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THE

ELEMENTARY PRINCIPLES

GRAPHIC STATICS

Th.

Elementary Principles

Graphic Statics

SPECIALLY PREPARED FOR STUDENTS OF SCIENCE
AND TECHNICAL SCHOOLS, AND THOSE ENTERING
FOR THE EXAMINATIONS OF THE BOARD OF EDU-
CATION IN BUILDING CONSTRUCTION, MACHINE
CONSTRUCTION, DRAWING, APPLIED MECHANICS,
AND FOR OTHER SIMILAR EXAMINATIONS

EDWARD HARDY

Teaclier of Building ConsU uction ; Certificates : — Honours in
Masonry and Brickwork ; Prizeman and Medallist in Masonry, etc.

ILLUSTRATED BY 192 DIAGRAMS

LONDON
B. T. BATSFORD 94 HIGH HOLBORN

1904

Butler & Tanner,

The selwood printing Works.

frome, and london.

270
H35

PREFACE

The following chapters are placed before students of
Building Construction, Applied Mechanics, and Machine
Construction and Drawing, in the hope that they may
be of service to those who desire aid in the study of the
" Statics " branch of these subjects.

It should be stated that, in the chapter on Graphic
Arithmetic, only such matter has been introduced as
is deemed necessary for the study of the succeeding
chapters.

The author desires to express his gratitude to Pro-
fessor Henry Adams, M.I.C.E., M.I.M.E., F.S.I., etc.,
for his kindness in reading through the MS., and for
his valuable help and advice.

EDWARD HARDY.

Saxatile, Merthyr Tydfil,
December, 1903.

CONTENTS

CHAPTER I

Graphic Arithmetic

PAGE

Graphic Representation of Quantities — Advantage of
Decimally-divided Scales — Addition — Subtraction —
Similar Triangles — Multiplication — Division — Pro-
portion— Examples ...... 9

CHAPTER II

Force

Definition, how Measured and how Represented —
Resultant — Equilibrium — Equilibrant — Parallel
Forces — Reaction — Moments and how Measured —
Point of Application of the Resultant of Parallel
Forces — The Three Orders of Levers — Solution of
Levers — Cranked or Bent Levers — Examples . 22

CHAPTER III

Centre of Gravity

Of a Parallelogram — Of a Triangle — Of a Trapezium
— Of any Quadrilateral Figure — A Door as a Lever —
Bow's Notation — Load — Stress — Strain — Examples . 44

CHAPTER IV

Non-Parallel Forces

Parallelogram of Forces — Triangle of Forces — Reso-
lution of Forces — Inclined Plane — Bent Levers —
Reaction of Door Hinges — Lean-to Roofs — Retaining
Walls for Water and Earth — Polygon of Forces —
Examples ........ 58

CHAPTER V

Funicular Polygon
Links, Pole, and Polar Lines or Vectors — Solution
of Parallel Forces — Reactions of the Supports of

CONTENTS

PAOK

Framed Structures — The Load Supported by a Roof
Truss, and how it is Conveyed to it — Centre of
Pressure of Irregular Masses — Examples . . 84

CHAPTER VI

Graphic Solution of Bending Moment

How to Obtain the B.M. Scale — Cantilevers Loaded
at Different Points — Beams with a Uniformly Dis-
tributed Load and Supported at Both Ends — How
to draw a Parabola — Cantilevers with a Uniformly
Distributed Load — Cantilevers and Beams Supported
at Both Ends with the B.M. Diagrams for Concentrated
and Uniformly Distributed Loads Combined —
Shearing Force — S.F. Diagrams: for Cantilevers
Loaded at Different Points — for Cantilevers with
Uniformly Distributed Loads — for Cantilevers with
Combined Concentrated and Uniformly Distributed
Loads — for Beams Supported at Both Ends with
Concentrated Loads, with Uniformly Distributed
Loads, and with Concentrated and Uniformly Dis-
tributed Loads Combined — Examples . . .106

