope  of  the  normal  to  the  beam,  or  the 
change  in  slope  of  the  tangent  to  the  beam,  in  the  length  Al. 

51.  Unit  Stress  in  the  Outer  Fibers.  —  The  extreme  upper 
and  lower  fibers  in  a  beam  which  is  bent  in  a  vertical  plane 
suffer  the  greatest  deformations  and  are  subjected  to  the  great- 
est stress.  The  allowable  unit  stress  in  these  fibers  determines 
the  load  which  the  beam  can  safely  carry. 

The  unit  stress  in  any  fiber  at  a  distance  v  from  the  neutral 
axis  is: 

•          — 


Z          V 

Substituting  in  equation  (5)  of  Article  50, 

M  =  —  ,  Formula  VII. 

v 

s  =        .  Formula  VIII. 


STRESSES   IN   BEAMS  89 

s  is  the  unit  stress  at  a  distance  v  from  the  neutral  axis.  In 
Fig.  77,  s  is  positive  (compression)  when  v  is  positive,  and  s 
is  negative  (tension)  when  v  is  negative.  When  v  becomes  v2, 
Formula  VIII  gives  the  unit  stress  in  the  outer  (top)  fibers. 
In  these  equations,  M  is  the  resisting  moment  of  all  the  fibers 
and  is  equal  to  the  external  moment.  Notice  in  Fig.  77  that 
the  compression  in  the  top  fibers  and  the  tension  in  the  bottom 
fibers  at  the  section  EFG  both  turn  counterclockwise  about  the 
axis  BB;  the  resisting  moment  is  negative  and  the  external  mo- 
ment is  positive. 

Notice  that  this  discussion  assumes  that  the  modulus  of  elas- 
ticity is  constant  for  all  stresses  used  and  is  the  same  in  com- 
pression and  in  tension. 

The  theory  of  Articles  50  and  51  may  be  given  more  briefly.  Let  k  be  the 
unit  stress  at  unit  distance  from  the  neutral  axis.  At  a  distance  v  from  the 
neutral  axis,  • 

s  =  kv.  (2) 

On  an  increment  of  area  dA, 

total  stress  =  kvdA.  (3) 

The  moment  of  this  total  stress  with  respect  to  the  neutral  axis  is 

dM  =  total  stress  on  dA  multiplied  by  v  =  kv2  dA,  (4) 

which  is  positive  for  all  values  of  v  when  k  is  positive  and  vice  versa. 


M  =      kiMA  =  kl.  (5) 

If  si  is  the  unit  stress  at  a  distance  vi  from  the  neutral  axis  si  =  kvi, 

*-*;  (6) 

Substituting  (6)  in  (5): 

M  =  —  -  Formula  VII. 

Vi 

While  Formula  VIII  may  be  obtained  at  once  from  VII,  they 
are  of  such  importance  that  both 
should  be  memorized. 

52.  Location  of  the  Neutral 
Axis.  —  We  have  yet  to  obtain 
the  position  of  the  neutral  axis. 
To  do  this,  we  make  use  of  the 
fact  that  the  total  tensile  stress 
across  the  section  EFG  below 
the  neutral  axis  is  equal  to 
the  total  compressive  stress  above  the  neutral  axis. 

Consider  Fig.   79,  a  horizontal  cantilever.     Take  the  forces 


90  STRENGTH  OF  MATERIALS 

which  act  on  the  portion  on  the  left  of  the  section  EFG,  and  re- 
solve horizontally.  Since  the  external  force  has  no  horizontal 
component,  the  only  forces  to  be  considered  are  the  internal 
forces  at  the  section.  The  portion  is  in  equilibrium,  hence  the 
total  compressive  force  across  the  section  below  the  neutral  axis 
must  equal  the  total  tensile  force  across  the  section  above  the 
axis. 

Stress  on  dA  =  kvdA,  (1) 

which  has  the  same  sign  as  k  when  v  is  positive  and  the  opposite 

sign  when  v  is  negative. 

/»f2 

Total  stress  on  entire  section  =  k   I     vdA  =  0;  (2) 

t/t>i 

k  is  not  zero  when  the  beam  is  bent,  hence 

•» 
v  dA  must  equal  zero.  (3) 

In  Mechanics  we  learned  that  the  position  of  the  center  of  gravity 
of  a  plane  area  is  given  by 

fydA 

;      g    '      i-2*-*-'  ,;;.,;•          . 

vA=fidA.  (4) 

The  second  member  of  equation  (4)  is  zero,  hence 

vA  =  0. 
Since  A  is  not  zero 

v  =.  0. 

The  neutral  axis  in  a  beam  of  any  section  passes  through  the 
center  of  gravity  of  the  section. 

This  is  on  the  assumption  that  the  modulus  of  elasticity  is  constant  for  all 
stresses  used,  and  is  the  same  in  compression  and  tension.  Where  the  beam 
is  stressed  beyond  the  elastic  limit,  this  is  not  strictly  true.  Since  working 
loads  should  be  well  within  the  elastic  limit,  these  results  are  strictly  correct 
in  well-designed  structures. 

PROBLEMS. 

1.  A  beam  4  inches  wide,  6  inches  deep,  and  10  feet  long  is  supported  at 
the  ends.  Find  the  unit  stress  in  the  top  and  bottom  fibers  at  the  middle  of 
its  length  due  to  a  load  of  400  pounds  at  the  middle. 

The  moment  at  the  middle  is  1000  foot  pounds.     The  moment  of  inertia 

of  the  cross  section  with  respect  to  the  neutral  axis  is  72  inches4.     -  equals 

24  inches3  for  both  top  and  bottom  fibers.     The  external  moment  and  -  must 

v 


STRESSES  IN   BEAMS  91 

be  expressed  in  the  same  units.  Since  we  express  stress  in  pounds  per  square 
inch,  we  reduce  the  moment  to  inch  pounds  to  use  in  Formula  VIII. 

Ans.  s  =  500  pounds  per  square  inch. 

2.  A  beam  6  inches  wide,  8  inches  high,  and  15  feet  long  is  supported  at 
the  ends  and  carries  a  uniformly  distributed  load,  including  its  own  weight, 
of  160  pounds  per  foot.     Find  the  maximum  moment  in  inch  pounds  and  the 
maximum  fiber  stress. 

Ans.  Maximum  fiber  stress,  844  pounds  per  square  inch. 

3.  A  10-inch  25-pound  I-beam  is  supported  at  two  points  13  feet  4  inches 
apart  and  carries  a  load  of  6000  pounds  midway  between  the  supports.     Neg- 
lecting the  weight  of  the  beam,  find  the  maximum  fiber  stress  under  the  load. 
(See  Cambria  for  the  moment  of  inertia  of  I-beams.) 

Ans.  9830  pounds  per  square  inch. 

53.   Section  Modulus.  —  The  expression  -  ,  where  v  is  the  dis- 

I  tance  to  the  extreme  outer  fiber,  is  used  so  often  that  it  is  con- 
venient to  give  it  a  name.  It  is  called  the  section  modulus,  or 
modulus  of  the  section.  Formula  VIII  becomes 

.,  .  „,  external   moment 

unit  stress  in  extreme  fibers  = —. —       — j— = 

section   modulus 

The  value  of  the  section  modulus  for  I-beams  and  channels  is 
given  in  Cambria  in  column  8  of  the  properties  of  these  shapes. 
It  is  convenient  to  know  the  section  modulus  of  a  rectangular 
section.  If  the  breadth  is  b  and  the  depth  is  d,  the  moment  of 

inertia  is  ^->  and  v  is  ~>  hence  the  modulus  of  the  section  is: 

/      bd* 

~  =  ~^r-  Formula  IX. 

v        6 


PROBLEMS. 

1.  How  much  stronger  is  a  beam  6  inches  wide  and  8  inches  deep,  than  a 
beam  of  the  same  material  4  inches  wide  and  6  inches  deep?  Ans.  8  :  3. 

2.  Compare  the  strength  of  a  4-inch  by  6-inch  beam  with  the  6-inch  side 
vertical,  with  that  of  the  same  beam  with  the  4-inch  side  vertical. 

3.  Show  that  the  ratio  of  the  strength  of  a  beam  of  rectangular  section 

with  the  side  b  vertical  to  its  strength  with  the  side  d  vertical  is  -3  • 

4.  A  4-inch  by  6-inch  cantilever  projects  6  feet  from  a  wall  and  carries  a 
distributed  load  of  40  pounds  per  foot  and  a  load,  P,  3  feet  from  the  free  end. 
What  is  the  maximum  value  of  P  if  the  fiber  stress  shall  not  exceed  1000 
pounds  per  square  inch?  Ans.  427  pounds. 

(In  giving  dimensions,  horizontal  are  given  first.     A  4-inch  by  6-inch  beam 
is  4  inches  wide  and  6  inches  deep.) 


92  STRENGTH  OF  MATERIALS 

5.  Find  the  I-beam  for  a  span  of  15  feet  with  a  uniformly  distributed  load, 
including  the  weight  of  the  beam,  of  600  pounds  per  foot,  and  a  load  of  4000 
pounds  at  the  middle,  with  allowable  stress  13,500. 

Ans.  The  section  modulus  is  28.3.  Use  a  10-inch  35-pound  I-beam, 
or  a  12-inch  31.5-pound  I-beam.  The  latter  is  cheaper  and  stronger,  and 
should  be  used  unless  the  2  inches  of  head  room  is  important. 

6.  A  15-inch  45-pound  I-beam,  supported  at  the  ends,  is  used  for  a  20-foot 
span  to  carry  a  uniformly  distributed  load,  including  its  own  weight,  of  500 
pounds  per  foot,  and  a  load  of  3000  pounds  4  feet  from  the  left  support. 
Find  the  maximum  fiber  stress.  Ans.  6189  pounds  per  square  inch. 

7.  Find  the  I-beam  for  a  span  of  20  feet  to  carry  a  uniformly  distributed 
load,  inclusive  of  its  own  weight,  of  600  pounds  per  foot  and  two  concentrated 
loads,  2000  pounds  3  feet  from  the  left  support  and  3000  pounds  4  feet  from 
the  right  support,  if  the  allowable  unit  stress  is  12,500  pounds  per  square  inch. 

Ans.  A  12-inch  35  pound  I-beam. 

8.  A  wooden  beam  is  used  for  a  span  of  12  feet  6  inches  to  carry  a  load  of 
4000  pounds  at  the  middle.     If  the  breadth  of  the  beam  is  8  inches  and  the 
allowable  working  stress  is  1000  pounds  per  square  inch,  what  is  the  minimum 
depth? 

9.  A  yellow-pine  beam  6  inches  wide,  10  inches  deep,  and  20  feet  long  is 
supported  at  the  left  end  and  4  feet  from  the  right  end  and  carries  a  distributed 
load,  including  its  own  weight,  of  120  pounds  per  foot  and  two  concentrated 
loads,  1200  pounds  2  feet  from  the  left  end  and  400  pounds  at  the  right  end. 
Construct  the  shear  diagram.     Find  the  moment  at  each  dangerous  section, 
and  the  maximum  fiber  stress. 

Ans.  Maximum  fiber  stress,  499  pounds  per  square  inch. 

10.  In  Problem  9  there  is  a  point  between  the  first  dangerous  section  and 
the  right  support  at  which  the  moment  is  zero.     Find  this  position  by  writing 
the  moment  equation  in  terms  of  the  distance  from  the  left  end  and  equating 
to  zero.     Solve  also  by  means  of  the  shear  diagram,  making  use  of  the  fact 
that  the  area  of  the  shear  diagram  between  the  point  to  be  found  and  the 
first  dangerous  section  is  equal  to  the  area  to  the  left  of  this  dangerous 
section. 

11.  Using  New  York  building  laws,  calculate  the  maximum  allowable  load, 
uniformly  distributed,  on  a  2-inch  by  12-inch  floor  joist  of  long-leaf  yellow  pine, 
of  16  feet  8  inches  span.  Ans.  2304  pounds,  138  pounds  per  foot. 

12.  If  the  weight  of  the  joist,  flooring,  and  plastering,  in  Problem  11,  is 
20  pounds  per  square  foot,  is  the  construction  allowable  for  a  school  building 
with  joists  spaced  12  inches  center  to  center? 

13.  Find  the  maximum  span,  according  to  New  York  building  laws,  for 
a  2-inch  by  10-inch  long-leaf  yellow-pine  joist  supporting  a  distributed  load, 
including  its  own  weight,  of  120  pounds  per  foot.  Ans.  14.9  feet. 

14.  What  is  the  maximum  span  for  a  20-inch  65-pound  I-beam  supporting 
a  uniformly  distributed  load,  including  its  own  weight,  of  1500  pounds  per  foot, 
if  the  allowable  unit  stress  is  13,500  pounds  per  square  inch? 

Ans.  26  feet  6  inches. 

15.  Determine  the  moment  of  inertia  of  a  circular  section  of  radius  a,  and 

show  that  the  section  modulus  is  —  • 

4 

16.  What  is  the  section  modulus  of  a  6-inch  square  with  the  diagonal  ver- 


STRESSES  IN  BEAMS 


93 


tical?     How  does  it  compare  with  the  section  modulus  of  the  same  section 
with  one  side  vertical?  Ans.  Section  modulus,  25.45;  ratio,  1  :  A/2. 

17.  A  square  section  with  diagonal  vertical  has  its  section  modulus  in- 
creased by  chamfering  the  top  and  bottom  corners.     What  must  be  the 
dimensions  of  the  triangular  sections  cut  away,  in  terms  of  the  side  of  the 
square,  to  make  the  section  modulus  a  maximum? 

18.  A  wooden  box  girder  is  made  of  two  2-inch  by  12-inch  planks  and  two 
2-inch  by  8-inch  planks.     What  is  the  section  modulus  when  these  are  united 
to  form  a  girder  of  square  section?  Ans.  231  inches3. 

19.  A  box  girder  is  made  of  two  10-inch  15-pound  channels  and  two  12- 
inch  by  Hnch  plates.     What  is  the  section  modulus  with  the  channels  verti- 
cal?    (Take  /  of  channels  from  table  in  Cambria.) 

20.  A  6-inch  by  8-inch  cantilever  10  feet  long  is  placed  with  its  8-inch 
faces  30  degrees  to  the  vertical  and  a  load  of  200  pounds  applied  at  the  free 
end.     Find  the  maximum  fiber  stress  at  the  corners. 

Ans.  574.7  pounds  per  square  inch. 

Resolve  the  force  or  the  moment  perpendicular  and  parallel  to  the  principal 
axes  of  inertia.  Find  unit  stress  separately  and  add. 

54.  Graphic  Representation  of  Stress  Distribution.  —  The 
unit  stress  in  a  beam,  provided  it  does  not  exceed  the  elastic 
limit,  varies  as  the  distance  from  the  neutral  axis.  It  may  be 
represented  by  the  straight  line  EF  of  Fig.  80.  This  straight 


H      F          C 


3 

\      >F= 

X^ 

^—  > 

II 


Fig.  80.  —  Variation  of  Stress  in  Sec-      Fig.  81.  —  Stress  Distribution  in  a 
tion.  Rectangular  Section. 

line  is  really  a  part  of  the  straight-line  portion  of  the  stress 
diagram  for  both  tension  and  compression,  while  the  vertical 
line  EC  is  the  X  axis.  If  the  stress  is  carried  beyond  the  elastic 
limit,  the  dotted  line  GH  represents  its  distribution  in  the  beam. 
In  a  beam  of  rectangular  section,  the  total  stress  on  any  area 
dA,  extending  across  the  section,  is  proportional  to  the  unit 
stress.  The  shaded  area  of  Fig.  80  may  represent  the  total 
stress  in  a  rectangular  section  as  well  as  the  unit  stress  in  a  sec- 
tion of  any  form.  It  is  often  convenient  to  represent  total  stress 


94  STRENGTH  OF  MATERIALS 

in  a  rectangular  section  by  a  figure  similar  to  the  shaded  area  in 
Fig.  81.  This  is  really  the  same  as  Fig.  80  with  oblique  axes. 
The  line  FC  represents  the  breadth  of  the  section  and  also  the 
total  stress  in  the  extreme  outer  fibers.  It  is  evident,  from  the 
similar  triangles,  that  t'he  total  stress  on  the  area  dA,  extending 
across  the  section  at  a  distance  v  from  the  neutral  axis,  will  be  to 
the  total  stress  at  the  top,  as  the  length  M  N  is  to  the  length  FC. 
The  actual  stress  over  the  entire  section  is  equal  to  a  uniform 
stress  of  intensity  equal  to  that  in  the  outer  fibers  over  the  shaded 
area.  If  the  cross  section  is  drawn  to  full  scale,  the  area  of  the 
shaded  triangle  OFC  gives  the  total  stress  above  the  neutral  sur- 
face when  the  maximum  stress  is  1  pound  per  square  inch.  Like- 
wise, the  area  OBG  gives  the  total  stress  below  the  neutral  surface. 
These  triangular  areas  are  equal  in  magnitude  and  opposite  in 
sign,  making  the  sum  of  the  total  stress  zero. 

The  section  modulus  of  a  rectangular  section  may  be  computed 

bd 
from  these  diagrams.     The  area  of  the  triangle  OFC  is'-j--    A 

uniform  stress  of  intensity  unity  over  the  entire  triangle  would 

bd 
have  a  resultant  -j-  at  the  center  of  gravity  of  the  triangle.     The 

moment  of  the  upper  half  with  respect  to  the  neutral  axis  is 

bd    d  =  bd* 
4*3       12 ' 

The  moment  of  the  lower  half  is  the  same  and  rotates  in  the  same 
direction  (since  both  stress  and  moment  arm  change  signs). 

bd2 

The  total  moment  is  -— • 
6 

When  the  unit  stress  in  the  outer  fibers  is  s  instead  of  unity 

sbd* 

total  moment  =  — 5 — 
o 

We  will  now  draw  the  stress  distribution  for  an  I-beam  section. 
This  diagram  for  the  rectangular  portion  of  the  flange  is  drawn 
like  Fig.  81.  Now  take  a  small  area  dA,  Fig.  82,  in  the  web. 
The  total  stress  in  this  area  is  to  the  total  stress  in  a  similar  area 
JK  at  the  top  as  the  distance  of  dA  from  the  neutral  axis  is  to  the 
distance  of  J K  from  this  axis.  To  get  the  total  stress  on  dA, 
we  first  project  its  length  on  the  upper  surface  to  locate  the  points 
JK.  Then  draw  straight  lines  from  the  center  0  to  J  and  K. 


STRESSES   IN   BEAMS 


95 


The  part  of  dA  between  these  lines  represents  the  total  stress. 
To  get  the  stress  on  the  triangular  portion  of  the  flange,  consider 
the  portion  ST  drawn  (for  convenience)  in  the  lower  flange. 
Project  S  and  T  on  the  lower  line  and  connect  the  center  0  with 
the  points  thus  found  by  means  of  the  dotted  lines.  The  portion 
S'T'  between  these  lines  measures  the  total  stress.  A  number 
of  these  lines  will  give  the  curved  area  required. 


j  K 


NEUTRAL  AXIS 


NEUTRAL  AXIS 


S'         T' 

Fig.  82.  —  Stress  Distribution 
in  an  I-beam. 


Fig.  83.  —  Distribution  in  a 
T-section. 


Fig.  83  is  the  stress  distribution  in  a  T  shaped  section.  The 
lower  portion  is  constructed  like  Fig.  81.  The  outer  fibers  at 
the  top  are  nearer  the  neutral  axis  than  those  at  the  bottom. 
Instead  of  projecting  on  the  upper  line  of  the  section,  we  project 
on  a  line  CF  whose  distance  from  the  neutral  axis  is  the  same  as 
the  lower  fibers.  The  total  stress  in  this  diagram  is  expressed 
in  terms  of  the  unit  stress  in  the  bottom  fibers.  We  might 
express  the  total  stress  in  terms  of  the  unit  stress  in  the  top 
fibers.  In  that  case  the  lower  part  of  the  diagram  would  extend 
beyond  the  section  to  the  right  and  left. 

PROBLEMS. 

1 .  Construct  the  stress-distribution  diagram  for  a  6-inch  by  4-inch  by  1-inch 
angle  section,  using  neutral  axis  parallel  to  shorter  leg. 

2.  A  beam  12  inches  deep  is  constructed  of  one  12-inch  by  £-inch  plate  and 
four  4-inch  by  3-inch  by  |-inch  angles  with  shorter  legs  parallel  to  the  plate. 
Construct  the  stress-distribution  diagram. 


96 


STRENGTH  OF  MATERIALS 


3.  Construct  the  stress-distribution  diagram  for  a  6-inch  by  8-inch  rectan- 
gular section  with  the  6-inch  side  horizontal.  Calculate  from  the  diagram 
the  total  stress  in  the  upper  strip  1  inch  high  and  6  inches  wide  as  compared 
with  the  total  stress  in  a  similar  strip  touching  the  neutral  axis. 

55.  Stress  Beyond  the  Elastic  Limit.  —  In  Fig.  85  the  shaded 
area  shows  the  distribution  of  stress  in  a  rectangular  section, 
when  the  stress  is  considerably  beyond  the  elastic  limit.  The 
actual  stress  in  the  outer  fibers  is  less  than  it  would  be  if  the 
modulus  were  constant  in  the  ratio  of  the  lengths  CH :  CF. 
The  moment  of  resistance  is  also  less. 


Fig.  84.  —  Distribu- 
tion in  an  Angle 
Section. 


E  G      E 

Fig.  86.  —  Rectangular  Section  Be- 
yond the  Elastic  Limit. 


Fig.  86  represents  the  stress  distribution  when  the  elastic 
limit  is  exceeded  as  compared  with  a  beam  of  constant  modulus 
having  the  same  resisting  moment.  The  moment  of  the  curved 
area  OMKHC  must  equal  the  moment  of  the  triangular  area 
OFC.  From  the  center  of  the  section  to  the  point  K  the  curve 
lies  outside  of  the  straight  line.  Beyond  K  it  is  inside.  The 
unit  stress  in  the  fibers  near  the  neutral  surface  is  greater  than 
if  the  modulus  were  constant;  and  the  unit  stress  in  the  outer 
fibers  is  Ies3.  The  moment  of  the  dotted  area  OMK  (or  the 
shaded  area  ONL)  is  equal  to  that  of  the  area  KFH  or  LGE. 

56.  Modulus  of  Rupture.  —  When  a  beam  is  broken  by  bend- 
ing, the  stress- distribution  diagram  for  a  rectangular  section, 
OMKH  (Fig.  86),  is  similar  to  the  complete  tension  or  compres- 


STRESSES  IN  BEAMS  97 

sion  curve  of  the  material.  The  actual  unit  stress  in  the  outer 
fibers  is  less  than  that  obtained  from  the  equation 

8  =  ^j-  Formula  VIII. 

in  the  ratio  of  CH:CF  (Fig.  86.)  The  calculated  value  of  the 
stress  in  the  outer  fibers  computed  from  Formula  VIII  is  called 
the  modulus  of  rupture,  or  the  transverse  ultimate  strength  of  the 
material.  It  is  also  called  the  extreme  fiber  stress  in  bending. 

While  the  modulus  of  rupture  does  not  give  the  actual  stress, 
it  enables  us  to  compare  stresses  in  similar  sections.  If  the 
modulus  of  rupture  is  obtained  from  the  test  of  beams  of  rec- 
tangular section,  this  figure  may  be  used  in  computing  the  ulti- 
mate transverse  load  in  other  beams  of  rectangular  section  made 
of  the  same  material.  The  results  may  also  be  used  with  little 
error  for  beams  of  other  shapes,  provided  they  are  symmetrical 
with  respect  to  the  neutral  axis.  With  unsymmetrical  sections, 
such  as  angles,  it  is  better  to  make  tests  and  obtain  the  modulus 
of  rupture  for  each  shape. 

The  student  will  remember,  however,  that  these  statements 
apply  to  the  stress  beyond  the  elastic  limit.  Since  allowable 
stresses  are  below  the  elastic  limit,  Formula  VIII  is  strictly 
correct  for  allowable  loads.  The  change  in  the  stress-distribu- 
tion diagram  when  the  stress  passes  the  elastic  limit  affects  the 
factor  of  safety  only. 

Ductile  materials,  such  as  soft  steel,  have  no  modulus  of 
rupture,  strictly  speaking,  since  beams  of  such  material  may  be 
bent  double  without  breaking. 

PROBLEMS. 

1.  A  white-pine  beam  1.78  inches  wide  and  1.25  inches  thick  was  supported 
at  two  points  12  inches  apart  and  broken  by  a  load  at  the  middle.     The  total 
load  at  rupture  was  1112  pounds.     Find  the  modulus  of  rupture. 

Ans.  7200  pounds  per  square  inch. 

2.  If  white  pine  of  quality  equal  to  that  of  Problem  1  is  used  in  the  form 
of  a  4-inch  by  4-inch  beam  to  carry  a  load  of  700  pounds  midway  between  two 
supports  6  feet  apart,  what  is  the  factor  of  safety? 

3.  A  rectangular  bar  of  cast  iron  1.04  inches  wide  and  0.80  inch  thick  is 
placed  on  two  supports  12  inches  apart  and  broken  by  a  load  of  1635  pounds 
at  the  middle.     What  should  be  the  allowable  working  stress  in  this  cast  iron 
in  beams  of  rectangular  section  with  a  factor  of  safety  of  10? 

Ans.  4400  pounds  per  square  inch. 


98  STRENGTH  OF  MATERIALS 

4.  A  beam  of  short-leaf  yellow  pine,  tested  by  Prof.  A.  N.  Talbot  at  the 
University  of  Illinois,  had  the  following  dimensions:  breadth,  7.12  inches; 
depth,  16.25  inches;  distance  between  supports,  13  feet  6  inches.     Two  equal 
loads  were  applied  at  points  4  feet  6  inches  from  the  supports,  making  the 
bending  moment  constant  and  the  shear  zero  between  these  points  (if  the 
weight  of  the  beam  is  neglected).     The  beam  broke  by  tension  in  the  outer 
fibers  between  the  loads  when  each  load  was  27,500  pounds.     Find  the  modulus 
of  rupture.  .      Ans.  4739  pounds  per  square  inch. 

5.  A  beam  of  long-leaf  yellow  pine  7.0  inches  wide  and  14.0  inches  deep, 
supported  and  loaded  as  in  Problem  4,  broke  under  a  total  load  of  37,300 
pounds.     What  was  the  ultimate  bending  strength  of  this  timber? 

Ans.  4400  pounds  per  square  inch. 

6.  A  second  beam  of  long-leaf  yellow  pine  7.0  inches  by  12.1  inches,  sup- 
ported and  loaded  as  above,  broke  under  a  total  load  of  52,900  pounds.     What 
was  the  ultimate  bending  strength  of  this  timber? 

Ans.  8362  pounds  per  square  inch. 

Problems  4,  5,  and  6  are  from  Bulletin  41  of  the  University  of  Illinois 
Engineering  Experiment  Station. 

7.  A  5-inch  by  6-inch  beam  of  1:2:4  concrete,  placed  on  supports  32 
inches  apart,  was  broken  by  a  load  of  1300  pounds  midway  between  the  sup- 
ports.    Neglecting  the  weight  of  the  beam,  find  the  modulus  of  rupture. 

Ans.  347  pounds  per  square  inch. 


CHAPTER  VII. 
DEFLECTION  IN  BEAMS. 

57.   Deflection  and  Moment.  —  In  Article  50,  equation  (5), 
we  have: 


dl 


(1) 


for  infinitesimal  lengths  dl,  measured  along  the  neutral  surface 
of  the  bent  beam.  The  angle  dd  is  the  change  in  slope  of  the 
tangent  to  the  neutral  surface  in  the  length  dl.  We  will  now 
determine  the  relation  existing  between  the  moment  and  the 
deflection  of  the  beam  from  its  original  form.  This  is  expecially 
easy  in  polar  coordinates.  The  lines  FG  and  MNj  of  Figs.  87 


Fig.  87.  —  Curvature  of  Beam. 

and  77,  make  an  angle  dd  with  each  other  (when  AZ  becomes  dl), 
and  intersect  at  some  point  beyond  the  drawing,  at  a  distance  p 
from  the  neutral  surface.     This  distance  p  is  the  radius  of  curva- 
ture of  the  neutral  surface. 
By  geometry: 

pde  =  dl,  (2) 


dl 


(3) 


Substituting  in  (1), 


M_  =  l       M  =  EI_ 
~EI~~p  p 

If  M  is  constant,  or  if  7  varies  as  M ,  p  is  constant,  and  the  curve 
of  the  beam  is  an  arc  of  a  circle  which  may  be  computed  by 
trigonometry. 


100  STRENGTH  OF  MATERIALS 

PROBLEMS. 

1.  A  3-inch  by  1-inch  steel  beam  10  feet  Igng  rests  on  two  supports  each 
30  inches  from  the  ends  and  carries  a  load  of  200  pounds  on  each  end.     Neg- 
lecting the  weight  of  the  beam,  what  is  the  bending  moment  at  the  supports? 
If  the  modulus  of  elasticity  is  30,000,000,  what,  is  the  radius  of  curvature? 
How  much  is  the  beam  deflected  upward  at  the  middle? 

{Moment,  6000  inch  pounds; 
Radius  of  curvature,  1250  inches; 
Deflection  at  the  middle,  0.36  inch. 

SUGGESTION.  —  With  the  radius  of  curvature  known,  calculate  the  angle  in 
radians  subtended  by  half  the  span.  The  versed  sine  of  this  angle  multiplied 
by  the  radius  of  curvature  is  the  deflection  at  the  middle.  As  ordinary  tables 
are  of  little  value  for  such  small  angles,  it  is  recommended  that  the  student  use 
the  first  two  terms  of  the  cosine  series  to  get  this  versed  sine.  (See  trigonom- 
etry for  series  or  develop  by  Maclaurin's  formula.) 

2.  In  Problem  5  of  Article  56,  if  E  is  1,500,000,  what  was  the  radius  of 
curvature  when  the  total  load  was  30,000  pounds? 


200*  \  '  200* 


Fig.  88. 

58.   Deflection  in  Rectangular  Coordinates.  —  To  express  the 

value  of  7^7  in  rectangular  coordinates,  we  must  determine  —^  in 
nil  al 

terms  of  x  and  y  and  their  derivatives.  Let  x  express  distance 
parallel  to  the  unbent  beam  and  y  express  deflection  of  the  beam 
from  its  original  position.  These  distances  are  usually  measured 
along  the  neutral  surface.  The  angle  6  may  be  measured  from 
any  fixed  line.  For  convenience  of  calculation,  we  will  measure 
0  from  a  line  parallel  to  the  X  axis.  The  angle  6  is  then  the 
angle  which  the  tangent  to  the  curved  beam  makes  with  the 
original  direction  of  the  beam.  It  is  the  angle  through  which 
the  tangent  to  the  beam  at  any  point  is  turned  when  the  beam  is 
bent. 

Fig.  89  shows  a  beam  supported  at  the  ends  and  bent.  The 
lower  figure  represents  the  neutral  axis  with  the  vertical  deflection 
exaggerated.  The  origin  is  taken  at  the  left  support,  and  x  is 


DEFLECTION  IN  BEAMS 


101 


taken  as  positive  to  the  right  and  y  as  positive  upwards,  as  is  the 
custom  in  mathematical  work. 
From  Fig.  89  (or  the  Calculus) : 


(1) 
(2) 


Differentiating  (2) : 


dO           dx2 

dx             /dy\2 

1 

\ 

1 

BEAM 

7 

T 

(3) 


;  COS  6  = 


From  Fig.  89,  dx  =  dl  cos  0,  which,  substituted  in  (3),  gives: 

d?y  d*y 

d0=       dx2  .  rfx2 

«-14; 


since 


sec  0  — 


Substituting  this  value  of  -    in  (1)  of  Article  57 


(4) 


M  =  El 


dx2 


(5) 


Equation  (5)  may  be  obtained  direct  from  (4)  of  Article  57  if  the 
student  remembers  the  expression  for  the  radius  of  curvature 
from  his  Calculus. 


102  STRENGTH  OF  MATERIALS 

In  beams,  as  used  in  engineering  practice,  the  deflections 
allowed  are  small,  and  -^  is  usually  so  small  compared  with  unity 
that  it  may  be  dropped  from  the  denominator  of  equation  (5). 
It  seldom  happens  that  ~  is  greater  than  0.01.  This  makes  f -r-j 

not  greater  than  0.0001,  and  the  error  in  dropping  it  entirely  from 
the  denominator  is  not  more  than  one  part  in  7000.  Equation 
(5)  then  becomes: 

M  =  El  -j-^  •  Formula  X. 

Formula  X  is  the  differential  equation  which  enables  us  to 
calculate  the  deflection  of  a  beam  or  column.  The  X  axis  must 

dii 
be  parallel  to  the  direction  of  the  unbent  beam  and  •—  must  be  so 

small  that  its  square  is  negligible  compared  with  unity. 

To  determine  the  deflection  of  a  beam,  we  must  solve  this  dif- 
ferential equation  for  y  in  terms  of  x  and  the  constants.  Where 
E  and  I  are  constant  this  means  that  we  must  express  M  in  terms 
of  x  and  y  and  solve.  When  alLthe  loads  are  vertical  and  the 
beam  horizontal,  or  in  general,  when  all  loads  and  reactions  are 
perpendicular  to  the  unbent  beam,  M  may  be  expressed  in  terms 
of  x  alone.  Our  problem  then  is  to  solve  a  differential  equation 
of  the  second  order  and  first  degree  of  the  form 

d2y 

-jgj  =  function  of  x. 

To  solve  this  we  merely  integrate  twice.  There  are  two  constants 
of  integration  which  must  be  determined  from  the  conditions  of 
the  problem.  There  are  two  things  to  be  done  in  solving  these 
problems.  The  first  is  to  write  an  expression  for  the  moment  at 
any  point  in  terms  of  x  and  the  known  loads  and  reactions.  The 
second  is  to  determine  two  conditions  from  which  the  two  inte- 
gration constants  may  be  evaluated. 

59.  Beam  Supported  at  Two  Points;  Moment  Constant.— 
This  is  the  case  shown  in  Fig.  90.  A  beam  rests  on  two  supports 
at  a  distance  I  apart,  overhangs  these  supports,  and  carries  loads 
on  the  free  ends  which  make  the  moment  at  each  support  the 
same  (Pc  =  Qe).  Let  M  be  the  value  of  this  moment.  If  the 
weight  of  the  beam  is  neglected,  the  moment  is  constant  through- 


DEFLECTION  IN  BEAMS  103 

out  the  span  of  length  /  from  one  support  to  the  other.  We  will 
consider  this  portion  only.  Remembering  that  the  sign  of  the 
moment  is  negative  and  substituting  in  Formula  X: 

nr^y  nf  /i\ 

El  -^  =  —M.  (1) 

Integrating : 

dx  =  ~     X  '       *' 
Integrating  again: 

EIy  =  -^j-  +  dx  +  C,.  (3) 

f  I  ff 

-c — 4< jc- J  |<—  e— JJ 


Fig.  90.  —  Beam  with  Constant  Moment. 

To  obtain  the  constants  d  and  C2,  we  have  the  condition  that 
y  =  0  at  the  left  support  where  x  =  0.  Substituting  in  equation 
(3),  we  get: 


Ely--       -  +  ClX.  (4) 

Equation  (4)  is  true  for  all  values  of  x  for  which  the  moment  is 
.  —  M.     It  is  true  at  the  right  support,  where  x  =  I  and  y  =  0. 
Substituting  in  (4)  : 

M12  +  ci  C       Ml 

-g-  +  Gi£,  Ci  =:  -g-- 

Substituting  this  value  of  Ci  in  equation  (4),  we  get: 


Equation  (6),  which  gives  the  value  of  y  in  terms  of  x  and  the 
constants  El  and  M ,  for  all  values  of  x  between  the  supports,  is 


104  STRENGTH  OF  MATERIALS 

called  the  equation  of  the  elastic  line  of  the  beam  between  these 

points.     To  find  the  position  of  maximum  deflection,  let  -p  =  0 

ax 

in  equation  (2).     After  substituting  the  value  of  Ci,  we  get: 


_^ 
dx~  EI\2 

~r  —  0    when    x  =  -> 
dx  2 

hence  the  point  of  maximum  deflection  is  at  the  middle  of  the 
span.  Substituting  in  (6),  to  get  the  maximum  deflection  at  the 
middle: 

Ml2  ,_ 

-- 


It  is  evident  that  if  the  loads  P  and  Q  are  equal  and  the  lengths  c  and 
e  are  equal,  the  beam  would  be  symmetrical  with  respect  to  the  middle  of 
the  span;  and  that  this  point  would  be  the  position  of  maximum  deflection. 

In  that  case  we  could  set  -—  equal  to  0  when  x  equals  -  in  equation  (2)  and 

solve  for  Ci  before  integrating  the  second  time.     If  P  and  Q  are  not  equal,  but 
the  products 

PC  =  Qe, 

the  symmetry  is  not  so  self-evident,  and  it  is  safer  to  obtain  the  constants  as 
we  have  done. 

PROBLEMS. 

1.  Show  that  the  deflection  for  all  parts  of  the  span  is  positive. 

2.  Apply  equation  (7)  to  Problem  1  of  Article  57. 

3.  Apply  equation  (6)  to  the  above  problem  to  find  the  deflection  at  10 
inches,  20  inches,  and  40  inches  from  the  left  support. 

Ans.          x  y 

10  inches.        0.20  inch. 
20      "  0.32      " 

40       "  0.32     " 

4.  In  the  above  problem  find  the  slope  of  the  tangent  at  either  support, 

fdu\2 
and  find  how  much  1  +  I  -~  ]  differs  from  unity. 


Ans.  0.024,  -0.024,  0.000576. 

5.  A  4-inch  by  4-inch  wooden  beam  12  feet  long  is  supported  4  feet  from 
the  left  end  and  3  feet  from  the  right  end.  A  load  of  150  pounds  is  placed  on 
the  left  end  and  a  load  of  200  pounds  on  the  right  end.  A  point  midway 
between  the  supports  rises  0.112  inch  when  the  loads  are  applied.  Find  E. 

Ans.  1,350,000  pounds  per  square  inch. 

60.  Cantilever  Beam  with  Load  on  the  Free  End.  —  Fig.  91 
represents  a  cantilever  beam  fixed  at  the  right  end  and  loaded 


DEFLECTION   IN  BEAMS 


105 


at  the  left  end.  The  origin  of  coordinates  is  taken  from  the  left 
end  before  the  load  was  applied.  The  moment  at  a  distance  x 
from  the  origin  is  —Px.  The  differential  equation  becomes: 


=  -Px. 
dx2 

dy          Px 

— 


(1 

'2) 


Vmaae 


Fig.  91.  —  Cantilever  with  Load  on  Free  End. 

At  the  wall,  where  x  =  Z,  the  beam  is  horizontal  and  -^- 

PI2 


=  0; 


dy 


Px*      PI* 


Px*     Pl*x 
EIy  =      ~          ^~ 


At  the  wall  x  =  I,  y  =  0; 


(3) 
(4) 
(5) 


c,  = 


PP 


Px3  ,  Pl"x      PI3 

=     ""   ~~  '' 


EI\S  "  2  "*"'«'" 

The  maximum  deflection  is  at  the  free  end,  where  x  =  0. 

PI3 


(6) 

(7) 
(8) 


2/max  = 


3  El 


Formula  XI. 


106 


STRENGTH  OF  MATERIALS 


PROBLEMS. 

1.  A  wooden  cantilever  6  inches  square  and  10  feet  long  is  deflected  0.64 
inch  at  the  end  by  a  load  of  180  pounds  at  the  end.     Find  E  and  maximum 
fiber  stress. 

Ans.  E,  1,500,000;  maximum  stress,  600  pounds  per  square  inch. 

2.  In  Problem  1  what  is  the  deflection  3  feet  from  the  free  end? 

Ans.  0.36  inch. 

3.  A  cantilever  beam  8  feet  4  inches  long  is  deflected  0.2  inch  at  the  end 
by  a  load  at  the  end.     What  is  the  deflection  30  inches  from  the  free  end? 

4.  A  4-inch  by  4-inch  wooden  cantilever  10  feet  long  is  deflected  1  inch  at 
the  end  by  a  load  at  the  end.     If  E  equals  1,200,000  pounds  per  square  inch, 
what  is  the  maximum  fiber  stress? 

5.  A  10-inch  I-beam,  as  a  cantilever  8  feet  long,  is  deflected  0.35  inch  at 
the  end  by  a  load  at  the  end.     If  E  is  29,000,000,  find  the  maximum  fiber 
stress.  Ans.  16,520  pounds  per  square  inch. 

6.  If  the  allowable  fiber  stress  in  a  wooden  beam  is  1000  pounds  per  square 
inch  and  the  modulus  of  elasticity  is  1,500,000,  derive  the  expression  for  the 
deflection  at  the  end  of  a  cantilever  of  length  I  and  depth  d  due  to  the  maximum 
allowable  load  at  the  end. 

7.  A  cantilever  of  length  a  +  c  has  a  load  P  at  a  distance  a  from  the  fixed 
end.     Find  the  deflection  under  the  load  and  at  the  free  end. 

8.  A  cantilever  of  length  a  +  c  has  a  load  P  at  the  free  end.     Find  the 
deflection  at  a  distance  c  from  the  free  end. 

61.   Cantilever  with  Uniformly  Distributed  Load.  —  Fig.  92 
represents  a  cantilever  with  a  uniformly  distributed  load.     We 


Fig.  92.  —  Cantilever  with  Uniformly  Distributed  Load. 

have  drawn  this  cantilever  fixed  at  the  left,  and  taken  the  origin 
at  the  wall.  To  determine  the  moment,  we  use  the  free  end  of 
length  I  —  x  to  the  right  of  the  section  at  a  distance  x  from  the 
origin.  If  w  is  the  load  per  unit  length,  the  weight  of  this  free 

portion  is  w  (I  —  x) .     Its  moment  arm  is  — ^— : 


M  =  —w 


DEFLECTION  IN  BEAMS  107 

The  sign  is  negative,  since  the  moment  with  which  the  left 
portion  tends  to  turn  the  right  portion  is  counterclockwise  (or 
the  beam  is  convex  upward). 

d*y  _       w(l-x}* 
hidtf-          ~2~ 

El      =  *^^  +  C,  .....  :  (2) 


At  the  wall,  where  x  =  0,  ~  =  0; 


24 

At  x  =  0,  y  =  0: 

' 


(7) 


The  maximum  deflection  is  at  the  point  where  x  =  I; 

1nu  w/3 

Wl  VV  L  -,-,  ,      ___ 

2/max  =  —  Q-^rf  =  ~~  ^^7'  Formula  XII. 

o  HiL  o  rjL 

where  W  =  wl,  the  total  distributed  load. 

We  stated  that  the  maximum  deflection  is  at  the  point  where  x  —  I.  This 
is  not  a  mathematical  maximum,  where  the  slope  of  the  tangent  is  zero.  It 
is  a  numerical  maximum  because  the  beam  ends  at  this  point.  The  curves 
considered  in  the  Calculus  are  of  indefinite  extent. 

PROBLEMS. 

1.  What  is  the  deflection  at  the  end  of  a  6-inch  by  6-inch  cantilever  12  feet 
long  due  to  a  distributed  load  of  40  pounds  per  foot,  if  E  is  1,200,000,  and 
what  is  the  maximum  fiber  stress? 

^       (  Deflection  at  free  end,  1.38  inches; 

'  1  Fiber  stress  at  wall,  960  pounds  per  square  inch. 

2.  If  the  allowable  fiber  stress  in  a  wooden  beam  is  1000  pounds  per  square 
inch  and  the  modulus  of  elasticity  is  1,500,000,  derive  the  expression  for  the 
deflection  at  the  end  of  a  cantilever  of  length  I  and  depth  d  due  to  uniformly 
distributed  load  which  develops  the  allowable  stress. 


108  STRENGTH  OF  MATERIALS 

3.  How  does  the  deflection  of  a  cantilever  with  a  uniformly  distributed 
load  compare  with  the  deflection  of  a  cantilever  with  a  concentrated  load  at 
the  free  end,  if  the  fiber  stress  is  the  same  in  both  cases? 

4.  Expand  equation  (1)  and  integrate  to  get  the  equation  of  the  elastic 
line.     Then  expand  equation  (7)  and  compare  the  results. 

5.  Take  a  cantilever  fixed  at  the  right  end  as  in  Fig.  91,  with  a  uniformly 
distributed  load  of  w  per  unit  length.     With  the  origin  at  the  free  end  as  in 
Fig.  91,  write  the  differential  equation  and  solve  for  the  equation  of  the  elastic 
line.     Compare  the  resulting  equation  with  (7)  for  points  at  the  free  end,  at 
the  middle,  and  at  a  distance  c  from  the  free  end. 

6.  What  is  the  total  deflection  of  a  wooden  beam  8  inches  wide,  6  inches 
deep,  as  a  cantilever  8  feet  4  inches  long,  due  to  a  distributed  load  of  84  pounds 
per  foot  and  a  load  of  350  pounds  at  the  free  end,  if  E  is  1,400,000?     What  is 
the  maximum  fiber  stress?     What  is  the  factor  of  safety  if  the  beam  is  white 
oak?  Ans.  Deflection  at  the  end,  1.01  inches. 

62.  Beam  Supported  at  the  Ends,  Uniformly  Loaded.  —  In  a 

beam  supported  at  the  ends  and  uniformly  loaded,  the  end  reac- 
tions are  each  equal  to  one-half  of  the  total  load, 

W      wl 


w  Pounds  Per  Unit  Length 

I  I'l  I  I  I  I  |  I  I  I  I  II  I  I  I  I  I  ITTTTT 


yt. 

^  _^_. 

2      2 


*       & 

Fig.  93.  —  Supports  at  Ends,  Load  Uniformly  Distributed 

The  moment  at  a  distance  x  from  the  left  support  is 

wlx      wx2 

and  the  differential  equation  becomes: 


wlx      w«, 

~o o~"  (1) 


EIdy  =  wfa2  _  wx* 
'    dx        4  6 

From  symmetry  we  see  that  the  tangent  is  horizontal  at  the 
middle, 

dy      A  I 

~r  =  Q    when     x  =  •= : 
dx 


£>  (3) 


DEFLECTION  IN  BEAMS  109 

-i-,rdy      wlx2      wx3      wl3  .  N 

EITx=~^r     "T   ~24* 

wl3x      r 

^ 

(6) 


When  x  =  -  the  deflection  is  a  maximum 

2 


5  rf 


ax          384^  384 

where  W  =  wl,  the  total  load. 

If  we  substitute  x  =  I  in  the  equation  of  the  elastic  line,  we  get 
y  =  0,  as  the  deflection  at  the  right  support.  We  might  have 
used  this  condition,  y  =  0,  when  x  =  I,  to  determine  Ci. 

PROBLEMS. 

1.  What  is  the  deflection  at  the  middle  of  a  2-inch  by  12-inch  floor  joist, 
15  feet  between  supports,  due  to  a  distributed  load  of  90  pounds  per  foot,  if 
E  is  1,350,000?  Ans.  0.264  inch. 

2.  A  15-inch  42-pound  I-beam  is  used  with  a  span  of  20  feet  to  carry  a 
load  of  800  pounds  per  foot.     If  E  is  29,000,000,  what  is  the  deflection?     What 
is  the  maximum  fiber  stress? 

(  Deflection  at  the  middle,  0.225  inch; 
'  I  Bending  stress,  8150  pounds  per  square  inch. 

3.  In  Problem  1  what  is  the  slope  of  the  tangent  to  the  beam  at  the  sup- 
ports?    If  the  beam  extends  4  feet  beyond  one  support,  how  much  is  this 
extension  elevated  at  the  end  when  the  load  is  applied  between  the  supports? 

4.  If  E  is  29,000,000  for  structural  steel,  what  is  the  greatest  deflection  in 
a  beam  of  length  I  between  supports  and  depth  d,  if  the  load  is  uniformly 
distributed  and  the  allowable  fiber  stress  is  15,000  pounds  per  square  inch? 
Apply  result  to  a  12-inch  beam  of  20-foot  span.  Ans.  0.517  inch. 

63.  Beam  Supported  at  the  Ends  with  a  Concentrated  Load 
at  the  Middle.  —  With  a  beam  supported  at  the  ends,  with  a 

p 
load  P  at  the  middle,  the  end  reactions  are  each  -^,  and  the 

Px 

moment  from  the  left  support  to  the  load  at  the  middle  is  -— 

This  moment  is  positive,  the  left  portion  at  any  section  tending 


110  STRENGTH  OF  MATERIALS 

to   turn   the   right   portion   clockwise.     (The   beam   is  concave 
upward.) 

**        Px 


dy  „  ,. 

EIte=~T     Cl' 

dii 
At  the  middle,  from  the  symmetry  of  the  sides,  -     =  0; 


p/2 

c>  =  -ir 

dy      Px*      PV 


, 


At  the  left  support,  where  x  =  0,  y  =  0; 

C2  =  0;  (6) 


F°rmula 


At  the  middle,  where  x  =  - 

« 


Formula  XIV  might  be  obtained  from  the  cantilever  with  a 
load  on  the  end.  A  beam  supported  at  the  ends  with  a  load 

at  the  middle  is  equivalent  to  two  cantilevers  of  length  -  each 

« 

p 

with  a  load  -^  at  the  end,  the  load  pushing  up  instead  of  down. 

P          I 
Substituting  -^  and  ^  for  P  and  I  in  Formula  XI,  we  get  Formula 

z         z 

XIV. 

Formula  XIV  is  much  used  to  determine  the  modulus  of 
elasticity.  Cantilevers  are  not  suitable  for  accurate  tests,  as  it 
is  difficult  to  so  fix  a  beam  at  one  end  that  the  tangent  will  not 
change  slightly  when  the  moment  is  applied.  It  is  easy  to  place 
a  beam  on  two  supports  and  measure  the  additional  deflection 
when  a  load  is  applied  at  the  middle. 


DEFLECTION  IN  BEAMS  111 


PROBLEMS. 

1.  A  selected  beam  of  red  oak,  1.75  inches  wide  and  1.25  inches  deep,  was 
placed  on  two  supports  12  inches  apart  and  a  load  applied  at  the  middle. 
When  an  addition  of  607  pounds  was  made  to  the  load  the  deflection  at  the 
middle  was  increased  0.050  inch.     Find  E. 

Ans.  1,534,000  pounds  per  square  inch. 

2.  In  the  beam  of  Problem  1  an  addition  of  721  pounds  produced  a  deflec- 
tion of  0.059  inch.     Find  E. 

3.  In  Problem  1  how  much  would  the  last  significant  figures  of  the  value 
of  E  be  changed  if  the  deflection  readings  were  incorrect  0.0005  inch?   if  the 
breadth  were  incorrect  0.005  inch?  if  the  depth  were  incorrect  0.005  inch?   if 
the  load  were  incorrect  1  pound?     How  much  would  E  be  changed  if  all  these 
errors  occurred  at  once  in  the  same  direction? 

Ans.  An  error  of  0.005  inch  in  the  depth  would  change  the  .result  1.2  per 
cent,  a  change  of  18  in  the  significant  figures. 

4.  The  beam  of  Problem  1  broke  under  a  load  of  2315  pounds.     Find  the 
fiber  stress  at  rupture. 

5.  What  is  the  deflection  at  the  middle  of  a  10-inch  25-pound  I-beam  of 
15  feet  span  due  to  a  load  of  6000  pounds  at  the  middle,  if  E  is  29,000,000? 

6.  If  a  rod  5  feet  long  is  clamped  to  one  end  of  the  beam  of  Problem  5, 
how  much  will  the  free  end  of  this  rod  be  elevated  when  the  load  is  applied 
to  the  beam?  Ans.  0.206  inch. 

7.  What  is  the  deflection  at  the  middle  of  a  12-inch  31.5-pound  I-beam  of 
20  feet  span  due  to  a  distributed  load  of  360  pounds  per  foot  and  a  load  of 
5000  pounds  at  the  middle,  if  E  is  29,000,000?     What  is  the  maximum  fiber 
stress  due  to  these  loads?  Ans.  Deflection  at  the  middle,  0.437  inch. 

8.  Substitute  x  —  I  in  equation  (7).     The  result  is  not  the  deflection  at 
the  right  support.     Why? 

64.  Beam  Supported  at  the  Ends,  Load  at  any  Point  between 
Supports.  —  In  the  case  of  a  beam  with  a  concentrated  load, 
the  moment  expression  changes  when  we  pass  this  load.  The 
differential  equation  (1)  and  the  equation  of  the  elastic  line  (7)  of 
the  preceding  article  apply  from  the  left  support  to  the  middle. 
Fortunately,  on  account  of  the  symmetry  of  the  two  ends  we 
could  assume  that  the  beam  was  horizontal  under  the  load  and 
thus  have  two  conditions  from  which  to  determine  the  two  arbi- 
trary constants.  When  the  load  is  not  at  the  middle,  ^  is  zero 

for  some  point  between  the  load  and  the  middle.  We  must  write 
the  differential  equation  for  both  sides  of  the  load.  The  solution 
of  these  two  differential  equations  gives  four  arbitrary  constants. 
Fig.  94  represents  a  beam  supported  at  the  ends  with  a  load  P 
at  a  distance  kl  from  the  left  support,  which  is  taken  as  the  origin. 
The  letter  k  represents  any  fraction  between  zero  and  unity,  and 


112 


STRENGTH  OF  MATERIALS 


is  constant  for  any  particular  problem.  The  left  reaction  is 
(1  —  k)  P  and  the  right  reaction  is  k P.  From  the  left  support 
to  the  load  the  moment  is 

P(l  -k)x; 
and  from  the  load  to  the  right  end  it  is 

P(l-k)x-P(x  -kl). 

It  is  convenient  to  write  the  differential  equations  for  both  por- 
tions of  the  beam  and  carry  the  integrations  through  together; 
remembering  that  the  one  set  of  equations  applies  from  x  =  0  to 
x  =  kl,  inclusive,  and  the  other  set  applies  from  x  =  kl  to  x  =  I, 
inclusive. 


-u- 


Fig.  94.  —  Supports 
at  Ends,  Load  at 
any  Point. 


For  all  points  from  x  =  0  to 


x  =  kl,  inclusive, 


(i) 


dy  =  P  (1  -  k)  x2 
dx  2 


+ 


For  all  points  from  x  =  kl  to 
x  =  I,  inclusive, 


EI% 

dx 


-P(x-kl);          (2) 

P(x-klY 
2 

(4) 


The  curve  of  the  beam  is  continuous  under  the  load  with  no 
sudden  change  of  direction,  and  -f  from  the  left  equation  (3)  is 

uX 

the  same  as  -^from  the  right  equation  (4)  at  the  common  point 

x  =  kl.      Accordingly,  when  x  =  kl,  the  second  members  of  equa- 
tions (3)  and  (4)  are  equal. 

Equating  these  second  members,  we  observe  that  the  first 
terms  of  the  two  expressions  are  the  same  when  x  =  kl,  and  that 
the  second  term  of  (4)  vanishes: 

Ci  =  C3. 


DEFLECTION  IN   BEAMS 


113 


Integrating  again,  remembering  that 


Ely 


P(l-k)x< 


.  (5) 


When  x  =  0,  y  =  0,  hence 


EIy  = 


C3: 
_P(l-k)x*     P(x-kl)* 


6 


(6) 


When  x  =  kl  the  values  of  y  for  (5)  and  (6)  are  the  same  and 
the  second  members  of  these  equations  are  equal;  from  which: 

0  =  C2  =  C4. 

When  x  =  I  in  (6),  y  =  0; 


Substituting  the  value  of 
from  (7)  in  (5),  we  get: 


.  (8) 


As  a  check,  apply  equation  (8)  to  find  the  deflection  under  the 

load  when  the  load  is  at  the  middle;  k  =  ^;  (1  —  k)  =  =  ;  x  =  ~; 

2  Zi  2i 

P      73        P  /1        1\73  P/3 

-  6  '  F6  +  f  (g  -  I)  l=  -  §•    (ComPare  Form"la  XIV'} 

The  point  of  maximum  deflection  is  given  by 


provided  that  k  >  J  so  that  equation  (3)  for  the  left  side  applies. 
It  may  be  shown  that  -  <  x  <  kl,  where  x  is  the  position  of 

Zi 

maximum  deflection  and  k  >  J;  and  consequently  the  point  of 
maximum  deflection  lies  between  the  load  and  the  middle  of  the 
beam. 

In  finding  the  point  of  maximum  deflection  and  the  deflection 
at  this  point,  it  is  convenient  to  use  the  simpler  equations  to  the 
left  of  the  load  (3)  and  (8).  In  case  the  load  is  to  the  left  of  the 
middle,  imagine  yourself  on  the  opposite  side  of  the  beam.  This 
will  make  k  greater  than  one-half. 


114  STRENGTH  OF  MATERIALS 

PROBLEMS. 

1.  If  k  =  0.6,  what  is  the  point  of  maximum  deflection?  Ans.  0.529  I. 

2.  Show  that  the  point  of  maximum  deflection  is  never  beyond  — r:  • 

V3 

3.  A  3-inch  by  2-inch  beam  10  feet  long  supports  a  load  of  45  pounds  6  feet 
from  the  left  end.     If  the  beam  is  supported  at  the  ends,  and  E  is  1,500,000, 
find  the  deflection  at  the  middle,  under  the  load,  and  the  maximum  deflection. 

Ans.  At  middle,  0.510  inch;  under  load,  0.498  inch. 

65.  Beam  Supported  at  Ends,  Two  Equal  Loads  Symmetrically 
Placed.  —  An  important  case  is  that  of  a  beam  supported  at  two 
points  with  two  equal  loads  at  equal  distances  from  the  supports. 
If  we  neglect  the  weight  of  the  beam,  the  shear  is  zero  and  the 
moment  is  constant  between  the  loads.  For  this  .reason  this 
method  of  loading  is  much  used  in  tests  of  beams,  for  it  enables 
the  experimenter  to  study  the  effect  of  moment  independent  of 
shear  between  the  loads,  and  the  effect  of  shear  and  moment 
combined  beyond  the  loads.  The  moment  and  the  resulting 
stress  in  any  horizontal  fiber  being  constant  between  the  loads, 
measurements  of  elongation  enable  him  to  locate  the  neutral  axis. 

p 


1                                                 (_ 

^                                  t 

I                                                                        I 

II 

a 

ill 

7  

Fig.  95.  —  Supports  at  Ends,  Two  Loads  Symmetrically  Placed. 

p 

In  Fig.  95,  the  load  at  each  point  is  represented  by  —     The 

2 

total  span  is  I  and  the  loads  are  at  a  distance  a  from  the  supports. 
As  usual,  the  left  support  is  taken  as  the  origin.  There  are  three 

cii] 
moment  equations.     Owing  to  the  symmetry,  we  see  that  -p  is 

zero  at  the  middle.  If  we  write  the  equations  for  the  left  portion 
and  for  that  between  the  loads,  we  shall  need  four  conditions  to 
determine  the  constants.  These  are  y  =  0  when  x  =  0  in  the 

first  portion ;  -p  has  the  same  value  for  both  equations  under  the 
first  load;  y  has  the  same  value  for  both  equations  under  this 

load;  and  -^  =  0  at  the  middle. 
dx 


DEFLECTION   IN   BEAMS 
Writing  the  equations  as  in  the  preceding  article: 


115 


From  x  =  0  to  x  =  a, 
d*y_Px 


Pa2 


(1) 

(3) 


From  x  =  a  to  x  —  (I  —  a) , 
rd2y_Pa 


dy  _  Pax 
dx~  ~2~ 


Pa2 


(2) 
(4) 


Pa2      Pal 


EIdy  =  Ps2      Pa2      Pal 
dx  ~     4  4  4 

Px3   ,  Pa2x      Palx 


C2  =  0. 


(5) 


I      dy 
When  a;  =  ->     -^  =  0; 

c  =-^?. 


EIy  = 


d?/  _  Pax  _  Pal 
dx~  ~^~      ~T 

Pax2      Palx 


^_  _  PaH  =  Pa3     Pa?l 
12   "      4  4  4  4 


(7) 


C4.      (6) 


Pa3 
12  ' 

_  Pao;2  _  Palx      .  w 
4  4       '  12  ' 

At  the  middle,  where  x  =  »  ' 
_  Pa  (<*_  _  ]?\ 

2/max  ~    p  r  I  i  o        n  /-  ]'  \y/ 


116  STRENGTH  OF  MATERIALS 

We  will  check  (9)  for  the  case  when  a  =  =  > 
_  PI  1 12       I2  \  _        PI* 

2/maX  77FT  I   OA  QO  / 


4SEI 

If  the  loads  are  placed  at  the  third  points,  the  deflection  at 
the  middle  is 

23  PI3 

2/max  1296^7" 


PROBLEMS. 

1.  If  a  is  equal  to  one-fourth  of  the  length  between  supports,  find  the 
maximum  deflection. 

2.  A  6-inch  by  8-inch  wooden  beam  supported  at  points  12  feet  apart  is 
loaded  with  two  equal  loads  of  800  pounds  each  4  feet  from  the  supports.     If 
E  is  1,200,000,  what  is  the  deflection  under  a  load  and  at  the  middle? 

Ans.  Under  a  load,  0.240  inch;  at  middle,  0.276  inch. 

3.  In  Problem  2,  two  vertical  lines  are  ruled  on  one  side  of  the  beam  20 
inches  on  the  right  and  left  of  the  middle.     When  the  load  is  applied,  what 
angles  will  these  lines  make  with  each  other? 


G  M 

Fig.  96. 

4.  In  Problem  3,  Fig.  96,  the  distance  FN  between  the  upper  ends  of  the 
lines  is  measured  with  a  delicate  extensometer.     How  much  is  this  distance 
diminished  when  the  loads  are  applied,  and  what  is  the  unit  deformation  in 
40  inches?  Ans.  Total,  0.0200  inch;  unit  0.00050  inch. 

5.  Compute  the  fiber  stress  in  the  upper  fibers  from  the  unit  deformation 
in  Problem  4  and  check  by  Formula  VIII. 

6.  Show  that  the  error  due  to  measuring  the  chord  instead  of  the  arc  in 
Problem  4  is  less  than  0.00004  inch,  and  that  the  relative  error  in  the  unit 
deformation  is  less  than  one  part  in  500.     (Use  the  first  two  terms  of  the  sine 
series  for  the  half-angle.) 

66.  Any  Beam  with  Two  Supports.  —  All  the  cases  of  deflec- 
tions so  far  considered  are  cases  of  two  supports.  (In  the  can- 
tilever one  of  these  supports  pushes  down  in  the  wall.)  In  all 
problems  of  this  sort,  the  reactions  may  be  computed  algebra- 
ically and  the  moment  equations  written  for  any  section.  At 
each  support  and  at  each  concentrated  load  the  equation  of 
moment  changes  and  a  different  differential  equation  must  be 


DEFLECTION   IN  BEAMS  117 

formed.  The  solution  of  each  differential  equation  of  the  second 
order  involves  two  integration  constants  which  must  be  deter- 
mined from  the  values  of  y  and  -p  at  points  to  which  the  equa- 
tions apply.  There  must  be  twice  as  many  of  these  known 
conditions  as  there  are  differential  equations. 

PROBLEMS. 

1.  A  beam  of  length  I  +  a,  uniformly  loaded  w  pounds  per  unit  length, 
is  supported  at  the  left  end  and  at  a  distance  a  from  the  right  end. 

Can  you  get  the  equation  of  the  elastic  line  of  the  portion  between  the 
supports  without  getting  the  equation  of  the  length  a  to  the  right  of  the  right 
support?  Why?  Can  you  get  the  equation  of  the  elastic  line  for  the  portion 
beyond  the  right  support  without  writing  the  differential  equation  for  the 
portion  between  the  supports?  Why? 

2.  A  6-inch  by  6-inch  wooden  beam  15  feet  long  is  supported  at  the  left 
end  and  5  feet  from  the  right  end  and  carries  a  distributed  load,  including  its 
own  weight,  of  60  pounds  per  foot.     The  modulus  of  elasticity  is  1,000,000. 

Draw  the  shear  diagram,  1  inch  equals  100  pounds.  Draw  the  moment 
diagram,  1  inch  equals  200  foot  pounds.  Draw  the  deflection  diagram,  1  inch 
equals  0.1  inch  deflection.  Find  the  point  of  zero  moment  from  the  moment 
equation  and  check  from  the  shear  diagram. 

3.  A  beam  of  length  I  +  a  is  supported  at  a  distance  a  from  the  right  end 
and  is  held  down  at  the  right  end.     Find  the  equation  of  the  elastic  line  when 
a  load  P  is  applied  at  the  left  end,  neglecting  the  weight  of  the  beam.     How 
does  this  differ  from  a  cantilever  with  a  load  on  the  end? 

4.  A  10-inch  25-pound  I-beam  20  feet  long  is  supported  3  feet  from  the 
left  end  and  5  feet  from  the  right  end.     What  is  the  maximum  allowable  load, 
including  its  own  weight,  which  may  be  uniformly  distributed  over  the  entire 
length,  if  the  allowable  fiber  stress  is  12,000  pounds  per  square  inch?     What 
is  the  deflection  at  the  middle,  the  maximum  deflection  between  the  supports, 
and  the  deflection  at  the  right  end?  Ans.  Load,  1952  pounds  per  foot. 

67.  Stiffness  of  Beams.  —  The  stiffness  of  a  beam  is  the  recip- 
rocal of  the  deflection.  The  stiffness  of  a  beam  may  be  defined 
as  the  load  which  will  produce  unit  deflection.  We  are  not 
accustomed  to  express  stiffness  in  this  way,  but  use  it  as  a  rela- 
tive term. 

In  the  expression  for  the  maximum  deflection  of  all  the  beams 
which  we  have  considered,  the  terms  E  and  7  occur  in  the  denom- 
inator. The  stiffness  of  a  beam  varies  directly  as  the  modulus 
of  elasticity  and  directly  as  the  moment  of  inertia  of  its  cross 
section.  The  moment  of  inertia  of  a  rectangular  section  varies 
as  the  cube  of  the  depth,  consequently  the  stiffness  of  a  rectan- 
gular section  varies  in  the  same  ratio. 


118  STRENGTH  OF  MATERIALS 

All  the  expressions  for  the  maximum  deflection  contained  the 
cube  of  the  length  in  the  numerator.  The  stiffness  of  beams  of 
the  same  cross  section  varies  inversely  as  the  cube  of  their  length. 

PROBLEMS. 

1.  How  does  the  stiffness  of  a  4-inch  by  6-inch  beam  compare  with  that  of 
a  4-inch  by  4-inch  beam  of  the  same  material? 

2.  How  does  the  stiffness  of  a  4-inch  by  6-inch  beam  with  the  6-inch  side 
vertical  compare  with  that  of  the  same  beam  with  the  4-inch  side  vertical? 

3.  How  does  the  stiffness  of  a  2-inch  by  12-inch  beam  15  feet  long  com- 
pare with  that  of  a  2-inch  by  8-inch  beam  10  feet  long?    Which  is  the  stronger? 


CHAPTER  VIII. 


BEAMS   WITH   MORE  THAN   TWO   SUPPORTS. 

68.  Relation  of  Deflection  to  Stress.  —  In  the  case  of  beams 
with  two  supports,  including  cantilevers,  we  are  able  to  compute 
the  moment  and  consequently  the  fiber  stress  without  making 
use  of  the  deflection.     From  this,  since  unit  stresses  are  more 
important  from  the  engineering  standpoint  than  deflections,  the 
student  may  get  the  impression  that  we  are  giving  too  much 
attention  to  the  elastic  curve.     However,  when  we  attempt  to 
get  the  stress  in  a  beam  with  more 

than  two  supports,  we  find  that  we 
must  write  the  differential  equations 
and  solve  for  the  elastic  curve  before 
we  can  get  even  the  reactions.  (In 
some  cases  it  is  only  necessary  to 
integrate  once  for  the  slope  of  the 
curve  and  determine  the  constants.) 
Consequently,  a  knowledge  of  these 
equations  of  deflection  is  indispen- 
sable for  the  calculation  of  stresses 
except  in  the  simplest  cases. 

69.  Cantilever  Supported  at  One 
End.  —  Fig.   97  represents  a  canti- 
lever fixed  at  the  right  end  and  sup- 
ported at  the  left  end,  and  carrying 

.»         T       j'lt*i_         11       i       f 
a  uniformly  distributed   load  of  w 

pounds  per  unit  length.     The  origin 
is  taken  at  the  left  support.     The  moment  at  a  section  at  a  dis- 
tance x  from  the  origin  is: 

wx2 
M  =  Rix ^~  * 

The  reaction  at  the  left  support,  Ri,  is  at  present  unknown. 
We  will  use  it  in  the  differential  equations  and  determine  its 
value  later.  ™  2 

'ZL.  (1) 


MOMENT 


g.  97. —  Beam  Fixed  at  One 
End  and  Supported  at  the 
Other. 


dx* 
dy 


wx' 


I    /"f 


(2) 


119 


120  STRENGTH  OF  MATERIALS 

When          x  =  l,         =  0,      hence     d  =         -;  (3) 


dy      R&*      RJ*      wx*      wl* 


~~     " 


When  x  =  0,  y  =  0,  hence  C2  =  0. 

We  still  have  one  condition  from  which  to  find  the  reaction  RI  : 
When  x  =  I,  y  =  0,  which,  substituted  in  equation  (5),  gives 


3TF  (. 

Rl  =  ~8~      T~' 

where  TF  is  the  total  distributed  load. 

tn 

y=-  ^j  (2x*  -  3  Zz3  +  ^)-  (7) 

With  RI  known  we  may  now  write  the  moment  equation 


PROBLEMS. 

1.  Draw  shear  and  moment  diagrams  for  beam  fixed  at  one  end  and  sup- 
ported at  the  other.     Find  the  moment  at  each  dangerous  section  from  the 
shear  diagram  and  compare  with  the  result  from  the  equation  of  moments. 

'  .-9  Wl        Wl 

Ans.  Moment  at  dangerous  sections,  -^r^  ,  --  —' 

iZo  o 

2.  How  does  the  greatest  moment,  numerically,  compare  with  that  of  a 
beam  supported  at  the  ends? 

3.  Find  the  position  of  maximum  deflection  and  the  value  of  this  maximum 
deflection. 

Ans.  Point  of  maximum  deflection  is  0.4215  I  from  the  left  support. 
Substituting  the  value  of  R  in  equation  (4),  or  differentiating  (7),  we  get 
for  the  points  where  the  beam  is  horizontal: 

Z3  =  0. 


Since  the  beam  is  horizontal  at  the  wall,  x  =  I  must  satisfy  this  cubic  equation. 
Dividing  by  the  corresponding  factor,  x  —  I,  we  get  a  quadratic.  Explain  the 
meaning  of  the  negative  root. 


BEAMS   WITH   MORE    THAN   TWO   SUPPORTS     121 

4.  A  4-inch  by  6-inch  wooden  beam  20  feet  long  is  supported  at  both  ends 
and  at  the  middle  and  carries  a  uniformly  distributed  load,  including  its  own 
weight,  of  120  pounds  per  foot.     What  is  the  maximum  fiber  stress?    Where 
is  it?     What  is  the  load  on  each  of  the  three  supports?     Would  the  beam  be 
stronger  or  weaker  if  it  were  cut  in  two  at  the  middle  and  simply  rested  on  the 
middle  support  instead  of  being  continuous  over  it? 

5.  Draw  the  shear  and  moment  diagrams  for  Problem  4:  1  inch  equals 
4  feet  length,  200  pounds  shear,  and  1000  foot-pounds  moment.     In  deriving 
the  theory  of  this  article,  we  assume  that  the  beam  is  absolutely  fixed  at  one 
end;  that  the  tangent  remains  horizontal  when  the  load  is  applied.     If  it  is 
not  perfectly  fixed,  as  in  the  case  of  a  beam  clamped  at  one  end,  the  reaction 
at  the  supported  end  will  be  greater  than  that  given  by  the  theory.     We  also 
assume  that  the  support  suffers  no  displacement  when  the  load  is  applied. 
Neither  of  these  conditions  is  fully  met  in  the  arrangement  of  Fig.  97.     Fig.  98 


U 


Fig.  98.  —  Continuous  Beam  of  Two  Equal  Spans. 

shows  a  beam  with  supports  at  the  ends  and  at  the  middle.  If  these  supports 
are  so  dimensioned  that  a  load  of  10  pounds  at  the  middle  will  produce  the 
same  deformation  as  a  load  of  3  pounds  at  either  of  the  others,  a  uniform  load 
over  the  length  will  leave  the  tangent  horizontal  over  the  middle  support, 
and  the  ends  will  not  be  deflected  with  reference  to  the  middle.  Either  half 
of  this  beam  will  now  fulfill  the  conditions  of  the  theory. 

6.  A  6-inch  by  8-inch  wooden  beam  is  supported  at  the  middle  and  15  feet 
on  each  side  of  the  middle  and  carries  a  load  of  150  pounds  per  foot  for  this 
30  feet  of  length.     The  end  posts  are  6  inches  by  6  inches  and  rest  on  concrete 
footings  1  foot  square.     What  should  be  the  dimensions  of  the  footing  for  the 
middle  post  if  the  settlement  of  all  shall  be  equal? 

7.  In  Problem  6  suppose  that  the  middle  post  settles  0.1  inch  and  that  E 
equals  1,200,000.     Find  the  reaction  on  each  support. 

8.  A  beam  is  absolutely  fixed  at  one  end  and  is  supported  at  the  other 
by  a  support  which  is  deformed  a  unit  distance  by  a  load  Q.     What  is  the 
reaction  at  this  support  due  to  a  uniform  load  W  over  the  entire  beam? 

Wl3 


3+-Q 

70.   Cantilever  Supported  at  One  End,  Load  Concentrated.  - 

Fig.  99  represents  a  beam  supported  at  the  left  end  and  fixed 
at  the  right  end,  with  a  load  P  at  a  distance  kl  from  the  left 
support.  We  will  carry  the  two  sets  of  equations  together  as  in 
Article  64. 


122 


STRENGTH  OF  MATERIALS 


Fig.  99. 


From  left  end  to  load, 


=  Rlx-P(x-kl).      (2) 


When  x  =  kl  the  curves  have  a  common  tangent,  from  which 

Ci=C8.  (5) 

When  x  =  I,  the  tangent  is  hori- 
zontal, 

=  0.    (6) 


We  will  not  substitute  the  value  of  C3  in  equations  (3)  and  (4) 
at  present. 

P(x  -  kl)s 


Ely  =  ^  +  C3x  +  C2.        (7) 
When  x  =  0,  y  =  0,  C2  =  0. 


Ely 


6  6 

C3x  +  C4. 


(8) 


When  x  =  kl,y  is  the  same  for  both  curves, 
0  =  C2  =  C4  =  0. 

When  x  =  I,  y  =  0. 


Combining  (6)  and  (9),  we  get: 
P 


^O.    (9) 


(10) 


PROBLEMS. 
1.   If  k  =  %,  show  that  the  end  reaction  is  &  P. 


BEAMS   WITH   MORE   THAN   TWO   SUPPORTS     123 

2.   If  k  =  \,  find  the  moment  at  each  dangerous  section  and  the  location 
of  the  point  of  inflection  where  the  moment  changes  sign. 


Fig.  100.  —  Beam  Fixed  at  One  End  and  Supported  at  the  Other  with 
Load  at  the  Middle. 


3.  A  2-inch  by  1-inch  wooden  beam  is  securely  clamped  so  that  8  feet 
projects  as  a  cantilever.  The  end  of  this  cantilever  rests  on  a  platform  scale. 
When  a  load  of  20  pounds  is  applied  3  feet  from  the  supported  end,  how  much 
should  be  the  increase  in  the  scale  reading?  Ans.  9.28  pounds. 

71.  General  Equations  of  Moment  and  Shear.  —  If  we  attempt 
to  derive  the  equations  for  a  beam  fixed  at  the  left  end  and  sup- 
ported at  the  right  end,  we  have  difficulty  in  forming  the  expres- 
sion for  the  moment.  It  is  therefore  advisable  to  derive  a  general 
equation  which  applies  to  all  cases.  In  Article  47,  we  have  this 
expression  for  moment: 

M  =R(c  +  x)-Q(b+x)-?.P(x-a)-^;        (1) 


M 


(2) 


Now  the  term  Re  —  Qb  represents  the  moment  at  the  origin  of 
all  the  forces  on  the  left  'of  the  origin.  R  —  Q  is  the  sum  of  the 
vertical  forces  to  the  left  of  the  origin.  Equation  (2)  may  be 
written : 


M 


o  +  -V  -  V  P  (x  -  a)  -  ==- ,     Formula  XV.     (3) 


where 


M  is  the  moment  at  any  section  at  a  distance  x  from  the 

origin; 
MQ  is  the  moment  at  the  origin; 


124  STRENGTH  OF  MATERIALS. 

S0  is  the  vertical  shear  at  the  origin; 

2P(x  —  a)  is  the  sum  of  the  moments  of  all  the  concentrated 
forces  between  the  origin  and  the  section; 

wy?1  \ 

~2~  is  the  moment  of  the  distributed  forces  between  the 

origin  and  the  section. 

PROBLEMS. 

1.  A  beam  20  feet  long  is  supported  at  the  right  end  and  5  feet  from  the 
left  end  and  carries  a  load,  including  its  own  weight,  of  120  pounds  per  foot 
and  a  load  of  2800  pounds  17  feet  from  the  left  end.     Calculate  the  moment 
at  the  left  support  and  write  the  moment  equation  for  the  portion  between 
the  supports,  using  a  point  infinitely  near  the  left  support  as  the  origin  of 
coordinates. 

Ans.  M  =  —1500  +  1560  x  —  60  x2,  from  x  =  0  to  x  =  12  feet. 

M  =  -1500  +  1560  x  -  60  x2  -  2800  (x  -  12),  from  x  =  12  feet  to 
x  =  15  feet. 

2.  Form  the  derivatives  of  the  moment  equations  above  and  solve  for 
the  position  of  maximum  moment.     Neither  gives  the  correct  result.     Why? 
What  is  the  meaning  of  the  results  obtained? 

3.  Solve  the  moment  equations  above  for  the  positions  of  zero  moment. 

Ans.  1  foot  and  15  feet. 

Each  of  these  quadratic  equations  has  two  solutions.  How  do  you  know 
which  one  to  use? 

72.  Point  of  Inflection.  —  A  point  of  inflection  in  a  beam  is 
a  point  where  the  moment  changes  sign,  and  where  the  center  of 
curvature  changes  from  one  side  of  the  beam  to  the  other.  A 
point  of  inflection  is  also  called  a  point  of  counterflexure.  From 
our  calculus  we  remember  that  a  point  of  inflection  occurs  when 

d2v  d*y 

-r~  equals  zero,  provided  -~  does  not  equal  zero  at  the  point. 

dx  dx 

In  the  case  of  an  elastic  curve, 

d?y=M^      d*y_   S 
dx2      El'    dx3      El' 

El  being  always  positive,  a  point  of  inflection  occurs  where  the 
moment  is  zero  and  the  shear  is  not  zero.  In  Problem  3  of  the 
preceding  article,  the  first  answer  gives  a  point  of  counterflexure. 
At  this  point  the  moment  is  zero  and  the  beam  could  be  cut  in 
two  and  one  portion  support  the  other  by  means  of  slight  projec- 
tion to  take  the  shearing  stress. 

Fig.  101  represents  the  beam  of  Problems  1  and  3  of  Article  71. 
The  left  portion  weighing  720  pounds  supports  the  right  portion 


BEAMS    WITH   MORE    THAN    TWO   SUPPORTS    125 

weighing  1680  pounds  and  the  load  of  2800  pounds.  In  the 
upper  figures  we  have  drawn  the  beam  as  made  of  two  parts 
united  by  a  pin  connection  at  the  point  of  inflection.  In  the 


,     ,  3040^ 

2160* 

Fig.  101.  —  Point  of  Counterflexure. 

lower  figure  the  beam  is  represented  as  cut  at  this  point  with 
one  end  resting  on  the  other.  To  find  the  force  with  which  the 
left  portion  lifts  the  right  portion,  take  moments  around  the  left 
support.  The  720  pounds  with  center  of  gravity  2  feet  to  the 
left  of  the  support  balances  1440  pounds  1  foot  to  the  right  of 
the  support.  Taking  moments  about  the  right  support: 

2800  X    3  =    8,400 

1680  X    7  =  11,760 

S  X  14  =  20,160 
S  =    1,440 
If  we  examine  the  shear  diagram,  we  find  the  shear  at  this  point 

is  1440  pounds. 

PROBLEMS. 

1.  A  beam  of  length  I  with  uniform  load  of  w  pounds  per  foot  is  supported 
one-fourth  the  length  from  the  left  end  and  one-fifth  the  length  from  the  right 
end.     Find  the  points  of  counterflexure. 

2.  From  the  shear  diagram  of  Article  69  determine  the  point  of  inflection 
of  a  beam  fixed  at  one  end  and  supported  at  the  other. 

73.  Cantilever  Fixed  at  Left  End,  Supported  at  Right  End, 
Load  Uniformly  Distributed.  —  Having  now  the  general  equation 
of  moment  (Formula  XV),  we  may  take  the  case  of  a  beam 
fixed  at  the  left  end.  With  a  uniform  load  only 


+  Ift-Q].'  (2) 


126  STRENGTH  OF  MATERIALS 

These  equations  so  far  apply  to  any  beam  fixed  at  the  left  end 
with  a  uniform  load.  Applying  the  results  to  a  beam  supported 
at  the  right  end: 

0  =  12  M0  +  4  S0l  -  wl2.  (4) 

We  have  used  our  three  conditions  for  determining  constants 
and  still  need  another  to  be  used  with  (4)  to  determine  M0  and 
SQ.  The  moment  at  the  right  support  where  x  =  Us  0.  Sub- 
stituting in  the  moment  equation: 

O-Afo'+SoJ-Tp-  (5) 

Combining  (5)  and  (4),  we  get: 

wl2          Wl 


_  5  wl  _  5  W 
~8~      ~8~J 

y=-  (2  &  -  5  Ix*  +  3  IV).  (6) 


PROBLEMS. 

1.  Substitute  I  —  x  f or  x  in  equation  (6)  and  compare  result  with  equation 
(7)  of  Article  69. 

2.  Find  the  position  of  maximum  deflection  and  compare  with  Article  69, 
Problem  3. 

3.  Draw  shear  and  moment  diagrams  and  compare  with  Fig.  97. 

4.  How  does  the  maximum  deflection  in  a  beam  of  this  kind  compare  with 
that  of  a  beam  supported  at  the  ends?  Ans.  41.6  per  cent  as  much. 

74.  Beam  Fixed  at  Both  Ends,  Uniformly  Loaded.  - 

"f '  (1) 

77T  J  ^U     "»  r  i       *-'U«*' 

0  =  6  M0  +  3  S0l  -  wl\  (3) 

(4) 

^  U  -  t 

0  =  12 Mo +  4 Sol -wl2;  (5) 

Q       wl      W 
S°-  2"==  2"' 

_wP=   _Wl 

"  12  "    "  12  " 
wz2 


BEAMS   WITH   MORE    THAN   TWO   SUPPORTS    127 

w  Pounds  Per  Unit  Length 


MOMENT 


Wl 


Fig.  102. 

PROBLEMS. 

1.  Show  that  the  moment  at  the  middle  is  one-half  that  at  the  ends. 

2.  Find  the  points  of  inflection. 


Ans.  0.21  II  and  0.789  I. 


3.   What  is  the  maximum  deflection  at  the  middle? 


Ans.   — 


Wl3 


384 


75.   Beam  Fixed  at  Both  Ends,  Concentrated  Load  kl  from 
Left  End.  - 

V/////////A 

Fig.  103. 


From  left  end  to  load, 


From  load  to  right  end, 


(1) 


=  0]. 


(3) 


+  [C2  =  0]. 


(6) 


(2) 


+  [C3  =  0].  (4) 


=  2M<t+Sttl-Pl(l-W.    (5) 

So*1     P(*-M)' 
~6~ 

(7) 

3.      (8) 


. 
+  [C4  =  0]. 


128  STRENGTH  OF  MATERIALS 

From  (5)  and  (8) : 

SQ  =  P{3(1  -A:)2 -2(1  -fc)*j; 
Mo  =  -  Pl\(l  _fc)2'_(i  -/c)3j. 

PROBLEMS. 

1.  Construct  the  moment  and  shear  diagrams  when  k  =  \. 

2.  When  k  =  \  show  that  the  deflection  at  the  middle  is  one-fourth  as 
great  as  it  is  in  a  beam  simply  supported  at  the  ends  with  a  load  at  the  middle 
and  that  the  unit  stress  is  one-half  as  great. 

3.  Show  that  the  beam  cut  off  as  in  Fig.  104  has  the  same  shear  at  walls, 
moment  at  walls,  moment  at  the  middle,  deflection  at  the  middle  and  quarter 
points,  as  the  beam  in  Fig.  103. 


PI 


MOMENT 


PI 


Fig.  104.  —  Points  of  Counterflexure  in  Beam  with  Ends  Fixed. 

76.  Theorem  of  Three  Moments.  —  The  methods  of  the  pre- 
ceding articles  may  be  applied  to  any  number  of  spans  or  to 
any  number  of  concentrated  loads.  However,  when  it  becomes 
necessary  to  write  more  than  two  moment  equations  and  solve 
for  the  corresponding  constants,  the  work  becomes  laborious. 
When,  as  is  usually  the  case,  it  is  desired  to  find  the  moments, 


BEAMS   WITH   MORE    THAN    TWO   SUPPORTS    129 

reactions,  and  shears,  without  getting  the  deflections,  the  theorem 
of  three  moments  is  of  great  use. 

The  theorem  of  three  moments  is  an  algebraic  equation  which 
expresses  the  relation  of  the  moments  at  three  successive  sup- 
ports of  a  continuous  beam  in  terms  of  the  length  of  the  inter- 
vening spans  and  the  loads  which  they  carry.  In  Fig.  105,  the 
moments  over  the  supports  are  represented  by  Ma,  Mb,  Mc. 
The  length  of  the  span  from  support  A  to  support  B  is  Zi,  and 
from  B  to  C  it  is  12.  Fig.  105  represents  a  uniformly  distributed 
load  of  Wi  pounds  per  unit  length  for  the  first  span  and  w2  pounds 
per  unit  length  for  the  second  span.  The  subscripts  a,  b,  c, 
represent  the  order  from  left  to  right  and  may  be  applied  to  any 
three  points  in  succession.  The  same  is  true  of  the  subscripts  1 
and  2  applied  to  the  spans  and  the  unit  loads. 

We  will  take  the  origin  of  coordinates  for  the  first  span  at  A 
and  for  the  second  span  at  B  and  write  a  differential  equation 
for  each  span,  remembering  that  the  origin  is  different  for  the 
two  equations. 

77.  Theorem  of  Three   Moments  for  Distributed  Loads.  — 


Ma       w-iPer  Unit 


MOMENT  M6Maftr  Pratt  Length  Me 


.ffl 

T 


tftrSHEAR      I 


Fig.  105. 


Span  AB. 


At  support  B}  x  —  li, 


Span  BC. 


=  Mb  +  Sbx-^f-  (1) 


EI^  = 
dx 


-=f-  +  c,. 


At  support  B,  x  =  0, 


dy  =  dy 
dx      dx 

i  +  3  Sall  -  wJi  +  6  Ci  =  6  C,. 


(2) 


(3) 


130 


STRENGTH  OF  MATERIALS 


2  6  24 

+  Ciz  +  [C,  =  0]. 

At  right  end  of  span, 


+  [C4  =  0].      (4) 

(5) 
At  right  end  of  span  BC, 


0 


0  = 

+  24  C8.  (6) 

Combining  equations  (6),  (5)  and  (3)  to  eliminate  Ci  and  C3: 
12  MJi  +  12  Mbl2  +  8  &g  +  4  &g  -  3  ^g  -  W2g  =  0.          (7) 
We  wish  to  eliminate  the  shear  and  bring  in  the  moment  at  C: 


=  M 


Substituting  these  values  of  the  shear  in  (7) : 
4  MJ!  +  8Mb  h  +       +  4 


~'         (8) 


=  0;  (9) 

oJS).          (10) 

Equation  (10)  is  called  the  theorem  of  three  moments  for  dis- 
tributed loads. 

If  the  spans  are  equal  and  the  loads  per  unit  length  in  the  two 
successive  spans  are  the  same,  the  equation  of  three  moments 
becomes : 

T£-         Formula  XVI. 


78.  Calculation  of  Moments.  —  The  theorem  of  three  moments 
applies  to  continuous  beams  with  any  number  of  supports.  We 
first  apply  the  theorem  for  three  consecutive  supports,  beginning 
with  the  first  one.  We  next  write  the  equation  beginning  with 
the  second  support  as  A.  This  is  continued  till  all  the  moments 
are  used.  If  there  are  four  supports,  two  equations  are  written; 
if  there  are  five  supports  three  equations  are  required.  Since 


BEAMS   WITH  MORE   THAN   TWO   SUPPORTS     131 

there  are  always  two  more  unknown  moments  than  there  are  equa- 
tions, we  must  know  two  of  the  moments  or  have  some  relations 
from  which  to  find  them.  Let  us  take  the  case  of  a  beam  with 
four  supports  and  three  equal  spans,  each  loaded  w  pounds  per 
unit  length.  The  simple  form  of  the  theorem  (Formula  XVI) 
applies. 

.Ml Ma Ms  MA 


k* 

Fig.  106.  —  Uniformly  Loaded  Beam  of  Three  Equal  Spans. 

This  is  represented  by  Fig.  106.  Representing  the  moments 
by  the  subscripts  1,  2,  3,  4,  as  they  refer  now  to  definite  moments 
and  not  merely  to  the  order  of  arrangement,  the  equations  are 

M2  +  4  M 3  -f  M 4  = ^ '  (2) 

If  the  beam  does  not  overhang  the  end  supports,  MI  =  0  and 
M4  =  0.     Solving  the  equations  for  this  case  we  get 

wl* 


PROBLEMS. 

1.   Find  the  moments  for  two  equal  spans,  with  uniform  loads  on  both,  with 
no  overhang  at  the  end  supports.  Ml  =  0,  M2  =  -  ?  ,  M3  =  0. 


2.  Find  the  moments  over  the  supports  for  four  equal  spans,  with  uniform 
loads  on  each  and  with  no  overhang. 

Ans.  MX  =  0,  M2  =  -  §^jt,  M3  =  -  y£»  M<  =  M2,  M5  =  0. 

3.  Find  the  moments  over  the  supports  with  two  equal  spans  of  length  I, 
and  overhangs  of  0.6  I  to  the  left  of  the  left  support  and  0.4  I  to  the  right  of 
the  right  support,  with  a  uniformly  distributed  load  throughout  the  entire 
length.  Ans.  M  i  =  -  0.18  wlz,  M2  =  -  0.06  wlz,  M3  =  -  0.08  wlz. 

4.  A  beam  30  feet  long,  weighing  w  pounds  per  foot,  rests  on  four  supports 
so  as  to  make  three  equal  spans  of  8  feet  each,  and  overhangs  the  left  support 
4  feet  and  the  right  support  2  feet.     Find  the  moment  over  each  support. 

Ans.   —  8  w,  —  4.4  w,  -  6.4  w,  —  2  w. 

5.  A  uniformly  loaded  beam  rests  on  three  supports  so  as  to  have  two 
equal  spans  with  equal  overhang  on  each  end.     What  must  be  the  ratio  of 
overhang  to  span  if  the  moments  at  all  supports  are  the  same? 

Ans.  Overhang  0.408  of  the  length  of  span. 


132  STRENGTH  OF  MATERIALS 

6.  A  beam  20  feet  long  with  a  uniformly  distributed  load  is  supported  at 
the  ends  and  12  feet  from  the  left  end.     What  is  the  moment  at  the  second 
support?  Ans.  —  14  w. 

7.  A  shaft  30  feet  long,  weighing  10  pounds  per  foot,  is  supported  by  three 
bearings  at  4  feet  from  the  left  end,  14  feet  from  the  left  end,  and  4  feet  from 
the  right  end.     It  carries  a  load  of  200  pounds  1  foot  from  the  left  end,  and  a 
load  of  300  pounds  2  feet  from  the  right  end.     Find  the  moment  at  each 
bearing.  Ans.   —  680,  185,  -  680  foot  pounds. 

79.  Calculation  of  Shear.  —  To  determine  the  shear  to  the 
right  of  any  support,  we  make  use  of  equation  (8)  of  Article  77. 

0  Mb  —  Ma     ,    Will  ^  ,      VT,TT 

Sa  = -, -f  -H-  *  Formula  XVII. 

li  2 

Sa  is  the  shear  just  to  the  right  of  any  support; 
Ma  is  the  moment  over  that  support; 
Mb  is  the  moment  over  the  next  support; 
Wi  is  the  load  per  unit  length  between  these  supports. 

Notice  that  the  last  term  of  the  formula  is  simply  half  the  load 
on  the  span. 

We  will  apply  this  formula  to  the  case  of  three  equal  spans  and 
four  supports  considered  in  Article  78. 

To  find  the  shear  at  the  right  of  the  first  support 


7  '      O 

i  fJ 

At  the  right  of  the  second  support, 

"  10       10   ,  wl 

S2  = j h  -o-  =  0.5  wl  =  0.5  W. 

In  the  same  way,  £3  =  0.6  wl  =  0.6  W. 

Fig.  107  gives  the  moments  over  the  supports  and  the  shears 
to  the  right  of  each  support  for  the  case  of  three  equal  spans. 

_  wlz                   wl2 
MOMENT     0  Jo ~-Jo 0 

SHEAR     \Lwi  \5wl  ULWI 


Fig.  107. 

PROBLEMS. 

1.   Calculate  the  shear  to  the  right  of  each  support  in  Problem  2  of  Article 
78  Am    S   =  Uwl    S   =  *5wl    S   =  13wl    S   =  17 ' wl 


BEAMS   WITH   MORE   THAN   TWO   SUPPORTS     133 

2.  Calculate  the  shear  at  the  right  of  each  support  in  Problem  3  of  Article 
78.  4ns.  0.62  wl,  0.48  wl. 

The  shear  at  the  left  of  any  support  is  obtained  from  the  shear 
at  the  right  of  the  preceding  support  by  subtracting  the  inter- 
vening load,  according  to  the  definition  of  vertical  shear.  The 
shear  at  the  left  of  the  second  support  in  Fig.  107  is  —  0.6  wl,  at 
the  left  of  the  third  support,  —  0.5  wl,  and  at  the  left  of  the  fourth 
support,  —  0.4  wl. 

The  reaction  at  any  support  is  computed  by  subtracting  the 
shear  at  the  left  from  the  shear  at  the  right  of  the  support. 

MOMENTA  ~™Wl*  ~™Wl*  "&Wl2  0 


SHEAR   \Uwl        -£wl\£wl       ~£wl<Mwl        -^Wl^wl        -iiirZ! 

-I >& Z- 


SHEAR 

Fig.  108. 

Fig.  108  gives  moments,  shears  and  reactions  for  a  beam  with 
four  equal  spans  uniformly  loaded  with  no  overhang. 

PROBLEMS. 

3.  Show  that  with  three  equal  spans,  uniformly  loaded,  the  reactions  are 
0.4  wl,  1.1  wl,  1.1  wl,  and  0.4  wl. 

4.  Calculate  the  reactions  at  the  supports  in  Problem  3,  Article  78. 

Ans.  1.22  wl,  0.86  wl,  0.92  wl. 

5.  In  Problem  4,  Article  78,  calculate  the  shears  and  find  the  reactions. 

Ans.  Reactions,  8.45  w,  7.30  w,  8.80  w,  5.45  w. 

6.  Draw  the  shear  diagrams  for  Problem  5.     From  this  diagram  find  all 
dangerous  sections  and  calculate  the  moment  at  each. 

Ans.  At  4.55  feet  to  the  right  of  the  third  support  M  =  3.95  w. 

7.  A  beam  carrying  a  uniformly  distributed  load  rests  on  three  supports 
spaced  10  feet  apart.     How  much  should  it  overhang  the  outer  supports  in 
order  that  the  reactions  at  all  the  supports  shall  be  the  same?     Ans.  4.4  feet. 


134 


STRENGTH  OF  MATERIALS 


So.  Theorem  of  Three  Moments  with  a  Single  Concentrated 
Load  on  each  Span.  —  Fig.  109  represents  a  continuous  beam 
with  a  load  PI  on  the  first  span  at  a  distance  ki  from  the  left  end 
and  with  a  similar  load^  at  a  distance  kzlz  from  the  left  end  of  the 
second  span,  etc.  Writing  the  equations  for  the  first  span  as  in 
Article  64, 


Fig.  109. 


sih— 4 


"*l* 4 


ax 


dx 


,  Sax* 

"1 n T 


EId£  =  Max  +  S-£ 


(z  - 


+ 


(3) 


=  C3. 


EIy=  ^+^  +  Ci* 

+  [C2=0].  (5) 


+ 


2  6 

Pi  (x  - 


6 
+  [Ct  =  0]. 


+ 


(7) 


+  6  d.  (9) 

From  the  general  equation  of  moments  when  there  is  no  dis- 
tributed load: 

~       Mb  —  Ma  ,    „  ,-       ,  x 


Substituting  in  (7)  : 

(2Ma  +  Mb)  h  +  Pi[(l  -  fa)  -  (1  - 


(11) 


g  +  6  Ci  =  0.     (13) 


A  similar  set  of  equations  may  be  written  for  the  second  span. 
We  will  give  these  the  even  numbers  and  represent  the  constants 
of  integration  by  K\,  etc. 

•n  T  dV  ,  ,  .      SbX2      .       rr 

El  ~Y~  =  Mbx  -\ H-  /LI.  (2) 

(2Mb  +  Mc)  k  +  P2[(l  -  fa)  -  (1  -  fa)3]  g  +  6 #!  =  0.     (14) 


BEAMS    WITH   MORE    THAN    TWO   SUPPORTS     135 

/•/7V 

At  support  B,  -j|  is  the  same  for  (3)  and  (2).     Substituting 

x  =  0  in  (2)  and  x  =  li  in  (3)  and  substituting  the  value  of  Sa  in 
the  latter,  we  get: 

(Ma  +  Mb)  h  +  Pl  [(I  -k)-(l-  A;)2]  II  +  2  d  =  2  K,.      (15) 

To  eliminate  the  two  unknowns  C  and  K  from  (13),  (14),  and 
(15),  multiply  (15)  by  3,  and  multiply  (13)  by  -  1  and  add  the 
results  to  (14). 

Mali  +  2  Mb  fa  +  Z2)  +  MC12  +  PJH  [2  (1  -  fci)  -  3  (1  -  fci)2 

-  /b2)  -  (1  -  /b2)3]  =  0.  (16) 


Equation  (16)  is  the  equation  of  three  moments  for  a  single 
concentrated  load  in  each  span. 

PROBLEMS. 

1.  A  beam  is  supported  at  the  ends  and  at  the  middle  and  carries  a  load  P 
at  the  middle  of  each  span.     Find  the  moment  at  the  middle  support  and  the 


reactions  at  each  support. 


Moment  at  the  middle,  -    —  . 


Ans. 

Reaction  at  end, 


Compare  these  results  with  a  beam  fixed  at  one  end,  supported  at  the  other, 
and  loaded  at  the  middle,  Article  70. 

2.  A  shaft  30  feet  long  is  supported  at  the  ends  and  the  middle.  A  load 
of  600  pounds  is  applied  at  the  middle  of  the  left  span  and  a  load  of  400  pounds 
at  the  middle  of  the  right  span.  Find  reactions,  the  moment  at  each  dan- 
gerous section,  and  the  points  of  inflection. 

81.  Superimposed  Loads.  —  In  many  cases  we  have  concen- 
trated loads  superimposed  on  distributed  loads.  In  the  case  of 
a  cantilever  we  get  the  deflection  and  stress  at  any  point  by 
adding  the  deflections  or  stresses  due  to  the  separate  loadings. 
The  same  is  true  of  a  beam  supported  at  the  ends.  In  the  case 
of  a  beam  supported  at  the  ends  with  a  concentrated  load  not 
at  the  middle  and  a  distributed  load,  the  sum  of  the  separate 
deflections  at  any  point  gives  the  resulting  deflection  at  that 
point,  but  the  point  of  maximum  deflection  is  between  the  posi- 
tion of  the  concentrated  load  and  the  middle  of  the  beam.  The 
dangerous  section  is  either  between  the  dangerous  sections  for 
the  separate  loads  or  under  the  concentrated  load. 

We  have  already  learned  to  locate  these  dangerous  sections 
by  means  of  the  shear  diagram.  It  is  seldom  necessary  to 


136  STRENGTH  OF  MATERIALS 

locate  exactly  the  point  of  maximum  deflection.  For  a  single 
concentrated  load  the  maximum'  deflection  is  between  the  point 
of  application  and  the  middle  of  the  beam  and  is  nearer  the 
latter.  For  a  uniformly  distributed  load  the  maximum  deflec- 
tion is  at  the  middle.  There  is  little  error  in  using  the  deflection 
at  the  middle  in  place  of  the  maximum.  If  the  maximum  deflec- 
tion due  to  the  concentrated  load  is  added  to  the  deflection  at 
the  middle  due  to  the  distributed  load  the  sum  will  be  a  trifle 
larger  than  the  actual  maximum  resulting  deflection. 

PROBLEMS. 

1.  A  6-inch  by  10-inch  beam  20  feet  long  is  supported  at  the  ends  and 
carries  a  distributed  load  including  its  own  weight  of  100  pounds  per  foot. 
If  E  is  1,000,000,  find  the  maximum  deflection  and  fiber  stress. 

Ans.  600  pounds  per  square  inch,  0.72  inch. 

2.  If  the  beam  of  Problem  1  carries  a  concentrated  load  of  800  pounds 
14  feet  from  the  left  end,  what  is  the  fiber  stress  under  the  load  and  at  the 
middle  due  to  this  load  alone?     What  is  the  deflection  under  the  load  and  at 
the  middle  and  what  is  the  maximum  deflection  due  to  the  concentrated  load 
alone? 

Ans.  Stress  under  load,  403.2;  at  middle,  288  pounds;  deflection  under 
load,  0.325  inch;  at  middle,  0.365  inch;  at  11  feet,  0.369  inch. 

3.  In  Problems  1  and  2,  if  both  loads  are  applied  at  the  same  time,  what 
is  the  maximum  fiber  stress?  Ans.  922.6  pounds  per  square  inch. 

In  Problem  3,  if  we  wish  to  find  the  exact  value  of  the  maxi- 
mum deflection  we  may  solve  the  differential  equation  for  dis- 
tributed and  concentrated  loads  combined.  The  equation  for 
finding  the  point  of  maximum  deflection  is  a  cubic  in  x.  The 
solution  of  this  cubic  for  Problem  3  gives  x  =  124  inches.  The 
maximum  deflection  is  1.086  inches  which  is  practically  the  sum 
of  the  separate  deflections  at  the  middle. 

PROBLEM. 

4.  Take  the  case  of  a  beam  fixed  at  the  left  end  and  supported  at  the  right 
end.     Solve  the  differential  equations  for  a  distributed  load,  a  load  concen- 
trated at  the  middle,  and  for  the  two  combined  and  compare  the  last  with  the 
sum  of  the  others. 

The  reactions,  moments,  shears  and  deflections  at  any  point 
due  to  a  combination  of  loads  may  be  obtained  by  taking  the 
sum  of  the  similar  quantities  for  the  loads  separately.  If  we 
want  the  maximum  moment  we  construct  the  shear  diagram  for 
the  combined  loads.  We  can  generally  locate  the  position  of 


BEAMS    WITH   MORE    THAN    TWO   SUPPORTS     137 

maximum  deflection  approximately  by  inspection  and  determine 
the  deflection  for  a  few  points.  In  Problem  3,  above,  the 
maximum  deflection  for  the  distributed  load  is  at  10  feet,  and 
for  the  concentrated  load  at  11  feet.  We  might  compute  the 
deflection  for  every  2  inches  between  these  points  and  plot  the 
curve  for  the  maximum. 

82.  Moments  in  Different  Planes.  —  It  frequently  happens 
that  the  forces  acting  upon  a  beam  are  not  all  in  the  same  plane. 
A  horizontal  shaft  may  be  subjected  to  a  vertical  load  due  to 
its  weight  and  the  weight  of  the  pulleys,  and  to  a  force  in  some 
other  direction  due  to  the  tension  on  belts.  If  a  beam  thus 
loaded  has  the  same  moment  of  inertia  in  all  directions  it  is  only 
necessary  to  find  the  resultant  moment  at  any  section,  making 
use  of  the  fact  that  moments  are  vector  quantities  and  may  be 
combined  like  forces  or  other  vectors. 

PROBLEMS. 

1.  A  shaft  3  inches  in  diameter  and  10  feet  long  weighs  24  pounds  per 
foot.     The  shaft  is  supported  at  the  left  end  and  2  feet  from  the  right  end. 
It  carries  a  pulley  weighing  64  pounds  1  foot  from  the  right  end,  and  a  pulley 
weighing  40  pounds  2  feet  from  the  left  end.     With  these  loads  only,  find 
the  dangerous  sections  and  the  moment  at  each. 

Ans.  3  feet  from  the  left  support,  188  foot  pounds;  at  right  support, 
112  foot  pounds. 

2.  If  there  is  a  horizontal  pull  of  80  pounds  on  the  right  pulley  perpendicular 
to  the  shaft,  what  is  the  ^resulting  moment  at  the  right  bearing? 

Ans.  137.6  foot  pounds. 

3.  In  Problem  2  what  is  the  resultant  moment  3  feet  from  the  left  support? 

Ans.  190.4  foot  pounds. 

The  maximum  resulting  moment  is  a  little  to  the  right  of  the  dangerous 
section  for  the  vertical  loads.  Its  exact  position  in  this  case  involves  the 
solution  of  a  cubic  equation. 

4.  Write  the  expression  for  the  square  of  the  resultant  moment  for  points 
between  the  left  pulley  and  the  right  bearing.     Find  the  expression  for  the 
position  of  the  resultant  dangerous  section  and  solve  by  method  of  trial  to 
two  significant  figures. 

5.  Solve  Problems  2  and  3  if  the  pull  of  80  pounds  is  30  degrees  below  the 
horizontal. 

Find  the  resultant  of  two  couples  at  60  degrees  with  each  other,  or  resolve 
the  80  pounds  into  its  vertical  and  horizontal  components,  adding  the  vertical 
component  to  the  vertical  loads  and  using  the  horizontal  component  as  you 
used  the  pull  when  it  was  horizontal. 

When  a  beam  is  subjected  to  forces  which  are  not  parallel  to 
one  of  the  principal  axes  of  inertia  (and  perpendicular  to  the 


138 


STRENGTH  OF  MATERIALS 


other),  it  is  necessary  to  resolve  these  forces  parallel  to  the  prin- 
cipal axes.  Then  find  the  unit  stress  at  any  point  in  the  section 
separately  for  both  sets  of  forces  and  add  the  results. 

PROBLEMS. 

6.  A  6-inch  by  8-inch  beam  15  feet  long  is  supported  at  the  ends  and  carries 
a  load  of  800  pounds  at  the  middle.  The  load  is  30  degrees  from  the  vertical 
in  a  plane  normal  to  the  length  of  the  beam.  Find  the  unit  stress  at  each 
corner  at  the  dangerous  section. 

The  principal  axes  of  inertia  are  horizontal  and  vertical.  The  vertical 
component  of  the  load  is  692.8  pounds  and  the  horizontal  component  is  400 
pounds.  Using  the  horizontal  component  as  applied  at  D  (Fig.  110, 1),  we 

800* 


--D 


ir 


Fig.  110. 


find  a  unit  stress  of  375  pounds  per  square  inch  which  is  tensile  on  the  left 
side,  1-4,  and  compressive  on  the  right  side.  Using  the  vertical  component  as 
applied  at  A,  with  CD  as  the  axis  of  inertia,  we  get  487  pounds  which  is  com- 
pressive at  the  top.  The  unit  stress  at  corner  2  is  862  pounds  per  square  inch 
compression.  At  corner  1  it  is  112  pounds  compression. 

7.  A  6-inch  by  6-inch  by  1-inch  standard  angle,  10  feet  long,  is  used  as  a 
beam  supported  at  the  ends.     The  angle  is  placed  with  legs  horizontal  and 
vertical  (Fig.  110,  II),  and  a  load  of  1000  pounds  is  applied  at  the  middle,  over 
the  center  of  gravity  of  the  section.     Find  stresses  at  corners/ 

Here  the  principal  axes  are  AB  for  which  /  is  14.78,  and  CD  for  which  7  is 
56.14.     The  external  moment  for  each  axis  is  15,000  V2- 

.  n       15,000  X  V2  X  1.86  X  ^2 
Unit  stress  at  C  =    — • TT78 =  pounds. 

The  unit  tensile  stress  at  3  due  to  the  moment  about  the  axis  AB  is  3329 
pounds.  The  tensile  stress  due  to  the  moment  about  CD  is  1336  pounds.  The 
total  tensile  stress  at  this  corner  is  4665  pounds.  What  is  the  total  unit  stress 
at  corner  2?  How  do  you  know  that  the  stress  at  4  is  less  than  at  3?  Compute 
the  stress  at  1. 

8.  Find  the  horizontal  and  vertical  components  of  the  deflection  at  the 
middle  in  Problem  7  if  E  is  30,000,000. 

Solve  for  each  axis  separately,  then  resolve  deflections  horizontally  and 
vertically  and  add. 


CHAPTER  IX. 
SHEAR  IN  BEAMS. 

83.  Direction  of  Shear.  —  The  total  vertical  shear  in  beams 
is  calculated  by  the  methods  of  Article  45.  We  have  learned 
nothing  so  far  in  regard  to  distribution  of  the  shearing  stress. 
In  Article  30,  we  learned  that  shearing  stresses  occur  in  pairs, 
that  a  small  block  subjected  to  a  shearing  stress  of  a  given  inten- 
sity along  two  parallel  vertical  faces  is  subjected  to  a  shearing 
stress  of  the  same  intensity  along  two  horizontal  faces. 

Fig.  Ill,  I,  represents  a  beam  made  by  placing  one  plank  on 
top  of  another.  Fig.  Ill,  II,  is  the  same  beam  under  load, 


provided  that  the  planks  are  held  from  slipping  with  reference 
to  each  other  by  being  glued  or  bolted  together  to  form  a  single 
beam.  If  the  planks  are  free  to  move,  they  take  the  form  III, 
in  which  the  upper  plank  is  moved  outward  over  the  lower  one 
at  each  end.  Consider  a  small  block  B  in  the  upper  portion  of 
the  lower  plank.  The  plank  above  this  block  has  been  displaced 
to  the  left.  If  they  were  glued  together,  the  upper  plank  would 
have  exerted  a  horizontal  shearing  stress  upon  the  upper  surface 
of  the  block.  To  prevent  rotation  there  must  be  a  vertical  shear 

139 


140 


STRENGTH  OF  MATERIALS 


upwards  at  the  left  side.  The  actual  shearing  stresses  upon  this 
block  from  the  surrounding  material,  if  the  upper  plank  were 
glued  to  the  lower,  would  take  the  directions  of  the  arrows. 

The  shear  at  the  left  of  the  block  is  vertically  upward,  which 
is  the  direction  of  the  external  shear.  If  we  used  a  small  block 
to  the  right  of  the  load  P  we  would  find  that  the  shear  on  its 
left  side  was  vertically  downward.  This  is  the  direction  of  the 
external  shear  in  this  half  of  the  beam.  One  of  the  planks  in 
Fig.  Ill  may  be  thicker  than  the  other,  which  shows  that  the 
results  apply  to  all  parts  of  a  vertical  section  in  so  far  as  they 
effect  the  direction  of  the  shearing  stress. 

84.  Intensity  of  Shearing  Stress.  —  To  determine  the  inten- 
sity of  the  shearing  stress  we  will  consider  Fig.  112.  This 


B    F 


NEUTRAL  AXIS  ^[dX 


-^ 

4. 

f        ' 

—  > 

—  dx- 

—  <- 

SHEAR 

1 

\ 

i 

NEUTRAL  AXIS  , 

lj 

FRONT  SIDE 

Fig.  112.  —  Horizontal  Shear  in  a  Beam. 

represents  a  part  of  a  beam.  Imagine  a  small  block  extending 
across  the  beam  between  planes  dx  apart  and  extending  from 
the  top  of  the  beam  to  a  plane  at  a  distance  v3  from  the  neutral 
surface.  This  block  is  in  equilibrium  under  the  action  of  the 
direct  stress,  tension  or  compression,  acting  on  the  ends  (the 
rectangles  whose  diagonals  are  CB  and  GF)  and  a  shearing  stress 


SHEAR   IN  BEAMS  141 

from  the  material  below  acting  on  the  bottom  (on  the  rectangle 
whose  diagonal  is  GE). 

Consider  a  small  area  dA  in  the  left  end  of  this  block.     The 

unit  tensile  (or  compressive)  stress  on  this  area  is  — =^-   where 

MI  is  the  bending  moment  at  the  section  and  7i  is  the  moment 
of  inertia  of  the  entire  cross  section  of  the  beam  with  respect  to 
the  neutral  axis.  The  total  tension  (or  compression)  on  the  left 
end  of  this  block  is  the  integral  of  the  unit  stress  over  the  left  end 
of  the  block: 

MI  Cv* 
Total  tension  on  left  end  =  -y-  I     vdA.  (1) 

*1    Jv» 

In  the  same  way,  the  total  tension  (or  compression)  on  the 
right  end  of  the  block  is: 

M2    CVt 

Total  tension  on  right  end  =  -^—   I     vdA.  (2) 

The  resultant  horizontal  pull  (or  push)  on  the.  block  in  the 
direction  of  the  length  of  the  beam  is  the  difference  of  these 
integrals  (1)  and  (2).  If  the  section  of  the  beam  is  uniform 
/i  =  72  and  vz  and  vz  are  the  same  for  both  expressions.  The 
resultant  horizontal  pull  (or  push)  becomes: 

Resultant  force  =  — ^-= -1  I     vdA.  (3) 

J-  *J  V3 

This  resultant  horizontal  force  must  be  balanced  by  the  hori- 
zontal shear  at  the  bottom  of  the  block.  If  the  breadth  CE  at 
the  bottom  of  the  block  is  6,  the  total  area  in  horizontal  shear 
is  b  dx,  and  the  total  shear  is  sab  dx.  Equating  these  forces : 


s,b  dx 


M2- 


-iM-J  J.TJ.  1         I  j     ,  /C\ 

s-  =  -f&^r'  '**•  (5) 


Since  Mz  —  MI  is  equal  to  dM , 

M%  —  MI  _  dM 
dx  dx 

where  S  is  the  total  vertical  shear. 


s.  =  4  f*vdA,    Formula  XVIII. 
lo  j Vt 


(7) 


142  STRENGTH  OF  MATERIALS 

where  s,  equals  the  unit  horizontal  shear  at  a  distance  v3  from  the 
neutral  axis  and  also  equals  the  unit  vertical  shear  at  the  same 

£v2 
v  dA  is  the  moment  of  the  area  of  the  end 
.4 

of  the  block  with  respect  to  the  neutral  axis. 


rv* 

I    vdA      f*«     ,.      -A 
-      J9.  .    I      vdA  =  vA. 

V  =  -  *—. >  Jv3 


When  the  area  and  location  of  the  center  of  gravity  of  the  por- 
tion of  the  plane  section  above  the  line  CE  are  known,  the  integral 
may  be  replaced  by  the  equivalent  expression  of  (8). 

PROBLEMS. 

1.  Find  the  unit  shearing  stress  in  a  4-inch  by  6-inch  beam  at  points  1  inch 
above  the  neutral  axis,  if  the  total  vertical  shear  is  1440  pounds.  Find  I  vdA 
both  ways  and  check. 

Ans.      *vdA  =  4'\dv  =  2  [v*]\  =  16; 


vA  =  2  X  8  =  16; 
1440 


72X4 


X  16  =  80  pounds  per  square  inch. 


2.  In  Problem  1,  find  the  unit  shearing  stress  at  the  neutral  surface. 
Also  find  the  average  shearing  stress  by  dividing  the  total  vertical  shear  by 
the  cross  section. 

Ans.  Unit  shear  at  the  neutral  surface,  90  pounds  per  square  inch. 
Average  shearing  stress  60  pounds  per  square  inch. 

3.  Show  algebraically  that  in  beams  of  rectangular  section  the  average 
unit  shear  is  two-thirds  as  great  as  the  shear  at  the  neutral  surface. 

4.  A  6-inch  by  8-inch  beam  10  feet  long,  supported  at  the  ends,  carries  a 
load  of  6000  pounds  4  feet  from  the  left  ond.     Find  the  unit  shearing  stress 
at  the  neutral  surface  just  to  the  right  of  the  left  support,  using  the  result  of 
Problem  3.  Ans.  112.5  pounds  per  square  inch. 

5.  In  Problem  4  find  the  unit  shearing  stress  1  inch  from  the  top  and 
2  inches  from  the  top  near  the  left  support. 

Ans.  49.21  and  84.37  pounds  per  square  inch. 

6.  In  Problem  4  find  the  unit  shearing  stress  3  inches  from  the  top  and 
five  inches  from  the  top. 

7.  In  a  beam  of  solid  circular  section  what  is  the  ratio  of  the  unit  stress  at 
the  neutral  surface  to  the  average  unit  shearing  stress?  Ans.  4  :  3. 

8.  What  is  the  maximum  load  which  can  be  placed  on  a  short  6-inch  by 
6-inch  beam  supported  at  the  ends,  if  the  allowable  unit  shearing  stress  parallel 
to  the  grain  is  150  pounds  per  square  inch? 

Ans.  7200  pounds  if  placed  at  the  middle;  3600  pounds  if  placed  near 
the  end. 


SHEAR  IN  BEAMS  143 

9.  Calculate  the  unit  shearing  stress  in  terms  of  the  total  shear  in  the  web 
of  a  10-inch  25-pound  I-beam  at  the  neutral  surface  and  at  the  bottom  of  the 
flange.  Ans.  ss  =  0.368  S  at  the  neutral  surface. 

ss  =  0.291  S  at  the  bottom  of  the  flange. 

10.  In  practice,  engineers  calculate  the  unit  shearing  stress  in  I-beams  by 
dividing  the  total  shear  by  the  area  of  cross  section  of  web  regarded  as  extend- 
ing the  entire  depth  of  the  beam  (see  Cambria).     What  is  the  average  unit 
stress  by  this  method  in  the  beam  of  Problem  9?  Ans.  0.323  S. 

11.*  A  7-inch  by  14-inch  beam  of  long-leaf  yellow  pine,  placed  on  supports 
13  feet  6  inches  apart,  was  subjected  to  equal  loads  at  points  4  feet  6  inches 
from  the  supports.  When  the  total  load  was  57,500  pounds,  the  beam  failed 
by  shear  at  the  neutral  axis  at  one  end.  Find  the  ultimate  shearing  strength 
of  this  timber  parallel  to  the  grain.  Compare  the  result  with  the  figures  given 
by  the  United  States  Department  of  Agriculture  (see  Cambria). 