CHAPTER VII

Explanation of Reciprocal or Stress Diagrams

Rules for Drawing Stress Diagrams — Span Roof —
Couple Close — Couple Close with a King-rod Added —
King-post Truss — Other Forms of Roof Trusses
— Framed Cantilevers — Apportioning Distributed
Loads — How to Obtain the Magnitude of the Stresses
of the Members of Framed Cantilevers and Girders
from the Stress Diagrams — Warren Girder with a
Concentrated Load on the Top Flange — Warren
Girder with a Uniformly Distributed Load on the
Top Flange — Warren Girder with a Uniformly
Distributed Load on the Bottom Flange — N" Girder
with a Uniformly Distributed Load on the Top Flange
— N Girder with a Concentrated Load on the Top
Flange — N Girder with Concentrated Loads on the
Bottom Flange — Lattice Girder Without Verticals —
Lattice Girder With Verticals — Examples . .127

Answers to Examples . . . .162

Errata.

Page 30. 6tli line, for " say |' = 1" "
read " say ^" — 1'."

Page 147. 8th line from bottom, /or
'"bcxce" read "bc+ce."

^\:'-'-m»-i-i

CHAPTER I

GRAPHIC ARITHMETIC

1. In ordinary arithmetic a number (unity) is chosen,
and all quantities expressed in multiples of that num-
ber : thus, 5 means that unity is taken 5 times, and
4-5 means that unity is taken 4-5 times.

Calculations are then made arithmetically.

2. Instead of expressing unity by a figure we can
express it by a line. All other quantities are then
represented by lines whose lengths are proportional
to the magnitudes they represent.

Let a line J'^ long represent unity, then 5 would be
expressed by a line five times as long, and 4-5 would
be expressed by a line four and one-half times as long.
Again, let a line f' long represent one article (or 1 yd.,
1 hr., etc.), then 12 articles (or 12 yds., 12 hrs., etc.)
would be shown by a line twelve times as long, i.e. by
a 9"" line.

It. will thus be seen that, after having decided upon
a unit length, any quantity, whether abstract or con-
crete, can be expressed by lines.

When the quantities are represented by lines, the
calculations are made by means of geometrical draw-
ings, i.e. " graphically."

3. This work, being intended for students who are

ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

already familiar with geometry, it is assumed that
scale drawing is understood.

A rule, with all tlie divisions continued to the edge
and with the inch and the subdivisions of the inch
ih i? h h 8' ^^^•) being divided into tenths, will be
found the most convenient. An ordinary flat rule
marked on both faces would contain 8 such scales,
while a triangular one would contain 12.

The measurements should be transferred to the
paper (or read off) by applying the edge of the rule
directly to the line, and not with the dividers.

^SC :-

■ ,.

Mil Ml!

\\\\\\\\\

30 20

\zz

IIIIMIII

1 1 [ i 1 M M

Fig. 1.

4. The numbers 1, 2, 3, etc., can also be read as 10,
20, 30, etc., when the tenths will become units and
the hundredths will become tenths. In a similar way
they may be read as hundreds, thousands, etc.
For example see Fig. 1, where : —

if f' = 1 unit, then A B = 115 units
„ f' = 10 units, then AB = 17-5 „
„r = ioo „ „ AB = n5- „
„ r = 1000 „ „ AB = 1750- „
and, if |'' = 1 yd.,1 lb., etc.,then C Z) = 2-4 yds.,lbs.,etc.
„ „ f' = 10 yds., lbs., etc., „ O i) = 24 „ „ „
„ „r = 100 „ „ „ „ CD = 2^0- „ „ „

5' 8

1000

CD = 2400

GRAPHIC ARITHMETIC

5. Sometimes the unit is given by means of a line
whose length is not stated. In order to find the
numerical value of a line when the unit is given thus,
it is advisable to divide the unit line into tenths, plot
off as many units as possible, and state the remaining
portion (if any) as decimals.