Ans.  440  pounds  per  square  inch. 

12.*  A  7-inch  by  16-inch  beam  of  Douglas  fir,  supported  near  the  ends  and 
loaded  at  the  third  points  with  equal  loads,  failed  by  shear  when  the  total 
load  was  45,000  pounds.  Find  the  ultimate  shearing  strength  of  this  timber 
parallel  to  the  grain.  Ans.  301  pounds  per  square  inch. 

13.  Timber  having  an  allowable  shearing  stress  of  100  pounds  and  an 
allowable  bending  stress  of  1000  pounds  is  used  for  beams  supported  near  the 
ends  with  uniformly  distributed  loads.     Below  what  length  will  the  shear 
determine  the  load  in  a  6-inch  by  6-inch  beam?  in  a  4-inch  by  8-inch  beam? 

Ans.  For  a  6-inch  by  6-inch  beam,  5  feet. 

For  a  4-inch  by  8-inch  beam,  6  feet  8  inches. 

14.  Using  the  allowable  stresses  of  Problem  13,  what  is  the  total  safe  load 
uniformly  distributed,  on  a  6-inch  by  6-inch  beam  supported  at  the  ends 
when  the  length  is  4  feet?  when  the  length  is  6  feet? 

Ans.  4800  pounds,  4000  pounds. 

The  total  horizontal  shear  at  any  horizontal  plane  in  a  beam 
is  proportional  to  the  area  of  the  stress  distribution  diagram 
(Article  54),  above  or  below  this  plane;  and  the  unit  shearing 
stress  is  proportional  to  the  quotient  of  this  area  divided  by  b. 
The  unit  stress  at  a  distance  v  from  the  neutral  axis  varies  as  v; 
the  stress  on  an  area  dA  is  kvdA ;  and  the  total  stress  on  an  area 
extending  from  a  plane  at  a  distance  vs  from  the  neutral  axis  to 

the  top  of  the  beam,  is  k   I     vdA. 

J  vz 

In  a  rectangular  section  (Fig.  81)  the  stress-distribution 
diagram  is  a  triangle.  If  b  is  the  breadth  and  v2  the  distance 

to  the  top  from  the  neutral  axis,  the  area  of  this  triangle  is  -^  ;  at 
a  point  half  way  to  the  top,  the  area  is  reduced  one-fourth  so  that 

*  Problems  11  and  12  are  from  tests  made  by  Professor  A.  N.  Talbot, 
described  in  Bulletin  41  of  the  Engineering  Experiment  Station  of  The  Uni- 
versity of  Illinois. 


144 


STRENGTH    OF  MATERIALS 


it  is  evident  that  the  unit  shearing  stress  at  the  latter  point  is  f 
as  great  as  at  the  neutral  axis.  In  an  I-beam  (Fig.  82),  most  of 
the  shaded  area  representing  the  stress  distribution  is  in  the 
flange.  The  small  shaded  area  in  the  web  measures  the  differ- 
ence between  the  shearing  stress  at  the  neutral  axis  and  that  at 
the  bottom  of  the  flange. 

PROBLEM. 

15.  A  beam  of  rectangular  section,  10  inches  deep,  has  a  unit  shearing 
stress  of  150  pounds  per  square  inch  at  the  neutral  surface.  Find  the  unit 
shearing  stress  at  each  inch  above  the  axis  by  means  of  the  stress-distribution 
triangle.  Ans.  144,  126,  96,  and  54  pounds  per  square  inch. 

85.  Resultant  of  Shearing  and  Tensile  Stress. —  Fig.  113  rep- 
resents a  block  of  breadth  dx,  height  dy,  and  length  /,  subjected 


Ssldy 


Fig.  113.  —  Combined  Shear  and  Tension. 

to  tensile  stresses  of  intensity  st  perpendicular  to  the  left  and 
right  vertical  faces,  to  shearing  stresses  of  intensity  sa  parallel  to 
these  faces,  and  to  shearing  stresses  of  equal  intensity  in  the  top 
and  bottom  faces.  The  shear  on  the  left  face  is  upward  and  on 
the  top  face  toward  the  left.  We  wish  to  find  the  shearing  stress 
parallel  to  the  diagonals  BG  or  CF  and  the  tensile  stress  normal 
to  the  plane  BCFG.  We  will  consider  the  block  as  divided  into 
two  equal  triangular  prisms  by  the  plane  BCFG  and  will  take  the 
prism  on  the  left  of  the  plane  as  the  free  body  in  equilibrium. 
The  forces  which  act  on  this  free  body  are  five  in  number: 

Total  tension  stldy,  towards  the  left,  applied  at  center  of 
BCED; 

Total  shear  saldy,  upward,  applied  at  center  of  BCED; 

Total  shear  saldx,  towards  the  left,  applied  at  center  of  DEFG; 

Total  shear  on  BCFG,  parallel  to  BG,  applied  at  center  of 
BCFG; 

Total  tension  normal  to  BCFG  at  its  center. 


SHEAR  IN  BEAMS  145 

We  will  represent  the  unknown  unit  shearing  stress  in  the  diagonal 
plane  by  s/,  and  the  unknown  tensile  stress  by  s/.  The  total 
shear  on  this  plane  is  then  sa'lds,  where  ds  is  the  length  of  the 
diagonal  BG.  The  total  tension  on  the  diagonal  plane  is  st'lds. 
We  will  determine  the  magnitude  of  these  unknown  forces  by 
resolving  parallel  to  BG  and  normal  to  the  plane  BCFG.  These 
five  forces  are  represented  in  a  single  plane  in  Figure  113,  II. 
Resolving  parallel  to  BG  and  dividing  by  I, 

stdy  cos  0  +  ssdx  cos  6  —  s8dy  sin  0  =  ss'ds,  (1) 

where  0  is  the  angle  between  the  plane  BCFG  and  the  horizontal. 

Dividing  by  ds  and  substituting  for  -j-  and  -~  : 

as          as 

ssf  =  st  sin  0  cos  0  +  s8  [cos2  0  —  sin2  0],  (2) 

s3'=st  ^y^  +  sacos  20.  (3) 

Resolving  normal  to  ds: 

stdy  sin  0  +  s8dx  sin  0  +  ssdy  cos  0  =  st'ds.  (4) 

s/  =  s*  sin2  0  +  2  s8  sin  0  cos  0,  (5) 

+  s.u*2e.       .    '  (6) 


These  equations  apply  when  the  external  shearing  stresses  in  the 
block  have  the  directions  of  Fig.  113.  If  the  shear  is  reversed 
some  of  the  signs  are  changed. 

PROBLEMS. 

1.  With  the  unit  shearing  stress  100  pounds  per  square  inch  and  the  unit 
tensile  stress  in  the  same  direction  400  pounds  per  square  inch  (Fig.  114),  find 
the  resultant  unit  shearing  stress  along  a  plane  making  an  angle  of  20  degrees 
with  the  direction  of  the  tension.     Also  find  the  unit  tensile  stress  normal  to 
this  plane.  Ans.  ss',  205;  st',  111  pounds  per  square  inch. 

2.  With  unit  shearing  stress  100  pounds  per  square  inch  and  unit  tensile 
stress  zero,  find  the  resultant  tensile  stress  and  shearing  stress  at  45  degrees. 

Ans.  str,  100;  s/,  0. 

86.  Maximum  Resultant  Stress.  —  To  find  the  direction  that 
the  plane  BCFG  should  have  in  order  that  the  shearing  stress 
along  it  shall  be  a  maximum,  differentiate  the  expression  for  s/, 
Article  85,  (3),  with  respect  to  0: 


146  STRENGTH  OF  MATERIALS 

-rr  (s8)  =  st  cos  2  8  —  2  s,  sin  2  0  =  0  for  maximum  or  minimum,  (1) 
do 

Tan  20.  =  ^--  (2) 

To  find  the  direction  of  the  plane  in  order  that  the  unit  tensile 
stress  across  it  shall  be  a  maximum  or  a  minimum: 

st  sin  2  B  +  2  sa  cos  2  0  =  0,  (3) 

Tan20<=-?p-  (4) 

The  angle  of  equation  (4)  is  normal  to  that  of  (2),  consequently 
the  direction  of  maximum  shear  is  45  degrees  from  the  direction 
of  maximum  tension. 

PROBLEMS. 
1.   In  Problem  1,  Article  85,  find  the  direction  of  the  maximum  tension  and 

tan  208  =  f{$  =  2. 

20,  =  63°  26'    or  243°  26'. 

sin  2  0,  =      0.8944  "  -      0.8944. 

cos  2  0a  =      0.4472  "  0.4472. 

^  sin  20  =  178.88     "       -178.88. 
sa  cos  2  0  =    44.72      "      -    44.72. 


s/  =  223.60  -  223.60. 

08  =  31°  43'     "      121°  43'. 

Fig.  114,  II,  shows  the  direction  of  the  maximum  resultant  shearing  stress. 
At  31°  43'  the  portion  below  the  line  exerts  a  shear  to  the  right  on  the  portion 
above.  At  121°  43'  the  portion  on  the  side  of  the  line  in  which  the  angle  is 
measured  exerts  a  negative  shear  on  the  other  side  and  the  arrow  representing 
positive  shear  (away  from  the  origin)  is  on  the  other  side.  Fig.  114,  III,  shows 
how  the  shears  act  on  the  element  of  volume. 

For  the  maximum  resultant  tensile  stress, 

tan  2  0  =  -  *. 


20  = 

153°  26'  or 

333°  26'. 

sin  2  0  = 

0.4472  " 

-  0.4472. 

cos  2  0  =  - 

0.8944  " 

0.8944. 

s,  sin  2  0  = 

44.72   " 

-  44.72. 

(1  -  cos  20)  =      378.88        "          21.12. 


Maxs/  =      423.60  -23.60. 

Fig.  114,  IV,  shows  the  direction  and  relative  magnitude  of  these  unit  stresses. 
Note  that  the  minimum  is  a  compressive  stress,  along  a  line  normal  to  the  max- 
imum tension, 


SHEAR   IN  BEAMS  147 


400* 


SHEAR  223.6* 


II 


IV 

Fig.  114.  —  Shear  and  Tension. 

2.  With  unit  compressive  stress  of  200  pounds  per  square  inch  and  unit 
shearing  stress  of  160  pounds  per  square  inch,  find  the  maximum  resultant 
shear  and  compression  (Fig.  115). 

tan  2  0a  =  ~|Q°  =  -  0.625. 

2  08  =      148°      or         328°. 
max*.'  =-  188.68   "          188.68. 
tan20f=  1.600. 

et  =      29°        or         119°. 
maxst' =      88.68      "      -288.68. 

Fig.  115,  II,  shows  the  shearing  stress.  The  shear  at  74  degrees  is  negative 
and  is  downwards  towards  the  origin  on  the  right  side  of  the  line.  At  164 
degrees  the  shear  is  positive,  away  from  the  origin  on  the  upper  side.  We 
have  drawn  the  shear  at  —  16  degrees,  instead  of  along  the  opposite  line. 
Fig.  115,  III,  shows  all  the  maximum  shearing  stresses  acting  on  a  block  in  the 
solid. 

The  maximum  stress  along  the  29-degree  line  is  compressive,  and  the  mini- 
mum along  the  line  at  right  angles  to  this  is  tensile  (Fig.  115,  IV). 

To  find  the  maximum  resultant  shearing  stress  without  calculating  the 
angle  2  0S,  substitute  in 

s8'  =  ?jf  sin  2  0  +  ss  cos  2  0, 
the  values  of  sin  2  0  and  cos  2  0  calculated  from  the  relation 

from  which  we  get 


max s.'  =  y  si  +  (-}'  Formula  XIX, 

In  a  similar  way  substitute  the  values  of  cos  2  0  and  sin  2  6  from 
tan20,  =-— * 


148  STRENGTH  OF  MATERIALS 

in  equation  (6)  of  Article  85,  and  get 

maxs*'  =  ^  ±  y  s2a  +  I  ^  J   =  ^  ±  max  «/. 


Formula  XX. 


Generally  it  is  best  to  calculate  the  value  of  max  s8'  from  equation  (3)  of 
Article  85  and  then  get  maximum  s/  from  the  second  relation  of  Formula  XX. 


SHEAR  188.7* 
II 


COMPRESSION 


/^          88.1*TENSION 

IV 
Fig.  115.  —  Resultant  of  Shear  and  Compression. 

3.  Find  the  maximum  resultant  shearing  and  tensile  stresses  due  to  a  hori- 
zontal tension  of  300  pounds  per  square  inch  and  a  shearing  stress  of  160 
pounds  per  square  inch. 

Ans.  Tan  2  08  =  0.9375;     0.  =    21°  35';    max  ssf  =  219.28  pounds; 

111°  35';  -219.28      " 

max  st'  =  369.28       " 

-    69.28      " 

To  find  the  direction  of  the  larger  tensile  stress  we  notice  that  the  tension 
produced  by  the  shear  alone'  (Fig.  116,  I)  takes  the  direction  OC.  The  re- 


160* 


Fig.  116. 


TENSION 


sultant  maximum  tension  lies  between  this  direction  and  the  horizontal.  To 
get  this  position  measure  backward  45  degrees  from  the  position  of  maximum 
shear. 

4.   A  6-inch  by  10-inch  beam  rests  on  two  supports  30  inches  apart  and 
carries  a  load  of  20,000  pounds  at  the  middle.     Find  the  tensile  and  shearing 


SHEAR   IN  BEAMS 


149 


stress  5  inches  from  the  left  support  and  3  inches  from  the  bottom  of  the  beam. 
Find  also  the  maximum  resultant  stresses. 

sa  =  210  pounds;  st  =  200  pounds; 
maxs/  =  232.6  pounds;  max  s/  =  332.6  pounds;  mins/  =—  132.6  pounds; 


68  =  12°  44' 


6t=-  32°  16' 


0t  =  57°  44'. 


87.  Resultant  Stress  in  Beams.  —  In  beams,  the  maximum 
tensile  stress  is  in  the  outer  fibers,  while  the  maximum  shearing 
stress  is  at  the  neutral  axis.  The  shear  is  a  maximum  where 
the  tension  is  zero  and  the  tension  is  a  maximum  where  the  shear 
is  zero.  For  these  reasons  it  is  not  generally  necessary  to  com- 
pute the  resultant  stress  when  the  beam  is  made  of  material 
which  is  equally  strong  in  all  directions.  There  are  conditions, 
however,  where  the  maximum  resultant  stress  becomes  an  im- 
portant factor.  We  will,  therefore,  work  out  a  problem  to  show 
the  ways  in  which  these  stresses  act. 

PROBLEM.  . 

1.  A  6-inch  by  10-inch  beam  is  supported  at  points  30  inches  apart  and 
supports  a  load  of  20,000  pounds  midway  between  the  supports.  Find  the 
magnitude  and  direction  of  the  maximum  resultant  tension,  shear,  and  com- 
pression, at  sections  5  inches  and  10  inches  from  the  left  support  at  points 
0,  1,  2,  3,  4,  and  5  inches  from  the  neutral  axis. 

TABLE  VI. 

RESULTANT  SHEAR  AND  TENSION  IN  A  BEAM. 


Shear. 

Ten- 

Maximum  shear. 

Maximum  tension. 

Maximum  com- 

Distance below 

sion. 

pression. 

axis. 

Pounds. 

Pounds. 

Pounds. 

Angle. 

Pounds. 

Angle. 

Pounds. 

Angle. 

A  4.    K 

0 

250 

0 

250.0 

0°    0' 

250.0 

-45°     0' 

250 

45°    0' 

At  O 

1 

240 

100 

245.2 

5°  53' 

295.2 

-39°  07' 

195.2 

50°  53' 

inches 
from    • 

2 

210 

200 

232.6 

12°  44' 

332.6 

-32°  16' 

132.6 

57°  44' 

3 

160 

300 

219.3 

21°  35' 

369.3 

-23°  25' 

69.3 

66°  35' 

end. 

4 

90 

400 

219.0 

32°  53' 

419.0 

-12°  07' 

19.0 

77°  53' 

5 

0 

500 

250.0 

45°    0' 

500.0 

0°    0' 

0 

90°    0' 

0 

250 

0 

250.0 

0°    0' 

250.0 

-45°    0' 

250.0 

45°    0' 

At  10 

1 

240 

200 

260.2 

11°  49' 

360.2 

-33°  11' 

160.2 

56°  49' 

inches 

2 

210 

400 

290.0 

21°  48' 

490.0 

-23°  12' 

90.0 

66°  48' 

from 

3 

160 

600 

341.0 

30°  58' 

641.0 

-14°    2' 

41.0 

75°  58' 

end. 

4 

90 

800 

410.0 

38°  40' 

841.0 

-  6°  20' 

10.0 

83°  40' 

5 

0 

1000 

500.0 

45°    0' 

1000.0 

0°    0' 

0 

90°    0' 

Above  the  neutral  axis  the  shear  is  the  same  as  below  and  the  tension  and 
compression  change  places.     The  angles  are  numerically  the  same  but  are  on 


150 


STRENGTH  OF  MATERIALS 


opposite  sides  of  the  horizontal.  Fig.  117  shows  the  direction  and  relative 
magnitude  of  the  maximum  and  minimum  stresses  for  this  problem.  Near 
the  bottom  where  the  maximum  compression  is  small  its  direction  is  shown 
by  the  dotted  lines.  In  the  same  way  the  direction  of  the  tension  is  indicated 
near  the  top. 


1000#TENSION 


5"   FROM  LEFT  SUPPORT     10  " 
,  20000* 


30"- 


Fig.  117.  —  Resultant  Stress  in  a  Beam  Section. 

88.  Failure  of  Beams.  —  The  nature  of  the  failure  in  a  beam 
depends  principally  upon  the  relative  ultimate  strength  of  the 
material  in  the  different  directions  and  the  value  of  the  different 
maximum  stresses.  In  a  beam  which  is  short  relative  to  its 
depth,  the  unit  tensile  and  compressive  stresses  at  the  danger- 
ous section  are  small  compared  with  the  unit  shearing  stress  at 
the  neutral  surface  at  the  ends.  Owing  to  the  fact  that  timber 
has  a  small  shearing  strength  parallel  to  the  grain  such  a  beam, 
if  made  of  timber,  will  usually  fail  by  shear.  Fig.  118  shows 
4  wooden  beams  each  about  40  inches  long.  The  upper  beam  is  a 
yellow  pine  beam  glued  to  a  white  pine  beam.  The  total  depth 
was  3.80  inches  and  breadth  1.57  inches.  The  beam  was  sup- 
ported at  points  36  inches  apart  and  loaded  at  the  third  points; 
this  beam  failed  by  longitudinal  shear  at  one  end  when  the  total 
load  was  1950  pounds.  The  failure  followed  the  glued  surface 
but  began  in  the  white  pine. 


SHEAR   IN   BEAMS 


151 


Fig.  118.  —  Failure  of  Timber  Beams. 


Fig.  119.  —  Failure  of  a  Reinforced  Concrete  Beam. 


152  STRENGTH  OF  MATERIALS 

The  second  beam  of  white  pine  3.81  inches  by  1.68  inches, 
loaded  in  the  same  way  with  the  large  dimension  horizontal, 
failed  by  tension  under  a  total  load  of  2467  pounds. 

The  lower  beam  is  hickory,  2.39  inches  by  3.79  inches.  It 
failed  by  tension  under  a  total  load  of  12,900  pounds.  The  beam 
next  to  the  bottom,  2.02  inches  by  2.16  inches  is  also  hickory. 
It  failed  under  a  load  of  5540  pounds. 

PROBLEMS. 

1.  Find  the  ultimate  shearing  strength  of  the  white  pine  of  the  upper 
beam.     What  was  the  unit  tensile  stress  when  it  failed? 

2.  Find  the  unit  tensile  stress  at  the  middle  and  the  unit  shearing  stress 
at  the  ends  for  the  white  pine  stick. 

3.  Find  the  ultimate  bending  strength  of  hickory  from  the  two  samples; 
also  find  the  lower  limit  for  the  shearing  strength. 

Fig.  119  shows  a  reinforced  concrete  beam  supported  near 
the  ends  at  points  12  feet  apart  and  loaded  at  the  third  point. 
The  diagonal  line  shows  the  initial  failure.  This  is  frequently 
called  a  shear  failure.  It  is  really  failure  by  tension.  The  line 
of  failure  is  about  normal  to  the  direction  of  the  maximum 

resultant  tensile  stress.  Of  course, 
as  the  direction  of  this  stress  is  de- 
termined by  the  magnitude  of  the 
shear  there  is  good  reason  for  calling 
it  a  shear  failure.  The  upper  part 
of  this  crack  does  not  follow  a  direc- 
tion normal  to  the  resultant  tensile 
Fis  120. 

stress,  but  "it  must  be  remembered 

that,  as  the  crack  extended  up  in  the  beam,  the  direction  of  the 
resultant  stress  changed.  In  a  beam  loaded  in  this  way,  when 
cracks  develop  in  the  portion  between  the  loads  where  the  shear 
is  zero,  the  line  of  fracture  is  vertical. 

Fig.  120  shows  failure  of  a  concrete  beam  by  compression. 


CHAPTER  X.  \  n 

y     *  «$> 

BEAMS   OF   SPECIAL  FORM. 

89.  Beams  of  Constant  Strength.  —  A  beam  of  "  constant 
strength  "  is  one  in  which  the  section  modulus  varies  as  the 
moment,  so  that  the  extreme  fiber  stress  in  all  sections  is  the 
same.     To  design  such  a  beam,  we  write  the  moment  expression 
and  solve  for  the  section  modulus  in  terms  of  this  expression  and 
the  allowable  unit  stress  in  the  outer  fibers. 

90.  Cantilever  with  a  Load  on  the  End.  —  If  we  regard  the 
cantilever  as  fixed  at  the  right  end  and  take  the  left  end  as 
the  origin  of  coordinates,  the  moment  at  a  distance  x  from  this 
origin  is  Px.     If  s  is  the  allowable  unit  stress, 

Px  =  s  multiplied  by  the  section  modulus. 

bd2 
For  a  rectangular  section  the  section  modulus  is  -^- ,  and 

sbd* 

p*  =  ir 

PROBLEMS. 

1.  A  cantilever  of  constant  strength  with  the  load  on  the  end  is  of  rec- 
tangular section  of  constant  depth  6  inches.  The  allowable  fiber  stress  is 
800  pounds  per  square  inch.  Find  the  equation  for  the  breadth. 

,         Px 
Ans.  b  = 


2.  A  cantilever  beam  of  constant  strength  and  rectangular  section  has  a 
constant  breadth  6.     If  the  allowable  unit  stress  is  s,  find  the  expression  for 

the  depth  for  a  load  on  the  free  end.  Ans.  d?  =  —^-  • 

3.  A  cantilever  of  constant  strength  with  load  of  600  pounds  at  the  free 
end  is  4  inches  wide  throughout.     The  section  is  rectangular.     The  allowable 
fiber  stress  is  1200  pounds  per  square  inch.     If  the  length  is  60  inches  from 
the  load  to  the  fixed  point,  find  the  depth  at  each  10  inches. 

( Position:       10        20        30        40        50        60    inches. 
Ans-\  Depth:       2.74    3.87    4.74    5.48    6.12    6.71  inches. 

4.  A  cantilever  5  feet  long  carries  a  load  of  800  pounds  at  the  free  end. 
The  section  is  a  rectangle  with  the  depth  twice  the  breadth.     The  allowable 
stress  is  1000  pounds  per  square  inch.     Find  the  depth  at  each  10  inches. 

(  Position:       10        20        30        40        50        60    inches. 
Ans'\  Depth:       4.58     5.77    6.60    7.27    7.83    8. 32  inches. 
153 


154 


STRENGTH  OF  MATERIALS 


5.  A  cantilever  of  constant  strength  6  feet  long  carries  a  load  at  the  free 
end.  The  depth  at  the  wall  is  8  inches  and  the  breadth  is  constant.  Find  the 
depth  at  each  foot.  Ans.  3.27,  4.62,  5.66,  6.53,  7.30,  8.00  inches. 

Fig.  121  shows  some  cantilevers  of  constant  strength  and 
rectangular  section.  Fig.  121,  I,  is  a  beam  of  constant  depth. 
The  breadth  varies  as  z  —  the  equation  of  a  straight  line.  The 
plan  is  a  triangle.  Fig.  121,  II,  represents  a  beam  with  breadth 
constant.  The  depth  varies  as  the  square  root  of  x  —  the  equa- 
tion of  a  parabola.  One  surface  may  be  plane  as  in  II  or  both 
may  be  curved  as  in  Fig.  121,  III.  In  any  case  the  equation 
gives  the  total  depth.  Fig.  121,  IV,  represents  a  cantilever  in 
which  both  depth  and  breadth  vary,  all  sections  being  similar 
rectangles.  The  equation  is  that  of  the  cubical  parabola. 


I  DEPTH 
CONSTANT 


PROBLEM. 

6.  A  circular  steel  post  20  inches  long  is  used  as  a  cantilever  with  a  load 
at  the  end.  The  allowable  unit  stress  is  12,000  pounds  per  square  inch.  Find 
the  diameter  at  each  5-inch  interval,  if  the  load  is  600  pounds. 

91.  Shearing  and  Bearing  Stresses  at  the  End.  —  In  Fig.  121, 
the  load  P  is  represented  at  the  extreme  ends  of  the  beams. 
Allowance  must  be  made  at  the  ends  for 
bearing  and  shearing  stresses.  For  in- 
stance, in  Problem  3  of  Article  90,  sup- 
pose the  allowable  unit  shearing  stress  to 
J,jt  l^^y  be  200  pounds  per  square  inch.  The  sec- 

^1  N.I    1    tion  should  never  be  less  than  3  square 

inches;  the  minimum  depth  should  be  f 
inch.  Suppose  also  that  the  allowable 
bearing  stress  is  300  pounds  per  square 
inch,  and  that  the  center  of  the  load  must 
be  5  feet  from  the  wall;  the  bearing  area 
must  be  at  least  2  square  inches.  If  the 
load  extends  the  en- 
tire width  of  the 
beam  the  bearing 
area  must  be  4  inches 
by  J  inch.  The  ac- 
tual beam 


4" 


IV  SECTIONS  SIMILAR^ 
RECTANGLES 


Fig.  121.  —  Cantilevers  of 
Constant  Strength. 


Fig.  122. — Requirements 
for  Shear  and  Bearing. 


tend  at  least  J  inch 
beyond  the  center  of  the  load.     Fig.  122  shows  the  details  for 
these  conditions.     The  dotted  lines  are  the  limits  for  the  beam 


BEAMS   OF  SPECIAL   FORM  155 

figured  for  bending  only.  The  solid  lines  show  the.  minimum 
dimensions  figured  for  all  stresses.  The  actual  beam  should  be 
somewhat  larger  at  the  end  than  shown,  as  a  great  increase  in 
safety  can  be  secured  here  with  practically  no  increase  in  cost 
and  weight.  Artistic  appearance  and  convenience  of  construc- 
tion may  cause  further  modifications  outside  of  the  minimum 
dimensions. 

PROBLEMS. 

1.  Design  a  cantilever  of  constant  strength  for  a  load  of  500  pounds  at  a 
distance  of  40  inches  from  a  wall:  the  maximum  bending  stress  to  be  800 
pounds;  the  maximum  shearing  stress,  100  pounds;  and  the  maximum  bear- 
ing stress,  200  pounds  per  square  inch.     The  depth  of  the  beam  is  constant, 
4  inches. 

2.  Design  the  same  cantilever  with  square  section,  all  other  conditions 
remaining  the  same  as  in  Problem  1. 

92.  Cantilever  with  Uniformly  Distributed  Load.  —  The  only 
difference  between  a  cantilever  with  uniformly  distributed  load 
and  one  with  a  concentrated  load  is  in  the  expression  for  the 
external  moment. 

PROBLEMS. 

1.  A  cantilever  of  constant  strength  has  a  rectangular  section  and  constant 
breadth  b.     The  load  is  uniformly  distributed  and  is  w  pounds  per  inch  of 
length.     If  s  is  the  allowable  unit  stress,  find  the  expression  for  the  depth. 

3  wx* 

Ans.  d?  =  — j— . 
sb 

2.  Draw  a  cantilever  of  constant  strength  and  constant  breadth  of  2 
inches  to  carry  a  load  of  180  pounds  per  foot  uniformly  distributed,  with  an 
allowable  unit  stress  of  1000  pounds  per  square  inch.     The  length  of  the 
cantilever  is  40  inches. 

3.  A  cantilever  of  constant  strength  of  rectangular  section  is  d  inches  deep. 
If  the  load  is  uniformly  distributed,  find  the  expression  for  the  breadth. 

,       3wx2 
Ans.  b  =  — =r-  • 
sd2 

4.  Draw  a  cantilever  to  satisfy  the  conditions  of  Problem  3.     The  breadth 
at  the  wall  shall  be  6  inches,  and  the  length  30  inches. 

5.  A  cantilever  of  constant  strength,  of  rectangular  section,  carries  a  dis- 
tributed load  of  60  pounds  per  foot.     The  depth  is  4  inches  and  the  allowable 
unit  stress  is  800  pounds  per  square  inch.    Find  the  breadth  for  every  10  inches 
for  a  length  of  5  feet. 

6.  A  cantilever  of  square  section  for  a  uniformly  distributed  load  is  6  inches 
square  at  the  wall  40  inches  from  the  free  end.     Find  the  dimensions  at  each 
10  inches.  Ans.  2.38,  3.78,  4.95  inches. 

7.  Design  and  draw  a  cantilever  of  constant  strength  and  constant  breadth 
of  2  inches  to  carry  a  distributed  load  of  100  pounds  per  foot  and  a  load  of 


156  STRENGTH  OF  MATERIALS 

400  pounds  2  feet  from  the  free  end.  The  length  of  the  cantilever  is  4  feet 
and  the  allowable  stress  is  600  pounds  per  square  inch. 

8.  A  cantilever  of  constant  strength  for  a  uniformly  distributed  load  has  all 
sections  similar  triangles.  At  50  inches  from  the  free  end  it  is  8  inches  wide 
and  6  inches  deep.  Find  the  dimensions  at  each  10-inch  interval. 

93.  Beams  of  Constant  Strength  Supported  at  the  Ends.  - 

The  methods  of  solution  are  the  same  as  for  cantilevers.  For 
a  single  load  at  the  middle  the  problem  is  exactly  the  same  as 
that  of  a  cantilever  of  one-half  the  length  with  a  load  at  the  end. 
A  beam  with  a  single  load  at  any  point  may  be  regarded  as  made 
of  two  cantilevers  fixed  at  that  point  with  loads  equal  to  the 
respective  end  reactions.  Beams  with  distributed  loads  are  not 
quite  so  simple.  Allowance  must  be  made  in  this  beam  for  shear 
and  bearing  at  the  supports,  which  was  not  necessary  in  the  case 
of  cantilevers  with  uniformly  distributed  loads. 

PROBLEMS. 

1.  A  box  girder  is  made  of  two  12-inch  35-pound  channels  riveted  to  a 
pair  of  10-inch  by  Hnch  plates  extending  the  entire  length  and  other  similar 
plates  extending  part  of  the  length  on  both  sides  of  the  middle.     The  load  is 
600  pounds  per  foot;  the  span  is  40  feet;  and  the  allowable  unit  stress  is  10,000 
pounds  per  square  inch-.     Find  the  minimum  length  of  these  plates,  making 
no  allowance  for  weakening  due  to  rivet  holes. 

Ans.  One  pair  of  plates  should  be  22  feet  1  inch  long. 

2.  Design  a  wooden  beam  of  constant  strength  supported  at  the  ends  for 
a  span  of  6  feet  with  a  load  of  600  pounds  in  the  middle  and  a  distributed  load 
of  120  pounds  per  foot.     The  allowable  fiber  stress  is  1000  pounds  per  square 
inch  and  the  beam  sections  square.    The  allowable  shearing  stress  is  100  pounds 
per  square  inch  and  the  bearing  stress  120  pounds  per  square  inch. 

3.  Solve  Problem  2  if  the  concentrated  load  is  4  feet  from  one  end. 

94.  Deflection  of  Beams  of  Constant  Strength.  —  The  prob- 
lem of  finding  the  deflection  of  a  beam  of  constant  strength 
differs  from  that  of  a  uniform  section  in  that  the  moment  of 
inertia  is  no  longer  constant  but  is  a  function  of  x.     In  beams 
symmetrical  with  respect  to  the  neutral  surface, 

2  si 


M  = 


d 


where  s  is  constant  throughout  the  length  and  d  may  be  con- 
stant or  variable.  In  the  following  discussions  when  the  depth 
is  constant  we  will  represent  it  by  the  capital  D,  and  when  the 
breadth  is  constant  we  will  use  the  capital  B.  This  will  enable 


BEAMS   OF  SPECIAL   FORM  157 

us  more  readily  to  distinguish  between  constants  and  variables 
in  integrating. 
95.   Cantilever  of  Constant  Depth  with  Load  on  the  End.  - 


Dividing  by  /: 


where  E,  s,  and  D  are  constants.     We  will  consider  the  beam  as 
fixed  at  the  right  end. 


2/ 


(l-xY+  (Ct  =  0).  (5 

sl2 


max 


We  have  now  the  maximum  deflection  in  terms  of  s  and  D. 
We  will  calculate  it  in  terms  of  the  dimensions  at  the  wall.  If 
Im  is  the  maximum  value  of  the  moment  of  inertia  (at  the  wall)  : 

PDl 


PI* 
2EDIm~      2  EIm' 


From  equation  (7)  we  see  that  the  deflection  of  a  cantilever 
of  constant  strength  and  constant  depth,  due  to  a  load  on  the 
free  end,  is  one  and  one-half  times  as  great  as  the  deflection  of  a 
cantilever  of  uniform  section  having  the  same  maximum  dimen- 
sions. 

96.   Cantilever  of  Constant  Breadth  with  Load  on  the  End.  - 

wd*y  -    2s  m 

*a?"   7' 

where  d  is  a  variable. 


158  STRENGTH  OF  MATERIALS 

where  B  is  the  constant  breadth.     Equation  (1)  becomes: 


4  si3    IsB 

-rVep 


Substituting  the  value  of  s  in  terms  of  the  breadth  and  depth  at 
the  wall: 

,KN 

which  is  twice  the  deflection  of  a  cantilever  of  uniform  section 
with  the  same  maximum  dimensions. 

PROBLEMS. 

1.  Find  the  expression  for  the  deflection  at  the  middle  for  a  beam  of  con- 
stant strength  and  constant  depth  due  to  a  load  at  the  middle  of  the  span, 

PI3 

the  beam  being  supported  at  the  ends.  Ans.   —  ort  „,    . 

62,  rjl-m 

2.  Find  the  expression  for  the  deflection  of  a  beam  of  constant  strength 
and  constant  breadth,  supported  at  the  ends  with  a  load  at  the  middle. 

PI3 


3.   In  the  case  of  a  cantilever  of  constant  depth  with  a  load  of  w  pounds  per 
inch  uniformly  distributed,  what  is  the  deflection  at  the  free  end? 


Ans-  ~ 

4.   Solve  the  case  of  a  cantilever  of  constant  strength  and  constant  breadth 
with  a  uniformly  distributed  load. 

Ans.  Ey  =  -  2sy/||7zlog|  -  z  +  l\ 
tflft  Wl* 


~      2EIm 

5.  How  do  the  deflections  at  the  ends  in  Problems  3  and  4  compare  with 
those  in  uniform  beams? 

6.  Find  the  expression  for  the  deflection  at  the  end  of  a  cantilever  of  con- 
stant strength,  with  sections  similar  rectangles,  due  to  a  load  at  the  free  end. 

97.  Cast-iron  Beams.  —  Parts  of  machines  are  frequently 
made  of  cast  iron.  Cast  iron  differs  from  most  structural  mate- 
rials *  in  that  the  ultimate  strength  in  tension  is  much  less  than 

*  There  is  a  greater  difference  in  concrete. 


BEAMS   OF   SPECIAL   FORM  159 

in  compression.  A  good  sample  of  cast  iron  may  have  an  ulti- 
mate tensile  strength  of  25,000  pounds  per  square  inch  and  a 
compressive  strength  of  as  much  as  100,000  pounds  per  square 
inch.  Working  stresses  in  cast-iron  members  subject  to  bending 
should  be  about  3000  pounds  per  square  inch  on  the  tension  side, 
and  may  be  10,000  pounds  per  square  inch  on  the  compression 
side.  The  ultimate  strength  in  compression  being  four  times  as 
much  as  in  tension  we  would  naturally  expect  that  this  ratio 
should  hold  in  designing  cast-iron  beams;  but  owing  to  the  fact 
that  the  neutral  axis  is  shifted  from  the  center  of  gravity  of 
the  cross  section  after  the  material  passes  the  elastic  limit,  a 
lower  ratio  should  be  used. 

The  cast-iron  beam  of  rectangular  section  in  Problem  3, 
Article  56,  showed  a  modulus  of  rupture  of  44,000  pounds  per 
square  inch.  The  same  bar  in  tension  broke  under  27,000  pounds 
per  square  inch,  the  shifting  of  the  neutral  axis  and  the  devia- 
tion of  the  tension  and  compression  stress-strain  diagrams  from 
straight  lines  accounting  for  this  difference. 

In  order  to  use  the  material  economically,  cast-iron  beams 
are  generally  made  of  sections  which  bring  the  center  of  gravity 
two  or  three  times  as  far  from  the  compression  side  as  it  is  from 
the  tension  side.  The  corresponding  stresses  in^the  outer  fibers 
bear  approximately  the  same  ratio. 

PROBLEM. 

1.  A  cast-iron  beam  40  inches  long  is  supported  at  the  ends  and  carries  a 
load  at  the  middle.  The  section  is  that  of  Fig.  123.  If  the  allowable  fiber 
stress  in  tension  is  3000  pounds  per  square  inch  and  the 
allowable  stress  in  compression  is  10,000  pounds  per  square 
inch,  what  is  the  safe  load  with  the  flange  at  the  top  and 
also  with  the  flange  at  the  bottom? 

Ans.  595  pounds,  1390  pounds. 

The  above  problem  has  been  solved  with  the  assump- 
tion that  the  stress-strain  diagram  is  a  straight  line  and 
that  the  modulus  of  elasticity  is  the  same  in  compression 
and  in  tension.  How  near  these  assumptions  are  true 
may  be  seen  from  Fig.  124.  This  figure  represents  the 
stress-strain  diagrams  for  cast  iron  from  the  same  heat, 
tested  at  the  Watertown  Arsenal.  (Tests  of  Metals, 
885,  pages  475-490.)  The  tension  curve  is  drawn  from  Fig.  123. 

the  mean  of  four  tests  and  the  compression  curve  from 
the  mean  of  twelve  tests.     (Only  a  part  of  the  compression  curve  is  drawn. 
The  ultimate  strength  of  this  iron  in  compression  was  apparently  52,000  pounds 
per  square  inch  for  bars  12  inches  long  and  1  square  inch  cross  section.    As 


160 


STRENGTH  OF  MATERIALS 


the  bars  bent  considerably  before  failure  the  actual  compressive  stress  was 
much  above  this  figure.) 

The  compressive  diagram  is  a  straight  line  up  to  13,000  pounds  per  square 
inch.  The  tension  diagram  is  slightly  curved  in  the  neighborhood  of  3000 
pounds.  The  modulus  of  elasticity  in  tension  is  slightly  greater  than  the 
modulus  in  compression.  The  curves  are  so  close  together,  however,  that 
little  error  is  made  in  using  the  above  assumptions  for  tensile  stresses  of  3000 
pounds  with  compressive  stresses  of  10,000  pounds. 


35000 
31000 
33000 
32000 
31000 
30000 

^^ 

^ 

x* 

^* 

** 

^x 

X 

^ 

^ 

1ESS  POUNDS  PER  SQUARE  INCH 

/ 

/ 

/ 

/ 

j 

',c 

v 

*** 

^~ 

7 

I 

^ 

+^ 

/ 

"^ 

^ 

/ 

^ 

/ 

V 

\* 

/C 

£ 

/ 

X 

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/ 

t 

/ 

1 

/ 

/ 

/ 

/ 

/ 

/ 

/ 

'  / 

"12000 
2  HOOO 

3  10000 

/ 

/ 

(/ 

/ 

8000 

/ 

7000 

A 

6000 

1 

5000 

y 

4000 

r 

3000 
2000 

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f 

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0 

y 

0001           .0005                .0010                .0015                .0020                .0025            .0030 
UNIT  DEFORMATIONS 

Fig.  124.  —  Stress-strain  Diagrams  for  Cast  Iron. 

When  it  comes  to  computing  the  unit  stress  at  rupture,  the  case  is  different. 

Let  us  consider  a  beam  of  rectangular  section  1  inch  wide  with  the  outer 
fibers  in  the  tension  side  1  inch  from  the  neutral  axis  and  determine  the  depth 
of  the  compression  side  in  order  that  the  total  compression  shall  equal  the  total 
tension  when  the  elongation  in  the  outer  fibers  in  tension  is  0.0025.  The 
unit  stress  in  these  outer  fibers  from  the  curve  is  found  to  be  24,800  pounds 
per  square  inch.  The  total  tension  is  obtained  from  the  average  ordinate  of 
the  curve.  To  get  this  average  ordinate  we  divide  the  area  measured  by  a  pla- 
nimeter  by  the  length  of  the  base.  Or  each  ordinate  may  be  regarded  as  the 
mean  altitude  of  a  strip  of  width  equal  to  the  distance  between  the  ordinates. 
For  instance,  the  ordinate  which  corresponds  to  the  unit  elongation  of  0.001 
is  16,100  pounds.  This  may  be  taken  as  the  altitude  of  a  strip  of  width  0.0001 
extending  from  elongation  0.00095  to  0.00105.  The  sum  of  the  ordinates 


BEAMS   OF   SPECIAL   FORM 


161 


r 

i 
I 

<     i-    > 

t 

.88 

1 

from  0.0001  to  0.0024  inclusive  multiplied  by  the  width  of  the  interval  will 
give  the  area  from  0.00005  to  0.00245.  The  half  intervals  from  0  to  0.00005 
and  from  0.00245  to  0.00250  may  be  computed  separately.  Calling  the  inter- 
val between  the  ordinates  unity  to  avoid  decimals, 

Area  from  0  to  0.5  ...................................  131 

Area  from  0.5  to  24.5  (Sum  of  24  ordinates)  .............  398,650 

Area  from  24.5  to  25  .................................  12,350 

Total  ......  411,131 

Dividing  by  25  ..........................  Mean  stress  .  .  16,445 

If  the  width  of  the  section  is  6  and  the  height  from  the  neutral  surface  is  v*, 
the  total  tension  is  the  product  of  the  area  bv% 
multiplied  by  the  mean  unit  stress. 

The  total  compression  below  the  neutral  surface 
must  equal  the  total  tension  above.     If  the  width  is 
unity  as  in  Fig.  125,  the  total  compression  is  the 
average  compressive  stress  multiplied  by  v\.    But  this 
product  is  the  area  of  the  compression  diagram.     It 
is  only  necessary,  then,  to  find  the  ordinate  which 
forms  the  right  boundary  of  an  area  below  the  com- 
pression curve  which  is  411,131  units.     The  curve   Fig.  125.  —  Neutral  Axis 
being  a  straight  line  to  the  deformation  0.00075,  at       in    Rectangular    Cast 
which  the  unit  stress  is  14,300  pounds,  we  get  this       Iron  Section. 
much  of  the  area  from  the  triangle. 

Area  from  0  to  7.5  (taken  as  a  triangle)  ................  53,625 

Area  from  7.5  to  20.5  (Sum  of  13  ordinates)  .............  309,570 

Total  to  20.5..  363,195 

^     Ordinate  at  21  ......................................  30,850 

Total  to  21.  5.  .  394,045 

Area  required  beyond  21.5  ............................  17,086 

Total.  .  411,131 

Dividing  17,086  by  31,400  which  is  approximately  the  mean  ordinate  for 
the  remainder  of  the  area,  we  get  0.56. 

21.5  +  0.56  =  22.06. 

The  compression  depth,  v1}  is  to  the  tension  depth,  v2,  as  22.06  to  25.  In  other 
words,  vi  is  practically  88  per  cent  of  v2  as  shown  in  Fig.%125. 

The  moment  of  the  total  tension  is  obtained  by  multiplying  each  ordinate 
by  its  abscissa  and  adding  the  products.  The  result  for  a  section  above  the 
neutral  axis  1  inch  square  is  10,506  inch  pounds.  In  the  same  way  the  moment 
of  the  compression  area  for  a  section  below  the  neutral  axis  1  inch  wide  and 
0.88  inch  deep  is  9541  inch  pounds.  The  total  moment  of  the  section  is  20,047 
inch  pounds. 

If  we  take  a  beam  1  inch  wide  and  1.88  inches  deep  and  compute  the  maxi- 


mum fiber  stress  from 


-j-  ,  we  get  34,000  pounds  per  square  inch  for  the 


value  of  s  corresponding  to  this  moment  instead  of  24,800  pounds  in  tension 
and  31,400  pounds  in  compression.  Fig.  126  shows  the  difference  between  a 
cast-iron  beam  of  rectangular  section  having  an  ultimate  tensile  stress  in  the 


162 


STRENGTH  OF  MATERIALS 


outer  fibers  of  24,800  pounds  per  square  inch  (II)  and  a  similar  beam  subjected 
to  the  same  moment  but  with  a  straight  line  curve  (I).  These  figures  show 
the  shifting  of  the  neutral  axis  and  show  why  the  modulus  of  rupture  does  not 
agree  with  the  maximum  tensile  strength. 

Fig.  127  shows  the  stress  distribution  in  a  cast-iron  T  section  with  a  rela- 
tively thick  stem  with  the  flange  in  tension. 


I  II 

Fig.  126.  —  Stress  Distribution  in  a 
Cast  Iron  Section. 


Fig.  127. 


In  the  case  of  cast-iron  beams,  where  it  is  not  practicable  to  test  full-size 
members,  it  is  advisable  to  make  small  beams  of  similar  sections  and  test 
them.  The  strength  of  the  large  beam  of  similar  cross  section  and  the  same 
length  may  be  computed  from  the  cube  of  the  like  dimensions. 

98.  Beams  of  Two  or  More  Materials.  —  Beams  are  fre- 
quently made  of  two  or  more  materials  having  different  moduli 
of  elasticity.  The  most  common  cases  are  beams  made  by  bolt- 
ing iron  or  steel  plates  to  wooden  beams,  and  reinforced  concrete 
beams,  in  which  steel  rods  are  embedded  in  the  concrete  in  the 
tension  side. 

Fig.  128  represents  a  beam  made  by  bolting  a  steel  plate  to 
one  side  of  a  wooden  beam.  In  calculating  the  section  modulus 
we  must  remember  that  the  modulus  of  elasticity  in  the  wood  is 
less  than  that  in  the  steel,  and  that  with  a  given  deformation  the 
unit  stress  in  the  two  materials  is  proportional  to  their  respective 
moduli.  If  the  modulus  of  the  steel  is  30,000,000  and  that  of 
the  wood  is  1,500,000  the  stress  in  the  steel  corresponding  to  a 
given  unit  elongation  is  20  times  as  great  as  the  stress  in  the 
wood.  To  find  the  neutral  axis  in  this  case  we  may  use  Fig.  128, 
II,  regarding  the  steel  area  A  BCD  as  having  a  density  20,  while 
the  wood  area  has  a  density  unity;  or  we  may  use  Fig.  128,  III, 


BEAMS   OF   SPECIAL   FORM  163 

regarding  the  steel  area  A  BCD  as  replaced  by  the  wood  area 
EFGH  twenty  times  as  wide  as  the  steel  and  of  the  same  thick- 
ness. 

PROBLEMS. 

(Use  E  for  steel,  20  times  E  for  wood  in  these  problems.) 

1.  A  6-inch  by  8-inch  wooden  beam  has  a  6-inch  by  Hnch  steel  plate 
fastened  to  the  lower  side.     Find  the  neutral  axis. 

Ans.  1.64  inches  from  the  bottom  of  the  wood. 

2.  A  4-inch  by  4-inch  wooden  beam  has  a  4-inch  by  |-inch  steel  plate  on 
the  lower  side  and  a  1-inch  by  1-inch  bar  embedded  in  the  upper  side,  the  top 
of  the  bar  being  flush  with  the  top  of  the  timber.     Find  the  neutral  axis. 

Ans.  1.18  inches  from  the  bottom  of  the  timber. 

To  calculate  the  stress  in  beams  of  this  kind  we  may  use  the 
timber  as  the  unit  and  regard  the  steel  as  expanded  into  the  equiv- 
alent amount  of  wood  as  in  Fig.  128,  III.  We  may  determine 


LWOOD 

WOOD  EQUIVALENT 

-STEEL         E  OF  STEEL 


H 


III 


Fig.  128. 


the  section  modulus  on  this  assumption  and  find  the  stress  in  the 
outer  fibers.  To  find  the  stress  for  steel  at  a  given  point,  multi- 
ply the  stress  calculated  for  wood  by  the  ratio  of  the  moduli  of 

elasticity. 

PROBLEMS. 

3.  In  Problem  1,  what  is  the  equivalent  moment  of  inertia  of  the  section? 

Ans.  738.5  inches4. 

4.  In  Problem  3,  the  beam  is  subjected  to  a  bending  moment  of  120,000 
inch  pounds.     What  is  the  maximum  fiber  stress  in  the  steel  and  in  the  wood? 

Ans.   1033  pounds  per  square  inch  in  the  wood. 
6955  pounds  per  square  inch  in  the  steel. 

5.  In  Problem  2,  the  beam  is  10  feet  long  and  supported  at  the  ends.     What 
is  the  greatest  load  which  may  be  applied  at  the  middle  if  the  allowable  unit 
stress  in  the  steel  is  15,000  pounds  per  square  inch?  Ans.  1938  pounds. 

6.  A  beam  10  feet  long,  supported  at  the  ends  and  loaded  at  the  middle, 
is  made  of  two  Hnch  by  8-inch  plates  of  steel  bolted  to  the  vertical  faces  of 
a  6-inch  by  8-inch  wooden  beam.     If  the  allowable  unit  stress  in  the  steel  is 
15,000  pounds  and  in  the  wood  is  1000  pounds,  what  is  the  maximum  safe 
load? 


164 


STRENGTH  OF  MATERIALS 


7.  In  a  beam  of  the  form  of  Problem  6  what  should  be  the  ratio  of  the 
depth  of  the  wooden  beam  to  that  of  the  steel  plates  if  both  reach  their  allow- 
able load  at  the  same  time? 

99.  Reinforced  Concrete  Beams.  —  Reinforced  concrete  rep- 
resents another  form  of  combination  beam.  A  reinforced  con- 
crete beam  has  steel  rods  embedded  in  the  concrete  near  the- 
surface  in  the  tension  side.  Sometimes  both  tension  and  com- 
pression sides  are  reinforced.  These  rods  may  be  ordinary 
round  or  square  steel  bars.  Usually  they  are  corrugated  in 
some  way  or  made  of  cable  or  twisted  square  bar  so  that  they 
will  not  slip  even  if  the  grip  of  the  concrete  should  be  weakened. 
Many  designers  use  steel  with  a  high  yield  point,  50,000  pounds 
per  square  inch  or  more.  Such  steel  may  safely  be  subjected  to 
stresses  of  18,000  pounds  per  square  inch,  provided  it  has  not 
been  subjected  to  rough  treatment  causing  short  bends  or  kinks. 

Fig.  129  represents  a  portion  of  a  reinforced  concrete  beam  8 
inches  by  11  inches  in  cross  section.  The  reinforcement  consists 


Fig.  129. 

of  three  rods  with  centers  1  inch  from  the  bottom  of  the  beam. 
The  photograph,  Fig.  119,  shows  a  beam  of  this  size  after  failure. 

In  working  out  the  theory  of  concrete  beams,  it  is  customary  to 
regard  the  steel  as  taking  all  the  tension.  If  the  unit  stresses 
are  kept  low,  the  concrete  on  the  tension  side  of  the  neutral  axis 
does  exert  some  tensile  stress,  but  at  loads  of  less  than  one  third 
of  the  ultimate  strength  of  the  beam  fine  cracks  form  in  the  ten- 
sion side  and  tests  show  that  the  steel  takes  practically  all  the 
tension  at  larger  loads. 

We  are  accustomed  to  speak  of  the  per  cent  of  reinforcement. 
This  is  obtained  by  dividing  the  area  of  the  steel  by  the  area  of 
the  beam  section  above  the  center  of  the  steel.  In  Fig.  129  the 
beam  is  regarded  as  an  8-inch  by  10-inch  section;  the  inch  of  con- 
crete is  considered  as  simply  protecting  the  steel.  If  the  rods  in 
Fig.  129  are  f  inch  round,  the  reinforcement  is  1.15  per  cent. 


BEAMS   OF   SPECIAL   FORM 


165 


More  or  less  elaborate  formulas  have  been  proposed  for  rein- 
forced concrete  beams.  Any  formula  depends  upon  the  form  of 
the  compression  curve  of  the  concrete.  This  curve  varies  greatly 
with  the  material,  the  proportions,  the  care  in  mixing,  the  age, 
and  the  stresses  to  which  it  has  been  subjected.  The  modulus 
of  elasticity  of  concrete  is  lowered  greatly  by  slight  overloads. 
For  these  reasons  there  is  little  use  for  great  refinement  of  calcu- 
lation unless  you  have  good  experimental 
data  in  regard  to  the  concrete  which  you 
are  using. 

The  portion  of  the  section  above  the 
neutral  axis  is  in  compression.  The  line 
OF}  Fig.  130,  represents  the  compression 
curve.  If  the  stresses  are  small,  this 
curve  is  a  straight  line.  If  they  are 
large,  it  is  somewhat  curved,  as  in  Fig. 
85.  (Fig.  10  is  the  compression  curve 
of  one  sample  of  concrete).  If  the  sec- 
tion of  the  beam  is  rectangular  above 
the  neutral  axis  the  total  compression 

may  be  represented  by  the  triangular  area  OFC.  If  the  angle  of 
bend  in  unit  length  of  the  beam  is  dd,  the  unit  stress  in  the  top 
fibers  at  a  distance  v2  from  the  neutral  surface  is 

v2Ec  dO, 

where  Ec  is  the  modulus  of  the  concrete  in  compression.  The 
total  stress  in  a  section  of  width  b  is  given  by 

6z|  Ec  dB 
Total  compression  in  concrete  = 

If  we  take  vi  as  the  distance  from  the  neutral  surface  to  the  center 
of  the  steel,  the  total  stress  in  the  steel  is  given  by 

Total  tension  in  steel  =  A8viE8d0, 

where  As   is  the  steel  area  and  E8  its  modulus  in  tension. 

Since  the  total  compression  in  the  concrete  equals  the  total 
tension  in  the  steel, 

AE8 


crete  Beam. 


bEc 


2 

Also  vz  +  vi  =  h, 

where  h  is  the  depth  from  the  top  of  the  beam  to  the  center  of 


CD 

(2) 


166  STRENGTH  OF  MATERIALS 

reinforcement.     These  equations  enable  us  to  locate  the  neutral 
axis  in  terms  of  h  for  any  per  cent  of  reinforcement. 

It  must  be  remembered  that  the  theory  above  is  on  the  simple 
assumption  that  the  concrete  curve  for  the  stresses  used  is  prac- 
tically a  straight  line. 

PROBLEMS. 

1.  If  the  modulus  of  elasticity  of  the  steel  is  fifteen  times  as  great  as  that 
of  the  concrete  and  the  area  of  the  steel  is  1  per  cent  of  the  area  bh,  find  v2 
in  terms  of  h.  Ans.  vz  =  0.42  h. 

2.  If  the  modulus  of  elasticity  of  the  steel  is  30,000,000,  and  that  of  the 
concrete  2,000,000,  locate  the  neutral  axis  for  reinforcements  of  1  per  cent, 
1.2  per  cent,  1.5  per  cent.  Ans.  vl  =  0.58  h,  0.55  A,  0.52  h. 

3.  In  Problem  1  what  is  the  average  unit  compressive  stress  in  the  concrete 
if  the  unit  tensile  stress  in  the  steel  is  10,000  pounds  per  square  inch,  and 
what  is  the  unit  compressive  stress  in  the  outer  fibers? 

Ans.  238,  476  pounds  per  square  inch. 

4.  In  Problem  2  what  is  the  compressive  stress  in  the  top  fibers  of  the 
concrete  when  the  unit  tensile  stress  in  the  steel  is  12,000  pounds  per  square 
inch?  Ans.  571,  640,  750  pounds  per  square  inch. 

The  unit  stresses  in  Problem  4  are  too  high  to  be  allowed.  The 
results  show  that  it  is  not  economical  to  use  too  great  a  percentage 
of  steel,  for  with  large  percentage  of  steel  the  concrete  reaches 
its  allowed  value  in  compression  while  the  stress  in  the  steel  is 
still  relatively  small. 

PROBLEMS. 

5.  With  one-half  of  1  per  cent  reinforcement,  and  with  the  modulus  of 
steel  fifteen  times  as  great  as  that  of  concrete,  what  is  the  maximum  com- 
pressive stress  in  the  concrete  when  the  tensile  stress  in  the  steel  is  12,000 
pounds  per  square  inch?  Ans.  375  pounds  per  square  inch. 

6.  With  0.8  per  cent  of  reinforcement  and  the  modulus  of  the  steel  fifteen 
times  that  of  the  concrete,  what  is  the  unit  stress  in  the  steel  when  the  maxi- 
mum stress  in  the  concrete  is  400  pounds  per  square  inch? 

100.  Resisting  Moment  in  Reinforced  Concrete.  —  The  com- 
pressive stress  in  the  concrete  above  the  neutral  axis  and  the  ten- 
sile stress  in  the  steel  form  a  couple.  The  moment  of  this  couple 
is  the  product  of  either  total  stress  multiplied  by  the  distance 
between  their  resultants.  The  resultant  stress  in  the  steel  is 
taken  as  applied  at  the  center  of  the  steel  section.  The  resultant 
of  the  compressive  stress  is  at  the  center  of  gravity  of  the  tri- 
angle OFC,  Fig.  130.  The  moment  arm  is 

Vi  +  §  V2, 

provided  the  compression  diagram  is  a  straight  line.     If  the 


BEAMS   OF  SPECIAL   FORM  167 

compression  diagram  is  curved,  the  moment  arm  is  slightly 
smaller.  Calculations  from  the  compression  curves  of  Fig.  10 
show  that  the  center  of  gravity  of  the  area  is  0.62  vz  at  the  ultimate 
stress  and  0.64  v%  when  the  stress  is  carried  to  400  pounds  per 
square  inch.  There  is,  therefore,  little  error  in  assuming  that 
the  resultant  of  the  compressive  stress  is  0.67  v2  above  the  neutral 
axis  in  all  cases. 

PROBLEMS. 

1.  Using   1   per  cent  reinforcement  and  Es  fifteen  times  Ec,  what  is  the 
effective  resisting-moment  arm?  Ans.  0.86  h. 

2.  In  Problem  1,  if  the  unit  stress  in  the  steel  is  10,000  pounds  per  square 
inch,  what  is  the  resisting  moment?  Ans.  86  bh2. 

3.  If  the  beam  in  Problem  2  is  8  inches  wide  and  10  inches  deep  from  top 
to  center  of  reinforcement,  what  is  the  resisting  moment? 

Ans.  68,800  inch  pounds. 

4.  In  Problem  3  the  beam  is  10  feet  long  and  supported  at  the  ends.     What 
load  at  the  middle  will  give  the  required  moment,  computing  the  concrete 
at  150  pounds  per  cubic  foot  and  neglecting  the  weight  of  the  steel  and  of 
the  concrete  below  the  reinforcement?  Ans.  1877  pounds. 

To  get  the  resisting  moment  of  a  reinforced  concrete  beam, 
we  compute  the  moment  of  the  couple  of  which  the  total  tension 
is  one  force  and  the  total  compression  the  other  force.  The 
total  tension  in  the  steel  is  easily  computed  by  multiplying  the 
area  by  the  allowable  unit  stress.  The  moment  arm  must  be 
less  than  h',  experiments  and  theory  show  that  it  is  about  0.8  h. 
To  get  the  resisting  moment,  then,  we  multiply  the  area  of  the 
steel  by  its  allowable  unit  stress  and  multiply  the  total  stress 
thus  obtained  by  0.8  of  the  distance  from  the  center  of  the  steel 
to  the  top  of  the  beam.  This  gives  a  simple  method  of  approx- 
imate calculation  for  a  beam  of  rectangular  section.  We  must 
not  apply  this,  however,  to  beams  with  too  great  a  per  cent  of 
reinforcement,  as  these  will  fail  by  compression  of  the  concrete 
before  the  steel  reaches  a  considerable  stress.  The  allowable 
amount  of  reinforcement  may  be  computed  for  beams  of  rec- 
tangular section  by  the  methods  of  the  preceding  article,  or  may 
be  determined  from  the  results  of  tests  of  large  concrete  beams. 

PROBLEMS. 

5.  Assuming  that  the  point  of  application  of  the  resultant  compression 
is  0.8  of  the  depth  from  the  steel  and  that  the  allowable  unit  stress  in  the  steel 
is  15,000  pounds,  find  the  total  safe  load  at  the  middle  of  an  8-inch  by  10-inch 
beam  10  feet  long,  supported  at  the  ends,  with  a  reinforcement  of  1  per  cent. 

Ans.  3200  pounds  if  the  weight  of  the  beam  is  neglected. 


168  STRENGTH  OF  MATERIALS 

6.  What  must  be  the  approximate  depth  of  a  rectangular  beam  12  inches 
wide  and  15  feet  long,  supported  at  the  ends,  to  carry  a  load  of  600  pounds  per 
foot,  including  its  own  weight,  if  the  reinforcement  is  0.9  per  cent  and  the 
allowable  stress  12,000  pounds  per  square  inch? 

i oi.  Resultant  Tensile  Stress.  —  In  Article  86  we  learned 
that  the  combined  effect  of  a  tensile  and  shearing  stress  produced 
a  resultant  tensile  stress  at  an  angle  with  the  direct  tension.  The 
tensile  strength  of  concrete  is  small  relatively  to  the  compressive 
strength  and  is  the  same  in  all  directions.  A  concrete  beam  is 
likely  to  fail  in  tension  along  a  line  perpendicular  to  the  direc- 
tion of  the  maximum  resultant  tensile  stress.  The  photograph, 
Fig.  119,  shows  a  failure  of  this  kind.  A  concrete  beam  should 
have  some  reinforcement  along  the  direction  of  the  resultant 

tensile  stress  as  well  as  longi- 

T  tf  ^^\ 1    tudinally    along    the    tension 

V ^.^          side.     Fig.  131  is  a  diagram 

D  .  ,.  showing  the  cracks  at  failure 

Fig.  131.  —  Failure  of  a  Reinforced 

Concrete  Beam.  m  a  beam  similar  to  Fig.  119 

and  loaded  in  the  same  way. 

Notice  that  the  cracks  in  the  middle  third  where  there  is  no 
shear  run  nearly  vertical,  while  the  crack  outside  of  the  load  is 
inclined.     This  beam  and  the 
one  in  Fig.  119  had  longitudi-    I  \\X\\  /////\ 

X   X    X   x.   X.  S  #    j?    JP    #       I 

nal  reinforcement  only.     Un-    Fr^ 

less    the    shear    is    relatively     Fig.  132.  _  shear  Bars  for  Reinforced 

small,  a  concrete  beam  should  Concrete. 

have  additional  reinforcement 

arranged  as  shown  in  Fig.  132.     These  inclined  bars  are  called 

shear  bars.     They  may  be  rigidly  attached  to  the  tension  bars  or 

may  be  separate. 

When  a  beam  is  continuous  over  several  supports,  it  follows, 
of  course,  that  the  reinforcement  must  be  placed  near  the  top  in 
the  portion  where  the  moment  is  negative. 


CHAPTER  XI. 


BENDING  COMBINED  WITH  TENSION  OR 
COMPRESSION. 

1 02.  Transverse  and  Longitudinal  Loading.  —  It  often  hap- 
pens'that  a  beam  is  subjected  to  a  direct  tension  or  compression 
in  the  direction  of  its  length  and  a  transverse  force  producing 
a  bending  moment.  The  unit  stress  at  any  point  in  a  given 
section  is  the  sum  of  the  direct  stress  and  the  bending  stress  at 
that  point.  For  example,  suppose  a  4-inch  by  4-inch  post  stands 
vertical  and  supports  a  load  of  4000  pounds  at  the  top.  The 
direct  compressive  stress  is  250  pounds  per  square  inch.  Sup- 
pose this  post  is  fixed  at  the  bottom  (Fig.  133),  and  that  a  hori- 
zontal push  of  200  pounds  is 
applied  2  feet  from  the  bot- 
tom. This  transverse  force  pro- 
duces a  tensile  stress  of  450 
pounds  per  square  inch  in  the 
outer  fibers  at  the  bottom  on 
the  side  of  the  push  and  a 
compressive  stress  of  the  same 
magnitude  in  the  opposite  side. 
The  resultant  stress  is  700 
pounds  per  square  inch  in  the 
one  side  and  200  pounds  per 
square  inch  in  the  other.  Fig. 
133,  IV,  shows  the  distribution 
of  the  stress,  compression  being 
represented  by  the  vertical  dis- 
tance downward.  In  Fig.  133, 


•200* 


COMPRESSION  250* 
DIRECT  STRESS 

TENSION  45 


^COMPRESSION  700^ 


Fig.  133.  —  Post  with  Compression 
and  Bending. 


II,  we  have  the  compression  alone  due  to  the  direct  load  of  4000 
pounds.  In  Fig.  133,  III,  we  have  the  stress  due  to  bending;  it 
is  450  pounds  compression  on  the  left  side  and  450  pounds  ten- 
sion on  the  right.  At  the  middle  of  the  section  it  is  zero.  In 
Fig.  133,  IV,  the  two  stresses  are  combined.  The  line  EF,  which 
is  the  zero  line  for  the  bending  stress,  is  placed  on  the  line  CD, 

169 


170  STRENGTH  OF  MATERIALS 

representing  the  compressive  stress  in  II.  At  a  point  f  inch  from 
the  right  of  the  diagram  the  resultant  stress  is  0. 

The  resultant  stress,  being  the  sum  of  the  direct  and  the 
bending  stress,  is  given  by  the  expression: 

Unit  stress  =  -^  +  -p,  Formula  XXI. 

where  P  is  the  total  load  parallel  to  the  length  of  the  beam  and 
M  is  the  bending  moment  from  any  source  whatever.  Since 
v  has  the  positive  sign  on  one  side  of  the  neutral  axis  and  the 
negative  sign  on  the  other  side,  the  second  term  may  be  positive 
or  negative,  according  to  the  position. 

PROBLEMS. 

1 .  A  wooden  post  6  inches  square  and  5  feet  high  is  fixed  at  the  lower  end 
and  carries  a  load  of  7200  pounds  at  the  top.     A  horizontal  push  of  180  pounds 
is  exerted  upon  the  north  side  of  the  post  50  inches  from  the  bottom.     Find 
the  maximum  tensile  and  compressive  stress  at  the  bottom. 

Ans.    50  pounds  per  square  inch  tension  on  the  north  side; 

450  pounds  per  square  inch  compression  on  the  south  side. 

2.  A  wooden  post  6  inches  by  8  inches  is  placed  vertical,  with  the  8-inch 
faces  in  the  meridian.     The  post  projects  fi  feet  from  the  ground.     A  load  of 
12,000  pounds  is  placed  on  the  top  and  a  horizontal  push  of  400  pounds  directed 
south  is  applied  to  the  north  face  1  foot  from  the  top.     Find  (a)  the  unit  stress 
at  the  bottom  on  the  north  side;  (6)  the  unit  stress  at  the  bottom  on  the  south 
side;  (c)  the  location  of  the  line  of  zero  stress  in  the  bottom  section;  (d)  the 
position  of  a  section  having  zero  stress  on  the  north  side. 

Ans.  (a)  125  pounds  tension;  (6)  625  pounds  compression;  (c)  If  inches 
from  the  north  side;  (d)  4  feet  4  inches  from  the  top. 

3.  A  2-inch  solid  shaft  supported  on  bearings  4  feet  apart  carries  a  load 
of  400  pounds  at  the  middle  and  is  subjected  to  a  horizontal  compression  of 
8000  pounds  parallel  to  its  length.     Find  the  maximum  and  minimum  unit 
stress,  neglecting  the  weight  of  the  shaft. 

Ans.  8662  pounds  compression;  3570  pounds  tension. 

4.  A  concrete  wall  10  feet  high  and  1  foot  thick  is  subjected  to  a  horizontal 
water  pressure  which  varies  as  the  depth  and  is  62.4  pounds  per  square  foot 
at  a  depth  of  1  foot.     The  concrete  weighs  150  pounds  per  cubic  foot.     Find 
the  maximum  tensile  and  compressive  stress  at  the  bottom. 

103.  Eccentric  Loading.  —  Let  a  rigid  bar  G,  Fig.  134,  be 
supported  by  three  equal  and  symmetrically  placed  rubber 
bands  (or  springs)  suspended  from  a  fixed  horizontal  support. 
Each  of  the  bands  will  be  stretched  equally  and  the  bar  will 
hang  in  a  horizontal  position  (Fig.  134,  I).  Now  attach  a  load 
P  at  the  middle  of  the  bar.  Each  rubber  band  will  receive  the 


BENDING  WITH  TENSION  OR  COMPRESSION    171 


same  elongation  and  the  bar  will  remain  horizontal  in  the  posi- 
tion of  Fig.  134,  II.  If  the  load  P  be  moved  to  the  right,  as  in 
Fig.  134,  III,  the  middle  band  will  receive  the  same  elongation 
as  in  the  preceding  case,  while  the  left  band  will  be  elongated  less 
and  the  right  band  more.  If  we  place  the  load  still  farther  to  the 


G, 

j 

G 

C 

G 

c 

r 

r* 

l 

£=£= 

I  II  III 

Fig.  134.  —  Eccentric  Loading  of  Rubber  Bands. 

right,  we  finally  reach  a  position  where  the  left  end  is  elevated 
above  the  position  which  it  occupied  before  the  load  was  applied, 
so  that  finally  no  load  whatever  comes  on  the  left  band.  If 
instead  of  the  rubber  bands,  we  use  helical  springs  of  relatively 
large  cross  section,  which  are  able  to  resist  compression  as  well 
as  tension,  we  may  secure  compressive  stress  in  the  spring  which 
is  on  the  side  of  the  center  away  from  the  load. 

Instead  of  the  rubber  bands  we  might  use  a  continuous  body, 
as  a  sheet  of  rubber  or  metal.  If  such  a  sheet  is  fastened  to  a 
rigid  body  at  the  top  and  bottom  and  a  load  is 
applied  considerably  to  one  side  of  the  center, 
there  will  be  an  elongation  on  that  side  and  a 
shortening  or  buckling  on  the  other.  A  similar 
result  obtains  when  a  "compressive  load  is  ap- 
plied to  a  body.  Fig.  135  shows  a  block  of  soft 
rubber  with  the  load  central,  and  Fig.  135,  II, 
shows  the  effect  of  moving  this  load  a  little  to 
one  side. 

If  we  consider  the  bar  G  of  Fig.  134,  III,  we 
see  that  the  effect  of  the  eccentric  load  is  a 
translation  downwards,  of  the  same  magnitude 
as  that  due  to  the  central  load  in  II,  together 
with  a  rotation  about  the  bottom  of  the  middle 
band  as  an  axis.  Taking  moments  about  this 
point  C,  we  find  that  equilibrium  occurs  when 
the  moment  of  the  load  P  with  respect  to  C  is  equal  to  the 
moment  of  the  excess  of  tension  in  the  right  band  plus  the 


IL 


Fig.  135.  —  Com- 
pression with  Di- 
rect and  with  Ec- 
centric Loading. 


172 


STRENGTH  OF  MATERIALS 


moment  of  the  deficiency  of  tension  in  the  left  band.  Suppose 
that  the  bands  are  4  inches  apart,  and  that  a  load  of  1  pound 
stretches  the  bands  0.4  inch.  One  pound  will  stretch  a  single 
band  1.2  inches.  Now  move  the  load  of  1  pound  2  inches  to 
the  right  of  the  middle.  The  moment  of  the  load  is  2  inch 
pounds,  which  is  balanced  by  a  force  of  0.5  pound  at  4  inches. 
The  tension  in  the  right  band  is  0.25  pound  more  than  that  in 
the  middle,  and  the  tension  in  the  left  band  is  0.25  pound  less. 
If  the  load  is  moved  more  than  3  inches  from  the  middle,  the 
tension  in  the  left  band  becomes  less  than  it  was  before  the  load 
was  applied. 

We  learned  in  Mechanics*  that  a  force  along  any  line  may 
be  replaced  by  an  equal  force  along  any  parallel  line  and  a 
couple  whose  moment  is  the  product  of  the  force  multiplied  by 
the  distance  between  the  lines.  In  Fig.  136  the  body  is  sub- 


C 

Y 

-> 

C 

\ef~B 
B 

^^ 

'^W// 

COUPLE  &Q=eP 


COUPLE  dR=eP 


Fig.  136.  —  Block  with  Eccentric  Loading. 

jected  to  a  compressive  force  of  P  pounds  along  a  line  through 
the  point  B.  The  force  is  parallel  to  the  axis  CC.  The  distance 
of  the  line  of  force,  BB,  from  the  axis  is  called  the  eccentricity 
of  the  load  and  is  represented  by  the  letter  e.  The  force  P  at 
B  may  be  replaced  by  a  force  P  at  C  and  a  couple  of  moment 
eP  tending  to  rotate  the  top  of  the  block  in  a  clockwise  direction. 
The  two  equal  and  oppositely  directed  forces  which  comprise 
this  couple  may  be  regarded  as  having  any  magnitude,  direction, 
and  position,  provided  only  that  the  product  of  the  magnitude  of 
either  force  multiplied  by  the  distance  between  them  is  equal  to 
moment  eP.  We  may  consider  the  couple  at  the  top  as  made  of 

*  Hoskins'  Theoretical  Mechanics,  Art.  94;  Maurer's  Technical  Mechanics, 
Art.  31. 


BENDING  WITH  TENSION  OR  COMPRESSION     173 


two  horizontal  forces  Q  at  a  distance  b  apart,  provided  bQ  =eP. 
In  the  same  way,  the  opposite  couple  at  the  bottom  may  be 
made  of  two  horizontal  forces  R  at  a  distance  d  apart,  with  a 
moment  dR  =  eP,  tending  to  produce  counterclockwise  rotation. 
It  is  evident,  then,  that  the  portion  of  the  body  between  the 
lower  force  Q  and  the  upper  force  R  is  subjected  to  a  bending 
moment  eP  and  a  direct  load  P  at  C.  The  forces  R  and  Q  may 
be  regarded  as  indefinitely  large  and  the  distances  6  and  d  in- 
definitely small,  so  that  the  entire  block  may  be  considered  as 
subjected  to  the  bending  moment. 

It  will  be  noticed  that  the  force  P  at  the  top  of  Fig.  136  is 
really  the  resultant  of  a  set  of  forces  distributed  over  the  entire 
top,  and  varying  uniformly  from  left  to  right.  The  cap  at  the 
top  and  the  support  at  the  bottom  are  supposed  to  be  relatively 
very  rigid.  If  the  load  is  applied  to  a  small  area  instead  of  over 
the  entire  end  of  the  body,  the  unit  stress  at  some  points  near  the 
end  will  be  larger  than  that  indicated  by  the  theory.  Near  the 
middle  of  the  length,  the  stress  will  be  practically  the  same  as 
if  the  force  were  applied  by  means  of  a  rigid  cap  and  support. 

Fig.  137  shows  large  eccentricity,  the  resultant  load  lying 
entirely  outside  of  the  section.  Here  the  existence  of  bending 


i  ii  in 

Fig.  137.  —  Large  Eccentricity. 

and  direct  stress  together  is  almost  self-evident.  Consider  the 
portion  above  any  section  G.  Resolving  vertically,  the  vertical 
reaction  at  the  section  is  equal  to  the  load  P  at  the  top.  Taking 
moments  about  an  axis  perpendicular  to  the  plane  of  the  paper 
through  the  center  of  the  section  G,  the  resisting  moment  of  the 
section  must  equal  the  moment  eP.  Fig.  137,  II,  shows  the  effect 
of  compression  and  Fig.  137,  III,  the  effect  of  tension. 


174 


STRENGTH  OF  MATERIALS 


Compression  combined  with  bending  is  shown  in  Fig.  138. 
The  forces  P  are  applied  to  the  wrenches  by  the  screw  clamp. 
The  wrenches  as  cantilevers  transmit  the  bending  moments  and 
direct  compression  to  the  bar.  The  experiment  may  easily  be 
performed  by  two  wrenches  and  a  steel  or  wooden  bar,  the  force 
being  applied  to  the  wrenches  by  hand.  The  bar  will  bend  as 
in  Fig.  137,  II,  if  the  forces  are  towards  each  other,  and  as  in 
Fig.  137,  III,  if  the  forces  are  from  each  other. 

The  clamp  of  Fig.  138  is  subjected  to  tension  and  bending. 
The  eccentricity  is  the  distance  from  the  center  of  the  screw  to 


Fig.  138. 

the  center  of  gravity  of  any  section.  In  a  hook  the  load  line 
joins  the  shank  with  the  point  which  is  immediately  below  it 
when  loaded.  This  point  is,  of  course,  the  point  in  concave 
portion  which  is  farthest  from  the  shank.  The  eccentricity  is 
the  distance  of  this  load  line  from  the  center  of  gravity  of  the 
section. 


BENDING   WITH   TENSION  OR  COMPRESSION     175 

PROBLEMS. 

1.  A  4-inch  by  6-inch  short  block  is  subjected  to  a  compressive  load  of 
6000  pounds,  the  line  of  the  resultant  load  being  5  inch  from  the  axis  of  the 
block  in  the  plane  parallel  to  the  6-inch  faces  which  passes  through  the  axis. 
Find  the  maximum  compressive  stress  and  the  minimum  stress. 

Ans.  Maximum,  375  pounds  per  square  inch; 

Minimum,  125  pounds  per  square  inch  compression. 

2.  In  Problem  1  solve  for  the  case  when  the  load  is  1  inch  from  the  axis. 

Ans.  500  pounds  per  square  inch,  0. 

3.  A  solid  circular  rod  1  inch  in  diameter  is  subjected  to  a  pull  of  4000 
pounds.    What  is  maximum  and  minimum  unit  stress  when  the  line  of  pull 
is  0.1  inch  from  the  axis? 

Ans.  9167  pounds  tension,  1019  pounds  tension. 

4.  Solve  Problem  3  if  the  line  of  pull  is  0.2  inch  from  the  axis. 

Ans.  13,241  pounds  tension,  3055  pounds  compression. 

5.  A  block  b  wide  and  d  thick,  of  rectangular  section,  has  the  load  so 
placed  that  the  unit  stress  in  the  outer  fibers  on  one  side  is  zero.     If  the  line 
of  load  is  in  the  plane  of  symmetry  parallel  to  the  faces  of  breadth  d,  what  is 

the  eccentricity?  Ans.  5. 

o 

6.  What  eccentricity  in  a  solid  circular  section  of  radius  a  will  make  the 

unit  stress  on  one  side  zero?  Ans.  e  =  -.  - 

4 

7.  A  hollow  circular  cylinder  of  outside  radius  a  and  inside  radius  b  is  so 
loaded  that  the  unit  stress  on  one  side  is  zero.    What  is  the  eccentricity? 

a2 +  62 

Ans.  e  =  —. — -  • 
4  a 

8.  In  a  hook  of  circular  section  the  distance  from  the  center  of  the  section 
to  the  line  of  the  load  is  3  inches.1    The  load  is  1000  pounds  and  the  diameter 
of  the  section  is  2  inches.     What  is  the  maximum  tensile  and  compressive  stress? 

Ans.  4138  pounds  tension,  3501  pounds  compression. 

9.  A  hook  of  circular  section  has  the  center  of  the  section  4  inches  from 
the  line  of  the  load.     What  must  be  its  diameter  to  carry  a  load  of  6000  pounds 
if  the  unit  stress  in  tension  shall  not  exceed  10,000  pounds  per  square  inch? 

Ans.  2.99  inches. 

10.  What  is  the  minimum  diameter  of  the  shank  in  Problem  9,  if  the  allow- 
able unit  stress  is  6000  pounds  per  square  inch? 

The  hooks  of  Problems  8  and  9  have  circular  sections  and  the 
unit  stress  on  the  tension  side  is  greater  than  on  the  compression 
side.  It  is  better  to  make  the  hook  section  of  the  form  of  Fig.  139, 
I,  so  that  the  tensile  stress  and  compressive  stress  are  about 
equal. 

In  the  case  of  cast-iron  hooks,  since  the  allowable  compressive 
stress  is  about  three  times  as  great  as  the  allowable  tensile  stress, 
it  is  customary  to  use  a  modified  T  section  (Fig.  139,  II),  which 
will  make  the  distance  from  the  neutral  axis  to  the  outer  fibers  in 
compression  much  larger  than  the  distance  to  the  tension  side. 


176 


STRENGTH  OF  MATERIALS 


PROBLEMS. 

11.  The  section  of  a  steel  hook  at  the  maximum  distance  from  the  load 
line  consists  of  two  semicircles  with  centers  1.5  inches  apart  joined  by  two 
straight  lines.     The  circle  nearest  the  load  line  is  1  inch  in  diameter  and  the 
other  is  0.8  inch  in  diameter.      The  load  line  passes  1.5  inches  from  the  center 
of  the  larger  circle.      Calculate  the  maximum  compressive  stress  when  the 
maximum  tensile  stress  is  6000  pounds  per  square  inch. 

12.  A  post  6  inches  square  and  5  feet  high  with  two  faces  in  the  meridian 
carries  a  load  of  1800  pounds  0.5  inch  west  of  the  middle  on  the  top,  and  is 
subjected  to  a  horizontal  push  of  72  pounds  from  west  toward  east  applied  at 
the  top.     What  is  the  unit  stress  at  the  bottom  on  the  east  and  west  sides? 


Fig.  139. 

104.  Maximum  Eccentricity  without  Reversing  Stress.  —  A 
brick  pier  laid  in  lime  mortar  has  practically  no  tensile  strength, 
and  the  tensile  strength  of  masonry  laid  in  cement  is  not  reliable. 
For  this  reason  the  load  on  a  masonry  pier  should  always  be 
placed  so  that  the  stress  over  the  entire  section  shall  be  com- 
pressive.  Problem  5  of  Article  103  showed  that  a  load  on  a 
rectangular  section  at  a  distance  from  the  center  greater  than 
one-sixth  of  the  dimension  of  the  section  in  this  direction  pro- 
duces a  negative  stress  in  the  outer  fibers  in  the  opposite  side. 
For  this  reason  it  is  a  rule  of  architects  and  engineers  that  the 
resultant  load  shall  not  fall  outside  of  the  middle  third  in  the  case 
of  rectangular  columns  or  piers.  From  Problem  6  we  see  that, 
with  round  piers  of  solid  section,  the  load  must  not  lie  outside 
of  a  circle  whose  diameter  is  one-fourth  that  of 
the  pier. 

The  statement  that  the  load  must  lie  inside 
the  middle  third  means  that  if  the  load  is  on 
the  line  BD,  Fig.  140,  through  the  center  of  the 
rectangle  parallel  to  the  side  d,  it  must  lie  in 
the  middle  third  of  this  line.  In  the  same  way, 


Fig.  140. 


if  the  load  falls  on  the  line  FG,  it  must  be  between  the  points 
CiCi  on  this  line.     CC  is  one-third  of  the  length  BD. 


BENDING  WITH  TENSION  OR  COMPRESSION     177 


PROBLEMS. 

1.  A  12-inch  square  brick  pier  carries  a  load  of  7200  pounds.     What  is 
the  unit  stress  when  the  resultant  load  lies  1  inch  from  the  center  of  the  section 
on  a  line  through  the  center  parallel  to  one  side? 

Ans.  75  pounds  per  square  inch. 

2.  Solve  Problem  1  when  the  load  is  1  inch  from  the  center  on  the  diagonal 
of  the  square.  Ans.  85.3  pounds  per  square  inch. 

3.  In  Problem  1  how  far  may  the  load  be  placed  from  the  center  and  not 
have  tensile  stress?  .Ans.  2  inches. 

4.  In  Problem  2  how  far  may  the  load  be  placed  from  the  center  and  not 
reverse  the  stress?  Ans.  1.41  inch. 

5.  Compare  the  strength  of  a  12-inch  by  12-inch  pier  with  the  load  in  the 
middle  with  that  of  a  12-inch  by  16-inch  pier  with  the  load  2  inches  from  the 
middle  on  a  line  through  the  middle  parallel  to  the  long  side. 

Ans.  The  square  pier  is  stronger  in  the  ratio  21  :  16. 

6.  A  square  section  of  side  6  has  the  resultant  load  at  a  point  C,  the  coordi- 
nates of  which  are  (x,  y),  Fig.  141, 1.     Show  that  when  the  unit  stress  at  F  is 
zero,  the  position  of  C  satisfies  the  equation 


Fig.  141.  —  Maximum  Eccentricity  without  Reversing  Steess. 

SUGGESTION.  —  The  moment  of  inertia  of  a  square  section  being  the  same  for 
all  axes  through  the  center,  the  rotation  will  be  about  the  axis  OE  perpendicular 
to  OC.  The  distance  of  the  extreme  fibers  at  F  from  this  axis  is  equal  to  EB. 


The  distance 


EB  =  ;r  (cos0  +  sin0). 


7  = 


b* 
12 


For  zero  stress  at  the  corner,  F, 

e  (cos  0  +  sin  0)  =  g  • 
b 


105.   Resultant  Load  not  on  a  Principal  Axis.  —  In  all  the 

problems  of  the  preceding  articles,  the  resultant  load  fell  on  one 
principal  axis  and  rotation  took  place  about  the  other  principal 
axis.  In  the  case  of  a  round  or  square  section,  the  moment  of 


178  STRENGTH  OF  MATERIALS 

inertia  is  the  same  for  all  axes  through  the  center  of  gravity,  and 
any  such  axis  may  be  regarded  as  a  principal  axis.  In  other  sec- 
tions, when  the  load  does  not  fall  on  a 
principal  axis,  the  axis  of  rotation  is  not 
the  line  OE  normal  to  OC,  but  is  some 
line  OG  (Fig.  142)  between  OE  and  the 
x  axis  for  which  7  is  a  minimum. 

To  solve  the  problem,  we  use  the 
•  method  of  resolving  the  couple  eP  into 
Fig.  142.  —  Eccentric  Load-  components  producing  rotation  about 
icipalAxis.  the  two  principal  axes.  These  compo- 
nents  for  Fig.  142  are  the  couple  Px  tending  to  produce  rotation 
about  the  Y  axis  and  the  couple  Py  tending  to  produce  rotation 
about  the  X  axis.  The  bending  stress  at  any  point  xi  in  the 
section  due  to  the  moment  Px  is 

Pxx1 


where  Iy  is  the  moment  of  inertia  with  respect  to  the  Y  axis.  In 
the  same  way  the  stress  due  to  the  couple  Py  may  be  calculated. 
The  resultant  stress  due  to  bending  is  the  sum  of  the  two  com- 
ponents. 

EXAMPLE.  —  A  rectangular  block  12  inches  long  measured  from  east  to 
west  and  10  inches  wide  is  subjected  to  a  load  of  3600  pounds  2  inches  from 
the  east  edge  and  2  inches  from  the  north  edge.  Find  the  unit  stress  at  each 
corner. 

The  bending  stress  at  the  north  and  south  edges  due  to  the  couple  of  10,800 
inch  pounds  is  54  pounds  per  square  inch.  The  bending  stress  at  the  east 
and  west  edges  due  to  the  couple  of  14,400  inch  pounds  is  60  pounds  per  square 
inch.  The  direct  compression  is  30  pounds  per  square  inch.  The  unit  stresses 
at  the  corners  are:  144  pounds  compression  at  the  northeast  corner;  24  pounds 
compression  at  the  northwest  corner;  84  pounds  tension  at  the  southwest 
corner;  36  pounds  compression  at  the  southeast  corner. 

PROBLEMS. 

1.  A  rectangle  8  inches  long  and  6  inches  wide  has  a  load  of  1200  pounds 
applied  2|  inches  from  one  8-inch  side,  and  3f  inches  from  one  6-inch  side. 
Find  the  unit  stress  at  each  corner. 

^Ans.  50,  25,  0,  25  pounds  per  square  inch. 

2.  A  rectangle  of  length  b  and  height  d  has  a  load  applied  at  a  point  whose 
coordinates  are  (x,  y)  with  respect  to  axes  through  the  center  parallel  to  the 

sides.  Show  that  if  this  point  lies  on  a  line  whose  intercepts  are^  and^,  the 
stress  will  be  zero  at  the  opposite  corner. 


BENDING  WITH  TENSION  OR  COMPRESSION     179 


3.  A  rectangular  post  6  inches  by  10  inches  has  the  10-inch  sides  in  the 
meridian.  A  load  of  2400  pounds  is  applied  1  inch  north  and  1  inch  east  of 
the  middle.  A  horizontal  push  of  200  pounds  is  20  inches  above  the  bottom 
directed  south  30  degrees  west,  the  line  of  the  push  passing  through  the  center 
of  the  section.  Find  the  unit  stress  at  each  corner  at  the  bottom. 

36.03  pounds  at  northeast  corner. 

Ans. 


22.69 
43.97 
57.31 


northwest 
southwest      " 
southeast      " 


CHAPTER  XII. 


COLUMNS. 

1 06.  Definition.  —  In  the  discussion  in  the  preceding  chapter, 
we  said  nothing  in  regard  to  the  deflection  of  the  body  considered, 
and  the  effect  of  this  deflection  in  changing  the  amount  of  eccen- 
tricity. In  tension,  the  deflection  is  in  the  direction  to  diminish 
the  eccentricity  (Fig.  137,  III).  In  compression,  on  the  other 
hand,  the  deflection  increases  the  eccentricity  and  consequently 
increases  the  unit  stress  (Fig.  137,  II).  A  yard  stick  may  be 
placed  with  one  end  on  the  floor  and  a  compressive  force  applied 
with  the  hand  to  the  other  end.  When  the  force  reaches  a  cer- 
tain amount,  the  stick  suddenly  bends  and  may 
deflect  several  inches  from  the  straight  line. 
The  original  eccentricity  of  possibly  0.01  inch 
is  increased  to  several  inches  and  the  unit  stress 
may  be  sufficient  to  cause  rupture.  If  the  stick 
is  placed  with  one  end  on  a  platform  scale,  as 
in  Fig.  143,  it  is  found  that  the  load  which  pro- 
duces a  deflection  of  2  inches  is  little,  if  any, 
greater  than  the  load  which  causes  a  deflection 
of  1  inch.  The  resisting  moment  has  been 
nearly  doubled,  but  the  external  moment  has 
likewise  been  doubled,  owing  to  the  increased 
=*  length  of  the  moment  arm. 
^^  A  compression  member  whose  length  is  sev- 

lg  Column  >Dg  era^  times  as  Srea"t  as  its  smallest  transverse 
dimension  is  called  a  column.  There  is  no  def- 
inite ratio  of  length  to  diameter  at  which  a  compression  member 
ceases  to  be  a  short  block  and  becomes  a  column.  We  find, 
however,  that  when  the  ratio  of  length  to  the  smallest  transverse 
dimension  is  less  than  10,  the  error  made  by  neglecting  the  de- 
flection is  so  small  that  it  may  ordinarily  be  neglected.  Some 
engineers  call  a  compression  member  of  length  less  than  15  diam- 
eters a  short  block  or  pier  and  calculate  it  by  the  methods  of  the 
preceding  chapter. 

180 


COLUMNS 


181 


Columns  may  be  vertical,  as  the  intermediate  posts  of  bridges, 
or  horizontal,  as  the  top  chords  of  a  bridge.  The  connecting  rod 
of  an  engine  is  a  column  during  the  forward  stroke.  When  a 
column  is  vertical,  the  only  bending  moment  is  that  due  to  the 
eccentricity  of  the  load  and  the  deflection.  When  a  column  is 
horizontal  or  inclined,  its  own  weight  applied  as  a  beam  becomes 
an  appreciable  factor.  The  rafters  supporting 
a  roof  act  as  columns  and  inclined  beams. 

A  compression  member  of  some  length  is 
frequently  called  a  strut. 

107.  Column  Theory.  —  Fig.  144  represents 
a  vertical  column  with  ends  free  to  turn  with- 
out friction  about  horizontal  axes  perpen- 
dicular to  the  plane  of  the  paper.  Fig.  144, 
I,  represents  the  actual  column  with  deflec- 
tion somewhat  exaggerated,  and  Fig.  144,  II, 
shows  the  central  axis  of  the  column  with  all 
horizontal  distances  enlarged.  The  X  axis  is 
taken  vertical  and  positive  upward.  (In  all 
beam  and  column  theory,  the  X  axis  is  taken 
parallel  to  the  original  length.)  The  Y  axis 
is  horizontal  and  positive  toward  the  left. 
We  might  take  the  column  horizontal  and 
neglect  the  bending  moment  due  to  its  weight 

with  the  same  result.     The  origin  of  coordi-  Fig'   *44'T.Column 

Deflection. 
nates  is  at  the  lower  end  at  the  center  of  the 

section.  At  a  section  at  a  distance  x  from  the  origin,  the 
moment  arm  with  respect  to  the  center  of  gravity  of  the  section 
is  y  +  e, 

and  the  moment  is 


Before  writing  the  differential  equation  we  must  determine 
which  sign  to  use  with  the  moment  expression.  In  Fig.  144,  y  is 
positive  throughout  the  entire  length  of  the  column.  The  first 

dv 
derivative,-^,  is  positive  at  the  beginning  and  negative  at  the 

d2y 
end;  consequently  the  second  derivative,  -r^,  is   negative.     In 

the  same  way,  if  the  eccentricity  were  on  the  other  side  of  the 
column  so  that  the  deflection  would  come  on  the  right  (negative), 


182  STRENGTH  OF  MATERIALS 

dzy 

-j-^  would  be  always  positive.     The  second  derivative  has  the 

negative  sign  when  y  and  e  are  positive  and  the  positive  sign 
when  y  and  e  are  negative. 

(The  direction  of  e  is  from  the  line  of  force  to  the  axis  of  the 
column.) 

Since  the  signs  of  the  second  derivative  and  the  moment  arm 
are  opposite,  we  write  the  differential  equation 

£/g  =  -P(2/  +  e).  (1) 

This  equation  may  be  written 

d?y     Py_      eP9  (2] 

dx2  ^  EI~       El 

which  is  a  differential  equation  of  the  second  order  and  first 
degree,  with  the  right-hand  member  constant.  The  student 
familiar  with  Differential  Equations  will  write  it: 


The  solution  of  (3)  is: 

FT  FT 

y  =  Ci  cosy-^yZ  +  C2siny  ^x  -  e.  (4) 

The  .student  who  is  not  familiar  with  Differential  Equations  may 
verify  (4)  by  performing  the  inverse  operations  to  get  (2).  A 
solution  of  a  differential  equation  may  always  be  proved  by 
differentiating  and  eliminating  the  constants  of  integration.  In 
this  case,  since  the  equation  is  of  the  second  order  requiring 
two  integrations,  we  must  differentiate  twice.  Differentiate  (4) 

W  T 
twice  and  multiply  by  -p-  .     The  result  obtained  added  to  the 

original  equation  (4)  gives  equation  (2),  which  proves  the  solu- 
tion. 

To  obtain  the  integration  constants,  the  conditions  are 

y  =  o  when  x  =  0,     and    y  =  0  when  x  =  I. 
From  the  first  condition: 

CiCosO  -  e  =  0; 

Ci  =  e;  (5) 

IT  I  P 

.  y  =  e  cos  y  -^jx  +  C2  sin  y  -^  x  -  e.  (6) 


COLUMNS  183 

Substituting  the  second  condition  in  (6) : 

0  =  e  *»\'-gft+  ^siny-J^Z  -  e;  (7) 


el  — 


sm 


V  =  «(c°s  V  ^jx  +  tan  y/J  1  sh  vjj*  -  l) 


(9) 


Equation  (9)  gives  the  deflection  at  any  point  of  a  column  with 
ends  free  to  turn  but  not  free  to  move  laterally.  It  is  a  sine 

curve.     To  find  the  point  of  maximum  deflection,  differentiate 

I 
and  set  the  first  derivative  equal  to  zero  and  find  that  x  =  - 

is  the  position  required.  We  might  have  assumed,  from  the 
symmetry  of  the  figure,  that  the  maximum  deflection  is  at  the 
middle  and  used  this  instead  of  the  second  condition  in  getting 
the  constant  C* 

To  get  the  maximum  deflection  at  the  middle: 


=  esec 


s^  \J  ^  |  +  **y£-f  |)-  e; 
=  6  (sec  y^  1  -  l)  ;  Formula  XXII.      (10) 


Maximum  moment  =  eP  sec 


/T~  Z 
y  ^j  „" 


P   .  4 

Maximum  stress  -  "T  T  ~y  sec  V  U  2  ' 


184  STRENGTH  OF  MATERIALS 


Maximum  stress  =  -j  (  1  +  —  sec  V 

^A  \         T  * 


-=-=.  - 
Hj  L  Z 


,... 

where  A  is  the  area  of  the  section  and  r  is  the  radius  of  gyration. 


PROBLEMS. 

1.  A  wooden  bar  1  inch  square  and  5  feet  long,  as  a  column  has  the  load 
0.1  inch  from  the  center  of  the  section  on  a  line  through  the  center  parallel  to 
one  side.  If  E  is  1,500,000,  what  is  the  deflection  at  the  middle  of  the  length 
due  to  a  load  of  200  pounds? 


.  V  ETT  o  = 

* 


Ans.       ETT  o  =  1-2  radian. 

£jl     2i 

2/max  =  0.176  inch. 

2.  Find  the  maximum  moment  and  fiber  stress  in  Problem  1. 

Ans.  Maximum  moment,  55.18  inch  pounds; 

Maximum  compressive  stress,  531  pounds  per  square  inch. 

3.  Solve  Problems  1  and  2  for  a  length  of  75  inches. 

Ans.  i/max  =  1.311  inches;  maximum  compressive  stress,  1894  pounds. 

4.  A  2-inch  round  steel  rod  5  feet  long  is  used  as  a  column  with  ends  free 
to  turn.     Find  the  deflection  at  the  middle,  and  the  maximum  fiber  stress  on 
the  concave  side  for  a  load  of  10,000  pounds,  if  the  eccentricity  is  0.1  inch,  and 
E  is  30,000,000. 

Ans.  t/max  =  0.0227  inch;   maximum  stress,  4744  pounds  per  square  inch. 

5.  Solve  Problem  4  if  the  eccentricity  is  0.01  inch;  also  for  0.5  inch. 

6.  Solve  Problem  4  for  loads  of  20,000  pounds,  30,000  pounds,  50,000 
pounds,  60,000  pounds,  and  70,000  pounds  for  eccentricities  of  0.01  inch  and 
0.1  inch. 

Ans.  Load:      20,000     30,000     50,000      60,000     70,000 
Stress:      6,763     10,340     19,295      32,483     Infinite. 
"         10,337     17,500    49,800     154,000 

Vr>      7 
=TT  x  =  90  degrees? 
I'j  I    4 

Ans.  64,570  pounds. 

If  we  observe  Problem  6,  we  find  that  a  load  of  50,000  pounds 
with  an  eccentricity  of  0.1  inch  produces  a  maximum  stress  of 
49,800  pounds  per  square  inch.  If  the  ultimate  strength  of  this 
steel  in  compression  is  50,000  pounds  per  square  inch,  this  is 
practically  the  ultimate  load.  On  the  other  hand,  a  load  of 
60,000  pounds  with  an  eccentricity  of  0.01  inch  produces  a  stress 
of  32,000  pounds,  so  that  with  the  smaller  eccentricity  the  load 
can  go  considerably  above  60,000  pounds.  A  load  of  64,570 


COLUMNS  185 

pounds  will  cause  failure  with  a  column  of  these  dimensions,  no 
matter  how  small  the  eccentricity;  for  this  load  makes  the  angle 

P      I  7T 

^j  ~  equal  to  ~,  an  angle  whose  secant  is  infinity.     A  shorter 

lit!    Z  Z 

column  of  the  same  dimensions  will  carry  a  greater  load.  A 
column  10  feet  long  and  2  inches  in  diameter  will  carry  less  than 
16,200  pounds.  A_long  column  with  the  load  exactly  central, 

*     /  P    I 

when  the  angle  V  ^j  ^  =  90  degrees,  is  in  a  condition  of  unstable 
»   ciL  A 

equilibrium;  the  least  vibration  will  start  it  to  bend,  and  it  will 
continue  to  bend  without  increase  of  load  till  it  fails. 

The  formulas  of  this  article  are  calculated  on  the  assumption 
that  E  is  constant.  This  is  the  case  below  the  true  elastic  limit 
only.  Beyond  this  limit  the  change  in  direction  of  the  stress- 
strain  diagram  is  practically  the  same  as  a  reduction  of  the 
value  of  E.  This  reduction  of  E  occurs  in  the  outer  fibers,  which 
are  subjected  to  the  greatest  stress,  so  that  it  causes  some  change 
in  the  location  of  the  neutral  axis  and  modifies  the  eccentricity. 
These  formulas  are,  therefore,  strictly  correct  only  to  the  true 
elastic  limit. 

Within  the  true  elastic  limit  these  formulas  are  theoretically 
and  experimentally  correct.  When  the  dimensions  of  the  col- 
umn are  given  and  the  eccentricity  is  known,  equation  (13)  gives 
the  unit  stress.  This  equation  may  be  used  to  determine  whether 
a  given  column  will  carry  a  given  load  with  safety. 

1 08.  Application  of  Column  Formulas.  —  When  it  comes  to 
computing  the  total  load  P  which  a  given  column  will  carry  with 
a  certain  allowable  unit  stress,  or  to  computing  the  size  of  column 
for  a  given  load,  these  equations  are  not  convenient,  since  neither 
of  these  quantities  is  expressed  explicitly.  Such  problems  may 
be  solved  by  the  method  of  trial.  Choose  some  column  and- 
calculate  the  unit  stress  by  equation  (13)  of  Article  107;  if  it 
comes  too  high,  choose  a  larger  column  until  you  get  one  for 
which  the  unit  stress  is  equal  to  the  allowable  stress,  or  a  little 
under  this  figure. 

Where  a  number  of  such  problems  are  to  be  solved,  it  is  a 
great  saving  of  time  to  represent  equation  (14)  by  means  of  a 

P          I 

curve  or  table.      To  determine  the  relative  values  of  -r  and  - 

.rL  T 

which  make  the  unit  stress  equal  to  the  ultimate  strength  of  the 


186  STRENGTH  OF  MATERIALS 

material  for  any  assumed  eccentricity  of  loading,  we  write  the 
second  form  of  equation  (14) : 

ev       .  I~P~    I       su 


where  su  is  the  ultimate  strength  of  the  material.     It  is  difficult 

P  I 

to  solve  for  -j  corresponding  to  a  given  value  of  -,  but  it  is  easy 

I  P 

to  solve  for  -  corresponding  to  a  given  -r-     The  table  below  gives 

T  A. 

most  of  the  work  for  steel,  for  which  E  =  29,000,000,  su  =  48,000, 
and  the  eccentricity  is  such  that 


This  table  is  computed  by  means  of  logarithms  : 
Radians  =  degrees  X  7^7:  ; 

loU 


the  log  of  V29/KXJ  =  0.77411. 

P 

When  -j  is  8000,  the  angle  is  87  degrees  42  minutes  =87.7  degrees; 

A. 

Log  87.7  1.94300 

.77411 


2.71711 
Log  VS  .45154 

2.26557 
j  =  184.3. 
PROBLEM. 

7         p 

Calculate  -  for  T  =  9000, 11,000, 14,000,  and  18,000  pounds  per  square  inch. 

T         A 


COLUMNS 


187 


TABLE  VII. 

ULTIMATE  UNIT  LOADS  ON  A  COLUMN  WITH  ROUND  ENDS. 

Calculated  from:  ultimate  strength,  48,000  pounds  per  square  inch;  E, 
29,000,000;  ^  equals  0.2. 

p 

-r  is  ultimate  unit  load  in  pounds  per  square  inch. 

A. 

-  is  the  ratio  of  the  length  of  the  column  to  its  radius  of  gyration. 


P 

A  ' 

"•'"Visfr 

-V££- 

& 

/ 

72r' 

l 
r 

Pounds. 

Degrees. 

Radians. 

1,000 

47 

235 

89°  46' 

1.566 

533 

2,000 

23 

115 

89  30 

1.562 

376 

3,000 

15 

75 

89  14 

1.557 

306 

4,000 

11 

55 

88  58 

1.552 

264 

5,000 

8.6 

43 

88  40 

1.547 

236 

6,000 

7.0 

35 

88  22 

.542 

214 

8,000 

5.0 

25 

87  42 

.531 

184 

10,000 

3.8 

19 

86  59 

.518 

164 

12,000 

3.0 

15 

86  11 

.504 

148 

16,000 

2.0 

10 

84  17 

.471 

125 

20,000 

1.4 

7.0 

81  47 

1.427 

108.8 

24,000 

1.0 

5.0 

78  41 

1.373 

94.4 

26,000 

.8462 

4.231 

76  20 

.332 

88.8 

28,000 

.7143 

3.571 

73  44 

.387 

82.8 

30,000 

.6000 

3.000 

70  32 

.231 

76.6 

32,000 

.5000 

2.500 

66  25 

.159 

69.8 

34,000 

.4119 

2.059 

60  57 

.064 

62.2 

36,000 

.3333 

1.667 

53  07 

.927 

51.4 

37,000 

.2973 

1.486 

47  43 

.833 

46.6 

38,000 

.2632 

1.316 

40  32 

.707 

39.0 

38,500 

.2468 

1.234 

35  51 

.626 

34.4 

39,000 

.2308 

1.154 

29  55 

.522 

28.6 

39,200 

-.2244 

1.122 

27  01 

.471 

25.6 

39,400 

.2182 

1.091 

23  36 

.412 

22.4 

39,600 

.2121 

1.061 

19  27 

.339 

18.4 

39,700 

.2090 

1.045 

16  57 

.296 

16.0 

39,800 

.2060 

1.030 

13  53 

.242 

13.0 

39,900 

.2030 

1.015 

9  52 

.172 

9.2 

40,000 

.2000 

1.000 

0 

.0 

0 

188 


.STRENGTH  OF  MATERIALS 


Fig.  145  shows  the  values  of  the  unit  load,  -j,  which  make  the 

unit  stress  in  the  outer  fibers  on  the  concave  side  of  a  steel  col- 
umn 48,000  pounds  per  square  inch  (provided  the  elastic  limit  is 
48,QOO  pounds,  as  in  the  case  of  cold-  rolled  steel,  for  instance). 
The  curves  are  calculated  for  a  modulus  of  29,000,000.  Curve  I 

0  2 


is  plotted  from  Table  VII  where  e  = 


Curve  II  is  plotted 


ev 


for  an   eccentricity  one-half  as  great,  —  =  0.1,  column   III  of 

Table  VIII.     Curve  III  is  plotted  from  zero  eccentricity,  the 
second  column  of  Table  VIII. 


20  40  60  80  100  120  140  160  180  200  220  240  260  280  300  320  340  360  380  400 

LENGTH-M_EAST  RADIUS  OF  GYRATION 


Fig.  145.  —  Column  Curves. 


We  notice  that  the  eccentricity  makes  a  large  difference  for 
values  of  -  less  than  100.  For  values  greater  than  160  the 
eccentricity  makes  little  difference. 

We  will  now  apply  one  of  these  curves  to  the  solution  of  a  problem.  Sup- 
pose we  want  a  solid  circular  column  80  inches  long  with  round  ends  to  carry 
a  load  of  40,000  pounds  with  a  factor  of  safety  of  4;  and  suppose  that  we. 
are  sure  that  the  eccentricity  of  the  load  is  not  greater  than  one-twentieth  of 
the  radius. 

If  a  is  the  radius, 

,=a;r=!;e=|j;2«o.2. 

Curve  I  may  be  used.  We  will  find  the  column  for  which  the  ultimate 
load  ig  160,000  pounds. 


COLUMNS 


189 


A  short  block  with  this  eccentricity  will  require  an  area  of  4  square  inches. 
The  diameter  must  be  at  least  2.2  inches  (use  table  in  Cambria  to  find  diameters 

from  areas);-  is  less  than  150  [80 -r-  0.55  =  145].     If  we  look  on  curve  I,  we 

p 

find  the  corresponding  value  of  -j  is  about  12,000  pounds  per  square  inch,  /e- 

quiring  a  section  of  about  13  square  inches.  The  area  required  is  between 
4  square  inches  and  13  square  inches.  As  the  next  step  in  this  trial  method, 
take  an  area  which  is  about  midway  between  4  and  13.  This  mean  area  has  a 

diameter  of  about  3.2  inches,  which  we  will  take  because  it  gives  -  equal  to  100. 
The  work  may  be  arranged  as  shown  below. 


Diameter  in 
inches. 

Radius  of  gy- 
ration. 

l 

r 

—  from  curve. 

Area. 

Total  load. 

Square  inches. 

Pounds. 

3.2 

.80 

100 

22,000 

8.04 

176,900 

3.0 

.75 

107 

20,500 

7.07 

145,000 

(Interpolate  for  the  diameter  which  gives  160,000.     Result,  3.09.) 

3.09 

.772 

104 

21,000 

7.50 

157,500 

3.10 

.775 

103 

21,200 

7.55 

160,060 

Note  that  this  calculation  gives  the  column  in  which  the  total  load  is  one- 
fourth  the  ultimate  load.  If  we  take  a  column  3.1  inches  in  diameter  and 
compute  the  maximum  stress  when  the  total  load  is  40,000  pounds,  we  find  it 
to  be  about  one-seventh  of  48,000  pounds  per  square  inch.  The  difference  is 
due  to  the  fact  that  the  secant  is  not  proportional  to  the  angle. 


PROBLEMS. 

8.  Using  curve  I,  find  the  diameter  of  a  steel  column  50  inches  long  to 
carry  a  load  of  10,000  pounds  with  a  factor  of  safety  of  3. 

9.  Using  curve  I,  find  the  Z-bar  column  15  feet  long  to  carry  a  load  of 
150,000  pounds  with  a  safety  factor  of  4. 

Ans.   A  column  made  of  four  5-inch  by  ^-inch  Z-bars  with  one  7-inch  by 
plate. 


l~p  I 
109.   Euler's  Formula.  —  As  any  column  will  fail  when  y  -^.^ 

becomes  90  degrees,  we  have,  as  the  upper  limit  of  possible  loading, 
the  condition: 


EI2      2' 


from  which 


P  = 
P 


Formula  XXIII 


190  STRENGTH  OF  MATERIALS 

Formula  XXIII  is  Euler's  formula.  It  may  be  derived  directly 
from  equation  (1)  of  Article  107  for  the  case  where  the  eccen- 
tricity is  0. 


Multiply  by  dy, 

(2) 


dzy  j        dy  d2y          ,      ,  dy 

-j^>  dy  =  •/•  -j2-  =  z  dz  where  z  =  -/•  • 
cfo2  dx  dx  da; 

E7T 

-p-zdz  =-ydy.  (3) 


(4) 

(6) 
(7) 


y  =  0  when  x  =  0,  hence  C2  =  0  (or  TITT) 
Using  ..         C2  =  0, 


El 
When  x  =  I,  y  =  0, 


jl  =  TT  (or  mr) ; 
p  =  <jr*EI          Formula  XXIII. 

From  equation  (8)  we  see  that  the  curve  of  the  column  is  a  sine 
curve.  Formula  XXIII  contains  7  but  does  not  include  the 
distance  to  the  outer  fiber.  From  this  we  conclude  that  when 


COLUMNS  191 

the  eccentricity  is  indefinitely  small,  and  the  length  compared' 
with  the  radius  of  gyration  is  sufficiently  great  for  the  column  to 
fail  by  bending,  the  value  of  the  ultimate  load  does  not  depend 
upon  the  form  of  the  column  except  in  so  far  as  the  form  changes 
the  value  of  I. 

Curve  III  of  Fig.  145  is  Euler's  curve  for  a  modulus  of  elasticity 
of  29,000,000.  As  a  mathematical  curve  it  is  of  infinite  length. 
As  an  engineering  curve  it  must  not  be  used  above  the  point  B, 
where  the  unit  load  is  the  ultimate  compressive  strength  of  the 
material. 

Since  Euler's  curve  is  derived  on  the  assumption  of  a  constant 
modulus  of  elasticity,  it  is  really  good  to  the  true  elastic  limit  only. 
For  ordinary  structural  steel,  the  true  elastic  limit  is  about 
30,000  pounds  per  square  inch.  Curve  III  of  Fig.  145  is  correct 

below  the  point  C  for  values  of  -  greater  than  98.     From  C  to  B 

it  gives  too  great  values  for  the  unit  load.  Above  B,  if  48,000 
is  the  ultimate  compressive  strength,  it  is  a  mathematical  fiction, 
"  out  of  bounds." 

We  will  see  later  that  it  is  better  to  use  Euler's  curve  only  for 

values  of  -  which  make  the  ultimate  load  one-third  of  the  ultimate 
r 

strength. 

PROBLEMS. 

1.  A  yard  stick,  with  the  ends  slightly  rounded,  was  placed  vertical  with 
the  lower  end  on  a  platform  scale  and  a  load  was  applied  to  the  upper  end 
(Fig.  143).  The  load  and  deflection  were  measured. 


Load  in  pounds. 

5.00  ............................  0.03 

6.00  ............................  0.20 

6.40  ............................  0.25 

6.48  ...........................  1.00  (Load  dropped  to  6.28.) 

6.28  ............................  2.50 

Calculate  El  from  the  last  two  readings  by  Euler's  formula.          Ans.  851,  825. 

2.  The  yard  stick  of  Problem  1,  supported  as  a  beam  at  points  34  inches 
apart,  was  deflected  f  $  inch  at  the  middle  by  a  load  of  1  pound  at  the  middle. 
Find  El  and  compare  the  result  with  Problem  1. 

3.  The  yard  stick  above  mentioned  was  1.06  inch  wide  and  0.18  inch  thick. 

Find  E  and  -  • 

4.  Find  the  total  load  with  a  factor  of  safety  of  4  on  a  round  steel  rod 
2  inches  in  diameter  for  lengths  of  20,  40,  60,  80,  and  100  inches,  if  the  ultimate 


192 


STRENGTH  OF  MATERIALS 


^compressive  strength  of  the  steel  is  50,000  pounds  per  square  inch  and  E  is 
30,000,000. 


Ans. 


f  Length. 
20  inches. 
40      "     .. 


Total  safe  load. 


16,148 
9,090 
5,813 


Ultimate  unit  load. 

185,060 

46,260 

60      "    20,560 

80      "    11,565 

100      "     .V  7,400 

Why  not  use  the  results  for  the  first  two  lengths? 

5.  Find  the  total  safe  load,  with  a  factor  of  safety  of  4,  on  a  4-inch  by 
4-inch  wooden  post,  12  feet  long,  if  E  is  1,500,000,  and  the  ultimate  strength 
is  5000  pounds  per  square  inch.  Ans.  3808  pounds. 

6.  Should  Euler's  formula  be  applied  to  the  post  of  Problem  5  if  the  length 
is  6  feet?  4  feet?    Why? 

TABLE  VIII. 

ULTIMATE  UNIT  LOADS  ON  A  COLUMN  WITH   ROUND  ENDS. 
Calculated  by  Euler's  formula  and  by  equation  (14)  of  Article  107  for 


three  values  of  su,  with  e~  =  0.1,  and  E 


29,000,000. 


Unit  load, 
P 
A' 

-  ,  Length  divided  by  least  radius  of  gyration. 

Euler's. 

stt  =48,000. 

stt  =40,000. 

SM  =32,000. 

Pounds. 

1,000 

535 

534 

534 

534 

2,000 

378 

377 

377 

377 

3,000 

309 

307 

307 

307 

4,000 

268 

266 

266 

266 

5,000 

239 

237 

237 

237 

6,000 

218 

216 

216 

216 

8,000 

189 

186 

186 

186 

10,000 

169 

166 

166 

166 

12,000 

154 

151 

150 

150 

14,000 

143 

139 

138 

138 

16,000 

134 

129 

128 

128 

18,000 

126 

121 

120 

116 

20,000 

120 

114 

112 

107 

22,000 

114 

108 

105 

98 

24,000 

109 

102 

99 

88 

26,000 

105 

97 

92 

77     .  . 

28,000 

101 

92 

86 

51 

29,000 

16 

30,000 

98 

87 

79 

32,000 

95 

82 

70 

34,000 

92 

77 

57 

36,000 

89 

72 

26 

38,000 

87 

65 

40,000 

85 

56 

42,000 

82 

42 

COLUMNS  193 

no.  Effect  of  Ultimate  Strength  of  Material  on  Strength  of 
Columns.  —  We  notice  from  Euler's  formula  that  the  ultimate 
load  on  a  long  column  depends  upon  the  modulus  of  elasticity 
and  the  moment  of  inertia  of  the  cross  section  and  is  independent 
of  the  ultimate  strength  of  the  material.  The  ultimate  strength 

of  the  material,  however,  determines  the  lower  limit  of  - ,  for  which 

Euler's  formula  may  be  used. 

Euler's  formula  assumes  zero  eccentricity.  From  the  curves 
of  Fig.  145  we  see  that  the  eccentricity  makes  little  difference  if 

-  is  greater  than  160.     The  effect  of  the  ultimate  strength  and 

small  eccentricity  together  is  shown  in  Table  VIII.     This  table, 
calculated  like  Table  VII,  gives  the  values  of  the  unit  load  for 

0.1  r2 
the  eccentricity  e  =  -  —  for  three  values  of  the  ultimate  strength. 

The  table  also  gives  the  results  of  Euler's  formula.     We  notice 
that  there  is  little  difference  when  -  is  140.     When  -  becomes  less 

than  100  the  difference  is  great. 

in.  Classification  of  Columns.  —  Columns  may  be  divided, 
according  to  the  nature  of  the  ends,  into  the  following  classes: 

I.  Both  ends  free  to  turn  about  horizontal  axes  but  not  free 
to  move  laterally,  Figs.  144  and  146,  I. 

II.  One  end  fixed  and  the  other  end  free  to  turn  and  free  to 
move  laterally,  Fig.  146,  II. 

III.  Both  ends  fixed  so  that  the  tangents  at  the  ends  do  not 
change,  Fig.  146,  III, 

IV.  One   end  fixed   and  the  other  end  free  to  turn  about 
one  or  more  horizontal  axes,  but  not  free  to   move  laterally, 
Fig.  146,  IV. 