Let J. be a line whose magnitude is required, and B
the unit line.

Divide B into tenths.

. A

Fig. 2.

It will be seen that B can be plotted 3 times along
A, and the remaining portion of A is equal to 7 tenths
of J5.

A, therefore, represents 3-7 units.

6. Addition. — ^Let it be required to add 1-7 yds. to
1-35 yds.

Take any convenient scale, as V — 1 yard. Set off
A B (Fig. 3) — 1-7, and adjoining this, and in a straight
line with it, set off J5 O = 1-35.

A B . C

Fig. 3.

It is now evident that A C equals the sum of ^ ^
and B C, and, by applying the scale to it, it is found
to be 3-05, i.e. the sum = 3-05 yds.

7. Find J he sum of 5-4 tons, 4-7 tons, and 3-2 tons.

Adopt a scale— say J'' = 1 ton. Set oE A B, B C

12 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

and C D (Fig. 4) in a straight line, and equal to 5-4,
4-7 and 3-2 respectively. Then AD = their sum, and,

A BCD

Fia. 4.

if measured, will be found to represent 13-3 tons.
8. Subtraction. — Take 42-5 lbs. from 73 lbs.

Fig. 5.

Let the scale be J^' = 10 lbs. Draw A B = 13 lbs.
(Fig. 5), and from B set back along B A, B C = ^25
lbs. A C is now the remainder, and by measurement
this is seen to represent 30-5 lbs.

9. It should be noted that all positive numbers, or
plus values, are set off in one direction, usually from
left to right, while the negative numbers, or minus
values, are measured back in the opposite direction.

10. Example. — Simplify (37 — 42 — 3 + 41) tons.
Take a scale such as f' = 10 tons.
Commencing at the point A, measure off A B = 31

tons to the right, as in Fig. 6. From B mark off B C
= 42 tons, but, since the 42 tons are to be subtracted,
B C must be taken in the opposite direction to A B.
From G measure C D = 3 tons. This again being
negative, it must be taken in the same direction as B C.
From D measure D E =: 4:1 tons. This being positive,
it must be taken in the same direction as A B.

D C A £ B

FiQ. 6.

GRAPHIC ARITHMETIC

The distance from A to E (i.e. the first and last
points) will give the answer. In the example given it
will be noticed that the point E comes on that side of
A towards which the positive quantities were taken,
therefore ^ ^ is positive. If E had come on the other
side of A, then the answer would have been negative.

II. Similar Triangles. — Before proceeding to
multiplication and division, it is necessary to study
the relationship between similar triangles. Similar
triangles are those whose angles are equal, each to
each— i.e., if the two triangles ABC and DE F (Fig. 7)
be similar, the angle ^ ^ C is equal to the angle D E F ;
the angle BAG — the angle E D F ; and the angle
AC B = the angle E F D.

Fig. 7.

If two angles of one triangle be equal to the two
angles of another triangle, each to each, then, since
the three angles of every triangle = 180°, the third
angle of the one is equal to the third angle of the other,
and the triangles are similar in every respect.

The particular point to be noted concerning similar
triangles is that the sides of the one bear the same
relation to each other as do the sides of the other tri-
angle, each to each ; that is, if A B (Fig. 7) be twice
B C, then D E in twice E F ; or, ii A C he H times
A B, then D F m l^ times D E.

14 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

This comparison is true of any two sides, providing
tlie sides chosen in the one triangle correspond with
those chosen in the other.

12. It is also true if the perpendicular height be taken

0 B

X £

Fig. 8.

as a side, because it can easily be shown that the tri-
angle A 0 C (Fig. 8) is similar to the triangle D X F,
or that the triangle A 0 B is similar to the triangle
DXE,

13. The relationship between the sides of similar
triangles is generally expressed as follows : —

^ i? is to ^ C as Z) ^ is to ^ J^ (see Fig. 7), and
written A B : B C : : D E : E F.