Case  I  is  the  only  one  so  far  considered.  If  L  is  the  total 
length  of  the  column,  and  I  is  the  length  of  the  sine  curve  ABC 
as  used  in  the  theory  of  Articles  107  and  109,  I  =  L  for  case  I. 

In  case  II,  the  entire  column  of  length  L  corresponds  to  the 
upper  half  AB  of  the  sine  curve.  Hence  for  case  II  we  use 
2  L  for  I  in  Formulas  XXII  and  XXIII. 

In  case  III,  -^  is  zero  at  each  end  and  at  the  middle.     The 
'  dx 

middle  half  ABC  corresponds  to  the  sine  curve  of  case  I.     This 


194 


STRENGTH  OF  MATERIALS 


portion  of  the  sine  curve  is  represented  by  I  in  the  formulas. 
If  L  is  the  entire  length  DF,  then  I  =  -~  -     A  column  with  both 

ends  rigidly  fixed  will  carry  as  great  a  load  as  a  column  of  half 
its  length  with  ends  free  to  turn, 


Fig.  146.  —  Types  of  Ideal  Columns. 

The  points  A  and  C,  at  one-fourth  the  length  from  the  ends, 
are  points  of  counterflexure.  The  portion  AD  is  one-half 
of  a  sine  curve.  If  revolved  180  degrees  in  the  plane  of  the 
paper  about  the  point  A,  the  curve  AD  will  coincide  with  A B. 
The  moment  is  zero  at  A  and  C.  Case  IV  is  fixed  at  one  end 
and  free  to  turn  at  the  other,  but  not  free  to  move  laterally. 
The  point  of  counterflexure  is  at  C.  As  there  is  no  moment  at 
C,  the  resultant  force  at  A  must  be  in  the  direction  AC.  The 
portion  ABC  forms  a  sine  curve  similar  to  the  preceding  cases 
with  the  line  AC  ^corresponding  to  the  X  axis  in  Fig.  144.  The 
lower  portion  CF  forms  a  part  of  a  sine  curve  as  far  as  the  plane 
of  the  body  which  holds  it.  Below  that  plane  it  is  straight. 
The  portion  CG  is  less  than  one-half  of  AC.  It  is  evident,  there- 
fore, that  AC  is  more  than  two-thirds  of  L.  The  solution  of  the 
differential  equation  shows  that  A  C  is  nearly  0.7  L. 
1  =  0.7  L  12  =  0.5  L2  nearly. 


COLUMNS  195 

In  Fig.  146,  V,  the  top  of  the  column  has  been  displaced 
laterally.  If  this  displacement  is  such  that  the  point  B  is  as 
far  from  the  line  AC  as  the  top  A  is  displaced  from  the  vertical 
line  through  F,  then  the  line  AC  from  the  end  to  the  point  of 
counterflexure  becomes  vertical.  In  this  position  AC  is  two- 
thirds  of  L,  there  is  no  horizontal  force  at  the  top,  and  the 
vertical  force  P  is  greater  than  in  Fig.  146,  IV.  The  position  is 
unstable  and  easily  changes  to  the  one  in  which  the  curves  are 
reversed,  with  C  and  B  deflected  to  the  right  of  the  vertical  line 
through  F,  in  which  position  the  load  P,  which  produces  a  large 
deflection,  is  less  than  in  case  IV. 

PROBLEMS. 

1.  A  yard  stick,  with  ends  rounded,  was  deflected  a  large  amount  by  a 
load  of  6.1  pounds  at  the  end,  as  in  case  I.     Find  El  by  Euler's  formula. 

2.  The  same  yard  stick  was  clamped  4  inches  from  one  end  and  the  load 
was  applied  as  in  Fig.  146,  IV.    It  was  found  that  a  load  of  15.42  pounds  pro- 
duced a  deflection  of  over  1.5  inches.     Find  El  by  Euler's  formula. 

3.  The  load  in  Problem  2  was  displaced  1  inch  south  of  the  vertical  line 
through  the  bottom.     The  vertical  component  of  this  load  was  17.12  pounds 
with  a  deflection  of  2  inches  south.     The  horizontal  component  was  found  to 
be  zero.     Find  El,  using  two-thirds  of  32  inches  as  I. 

4.  A  solid  circular  steel  rod  stands  in  a  vertical  position  with  the  lower 
end  fixed.     A  load  of  100,000  pounds  is  applied  at  the  free  upper  end  at  a 
distance  of  1  inch  from  the  center.      The  diameter  of  the  rod  is  6  inches, 
and  its  length  from  the  fixed  point  is  15  feet.     If  E  is  30,000,000,  find  the 
deflection  at  the  end  and  the  maximum  fiber  stress  by  the  formulas  of  Article 
107.  Ans.  Maximum  stress,  21,320  pounds  per  square  inch. 

ii2.  End  Conditions  in  Actual  Columns.  —  The  classification 
of  Article  111  represents  ideal  conditions,  which  are  only  approx- 
imated in  practice.  The  columns  in  actual  use  are: 

Round-end  columns,  which  end  with  spherical  or  cylindrical 
surfaces.  They  sometimes  end  with  knife-edges,  which  may  be 
regarded  as  cylinders  of  small  radius.  The  round  surfaces  roll 
on  plane  surfaces  with  practically  no  friction.  Round-end  columns 
are  not  used  in  structures  and  are  rarely  used  in  machines. 
They  are  used  in  tests  to  check  the  accuracy  of  theory,  as  they 
fulfill  very  closely  the  conditions  of  case  I  of  ends  free  to  turn. 

A  pin-end  or  hinged-end  column  ends  with  cylindrical  surfaces 
which  turn  in  cylindrical  bearings  (Fig.  147).  Fig.  148  shows  one 
end  of  a  pin-connected  column  made  of  two  channels  latticed 
together.  This  form  of  connection  is  commonly  used  in  bridges. 


196 


STRENGTH  OF  MATERIALS 


A  column  which  ends  with  a  ball  and  socket  is  practically  the 
same  as  a  hinged-end  column,  except  that  it  is  free  to  turn  in 
any  plane  instead  of  in  the  single  plane  normal  to  the  axis  of 
the  hinge. 


© 

4 

4 

Q, 

Fig.  : 
en 

f 

L47.— 
dColu 

A 

mn 

f 

Pin- 

Fig.  148.  —  End  of  Pin-connected  Bridge  Post. 


If  the  pin  of  a  hinged-end  column  rolled  on  a  plane  surface, 
there  would  be  little  friction,  and  the  case  would  be  the  same  as 
that  of  the  round-end  columns.  Usually  the  pin  turns  in  a  close- 
fitting  bearing,  so  that  the  friction  is  considerable.  A  hinged- 
end  column  may  be  anywhere  between  case  I,  with  the  ends  free 
to  turn,  and  case  III,  with  the  ends  fixed.  If  the  pin  is  small, 
the  moment  arm  of  the  friction  is  small,  and  a  slight  eccentricity 
will  cause  it  to  turn.  If  the  pin  is  large,  the  opposite  is  true. 
In  the  moving  parts  of  machinery,  the  pin  connections  are  lubri- 
cated so  that  they  turn  easily.  The  connecting  rod  of  an  engine 
is  an  example. 

Fig.  149  shows,  diagrammatically,  the  behavior  of  a  pin-end 
column.  At  first  it  acts  as  a  column  with  fixed  ends  (Fig.  149, 1). 
When  the  moment  at  the  end  becomes  greater  than  the  product 
of  the  starting  friction  at  the  surface  of  the  pin  multiplied  by 
its  radius,  the  column  turns  at  the  end  to  some  position  similar 
to  Fig.  149,  II.  In  this  position,  the  points  of  counterflexure  A 


COLUMNS 


197 


i         ii       in 

Fig.  149.— Deflection  of 
Hinged-end  Column. 


and  C  are  nearer  the  ends,  and  the  moment  on  the  pins  is  less. 
The  column  may  finally  change  to  the  position  of  Fig.  149,  III, 
which  is  that  of  case  I.  In  this  last  position,  it  will  support  a 
load  much  smaller  than  in  the  first  position.  If  the  ratio  of  the 
length  to  least  radius  of  gyration  is  200  or  more,  so  that  Euler's 
formula  applies  to  both  the  whole  and  the  half  length,  the 
column  in  the  last  position  will  carry  a  load 
only  one-fourth  as  great  as  in  the  first  one. 

Some  interesting  tests  of  columns  were 
made  at  the  Pencoyd  Iron  Works  in  1883 
by  James  Christie.*  In  these  tests,  some 
of  the  so-called  hinged-end  columns  were 
fitted  with  hemispherical  balls  turning  in 
sockets.  The  balls  were  located  as  nearly 
as  possible  in  the  line  of  the  axis  of  the 
column  by  careful  measurements.  Owing 
to  the  fact  that  no  column  is  absolutely 
straight  and  perfectly  uniform  in  section 
and  homogeneous  in  structure  throughout 
its  entire  length,  this  method  did  not  always 
put  the  centers  of  the  hemispheres  exactly 
on  the  axis  of  the  column.  The  final  adjustment  was  deter- 
mined by  trial  in  the  testing  machine;  a  small  load  was  ap- 
plied and  the  deflection  measured.  The  hemispheres  were  then 
moved  a  little  and  the  test  repeated,  until  a  position  was  found 
where  a  considerable  load  caused  no  appreciable  deflection.  The 
column  was  then  loaded  to  failure. 

t  "  When  the  point  of  greatest  strength  was  reached,  the 
behavior  of  the  specimen  was  peculiar.  Under  ordinary  cir- 
cumstances the  bar,  while  bending  under  strain,  rotated  from  the 
start  on  its  hinged  ends.  When  correctly  centered,  no  such 
rotation  occurred  at  the  beginning  of  the  deflection,  but  the  bar 
bent  like  a  flat-ended  strut,  till  the  point  of  failure  was  reached, 
when  it  rotated  on  its  ends  suddenly,  as  sometimes  to  spring 
from  the  machine."  These  results  could  not  be  secured  when  the 
balls  or  pins  rolled  on  plane  surfaces,  and  were  difficult  to  get 
when  the  pins  were  small. 

The  effect  of  the  size  of  the  pin  was  shown  in  these  experi- 

*  Transactions  of  the  American  Society  of  Civil  Engineers,  1883,  pages 
85-122. 

f  Ibid.,  page  87. 


198 


STRENGTH  OF  MATERIALS 


ments.  Two  angles  of  the  same  length  were  cut  from  the 
same  bar.  One  of  these  tested  with  a  2-inch  ball  and  socket 
failed  at  36,500  pounds  per  square  inch;  the  other  tested  with  a 
1-inch  ball  and  socket  failed  at  24,010  pounds.  Similar  results 
were  obtained  in  other  experiments. 

These  tests  and  many  others  show  that  the  friction  at  the 
ends  of  a  hinged-end  column  partially  fixes  the  ends  and  increases 
the  ultimate  strength.  It  must  be  remem- 
bered, however,  that  in  the  testing  machine 
the  loads  are  applied  with  little  vibration. 
In  structures  such  as  railway  bridges, 
where  there  is  large  vibration,  it  is  probable 
that  the  friction  of  the  pins  gives  little  help, 
and  it  is  safest  to  regard  hinged-end  columns 
as  equivalent  to  round-end  columns. 

Square-end  or  flat-end  columns  end  with 
plane  surfaces  in  contact  with  plane  surfaces. 
They  must  be  accurately  fitted  if  eccentric 
loading  is  to  be  avoided.  If  a  beam  rest- 
ing on  a  square-end  column  bends  under  its 
load  (Fig.  150,  II),  the  load  on  the  column 
Footings  which  support  columns  often  settle 


//  ///#//////////, 
i  ii 

Fig.  150.  —  Square- 
end  Columns. 


becomes  eccentric. 

unevenly  and  cause  large  eccentricity. 

Pin-end  columns  are  practically  square-end  with  respect  to 
the  axis  of  the  pin. 

A  column  with  a  pin  connection  at  one  end  and  a  square 
connection  at  the  other  is  called  a  pin-smd-square  column.  This 
term  includes  columns  with  one  end  fixed  and  the  other  hinged. 
This  column  approximates  the  conditions  of  case  IV., 

Fixed-end  columns  are  riveted  to  the  remainder  of  the  struc- 
ture in  buildings  and  bridges.  In  machines  they  are  fastened 
in  various  ways.  The  connection  can  never  be  absolutely  rigid, 
and  the  member  to  which  the  column  is  fixed  must  suffer  some 
deflection,  so  that  there  is  always  some  change  in  the  slope  of 
the  tangent  at  the  "  fixed  "  points.  When  the  column  is  very 
flexible  compared  with  the  body  to  which  it  is  fixed,  it  may  then 
be  regarded  as  an  example  of  the  ideal  case  and  I  may  be  taken 
as  equal  to  half  the  length  L.  In  the  case  of  the  yard  stick 
described  in  Problem  2  of  Article  111,  the  column  was  firmly 

L 
clamped  to  the  2-inch  by  4-inch  post  and  the  value  of  --  was 


COLUMNS  199 

over  800,  making  it  relatively  very  flexible,  so  that  this  gave 
consistent  results  when  treated  as  an  example  of  a  column 
fixed  at  one  end.  In  machines  this  condition  is  sometimes  met, 
but  it  never  occurs  in  structures. 

113.  Some  Experiments  Showing  Effect  of  End  Conditions.  — 
It  is  evident  that  the  value  of  I  to  be  used  with  "  square-  "  and 
"  fixed-  "  end  columns  in  calculating  the  unit  load  is  greater  than 

-~  and  less  than  L.     In  the  case  of  the  "  pin-and-square  "  column, 

it  is  less  than  L  and  (if  the  friction  of  the  pin  is  small)  greater 
than  0.7  L.  The  best  values  to  be  used  should  be  determined 
from  tests  of  full-size  columns  under  a  wide  range  of  conditions. 

The  experiments  of  Christie,  previously  mentioned,  are  in- 
structive in  this  regard.  In  Table  IX  are  the  results  of  these 
experiments  for  angle  and  tee  sections.* 

In  this  table  L  is  the  total  length  of  the  column  as  in  Fig.  146, 
and  r  is  the  least  radius  of  gyration.  It  was  found  that  the 
columns  failed  in  the  direction  for  which  radius  of  gyration  was 
the  minimum. 

The  figures  in  this  table  give  us  some  idea  of  the  relative  value 
of  hinged,  flat,  and  fixed  ends,  as  compared  with  round  ends. 
In  the  case  of  the  hinged  ends,  owing  to  the  lack  of  vibration, 
the  load  was  probably  greater  than  would  be  found  under  the 
conditions  of  railway  bridges  subjected  to  the  jar  of  fast  trains. 
The  fixed  ends  were  clamped  to  the  testing  machine,  which  was 
relatively  rigid.  Notice  that  with  the  short  lengths,  where  the 
columns  were  relatively  stiff,  the  fixed-end  columns  were  inferior 
to  the  flat  ends  and  only  a  little  better  than  the. hinged  ends. 

With  —  greater  than  100  the  superiority  of  the  fixed  ends  becomes 

marked. 

p 
If  we  consider  the  table,  we  find  that  with  -r  equal  to  25,000 

A. 

pounds  per  square  inch,  —  is  80  for  round  ends.     For  flat  ends 

P  L 

this  value  of  -r  lies  between  the  values  120  and  140  for—.      If 
A  r 

*  These  figures  are  averages  of  the  results  for  angles  and  tees  from  the 
table  on  page  116  of  the  Transactions  of  the  American  Society  of  Civil  Engi- 
neers for  1883.  The  results  for  channels  and  I-beams  are  not  included  in 
these  averages,  as  these  were  used  with  the  flat-end  condition  only. 


200 


STRENGTH  OF  MATERIALS 


we  interpolate  between  26,500  and  23,250  we  get  -  =  129  for 

p 

-r  =  25,000.     As  far  as  this  experiment  goes,  it  indicates  that 

A 

in  calculating  a  flat-end  column  the  value  of  I  in  the  formulas 

orv    T 

should  be  taken  as  -j^  =  0.62  L.     In  the  same  way  for  fixed 
.  LZ\J 

T  p 

ends  we  find  that  —  =  144  gives  -j  equal  to  25,000.     This  makes 
r  ,A 

I  =  0.56  L  for  this  particular  case. 


TABLE  IX. 

PENCOYD  TESTS  OF  WROUGHT-IRON  STRUTS. 

Average  Results  for  Angles  and  Tees. 


L 

r 

P 

—  ,  Ultimate  unit  load  in  pounds  per  square  inch. 
A. 

Round  ends. 

Hinged  ends. 

Flat  ends. 

Fixed  ends. 

20 

44,000 

46,000 

49,000 

45,000 

40 

36,500 

40,500 

41,000 

38,000 

60 

30,500 

36,000 

36,500 

34,000 

80 

25,000 

31,500 

33,500 

32,000 

100 

20,500 

28,000 

30,250 

30,000 

120 

16,500 

24,250 

26,500 

28,000 

140 

12,800 

20,250 

23,250 

25,500 

160 

9,500 

16,350 

20,500 

23,000 

180 

7,500 

12,750 

18,000 

20,000 

200 

6,000 

10,750 

15,250 

17,500 

220 

5,000 

8,750 

13,000 

15,000 

240 

4,300 

7,500 

11,500 

13,000 

.260 

3,800 

6,500 

10,250 

11,000 

280 

3,200 

5,750 

8,750 

10,000 

300 

2,800 

5,000 

7,350 

9,000 

320 

2,500 

4,500 

5,750 

8,000 

340 

2,100 

4,000 

4,650 

7,000 

360 

1,900 

3,500 

3,900 

6,500 

380 

1,700 

3,000 

3,350 

5,800 

400 

1,500 

2,500 

2,950 

5,200 

420 

1,300 

2,250 

2,500 

4,800 

440 

2,100 

2,200 

4,300 

460 

1,900 

2,000 

3,800 

480 

1,700 

1,900 

COLUMNS  201 

PROBLEMS. 

1.  Using  —  equals  60  for  round  ends,  find  the  equivalent  lengths  of  hinged, 

flat,  and  fixed  ends,  and  the  corresponding  values  of  I  in  terms  of  the  entire 
length  L.  Ans.  I  =  0.70  L,  I  =  0.61  L,  I  =  0.63  L. 

2.  Using  —  equals  100  for  round-end  columns,  find  the  corresponding  values 
for  hinged,  flat,  and  fixed  ends,  and  values  of  I  in  terms  of  L. 

.      =  138,160,177. 


If  we  take  all  the  values  for  round  ends  from  40  to  200  inclu- 
sive and  determine  the  values  of  —  which  give  the  same  unit  load 
for  the  other  end  conditions,  we  get  the  following  ratios  : 

Hinged.  Flat.  Fixed. 

Minimum  .....................     1.29  1.50  1.25 

Maximum  .....................     1.45  1.69  1.87 

Meanofall  ....................     1.37  1.60  1.72 

In  the  case  of  the  fixed  ends,  only  one  value  was  below  1.50. 

As  far  as  these  figures  go,  they  indicate  that  a  flat-end  column 
16  feet  long,  a  fixed-end  column  17.2  feet  long,  or  a  hinged-end 
column  13.7  feet  long,  will  carry  the  same  total  load  as  a  round- 
end  column  10  feet  long  of  the  same  cross  section. 


CHAPTER  XIII. 

COLUMN  FORMULAS  USED   BY  ENGINEERS. 
114.   Straight-line  Formulas.  —  We  haveifound  that  for  large 
values  of  - ,  Euler's  formula  may  be  used,  and  a  considerable 

eccentricity  makes  little  difference.      For  smaller  values  of  -, 

Euler's  formula  must  not  be  used,  and  a  small  eccentricity 
makes  a  large  difference  with  the  results  of  the  secant  formula 
of  Article  107.  In  structures  there  is  generally  considerable 
uncertainty  in  regard  to  the  amount  of  eccentricity.  This  is 
especially  true,  for  flat-  or  fixed-end  columns.  It  is,  therefore, 
not  worth  while  to  go  through  the  labor  of  calculating  with  these 
formulas  (except  in  the  case  of  a  column  with  a  relatively  large 
known  eccentricity).  Engineers  make  use  of  simpler  working 
formulas.  Of  these,  Rankine's  formula  was  formerly  most  used. 
At  present,  the  straight-line  formulas  are  preferred. 

A  straight-line  column  formula  for  the  ultimate  unit  load 
has  the  form: 

£  =  Su  -  k  l-r  -  Formula  XXIV. 

The  student  will  recognize  this  as  the  equation  of  a  straight 
line  through  the  point  (0,  su)  and  sloping  downwards.  If  we 
draw  a  straight  line  through  the  point  (0,  su)  of  Fig.  145,  and 
tangent  to  Euler's  curve  III,  we  find  that  this  straight  line  does 
not  deviate  far  from  curves  I  and  II.  Except  for  small  values 

of  - ,  a  small  change  in  the  eccentricity  will  cause  the  curve  to 

change  from  one  side  of  this  straight  line  to  the  other.  Such  a 
straight  line,  then,  will  give  a  fair  value  of  the  unit  load  for  the 
uncertain  eccentricities  which  occur  in  practice,  for  all  values  of 

-to  the  left  of  the  point  of  tangency  except  extremely  small  ones. 

Fig.  151  shows  the  method  of  obtaining  the  constant  of  a 
straight-line  formula  graphically.  Curve  I  is  Euler's  curve  for 
steel  with  E  equal  to  30,000,000,  and  su,  50,000.  The  straight  line 

202 


COLUMN  FORMULAS  USED  BY  ENGINEERS     203 


II  is  drawn  from  the  point  sU9  whose  coordinates  are  (0,  50,000) , 
tangent  to  curve  I.  The  coordinates  of  the  point  of  tangency, 
D,  are  133  and  16,700.  To  get  the  equation  of  the  straight  line, 


UNIT  LOAD  IN  POUNDS  PER  SQUARE  INCH 

ll 

Sn 

s 

I  EULER'S  CURVEj-^|^ 

\ 

TT 

I1 

II 

STRAIGHT  LINE 
-£-50,000-250^ 

[  STRAIGHT  LINE  WITH 
FACTOR  OF  SAFETY  4 
£=  12,500-62.5  -i 

FOR  STEEL 

£=30,000,000 
MPRESSIVE  STRENGTH 
000    PER  SO.  INCH 

\ 

% 

BE 

N^ 

if! 

11J 

CO 
50 

T 

\ 

\C 

^ 

^ 

1 

•p-~. 

1 
•^^ 

f'0f 

s 

~***«^ 

*"* 

n 

^ 

f; 

— 

^ 

^ 

**«^ 
H 

^  ^ 

s=i; 

•  • 

•    • 

-r_ 

20    40    60   80  100  120  110  160  180  200  220  240  260  280  300  320  340  360  380  400 
LENGTH-^RADIUS  OF  GYRATION 

Fig.  151.  —  Straight  Line  and  Euler's  Curves. 

we  notice  that  its  ordinate  is  25,000  when  its  abscissa  is  100; 
that  is,  it  drops  25,000  in  100.     The  equation  is  then 


^  =  50,000-250- 

A.  T 


(2) 


Beyond  D  the  straight  line  falls  below  Euler's  curve.     It 
should  not  be  used  for  values  of  -  greater  than  140,  as  the  results 

beyond  this  limit  are  unnecessarily  small. 

The  straight  line  should  be  used  to  the  point  of  tangency  (or  a 

little  farther)  and  Euler's  equation  beyond  that  point. 

p 
Curve  III  is  the  straight  line  for  allowable  values  of  -r  for  a 

factor  of  safety  of  4.     It  stops  at  G  where  -  is  133.     Beyond  that 
point  use  Euler's  and  divide  the  result  by  4. 

PROBLEMS. 

1.  Using  equation  (2),  solve  Problem  4  of  Article  109  for  the  lengths  of 
20  inches,  40  inches,  and  60  inches  with  a  safety  factor  of  4. 

Ans.  Total  safe  loads,  31,400,  23,560,  and  15,700  pounds. 

2.  Find  the  total  load  with  a  factor  of  safety  of  4  on  a  4-inch  by  4-inch  by 
Hnch  angle  with  ends  free  to  turn,  for  lengths  of  5  feet,  10  feet,  and  15  feet. 

Ans.  28,850,  11,730,  and  5210  pounds. 

3.  Plot  Euler's  curve  for  timber  for  E  =  1,500,000.     Draw  straight  line 
if  ultimate  strength  is  5000  pounds  per  square  inch.    Derive  a  working  straight- 
line  equation  with  a  factor  of  safety  of  5. 


204  STRENGTH  OF  MATERIALS 

115.  Algebraic  Derivation  of  the  Straight-line  Formulas.  - 
While  a  straight-line  formula  may  always  be  derived  graphically 
by  plotting  Euler's  curve  and  drawing  the  tangent,  it  may  also 
be  derived  by  the  methods  of  the  Calculus,  being  the  problem 
of  drawing  a  straight  line  through  a  given  point  tangent  to  a 
given  curve.  We  may  write  Euler's  formula: 


p         i 

where  y  =  -j,     x  =  -,     and    a  = 

A.  T 

Our  problem  is  to  draw  a  tangent  to  the  curve  (1)  which  shall  pass 
through  the  point  (0,  SM).     The  equation  of  this  tangent  line  is: 

y  =  --^rx  +  su,  (2) 

where  x\  is  the  abscissa  of  the  point  of  tangency. 

Since  the  straight  line  (2)  passes  through  the  point  of  tangency 
whose  coordinates  are  fe,  2/i)>  these  coordinates  satisfy  the  equa- 
tion of  the  line;  hence 

!h=-|r  +  *,.  (3) 

Also,  since  the  point  of  tangency  is  on  the  curve,  these  coordi- 
nates satisfy  equation  (1);  and 


Combining  (3)  and  (4)  for  the  coordinates  of  this  point  of  con- 
tact: 

yi  =  jf>  Formula  XXV.     (5) 

#-¥•  '      '      .  (6) 

su 

We  may  substitute  the  value  of  x\  in  equation  (2)  and  get  the 
desired  straight-line  equation.  It  is  better  simply  to  use  the 
easily  remembered  fact  that  the  ordinate  of  the  point  of  tangency 
to  Euler's  curve  is  one-third  of  the  Y  intercept  of  the  straight 
line  (Formula  XXV).  Substitute  this  value  in  Euler's  equation 
and  get  the  abscissa  of  the  point  of  tangency.  This  gives  the 
coordinates  of  two  points  on  the  straight  line  from  which  to  write 
its  equation. 


COLUMN  FORMULAS  USED  BY  ENGINEERS     205 

The  abscissa  of  the  point  of  contact  is  the  lower  limit  of  -  for 

Euler's  formula.  It  is  also  the  upper  limit  for  the  straight-line 
formula,  although,  if  it  is  exceeded  by  a  small  amount,  the  error  is 
small  and  on  the  safe  side. 

Referring  to  Fig.  151,  Euler's  curve  might  be  used,  if  there  is 
no  eccentricity,  up  to  the  point  C,  the  true  elastic  limit;  but  with 
the  uncertain  eccentricity  which  occurs  in  practice  it  is  best  to 
use  it  only  to  D.  When  you  solve  by  Euler's  formula  for  the  ulti- 
mate unit  load  and  get  a  result  greater  than  one-third  the  ultimate 
strength  of  the  material,  discard  your  work  and  solve  by  a  straight- 

p 
line  formula  (or  by  Rankine's  formula).      If  the  value  of  -r  by 

A. 

Euler's  formula  comes  out  less  than  one-third  of  su,  it  may  be 
used  when  divided  by  suitable  factor  of  safety. 

PROBLEMS. 

1.  Find  a  straight-line  equation  with  a  factor  of  safety  of  5  for  long-leaf 
yellow  pine  for  which  the  ultimate  compressive  strength  is  5000  pounds  per 
square  inch  and  E  is  1,500,000. 


P        ^000     7 

When^-  is  2™,  tin  Euler's  formula  is  30  TT  =  94.2. 
A.         or 

The  slope  of  the  line  is  3333  divided  by  94.2  =  35.4. 
For  the  ultimate  load: 

^=5000-35.4-- 

A.  T 

With  a  factor  of  safety  of  5: 

£  =  1000-7-- 
A  r 

2.  With  the  data  of  Problem  1,  find  the  total  safe  load,  with  a  factor  of 
safety  of  5,  on  a  6-inch  by  6-inch  long-leaf  yellow-pine  post  for  lengths  of 
10  feet,  15  feet,  and  20  feet.  Ans.  18,500,  9870,  and  5550  pounds. 

3.  Using  timber  having  an  ultimate  strength  of  5000  pounds  per  square 
inch  and  a  modulus  of  elasticity  of  1,200,000,  derive  a  working  straight-line 
formula  with  a  factor  of  safety  of  4. 

Ans.  j  =  1250  -  10  -,  for  values  of  -  up  to  85. 

4.  Derive  a  straight-line  formula  for  cast  iron  for  which  E  is  15,000,000 
and  the  ultimate  compressive  strength  is  50,000,  with  a  factor  of  safety  of  5. 

Ans.  y  =  10,000  -  70  -,  for  values  of-  up  to  95. 

A.  T  T 

5.  Calculate  the  total  safe  load  on  a  hollow  cast-iron  column  8  inches 
diameter  and  1  inch  thick  for  lengths  of  10  feet  and  15  feet. 

Ans.  146,000,  109,100  pounds. 


206 


STRENGTH  OF  MATERIALS 


6.   Using  E  =  29,000,000  and  su  =  45,000,  find  the  straight-line  formula 
with  a  safety  factor  of  3  for  structural  steel  columns  with  round  ends. 

15,000  -  72  -,  up  to  about  140. 


Ans.  -r 
A 


116.  The  Ultimate  Strength.  —  The  straight-line  formulas  de- 
pend upon  su,  the  ultimate  compressive  strength  of  the  material 
in  a  short  block.  In  the  Pencoyd  tests  of  Table  IX,  this  figure 
seems  to  be  about  49,000  pounds  per  square  inch  for  wrought 
iron.  This  is  considerably  above  the  yield  point  of  the  iron  used. 

These  tests  were  made  slowly,  so  that 
there  was  ample  time  for  the  raising 
of  the  elastic  limit,  which  occurs  when 
wrought  iron  and  soft  steel  are  loaded 
beyond  the  yield  point.  Also  the 
columns  used  in  these  tests  were  each 
made  of  a  single  piece,  so  that  there  was 
not  that  opportunity  for  local  failure 
which  exists  in  columns  built  up  of 
several  pieces  riveted  together. 

For-  built-up  columns,  the  ultimate 
strength  is  the  yield  point  of  the  material. 
Fig.  152  shows  one  of  a  set  of  wrought- 
iron  columns  tested  at  Watertown  Ar- 
senal in  1884  ("  Tests  of  Metals,"  1884, 
page  17).  The  column  was  tested  with 
3.5-inch  pins.  The  length  center  to 
center  of  pins  was  20  feet,  and  the  def- 
ormation was  measured  in  a  gauged 
length  of  200  inches.  The  average  cross 
section  of  channels  and  plates  was  de- 
termined from  the  weight  and  specific  gravity.  The  cross  sections 
were: 

Square  inches. 

Channel  A  ...................................       3.00 

Channel  B  ...................................       3.05 

Plate  C  ......................................       2.66 

Plate  D  ..............  :  .......................       2.60 

11.31 

The  initial  load  was  5000  pounds.  The  set  was  determined  by 
returning  to  the  initial  load  after  each  50,000  pounds  increment. 
The  deflection  at  the  middle  was  measured  perpendicular  and 
parallel  to  the  pins.  Some  of  the  readings  are  given  in  Table  X. 


w, 
Watertown  Arsenal. 


COLUMN  FORMULAS  USED  BY  ENGINEERS     207 


TABLE  X. 

TEST  OF  WROUGHT-IRON  PLATE  AND  CHANNEL  COLUMN  AT 
WATERTOWN  ARSENAL. 


Total   load. 

Compression  in  gauged 
length  of  200  inches. 

Deflection  at  the  middle. 

Perpendicular  to  pins. 

Parallel  to  pins. 

Pounds. 

Inch. 

Inch. 

Inch. 

5,000 

.0 

.0 

.0 

30,000 

.0169 

.0 

.0 

50,000 

.0293 

.01 

.01 

5,000 

.0 

.0 

.0 

80,000 

.0482 

.01 

.01 

100,000 

.0610 

.02 

.01 

5,000 

.0 

.0 

.0 

130,000 

.0804 

.03 

.02" 

150,000 

.0931 

.03 

•    .02 

5,000 

.0010 

.0 

.01 

180,000 

.1118 

.04 

.03 

200,000 

.1247 

.04 

.03 

5,000 

.0016 

.0 

.02 

230,000 

.1444 

.06 

.03 

250,000 

.1580 

.07 

.03 

5,000 

.0041 

.0 

.03 

260,000 

.1651 

.09 

.03 

270,000 

.1725 

.10 

.03 

280,000 

.1797 

.12 

.03 

290,000 

.1870 

.13 

.03 

300,000 

.1954 

.17 

.03 

5,000 

.0110 

.03 

.03 

310,000 

(Micrometer 

.20 

.03 

320,000 

removed) 

.27 

.03 

325,000 

.... 

.32 

.03 

330,000 

.45 

.03 

330,100 

.      .... 

.48 

.03 

Failed  by  deflection  perpendicular  to  the  plane  of  the  pins;  with  plate  C 
on  the  convex  side. 

"After  reaching  the  maximum  load,  the  deflection  increased  slowly  till  it 
reached  0.75  inch,  the  load  at  the  time  being  320,000  pounds.  From  this 
point  the  rate  of  deflection  accelerated  till  it  reached  1.80  inches  under  310,000 
pounds  load,  when  sudden  springing  occurred,  increasing  the  deflection  to 
3.35  inches,  while  the  pressure  fell  to  155,000  pounds. 

"  Released  to  the  initial  load,  the  deflection  was  2.08  inches. 

"A  sharp  bend  was  found  20  inches  from  the  middle  of  the  post;  the  plate 
on  the  concave  side  buckled  between  the  riveting.  Pinholes  elongated 
0.01  inch." 


208  STRENGTH  OF  MATERIALS 

In  this  test  the  maximum  load  was  nearly  29,200  pounds 
per  square  inch.  The  unit  stress  on  the  concave  side  was  this 
figure  plus  the  stress  due  to  bending.  We  get  the  maximum 
bending  moment  by  multiplying  the  load  by  the  deflection  0.48 
inch.  The  moment  of  inertia  of  the  section  with  respect  to  an 
axis  parallel  to  the  pins  is  about  86.  This  gives  a  bending  stress 
of  6100  pounds,  making  the  total  compressive  stress  at  the 
beginning  of  failure  35,300  pounds  per  square  inch. 

It  is  probable  that  at  the  beginning  of  failure  the  column  was 
in  condition  I  of  Fig.  149,  the  friction  causing  it  to  act  as  if  the 
ends  were  fixed.  In  this  case  the  moment  arm  is  only  half  the 
deflection  at  the  middle  and  the  actual  maximum  stress  is  only 
32,000  pounds  per  square  inch. 

Other  columns  of  the  same  set  gave  similar  results.  Other 
sets  of  tests,  notably  those  of  Buchanan*  for  the  Pennsylvania 
Railroad,  agree  in  indicating  that  the  value  of  su  should  not 
exceed  35,000  for  wrought  iron  and  40,000  for  structural  steel. 

PROBLEMS. 

1.  Using  E  =  27,500,000  and  su  =  35,000,  derive  a  straight-line  formula 
for  the  ultimate  load  on  wrought-iron  columns,  also  a  working  formula  with 
a  factor  of  safety  of  2.5. 

P  I 

-r  =  35,000  -  153  -  ;  point  of  tangency  at  152.5; 

A.  T 

P  1 

^  =  14,000  -  61  -  • 
A  r 

2.  Using  E  =  29,000,000  and  s«  =  40,000  for  structural  steel,  derive  a 
working  straight-line  formula  with  a  factor  of  safety  of  2.5. 

Ana.  ^  =  16,000  -  73  -  to  about  150. 

J\.  T 

The  American  Railway  Engineering  and  Maintenance  of  Way 
Association  has  adopted  for  structural  steel 

£  =  16,000  -  70->  Formula  XXVI. 

A.  T 

with  a  maximum  of  14,000. 

This  formula,  which  we  will  call  the  American  railway  formula, 
is  practically  the  same  as  the  results  of  Problem  2. 

Fig.  153  shows  the  reason  for  having  a  maximum  value  and 

not  carrying  the  straight  line  entirely  back  to  the  Y  axis.     In 

p 
this  figure,  curve  I  gives  the  ultimate  values  of  -r  which  make 

•A  v 

*  Engineering  News,  Dec.  26,  1907. 


Ans. 


COLUMN  FORMULAS  USED  BY  ENGINEERS  209 
the  unit  stress  40,000  pounds  per  square  inch  calculated  in  the 
same  way  as  the  curves  of  Fig.  149  for  —2  =  0.1  (column  IV  of 

Table  VIII).  This  curve  is  nearly  horizontal  at  first,  with  a 
maximum  value  of  36,360  pounds  per  square  inch.  Curve  II  is 
the  American  railway  formula  multiplied  by  2.5,  which  makes 


50000 


£40000 
§35000 
£30000 
"25000 
|  20000 
o  15000 

S  loooo 


STEEL  COLUMNS  WITH 

ROUND  ENDS; 

I,    £.=29,000,000; 


II,  STRAIGHT  LINE 
-j=40, 000-175  j;. 


20    40    CO     80    100  120  140  160  180  200  220  240  260  280  300  320  340  360  380  400 
LENGTH-^- LEAST  RADIUS  OF  GYRATION 

Fig.  153.  —  Straight  Line  and  Secant  Curves. 


it  pass  through  40,000  on  the  Y  axis.  The  horizontal  line  35,000 
is  a  little  below  curve  I.  (With  a  little  greater  eccentricity 
curve  I  would  fall  below  35,000.)  With  the  eccentricity  used, 


curve  I  falls  above  curve  II  for  -  less  than  110. 
values  of -curve  I  is  slightly  below  curve  II. 


For  greater 


PROBLEMS. 

3.  Calculate  the  total  safe  load  by  the  American  railway  formula  for  a 
steel  post  20  feet  long  made  of  one  8-inch  by  f-inch  plate  and  four  4-inch  by 
3-inch  by  f-inch  angles.     (See  Cambria.)  Ans.  210,000  pounds. 

4.  Calculate  the  total  safe  load  on  a  Z-bar  column  16  feet  long  made  of 
four  4-inch  by  4-inch  by  Hnch  Z-bars  and  one  6f-inch  by  Hnch  web  plate. 

5.  Using  the  result  of  Problem  1  of  Article  115  for  safe  loads,  design  a 
square  yellow-pine  post  10  feet  long  to  carry  a  total  load  of  18,000  pounds. 

Ans.  5.94  inches  square;  use  6-inch  by  6-inch. 

6.  Using  the  same  formula,  find  the  thickness  of  a  yellow-pine  post  8 
inches  wide  and  10  feet  long  to  carry  a  total  load  of  28,000  pounds. 

Ans.  6.4  inches. 

7.  Solve  Problem  6  for  a  load  of  60,000  pounds.  Ans.  11.8  inches. 

8.  Solve  Problem  6  for  a  load  of  1200  pounds,  using  the  constants  of 
Problem  1,  Article  115.  Ans.  2.06  inches  thick. 


210  STRENGTH  OF  MATERIALS 

9.  Using  the  American  railway  formula,  design  a  solid  circular  steel  column 
to  carry  a  total  load  of  60,000,  the  length  of  the  column  being  80  inches. 

Ans.  Diameter,  3  inches  nearly. 

10.  What  formula  would  you  use  for  Problem  9  if  the  length  were  doubled? 

117.   Straight-line   Formulas  for  Square  or  Fixed  Ends. — 

In  applying  straight-line  formulas  to  columns  with  square  or 
fixed  ends,  it  is  customary  to  modify  the  constant  k  and  use  the 
entire  length  of  the  column  as  I  in  the  formula.  The  American 
Railway  Engineering  and  Maintenance  of  Way  Association 
uses  the  one  constant  (70)  for  all  cases,  treating  the  so-called 
fixed  and  square  ends  as  no  better  than  hinged  ends.  This  is 
good  practice  for  bridges  and  similar  structures.  When  a 
bridge  post  is  riveted  to  the  floor  beam,  experiments  show  that 
the  deflection  of  the  beam  often  produces  a  bending  stress  in 
the  post  which  is  equivalent  to  a  large  eccentricity.  In  pin- 
connected  bridges,  a  slight  difference  in  the  length  of  the  eye- 
bars  which  form  the  diagonals  of  the  truss  sometimes  causes 
such  concentration  of  stress  in  one  side  of  the  post  that  it  is 
weaker  in  the  plane  of  the  pins  than  perpendicular  to  that  plane. 
In  buildings,  where  the  floor  beams  are  riveted  to  the  posts, 
there  is  likely  to  be  considerable  eccentricity  in  the  end  posts. 
At  intermediate  posts  with  beams  on  both  sides  the  eccentricity 
is  less.* 

The  American  railway  formula  may  well  be  used  for  all  struc- 
tures built  of  structural  steel,  provided,  of  course,  due  allowance 
is  made  for  live  loads  and  impact  in  computing  the  total  load. 

The  building  laws  of  New  York  city  require  for  structural 
steel  columns  with  square  or  fixed  ends : 

•jj -.15,200- 58 1-  (1) 

If  we  regard  the  length  of  the  sine  curve  as  0.8  of  the  total 
length  of  the  column,  we  get  from  the  American  railway  formula 
0.8  X  70  =  56.  If  k  is  70  for  round  ends,  and  the  effect  of 
square  or  fixed  ends  is  sufficient  to  make  a  10-foot  column 
equal  in  strength  to  an  8-foot  round-end  column,  the  constant 
58  may  be  taken  as  practically  correct  for  the  square  and  fixed 
ends,  if  the  total  length  of  the  column  is  taken  as  L  (see  experi- 
ments, Table  IX). 

*  See  paper  by  C.  T.  Morris,  Engineering  News,  Nov.  2,  1911,  p.  530. 


COLUMN  FORMULAS  USED  BY  ENGINEERS     211 
In  Problem  1  of  Article  115,  we  found  the  formula 

J  =  1000-7^i  (2) 

for  round-end  columns  of  long-leaf  yellow  pine  with  a  modulus 
of  1,500,000,  an  ultimate  strength  of  5000,  with  a  factor  of  safety 
of  5.  The  New  York  building  laws  require 

J=1000-18j-  (3) 

For  solid  columns  of  circular  section  D  =  4  r,  so  that  equation  (2) 
becomes 

j  =  1000  -  28  ~  (3) 

If  the  ends  of  the  square-end  columns  used  in  buildings  are  suffic- 
iently well  fixed  that  9  L  is  equal  to  14  I,  equations  (2)  and  (3) 
are  consistent. 

The  New  York  building  formulas  for  timber  posts  are  recom- 
mended for  square-  or  fixed-end  conditions.  They  may  be  applied 
to  rectangular  posts  if  D  is  taken  as  the  least  dimension. 

PROBLEMS. 

1.  If  cast  iron  has  an  ultimate  strength  in  compression  of  80,000  pounds 
per  square  inch  and  a  modulus  of  15,000,000,  derive  a  formula  for  square-end 
cast-iron  columns  with  a  factor  of  safety  of  8,  assuming  that  the  ends  are  so 
fixed  that  3  L  is  equivalent  to  4  L     (Compare  with  Cambria.) 

2.  Find  the  total  safe  load  by  the  New  York  building  laws  for  a  12-inch 
31.5-pound  I-beam  as  a  column  with  square  ends,  for  lengths  of  10  feet  and 
15  feet.  Ans.  76,940  pounds  and  45,040  pounds. 

3.  Find  the  total  safe  load  on  a  6-inch  by  6-inch  yellow-pine  post  10  feet 
long  by  the  New  York  building  laws.  Ans.  23,040  pounds. 

4.  Find  the  size  of  a  square  post  of  long-leaf  yellow  pine  10  feet  in  length, 
to  carry  a  load  of  40,000  in  accordance  with  the  requirements  of  the  New  York 
building  laws.  Ans.  7.49  inches  square;  use  an  8-inch  by  8-inch. 

5.  Solve  Problem  4  for  hemlock. 

6.  Find  the  total  safe  load  by  New  York  building  laws  on  a  column  of 
medium  steel  made  of  two  9-inch  15-pound  channels  and  two  13-inch  by  5- 
Lnch  plates,  for  lengths  of  20  feet  and  30  feet  (see  Cambria  for  constants). 

Ans.  255,900  pounds  and  218,000  pounds. 

7.  Solve  Problem  6  by  the  American  railway  formula. 

8.  Select  a  plate  and  channel  column  of  medium  steel,  20  feet  in  length, 
to  carry  a  load  of  100,000  in  accordance  with  the  New  York  building  laws. 
(Use  Cambria  tables  to  get  the  column  approximately  and  then  apply  the 
formula  to  see  if  the  column  thus  selected  will  do.) 

9.  Solve  Problem  8  by  the  American  railway  formula. 


212  STRENGTH  OF  MATERIALS 

10.  Select  a  plate  and  angle  column  20  feet  long,  to  carry  300,000  pounds 
safely  in  accordance  with  the  American  railway  formula. 

Ans.  One  12-inch  by  ^-inch  plate  with  four  6-inch  by  3^-inch  by  ^- 
inch  angles  carries  316,000  pounds. 

118.  Rankine's  or  Gordon's  Formulas.  —  While  the  straight- 
line  formulas  are  coming  into  general  use  among  engineers,  on 
account  of  the  ease  of  application  and  the  fact  that  they  agree 
as  well  with  the  results  of  tests  as  the  more  complicated  expres- 
sions, another  type  of  formula  formerly  had  the  preference  and 
is  still  used  considerably.  This  type  is  called  Gordon's  or 
Rankine's  formula.  It  is  an  empirical  formula  "which  was  first 
derived  from  the  results  of  tests.  It  has  the  advantage  that  it 
applies  to  columns  of  any  length.  It  is: 

S  Formula  XXVII. 


where  su  is  the  ultimate  strength  in  compression  in  the  case 
of  a  short  block,  and  q  is  a  coefficient  determined  from  experi- 
ments with  columns  of  various  lengths.  To  use  the  formula 
with  any  given  factor  of  safety,  simply  divide  the  numerator 
by  the  factor;  that  is,  use  the  allowable  unit  stress  instead  as 
the  ultimate  strength. 

The  figures  for  structural  steel  columns  with  flat  ends,  given 
in  Cambria,  were  computed  by  this  formula,  using  50,000  as  su 

for  medium  steel  and  for  q.     (This  value  of  q  is  the-  one 

OO,UUU 

recommended  by  Rankine.) 

PROBLEMS. 

1.   Find  the  total  load,  with  a  factor  of  safety  of  4,  on  a  12-inch  40-pound 
I-beam  of  medium  steel,  12  feet  in  length,  as  a  column  with  square  ends. 

Ans.  90,460  pounds. 

_  0.625;        ?=     12'5°° 


36,000  \rj  A       1  +  0.625 

Compare  the  results  with  Cambria,  "Safe  Loads  for  I-Beams  used  as  Columns.'.' 

Using  su  =  50,000  and  q  =  ,  calculate  the  total  load,  with  a  safety 


factor  of  4,  on  the  following  square-end  columns  ef  medium  steel. 

2.  A  10-inch  25-pound  I-beam  15  feet  long. 

3.  A  12-inch  31.5-pound  I-beam  16  feet  long. 

4.  A  12-inch  31.5-pound  I-beam  32  feet  long. 

5.  A  6-inch  by  4-inch  by  1-inch  angle  12  feet  long. 


COLUMN  FORMULAS  USED  BY  ENGINEERS     213 

6.  A  post  made  of  two  10-inch  20-pound  channels,  latticed  together  6  inches 
back  to  back,  for  lengths  of  15  feet  and  20  feet. 

7.  A  column  made  of  two  10-inch  15-pound  channels,  9  inches  back  to  back, 
and  two  15-inch  by  j-inch  plates,  for  lengths  of  20  feet  and  40  feet. 

119.  Ritter's  Rational  Constant  for  Rankine's  Formula.  - 

While  the  constant  q  was  originally  derived  from  a  few  tests  of 
columns,  it  may  be  obtained  from  the  constants  of  the  materials. 
We  know  from  experiments  and  theory  that  Euler's  formula 
gives  the  ultimate  load  when  the  load  is  exactly  central,  the  ends 
either  perfectly  free  to  turn  or  absolutely  fixed,  and  the  value  of 

I  P 

~  so  great  that  the  computed  —r  is  below  the  true  elastic  limit  of 

T  A. 

the  material.  Any  curve  which  is  to  be  used  with  all  lengths 
must  coincide  with  Euler's  curve  when  -  becomes  indefinitely 

large,  and  must  also  pass  through  the  point  su  when  -  =  0. 

I  P 

We  see  that  when  -  equals  zero,  -r  equals  su'}  Rankine's  for- 

T  A. 

mula  satisfies  the  second  of  the  above-mentioned  conditions.    To 

make  it  satisfy  the  first  condition,  we  must  find  some  value  of 

p 
q  which  will  make  -r  the  same  in  Rankine's  and  Euler's  formulas 

for  large  values  of  - : 


!+<?(- 

For  large  values  of  -  the  second  term  in  the  denominator  of  Ran- 
kine's formula  is  so  large  relatively  that  the  first  term  (unity) 
may  be  dropped.  Then 


This  value  of  q  is  Ritter's  rational  constant. 


214  STRENGTH  OF  MATERIALS 

PROBLEMS. 

1.  Find  the  value  of  q  for  steel  having  a  modulus  of  29,000,000  and  ultimate 

strength,  in  compression,  of  40,000  pounds  per  square  inch.  Ans.   =—— . 

vloo 

2.  Using  the  ultimate  strength  of  steel  in  compression  as  40,000  pounds 
per  square  inch  and  the  result  of  Problem  1  as  q}  find  the  ultimate  loads  in 

pounds  per  square  inch  for  round-end  columns  for  values  of  -  ,  differing  by  20, 

20         40         60         80         100       120         140       160     180     200 
Ans. 

1^-,  37,900  32,700  26,600  21,100  16,700  13,300  10,700  8700  7200  6100 

\A 

P 

3.  Solve  for  -r  in  Problem  2  by  Euler's  formula. 

A 

4.  Using  the  constants  of  Problem  1,  find  the  total  safe  load  with  a  factor 
of  safety  of  4  on  a  round-end  solid  circular  steel  column  2  inches  in  diameter 
and  40  inches  long.  Ans.  16,600  pounds. 

5.  Using  the  constants  of  Problem  1,  find  the  total  load  with  a  factor  of 
safety  of  5  which  may  be  placed  on  a  10-inch  25-pound  I-beam  10  feet  long, 
used  as  a  column  with  round  ends.  Ans.  18,800  pounds. 

For  the  wrought  iron  used  in  Ritchie's  experiments,  Table  IX, 
the  modulus  of  elasticity  was  about  25,000,000  and  the  ultimate 
strength  about  49,000.  If  we  take  E  as  24,800,000,  making 

TT*E=  245,000,000,  we  get  ^-^  as  Ritter's  constant  for  these  ex- 


periments. 

P  49,000 


(4) 
A       1  + 


5000  Vv 

gives  the  ultimate  strength  of  wrought-iron  struts  with  ends  free 
to  turn. 

PROBLEMS. 

P  I 

6.  Calculate  -r  for  all  values  of  -  from  0  to  400  at  intervals  of  20  by  means 

A  r 

of  equation  (4),  and  plot  curve  with  abscissas  1  inch  =  -  =  40;  ordinates 

1  inch  =  ^  =  10,000. 

Also  on  the  same  sheet  with  the  same  origin  and  coordinates,  plot  the  results 
of  Table  IX  for  round  ends. 

7.  With  E  =  24,800,000,  calculate  -T  by  Euler's  formula  from  -  =  60  to  400 

A  T 

at  intervals  of  20,  and  plot  the  results  with  those  of  Problem  6. 


COLUMN  FORMULAS  USED  BY  ENGINEERS     215 

8.  Draw  straight  line  tangent  to  Euler's  curve  through  the  point  (0,  49,000) 
and  determine  its  equation.     Compare  this  straight  line  with  the  results  of 
the  test. 

9.  With  the  results  of  Problems  6,  7,  and  8  as  a  basis,  write  your  conclusions 
as  to  the  merits  of  Rankine's,  Euler's,  and  the  straight-line  formulas. 

Curve  I  of  Fig.  154  is  drawn  from  Rankine's  formula  with 
Hitter's  constant  calculated  from  su  =  40,000  and  E  =  29,000,000, 

which   makes   q  =  ^r^v     Curve   II   is   Euler's.     We   see   that 
71oo 

curve  I  gradually  approaches  Euler's  curve  and  is  always  on  the 


II      EULER'S  CURVE, 

E-29, 000,000; 
III,  SECANT  CURVE  FOR 

iv,     -"•" 


20    40    60    SO    100  120  140  160  180  200  220  240  2t>0  230  300  320  340  380  380  400 
LENGTHS  RADIUS  OF  GYRATION 


Fig.  154.  —  Column  Curves. 


safe  side.     Curve  III  is  drawn  from  the  secant  formula  for 

cv 

-5  =  0.1  (curve  I  of  Fig.  145).  We  notice  that  curve  I  is  de- 
cidedly below  curve  III,  especially  for  the  values  of  -  between  40 

and  120,  which  values  are  most  used  in  actual  columns.  Ran- 
kine's formula  with  Ritter's  constant  gives  correct  values  for 
very  long  columns  and  for  columns  of  zero  length  with  no  eccen- 
tricity (these  being  the  conditions  under  which  the  constants  are 
determined),  but  for  columns  of  the  usual  length  is  not  so  good 
as  a  straight  line  tangent  to  Euler's. 

Curve  IV  is  computed  by  means  of  Rankine's  experimental 

constant  ^5-7^-     It  is  not  far  off  for  short  columns  especially 
lo,U(Ju 

if  the  eccentricity  is  very  small,  but  it  is  too  high  with  columns 
which  are  relatively  long. 

Since  the   only  advantage  possessed  by   Rankine's  formula 
over  a  straight-line  formula  is  the  fact  that  the  former  may  be 


216  STRENGTH  OF  MATERIALS 

used  with  columns  of  all  lengths,  it  follows,  that  if  it  is  used  at  all 
it  should  be  used  with  Hitter's  constant,  so  that  the  errors  will 
be  always  on  the  safe  side  (except  for  very  short  posts).  The 

experimental  constants  ^  7^  for  hinged  ends  and  oa  -__  for  fixed 
lo,UUU  o 


ends  which  were  recommended  by  Rankine  are  based  on  a  limited 
number  of  tests.  While  they  are  not  far  off  for  short  columns, 
these  constants  give  unsafe  results  in  relatively  long  ones. 

The  column  tables  in  Cambria  are  computed  with  these  con- 
stants. However,  a  larger  factor  of  safety  is  taken  than  is  used 
by  many  engineers,  12,500  being  employed  as  the  numerator  for 
buildings  and  10,000  recommended  for  bridges.  On  this  account 

the  results  given  are  safe,  and  especially  for  values  of  -  less  than 

100. 

In  adapting  Rankine's  formula  with  Ritter's  constant  to  fixed 
and  square  ends,  the  entire  length  L  is  used  and  q  is  modified. 
If  the  column  were  absolutely  fixed  at  both  ends,  the  q  of  Ritter's 

expression  would  be  divided  by  4  ;  if  q  were  ^        for  round  ends, 

1  000 

it  would  be  ^5-7^7;  for  ends  rigidly  fixed.     In  actual  columns,  since 


the  ends  are  not  perfectly  fixed,  we  must  use  a  ratio  less  than  4. 
Suppose  that  I  is  taken  as  0.8  L  (Table  IX),  then  I2  =  0.64  L2. 

If  we  take  q  for  structural  steel  as  =r^  (Problem  1), 


I2  L2 

nearly. 


7155      11,200 
The  Philadelphia  laws  use,  for  medium  steel: 
P_  16,250 

A  =  i+    L* 


11,000  H 

A  few  years  ago  several  railroads  used  the  same  formula  with 
17,000  as  the  numerator  instead  of  16,250.  Both  are  rather 
high;  15,000  is  better.  This  value  of  q  is  good  for  square  ends 
where  the  eccentricity  is  small,  but  should  not  be  used  with  pin 
ends  except  with  a  large  factor  of  safety. 


COLUMN  FORMULAS  USED  BY  ENGINEERS     217 

PROBLEMS. 

10.  Calculate  the  total  safe  load  on  a  4-inch  by  4-inch  by  £-inch  angle  of 
medium  steel  for  lengths  of  10  feet  and  15  feet  by  the  Philadelphia  formula. 

11.  Compute  the  total  safe  load  on  a  strut  with  flat  ends,  made  of  a  6-inch 
by  6-inch  by  1-inch  angle  of  medium  steel,  by  the  New  York  building  laws, 
the  Philadelphia  building  laws,  and  the  American  railway  formula,  for  lengths 
of  10  feet  and  15  feet. 

12.  Find  the  Z-bar  column  of  medium  steel  20  feet  long  with  flat  ends  to 
carry  a  load  of  400,000  pounds,  by  each  of  the  formulas  above  mentioned. 

13.  Solve  Problem  2  of  Article  117  by  the  Philadelphia  formula. 

P  I 

14.  Find  -r  for  medium  steel  for  values  of  -  from  20  to  140  at  intervals  of 

A.  T 

20,  by  the  American  railway  formula,  by  the  New  York  formula,  and  by  the 
Philadelphia  formula.  Compare  the  results  with  those  of  Cambria  for  square 
ends  with  a  factor  of  safety  of  4. 

15.  Solve  Problem  14  from  160  to  300  inclusive. 

120.  General  Conclusions.  —  The  calculation  of  columns  is  not 
as  satisfactory  as  that  of  beams.  This  is  due  to  two  reasons: 
the  location  of  the  load,  and  the  relative  freedom  of  the  ends. 
In  a  beam,  the  location  of  the  load  is  known  with  a  large  relative 
accuracy.  A  1-inch  displacement  of  the  load  in  a  horizontal 
beam  10  feet  long  produces  a  very  small  effect  upon  the  unit 
stress;  an  equal  displacement  of  the  load  at  the  end  of  a  block  6 
inches  square  will  double  the  maximum  stress  on  one  face.  Again, 
we  generally  deal  with  beams  entirely  free  to  turn  at  the  supports 
or  with  cantilevers  which  are  entirely  free  to  move  and  turn  at  one 
end  and  which  are  perfectly  fixed  at  the  other,  so  far  as  concerns 
the  moment  arms.  The  results  which  we  get  in  calculating  beams 
are  correct  inside  the  true  elastic  limit  and  approximately  true 
beyond  that  limit.  If  we  take  a  column  perfectly  free  to  turn 
at  both  ends  and  know  the  position  of  the  load  with  the  same 
relative  certainty  as  in  the  case  of  the  beam  supported  at  the  ends, 
we  may  calculate  the  unit  stress  with  the  aid  of  Formula  XXII  as 
accurately  as  we  can  compute  it  in  the  beam  by  the  use  of  For- 
mula VIII.  There  is  this  apparent  difference:  in  the  beam  the 
unit  stress  varies  as  the  load;  in  the  column  it  increases  more 
rapidly.  Again,  a  column  fixed  at  one  end  and  free  at  the  other 
(case  II,  Fig.  146)  can  be  calculated  with  the  same  accuracy  as 
a  cantilever  with  one  end  free,  provided  the  load  is  located  with 
the  same  relative  accuracy  and  the  end  is  so  well  fixed  that  the 
relative  change  in  moment  due  to  change  in  tangent  at  the  "  fixed 
end  "  is  the  same  in  both  cases.  The  change  in  moment  due  to 


218  STRENGTH  OF  MATERIALS 

change  in  the  tangent  at  the  ends  is  proportional  to  the  rate  of 
change  of  the  cosine  of  a  small  angle  in  the  case  of  a  beam  and  to 
the  rate  of  change  of  the  sine  of  a  small  angle  in  the  column. 
The  loads  being  much  greater  in  a  column  than  in  a  beam  of  the 
same  section,  the  effect  of  friction  in  partially  fixing  the 'ends  is 
greater  in  the  column. 