But this is proportion, and the product of the ex-
tremes is equal to the product of the means :
therefore ABxEF = BGxDE,
BGxDE

and AB =z

(Similarly any one side can be found in terms of the
others.

14. Multiplication, Division and Proportion can each
be worked by means of similar triangles.

Multiplication. — Let it be required to find a line
2-3 times as long as a given line A.

GRAPHIC ARITHMETIC

Take a scale such as V = 1 unit. Draw B G = I
unit, and BD = 23 units (Fig. 9).

I I I I 1 ITI I I I

JZS

Scale of Uruts

Fig. 9.

From G erect a perpendicular G E equal to the given
line A. Join BE and produce it until it meets a
perpendicular from D at F.

Then D F = 2-^ times G E or 2-3 times as long as
the given line A.

15. Division. — Divide a given line A by 2-3.
Draw a line B G = I unit, and 5 i> = 23 units

(Fig, 10). From D erect a perpendicular Z) i^ equal to
the given line A. Join B F, and from G erect the
perpendicular C E, meeting it at E.

Then G E represents the quotient of D F,ov A divided
by 2-3.

16. Proofs. — Suice the triangles EBG and FBD
(Figs. 9 and 10) have the angle FBD common to both,
and the angle E G B = the angle F D B, both being
right angles, then the two triangles are similar in every
respect (§ 11).

16 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

(zr

I I I I I I

I_L

xz:^

Scale oF Units

Fig. 10.

Therefore D F : B D : : C E : B G

GE X BD
and i)i^= TTT^ , (1)

also G E
But BG
therefore D F =

BG
DF X BG

-. (2)

1 unit and jB D = 2-3 units,

0^x23 ^„ ^^
-^ =GE X 2-3,

and G E

DF X 1
2¥~

= DF -r 2-3.

17. Proportion. — It should be noticed that multipH-
cation and division are simply proportions where one
of the quantities of the known ratio is unity.

The equations 1 and 2 are true whatever values are
given to ^ O and B D, provided that they are properly
set out to scale, so the construction for a problem in
proportion is similar to that for multiplication and
division.

GRAPHIC ARITHMETIC

Notes : —

\. B C and B D must be drawn to the same scale.

2. DF must be measured by the same scale as

that by which C E is drawn, and vice versa.

3. C E (Fig. 9) and D F (Fig. 10) were each drawn

equal to a given line, but they could have been
drawn to scale equal to any known quantity.

4. A different scale may be used for B C and B D

to that used for C E and D F.

5. The perpendicular representing the known

quantity must be erected at the end of the
line shown in the denominator of the equations
1 and 2.
1 8. Examples. — Multiply 350 lbs. by 1-7.
Let the scale for the multiplicand be f' = 100 lbs.,
and the scale for the multiplier be 2'" = 1 unit.

By the second scale set off ^ ^ = 1 unit, and A G
= 1-7 units (Fig. 11). £

Fig. 11.

ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

b'roni B draw the perpendicular B D = 350 lbs. by
the first scale.

Join A D, and produce it to meet the perpendicular
C E. By applying the first scale, C E will be found to
represent 595 lbs.

19. Find the product of the lines A and B if the line
C be the unit.

Draw DE = C, Sind DF = B (Fig. 12).

At E erect the perpendicular E G = A. Join D G,
and produce it to meet the perpendicular F H at //,

Then F H is the product of A and B.

N.B. — The result would have been the same \i D F
were made equal to A, and E G equal to B.

20. Divide 42-5 yards by 3-4.

Make A B and A C equal to 1 and 3-4 units respec-
tively. At C set up O Z) equal to 42-5 yards by scale.

Join A D and erect the perpendicular BE. BE
measured to scale gives 12-5 yards.

21. Find five-sevenths of a given line. This may
be stated as a proportion, thus :

GRAPHIC ARITHMETIC

^ivou line : required part.

given line x

therefore the required part =

Taking a suitable scale, make Q

AB = 5^ndAC = 7 (Fig. 14).
7 is the denominator in the
above equation, and it is repre-
sented by A C. Therefore from
G erect the perpendicular C D
equal to the given line. Join

A D, and from B erect the per- on

pendicular B E meeting it at ^. A D 0

Then BE = fCD. ^'^- ^^^

22. If a rod of iron 3-2 yds. long weighs 12-6 lbs.
what would be its weight if it were 7-5 yds. long ?

The proportion is 3-2 yds. : 7-5 yds. : : 12-6 lbs. :
X lbs.

12-6 lbs. X 7-5
therefore x = ^ •

Draw ^ J5 and ^ C (Fig. 15) equal to 3-2 yds. and

nil

\ /

W 0

lllllllllll

Scales
Fia. 15.

yd6

20 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

7-5 yds. respectively. From B erect the perpendicular

5 Z) = 12-6 lbs. Join A D, and produce it until it

meets the perpendicular C E. Then C E = 295 lbs.

23. The preceding examples, 18 to 22, could have

been solved equally well by using another diagram.

12-6 lbs. X 7-5
In the last exercise x = ~

Draw A B Sind AC (Fig. 16) equal to 3-2 and 7-5
respectively.

At A erect the perpendicular A D equal to 12-6
lbs.

Since A B represents the
denominator in the above
equation, join B to D.

From C draw C E parallel
to B D until it meets the
perpendicular from A at E.

Then A E = x = 29-5
lbs.

Examples to Chapter I.

Ex. Ch. I.— Fig. 1.

1. What does A B (Fig. 1) show :—

(a) With a scale of J'' = 1 unit.

(b) With a scale of f '' = 1 yard.

2. Taking a convenient scale, graphically determine
the following : —

(a) The diagonal of a square whose sides are 36 ft.

EXAMPLES TO CHAPTER I

(6) The perpendicular height of an equilateral

triangle with 15'' sides,
(c) The height of a wall, if a 24 ft. ladder leaning

against it, with the foot 7-5 ft. from the wall,

just reaches the top.
{d) The length of the different members in Fig. 2.

<- - - 26.0' - - ^

Ex. Ch. I.— Fig. 2.

3. Draw a line 2'' long, and find |. of it.

4. Find the sum of A and B (Fig. 3) if the unit line
measures J".

5. Find the product of A
and B (Fig. 3) if J'' = 1 unit.

6. Draw two lines, A 2-3
inches long, and B 15 inches
long.

If the scale be J'' = 1 unit, graphically determine

A
B

Ex. Ch. I.— Fig. 3.

^+B.

Chapter II
FORCE

24. Force is (a) that which tends to move a body,

or (6) that which tends to stop a body

when it is moving,

or (c) that which tends to change the

direction of a body when it is moving.

In this work it is only intended to deal with force as

defined in (a). No reference will be made to velocity, and

only bodies which are in a state of rest relatively to

neighbouring bodies will be treated upon,

25. Force is measured in units of lbs., cwts., or tons.

26. We have already seen that lbs., etc., can be repre-
sented by lines drawn to scale. Hence, if the magni-
tude of a force be known, a line may be drawn whose
length will be proportional to the force.

27. Force must be exerted in a certain direction ;
the line representing it must, therefore, be drawn in that
direction.

28. An arrow can be placed on the line indicating
the sense of the force, that is, showing in which direction
along the line the force is acting.
^^ Example. — If V — 10

lbs., then A represents a
forc(^= 32 lbs. acting from
left to right, and B a
downward force = 15 lbs.
29. The j)oint of appli-
cation (i.e. the place at

Example.
which the force is applied) must be known.

RESULTANT — EQUILIBRIUM — EQUILIBR^VNT 23

30. When the magnitude, direction, sense, and point
of apphcation of a force are known, the force is said to
be known.

These four points should be clearly understood, and
always kept in mind. In determining a force the
student must see that he finds all four.