Beams  fixed  at  both  ends  or  fixed  at  one  end  and  supported 
at  the  other  are  indefinite,  because  it  is  not  possible  to  fix  the 
beam  perfectly  so  that  it  will  not  turn,  or  support  it  so  that  it 
will  not  move.  For  these  reasons  the  calculation  of  the  unit 
stresses  in  relatively  stiff  beams  of  these  kinds  is  always  open  to 
question.  The  same  is  true  of  columns  fixed  at  both  ends,  or 
fixed  at  one  end  with  a  hinge  connection  at  the  other. 

Euler's  formula  gives  the  ultimate  load  which  will  cause  a 
column  with  practically  no  eccentricity  to  deflect  without  limit. 
Unless  the  column  is  relatively  long,  it  will  fail  by  crushing 

before  this  load  is  reached  (see  Fig.  145).     Euler's  formula  must 

p 
never  be  used  if  the  value  of -r   comes  out  greater  than  the  elastic 

limit  of  the  material.     It  is  better  to  use  the  results  of  Euler's 

p 

formula  only  when  they  give  -r  less  than  one-third  of  the  ulti- 
mate strength.  In  such  case,  divide  the  results  by  a  suitable 
factor  of  safety  to  get  the  allowed  load. 

For  shorter  columns,  draw  a  straight  line  tangent  to  Euler's 
curve  and  passing  through  the  ultimate  strength  of  the  material 
on  the  Y  axis.  Divide  the  equation  of  this  straight  line  by  a 
suitable  factor  of  safety  for  a  working  formula.  Use  this  straight 
line  to  the  ordinate  of  the  point  of  tangency  and  use  Euler's 
equation  for  the  longer  columns. 

With  hinged  end  columns  the  I  of  these  formulas  is  the  entire 
length  between  hinges.  With  case  II,  Us  twice  the  length  of  the 
column.  With  fixed  ends  or  with  pin-and-square  ends,  it  is  also 
safest  to  take  I  as  the  entire  length  of  the  column.  If  any  allow- 
ance is  made,  it  should  never  be  as  great  as  that  of  the  ideal 
cases,  which  are  never  met  in  practice.  The  amount  of  allow- 
ance depends  upon  the  relative  dimensions  of  the  column  and 
the  beams  to  which  it  is  attached  and  the  method  of  attach- 
ment. 

The  effect  of  eccentricity  is  taken  into  account  by  using  a 


COLUMN  FORMULAS  USED  BY  ENGINEERS     219 

limiting  stress  for  short  columns,  as  in  the  case  of  the  American 
railway  formula,  and  by  the  use  of  a  large  factor  of  safety  (well 
called  a  factor  of  ignorance)  to  take  care  of  any  uncertainties 
in  this  respect.  (The  real  factor  of  safety  in  many  columns 
which  are  standing  is  probably  much  less  than  figured  by  the 
designer.) 

Rankine's  formula  is  used  by  some  engineers.  With  Hitter's 
constants  it  is  always  safe  —  unnecessarily  safe  for  columns  of 
moderate  length.  With  Rankine's  constants  it  should  not  be 
used  for  long  columns. 

,  If  the  eccentricity  of  the  load  were  sufficiently  well  known, 
the  secant  formulas  of  Article  107  are  strictly  correct  for  round 
ends,  below  the  true  elastic  limit.  A  set  of  curves  like  I  and 
II  of  Fig.  145,  for  the  different  relative  eccentricities,  could  be 
used  for  all  cases  of  this  kind. 

121.  Failure  of  Beams  Due  to  Flexure  on  the  Compression 
Side.  —  The  compression  flange  of  a  beam  is  really  a  column, 
and  may  fail  by  buckling  laterally.  Fig.  155,  II,  is  the  plan  of 


Uij 

I 


BOTTOM  FLANGE 


TOP  FLANGE 

II  PLAN 


Fig.  155. 


an  I-beam  supported  at  the  ends.  The  top  flange,  being  in 
compression,  may  buckle  as  shown,  while  the  bottom  flange  re- 
mains straight.  The  unit  stress  in  the  top  fibers  computed  as  a 

p 
beam  is  the  maximum  value  of  the  unit  load  -j,  of  the  column 

A. 

theory.  As  only  the  very  top  fibers  reach  the  maximum  value 
of  the  stress,  and  as  these  fibers  are  held  from  buckling  laterally 
by  the  fibers  below,  this  unit  load  may  be  taken  somewhat  larger 
than  in  simple  columns.  Suppose  we  use  18,000  pounds  per 
square  inch  as  this  maximum  stress  and  use  Rankine's  formula 

with  the  constant  oa  nAA  for  fixed  ends.     If  6  is  the  breadth  of 

ob,UUU 

b2 
the  flange,  the  square  of  the  radius  of  gyration  is  j^ 

P_         18,000  18,000 

A  =  I2  I2 

-i-      |      o/^   f\f\f\      9  I 


36,000  r2          '  3000  65 


220 


STRENGTH  OF  MATERIALS 


PROBLEMS. 

1.  If  I  is  the  length  of  the  beam  (or  the  distance  between  stiff eners)  and 
16,000  per  square  inch  is  the  maximum  compressive  stress  due  to  bending, 
find  the  maximum  ratio  of  I  to  b  in  order  that  the  flange  will  not  buckle. 

2.  Solve  Problem  1  if  the  maximum  compressive  stress  due  to  bending 
is  12,000  pounds  per  square  inch. 

3.  What  is  the  maximum  compressive  stress  allowable  by  the  above  formula 
if  the  stiff  eners  are  placed  at  intervals  of  60  times  the  breadth  of  the  flange? 

Compare  the  results  of  these  problems  with  Cambria  under  "  Lateral 
Strength  of  Beams  without  Lateral  Support." 

4.  Solve  Problem  2  by  means  of  the  American  railway  formula  for  columns 
instead  of  the  one  given  above. 

5.  Using  the  New  York  formula,  find  the  maximum  compressive  stress  in 
a.  15-inch  42-pound  I-beam  10  feet  long  without  stiffeners  or  other  lateral 
supports.  Ans.  10,840  pounds  per  square  inch. 

6.  Using  the  American  railway  formula,   find    the    maximum    distance 
between  stiffeners  in  an  18-inch  55-pound  I-beam,  when  the  maximum  com- 
pressive stress  due  to  bending  is  13,000  pounds  per  square  inch. 

122.  Failure  Due  to  Buckling  of  the  Web.  —  We  learned  in 
Article  31,  that  a  vertical  shear  produces  compression  at  45 


Fig.  156.  —  Web  of  I-Beam  as  a  Column. 


degrees  to  the  vertical,  and  that  the  intensity  of  this  compres- 
sive stress  is  equal  to  the  unit  vertical  or  horizontal  shearing 
stress.  All  parts  of  the  web  of  an  I-beam  subjected  to  vertical 
shear  may  be  regarded  as  made  up  of  a  series  of  columns  with 
fixed  ends  placed  45  degrees  to  the  vertical.  In  Fig.  156,  FG 
represents  one  such  imaginary  column.  Small  trusses  are  made 
similar  to  Fig.  157.  The  riveted  diagonals  such  as  CD  and  FG 
transmit  the  shear  from  the  top  to  the  bottom  chord.  FG  is 
in  compression  and  acts  as  a  column.  Instead  of  single  bars 


COLUMN  FORMULAS  USED  BY  ENGINEERS     221 

riveted  at  their  intersections,  a  plate  with  part  of  the  material 
cut  away  might  be  used  to  connect  the  top  and  bottom  chords. 
Finally,  if  this  plate  is  continuous  it 
becomes  a  plate  girder  or  I-beam. 

Considering  the  column  FG  in  Fig. 
156,  its  thickness  is  t,  the  thickness 
of  the  web,  and  its  length  is  \/2  c, 
where  c  is  the  distance  between  the 
flanges  (represented  by  I  in  the  Cam- 
bria sketches  of  I-beam  sections).  The 
unit  shearing  stress  in  the  web  varies 
slightly  (see  Article  84,  Problems  9  and  10).  It  is  customary 
to  find  the  mean  vertical  shear  in  the  web  of  an  I-beam  by 
dividing  the  total  vertical  shear  by  the  area  td,  where  t  is  the 
thickness  of  the  web  and  d  is  the  entire  depth  of  the  beam. 
Since  unit  compressive  stress  is  equal  to  the  unit  shearing  stress, 


Fig.  157. 


S 


A      td' 


(D 


P  . 


where  -j  is  the  unit  load  in  the  column  formula  and  S  is  the 

total  vertical  shear.     To  find  the  safe  value  for  S,  it  is  only 
p 

necessary  to  solve  -j  by  any  column  formula,  remembering  that 
A. 

t2 
r2  =  TI-X   for  a  rectangular  section  of  thickness   t.     Using   first 

\£i 

Rankine's  formula  with  12,000  as  the  numerator  and  q  equal 
to  -^-^x  in  order  to  compare  with  Cambria,  we  get: 


36,000 


P 

12,000 

A  ~ 

P 
A~ 

I2 
i    \ 

1  36,000  r2 
12,000 

1     /c\» 

(2) 


(3) 


1500 


PROBLEMS. 

1.  Find  the  maximum  value  of  the  unit  shear,  the  total  vertical  shear, 
and  the  total  load  uniformly  distributed,  on  a  12-inch  31.5-pound  I-beam,  by 
means  of  the  above  formula. 

Ans.  7488  pounds  per  square  inch,  31,450  pounds,  and  62,900  pounds. 


222  STRENGTH  OF  MATERIALS 

2.  Solve  Problem  1  for  a  15-inch  42-pound  I-beam.     Compare  results  with 
Cambria  under  "Maximum  Loads  of  I-Beams  and  Channels  Due  to  Crippling 
the  Web." 

3.  If  the  allowable  unit  stress  due  to  bending  is  16,000  pounds  per  square 
inch,  what  is  the  minimum  length  for  which  the  full  bending  stress  may  be 
developed  by  a  uniformly  distributed  load  without  producing  excessive  buck- 
ling stresses  in  a  12-inch  31.5-pound  I-beam? 

4.  Solve  Problem  3  for  a  20-inch  65-pound  I-beam  for  the  maximum  load 
and  minimum  span  without  crippling  the  web  if  the  load  is  concentrated  at 
the  middle. 

5.  Solve  Problem  1  by  the  American  railway  formula. 

Ans.  Total  load,  47,780  pounds. 

6.  Solve  Problem  1  by  the  New  York  building  laws. 

Ans.  56,000  pounds. 

7.  Solve  Problem  1  by  the  Philadelphia  laws  for  medium  steel. 

Ans.  45,800  pounds. 


CHAPTER  XIV 


TORSION. 

123.  Torque.  —  A  shaft  or  rod  subjected  to  a  pair  of  equal 
and  opposite  couples  in  parallel  planes  at  right  angles  to  its 
length  is  in  torsion  between  these  planes.  In  Fig.  158  we  have 
a  rope  wound  around  a  shaft  and  carrying  a 
weight.  Attached  to  the  shaft  is  a  pulley 
upon  which  runs  a  belt.  The  tension  on  the 
rope  and  the  reactions  of  the  bearings  form 
a  counterclockwise  couple,  while  the  tension 
on  the  belt  and  the  reactions  form  a  clock- 
wise couple.  If  there  is  no  friction  at  the 
bearings,  these  couples  are  equal,  provided 
the  shaft  is  stationary  or  moving  in  either 
direction  with  constant  speed.  The  moment 
of  either  couple  is  the  twisting  moment,  or 
torque,  in  the  portion  of  the  shaft  between  the  pulley  and  the 
rope.  We  will  represent  torque  by  M t. 


Fig.  158.  —  Shaft  in 
Torsion. 


PROBLEMS. 

1.  An  axle  8  inches  in  diameter  is  used  to  lift  a  load  of  400  pounds  which 
is  carried  by  a  1-inch  rope.     If  the  axle  is  turned  by  a  crank  at  one  end,  what 
is  the  torque?  Ans.  1800  inch  pounds. 

2.  A  shaft  carries  a  pulley  3  feet  in  diameter  and  a  second  pulley  2  feet  in 
diameter.     At  the  2-foot  pulley  the  belt  runs  horizontally  to  the  right,  with 
tension  of  600  pounds  in  the  upper  portion  and  100  pounds  in  the  lower  portion. 
The  belt  on  the  3-foot  pulley  runs  vertically  downward,  with  tension  of  120 
pounds  on  the  right  portion.     Find  the  torque  between  the  pulleys  and  the 
tension  on  the  left  portion  of  the  vertical  belt. 

-4ns.  500  foot  pounds,  453.3  pounds. 

3.  A  bolt  is  turned  by  means  of  a  wrench.     The  pull  on  the  wrench  is 
applied  16  inches  from  the  axis  of  the  bolt.     Find  the  torque  when  the  pull  is 
80  pounds.  Ans.  1280  inch  pounds. 

124.  Shearing  Stresses  in  a  Shaft.  —  Let  us  consider  a  por- 
tion of  the  shaft  of  Fig.  158,  between  the  pulley  and  the  rope. 
Fig.  159  represents  such  a  portion  with  axis  vertical.  We  will 

223 


224 


STRENGTH  OF  MATERIALS 


suppose  the  lower  end  to  be  stationary  with  respect  to  the  upper 
end.     When  the  shaft  is  twisted  in  a  counterclockwise  direc- 


L  L  F-f'dA 


L2L'2 


II 


Fig.  159.  —  Portion  of  Shaft  in  Torsion. 


tion,  the  plane  OBDC  becomes  the  surface  OB'DC,  and  the 
radius  OFB  remains  a  straight  line  OF'B'  (in  case  the  section 
is  circular).  The  displacement  FF'  of  any  point  is  proportional 
to  its  distance  from  the  axis,  and  also  proportional  to  its  dis- 
tance from  the  base  regarded  as  stationary.  Consider  a  hollow 
cylinder  of  radius  OF  =  r,  the  upper  base  of  which  is  the  ring 
LFML.  This  cylinder  developed  gives  the  rectangle  L  GGZ  L2  of 
Fig.  159,  II.  When  the  shaft  is  twisted  this  developed  cylinder 
becomes  the  parallelogram  L'GG^Lz'.  Every  filament  as  EF  in 
the  cylinder  has  suffered  a  shearing  displacement  FFf.  Since 
OF'B'  remains  straight,  this  displacement  varies  as  r,  and  the 
unit  shearing  stress  may  be  represented  by  kr,  where  k  is  the 
unit  shearing  stress  at  unit  distance  from  the  axis.  On  a  fila- 
ment of  area  dA,  the  total  shearing  stress  is  the  area  times  the 

unit  stress: 

Shear  on  dA  =  krdA.  (1) 

The  resisting  moment  of  the  shear  on  the  filament  EF'  of  area 
dA  with  respect  to  the  axis  of  the  cylinder  OC  is  the  product  of 
the  shear  on  the  area  multiplied  by  the  radius : 

Resisting  moment  of  filament  =  krzdA.  (2) 

The  total  resisting  moment  of  the  entire  portion,  being  the  sum 
of  the  moments  of  all  the  filaments,  is  the  integral  of  (2)  over 


TORSION  225 

the  entire  area  of  cross  section,  and  is  equal  to  the  external 
torque. 

Torque  =  k  Cr2dA  =  kJ,  (3) 

where  J  is  the  polar  moment  of  inertia. 
At  a  distance  r  from  the  axis, 


s  J 
Torque  =  k  J  =  -—  J 

_         torque  X  radius 
polar  moment  of  inertia  * 

If  a  is  the  radius  of  the  cylinder,  the  shearing  stress  in  the  outer 
fibers  is  given  by 

Mta 


Formula  XXVIII. 


/n  Mv\ 

(Compare  with  s  =  — *-•] 


PROBLEMS. 

1.  A  2-inch  solid  shaft  is  twisted  by  means  of  a  pipe  wrench.     The  force 
is  200  pounds  applied  3  feet  from  the  axis  of  the  shaft.     Find  the  maximum 
shearing  stress.  Ans.  4583  pounds  per  square  inch. 

2.  A  6-foot  flywheel  is  driven  by  a  4-inch  solid  shaft.     The  tension  on  the 
upper  part  of  the  belt  running  from  the  wheel  is  1800  pounds  and  on  the  lower 
part  it  is  300  pounds.     Find  the  torque  and  the  maximum  unit  stress. 

Ans.  s8  =  4298  pounds  per  square  inch. 

3.  Calculate  the  unit  stress  in  a  2-inch  solid  shaft  which  is  used  to  turn  a 
drum  30  inches  in  diameter  which  is  lifting  800  pounds  by  means  of  a  1-inch 
rope.  Ans.  7894  pounds  per  square  inch. 

4.  In  Problem  2  what  would  be  the  unit  stress  if  the  shaft  were  hollow, 
inside  diameter  2  inches,  outside  diameter  4  inches? 

^ins.  4583  pounds  per  square  inch. 

5.  Solve  Problem  2  with  the  shaft  hollow,  inside  diameter  2  inches,  and 
outside  diameter  such  that  the  weight  per  foot  equals  that  of  a  4-inch  solid 
shaft. 

6.  In  Problem  2,  if  the  wheel  makes  240  revolutions  per  minute,  what  is 
the  horse  power  transmitted  by  the  belt?  Ans.  206  horse  power  nearly. 

7.  A  3-inch  solid  shaft  running  250  revolutions  per  minute  drives  a  wheel 
4  feet  in  diameter.     What  is  the  difference  in  tension  in  the  two  parts  of  the 


226  STRENGTH  OF  MATERIALS 

belt  when  the  unit  shearing  stress  is  4000  pounds  per  square  inch?     What  is 
the  horse  power  transmitted?  Ans.  883  pounds;  85  horse  power. 

8.  A  key  f  inch  wide  unites  a  6-inch  shaft  to  a  flywheel.  What  must  be 
the  length  of  this  key  and  the  length  of  the  hub  of  the  wheel  if  the  shearing 
stress  in  the  key  is  equal  to  the  maximum  shearing  stress  in  the  shaft,  no 
allowance  being  made  for  friction  between  hub  and  shaft? 

Ans-.  18.8  inches. 

125.  Relation  of  Torque  to  Angle  of  Twist.  —  In  Fig.  159 
the  displacement  FF'  at  a  distance  r  from  the  axis  of  the  shaft 
is  r0,  where  0  is  the  angle  FOF'.  This  angle  6  is  the  displace- 
ment of  the  top  plane  with  reference  to  the  bottom.  The  unit 
displacement  is  the  total  displacement  divided  by  the  length  of 
the  portion: 

ffi 
Unit  displacement  =  y>  (1) 

where  I  is  the  length  of  the  portion  considered.     If  Ea  is  the 
modulus  of  elasticity  in  shear, 

Unit  shearing  stress  =  —  j—  •  (2) 

Resisting  moment  =  —  -,  —  d  A  .  (3) 

Total  resisting  is  the  integral  of   (3)  over  the  entire  section, 
remembering  that  0  is  constant  for  any  given  value  of  I: 


M 
t 


80       I        9    J    A  E80J  f      . 

-J  r2dA  =   -y-;  (4) 

0  =  ^4  •  Formula  XXIX.     (5) 


This  theory  applies  rigidly  to  circular  shafts  only.  For  other 
sections  the  deformations  and  stresses  are  not  exactly  propor- 
tional to  the  distance  from  the  axis. 

PROBLEMS. 

1.  A  4-inch  solid  steel  shaft  is  twisted  2  degrees  in  a  length  of  12  feet.     If 
the  modulus  of  shearing  elasticity  is  12,000,000,  what  is  the  torque? 

Ans.  73,100  inch  pounds. 

2.  In  Problem  1  what  is  the  unit  shearing  displacement  of  the  extreme 
fibers  and  what  is  the  unit  stress? 

Ans.  ss  =  5818  pounds  per  square  inch. 

This  problem  should  be  solved  from  the  geometry  of  the  shaft  without 
using  the  formulas  of  this  and  the  preceding  articles. 


TORSION  227 

3.  A  2-inch  solid  shaft  is  twisted  3  degrees  in  a  length  of  5  feet  by  a  force 
of  300  pounds  applied  at  the  ends  of  a  lever  48  inches  long.     Find  E8. 

4.  When  a  shaft  of  radius  a  and  length  I  is  twisted  an  angle  of  6  radians, 
show  from  Fig.  159,  without  integrating,  that 

=  E8ad 

and  from  this  result  by  means  of  Formula  XXVIII  derive  Formula  XXIX. 

126.  Relation  of  Torque  to  Work.  —  It  is  often  convenient 
to  use  the  relation  of  torque  to  foot  pounds  of  work  per  revolu- 
tion.    If  a  force  P  is  applied  at  the  end  of  an  arm  R  feet  in  length 
measured  from  the  axis  of  the  shaft,  the  work  per  revolution 
is  2  irRP  foot  pounds,  since  the  force  P  displaces  its  point  of 
application  a  distance  2  irR.     But  since  PR  is  the  torque,  the 
work  per  revolution  is  2  irM t. 

In  all  problems  involving  work  done  by  a  shaft,  solve  first  for 
the  torque.  When  the  torque  is  obtained  in  a  numerical  or 
literal  equation,  it  may  be  used  in  Formulas  XXVIII  or  XXIX. 

PROBLEMS. 

1 .  A  shaft  transmits  300  horse  power  at  200  revolutions  per  minute.     What 
is  the  torque  in  foot  pounds?  Ans.  7875  foot  pounds. 

2.  In  Problem  1  what  must  be  the  diameter  of  the  shaft  if  the  maximum 
unit  shearing  stress  is  4000  pounds  per  square  inch?  Ans.  4.93  inches. 

3.  Find  the  diameter  of  a  solid  shaft  to  transmit  1200  horse  power  at  150 
revolutions  per  minute  with  a  maximum  shearing  stress  of  6000  pounds  per 
square  inch.  Ans.  7.54  inches. 

4.  A  shaft  coupling  is  made  of  two  disks  fastened  together  by  six  1-inch 
bolts.     The  axis  of  each  bolt  is  8  inches  from  the  axis  of  the  shaft.     What 
horse  power  will  the  coupling  transmit  if  the  allowable  unit  shearing  stress  in 
the  bolts  is  9000  pounds  per  square  inch? 

Ans.  323  times  the  number  of  revolutions  per  second. 

127.  Torsion  Combined  with  Bending  or  Tension.  —  In  beams 
the  maximum  bending  stress  exists  in  the  upper  and  lower  fibers, 
while  the  maximum  shearing  stress  is  at  the  neutral  surface, 
so  that  the  resultant  is  seldom  much  greater  than  the  maximum 
values  of  the  separate  stresses.     On  the  other  hand,  in  shafting 
subjected  to  bending  and  twisting,  the  maximum  shearing  stress 
is  in  the  outer  fibers,  while  the  maximum  tensile  stress  due  to 
bending  comes  in  some  of  these  same  fibers,  so  that  the  resultant 
is  much  larger  than  either  stress.     A  similar  condition  exists 
when  torsion  is  combined  with  direct  tension  or  compression. 


228 


STRENGTH  OF  MATERIALS 


PROBLEMS. 

1.  A  1-inch  round  rod  projects  from  a  vise.  A  wrench  is  applied  to  the 
rod  8  inches  from  the  vise,  and  a  force  of  60  pounds  is  applied  to  the  wrench 
12  inches  from  the  axis  of  the  rod.  Find  the  unit  tensile  stress  due  to  bending 
at  the  vise  (Fig.  160).  Find  the  unit  shearing  stress  due  to  twisting  in  all 
the  outer  fibers. 

3840  X  4  .     , 

=  4889  pounds  per  square  inch; 


Ans. 


7T 

3840  X  3 


ss  =  —  —  =  3667  pounds  per  square  inch. 

2.   In  Problem  1  find  the  maximum  resulting  tensile  and  shearing  stress. 

Ans.  sa'  =  1222  V32  +  22  =  4407  pounds  per  square  inch; 

at   =  2444  +  4407  =  6851  pounds  per  square  inch. 

Observe  since  the  section  modulus  used  in  torsion  is  twice  that  used  in 
bending  and  the  load  is  the  same  for  both,  there  is  always  a  large  common 

factor  which  may  be  taken  out  to 
reduce  the  work. 

3.  If  the  rod  in  Problem  1  were 
twisted  by  two  wrenches  extending  in 
opposite  directions  with  equal  forces 
of  30  pounds  on  each,  what  would  be 
the  maximum  resultant  stress  ? 

4.  A   3-inch    solid   shaft    is    sup- 
ported on  bearings  4  feet  apart.     It 
carries  a  3-foot  pulley  1  foot  from  the 
left  bearing  and  a  2-foot  pulley  1  foot 
from  the  right  bearing.      The  3-foot 
pulley  weighs  200  pounds   and  the 
2-foot  pulley  weighs  150  pounds.    The 
belts  run  vertically  downward.     The 
tension  on  the  front  on  the  3-foot 
pulley  is  500  pounds  and  on  the  back 

120  pounds.  The  tension  on  the  front  on  the  2-foot  pulley  is  160  pounds. 
Find  the  tension  on  the  back  on  the  2-foot  pulley  and  the  horse  power  trans- 
mitted at  300  revolutions  per  minute,  assuming  that  there  is  no  friction 

Ans.  32.6  horse  power. 

5.  In  Problem  4  find  the  maximum  resultant  shearing  and  tensile  stress 
at  the  middle  of  the  span,  neglecting  the  weight  of  the  shaft. 

Ans.  2469  pounds  per  square  inch;  4574  pounds  per  square  inch. 

6.  In  Problem  4  find  the  resultant  tensile  and  shearing  stress  18  inches 
from  the  left  bearing. 

7.  Solve  Problem  5  if  the  belts  on  the  3-foot  pulley  run  horizontally  back- 
wards, all  other  factors  remaining  the  same. 

8.  A  4-inch  vertical  shaft  is  subjected  to  a  direct  compression  of  6000 
pounds  and  a  twisting  moment  due  to  a  force  of  600  pounds  at  the  end  of  a 
4-foot  lever.     What  is  the  resultant  compressive  stress  in  the  outer  fibers? 

Ans.  2543  pounds  per  square  inch. 

128.   Helical  Springs.  —  An  interesting  example  of  torsion  is 
the  stretching  or  compression  of  a  helical  spring,  such  as  is  shown 


Fig.  160.  —  Torsion  and  Bending. 


TORSION 


229 


in  Fig.  161.     A  helical  spring  is  made  by  winding  a  wire  or  rod 
on  a  cylinder  (in  a  single  layer,  usually).     The  radius  of  the  coil 
of  the  spring  is  the  sum  of  the  radii  of  wire  and  the  cylinder 
about  which  it  is  wound.     The  ends  of  the  wire, 
when  the  spring  is  to  be  used  in  tension,  are  turned 
in  to  the  center  so  as  to  apply  the  force  in  the  line  of 
the  axis. 

The  greater  part  of  the  elongation  of  a  helical 
spring  is  due  to  torsion.  If  we  consider  a  section 
at  0,  Fig.  161,  II,  we  find  that  there  is  a  twisting 
moment  PR  due  to  the  load  P  at  the  axis.  (Fig. 
161,  II,  is  a  plan  of  the  lower  turn;  the  force  P  is 
normal  to  the  plane  of  the  paper.)  The  point  C  at 
which  the  load  P  is  applied  being  at  the  center  of 
the  circle,  it  lies  in  the  plane  of  the  section  and 
therefore  there  is  no  bending  moment.  The  effect 
of  a  force  acting  on  an  arm  CBO  is  independent 
of  the  form  of  the  arm.  As  far  as  concerns  the 
stresses  at  the  section,  CBO  might  be  a  straight  rod 
from  C  to  0.  The  point  0  is  any  point  in  the  spring 
except  the  portion  CB  and  the  similar  portion  at 
the  top.  With  these  exceptions  the  entire  spring 
is  subjected  to  a  twisting  moment  PR.  In  addi- 
tion to  this  torsion,  there  is  a  constant  total  shear  P.  jj^^j  g  ^rig 
Also,  since  the  turns  of  the  coils  are  not  exactly 
horizontal,  there  is  another  slight  correction.  Both  of  these  are 
neglected  in  ordinary  calculations. 

To  get  the  elongation  of  a  helical  spring  due  to  a  given  load, 
multiply  the  angle  of  twist  in  the  entire  spring  by  the  radius  R. 


PROBLEMS. 

1.  A  rod  0.4  inch  in  diameter  is  used  to  make  a  helical  spring  of  20  turns. 
The  radius  from  axis  of  coil  to  center  of  all  sections  is  2  inches.      If  Es  is 
12,000,000,  find  the  elongation  due  to  a  load  of  10  pounds.  Ans.  f  inch. 

2.  In  Problem  1  what  is  the  unit  shearing  stress  in  the  outer  fibers? 

Ans.  1591  pounds  per  square  inch. 

3.  If  the  same  length  of  rod  were  used  to  make  40  turns  of  1-inch  radius, 
what  would  be  the  elongation  due  to  a  load  of  10  pounds?          Ans.  ^  inch. 

4.  What  is  the  unit  stress  in  Problem  3? 

5.  A  wire  0.1  inch  in  diameter  is  bent  into  40  turns  of  a  helical  spring  of 
0.5-inch  radius.     If  E8  is  12,000,000,  how  much  will  a  load  of  1  pound  elongate 
this  spring?  Ans.  •£$  inch. 


230  STRENGTH  OF  MATERIALS 

6.  How  many  turns  of  wire  are  required  to  make  a  spring  similar  to  that 
of  Problem  5  which  a  load  of  1  pound  will  elongate  1  inch? 

7.  If  R  is  the  radius  of  the  helix,  r  the  radius  of  the  rod,  P  the  load,  Es 
the  modulus  of  elasticity  in  shear,  and  n  the  number  of  turns,  prove  that  the 
elongation  is 


8.  At  Watertown  Arsenal,  a  steel  rod  1.24  inches  in  diameter  and  about 
241  inches  long  was  formed  into  a  helical  spring  7.64  inches  outside  diameter. 
A  load  of  5000  pounds  shortened  this  spring  4.64  inches.     Find  the  modulus 
of  shearing  elasticity.  Ans.  11,460,000. 

9.  Derive  an  expression  for  the  elongation  of  a  helical  spring  which  shall 
contain  the  total  length  of  the  rod  or  wire  instead  of  the  number  of  turns. 


CHAPTER  XV. 
RESILIENCE  IN  BENDING  AND  TORSION. 

129.   Resilience  in  Beams.  —  In  Article  12  we  learned  that  the 

s2 
elastic  resilience  per  cubic  inch  is  =-=,  and  that  the  total  energy 

L  til 

is  this  quantity  multiplied  by  the  volume.  In  beams  the  same 
relation  holds,  but  the  stress  varies  from  the  neutral  axis  to  the 
outer  fibers,  and  also  varies  with  the  moment  from  one  end  to  the 
other. 

We  may  calculate  the  total  energy  in  either  of  two  ways:  we 
may  find  the  total  work  done  by  the  external  forces,  or  we  may 


^ 

55i 

^\^1 

\ 

^H^ 

•dx 

\ 

"  J-xx 

derive  an  expression  for  the  internal  work  in  each  increment  of 
volume  and  integrate  over  the  entire  volume  of  the  beam. 

By  external  work,  if  a  load  P  causes  a  deflection  i/max  at  its 


point  of  application,  the  work  is 


(Fig.  162,  II).     In  a  beam 


with  a  single  concentrated  load,  if  we  find  the  deflection  under 
the  load  for  any  particular  case,  the  total  work  is  easily  cal- 
culated. In  a  cantilever  with  a  load  on  the  end,  the  deflection  at 

the  end  is 

PI* 
2/max-  3EI> 

231 


232  STRENGTH  OF  MATERIALS 

and  the  work  done  by  the  load  P  is 

P273 

Work^-  (1) 

Where  the  load  is  uniformly  distributed,  w  pounds  per  unit 
length,  each  increment  of  load  wdx  on  a  length  dx  does  work 

amounting  to     ~     >  where  y  is  the  deflection  of  the  particular 

part  of  the  beam  on  which  the  increment  rests.     In  Fig.  163,  II, 
one  increment  wdx  is  deflected  a  distance  yit  another,  y2,  etc. 

% 

w  Pounds  Per  Unit  Length 


The  different  values  of  y  are  determined  from  the  equation  of  the 
elastic  line.  The  total  work  is  the  sum  of  these  increments  of 
work. 

Total  work 


=  |  fydx,  (2) 


with  the  ends  of  the  beam  as  the  limits. 

130.  Expression  for  Internal  Work.  —  In  a  beam  the  unit 

stress  at  a  distance  v  from  the  neutral  axis  is  -j--    In  Fig.  162,  1, 

there  is  an  element  of  volume  of  cross  section  dA  and  length  dx 
at  a  distance  v  from  the  neutral  axis.  The  energy  dW  in  this 
element  of  volume  dAdx  is 


dW  =  ~  dAdx  =  dAdx.  (1) 


r  r  MZ 

=    /    /  ^-^j^v2dAdx.  (2) 


Total  work  in  beam 


Integrating  first  with  respect  to  v  gives  the  work  done  upon  the 
volume  of  length  dx  between  two  vertical  planes.     Throughout 


RESILIENCE  IN  BENDING   AND   TORSION     233 

this  volume  x,  M,  and  /  are  constant.     The  integral  of  v2dA  across 
the  beam  from  the  bottom  to  the  top  is  /. 


/M2 
f^jdx.  (3) 


Equation  (3)  enables  us  to  calculate  the  total  work  for  any  beam. 
M  and  /  are  expressed  in  terms  of  x  before  integrating. 

131.  Cantilever  with  Uniformly  Distributed  Load.  —  The  mo- 
ment is  -=-,  if  the  origin  is  taken  at  the  free  end. 

z 

Total  work  = 
For  a  beam  of  constant  cross  section  for  which  I  is  constant, 

Total  work  =  *%  p>T  =  40  El'  ® 

In  a  rectangular  section  the  maximum  fiber  stress  at  the  wall  is 
given  by 

Wl      2  si 


Substituting  in  (2)  : 

Work-48'" 
~ 


Work  per  unit  volume  =  ^-^-  (4) 

oU  Hi 

The  total  energy  in  a  cantilever  of  uniform  rectangular  section 
is  only  one-fifteenth  as  much  as  that  in  a  block  of  the  same  volume 
with  constant  compressive  stress  throughout. 


PROBLEMS. 

1.  Find  the  expression  for  the  work  per  unit  volume  in  a  solid  circular 
cantilever  with  a  uniformly  distributed  load,  in  terms  of  the  maximum  fiber 

s2 

stress.  Arts.    .n  _.> 

40  h 

2.  Find  the  total  work  in  a  cantilever  of  uniform  section  by  means  of  the 
external  work,  using  the  expression  of  Article  129. 


234 


STRENGTH  OF  MATERIALS 


3.  By  the  method  of  internal  work  find  the  total  work  done  by  a  load  at 
the  end  of  a  cantilever.     Compare  the  result  with  equation  (1),  Article  129. 

4.  Find  the  work  per  unit  volume  in  a  cantilever  with  a  load  on  the  end, 

s2 
if  the  section  is  rectangular.  Ans.  r^-^,  - 

lo  tii 

5.  Find  the  energy  per  unit  volume  in  a  cantilever  of  uniform  circular  sec- 

2 
tion  with  load  on  the  end.  Ans. 


6.  Show  that  the  energy  per  unit  volume  in  the  case  of  a  beam  supported 
at  the  ends  with  a  load  at  the  middle  is  the  same  as  that  of  a  cantilever  with  a 
load  on  the  end. 

7.  Find  the  work  per  unit  volume  in  a  beam  of  rectangular  section  sup- 
ported at  the  ends  and  uniformly  loaded. 


Ans.  Total  work  = 


for  any  section; 


240  #/ 
Work  per  unit  volume  for  rectangular  section  =  4 


132.  Beams  of  Variable  Section.  —  In  beams  of  variable  sec- 
tion the  moment  of  inertia  is  a  variable.  Take  the  case  of  a  can- 
tilever beam  of  constant  strength 
and  constant  depth  d  with  a  load 
on  the  free  end.  If  B  is  the 
breadth  at  the  wall,  Fig.  164,  the 
breadth  at  a  distance  x  from  the 


Fig.  164. 


,  ,  .    Ex 

free  end  is  -=-• 

Bxd* 
121  ' 


M  =  Px 
Total  work 


-T 

Jo 


Since 


EBd* 
=  sBd2, 


, 
dx  = 


EBd* 


s2Bdl       s2 
Total  work  =  =        volume. 


(D 


(2) 


(3) 


PROBLEMS. 

1.  From  equation  (2)  find  the  deflection  at  the  end  of  a  cantilever  of  con- 
stant strength  and  constant  depth  with  a  load  at  the  end.  Compare  result 
with  Article  95. 


RESILIENCE  IN   BENDING  AND    TORSION     235 

2.   Find  the  total  resilience  and  the  energy  per  unit  volume  in  a  cantilever 
of  constant  breadth  and  constant  strength  with  a  load  on  the  free  end. 


Ans.  Total  work  = 


Work  per  unit  volume  =  r^  . 

6  .Cr 

3.  By  means  of  the  external  work  and  the  total  work  of  Problem  2,  find 
the  deflection  at  the  end  of  a  cantilever  of  constant  strength  and  constant 
breadth  due  to  a  load  at  the  end. 


i 

'• 


133*   Leaf  Springs.  —  The  common  leaf  springs  as  used  in 
vehicles  are  beams  of  constant  strength  made  of  several  parts  or 
leaves.     Fig.   165,    II,   shows  a 
spring  of  this  kind.     Fig.  165,  I, 
represents    the    leaves    straight 
with   a  little   distance   between 
them.    In  the  upper  beam  the 
moment  is  constant  from  A  to 
B.     In  the  second  beam  it  is 
constant  from  C  to  D. 

With  a  constant  moment  M 
the  energy  is 


M' 


MH 


if  I  is  the  length  from  the  points  of  support  (A  and  B  for  the  top 
leaf). 

2 


for  a  rectangular  section. 

Since  bdl  is  the  volume  between  supports,  the  energy  for  the 


portion  having  constant  moment  is  £-=  per  unit  volume. 

b  £j 


This 


is  the  same  as  in  the  rectangular  beams  of  constant  strength  in 
the  preceding  article. 

The  portions  of  the  top  leaf  to  the  left  of  A  and  to  the  right  of 
B  are  cantilevers  with  load  on  the  end,  and  store  only  one-third 
as  much  energy  per  cubic  inch  as  the  portion  between  A  and  B. 
In  the  actual  leaf  spring  as  shown  by  Fig.  165,  II,  the  contact 


236  STRENGTH  OF  MATERIALS 

takes  place  over  a  considerable  area  and  the  stresses  are  some- 
what modified  by  the  friction. 

134.  Resilience  in  Torsion.  —  The  work  done  by  a  force  P  at 
the  end  of  an  arm  of  length  R  when  the  arm  is  turned  through  an 
angle  0  is 

PR8      Mt6 
~2~      -2" 

provided  the  force  is  applied  gradually. 

If  we  substitute  in  (1)  the  values  of  M  t  and  6  from  Formulas 
XXVIII  and  XXIX,  we  get  for  a  rod  of  length  I  and  radius  r, 

s  ZJI         s 
Total  work  of  torsion  =     *     ,  =  -r^-  volume.  (2) 


The  modulus  of  elasticity  in  shear  is  about  two-fifths  as  great  as 
in  tension  or  compression,  so  that  for  the  same  value  of  the  unit 
stress  the  total  energy  of  a  rod  in  torsion  is  one-fourth  greater 
than  that  of  the  same  rod  in  tension.  However,  since  the  elastic 
limit  of  steel  and  similar  materials  is  less  in  shear  than  in  ten- 
sion, the  total  energy  which  can  be  stored  is  about  the  same  in 
both  cases. 

Since  torsion  gives  so  much  more  energy  than  bending,  it  is  to 
be  preferred  in  the  design  of  springs.  A  helical  spring  in  which 
the  stresses  are  chiefly  torsional  is  the  most  convenient  form  to 
take  up  energy. 

PROBLEMS. 

1.  In  Problem  8  of  Article  128  find  the  total  work  in  compressing  the  spring, 
and  the  energy  per  cubic  inch  and  per  pound. 

Ans.  Total  energy,  966.7  foot  pounds. 

2.  A  spring  at  the  Watertown  Arsenal  was  made  of  36  pounds  of  steel  rod 
1.02  inches  in  diameter.     The  outside  diameter  of  the  coil  was  4.30  inches. 
A  load  of  11,000  pounds  changed  the  length  of  this  spring  from  20.63  inches  to 
16.67  inches.     After  the  load  was  removed  the  spring  returned  to  its  original 
length  to  within  0.02  inch.     Find  the  energy  per  cubic  inch  and  the  energy 
per  pound.  Ans.  50.4  foot  pounds  per  pound. 

3.  In  Problem  2  what  was  the  maximum  shearing  stress  due  to  torsion? 

Ans.  86,580  pounds  per  square  inch. 

135.  Sections  of  Maximum  Resilience.  —  To  obtain  the  maxi- 
mum resilience  per  unit  volume,  the  stress  in  all  portions  of  the 
solid  should  be  the  maximum  allowable  stress.  We  can  only 
secure  this  condition  of  perfect  efficiency  when  the  material  is 
used  in  direct  tension  or  compression,  which  is  not  practicable 


RESILIENCE  IN  BENDING  AND    TORSION     237 

in  the  construction  of  springs  on  account  of  the  small  displace- 
ment secured  and  the  large  force  required  (except  in  the  case  of 
soft  rubber). 

In  bending  and  torsion  only  the  outer  fibers  are  subjected  to 
the  maximum  stress,  so  that  the  energy  per  unit  volume  is  always 

s2 
less  than  -^  • 

In  any  section  subjected  to  bending  the  unit  stress  =  kv,  and 
the  total  energy  in  a  portion  of  length  dx,  extending  from  the 
neutral  surface  to  the  outer  fibers  at  a  distance  vi  from  this 
surface,  is 

k2dx   £V-.-,-i 

For  a  rectangular  section  of  breadth  b  this  becomes 

mdx 


Since  kvi  =  s  and  bvi  dx  =  the  volume,  we  get 

s2 
Energy  per  unit  volume  =  -^=  -  • 

\)JtL 

In  a  solid  rectangular  section  used  as  a  beam  of  constant 
strength  or  as  a  beam  of  uniform  section  with  a  constant  moment, 
the  efficiency  as  a  spring  is  one-third. 

In  an  I-beam  section  a  relatively  large  portion  of  the  section 
is  in  the  flange,  where  the  unit  stress  approximates  the  maximum, 
so  that  the  energy  per  unit  volume  is  greater  than  in  the  rec- 
tangular section. 

In  a  circular  section  which  is  twisted  an  angle  8  in  a  length  I 

/j 

the  unit  deformation  at  a  distance  r  from  the  axis  is  y ,  and  the 
shearing  force  on  a  circular  element  of  radius  r  and  thickness  dr 


is     w    ^r  —  t  .     The  energy,  being  one-half  the  product  of  the 
force  by  the  deformation,  is 


8,  8     (a*  -  V) 

Total  energy  =  -  -  ,  (3) 


238  STRENGTH  OF  MATERIALS 

where  b  is  the  inside  radius  and  a  is  the  outside  radius.     At  the 

77f         f\ 

outer  fibers  ss  =  —  y  —  ,  which  substituted  in  (3)  gives 
i 

s87rl  (a2  -  fc2)  (a2  +  62)       (a2  +  62)  si  . 
Total  work  =  L__f  x  volume.    (4) 


PROBLEMS. 

1.  Show  that  (4)  reduces  to  (2)  of  Article  134  when  6  =  0. 

2.  What  is  the  energy  per  unit  volume  in  a  hollow  cylinder  whose  inside 

diameter  is  one-half  its  outside  diameter?  Ans.    -  ^r- 

ID  AS 

3.  A  hollow  rectangular  beam  is  6  inches  by  8  inches  outside,  and  4  inches 
by  4  inches  inside.    Find  its  energy  per  unit  volume  if  the  external  moment 

11s2 
is  constant  throughout  its  length.  Ans. 


CHAPTER  XVI. 
CENTER  OF  GRAVITY. 

136.  Center  of  Gravity.  —  When  each  of  the  particles  which 
compose  a  body  or  system  of  bodies  is  subjected  to  a  force  which 
is  proportional  in  magnitude  to  the  mass  of  the  particle  and 
parallel  to  the  similar  forces  in  every  other  particle,  the  line  of 
application  of  the  resultant  of  these  forces  passes  through  the 
center  of  gravity  of  the  body  or  system. 

The  location  of  the  center  of  gravity  is  determined  from  the 
intersection  of  two  such  resultants. 

Fig.  166  represents  three  particles  of  relative  masses  2,  3, 
and  4,  united  by  weightless  rods  to  form  a  single  body.  In 


£    c 


Fig.  166,  I,  these  particles  are  subjected  to  forces  directed  ver- 
tically downwards.  The  resultant  of  these  forces  is  a  force  of 
9  units  along  the  line  CD.  The  center  of  gravity  is  located  at 
some  point  on  this  line.  In  Fig.  166,  II,  the  forces  are  horizontal, 
and  their  resultant  is  a  horizontal  force  of  9  units  along  the 
line  EF.  The  point  0  at  the  .intersection  of  EF  with  CD  is  the 
center  of  gravity. 

The  center  of  gravity  is  also  called  the  center  of  mass. 

137.  Determination  of  the  Center  of  Gravity  by  Balancing.  - 
The  force  with  which  the  earth  attracts  the  particles  of  a  body 
is  proportional  *  to  the  mass  of  each  particle.     These  forces  are 

*  There  is  a  difference  in  the  attraction  of  the  earth  due  to  difference  in 
the  distance  of  the  various  particles  from  the  center  of  the  earth  amounting 

239 


240 


STRENGTH  OF  MATERIALS 


directed  toward  the  center  of  the  earth,  so  that  for  bodies  of 
ordinary  dimensions  they  may  be  regarded  as  parallel,  within 
the  limits  of  accuracy  of  our  measurement.  The  resultant 
force  of  gravity  on  any  body  passes  through  the  center  of  gravity. 
A  body  may  be  held  in  equilibrium  by  a  single  force  provided 
that  force  is  along  the  line  of  the  resultant  of  all  the  other  forces. 
When  a  body  is  supported  by  a  flexible 
cord  or  by  a  point  about  which  it  is  free  to 
turn  without  friction,  the  center  of  gravity 
must  be  on  the  vertical  line  through  the 
point  of  application  of  the  cord  or  point 
(provided,  of  course,  no  forces  are  acting 
except  gravity  and  the  cord  or  point). 

Fig.  167  shows  the  same  body  as  Fig.  166. 
In  Fig.  167,  I,  it  is  supported  at  C  by  a 
cord.  A  plumb  line  let  fall  from  C  passes 
through  the  center  of  gravity.  In  Fig.  167, 
II,  it  is  supported  on  a  point  or  knife  edge 
at  E,  and  turns  under  the  action  of  gravity 
until  its  center  of  gravity  comes  directly 
below  the  point  of  support.  The  intersec- 
tion  of  the  plumb  line  from  E  with  the  line 
CD  (previously  marked  in  any  convenient  way)  gives  the  center 
of  gravity  0. 

This  method  of  finding  the  location  of  the  center  of  gravity 
is  of  little  practical  use,  owing  to  the  fact  that  the  point  to  be 

found   is   usually  surrounded   by  solid 

material,  making  it  necessary  to  find  c  — —fir — 

the  intersection  of  three  planes  instead  _ 

£  , ,       .    ,  ,.          £,         r  T,  .     Fig.  168.  — Center  of  Gravity 

of  the  intersection  of  two  lines.     It  is  b    Balancing 

useful  in  relatively  long  bodies,  espe- 
cially  if   there   are   some   plane   surfaces   to   use    as   planes  of 
reference.     Fig.    168   represents   a   beam   balanced  on   a   knife 
edge.     The  center  of  gravity  is  in  the  vertical  plane  of  the  knife 
edge. 

138.  Center  of  Gravity  by  Moments.  —  In  theoretical  dis- 
cussions the  center  of  gravity  is  usually  located  by  moments. 
The  plane  of  application  of  the  resultant  of  any  set  of  forces 


to  about  one  part  in  ten  million  for  a  difference  of  one  foot.     This  is  negligible 
for  ordinary  bodies.     It  would  not  be  negligible  in  the  case  of  a  mountain. 


CENTER   OF   GRAVITY 


241 


can  be  determined  by  dividing  the  sum  of  the  moments  of  all  the 
forces  with  respect  to  any  axis  by  the  magnitude  of  the  resultant 
force.  This  comes  from  the  proposition  of  Mechanics  that  the 
sum  of  the  moments  of  any  set  of  forces  with  respect  to  any 
axis  is  equal  to  the  moment  of  their  resultant  with  respect  to 
that  axis. 

Fig.  169  represents  a  body  made  up  of  four  particles  of  masses 
mi,  m2,  etc.,  which  are  not  all  in  one  plane.  The  X  axis  is  taken 
in  the  usual  way;  the  Y  axis  is  vertical  as  in  the  analytics  of  two 
dimensions;  and  the  Z  axis  is  horizontal  with  the  positive  direc- 


Fig.  169.  —  Center  of  Gravity  by  Moments. 

tion  toward  the  front.  In  Fig.  169,  I,  the  coordinates  of  the 
particles  in  the  direction  of  the  Z  axis  are  drawn  in  the  YZ 
plane;  the  Y  coordinates  are  in  the  Y  axis;  the  X  coordinates  are 
perpendiculars  let  fall  from  the  points  mi,  w2,  etc.,  upon  the 
YZ  plane. 

We  will  first  take  moments  about  the  Z  axis,  and  we  will  con- 
sider that  the  letters  mi,  w2,  etc.,  represent  relative  masses  as 
well  as  the  positions  of  the  particles.  If  the  forces  which  act 
on  these  particles  are  vertically  downward,  the  moment  arm  of 
each  force  is  the  x  of  the  particle.  In  the  case  of  w4,  for  instance, 
the  force  is  directed  toward  the  point  E±  in  the  XZ  plane;  its 
moment  arm  is  the  line  F4E^  which  is  equal  in  length  to  z4.  The 
total  moment  of  all  these  vertical  forces  is  given  by 

Total  moment  =  miXi  -f  m2x2  +  m3xs  +  ra4z4.  (1) 

The  total  force  is  the  sum  of  the  separate  forces,  since  they 
are  all  parallel. 


242  STRENGTH  OF  MATERIALS 

The  coordinate  of  the  center  of  gravity,  represented  by  x,  is 

.  . 
i         2         3         4 

Since  the  forces  are  parallel  to  the  YZ  plane,  if  we  used  any 
axis  in  that  plane  parallel  to  the  Z  axis,  the  moment  arms  would 
remain  the  same,  consequently  we  are  accustomed  to  speak  of 
the  moments  in  (1)  as  the  moments  with  respect  to  the  YZ  plane. 

Equation  (2)  is  the  equation  of  a  plane  normal  to  the  X  axis, 
which  passes  through  the  center  of  gravity.  To  locate  the  point 
we  must  get  the  equations  of  two  other  planes. 

If  we  were  endeavoring  to  measure  y  (the  distance  of  the 
center  of  gravity  above  the  XZ  plane)  experimentally,  it  would 
be  necessary  to  rotate  the  entire  system  90  degrees,  so  as  to  make 
the  Y  axis  horizontal  and  give  an  effective  moment  arm  to  the 
forces.  In  a  theoretical  calculation,  we  may  imagine  the  forces 
turned  90  degrees.  In  Fig.  169,  II,  the  force  upon  nil  is  drawn 
parallel  to  the  X  axis  and  we  will  suppose  that  all  the  other  forces 
are  in  the  same  direction.  The  moment  arm  of  the  force  nil 
with  respect  to  the  Z  axis  is  the  line  m\E\  which  is  y\.  Instead 
of  using  the  Z  axis  we  might  use  any  line  in  the  XZ  plane  parallel 
to  the  Z  axis  as  the  axis  of  moments.  It  is  not  necessary  that 
the  forces  be  parallel  to  the  X  axis  as  they  may  be  parallel  to 
the  Z  axis  shown  in  Fig.  169,  II,  at  w2,  in  which  case  the  axis  of 
moments  is  some  line  in  the  XZ  plane  parallel  to  the  X  axis; 
or  the  forces  may  be  in  any  direction  parallel  to  the  XZ  plane. 
When  we  think  of  a  moment  with  respect  to  a  plane  we  do  not 
generally  consider  the  direction  of  the  forces,  but  concern  our- 
selves only  with  their  magnitude  and  the  distance  from  the  plane 
of  reference. 


-  = 

mi  -f-  W2  +  W3  +  W4 

A  similar  expression  gives  z,  the  moments  being  taken  with  re- 
spect to  the  XY  plane. 

To  get  the  distance  of  the  center  of  gravity  of  a  body  from  any 
plane,  divide  the  sum  of  the  moments  of  all  the  particles  which 
compose  the  body  with  respect  to  the  plane,  by  the  total  mass 
of  the  body. 


CENTER   OF   GRAVITY  243 

PROBLEMS. 
1.  A  body  is  composed  of  three  particles  in  the  same  plane 

Mass  x  y 

3  57 

2  48 

5  36 

Find  x  and  y. 

_  3  X5  +  2X4  +  5X3      38 

=       =  3<8' 


10 


3  X  7  =  21 
2  X  8  =  16 
5  X  6  =  30 


Wy     =67 
y     =6.7 

The  second  form  of  solution  is  preferable,  especially  when  the  numbers  are 
large.  It  is  still  better  to  arrange  the  data  and  results  in  a  single  table  omitting 
the  multiplication  and  equality  signs. 

2.   A  body  is  composed  of  four  particles  of  the  following  masses  and  coordi- 
nates. 

mx  my  mz 


Mass 

x 

y 

z 

3 

5 

4 

3 

5 

2 

6 

5 

6 

4 

-2 

8 

2 

3 

9 

4 

Ans.  y  =  3,  etc. 

139.  Center  of  Gravity  of  Continuous  Bodies.  —  When  a 
body  is  made  up  of  a  great  number  of  particles  m\,  m^  etc.,  at 
distances  x\,  xz  from  the  YZ  plane,  the  expression  for  the  moment 
with  respect  to  the  plane  is  written  2mx,  and  the  position  of 
the  center  of  gravity  is  given  by 


,  /1  , 

*=  ZST    -M- 

When  the  number  of  particles  is  indefinitely  great  and  their  mag- 
nitudes indefinitely  small  we  write 


Formula  XXX. 
M 


244 


STRENGTH  OF  MATERIALS 


If  dV  is  an  element  of  volume  and  p  is  the  density,  dM  =  p  dV. 
If  p  is  constant,  Formula  XXX  becomes 


p  CxdV       CxdV       Cx 


V 


(3) 


In  these  expressions,  x  is  the  distance  of  the  center  of  gravity 
of  the  element  of  mass  from  the  YZ  plane.  If  the  dimension  of 
the  element  of  volume  parallel  to  the  X  axis  is  infinitesimal  (dx) 
and  the  origin  of  moments  is  taken  in  the  YZ  plane,  the  "x" 
in  the  expressions  above  is  the  x  of  the  Cartesian  coordinates. 
Where  the  element  of  volume  has  a  finite  length  in  the  direction 
parallel  to  the  X  axis,  this  is  not  the  case,  as  will  be  seen  in  some 
of  the  examples. 

Y 


in 


Fig.  170.  —  Center  of  Gravity  of  a  Prism. 

PROBLEMS. 

1.   Find  the  center  of  gravity  of  a  rod  of  uniform  section  and  density,  using 
the  plane  of  one  end  as  the  YZ  plane. 

If  A  is  the  area  of  any  cross  section  parallel  to  the  ends,  Fig.  170,  I, 
dV  =  A  dx, 


xdx 


fdx 


A|2_l.       T 

17 


CENTER   OF   GRAVITY 

The  constant  term  A  might  have  been  canceled  before  integrating;  but  it 
is  best  to  retain  all  such  constants  until  the  limits  are  put  into  the  integrals. 
The  denominator  of  the  fraction  is  the  volume  of  the  solid,  which  we  know  in 
many  cases,  so  that  if  no  constant  terms  are  dropped  we  are  enabled  to  check 
that  much  of  our  work. 

Fig.  170,  1,  is  a  rectangular  parallelepiped,  but  there  is  nothing  in  the  work 
which  limits  us  to  a  rectangular  section.  The  results  apply  to  right  prisms 
of  any  section  whatever  (Fig.  170,  II).  Nor  are  we  limited  to  right  prisms. 
Fig.  170,  III,  shows  an  oblique  prism  with  one  end  in  the  YZ  plane  and  the 
other  end  parallel  to  this  plane  and  with  all  sections  alike.  The  center  of 
gravity  of  this  prism  lies  in  a  vertical  plane  midway  between  the  planes  of  the 
ends. 

2.  Solve  Problem  1  with  the  XZ  plane  (and  the  origin  of  coordinates)  at 
one-fourth  the  length  of  the  prism  from  the  left  end. 

140.  Center  of  Gravity  of  Plane  Areas.  —  A  plane  area  may 
be  regarded  as  a  plate  of  uniform  thickness  t  and  uniform  density. 
If  the  faces  of  this  plate  are  parallel  to  the  XY  plane 

t  CxdA       CxdA 


CxdA       C 

- 


where  dA  is  an  element  of  area  and  A  is  the  entire  area  of  the 
surface. 

(yd  A 

--.-./_ (2) 

dA 


r 

f 


PROBLEM. 

1.  Find  the  center  of  gravity  of  a  triangular  area,  of  base  6  and  altitude  h, 
by  integration.  We  will  take  the  origin  at  the  vertex  C  and  so  place  the  base 
ED  of  length  b  that  it  shall  be  parallel  to  the  Y  axis.  The  element  of  area 
is  a  strip  of  width  dx.  From  similar  triangles,  the  length  of  this  strip  FF'  is 
bx 
h  ' 

«4>5*j 

w 

z  =  FF7  =  ^f=i'  (3) 

hJXX      ftl_2jo         2 
2-|*. 

We  recognize  the  denominator  of  the  last  term  of  (3)  as  the  area  of  the  triangle, 
which  shows  that  our  increment  of  area  was  taken  correctly  and  the  proper 
limits  were  used. 


246 


STRENGTH  OF  MATERIALS 


The  center  of  gravity  of  a  triangular  area  falls  on  the  line  JK  parallel  to 
the  base  and  at  a  distance  from  the  base  equal  to  one-third  the  altitude. 

The  triangle  may  be  turned  and  the  origin  taken  at  E  or  D  and  another  line 
at  one-third  the  altitude  from  the  new  base  may  be  found. 