31. Resultant. — If a number of forces (whether
parallel or otherwise) act on a body, and move it in a
certain direction, it is evident that another force could
be found, which, acting in that direction, would do
the same work.

This force is called the resultant of the others.

32. Equilibrium. — If the resultant of a number of
forces be zero, then they are said to be in equilibrium.
If thes3 forces be applied to a body in a state of rest,
then it will still remain -at rest.

33. Equilibrant. — When a body is not in equilib-
rium it moves in a certain direction with a force which
has a resultant. Another force equal to the resultant
in magnitude, acting in the same line and opposite in
sense, would produce equilibrium.

This force is called the equilibrant.

34. The equilibrant and the resultant of a system of
forces are always equal in magnitude, act in the same
line of direction, and are of opposite sense.

Note. — By old writers the word direction meant line
of action and sense together, and in common language
direction is still used in this way.

35. If a number of forces be in equilibrium, any one
of them is the equilibrant of the others, and if the sense
be reversed, it will represent their resultant.

For, if a system of forces be in equilibrium, each force
helps to maintain it, and the removal of any one of them

24 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

would cause the remainder to move along the line on
which it was acting, and in the opposite direction.

36. Parallel Forces. — Forces are said to be parallel
when the lines along which they act are parallel.

Suppose forces equal to 5, 3, and 8 lbs. to be acting in
one direction, and forces equal to 4, 7, and 2 lbs. in the
opposite direction. In the one direction a force equal
to 16 lbs. would be acting, whilst against that a force
equal to 13 lbs. would be exerted. The whole system
will have a resultant of 3 lbs. acting in the first direction.
The 3 lbs. represents the resultant of the 6 forces, and
acts in the direction of the 16 lbs.

If a force equal to 3 lbs. be added to the second set
of forces, or taken from the first, then the whole system
would be in equilibrium.

37. In dealing with parallel forces those acting in
one direction are taken as positive, and those acting
in the opposite direction as negative. The algebraical
sum of all the forces will represent the resultant, and
it will act in the direction of those whose sum is the
greater.

38. If this sum be zero, then all the forces
g are in equilibrium, and, conversely, if a

system of parallel forces be in equilibrium,
then the algebraical sum = 0.

Fig. 17 represents a force A B acting on a
point A. If only this force were acting on it,
it would move in the direction of A to B, and
to keep it in equilibrium, another force A C
equal to A B, and pulling it in the opposite
direction, must be introduced.

Figs. 18 and 19 show how by means of two

Fig. 17. spring balances the student can prove for

EEACTION — MOMENTS

Fig. 1!

bal-

himself that the sum of all the forces acting
downwards is equal to the sum of the forces
supporting them. Of course, the weight of
the pulley (Fig. 18) or the bar (Fig. 19) must
be added, as it

exerts a down- 5? X

ward pressure.
Fig. 20 repre-
sents a beam
carrying three
loads. On ex-
amination it
will be seen that

the two walls act in just the same way as the

ances in Fig. 19. Hence

the force exerted by the

two walls together must

equal the sum of all the

forces acting downwards,

together with the weight of

the beam,

39. Reaction. — When a force acts on a body it
produces a resisting force from that body. This second
force is always equal to and opposite to the first.

The beam with its load (Fig. 20) exerts a force on
the walls, and this produces a resistance from each
wall equal to the portion of the weight it has to carry.
These resistances are called the reactions of the wall.

40. Moments. — It now becomes necessary to ascer-
tain why the point of application of a force should be
known. Place a book or similar object at A B C D
(Fig. 21) on a table. Apply a force, as shown at P.
This will cause the book to rotate clockwise, i.e. in

Fig. 19.

^ I ^

Fig. 20.

26 ELEMENTAliY TIUNCIPLES OF GUAPHIC STATICS

the same direction as the hands of a clock. If the
force be appUed near A, it will rotate in the opposite
direction, or anti-clockwise. By applying the force at
different points, the student will find that to move
the book forward he must apply a force in a direction
which, if produced, would pass through the point G.

Again, let him take a lath, holding it horizontally
by one end. Place a 1 lb. weight 1 ft. from the hand.
He will find that the weight causes the lath to try to
rotate with his hand as the centre of rotation, and he
also experiences a difficulty in counteracting this rota-
iP tion. If he moves the weight
2 ft. from the hand, he will
find the tendency to rotate
twice as great, and that it is
twice as difficult to keep the
weight in position.
Fig. 21. ^j^ From this it will be seen

that the greater the distance the weight or force is
from a certain point, the greater is its tendency to
produce rotation round that point.

42. This tendency of a force to produce rotation is
called the Moment of the force, or Bending Moment.

43. If a force passes through the point chosen, and
the point is supported, the force cannot produce rota-
tion, hence there is no Moment. It acts simply as a
force.

44. Referring again to the weight supported on the
lath 2 ft. from the hand, it is wrong to say that the
strain on the hand is 2 lbs. because only 1 lb. is sup-
ported. How, then, shall the moment be measured ?
It is now clear that it will depend directly upon the
magnitude of the force, and upon the perpendicular

MOMENTS 27

distance of the line of action of tlie force from the
point on which it is supposed to rotate.

Hence, in the last example the strain on the hand
is 1 lb. X 2 ft. = 2 ft.-lbs.

As before explained, the unit of force is generally
expressed in lbs., cwts., or tons, and the unit of length
in inches or feet, so the moment of a force would be
expressed as — inch-lbs., ft.-lbs., ft.-cwts., etc.

45. The moment of a force which produces clock-
wise rotation is generally taken as negative, and that
producing rotation in the opposite direction as positive.

We will now proceed to find

lbs Ih.

the moments in an example. a^^l {if

The student should again use |- '^ ^ |

his spring-balances, and arrange Y /\ /^^ -S

them as in Fig. 22. To prevent '^(3

the arrangement being cumber-
some, he can take an inch for his unit of length and
mark off the inches on the bar.

In the example shown the bar is 8"', and the weight,
which is 6 lbs., is shown 3"' from one end. The balance
nearer the weight now registers 3| lbs., and the other
21 lbs.

The system shows three parallel forces in equilibrium,
and the force acting downwards is equal to the sum
of those acting m the opposite direction.

Any point can now be selected, and the moments of
all the forces about that point ascertained, care being
taken to prefix the + (plus) or — (minus) signs, as
already explained.

First let the centre of the beam be taken as the point.

This must now be considered as a pivot on which
the beam can turn.

28 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

The 6 lbs. would cause it to turn clockwise, and is
V from the point, so the moment is — 6 inch-lbs.

The 3| lbs. would cause it to revolve in the opposite
direction, and is 4^' from the point, so the moment is
+ 15 inch-lbs.

The 2J lbs. is 4'' away, and would cause clockwise
motion, so the moment is — 9 inch-lbs.

The algebraical sum of the moments of all the forces
about this point is ( — 6 + 15 — 9) inch-lbs. = o.

Take the point y 2"' from A as shown (Fig. 15).

The moments are + (21 lbs. x 2'') - (6 lbs. x T)
+ (3| lbs. X W) = (41 - 42 + 37i) in.-lbs. = o.

Lastly, take a point through which one of the forces
acts. Then by § 43 the moment of that force is nil.
Taking the point B, the only forces producing rotation
are the 2 J lbs. and the 6 lbs.

The moments are (2 J x 8 — 6x3) in. lbs. =
(18-18) in. lbs. = o.

46. From the foregoing example it will be seen that —
If a system of forces be in equilibrium, the algebraical

sum of the moments of all the forces about any point
is zero.

47. The converse of this is also true, so —

If the algebraical sum of the moments of all the forces
in a system be zero, then the system is in equilibrium.