In  Fig.  171,  II,  the  line  CL,  drawn  from  the  vertex  C  to  the 
middle  point  of  the  base,  passes  through  the  center  of  gravity. 
If  we  divide  the  triangle  into  narrow  strips 
such  as  FF'  by  lines  parallel  to  the  base, 
this  median  line  bisects  each  of  these  strips. 
The  line  CL  passes  through  the  center  of 
gravity  of  each  strip  and  each  strip  could 
be  balanced  on  the  line  as  a  knife  edge 
support.  'This  is  also  true  of  median  lines 
from  D  and  E.  The  center  of  gravity  of  a 
triangular  area  is  at  the  intersection  of  the 
medians.  We  know  from  geometry  that 
the  intersection  of  the  medians  of  a  tri- 
angle is  two-thirds  the  length  of  any 
median  from  its  vertex.  The  center  of 
gravity  of  the  triangle  of  Fig.  171,  II,  is  at 
G  on  the  median  CL  at  a  distance  from 
L  equal  to  one-third  of  CL. 

If  any  plane  area  has  a  line  of  symme- 
try, the  area  could  be  balanced  on  this  line;  consequently  the 
line  of  symmetry  passes  through  the  center  of  gravity  of  the  area. 
If  a  solid  is  symmetrical  with  respect  to  a  plane,  this  plane  passes 
through  the  center  of  gravity  of  the  solid.  These  facts  enable 
us  to  locate  the  center  of  gravity  of  many  areas  and  volumes 
without  integrating. 

PROBLEMS. 

2.  A  triangle  of  base  6  inches  and  altitude  9  inches  has  its 
base  on  the  upper  edge  of  a  6-inch  square.      Knowing  the  loca- 
tion of  the  center  of  gravity  of  the  triangle  and  square,  find  the 
distance  of  the  center  of  gravity  of  the  combined  area  from  the 
base  of  the  triangle.     Prove  the  result  by  finding  the  distance 
from  the  base  of  the  square  used  as  the  origin. 

Ans.  y  =  j  inch  when  the  origin  is  on  ED  of  Fig.  172. 

3.  If  the  median  line  from  C  (Fig.  172)  makes  an  angle  of 
60  degrees  with  the  horizontal,  find  x,  using  two  origins,  and 
compare  the  results. 

4.  A  circular  area  4  inches  in  diameter  is  tangent  to  a  rectangle  4 


Fig.  171.  — Center  of 
Gravity  of  a  Trian- 
gular Area. 


CENTER   OF   GRAVITY 


247 


wide  and  5  inches  high  at  a  point  on  the  right  side  of  the  rectangle  2  inches 
above  the  lower  right  corner.     Locate  the  center  of  gravity. 

Ans.  x  =  3.54  inches  from  the  left  side  of  rectangle; 
y  =  ? 

5.  A  rectangular  board  12  inches  wide  and  16  inches  high  has  a  circular 
hole  cut  out.     The  center  of  the  hole  is  5  inches  from  the  right  edge  and 
5  inches  from  the  bottom.     Its  diameter  is  3  inches.     Locate  the  center  of 
gravity  of  the  remainder. 

This  is  easiest  solved  by  subtracting  from  the  moment  of  the  entire  board 
the  moment  of  the  area  cut  away. 

6.  Find  the  distance  from  the  Y  axis  of  the  center  of  gravity  of  the  plane 
area  bounded  by  the  X  axis,  the  parabola  y2  =  4  x,  and  the  ordinate  x  —  9 
(Fig   173,1). 


fdx 


dy 


II 


III  IV 

Fig.  173.  —  Center  of  Gravity  of  Area  Bounded  by  a  Parabola. 

The  element  of  area  is  ydx,  and  the  moment  arm  is  x  of  the  curve 


(1) 


Since  there  are  two  variables  in  the  integrals  of  (1),  we  must  eliminate  one 
of  these  by  substituting  its  value  in  terms  of  the  other  from  the  equation  of 
the  curve.  Solve  first  by  eliminating  y  and  integrating  between  the  proper 
limits  for  x;  then  solve  by  eliminating  x  and  dx  and  integrating  between  the 
proper  limits  for  y.  Compare  results.  The  result  should  be  greater  than 
one-half  of  9  and  should  be  less  than  6 .  Why? 

7.  Solve  Problem  6  for  y,  using  the  element  of  area  of  Fig.  173,  II.  The 
result  should  be  greater  than  2  and  less  than  3.  Why? 


248  STRENGTH  OF  MATERIALS 

8.  Solve  Problem  7  for  y,  using  the  element  of  area  of  Fig.  173,  1. 

9.  Solve  Problem  6  for  x,  using  the  horizontal  element  of  Fig.  173,  II. 

10.  Solve  Problem  6  for  x  by  double  integration,  using  the  element  of  area 
dx  dy.     Integrate  first  with  respect  to  y  and  compare  the  result  of  this  inte- 
gration with  the  original  integral  of  Problem  6. 

11.  Solve  Problem  6  for  x  by  double  integration,  integrating  first  with 
respect  to  x.     Compare  result  of  this  integration  with  the  integral  of  Prob- 
lem 9. 

12.  Solve  for  y  by  double  integration,  integrating  first  with  respect  to  y 
so  as  to  build  up  vertical  strips  extending  from  the  X  axis  to  the  curve  (Fig. 
173,  III).     Compare  the  result  of  this  first  integration  with  the  integral  of 
Problem  8. 

13.  Solve  for  y  by  double  integration,  integrating  first  with  respect  to  x 
(Fig.  173,  IV). 

14.  Find  x  of  the  area  entirely  bounded  by  the  curve  y2  =  4  x  and  the 
straight  line  3  y  =  2  x,  by  a  single  integration  with  element  of  area  parallel 
to  the  Y  axis.  Ans.  x  =  3.6. 

15.  Solve  Problem  14  by  subtracting  from  the  moment  and  area  of  Problem 
6  the  moment  and  area  of  the  triangle  bounded  by  OD  and  DC  and  the  straight 
line  from  0  to  C  in  Fig.  173,  1. 

16.  Find  x  of  the  area  bounded  by  the  Y  axis,  the  line  y  =  6,  the  hyperbola 
xy  =  12,  the  line  x  =  12,  and  the  X  axis,  using  a  vertical  strip  as  the  increment 
of  area. 


12+  fxydx       12  +  12  fdx  ,„ 

J  J  \z-\-L4\x\z 

=  "' 


_ 

x  = 


17.  Solve  Problem  16,  using  the  element  of  area  in  the  form  of  horizontal 
strips. 

18.  Find  x  of  a  60-degree  sector  of  a  circle  of  radius  a  with  the  X  axis  as 
one  of  the  bounding  lines  (Fig.  174).     Solve  by  polar  coordinates,  integrating 

first  with  respect  to  r.  (The  order  of  integration  is 
immaterial  in  this  case,  as  the  limits  of  one  variable 
are  independent  of  the  other  variable.  Where  this  is 
not  the  case,  integrate  first  with  respect  to  r.) 

V3  a      n  __, 
Ans.  x  =  -    -  =0.551  a. 

7T 

19.   Using  the  value  of  x  from  Problem  18,  find  y 
without  integrating. 

^  _  -  _  _  a 

Fig.  174.  —  Center  of  "»! 

Gravity  of  a  Sector        20.   Solve  Problem  18  for  x  if  the  sector  is  so  placed 
of  a  Circle.  that  the  X  axis  bisects  it.    Compare  with  results  of  the 

preceding  problems. 

21.  Find  the  center  of  gravity  of  segment  of  a  circle  of  radius  10  bounded 
on  one  side  by  a  straight  line  at  a  distance  5  from  the  center  of  the  circle. 
Solve  by  rectangular  coordinates,  using  strips  parallel  to  the  boundary  line  as 
increments  of  area.  Ant,  x  —  7.05. 


CENTER   OF   GRAVITY 

Using  only  the  half  above  the  X  axis  and  calling  the  radius  a: 
^fxydx_f(a*-x*)*xdx 
ydx        -a?  f  sin2  6d8 


a3\/3 


249 


V3 


The  independent  variable  is  changed  in  the  denominator  and  might  also  be 
changed  in  the  numerator.      Why  is  the  upper  limit  in  the  denominator  0 

and  not  J  ?    Explain  the  geometric  meaning  of  each  term  in  the  denominator. 
o 

22.   Solve  Problem  21  by  double  integration  with  polar  coordinates  (Fig. 
175,  II). 


r  ii 

Fig.  175.  —  Center  of  Gravity  of  a  Seg- 
ment of  a  Circle. 


Fig.  176.  —  An  Angle 
Section. 


23.  Find  the  center  of  gravity  of  a  semicircular  area  of  radius  a. 

Ans.  x=~  =  0.4244  a. 
Sir 

24.  A  regular  hexagon  is  bisected  by  a  line  joining  opposite  angles.     Find 
the  distance  of  the  center  of  gravity  of  the  half  hexagon  from  this  line,  without 
integrating. 

25.  Fig.  176  represents  a  6-inch  by  4-inch  by  1-inch  standard  angle.     Find 
the  distance  of  the  center  of  gravity  from  the  back  of  each  leg,  and  compare 
the  results  with  the  tables  in  Cambria. 


250 


STRENGTH  OF  MATERIALS 


26.  Fig.    177  represents   a  standard   10-inch   15-pound  channel  section. 
Find  the  distance  of  the  center  of  gravity  of  the  section  from  the  back  of  the 
web,  and  compare  with  Cambria  under  "  Properties  of  Standard  Channels." 

27.  Look  up  dimensions  of  a  12-inch  30-pound  channel  and  calculate  the 
area  and  the  location  of  the  center  of  gravity. 

28.  A  section  is  made  of  two  10-inch  15-pound  channels  and  one  12-inch 
by  ^-inch  plate.     How  far  is  the  center  of  gravity  of  the  section  from  the  center 
of  the  channels  (see  Fig.  178)? 


Fig.  177.  —  A  Channel  Section. 


Fig.  178.  —  A  Plate  and  Chan- 
nel Column  Section. 


29.  Find  the  center  of  gravity  of  an  area  inclosed  by  the  X  axis,  a  circle 
of  10  inches  radius  with  center  at  the  origin,  the  straight  line  y  =  2  (x  —  3), 
and  the  circle  x2  +  yz  =  25. 

30.  Find  the  center  of  gravity  of  the  area  between  the  circle  with  center 
at  the  origin  and  radius  6  and  the  rectangular  hyperbola  xy  =  12. 

31.  A  circular  board  20  inches  in  diameter  has  a  hole  5  inches  in  diameter 
cut  out,  the  center  of  the  hole  being  4  inches  from  the  center  of  the  board. 
Find  the  center  of  gravity  of  the  remainder. 

32.  Solve  Problem  31  if  the  center  of  the  hole  is  8  inches  from  the  center 
of  the  board. 


CHAPTER  XVII. 
MOMENT  OF  INERTIA. 

141.  Definition. —  The  moment  of  inertia  of  a  body  with 
respect  to  an  axis  is  the  sum  of  the  products  obtained  by  multi- 
plying the  mass  of  each  particle  of  the  body  by  the  square  of  its 
distance  from  the  axis.  If  m  is  the  mass  of  any  particle,  and  r 
is  its  distance  from  the  axis, 

7  =  Swr2. 

For  a  continuous  body,  the  definition  expressed  mathematically 
is 

/  =fr*dM,  Formula  XXXI. 

where  /  is  the  moment  of  inertia  and  dM  is  any  element  of  mass 
(finite  or  infinitesimal),  all  parts  of  which  are  at  a  distance  r  from 
the  axis. 

In  Fig.  179,  the  Z  axis  is  taken  as  the  axis  of  inertia  for  the 
solid.  The  element  BBr  extending  entirely  through  the  body 
parallel  to  the  Z  axis  is  the  element  of 
mass  of  Formula  XXXI.  The  element 
of  mass  might  have  the  form  of  a  hollow 
cylinder  of  radius  r,  and  thickness  dr. 
It  could,  of  course,  be  of  infinitesimal 
dimensions  in  three  directions,  in  which 
case  the  volume  would  be  represented  by 
dx  dy  dz  in  rectangular  coordinates,  by 
rdOdrdz  in  cylindrical  or  mixed  coordi- 
Fig.  179.  nates,  and  by  r2  sin  6  d&  d<f>  dr  in  spherical 

coordinates  (with  the  Z  axis  as  the  axis  of  the  sphere  from  which 
6  is  measured).  When  taking  moment  of  inertia  with  respect 
to  the  Z  axis  the  values  of  r2  of  Formula  XXXI  in  terms  of 
the  coordinates  of  the  element  are 

x2  +  y2  for  rectangular  coordinates, 
r2  for  cylindrical  coordinates, 

r2  sin2  6  for  spherical  coordinates. 
251 


/ 
x 


252 


STRENGTH  OF  MATERIALS 


The  element  BBf  of  Fig.  179  may  be  regarded  as  an  example 
of  rectangular  coordinates  after  integration  with  respect  to  Z. 
If  its  cross  section  were  of  the  form  of  an  element  of  area  in  polar 
coordinates  it  would  be  an  example  of  the  cylindrical  element 
of  volume  after  one  integration.  A  second  integration  of  this 
element  of  volume  with  respect  to  0  would  give  the  hollow 
cylinder  of  thickness  dr. 

PROBLEMS. 

1.   Find  the  moment  of  inertia  of  a  rectangular  parallelepiped  of  width  6, 
height  d,  and  length  I,  with  respect  to  an  edge  parallel  to  its  length  (Fig.  180) 
by  double  integration. 

Ans.  I  =  —5-  (i 


Fig.  180. —  Moment 
of  Inertia  of  Paral- 
lelopiped. 


where  p  is  the  mass  per  unit  volume  and  M  is  the  total 
mass. 

2.   Find  /  of  a  homo- 

I     geneous  solid   cylinder    of 

^length  I  and  radius  a  with 
respect  to  the  axis  of  revo- 
lution- (00',  Fig.  181). 

Ans.  7 


rde 


2  2 

This  is  a  case  of  cylin- 
drical coordinates.  The  element  of  volume  for  double 
integration   has   a  length  I  and  a  cross  section 
r  dd  dr.     Integrating  first  with  respect  to  r  gives 
a  wedge-shaped  element  between  the  planes  whose 
traces  on  the  front  are  the  dotted  lines  OE  and  OF.    .— 
The  second  integration  builds  up  the  cylinder  of    *lgJ   l          '  Moment  of 
a  series  of  such  wedges.  Inertia  of  Cylinder- 

If  6  be  integrated  first  between  the  limits  0  and  2  TT,  we  get  a  hollow  cylinder 

of  radius  r  and  thickness  dr.  The  vol- 
ume of  this  hollow  cylinder  is  2  irrl  dr, 
and  its  moment  of  inertia  with  respect 
to  the  axis  00'  is  2  irplr3  dr,  which  might 
have  been  obtained  directly  without 
integrating. 

3.  Find  the  moment  of  inertia  of 
a  right  cone  of  height  h  and  radius  of 
base  a  with  respect  to  the  axis  of  rev- 
olution, by  a  single  integration  using 
a  hollow  cylinder  as  the  element  of 
volume. 

.  _  Trpa*h  =  3  Ma2 
10  10 


Fig.  182.  —  Moment  of  Inertia  of 
Right  Cone. 


Ans.  1=2  irpr^z  dr  =  2  irph 


4.  Find  the  moment  of  inertia  of  a  homogeneous  solid  sphere  of  radius  a 
with  respect  to  a  diameter  by  double  integration.     Use  as  the  element  of 


MOMENT   OF   INERTIA  253 

volume  a  ring  of  radius  r  and  cross  section  dr  dz  and  integrate  first  with  respect 
to  z.  Ans.  7  =  |  Ma2. 

5.  Solve  Problem  4,  integrating  first  with  respect  to  r.     Show  that  this 
is  the  same  as  a  single  integration  using  a  disk  or  short  cylinder  of  length  dx 
as  the  element  of  volume,  and  applying  the  results  of  Problem  2. 

6.  Solve  Problem  3  by  a  single  integration  building  the  cone  of  flat  disks 
parallel  to  the  base,  the  moment  of  inertia  of  each  disk  being  from  Problem  2, 

—  jy-  dx,  where  r  is  measured  from  the  axis  to  the  surface. 

142.  Radius  of  Gyration.  —  The  radius  of  gyration  may  be 
denned  algebraically  by  the  equations: 

Mk*  =  I,  (1) 

k*  =  W  (2) 

where  k  is  the  radius  of  gyration.* 

The  radius  of  gyration  is  the  distance  from  the  axis  at  which 
the  entire  mass  could  be  concentrated  and  leave  the  moment  of 
inertia  unchanged.  In  the  case  of  a  homogeneous  solid  cylinder 
with  respect  to  the  axis  of  revolution, 

_Maz 
2    ' 


k=-=.  =  0.7071  a. 

A/2 

If  the  entire  mass  of  a  solid  cylinder  of  radius  a  were  condensed 
into  a  hollow  cylinder  of  radius  0.707  a  and  negligible  thickness, 
or  into  a  single  filament  at  a  distance  of  0.707  a  from  the  axis, 
the  moment  of  inertia  in  each  case  would  be  the  same  as  that  of 
the  solid  cylinder. 

PROBLEMS. 

1.  Find  the  square  of  the  radius  of  gyration  of  a  homogeneous  solid  cylinder 
of  12  inches  radius.  Ans.  k*  =  72. 

2.  Find  the  radius  of  gyration  of  a  parallelepiped  of  breadth  8  inches  and 
depth  10  inches  with  respect  to  an  axis  parallel  to  the  length  along  one  edge 
(Problem  1,  Article  141).  Ans.  k  =  7.39  inches. 

3.  A  solid  cylinder  of  6  inches  radius  weighing  40  pounds,  is  coaxial  with 
a  solid  sphere  of  8  inches  radius  weighing  60  pounds.     Find  the  radius  of 
gyration  of  the  combination.  Ans.  k  =  4.75  inches. 

*  In  this  chapter  we  shall  represent  radius  of  gyration  by  k  to  avoid  con- 
fusion. In  the  chapter  on  columns  it  is  represented  by  r  which  is  customary 
in  books  on  this  subject. 


254 


STRENGTH  OF  MATERIALS 


4.   Find  the  radius  of  gyration  of  a  homogeneous  hollow  cylinder  of  outside 
radius  a  and  inside  radius  b  with  respect  to  the  axis  of  revolution. 

Ans.  k 


5.  By  integration,  find  the  moment  of  inertia  of  a  homogeneous  solid 
cylinder  with  respect  to  an  element  of  the  curved  surface  as  an  axis  (CC', 
Fig.  181).  Ans.  I  =  f  Ma2. 

143.  Transfer  of  Axis.  —  When  it  is  necessary  to  find  the  mo- 
ment of  inertia  with  respect  to  some  axis  for  which  the  equation 
of  the  solid  is  complicated  the  integration  becomes  laborious. 
Usually  it  is  best  to  first  find  the  moment  of  inertia  with  respect 
to  an  axis  giving  the  simplest  expression  for  the  equation  of  the 
solid  and  then  transfer  to  the  new  axis.  If  CC'  is  an  axis  passing 
through  the  center  of  gravity  of  a  solid  and  00'  is  a  parallel  axis 
at  a  distance  d  from  it,  we  will  prove  that 

/  =  /o  +  Md\  Formula  XXXII. 


where  I  is  the  moment  of  inertia  with  respect  to  00'  and  70  is  the 
moment  of  inertia  with  respect  to  CC'  . 


Fig.  183.  —  Transfer  of  Axis. 


Let  BB'  (Fig.  183)  be  an  element  of  mass  parallel  to  the  axes. 
Its  coordinates  with  respect  to  the  axis  CC'  are  (x,  y).  Let  the 
coordinates  of  the  center  of  gravity  with  respect  to  the  axis  00' 


be  (a,  6)  so  that  d  -- 
of  r2  in  the  expression 


With  respect  to  00'  the  value 


is 


MOMENT  OF  INERTIA  255 

With  respect  to  00'  the  expression  for  the  moment  of  inertia  is 
/  =  f  (a2  +  2ax  +  x2  +  62  +  2  by  +  y*)  dM.  (1) 

ydM.  (2) 


C 


We  recognize  the  first  term  of  the  second  member  of  (2)  as  the 
moment  of  inertia  with  respect  to  CC".  The  second  term  is 
(a2  +  62)  M,  which  is  M  d2. 

The  third  term,  2  a  I  x  dM,  is  zero  ;  x  dM  being  the  moment  of 

dM  with  respect  to  a  vertical  plane  through  CC'  ',  and  the  sum  of 
these  moments  is  zero  when  the  center  of  gravity  falls  in  this 
vertical  plane. 


When  y  is   measured  from  the   center  of  gravity  y  =  0  and 

/ydM 
,..    =  0,  consequently  the  last  term  of  (2)  is  zero  and  equa- 

tion (2)  becomes  Formula  XXXII. 

PROBLEMS. 

1.  Solve  Problem  5  of  Article  142  by  means  of  Formula  XXXII  and  the 
answer  of  Problem  2  of  Article  141. 

2.  Find  the  moment  of  inertia  of  a  homogeneous  solid  sphere  of  radius  a 
with  respect  to  a  tangent,  and  compute  the  square  of  the  radius  of  gyration. 

Ans.  jfc*  =  1.4  a2. 

3.  Find  the  radius  of  gyration  of  a  homogeneous  solid  cylinder  of  6  inches 
radius  with  respect  to  an  axis  10  inches  from  the  axis  of  revolution  and  parallel 
to  it.  Ans.  k  =  10.86  inches. 

4.  Find  the  radius  of  gyration  of  a  parallelepiped  of  breadth  Synches,  and 
depth  10  inches  with  respect  to  an  axis  parallel  to  its  length  through  the  center 
of  gravity,  by  means  of  the  result  of  Problem  2,  Article  142,  and  Formula 
XXXII. 


144.  Moment  of  Inertia  of  a  Thin  Plate. 
-  In  Fig.  184,  if  we  take  the  moment  of  in- 
ertia with  respect  to  the  vertical  axis  FF', 
we  get  from  Problem  1  of  Article  141 


F° 


Fig.  184. 


256 


STRENGTH  OF  MATERIALS 


If  I  is  small  relatively  to  b,  I2  may  be  neglected  and 

Mb* 
3 

In  a  similar  manner  in  respect  to  an  axis  through  the  center  if  6 
is  small  compared  with  Z, 

Ml2 


1  = 


12 


PROBLEMS. 

1.  A  rectangular  board  40  inches  long  and  1  inch  thick  weighs  6  pounds. 
Find  its  moment  of  inertia  with  respect  to  an  axis  parallel  to  its  breadth 
through  its  center  of  gravity.  Ans.  I  =  800.5. 

2.  In  Problem  1  what  is  the  moment  of  inertia  if  the  thickness  is  neglected? 

3.  A  rectangular  rod  60  inches  long  is  1  inch  square.     Find  its  radius  of 
gyration  with  respect  to  an  axis  through  the  middle  of  one  end  perpendicular 
to  its  length.     What  difference  does  it  make  if  the  thickness  is  neglected? 

Fig.  185  represents  a  plate  of  uniform  thickness  t  and  of  any 
form  whatever.     The  axis  00'  is  parallel  to  the  plane  surfaces. 

The  element  of  volume  BB'  has 
its  long  dimension  parallel  to 
00'.  If  the  thickness  of  the 
plate  is  so  small  relatively  that 
the  radius  r  drawn  from  any 
point  in  it  to  the  axis  00'  is 
not  appreciably  greater  than  the 
smallest  radius  y,  then  the  entire 
element  may  be  used  as  the 


7 

Fig.  185. 


— x  AXIS  mass  element  in  /  r2  dM,  which 

becomes    /  y2  dM  for  this  figure. 

If  dA  is  the  cross  section  of 
the  element  parallel  to  the  YZ  plane  its  volume  is  id  A.  The 
moment  of  inertia  of  a  thin  plate  with  respect  to  an  axis  in  one 
of  its  surfaces  or  an  axis  between  these  surfaces  is 

I  =  Ptfy2dA,  (1) 

where  dA  is  an  element  of  the  surface. 


MOMENT   OF  INERTIA  257 

145.  Moment  of  Inertia  of  a  Plane  Area.*  —  If  dA  is  an  ele- 
ment of  an  area  at  a  distance  r  from  some  axis,  /  r2  dA  is  called 

the  moment  of  inertia  of  the  area  with  respect  to  the  axis.  If 
the  axis  lies  in  the  plane  of  the  area,  the  moment  of  inertia  of  the 
area  is  the  same  as  that  of  a  thin  plate  of  such  thickness  and 
density  that  its  mass  is  unity  per  unit  area.  In  equation  (1)  of 
the  preceding  article,  if  the  product  pt  is  unity,  the  expression 
gives  the  moment  of  inertia  of  a  plane  area  with  respect  to  a 
horizontal  axis. 

The  moment  of  inertia  of  an  area  with  respect  to  an  axis  in 
its  plane  is  a  factor  in  all  problems  concerning  the  strength  and 
deflection  of  beams  and  columns. 

The  moment  of  inertia  of  a  surface  with  respect  to  an  axis 
normal  to  its  plane  is  called  the  polar  moment  of  inertia  of  the 
surface.  The  polar  moment  of  inertia  of  a  surface  is  equivalent 
to  the  moment  of  inertia  of  a  solid  plate  of  the  same  dimensions 
as  the  surface  and  of  such  thickness  and  density  that  its  mass 
is  unity  per  unit  area. 

In  Problem  2  of  Article  141  we  found  the  moment  of  inertia 
of  a  solid  cylinder  to  be 

,  _  TrpZa4 
2 

If  pi  is  unity  this  gives  for  the  polar  moment  of  inertia  of  a 
circle  with  respect  to  an  axis  through  its  center 

ira4      Aa2 


where  J  represents  the  polar  moment  of  inertia. 

In  a  similar  way  we  find  the  polar  moment  of  inertia  of  a 
rectangle  of  sides  b  and  d,  with  respect  to  an  axis  through  one 
corner,  to  be 


Formula  XXXII  holds  for  moment  of  inertia  of  areas. 

*  In  reality  this  is  not  a  true  moment  of  inertia  in  the  physical  sense  of  the 
term,  as  an  area  has  no  mass,  but  as  the  mathematical  expression  is  similar 
in  form  to  a  true  moment  of  inertia,  it  is  convenient  and  customary  to  call  it 
the  moment  of  inertia  of  the  area. 


258  STRENGTH  OF  MATERIALS 


PROBLEMS. 

1.  By  integration  find  the  moment  of  inertia  of  a  rectangle  of  breadth  b 

bd3 
and  depth  d  with  respect  to  the  side  b.  Ans.  I  =  - 

o 

2.  By  transfer  of  axis  find  the  moment  of  inertia  of  a  rectangle  of  sides 
b  and  d  with  respect  to  an  axis  in  the  plane  of  the  area  parallel  to  b  and  passing 

through  the  center  of  the  rectangle.  Ans.  /  =  —  • 

3.  By  integration  find  the  moment  of  inertia  of  a  circular  area  of  radius  a 
with  respect  to  a  diameter.  Ans.  I  =~  - 

The  results  of  Problems  1,  2,  and  3  should  be  memorized  on  account  of 
their  importance  in  the  theory  of  beams  and  columns. 

4.  What  is  the  radius  of  gyration  of  a  circular  area  with  respect  to  a 
diameter? 

5.  Find  the  polar  moment  of  inertia  of  a  square  with  respect  to  one  corner. 
How  does  it  compare  with  the  moment  of  inertia  with  respect  to  a  side  of  the 
square?  Ans.  J  =  21. 

6.  Compare  the  polar  moment  of  a  circle  with  respect  to  an  axis  through 
the  center  with  the  moment  of  inertia  with  respect  to  a  diameter. 

7.  Find  the  moment  of  inertia  of  a  triangle  of  base  6  and  altitude  h  with 
respect  to  a  line  through  the  vertex  parallel  to  the  base.     From  this  result  by 
transfer  of  axes  find  the  moment  of  inertia  of  the  triangle  with  respect  to  the 

bh3 
base.  Ans.  /  =  -7^  w^tn  respect  to  the  base. 

LA 

8.  Find  the  moment  of  inertia  of  a  6-inch  by  4-inch  by  1-inch  'angle  section 
(Fig.  186)  with  respect  to  an  axis  (1-1)  through  the  center  of  gravity  parallel 

to  the  longer  leg.  Divide  the  section  into 
two  rectangles.  Find  the  moment  of  inertia 
of  each  with  respect  to  a  line  through  its 
center  of  gravity,  then  transfer  to  axis  (1-1) 
and  add.  As  a  check  find  the  moment  of 
inertia  of  the  entire  figure  with  respect  to 
*  some  axis  which  is  a  common  base  of  both 
rectangles  (as  BC,  or  a  horizontal  line 
through  D),  then  transfer  to  the  center  of 

Fig.  186. -Moment  of  Inertia    g™^    ComPare    result    with    table   in 
of  Angle  Section.  ^am  ^  ^  moment  Q£  inertia  of  the 

above  angle  section  with  respect  to  an  axis  through  the  center  of  gravity 
parallel  to  the  shorter  leg. 

10.  A  "plate-and-angle  column"  (see  Cambria)  is  made  of  four  4-inch 
by  3-inch  by  £-mch  angles  and  one  12-inch  by  £-inch  plate.     The  angles  are 
riveted  to  the  plate,  the  back  of  the  longer  legs  being  one-eighth  inch  above  and 
below  the  edges  of  the  plate.     Taking  the  moments  of  inertia  and  location  of 
centers  of  gravity  from  Cambria  tables  of  angle  sections,   find  the  moment 
of  inertia  of  the  entire  section  with  respect  to  the  two  lines  of  symmetry. 

11.  Look  up  in  Cambria  the  "Diagram  for  Minimum  Standard  Channels" 
and  derive  the  formula  for  the  moment  of  inertia  there  given. 


MOMENT   OF  INERTIA 


259 


12.  From  the  dimensions  given  in  Cambria  find  the  moment  of  inertia  of 
a  15-inch  33-pound  standard  channel  with  respect  to  an  axis  through  the 
center  of  gravity  of  the  section  perpendicular  to  the  web. 

13.  A   "plate-and-channel  column"  is  made    of  two   15-inch    40-pound 
channels  placed  12|  inches  back  to  back  (with  toes  out),  and  two  20-inch  by 
f-inch  plates.     Taking  the  moments  of  inertia  of  the  channels  from  the  table 
of  the  properties  of  channel  sections,  find  the  moment  of  inertia  of  the  section 
with  respect  to  an  axis  parallel  to  the  channel  webs  and  midway  between  them 
and  also  with  respect  to  an  axis  parallel  to  the  plates  through  the  centers  of 
the  channels. 

14.  Find  the  moment  of  inertia  of  an  8-inch  by  3-inch  by  Hnch  Z-bar 
section  with  respect  to  axes  through  the  center  of  gravity  parallel  and  per- 
pendicular to  the  web. 

15.  Find  the  two  principal  moments  of  inertia  of  a  20-inch  65-pound  stand- 
ard I-beam  section.     Derive  the  formula  used. 


146.  Change  of  Direction  of  Axis.  —  Formula  XXXII  en- 
ables us  to  transfer  moment  of  inertia  from  one  axis  to  a  parallel 
axis.  It  is  frequently  necessary 
to  transform  to  an  axis  at  an 
angle  with  the  original  axis. 

Fig.  187  represents  an  area  in 
the  XY  plane.  The  moment  of 
inertia  of  this  area  with  respect  to 
the  X  axis  OX  we  will  call  Ix,  and 
the  moment  of  inertia  with  respect 
to  the  Y  axis  we  will  call  7W: 


Fig.  187.  —  Change  of  Direction 
of  Axis. 


,-fy*dA; 


2dA. 


Let  OX',  OYr  be  new  axes  making  an  angle  0  with  the  X  and 
Y  axes  respectively.  The  coordinates  of  the  element  of  area  dA 
with  respect  to  these  new  axes  are  (xf,  y'). 

The  moment  of  inertia  of  the  area  with  respect  to  OX'  is 


(1) 

(2) 
(3) 


.  2dA. 

From  the  geometry  of  the  figure 

y'  =  y  cos  6  —  x  sin  6. 


I  =  /  (?/2  COS2  0  —2xycosB  sin  0  +  z2  sin2  0)  dA. 


260  STRENGTH  OF  MATERIALS 

1=  Ix  cos2  0  +  Iy  sin2  8  -  sin  2  0  I  xy  dA.  (4) 

/=  Ix  +  7"  +  7*  ~  /ycos2  0  -  sin2  0  CxydA.  (5) 

44  j 

PROBLEMS. 

1.  Find  the  moment  of  inertia  of  a  rectangle  4  inches  wide  and  3  inches  high 
with  respect  to  an  axis  in  its  plane  which  passes  through  the  lower  left  corner 
and  makes  an  angle  of  20  degrees  with  the  horizontal. 

Ix  =  36  inches4, 
Iv  =  64  inches4. 


< — x — SdA  ^^  r  r*  rz 

I  xydA  =   \     \    xydxdy  =  36  inches4. 

7  =  50  -  14  cos  40°  -  36  sin  40° 
=  16.14  inches4. 
Fig.  188.  —  Moment  of  Inertia 

with  Respect  to  OX'.  ,  2'  f '™  *»H«™  J  >f  the  *™  !»"" 

through  the  lower  left  corner  and  makes 

a  negative  angle  of  20  degrees  with  the  horizontal. 

Ans.  1=  62.42  inches4. 

3.  Find  the  moment  of  inertia  of  a  3-inch  by  4-inch  rectangle  with  respect 
to  a  diagonal  by  means  of  equation  (5)  and  check  by  the  moment  of  inertia 
of  two  triangles  with  respect  to  the  diagonal  as  a  common  base  (Problem  7, 
Article  145). 

147.  Product  of  Inertia.  —  The  expression  /  xy  dA  is  called  the 

product  of  inertia  of  the  area.     It  is  represented  algebraically  by 
the  letter  H . 

If  an  area  is  symmetrical  with  respect  to 
either  one  of  a  pair  of  rectangular  axes,  its 
product  of  inertia  with  respect  to  that  pair 
of  axes  is  zero.  Fig.  189  represents  an  area 
symmetrical  with  respect  to  the  Y  axis.  If 
we  integrate  first  with  respect  to  x, 


H  = 


>  dy  =  *  J[*5  -  xl]  y  dy. 


Fig.  189. 


If  the  area  is  symmetrical  with  respect  to  the  Y  axis,  the  lower 
limit  Xi  is  numerically  equal  and'  opposite  in  sign  to  the  upper 
limit  x2,  and  the  squares  are  the  same  in  magnitude  and  sign; 
consequently  the  term  in  the  brackets  vanishes  and 

#=0. 


MOMENT   OF   INERTIA  261 

When  the  product  of  inertia  is  known  with  respect  to  a  pair 
of  rectangular  axes  through  the  center  of  gravity  of  an  area,  it 
may  be  calculated  for  a  second  pair  of 
parallel  axes  in  the  plane  of  the  area  by  a 
formula  similar  to  XXXII  for  the  transfer 
of  moments  of  inertia. 

Let  OX,  OF,  Fig.  190,  be  the  original 
pair  of  axes  through  the  center  of  gravity, 
and  let  (x,  y)  be  the  coordinates  of  an  ele- 
ment  dA  with   reference   to   these   axes.   Fig.  190.  —  Transfer  of 
Let  O'X',  O'Y'  be  a  new  pair  of  parallel      Axes    for   Product   of 
axes.      Let   (a,  6)   be  the  coordinates  of      Inertia- 
the  center  of  gravity  of  the  area  with  respect  to  the  new  axes. 

If  H  is  the  product  of  inertia  with  respect  to  the  new  axes, 


H  =(a  +  x)(b  +  y)dA.  (1) 

H  =  ab  CdA+b  CxdA+a   CydA  +  Cxy  dA.  (2) 

H  =  abA  +  0  +  0  +  #0,  (3) 

where  HQ  is  the  product  of  inertia  with  respect  to  the  axes 
through  the  center  of  gravity.  Equation  (3)  is  easily  remembered 
from  Formula  XXXII,  replacing  the  square  by  the  product. 

If  the  center  of  gravity  of  an  area  falls  in  the  first  or  third 
quadrant  with  respect  to  the  axes  for  which  its  product  of  iner- 
tia is  taken,  H  is  positive;  if  the  center  of  gravity  falls  in  the 
second  or  fourth  quadrant,  H  is  negative. 

PROBLEMS. 

1.  Find  the  product  of  inertia  of  a  rectangle  6  inches  wide  and  4  inches  high 
with  respect  to  the  lower  and  left  edges  as  axes.  Ans.  H  =  144  inches4. 

2.  Find  the  product  of  inertia  of  a  rectangle  5  inches  wide  and  4  inches  high 
with  respect  to  the  lower  edge  and  a  vertical  line  1  inch  to  the  left  of  the  right 
edge.  Ans.  H  =  -60. 

3.  Find  the  product  of  inertia  of  the  6-inch  by  4-inch  by  1-inch  angle 
section  of  Fig.  186  with  respect  to  the  axes  1-1  and  2-2. 

148.  Transformation  of  Direction  of  Axes  for  Product  of 
Inertia.  —  To  get  the  product  of  inertia  for  the  axes  OX'  ',  OF'  of 
Fig.  187,  we  have: 

(1) 


262  STRENGTH  OF  MATERIALS 

H'  =  I  (x  cos  B  -f-  y  sin  B)  (y  cos  6  -  x  sin  0)  dA,  (2) 

Hr  =  (cos2  6  -  sin2  0)  f  z?/  dA  +  cos  0  sin  0  C(yz  -  x2)  dA,    (3) 

H'=H  cos  2  0  +  /x  ~/vsin20.  (4) 

H'  becomes  zero  when  the  right  member  of  (4)  =  0,  when 

2  77 

tan  2  e  =  Y^  ^j-  •  (5) 

lv  —  LX 

PROBLEMS. 

1.  In  the  4-inch  by  3-inch  rectangle  of  Fig.  188  what  will  be  the  angle 
between  OX'  and  the  4-inch  edge  if  the  product  of  inertia  with  respect  to  OX' 
and  the  axis  through  0  normal  to  it  is  zero?  Ans.  0  =  34°  22'. 

2.  Find  the  direction  of  the  pair  of  axes  through  the  center  of  gravity  of 
the  6-inch  by  4-inch  by  1-inch  angle  section  of  Fig.  186  for  which  the  product 
of  inertia  is  zero. 

149.  Direction  of  Axis  for  Maximum  Moment  of  Inertia.  - 

Equation  (5)  of  Article  146  is  written: 

/  =  Ix  +  ly  +  ^-=-^  cos  2  0  -  #  sin  26.  (1) 

-  z 

Differentiating  with  respect  to  0, 

^  =  (/„  -  Ix)  sin  2  B  -  2H  cos  2  0,  (2) 

from  which  the  condition  of  maximum  or  minimum  is 

-  (3) 


y    —        x 

Comparing  (3)  with  (5)  of  Article  148,  we  find  that  the  condition 
of  maximum  or  minimum  moment  of  inertia  is  the  condition 
which  gives  zero  product  of  inertia.  There  are  two  solutions  for 
(3),  which  give  values  of  2  6  differing  by  180  degrees  and  values  of 
B  differing  by  90  degrees.  One  of  these  positions  is  that  of  maxi- 
mum moment  of  inertia  and  the  other  is  that  of  minimum  moment 
of  inertia. 

PROBLEMS. 

1.  In  the  rectangle  of  Problem  1  of  Article  148  find  the  maximum  and 
minimum  moment  of  inertia  for  axes  through  one  corner. 

2.  In  Problem  2  of  Article  148  find  the  maximum  and  minimum  moment 
of  inertia  and  compare  the  minimum  with  the  table  in  Cambria. 


MOMENT  OF  INERTIA  263 

3.  What  is  the  direction  of  the  axis  for  which  the  moment  of  inertia  is  a 
minimum  for  an  angle  section  with  equal  legs?     Why? 

4.  In  the  case  of  an  8-inch  by  8-inch  by  1-inch  angle  section  calculate  the 
distance  of  the  center  of  gravity  from  the  back  of  the  leg.     Find  the  moment 
of  inertia  with  respect  to  an  axis  through  the  center  of  gravity  parallel  to  the 
leg.     Find  the  minimum  and  maximum  moment  of  inertia  for  axes  through 
the  center  of  gravity.     When  the  work  is  complete  compare  with  Cambria. 

5.  Solve  the  8-inch  by  3-inch  by  £-inch  Z-bar  section  completely  for  the 
least  radius  of  gyration. 

The  maximum  and  minimum  moments  of  inertia  of  an  area  for 
axes  through  a  given  point  are  called  the  principal  moments  of 
inertia,  and  the  corresponding  axes  are  the  principal  axes. 

If  the  minimum  moment  of  inertia  is  known,  it  is  generally  easy 
to  find  the  maximum  by  means  of  a  simple  relation 

•L  max    r    *  min  =   •*  x  T~  ly  =  «/  • 

The  sum  of  the  moments  of  inertia  of  a  plane  area  for  any  pair 
of  rectangular  axes  in  the  plane  is  equal  to  the  polar  moment  of 
inertia  for  their  point  of  intersection. 

Let  one  of  these  axes  be  used  as  the  X  axis. 


If  the  other  rectangular  axis  is  used  as  the  Y  axis, 

/„  =  Cx2dA. 
For  the  polar  moment  of  inertia  r2  =  x2  +  yz, 

J  =  f(x2  +  yt)  dA  =  Cx*dA  +  Cy*dA. 

PROBLEMS. 

6.  Using  all  the  data  you  can  find  in  Cambria,  calculate  the  maximum 
moment  of  inertia  of  a  5-inch  by  3-inch  by  |-inch  angle  section. 

7.  Find  the  least  and  greatest  moment  of  inertia  and  radius  of  gyration  of 
a  semicircular  area  of  radius  a  with  respect  to  axes  in  its  plane  passing  through 
the  end  of  the  diameter  which  bounds  it. 

150.  The  Moment  of  Inertia  of  a  Prism  or  Pyramid.  —  The 

moment  of  inertia  of  any  solid  may  be  found  by  triple  integration 
with  an  element  which  is  infinitesimal  in  each  direction,  or  by 
double  integration  with  an  element  which  is  infinitesimal  in  two 


264 


STRENGTH  OF  MATERIALS 


directions  and  extends  entirely  through  the  mass  in  the  direction 
of  the  axis. 

It  is  often  easier  to  use  a  thin  plate  or  disk  which  is  infinitesimal 
in  one  direction  only  as  the  element  of  volume,  provided  the 
moment  of  inertia  of  this  element  is  known  with  respect  to  an 
axis  through  its  center  of  gravity  parallel  to  the  axis  of  inertia. 

PROBLEMS. 

1.  Find  the  moment  of  inertia  of  a  right 
pyramid  of  height  h,  with  a  square  base  of 
side  6,  with  respect  to  an  axis  through  the 
vertex  perpendicular  to  the  base. 

The  element  of  volume  is  the  square  plate 
of  thickness  dx.     Its  volume  is  A  dx,  where 
A  is  the  area  of  the  section.     From  similar 
solids  (Fig.  191), 
Fig.  191.  A  =  —  • 

As  each  side  is  -j- ,  its  polar  moment  of  inertia  with  respect  to  the  X  axis  is 
^-ry  dx.  The  total  moment  of  inertia  is  the  sum  of  that  of  the  several  plates. 
/  = 


3(M4  ~    30 


.10. 


2.  Find  the  moment  of  inertia  of  a  right  pyramid,  the  base  of  which  is  a 
hexagon  of  side  a,  with  respect  to  an  axis  through  the  vertex  perpendicular 
to  the  base. 

If  in  Fig.  191  we  wished  to  find  the  moment  of  inertia  with 
respect  to  the  Z  axis  OZ,  we  could  find  the  moment  of  inertia  of 
the  plate  with  respect  to  the  parallel  axis  CC'  and  then  transfer 
to  the  Z  axis.  The  moment  of  inertia  of  the  plate  is  the  same  as 
that  of  the  area  of  the  plate  with  respect  to  a  line  in  its  plane 
multiplied  by  the  thickness  dx  and  the  density.  The  moment  of 
inertia  of  the  plate  about  CC'  is 


where  /o  and  Md2  have  the  meaning  of  Formula  XXXII. 


MOMENT  OF  INERTIA 


265 


PROBLEMS. 

3.  Find  the  square  of  the  radius  of  gyration  of  a  right  pyramid  24  inches 
high  with  base  12  inches  square  with  reference  to  an  axis  through  the  vertex 
parallel  to  the  base.  Ans.  k2  =  352.8  inches2. 

4.  Find  the  moment  of  inertia  of  a  right  cylinder  of  radius  a  and  length  I 
with  respect  to  an  axis  perpendicular  to  the  axis  of  the  cylinder  through  the 

ll 

3    '    4 


center  of  one  end. 


Ans.  tf -L-HlL 


Observing  the  answer  of  Problem  4,  we  see  that  the  square  of 
the  radius  of  gyration  is  made  of  two  terms,  the  first  of  which  is 
k*  for  a  long  thin  rod  with  respect  to  an  axis  through  one  end 
perpendicular  to  its  length,  and  the  other  is  k2  for  a  circular  area 
with  respect  to  a  diameter.  The  moment  of  inertia  of  any  solid 
with  a  constant  cross  section  and  ending  with  parallel  planes 
normal  to  its  length  (any  right  prism  or  cylinder)  may  be  calcu- 
lated in  the  same  way.  Expressed  algebraically, 

where  ki  is  the  radius  of  gyration  of  the  prism  regarded  as  a  thin 
rod  and  kA  is  the  radius  of  gyration  of  a  cross  section.     Fig.  192 


Fig.  192. 


represents  a  triangular  prism  with  its  axis  parallel  to  the  X  axis. 
It  is  desired  to  find  its  moment  of  inertia  with  respect  to  the  Z 
axis. 

Using  the  element  BE'  (Fig.  192)  of  cross  section  dxdy  ex- 
tending entirely  through  the  body  in  the  direction  of  the  Z  axis : 


/  = 


(2) 


266  STRENGTH  OF  MATERIALS 

where  dA  is  an  area  of  length  BB'  and  height  dy. 

I  =  P  C  fx*dxdA+P    C  Cy*dxdA.  (3) 

When  we  integrate  with  respect  to  dA,  x  remains  unchanged,  as 
we  simply  pile  up  elements  of  the  form  of  BB'  from  the  bottom 
of  the  top  of  the  section  EFG  between  vertical  planes  at  a  dis- 
tance dx  apart: 

/  =  p  Cx*A  dx  +  P  C IAdx,  (4) 

where  A  is  the  area  of  the  section,  and  I  A  is  the  moment  of 
inertia  of  the  plane  area  with  respect  to  the  axis  CC'  in  the  XZ 
plane  parallel  to  the  Z  axis. 

Equation  (4)  applies  to  a  solid  of  any  form  whatever,  and  is 
not  limited  to  a  prism  as  shown  in  the  figure.  If  the  line  OX 
passed  through  the  center  of  gravity  of  all  the  sections,  we  would 
have  an  example  of  Formula  XXXII  as  in  Problems  3  and  4. 

If  the  solid  is  a  prism  or  cylinder  with  the  axis  parallel  to  the 
X  axis,  A  is  constant  and  I  A  is  constant;  then 

/  =  PA   Cx2  dx  +  PI A   Cdx.  (5) 

The  first  term  of  the  last  member  of  (5)  is  the  moment  of  inertia 
of  a  thin  rod  with  respect  to  an  axis  perpendicular  to  its  length. 
The  second  member  is  equal  to  plAk%  =  MkA,  which  proves 
equation  (1). 

PROBLEMS. 

5.  Find  the  moment  of  inertia  of  a  right  cylinder  18  inches  long  and  12 
inches  in  diameter  with  respect  to  an  axis  in  the  plane  of  one  end  and  tangent 
to  the  cylinder.  Ans.  I  =  153  M . 

6.  Find  the  moment  of  inertia  of  a  prism  6  inches  square  and  24  inches 
long  with  respect  to  an  axis  in  the  plane  of  one  end  perpendicular  to  the  end 
of  a  diagonal.  Ans.  7  =  M  (192  +  21). 

MISCELLANEOUS  PROBLEMS. 

1.  A  homogeneous  solid  sphere  6  inches  in  diameter  is  placed  at  the  end 
of  a  homogeneous  solid  cylinder  2  inches  in  diameter  and  40  inches  long,  the 
center  of  the  sphere  being  in  the  line  of  the  axis  of  the  cylinder.     If  the  cylinder 
and  sphere  have  the  same  density,  locate  the  center  of  gravity  of  the  combina- 
tion. Ans.  9.105  inches  from  the  point  of  contact. 

2.  In  Problem  1  what  is  the  radius  of  gyration  with  respect  to  a  diameter 
of  the  cylinder  in  the  end  opposite  the  sphere?  Ans.  34  inches  nearly. 


MOMENT  OF  INERTIA  267 

3.  The  section  of  a  right  prism  is  an  equilateral  triangle  4  inches  on  each 
side.     Its  length  is  30  Inches.     What  is  its  radius  of  gyration  with  respect  to 
the  line  of  intersection  of  a  side  and  an  end?  Ans.  k  =  17.38  inches. 

4.  A  block  20  inches  long,  16  inches  wide,  and  12  inches  thick  has  a  round 
hole  10  inches  in  diameter  and  6  inches  deep  in  the  middle  of  the  left  end. 
Find  the  center  of  gravity  of  the  remainder. 

5.  In  Problem  4  find  the  moment  of  inertia  and  the  radius  of  gyration  with 
respect  to  an  axis  through  the  middle  of  the  right  end  parallel  to  the  12-inch 
faces. 

6.  Show  that  the  radius  of  gyration  of  a  square  area  with  respect  to  any 
axis  in  its  plane  through  the  center  of  gravity  is  the  same. 


CHAPTER  XVIII. 
COMPUTATION   WITHOUT   INTEGRALS. 

151.  Areas  Which  Cannot  be  Integrated.  —  In  order  to  inte- 
grate a  plane  surface  for  area,  center  of  gravity,  or  moment  of 
inertia,  we  must  first  know  the  equations  of  the  curves  which 
form  its  boundary  lines.  It  sometimes  happens  that  sections 
are  used  which  are  not  conveniently  expressed  in  simple  equations. 
More  frequently  the  engineer  is  called  upon  to  compute  the  con- 
stants of  some  section  designed  in  accordance  with  some  more 
or  less  simple  expressions  which  cannot  easily  be  determined  from 
measurements  of  the  finished  product. 

When  a  section  is  bounded  by  several  curves  having  different 
equations,  it  requires  considerable  labor  to  make  the  several 
integrations  and  put  in  the  appropriate  limits. 

For  these  reasons  it  is  occasionally  necessary,  and  often  con- 
venient, to  determine  center  of  gravity  and  moment  of  inertia 
by  approximate  computations  or  physical  experiments. 


2425 


Fig.  193.  —  Finite  Increments  of  Area. 

152.  Approximate  Computation.  —  For  approximate  compu- 
tations it  is  usually  best  to  make  an  accurate  drawing  of  the 
section,  preferably  on  squared  paper.  If  the  section  is  symmet- 
rical with  respect  to  any  line,  it  is  necessary  to  draw  and  calcu- 
late only  one-half,  as  in  Fig.  193. 

Each  ordinate  in  Fig.  193  may  be  taken  as  approximately  the 
mean  altitude  of  a  strip  one  unit  wide  extending  one-half  unit 

268 


COMPUTATION  WITHOUT  INTEGRALS         269 

on  each  side.  The  ordinate  3  is  the  mean  altitude  of  the  strip, 
extending  from  2.5  to  3.5  between  the  dotted  lines.  To  get 
the  area,  simply  measure  each  ordinate  (from  1  to  24  inclusive 
in  Fig.  193),  and  multiply  the  sum  of  these  by  the  common 
interval.  This  gives  the  entire  area  from  0.5  to  24.5.  The 
small  triangles  outside  these  limits  may  be  computed  separately. 
To  get  the  moment  of  the  area  in  order  to  calculate  the  center 
of  gravity,  multiply  each  ordinate  by  its  abscissa  and  add. ! 

For  the  moment  of  inertia  with  respect  to  the  origin,  multiply 
each  ordinate  by  the  square  of  its  abscissa  and  add  the  results. 
To  this  sum  add  one-twelfth  of  the  product  obtained  by  multi- 
plying the  sum  of  the  ordinates  by  the  square  of  the  horizontal 
interval.  Also  add  in  the  moment  of  inertia  of  the  limiting 
triangles. 

To  get  the  moment  of  inertia  with  respect  to  a  line  through 
the  center  of  gravity,  transfer  by  Formula  XXXII. 

At  0.5  the  ordinate  is  2,  making  the  area  of  the  triangle  0.5 
square  unit.  The  moment  arm  of  this  triangle  is  ^,  so  that  the 
moment  and  moment  of  inertia  are  nearly  negligible.  At  24.5 
the  ordinate  is  1;  the  area  is  0.25;  the  moment  arm  is  24.67. 

The  quantity  8.9  which  is  added  to  the  moment  of  inertia  is 
the  sum  of  the  moments  of  inertia  of  the  strips  about  their 
centers  of  gravity. 


PROBLEMS. 

1.  In  the  figure  of  Table  XI,  how  far  is  the  center  of  gravity  from  the 
origin?  Ans.  13.66  units. 

2.  If  each  unit  represents  1  inch,  what  is  the  moment  of  inertia  of  the 
half  of  the  figure  above  the  X  axis  with  respect  to  a  vertical  line  through  the 
center  of  gravity?  Ans.  5023.8  inches4. 

3.  If  each  unit  represents  |  inch,  what  is  this  moment  of  inertia? 

Ans.  314  inches4. 

4.  Using  the  ordinates  of  Table  XI,  with  the  addition  that  24  extends  from 
2.5  to  4.4,  and  24.5  extends  from  3.0  to  4.0,  find  the  center  of  gravity  of  the 
upper  half  of  Fig.  193  from  the  X  axis,  using  the  vertical  strips. 

5.  Find  the  moment  of  inertia  of  the  upper  half  of  the  figure  with  respect 
to  the  X  axis. 

6.  Plot  the  curve  yz  =  4x.     Find  the  center  of  gravity  of  the  area  between 
the  positive  part  of  this  curve,  the  X  axis,  and  the  ordinate  x  =  4  by  the 
approximate  method.     Use  paper  ruled  10  lines  to  the  inch  and  take  the  first, 
third,  fifth  ordinate,  etc.     Compare  result  with  that  obtained  by  integration. 

7.  Solve  Problem  6  for  moment  of  inertia  with  respect  to  axes  through  the 
center  of  gravity  parallel  to  the  X  and  Y  axes. 


270 


STRENGTH  OF  MATERIALS 


The  calculation  of  the  area  of  Fig.  193  is  shown  by  Table  XI, 
taking  each  division  as  unity. 

TABLE  XI. 

COMPUTATION  OF  AREA,  CENTER  OF  GRAVITY,  AND  MOMENT  OF  INERTIA 

OF  FIGURE  193. 


X 

y 

w 

x*y 

1 

2.8 

2.8 

2.8 

2 

3.6 

7.2 

14.4 

3 

3.9 

11.7 

35.1 

4 

4.0 

16.0 

64.0 

5 

3.9 

19.5 

97.5 

6 

3.6 

21.6 

127.6 

7 

3.4 

23.8 

166.6 

8 

3.4 

27.2 

217.6 

9 

3.4 

30.6 

275.4 

10 

3.4 

34.0 

340.0 

11 

3.4 

37.4 

.  411.4 

12 

3.4 

40.8 

489.6 

13 

3.5 

45.5 

591.5 

14 

4.0 

56.0 

784.0 

15 

5.0 

75.0 

1,125.0 

16 

5.5 

88.0 

1,408.0 

17 

6.1 

103.7 

1,762.9 

18 

6.45 

116.1 

2,089.8 

19 

6.55 

124.4 

2,364.5 

20 

6.5 

130.0 

2,600.0 

21 

6.25 

131.3 

2,756.3 

22 

6.0 

132.0 

2,904.0 

23 

5.3 

121.9 

2,803.7 

24 

1.9 

45.6 

1,094.4 

105.25 

1,442.1 

24,526.1 

24.5 

.25* 

6.1 

152.1 

.5 

.5 

106.0 

1,448.2 

24,678.2 

8.9 

24,687.1 

*  The  last  two  figures  are  areas  of  triangles. 

153.  Center  of  Gravity  by  Experiment.  —  The  center  of  grav- 
ity of  a  section  may  easily  be  determined  by  cutting  it  out  of 
uniform  cardboard  or  sheet  metal,  and  balancing  on  a  knife 
edge.  A  better  method  of  finding  the  center  of  gravity  is  that 
of  balancing  it  on  the  beam  of  a  platform  scale  or  similar  device. 


COMPUTATION   WITHOUT   INTEGRALS         271 

Fig.  194,  I,  represents  a  body  on  a  beam  balanced  by  the  poise 
in  the  position  shown.  If  the  body  is  turned  end  for  end  on 
the  beam  with  the  edge  0  not  changed,  so  as  to  come  into  the 
position  shown  by  the  dotted  lines,  the  center  of  gravity  is  moved 
a  distance  GG' ',  which  is  twice  its  distance  from  the  0.  To  get 
a  balance,  the  poise  P  must  be  moved  to  the  dotted  position  P'. 
If  the  mass  of  the  body  is  known  in  terms  of  the  poise,  the  dis- 


Fig.  194.  —  Center  of  Gravity  by  Balancing. 

tance  GG'  may  easily  be  calculated.  Instead  of  rotating  about 
the  end  0,  any  vertical  line  may  be  used  as  the  line  of  reference 
whose  position  on  the  beam  is  not  changed. 

PROBLEM. 

1.  A  body  weighing  4.50  pounds  is  balanced  on  a  scale  beam.  When 
turned  about  a  vertical  line  through  the  end  nearest  the  knife  edge,  the  apparent 
change  in  weight  is  0.576  pound.  The  distance  from  the  central  knife  edge 
to  the  end  knife  edges  is  10  inches.  How  far  is  the  center  of  gravity  from  the 
line  about  which  it  turned?  Am.  0.64  inch. 

The  method  just  given  requires  that  we  know  the  weight  of 
the  poise  and  the  value  in  inches  of  a  division  on  the  scale,  or 
that  we  know  the  distance  from  the  central  knife  edge  to  the 
knife  edge,  upon  which  1  pound  weighs  1  pound.  Another 
method  is  that  shown  by  Fig.  194,  II.  The  body  is  placed  on 
the  beam  as  before  and  moved  till  equilibrium  is  secured  with 
some  convenient  weight  on  the  opposite  end.  It  is  then  turned 
end  for  end  and  moved  along  the  scale  beam  until  the  same 
balance  is  secured,  with  all  other  weights  unchanged.  The 
center  of  gravity  is  now  in  the  same  position  which  it  occupied 
before  turning.  If  the  position  of  any  point  such  as  A  is  noted 
before  turning  and  again  after  turning,  the  difference  of  these 


272 


STRENGTH  OF  MATERIALS 


two  positions  is  twice  the  horizontal  distance  of  A  from  the 
center  of  gravity.  This  may  be  done  with  great  accuracy  on 
the  beam  of  a  platform  scale  by  clamping  to  the  beam  a  small 
steel  scale  for  determination  of  the  displacement  of  the  points  as 
shown  in  the  figure. 

PROBLEM. 

2.  A  body  is  balanced  on  a  scale  beam.  When  turned  around  and  again 
balanced,  it  is  found  that  the  point  originally  at  the  left  end  is  displaced  3.32 
inches.  How  far  is  the  center  of  gravity  from  this  end?  Ans.  1.66  inches. 

154.  Moment  of  Inertia  by  Experiment.  —  A  common  method 
of  finding  the  moment  of  inertia  of  an  irregular  body  is  that  of 
determining  its  effect  upon  the  time  of  vibration  of  a  torsion 
pendulum.  The  time  of  vibration  of  a  torsion  pendulum  varies 
as  the  square  root  of  the  moment  of  inertia  of  its  mass  with 
respect  to  the  vertical  line  which  is  the  axis  of  the  supporting 
wire.  This  relation  may  be  expressed  briefly: 

T*  =  KI, 

where  T  may  be  the  time  of  a  complete  period  or  of  a  single 
vibration  (with  K  varying  accordingly),  and  K  is  a  constant 


---B 


rrr 


Fig.  195.  —  Moment  of  Inertia  by  Torsional  Vibration. 

which  depends  upon  the  length,  diameter,  and  modulus  of  shear- 
ing elasticity  of  the  supporting  wire.  The  factors  which  make 
up  K  need  not  be  determined  separately,  as  the  entire  term 
may  be  obtained  by  substitution  from  the  time  of  vibration  of  a 
mass  of  known  moment  of  inertia. 

Fig.  195,  I  and  II,  shows  a  uniform  solid  circular  steel  or  brass 
rod  in  a  horizontal  position  on  a  light  support  suspended  by  a 
single  steel  wire.  Fig.  195,  III  and  IV,  shows  a  second  body 
on  the  same  support.  It  is  desired  to  find  the  moment  of  inertia 


COMPUTATION   WITHOUT  INTEGRALS        273 

of  this  second  body  with  respect  to  an  axis  through  its  center 
of  gravity  perpendicular  to  the  line  AB.  If  the  body  can  be  so 
supported  that  A  B  is  horizontal,  it  will  rotate  about  the  desired 
axis;  for  the  center  of  gravity  of  the  combined  body  and  support 
must  fall  directly  under  the  axis  of  the  wire,  and  if  .the  support 
is  small  relatively  this  combined  center  of  gravity  will  practi- 
cally coincide  with  that  of  the  body,  even  if  the  support  does 
not  hang  in  exactly  the  position  which  it  occupies  when  it  is 
not  loaded. 

When  the  moment  of  inertia  of  the  support  is  small,  as  in  Fig. 
195,  the  unknown  moment  of  inertia  is  calculated  from 


T  rji  2  ' 

Ic       Ic 

where  the  subscript  A  refers  to  the  body  and  the  subscript  C  to 
the  cylinder. 

Generally  it  is  not  practicable  to  use  a 
very  light  support  and  get  the  body  in  the 
desired  position.  Fig.  196  shows  a  relatively 
large  support  carrying  the  unknown  body 
in  the  side  elevation  of  Fig.  196,  I,  and  the 
known  cylinder  in  the  other.  In  this  case 
we  get  the  time  of  vibration  with  the  sup- 
port alone;  and  then  with  the  support  and 
each  load  separately. 


i  ii 

Fig.  196.  —  Support  for 
Torsional  Vibration. 


KIB, 


TA=K  (I A  +  IB}, 

where  Tc  is  the  time  with  support  and  cylinder;  TB,  with  the 
support  alone,  etc. 

T        (Tj  -  Tg)  Ic 

•*•  A  rriz  rpz 

1C  ~    L  B 

PROBLEMS. 

1.  The  time  of  vibration  of  a  given  torsion  pendulum  with  the  support 
alone  is  0.46  second;  with  the  support  loaded  with  a  cylinder  10  inches  long 
and  \  inch  in  diameter  it  is  0.87  second;  with  an  unknown  body  in  place  of 
the  cylinder  it  is  0.94  second.  The  cylinder  weighs  0.556  pound  and  the  body 
1  25  pounds.  Find  the  moment  of  inertia  and  radius  of  gyration  of  the  body. 

Ans.  k  =  2.14  inches. 


274  STRENGTH  OF  MATERIALS 

2.  Under  what  conditions  may  the  unknown  moment  of  inertia  be  accu- 
rately determined  without  getting  the  time  of  vibration  of  the  support? 

3.  If  any  clamp  screws  are  used  in  the  support,  they  should  be  vertical. 
Why? 

155.  The  Moment  of  Inertia  of  a  Plane  Section.  —  The  method 
of  the  preceding  article  affords  a  method  of  obtaining  the  moment 
of  inertia  of  any  plane  section  when  the  material  can  be  cut  up 
into  pieces.  Suppose  we  have  a  beam  of  any  irregular  section. 
Cut  out  a  piece  of  some  convenient  length  and  finish  the  ends  to 
parallel  planes  perpendicular  to  the  length  of  the  beam.  A  con- 
venient length  for  the  finished  piece  is  1  inch.  Determine  the 
area  of  cross  section  by  calculation  from  the  weight  and  the 
specific  gravity.  Get  the  center  of  gravity  by  the  method  of 
Article  153.  Suspend,  and  compute  the  moment  of  inertia. 
Divide  by  the  weight  for  k2.  This  k2  is  the  square  of  the  radius 
of  gyration  of  the  solid  prism  1  inch  long. 

From  equation  (1)  of  Article  150  we  know  that  the  square  of 
the  radius  of  gyration  of  a  prism  is  equal  to  the  sum  of  the  squares 
of  the  radius  of  gyration  as  a  thin  section  and  the  radius  of  gyra- 
tion as  a  rod.  In  this  case  the  square  of  the  radius  of  gyration  as 
a  rod  is  one-twelfth  of  the  square  of  the  length. 

PROBLEMS. 

1.  In  the  case  of  the  unknown  section  of  Fig.  195,  III,  and  IV,  the  length  I 
is  1  inch,  the  weight  in  air  1.524  pounds,  the  weight  in  water  1.326  pounds. 
The  water  was  at  the  temperature  at  which  the  density  is  62.2  pounds  per 
cubic  foot.     What  is  the  area  of  cross  section?  Ans.  5.50  square  inches. 

2.  On  a  torsion  pendulum  with  a  light  support  the  body  in  the  position 
shown  made  100  vibrations  in  83.2  seconds.     A  rod  \  inch  in  diameter  and 
12  inches  long  weighing  0.668  pound  makes  100  vibrations  in  163.8  seconds. 
What  is  the  radius  of  gyration  of  the  body  and  of  its  cross  section? 

Ans.  k  of  cross  section  is  1.127  inches. 

3.  In  Problem  2  what  is  the  moment  of  inertia  of  the  cross  section? 

Ans.  6.99  inches4. 

4.  The  center  of  gravity  of  the  section  of  Fig.  194,  III,  is  2.48  inches  from 
A.    What  is  the  section  modulus? 


CHAPTER  XIX. 


REPEATED   STRESSES. 

156.  Lag  of  Deformation  behind  Stress.  —  When  force  is  ap- 
plied to  an  elastic  solid  the  deformation  lags  ia  little  behind  the 
stress.  In  the  test  of  a  rod  in  tension,  if  the  machine  is  stopped 
with  the  beam  balanced,  after  a  short  time  it  is  found  to  be  no 
longer  in  equilibrium.  If  the  load  is  increasing  when  the  ma- 
chine is  stopped,  the  beam  drops,  showing  a  continued  elongation 
with  a  slightly  diminished  stress.  If  the  stress  is  below  the  yield 
point,  the  beam  drops  slowly;  if  above  the  yield  point,  it  comes 
down  quickly.  If  a  little  more  load  is  applied  so  as  to  raise  the 
beam,  it  comes  down  more  slowly  the  next  time.  In  any  case,  an 
equilibrium  can  finally  be  found  at  which  there  is  no  change. 


eoooo 

55000 
50000 
45000 
40000 


3  35000 

o 

10  30000 


15000 
10000 
5000 


f- 


0      .02    .04     .06 


.10     .12     .14    .16     .18    .20    .22     .24 
UNIT  ELONGATION 


.28    .30     .32    .34    .30    .38   .40 


Fig.  197.  —  Effect  of  Time  on  Stress-strain  Diagram. 

As  a  result  of  this  lag  of  deformation  there  is  a  considerable 
variation  in  the  form  of  the  stress-strain  diagram  of  any  material 
depending  upon  the  rate  of  speed  of  the  testing  machine.  If  the 
machine  runs  rapidly  the  diagram  will  be  higher  than  if  it  goes 
slowly.  Frequently  the  machine  is  run  at  a  rather  high  speed 
until  a  given  load  is  reached  and  then  stopped  to  read  the  exten- 
someters.  Table  XII  gives  the  results  for  a  bar  of  soft  steel 

275 


276 


STRENGTH  OF  MATERIALS 


tested  in  this  way.  The  poise  was  set  at  the  load  given  in  the 
first  column  and  the  rod  was  stretched  till  equilibrium  was  secured. 
It  was  then  stopped  for  one  minute,  at  the  end  of  which  time 
the  poise  was  moved  to  get  a  new  balance. 

Fig.  197  is  plotted  from  Table  XII  (together  with  some  inter- 
mediate points  omitted  from  the  table).  If  the  steel  were  tested 
by  means  of  an  autographic  machine  which  draws  the  stress- 
strain  diagram,  the  curve  would  be  still  higher  than  the  upper 
one  of  the  figure,  depending  upon  the  speed  of  the  machine. 

TABLE  XII. 

TEST  OF  SOFT  STEEL  IN  TENSION. 
Area  of  section,  0.600  square  inch. 


Total  load. 

Unit  stress  per  square  inch. 

Elongation  when  machine 
stopped. 

When  machine 
stopped. 

After  one  min- 
ute. 

When  machine 
stopped. 

After  one  min- 
ute. 

In  8  inches. 

Unit. 

Pounds. 

Pounds. 

Pounds. 

Pounds. 

Inches. 

Inch. 

30 

30 

50 

50 

0 

0 

6,150 

6,000 

10,250 

10,000 

.0027 

.00034 

9,000 

8,650 

15,000 

14,420 

.00405 

.00051 

12,200 

11,800 

20,330 

19,670 

.0055 

.00069 

15,000 

14,400 

25,000 

24,000 

.00675 

.00084 

18,000 

17,100 

30,000 

28,500 

.0082 

.00102 

19,200 

18,250 

32,000 

30,420 

.0085 

.00106 

19,800 

18,600 

33,000 

31,000 

.0087 

.00109 

20,400 

19,150 

34,000 

31,920 

.0092 

.00115 

21,000 

19,200 

35,000 

32,000 

.0121 

.00151 

20,600 

19,350 

34,330 

32,250 

.0450 

.00562 

20,600 

20,000 

34,330 

33,330 

.0530 

.00662 

21,000 

19,000 

35,000 

31,670 

.0733 

.00916 

21,000 

19,200 

35,000 

32,000 

.1740 

.02175 

21,600 

20,900 

36,000 

34,830 

.2121 

.02651 

22,800 

21,750 

38,000 

36,250 

.2393 

.02991 

24,000 

22,900 

40,000 

38,160 

.2773 

.03466 

26,400 

25,050 

44,000 

41,750 

.3873 

.04841 

28,800 

26,900 

48,000 

44,830 

.5506 

.06882 

31,200 

29,050 

52,000 

48,520 

.83 

.104 

32,400 

30,600 

54,000 

51,000 

1.30 

.162 

32,800 

30,850 

54,670 

51,420 

1.62 

.202 

32,850 

31,200 

54,750 

52,000 

1.89 

.236- 

32,750 

30,950 

54,580 

51,580 

2.08 

.260 

32,600 

30,900 

54,330 

51,500 

2.52 

.315 

23,400 

39,000 

2.98 

.372 

Broke  at  23,400  pounds.     The  area  of  the  neck  was  0.196  square  inch. 


REPEATED   STRESSES 


277 


The  static  load  which  would  produce  a  given  deflection  would 
come  below  the  lower  curve.  The  apparent  ultimate  strength 
of  this  steel  is  54,750  pounds  per  square  inch.  The  actual  ulti- 
mate strength  in  terms  of  the  original  area  is  less  than  52,000 
pounds  per  square  inch. 

From  this  table  and  curve  we  see  that  in  order  to  compare  the 
results  of  tests  of  materials  the  speed  of  the  test  is  an  important 
factor.  This  is  especially  the  case  with  very  ductile  materials. 
The  modulus  of  elasticity  taken  rapidly  is  much  larger  than  if 
taken  slowly. 

The  form  of  the  stress-strain  diagram  and  the  amount  of  lag 
of  deformation  depend  somewhat  upon  the  previous  treatment 
of  the  material.  A  bar  of  hot-rolled  steel  when  tested  for  the 
first  time  will  generally  show  some  permanent  set  at  loads  below 
the  true  elastic  limit.  If  it  is  raised  nearly  to  the  yield  point  and 
then  reduced  to  zero  and  again  raised,  the  second  test  will  show 
no  set  and  will  generally  give  a  better  curve  than  the  first  one. 
If  steel  or  wrought  iron  is  carried  beyond  the  yield  point  and  the 
load  released,  the  second  test  will  show  a  yield  point  above  the 
previous  maximum  stress. 

TABLE  XIII. 

SOFT  STEEL  BEYOND  THE  YIELD  POINT. 
Original  area,  0.600  square  inch. 


Total  load. 

Unit  stress  per  square  inch. 

Elongation  when  machine 
stopped. 

When  machine 
stopped. 

After  one  min- 
ute. 

When  machine 
stopped. 

After  one  min- 
ute. 

In  8  inches. 

Unit. 

Pounds. 

Pounds. 

Pounds. 

Pounds. 

Inch. 

Inch. 

23,400 

39,000 

-.0576 



24  000 

23  300 

40  000 

38,830 

-.0441 

24,600 

23,500 

41,000 

39,170 

-.0148 

25,200 

.24,100 

42,000 

40,170 

.01425 

!  00178 

21,000 

21,075 

35,000 

35,125 

.01275 

.00159 

18,000 

18,025 

30,000 

30,040 

.0111 

.00139 

15,000 

15,125 

25,000 

25,210 

.0094 

.00117 

12,000 

12,150 

20,000 

20,250 

.00755 

.00096 

9,000 

9,325 

15,000 

15,540 

.00505 

.00076 

6,000 

6,250 

10,000 

10,420 

.00415 

.00052 

3,000 

3,175 

5,000 

5,290 

.0021 

.00026 

30 

125 

50 

210 

.0001 

30 

30 

50 

50 

.0 

'.6" 

3,000 

3,000 

5,000 

5,000 

.00105 

.00013 

6,000 

6,000 

10,000 

10,000 

.0026 

.00032 

278 


STRENGTH  OF  MATERIALS 


TABLE  XIII    (Continued-). 


Total  load. 

Unit  stress  per  square  inch. 

Elongation  when  machine 
stopped. 

When  machine 
stopped. 

After  one  min- 
ute. 

When  machine 
stopped. 

After  one  min- 
ute. 

In  8  inches. 

Unit. 

Pounds. 

Pounds. 

Pounds. 

Pounds. 

Inch. 

Inch. 

9,000 

8,900 

15,000 

14,830 

.00435 

.00054 

12,000 

11,900 

20,000 

19,830 

.0061 

.00076 

15,000 

14,950 

25,000 

24,920 

.00785 

.00098 

18,000 

17,825 

30,000 

29,710 

.00965 

.00121 

21,000 

20,900 

35,000 

34,830 

.0116 

.00145 

24,000 

23,600 

40,000 

39,330 

.01355 

.00169 

24,600 

24,200 

41,000 

40,330 

.0154 

.00192 

25,200 

24,550 

42,000 

40,920 

.01765 

.00221 

24,000 

23,900 

40,000 

39,830 

.01735 

.00217 

21,000 

21,150 

35,000 

35,250 

.0158 

.00197 

18,000 

18,100 

30,000 

30,170 

.0142 

.00177 

15,000 

15,075 

25,000 

25,125 

.0123 

.00154 

12,000 

12,225 

20,000 

20,375 

.0108 

.00135 

9,000 

9,300 

15,000 

15,500 

.0090 

.00112 

6,000 

6,225 

10,000 

10,375 

.0073 

.00091 

3,000 

3,125 

5,000 

5,210 

.0052 

.00065 

30 

120 

50 

200 

.0035 

30 

30 

50 

50 

.0033 

'.0004i 

Test  bar  rested  40  hours  without  load. 

30 

30 

50 

50 

.0033 

.00041 

3,000 

2,970 

5,000 

4,950 

.00455 

.00057 

6,000 

6,000 

10,000 

10,000 

.0059 

.00074 

9,000 

8,900 

15,000 

14,830 

.00745 

.00093 

12,000 

11,850 

20,000 

19,750 

.00895 

.00112 

15,000 

14,825 

25,000 

24,710 

.01035 

.00129 

18,000 

17,850 

30,000 

29,750 

.01185 

.00148 

21,000 

20,800 

35,000 

34,670 

.0134 

.00167 

24,000 

23,850 

40,000 

39,750 

.01505 

.00188 

21,000 

21,000 

35,000 

35,000 

.01365 

.00171 

18,000 

18,100 

30,000 

30,170 

.01215 

.00152 

15,000 

15,100 

25,000 

25,170 

.01075 

.00134 

12,000 

12,100 

20,000 

20,170 

.00925 

.00116 

9,000 

9,050 

15,000 

15,080 

.00785 

.00098. 

6,000 

6,175 

10,000 

10,290 

.00635 

.00079 

3,000 

3,175 

5,000 

5,290 

.00475 

.00059 

30 

30 

50 

50 

.0033 

.00041 

3,000 

3,000 

5,000 

5,000 

.00445 

.00056 

6,000 

5,950 

10,000 

9,920 

.00575 

.00072 

9,000 

8,850 

15,000 

14,750 

.00735 

.00092 

12,000 

11,850 

20,000 

19,750 

.0089 

.00111 

15,000 

14,900 

25,000 

28,830 

.01045 

.00131 

18,000 

17,600 

30,000 

29,330 

.0119 

.00149 

24,000 

23,800 

40,000 

39,670 

.0151 

.00189 

24,600 

24,500 

41,000 

40,830 

.0155 

.00194 

REPEATED   STRESSES 


279 


TABLE  XIII     (Concluded). 


Total  load. 

Unit  stress  per  square  inch. 

Elongation  when  machine 
stopped. 

When  machine 
stopped. 

After  one  min- 
ute. 

When  machine 
stopped. 

After  one  min- 
ute. 

In  8  inches. 

Unit. 

Pounds. 
25,200 

25,800 
26,400 
27,000 
27,600 

Pounds. 

Pounds. 

42,000 
43,000 
.  44,000 
45,000 
46,000 

Pounds. 

42,330 
43,580 
44,170 
44,170 

Inch. 

.01585 
.0162 
.0169 
.0174 
.01865 

Inch. 
.00198 
.00202 
.00212 
.00217 
t  .00233 

25,400 

26,150 
26,500 
26,500 

The  unit  stress  was  calculated  from  the  original  area  of  0.600  square  inch. 
The  actual  area  was  0.578  square  inch.  The  actual  length  of  the  gauged  por- 
tion was  8.3  inches  at  zero  elongation  as  read. 

Table  XIII  gives  some  of  the  results  for  a  test  piece  from  the 
same  rod  as  Table  XII.  This  test  piece  was  loaded  to  36,000 
pounds  per  square  inch,  producing  an  elongation  of  about  0.3 
inch  in  a  gauged  length  of  8  inches.  The  zero  of  the  extensom- 
eter  corresponds  to  0.3  inch  total  elongation. 


(35000 
60000 
55000 
50000 
45000 
40000 
35000 
30000 
25000 
20000 
15000 
10000 
5000 
0 

( 

H 

c 

D    , 

r*-""' 

F 

I 

B 

II 

A 

'fa 

/G 

f 

/ 

/ 

/ 

A 

A, 

/\ 

^ 

/ 

^ 

//* 

/ 

v 

' 

/ 

/* 

o/ 

/ 

y 

/ 

/ 

/ 

/£/ 

/ 

/ 

, 

/x 

' 

// 

' 

/ 

fr 

F 

/" 

G' 

.0002        .0000         .0010       .0014  .0375       .0379        .0383        .0387        .0391         .0395       .039 

UNIT  ELONGATION 

Fig.  198.  —  Stress-strain  Cycles. 

Curve  I  of  Fig.  198  is  plotted  from  Table  XII.  Curve  II  is 
plotted  from  Table  XIII.  The  test  of  the  piece  used  for  Table 
XIII  showed  that  the  curve  up  to  the  yield  point  was  practically 
the  same  as  that  of  Table  XII.  The  abscissas  of  curve  II 
begin  with  0.0375,  which  corresponds  with  the  0  of  Table  XIII. 

After  passing  the  point  D  on  curve  II  we  notice  that  the 
descending  portion  is  concave  toward  the  left.  The  same  is 
true  of  the  descending  portion  from  E  to  F.  All  of  these  lines 


280  STRENGTH  OF  MATERIALS 

are  drawn  from  the  readings  at  the  time  the  machine  stopped. 
If  we  were  to  draw  the  curve  for  the  end  of  one  minute  rest,  the 
descending  portions  from  D  and  E  would  lie  to  the  left  of 
the  curves  as  drawn  and  would  be  less  concave.  If  we  observe 
the  table  we  notice  that  in  going  down  from  D  the  unit  stress 
with  a  given  elongation  increases  after  a  short  rest.  The  maxi- 
mum increase  occurs  at  the  unit  stress  of  9000  pounds.  When 
soft  steel  is  stretched  it  returns  slowly  to  its  original  length  after 
the  load  is  removed.  The  deformation  lags  behind  the  stress 
in  both  the  ascending  and  descending  portions. 

In  coming  from  C  to  D,  the  maximum  load  at  stopping  the 
machine  was  42,000  pounds  per  square  inch,  but  the  maximum 
load  after  one  minute  was  only  40,170.  We  notice  that  the 
curve  coming  up  from  zero  load  is  practically  straight  (except 
at  the  lower  end)  until  40,000  pounds  is  reached,  where  its  slope 
drops  suddenly.  There  is  a  considerable  permanent  deforma- 
tion in  passing  from  40,000  to  42,000  pounds  at  E.  While  the 
curve  has  the  same  ordinates  at  D  and  E  the  permanent  stress 
is  750  pounds  greater  at  E.  Below  the  35,000-pound  ordinate 
the  descending  curves  from  D  and  E  are  the  same. 

The  ascending  curve  is  nearly  straight  up  to  40,000  pounds 
while  the  original  yield  point  was  less  than  35,000  pounds.  This 
is  an  illustration  of  the  fact,  mentioned  in  Article  25,  that  stress 
beyond  the  yield  point  raises  the  yield  point  of  soft  steel.  The 
new  yield  point  is  found  at  the  permanent  unit  stress  which  was 
previously  reached. 

From  F  to  G  and  back  to  F  is  the  cycle  after  40  hours'  rest 
without  load.  These  curves  are  more  nearly  straight  than  the 
others.  The  descending  and  ascending  portions  are  close  to- 
gether, and  the  lag,  as  shown  from  the  tables,  is  smaller.  The 
descending  curve  returns  to  zero,  showing  that  the  true  elastic 
limit  is  not  necessarily  the  limit  of  zero  set.  A  stress-strain 
diagram  may  be  considerably  curved  and  the  material  show  no 
permanent  set. 

Ascending  from  F  to  G  the  second  time  the  curves  coincide. 
The  effect  of  the  rest  is  shown  by  the  fact  that  the  curve  from  G 
to  H  lies  considerably  above  the  original  curve  to  E. 

The  ascending  portion  of  the  first  curve  has  a  less  slope  than 
the  similar  curve  after  rest,  showing  the  effect  of  rest  in  raising 
the  modulus  of  elasticity.  The  curve  FG  has  a  slightly  smaller 


REPEATED   STRESSES 


281 


slope  than  curve  I,  showing  an  apparent  decrease  in  the  mod- 
ulus due  to  overstraining.  The  unit  stresses  for  curve  II  are 
computed  from  the  original  area  of  0.600  square  inch,  while 
the  actual  area  at  the  new  zero  elongation  was  0.578  square 
inch.  Also  the  unit  elongations  are  computed  from  the  original 
length  of  8  inches.  If  we  use  the  actual  area  in  computing  the 
stress  and  the  actual  length  of  8.3  inches  in  calculating  the  unit 
deformations  the  modulus  is  practically  the  same  for  curve  I 
and  for  curve  II  from  F. 

PROBLEMS. 

1.  From  Table  XII  calculate  E  for  at  least  three  readings. 

2.  From  Table  XIII  calculate  E  for  at  least  two  readings  beginning  with 
the  initial  load,  and  for  two  readings  beginning  with  3000  pounds,  with  the 
actual  area  and  original  length. 

157.   Watertown  Arsenal  Tests  of  Eyebars.  —  An  interesting 
set  of  tests  illustrating  the  behavior  of  steel  and  wrought  iron 


UNIT  STRESS  IN  POUNDS  PER  SQUARE  INCH 

D 

a 

4 

/ 

ii 

/ 

y 

? 

7 

I 

c 

/ 

'/ 

t 

/ 

// 

// 

S 

// 

// 

Y 

> 

y 

// 

/ 

> 

'/ 

/f 

/ 

yi 

// 

/ 

f 

/ 

/ 

.0002         .0006          .0010         .0014    0    .0002         .0006         .0010        .0014          .0018         .0022 
UNIT  ELONGATION 

Fig.  199. 

was  made  at  the  Watertown  Arsenal  in  1886  (Tests  of  Metals, 
1886,  Part  2,  pp.  1571-1617).  These  tests  were  made  on  eye- 
bars  about  25  feet  long.  The  gauged  length  was  260  inches, 
enabling  the  unit  deformation  to  be  read  with  great  accuracy. 
Table  XIV  gives  some  of  the  results  for  the  first  loading  of  test 
piece  No.  4136  (p.  1578).  The  initial  load  (as  is  customary  in 
the  Watertown  tests)  was  1000  pounds  per  square  inch.  After 


282 


STRENGTH  OF  MATERIALS 


each  load  the  machine  was  reversed  to  the  initial  load  to  get  the 
permanent  elongation  which  is  given  in  the  table  as  "  Set  under 

TABLE  XIV. 

WATERTOWN  TESTS  OF  STEEL  EYEBAR. 


Total  load. 

Unit  stress 
per  square 
inch. 

Elongation. 

Set  under 
initial  load. 

Net  unit 
elongation. 

E 

In  260 
inches. 

Unit. 

Pounds. 

Pounds. 

Inches. 

Inch. 

Inches. 

Inch. 

5,250 

1,000 

0 

0 

26,250 

5,000 

.0456 

.000175 

.0065 

.000150 

26,670,000 

52,500 

10,000 

.0915 

.000352 

.0085 

.000319 

28,210,000 

78,750 

15,000 

»  .1369 

.000526 

.0089 

.000485 

28,860,000 

105,000 

20,000 

.1815 

.000698 

.0096 

.000661 

28,750,000 

131,250 

25,000 

.2264 

.000871 

.0101 

.000832 

28,850,000 

157,500 

30,000 

.2720 

.001046 

.0109 

.001004 

28,880,000 

183,750 

35,000 

.3194 

.001229 

.0147 

.001172 

29,010,000 

196,000 

37,330 

.3459 

.001330 

198,000 

37,710 

.3700 

.001423 

200,000 

38,090 

.5665 

.002179 

204,750 

39,000 

1.07 

.0041 

210,000 

40,000 

2.35 

.0090 

369,000 

70,286 

30.42 

The  bar  was  again  tested  to  a  total  load  of  262,500  pounds,  or  50,000  pounds 
per  square  inch  in  terms  of  the  original  area.  After  this  load  the  set  was 
negative,  and  equal  to  10  divisions.  After  a  further  rest  of  one  hour  at  the 
initial  load  the  reading  was  —  0.0150  inch. 

The  micrometer  was  reset  to  zero  and  the  bar  was  again  loaded  to  262,250 
pounds.  The  elongation  under  this  load  was  0.6263  inch  in  the  original  length 
of  260  inches.  The  immediate  set  was  0.0028  inch.  After  10  minutes  the  set 
was  0.0005  inch,  and  after  12  minutes  it  was  zero. 

The  micrometer  was  reset  to  zero  and  the  bar  was  tested  the  fourth  time 
to  a  load  of  341,250  pounds  and  back  at  intervals  of  5000  pounds  per  square 
inch  in  terms  of  the  original  area.  The  elongation  was  0.8660  inch.  The 
immediate  set  was  0.0301  inch.  After  20  minutes  the  set  was  0.0215  inch. 

The  micrometer  was  reset  and  the  bar  was  tested  to  262,500  pounds.  The 
elongation  in  the  gauged  length  was  0.6257  inch.  The  immediate  set  was 
0.0041  inch  and  the  set  after  6  minutes  was  0. 

>The  micrometer  was  reset  to  zero  and  the  bar  again  tested  to  262,500 
pounds.  At  each  load  the  elongation  was  taken  immediately  and  again  after 
an  interval  of  3  minutes.  The  results  are  given  in  Table  XV,  which  shows 
the  change  in  deformation  with  constant  stress,  while  Tables  XII  and  XIII 
show  the  change  in  stress  at  nearly  constant  elongation.  In  computing  the 
unit  stresses  for  this  table  the  permanent  area  at  the  end  of  the  first  test  is 
used  instead  of  the  original  area.  This  permanent  area  is  given  as  4.70  square 
inches.  Also  in  computing  the  unit  elongations  the  length  is  taken  as  290.4 
inches. 


REPEATED   STRESSES 


283 


TABLE  XV. 

WATERTOWN  TEST  OP  STEEL  EYEBAR. 

Area,  4.70  square  inches.     Gauged  length  290.4  inches. 

Bar  previously  stretched  from  260  inches  by  a  load  of  369,000  pounds. 


Total  load. 

Unit  stress  per 
square  inch. 

Elongation  in  290.4  inches. 

Immediate  unit 
elongation. 

Immediate. 

After  3  min- 
utes. 

Pounds. 

Pounds. 

Inch. 

Inch. 

Inch. 

5,250 

1,117 

0 

26,250 

5,585 

.0503 

.0504 

.000173 

52,500 

11,170 

.1093 

.1099 

.000376 

78,750 

16,755 

.1685 

.1690 

.000582 

105,000 

22,340 

.2299 

.2309 

.000792 

131,250 

27,925 

.2922 

.2935 

.001006 

157,500 

33,510 

.3562 

.3575 

.001226 

183,750 

39,095 

.4209 

.4222 

.001449 

210,000 

44,680 

.4868 

.4885 

.001676 

236,250 

50,265 

.5533 

.5549 

.001905 

262,500 

55,850 

.6209 

.6230 

.002148 

236,250 

50,265 

.5660 

.5659 

.001949 

210,000 

44,680 

.5082 

.5080 

.001750 

183,750 

39,095 

.4491 

.4489 

.001546 

157,500 

33,510 

.3886 

.3880 

.001338 

131,250 

27,925 

.3262 

.3252 

.001123 

105,000 

22,340 

.2631 

.2620 

.000906 

78,750 

16,755 

.1980 

.1971 

.000682 

52,500 

11,170 

.1331 

.1319 

.000455 

26,250 

5,585 

.0655 

.0627 

.000225 

5,250 

1,117 

.0048 

.0030 

.000016 

Bar  rested  15  hours  under  initial  load. 

5,250 



-.0054 

5,250 

1,117 

0 

(Micrometer 

26,250 

5,585 

.0504 

reset  to  0) 

.000173 

52,500 

11,170 

.1084 

.000373 

78,750 

16,755 

.1660 

.000572 

105,000 

22,340 

.2250 

.000775 

131,250 

27,925 

.2870 

.000988 

157,500 

33,510 

.3510 

.001209 

183,750 

39,095 

.4157 

.001431 

210,000 

44,680 

.4839 

.001666 

236,250 

50,265 

.5493 

.001891 

262,500 

55,850 

.6180 

.002128 

236,250 

50,265 

.5621 

.001935 

210,000 

44,680 

.5047 

.001739 

183,750 

39,095 

4459 

001535 

157,500 

33,510 

.3857 

.001328 

131,250 

27,395 

.3241 

.001082 

1 

284 


STRENGTH  OF  MATERIALS 


TABLE  XV  (Continued). 


Total  load. 

Unit  stress  per  square 
inch. 

Elongation  in  290.4 
inches. 

Unit  elongation. 

Pounds. 

Pounds. 

Inch. 

Inch. 

105,000 

22,340 

.2609 

.000898 

78,750 

16,755 

.1973 

.000679 

52,500 

11,170 

.1321 

.000455 

26,250 

5,585 

.0644 

.000222 

5,250 

1,117 

.0065 

.000023 

initial  load."  To  get  the  net  elongation  we  subtract  this  set 
from  the  total  elongation  before  dividing  by  the  original  length. 
The  results  are  given  graphically  by  curve  I  of  Fig.  199.  The 
net  unit  elongations  are  represented  by  the  circles  and  the  gross 
unit  elongations  by  the  crosses. 

Curve  II  of  Fig.  199  represents  the  first  portion  of  Table  XV. 
The  curve  is  plotted  from  the  immediate  unit  elongation.  The 
curve  is  practically  straight  from  K  to  D.  (D  is  considerably 
below  the  maximum  stress  of  Table  XIV.)  From  0  to  K  the 
curve  is  nearly  a  straight  line.  A  consideration  of  the  table 
shows  that  the  bar  continued  to  stretch  at  a  constant  stress  when 
the  load  was  ascending  and  shortened  under  a  constant  stress 
on  the  descending  curve.  The  greatest  difference  in  3  minutes 
in  the  descending  portion  occurred  at  the  apparent  unit  stress 
of  5000  pounds.  The  effect  of  the  lag  in  deformation  is  further 
shown  by  the  fact  that  the  zero  fell  0.0102  inch  in  15  hours 
of  rest. 

PROBLEMS. 

1.  From  the  curve  of  unit  stress  and  net  unit  elongation  of  Fig.  198, 1, 
determine  the  modulus  of  elasticity  of  this  steel. 

•  2.  Plot  the  stress-strain  diagram  from  the  second  part  of  Table  XV  which 
gives  the  results  after  15  hours  rest.  Calculate  E. 

3.  Using  the  original  area  and  the  original  length  calculate  E  from  the  last 
part  of  Table  XV. 


158.  Failure  under  Repeated  Stress.  —  There  is  a  consider- 
able area  between  the  ascending  and  descending  portions  of  the 
stress-strain  diagram  in  Figs.  198  and  199.  The  greater  the 
limits  of  stress  the  greater  this  area.  This  inclosed  area  is  a 
measure  of  the  work  expended  in  stretching  the  bar  which  is 
not  recovered  as  mechanical  work  when  the  load  is  released. 


REPEATED   STRESSES  285 

This  energy  is  lost  as  mechanical  energy,  being  transformed  into 
heat  or  expended  in  changing  the  molecular  condition  of  the 
material. 

PROBLEMS. 

1.  From  the  readings  of  Table  XIII  find  the  lost  energy  in  a  portion  of 
the  bar  one  inch  in  length  in  passing  through  the  cycle  from  F  to  G  and  back 
to  F  (Fig.  198,  II).  Ans.  0.8  inch  pound. 

2.  From  Table  XV  find  the  total  energy  expended  in  the  entire  gauged 
length  during  one  cycle.     Also  solve  after  15  hours'  rest. 

Since  energy  is  lost  in  a  cycle  of  this  kind  it  is  natural  to  expect 
that  a  great  number  of  repetitions  of  stress  would  cause  failure 
at  a  maximum  stress  less  than  the  ultimate  strength  of  the 
material.  The  experiments  of  Wohler  and  others  show  that  this 
is  the  case.* 

When  the  stress  varied  from  zero  to  a  maximum  it  was  found 
that  if  this  maximum  was  less  than  one-half  the  ultimate  strength 
the  piece  would  fail  under  a  great  number  of  repetitions  of  load. 
If  a  steel  bar  having  an  ultimate  strength  of  60,000  pounds  per 
square  inch  is  loaded  from  0  to  40,000  pounds  it  will  probably 
break  after  a  few  thousand  applications.  If  loaded  from  0  to 
35,000  it  will  last  much  longer.  If  from  0  to  30,000  it  may  fail 
after  several  million  repetitions.  If  loaded  from  0  to  25,000  it 
will  last  indefinitely.! 

When  the  stress  does  not  return  entirely  to  zero  the  area  of  the 
figure  representing  the  lost  energy  is  less.  The  experiments  with 
repeated  stresses  show  that  the  maximum  stress  can  be  greater 
without  failure  under  an  indefinite  number  of  repetitions.  Steel 
having  an  ultimate  strength  of  60,000  pounds  per  square  inch 
will  stand  a  stress  varying  from  25,000  to  40,000  pounds,  or 
from  20,000  to  35,000  without  failure. 

If  the  stress  changes  from  compression  to  tension  the  maxi- 
mum is  less  than  for  the  case  of  one  kind  of  stress  only.  Experi- 

*  See  Goodman's  "Mechanics  Applied  to  Engineering,"  under  the  head 
"Wohler's  Experiments."  Unwin's  "The  Testing  of  Materials  of  Construc- 
tion," pages  356-394,  gives  an  excellent  discussion  of  this  subject.  Also  see 
paper  by  Henry  B.  Seaman,  Transactions  of  the  American  Society  of  Civil 
Engineers,  Vol.  XLVI  (1899),  pages  141  to  150,  and  discussion  on  pages  166 
to  257. 

f  See  paper  by  J.  H.  Smith  entitled  "Some  Experiments  on  Fatigue  of 
Metals."  The  Journal  of  the  Iron  and  Steel  Institute,  1910,  Vol.  II,  pp.  246- 
318. 


286  STRENGTH  OF  MATERIALS 

ments  show  that  when  a  bar  is  tested  in  one  direction  its  elastic 
limit  in  the  other  is  lowered,  so  that  the  raising  of  the  elastic 
limit  which  occurs  when  a  bar  of  ductile  material  is  overstrained 
in  one  direction  is  lost  when  the  reverse  stress  is  applied.  The 
experiments  of  Wohler  show  that  steel  having  an  ultimate 
strength  of  60,000  pounds  per  square  inch  when  tested  in  tension 
will  fail  under  a  stress  which  changes  from  16,000  compression 
to  16,000  pounds  tension.  If  the  stress  changes  from  14,000 
tension  to  14,000  compression  the  piece  will  probably  stand  an 
indefinite  number  of  repetitions. 

159.  Design  for  Varying  Stresses.  —  A  number  of  methods 
have  been  proposed  for  designing  members  subjected  to  repeated 
stresses.  This  may  be  done  by  lowering  the  allowable  unit  stress 
or  adding  a  suitable  increment  to  the  applied  load. 

The  formula  of  Launhardt  is  an  empirical  formula  based  on 
Wohler's  experiments,  which  until  recently  was  considerably 
used  for  calculating  the  allowable  working  stress  for  varying 
loads.  This  formula  contains  a  factor  depending  upon  the  ratio 
of  the  ultimate  static  strength  to  the  ultimate  repetition  strength 
when  the  load  varies  from  0  to  the  maximum.  If  we  take  this 
ratio  as  2  which  coincides  reasonably  well  with  the  results  of  the 
tests,  the  formula  may  be  written 

_  sw/         minimum  load\ 
2  V         maximum  load/ 
where 

Sw  =  static  allowable  unit  stress, 

sv  =  maximum  allowable  unit  stress  with  varying  load. 

Q 

When  the  minimum  load  is  0,  sv  =  ^» 

z 

When  the  minimum  load  equals  the  maximum  load,  sv  =  sw. 


PROBLEMS. 

1.  If  the  allowable  unit  stress  for  a  given  steel  for  a  static  load  is  15,000 
pounds  per  square  inch,  what  is  the  maximum  allowable  unit  load  and  the 
required  area  of  cross  section  when  the  load  varies  from  20,000  to  30,000? 

Ans.  12,500  pounds  per  square  inch,  2.4  square  inches. 

2.  Find  the  area  of  cross  section  to  carry  safely  a  load  which  varies  from 
120,000  to  360,000  pounds  if  the  allowable  static  unit  stress  is  15,000  pounds 
per  square  inch.  Ans.  36  square  inches. 


REPEATED   STRESSES  287 

Launhardt's  Formula  applies  to  stresses  in  one  direction  only. 
Goodman  *  recommends  a  simple  rule  which  is  easy  to  remember 
and  convenient  to  apply.  Add  to  the  maximum  load  the  difference 
between  the  maximum  and  minimum  load  and  treat  the  sum  as  a 
static  load. 

PROBLEMS. 

3.  Solve  Problem  1  by  Goodman's  "dynamic"  rule. 

Ans.  2.67  square  inches. 

4.  What  is  the  cross  section  required  to  carry  a  load  which  varies  from 
30,000  pounds  compression  to  60,000  pounds  tension  if  the  allowable  static 
unit  stress  is  12,000  pounds  per  square  inch?  Ans.  12.5  square  inches. 

5.  A  shaft  is  supported  between  bearings  4  feet  apart  and  carries  a  load 
of  600  pounds  at  the  middle.     If  the  allowable  static  unit  stress  is  12,000 
pounds,  what  is  the  minimum  diameter  of  the  shaft  to  allow  for  the  alternate 
tension  and  compression  as  the  shaft  rolls  over? 

If  the  shaft  makes  100  revolutions  per  minute  in  what  time  will  the  stress 
reverse  one  million  times? 

1 60.  Impact  Stresses.  —  In  order  to  determine  the  magnitude 
of  the  stress  due  to  a  given  load  it  is  necessary  to  know  how  the 
load  is  applied. 

Loads  which  are  fixed  in  position  and  constant  in  magnitude 
are  dead  loads.  The  weight  of  a  structure  is  a  dead  load.  A 
load  which  is  applied  gradually,  as  the  weight  of  falling  snow,  is 
treated  as  a  dead  load.  A  load  which  varies  in  position,  such  as 
the  weight  of  a  moving  train  on  a  bridge,  is  a  live  load.  Any 
load  which  changes  in  position  or  magnitude  will  produce  impact 
stresses.  The  magnitude  of  this  impact  factor  depends  upon  the 
speed  of  application. 

Fig.  200,  I,  represents  a  suspended  spring.  In  II  a  load  is 
attached  to  the  spring  but  supported  by  B  so  that  it  exerts  no 
pull  on  the  spring.  In  III  the  support  B  is  lowered  gradually;  a 
part  of  the  weight  W  is  carried  by  B  and  the  remainder  by  the 
spring  S.  In  IV  the  spring  supports  the  entire  weight.  The 
upward  pull  exerted  by  the  spring  is  equal  to  the  weight.  If  P 
is  the  pull  required  to  stretch  the  spring  unit  distance,  the  total 
pull  in  position  IV  is  Pyi  where  y\  is  the  total  elongation.  If 
the  support  B  is  lowered  gradually  so  that  the  average  value  of 

*  Goodman's  "Mechanics  Applied  to  Engineering,"  page  535.  For  a 
discussion  of  the  various  formulas  see  Johnson's  "Materials  of  Construction," 
Article  389. 


288  STRENGTH  OF  MATERIALS 

the  sum  of  the  pull  of  the  spring  and  the  reaction  of  the  support 
is  equal  to  TF,  the  body  will  come  to  rest  without  vibration. 


W 

2/1  =  -p- 

The  energy  stored  in  the  spring  is  —  ^  ,  which  is  only  one-half 

& 

of  the  work  of  gravity  on  the  mass  W.     The  remaining  half  of 
the  energy  is  expended  on  the  support  B  which  moves  a  distance 

Pv 
y  with  an  average  push  of  —^  - 

If  the  support  B  is  suddenly  removed  from  W  in  the  position 
II,  the  entire  force  of  gravity  is  effective  throughout  the  whole 
distance.  At  first  the  spring  offers  no  resistance  and  the  entire 
load  goes  to  accelerate  the  mass  (provided  the  mass  of  the  spring 
is  negligible).  As  it  is  stretched,  the  resistance  of  the  spring 
increases.  At  the  position  IV  the  pull  of  the  spring  is  equal  to 
the  weight  and  the  acceleration  is  zero.  The  mass  has  its  highest 
velocity  at  the  point  where  it  would  come  to  rest  under  a  gradu- 
ally applied  load.  Beyond  this  point,  represented  by  IV.  the 
upward  pull  of  the  spring  is  greater  than  the  weight  and  the  body 
is  negatively  accelerated.  It  finally  stops  at  the  position  of 
Fig.  200,  V,  with  an  elongation  of  the  spring  yz.  To  calculate 
this  elongation,  we  have 


, 

2W 

2/2  =  -p-  =  2  t/i, 

=  2W. 

The  deflection  due  to  a  suddenly  applied  load  is  twice  as  great 
as  when  the  load  is  gradually  applied,  and  the  maximum  force 
is  twice  the  load.  After  reaching  the  maximum  elongation  the 
body  vibrates  back  to  its  original  starting  point  (provided  the 
spring  is  perfectly  elastic). 

Fig.  200,  VI,  shows  the  mass  W  lifted  a  distance  h  above  the 
position  of  II,  in  which  it  exerts  no  pull  on  the  spring.  If  released 
suddenly,  it  falls  this  distance  before  it  begins  to  stretch  the 


REPEATED   STRESSES 


289 


spring.  The  total  work  done  by  gravity  is  the  weight  multiplied 
by  the  total  distance  h  -f-  y.  At  the  lowest  position  VII  this 
work  has  been  transformed  to  energy  of  the  spring. 

Py2 

-tr)^-f  • 


s  i 


1 


VI 


lw\ 

VII 


Fig.  200.  —  Effect  of  Sudden  Loads  and  Impact. 


PROBLEMS. 

1.  A  force  of  6  pounds  stretches  a  given  spring  1  foot.     A  4-pound  mass 
is  placed  on  the  spring  and  gradually  lowered.     What  is  the  elongation  of  the 
spring  when  it  comes  to  rest?  Ans.  8  inches. 

2.  In  Problem  1  the  load  is  applied  suddenly.     What  is  the  elongation  of 
the  spring  and  the  maximum  pull?  Ans.  16  inches;  8  pounds. 

3.  In  Problem  1  the  load  is  lifted  1  foot  and  then  released  suddenly.     How 
much  does  it  stretch  the  spring,  and  what  is  the  maximum  tension? 

Ans.  2  feet;  12  pounds. 

4.  A  springboard  is  made  of  a  plank  12  inches  wide  and  2  inches  thick  as 
a  cantilever  10  feet  long.     What  is  the  maximum  fiber  stress  when  a  boy 
weighing  60  pounds  steps  on  it  suddenly  from  a  point  at  the  same  level  as  the 
end?  Ans.  1800  pounds  per  square  inch. 

5.  What  is  the  maximum  fiber  stress  in  Problem  4,  if  the  boy  jumps  down 
from  a  point  6  inches  above  the  end  of  the  plank,  if  the  modulus  of  the  timber 
is  1,200,000?  Ans.  5545  pounds  per  square  inch. 

In  most  cases  a  varying  load  requires  some  time  for  its  applica- 
tion, so  that  the  stress  produced  by  a  live  load  is  something  less 
than  twice  that  of  an  equal  static  load.  When  a  locomotive 
runs  on  a  bridge,  the  effective  stress  produced  may  be  50  per  cent 
greater  than  that  due  to  its  weight  alone.  We  say  then  that  an 
impact  factor  of  50  per  cent  should  be  added  to  the  live-load 
stress.  If  the  speed  is  reduced,  the  impact  factor  is  smaller. 


290  STRENGTH  OF  MATERIALS 

For  the  impact  factor  which  should  be  added  to  the  live-load 
stress  in  the  case  of  bridges,  consult  the  specifications  of  the 
American  Railway  Engineering  and  Maintenance  of  Way  Asso- 
ciation.* 

*  The  formula  used  is  an  empirical  one  based  on  experiments.  For  a 
description  of  the  experiments  upon  which  the  formula  is  based,  see  the  paper 
of  F.  E.  Turneaure  in  the  Transactions  of  the  American  Society  of  Civil 
Engineers,  Vol.  XLI,  pages  410-466. 


INDEX. 


A. 

Actual  unit  stress,  24. 
Allowable  unit  stress,  4,  5. 
American  Railway  Engineering  and 
Maintenance   of   Way   Associa- 
tion: 

allowable  stress,  5. 
column  formula,  208-211. 
impact  rule,  290. 

American  Society  of  Civil  Engineers: 
reference  to  transactions,  197,  199, 

285,  290. 

American  Society  for  Testing   Ma- 
terials: 

reference  to  transactions,  23,  32. 
Apparent  elastic  limit,  21. 
Apparent  unit  stress,  24. 
Approximate  computations,  161,  268, 

269,  270. 

Axes,  principal,  177,  178,  263. 
Axis,  neutral,  85,  89,  90,  162. 


B. 


Beams,  66-168. 

cantilever,  66,  104  et  seq. 

cast-iron,  158-162. 

constant   moment,    100,    102-104, 

114-116. 

constant  strength,  153-158. 
continuous,  66,  128-135. 
counterflexure  in,  124-128. 
dangerous  section,  83. 
deflection,  99-138,  156-158. 
deformation,  86. 
differential  equation,  101-102. 
distribution  of  stress,  93-96. 
external  moment,  69. 
failure  of,  150,  151,  168,  219,  220. 
modulus  of  rupture,  96-98. 


Beams,  modulus  of  section,  91-94. 

moment  diagrams,  77  et  seq. 

points  of  inflection,  124-128. 

radius  of  curvature,  99. 

reinforced  concrete,  151,  164-168. 

resilience  of,  231-235. 

shear  diagrams,  74-77,  80. 

shear  in,  68,  132,  139-150. 

stiffness  of,  117,  118. 

stresses  in,  67,  85,  87-96. 
Bearing  stress,  3,  49-51,  55-57,  154. 
Breaking  strength,  22. 
Building  laws,  5,  92,  210,  216. 

butt  joint,  58. 


C. 


Cambria  Steel  handbook,  4,  5,  23,  91, 
93,  143,  217,  222,  249,  250,  258. 
Cantilevers,  66,  104  et  seq. 

constant  strength,  153-156. 

deflection  of,  104-108,  156-158. 

supported  at  end,  119,  121. 
Cast  iron: 

beams,  158-162. 

stress-strain  diagrams,  25,  160. 

test,  data  of,  27. 
Center  of  gravity,  239-250,  270-271. 

by  experiment,  270-271. 

by  moments,  240-243. 

of  continuous  bodies,  243-245. 

of  plane  areas,  245-250. 
Cold-rolled  steel,  34-36. 

test  data,  35. 
Columns,  180-222. 

American  railway  formula,  208. 

classification  of,  193,  194. 

curves  for,  188,  203,  209,  215. 

definition,  180. 

differential  equation,  182. 


291 


292 


INDEX 


Columns,  end  conditions,  195-201. 

Euler's  formula,  189-192,  218. 

Pencoyd  tests,  197,  199-201. 

Rankine's  formula,  212-217. 

Hitter's  constant,  213,  214. 

Secant  formula,  183-188. 

straight-line  formulas,  202-211. 

theory,  181  et  seq. 

Watertown  tests,  207. 
Compression,  2,  48-50. 
Compressive  stress,  3,  161. 

caused  by  shear,  43,  44,  144-150. 
Concrete,  29,  30,  48. 

reinforced,  151,  164-168. 

stress-strain  diagram,  29. 

test  data,  30. 

Counterflexure,  124-128,  194. 
Cover  plates,  59. 


D. 


Dangerous  section,  83. 

Deflection  in  beams,  99-138,  156-158. 

cantilevers,  104-108. 

constant  strength,  156-158. 

fixed  at  ends,  119-128. 

supported  at  ends,  108-116. 
Deflection  in  columns,  183. 
Deformation,  5,  6,  39. 

lag  behind  stress,  275. 

of  beams,  86. 

shearing,  39. 

unit,  5,  6,  225. 
Density  of  materials,  4. 
Department  of  Agriculture,  4,  32. 

timber  tests,  5,  143. 

E. 

Eccentric  loads,  170-179. 
Efficiency  of  riveted  joint,  63. 
Elasticity : 

modulus,  7. 

shearing  modulus,  39. 
Elastic  limit,  6,  18. 

Johnson's  apparent,  21. 

proportional,  20. 

true,  18-20. 
Elastic  line,  104. 


Elongation,  15. 

per  cent,  23. 

Engineering  News,  208,  210. 
Euler's  formula,  189-192  et  seq. 

curves  for,  188,  203. 

derivation,  190. 

limits  to  use,  191,  218. 

relation  to  Rankine's  formula,  213. 

relation  to  straight  line,  202-205. 
Experimental  tables,  17,  27,  28,  30, 
35,  200,  207,  276,  277,  282,  283. 
Eyebar  tests,  282-284. 


Factor  of  safety,  29,  31. 

in  columns,  18$  et  seq. 
Failure,  46-53. 

by  cutting,  49-51. 

in  beams,  150,  151,  168,  219. 

in  compression,  48,  49,  50. 

of  riveted  joint,  60,  64. 

under  repeated  loads,  285. 
Fatigue,  285,  286. 
Fixed-end  beams,  120-128. 
Fixed-end  columns,  198. 
Flange  failure,  219. 
Franklin  Institute,  5. 

G. 

Graphic  representation  of  stress  dis- 
tribution, 93-96. 

Goodman's  impact  rule,  287. 

Goodman's  Mechanics,  reference, 
285-287. 

Gyration,  radius  of,  253. 

H. 

Helical  springs,  228-230. 
Hickory  beams,  151,  152. 
Hinged-end  columns,  195  et  seq. 
Hooks,  174-176. 
Horizontal  shear,  140-143. 
Horse  power,  225. 
Hoskins'  Mechanics,  172. 


INDEX 


293 


I. 


Illinois,  University  of: 

timber  tests,  21,  98,  143. 
Impact  stress,  287-290. 
Inflection,  points  of,  124-128. 
Intensity  of  stress,  3,  140. 
Internal  work,  232. 
Iron  and  Steel  Institute,  reference, 
285. 

J. 

Johnson's  apparent  elastic  limit,  21. 
Johnson's    "Materials   of   Construc- 
tion," 16,  21,  287. 


Lap  joint,  57. 
Launhardt's  formula,  286. 
Limit,  elastic,  6,  18. 

M. 

Maurer's  Mechanics,  172. 
Modulus  of  elasticity,  7. 

calculation  of,  21. 

in  shear,  39,  46. 

of  volume,  14. 
Modulus  of  resilience,  10. 
Modulus  of  rupture,  96-98. 
Modulus  of  section,  91-94. 
Moment,  69  et  seq. 

diagrams,  77. 

due  to  eccentric  loading,  172,  173. 

experimental  illustrations  of,  71-73. 

external,  69. 

general  equation  in  beams,  122, 124. 

positive  direction  of,  70. 

relation  to  shear  diagram,  79, 81-83. 

relation  to  stress,  87. 

resisting,  69,  70,  166. 

resultant,  137. 
Moment  of  inertia,  251  et  seq. 

axis  for  maximum,  262. 

by  experiment,  272-274. 

change  of  direction  of  axis,  259. 

definition,  251. 

in  thin  plate,  255. 


Moment  of  inertia,   of  plane  area, 

257-274. 
polar,  257. 
principal,  138,  263. 
transfer  of  axis,  254. 

N. ' 

Neutral  axis,  85. 

location  of,  89-90,  161-162. 
Neutral  surface,  85. 
New  York  building  laws,  5,  92,  210, 

211,  217,  222. 
Notation,  xi-xii. 

P. 

Pencoyd  column  tests,  197-200. 
Philadelphia  building  laws,  216,  217, 

222. 

Pin-end  connections,  195  et  seq. 
Poisson's  ratio,  11. 
Polar  moment  of  inertia,  257. 
Product  of  inertia,  260-262. 
Punching,  52. 

R. 

Radius  of  curvature,  99. 
Radius  of  gyration,  253. 
Railway   Engineering    and    Mainte- 
nance of   Way   Association,   5, 
208-211,  290. 

Rankine's  formula,  212-217. 
Reduction  of  area,  23. 
Reinforced  concrete,  151,  164-168. 
Repeated  stress,  275-287. 
Resilience,  9,  10,  231-238. 

in  beams,  234-238. 

modulus  of,  10. 

sections  for  maximum,  237. 

torsion,  237. 
Ritter's  constant,  213. 
Riveted  joints,  54-65. 

butt  joints,  58,  64. 

efficiency,  63. 

lap  joints,  57. 

tests,  59-61. 
Round-end  columns,  195  et  seq. 


294 


INDEX 


S. 


Section  modulus,  91. 

Shear,  37-47,  68,  74,  139-150. 

caused  by  tension,  41. 

deformation  in,  39,  45. 

forces  in  pairs,  42,  43. 

in   beams,   68,    74,    132,    139-150, 
154. 

maximum  resultant,  145. 

modulus  of  elasticity,  39. 

stress  caused  by  torque,  223. 
Shear  diagrams,  74  et  seq. 

relation  of  area  to   moment,   81, 

83. 
Shearing  stress,  37. 

in  beams,  69,  140,  144. 

in  shafting,  223-225. 
Society  of  Civil  Engineers,  197,  199, 

285,  290. 
Society   for  Testing    Materials,    23, 

33. 

Specifications,  Manufacturers',  23. 
Springs,  229,  234,  235,  289. 
Square-end  columns,  198. 
Steel  tests,  17,  35,  59-64,  276,  277, 
282. 

stress-strain  diagrams,  18,  19,  275, 

279,  281. 
Stress,  2. 

allowable  unit,  4. 

bearing,  3,  55-57. 

beyond  elastic  limit,  96. 

beyond  yield  point,  15,  33. 

compressive,  3. 

in  beams,  67,  85  et  seq. 

in  riveted  joints,  54. 

resultant,  144-150,  168. 

shearing,  37,  139-150,  224. 

tensile,  3. 

unit,  3,  88. 

working,  4. 
Stress-strain  diagrams,  15. 

cast  iron,  25,  160. 

concrete,  29. 

'steel,  18,  19,  279,  281. 

yellow  pine,  25,  29. 
Strut,  181. 


T. 

Tables,  calculated: 

VI.  resultant  stress  in  beams,  149. 

VII.  eccentric  loaded  columns,  187. 

VIII.  Euler's  formula,  192. 

XL  computation  of  center  of  grav- 
ity, etc.,  270. 

Tables,  experimental: 

I.  machine  steel,  17. 

II.  cast  iron,  27. 

III.  yellow  pine,  28. 

IV.  concrete,  30. 

V.  cold-rolled  steel,  35. 

IX.  Christie's  column  tests,  200. 

X.  a  wrought-iron  column,  207. 

XII.  soft  steel,  276. 

XIII.  soft  steel,  277,  278. 

XIV.  structural  steel,  282. 

XV.  structural  steel,  283,  284. 
Tensile  stress,  3. 

caused  by  shear,  43,  144,  168. 
Tension,  1,  12,  47,  55. 
Testing  Materials,  Society,  23,  32. 
Theorem  of  three  moments,  128-135. 
Torque,  223-230. 

relation  to  angle  of  twist,  225,  226. 

relation  to  stress,  224,  225. 

relation  to  work,  227. 
Torsion,  223  et  seq. 

combined  with  bending,  227,  228. 
Total  stress,  2. 

U. 

Ultimate  strength,  22. 

effect  of  form  on,  32,  33. 
Unit  stress,  3. 

actual,  24. 

apparent,  24. 

in  beams,  88. 

United  States  Department  of  Agri- 
culture, 4,  5,  32,  143. 
Unwin's  "Testing  of  Materials,"  285. 

V. 

Volume  change,  12. 


INDEX  295 

W.  Wrought  iron,  49. 

Wrought-iron  columns,  200,  206. 
Watertown  Arsenal  tests,  7,  13,  26, 

28,  59-64,  159,  206,  230,  281.  y 
Web  failure,  220. 

White  oak,  48,  50.  Yellow  pine,  3,  28,  47,  50,  152. 

White  pine,  151,  152.  column  formula,  211. 

Wohler's  experiments,  285.  Yield  point,  18-20,  23. 

Working  stress,  4.  Young's  modulus,  6.