48. Further —

The moment of the resultant of any number of forces
about any point is equal to the algebraical sum of the
moments of all those forces about the same point.

For, taking Fig. 22, the force at A may be considered
the equilibrant of the other two, and, if the sense be
reversed, it will represent the resultant of the same
two, (§ 35).

MOMENTS

Take the moments about the centre.

The force at A now acts downwards, so the moment
of this force is 2^ lbs. x 4"' = + 9 in. -lbs.

The moments of the other two forces are (15 — 6)
in. -lbs. = + 9 in.-lbs.

49. If, then, it be necessary to find the moments
of a number of forces about a point, it is sufficient to
find the moment of the resultant of those forces about
that point.

50. The knowledge we have gained of these moments
is of great assistance in determining forces.

Fig. 23 represents a beam rest-
ing on two walls, A and B, 12 ft.
apart, and carrying a load of 6
tons 4 ft. from B.

It is necessary to find what
portion of the load each wall
carries.

The reactions of ^ + B z= 6 tons.

There are two unknown forces, so we will take the
moments about the point B to eliminate the moment
of that force (§ 43).

Let ^'s share of the load be called x tons.

Then 6 tons x 4' - a: tons x 12' = o (§ 46)

, tons

»^

■*(-L0

Fig. 23.

fore X tons x 12' = 6 tons x 4', and x tons :

= 2 tons = ^'s share. But A+ B = Q
fore -B's portion is 4 tons.

51. The student will recognize x tons =

there-
6 tons X 4'

"" r2^

tons ; there-

6 tons X 4 ft.
12 ft.

as a proportion (§ 13), hence x may be found graphically.
In the above expression there are feet and tons,
hence two different scales are necessary, the one to set

ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

out the length of the beam and the position of the load,
and the other by which to draw the magnitude of the
load.

The former is an ordinary lineal scale, and the latter
is known as the force scale.

Set out the beam to scale (say ^ = V), and mark
the position of the load (Fig. 24).

To find ^'s load, set up at A the perpendicular
A D = 6 tons by scale (say J"' = 1 ton).

Join B D, and erect the perpendicular C E to meet it.

Fig. 24.

Then C E= x (§ 13)=J's load drawn to scale = 2 tons.

To find i?'s load, draw from E the line E F parallel
to A B. Since A F = C E = A'^ load, and AD = Q
tons, therefore F D = B's load = 4 tons.

52. Fig. 25 represents a beam supporting two loads.

Find the reactions of the supports.

Let A again = x tons, and take the moments about B.

Then 4 tons x 8 ft. + 5 tons x 12 ft. — x tons
X 18 ft. = 0,

TAEAT.LEL FORCES

) I tons 4 I

tons

6'o"

-L.0- ^
Fig. 25.

i.O

so X tons X 18 ft. — (32 + 60) ft.-tons,

and X tons =

(32 + 60) ft.-tons

Therefore B's load

18 ft.

= 9 tons

A's load.

5.V tons = 3# tons.

To find these reactions, as shown in Fig. 24, would
necessitate two figures and the answers added together.
In Chapter V a better method will be shown for graphi-
cally finding the reactions when there is more than one
load.

53. If the beam carries a uniformly distributed load
over its whole length, then each support carries one-
half the total load.

54. To ascertain this arithmetically or graphically
the whole load must be considered as concentrated
at its centre of gravity, which will be over the centre
of the beam.

55. If the load be uniformly distributed over a
portion of the length of the beam, the load must be
treated as acting at its centre of gravity.

56. Point of Application of the Resultant of
Parallel Forces. — It was pointed out (in § 31) that
the resultant of a number of forces is the force which
can be substituted for them, and, in the case of parallel
forces, it is equal to their algebraical sum.

It is now necessary to find the point of application

32 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

of the resultant, that is, a point in the line along wliich
the resultant would act to do the same work as the
forces for which it is substituted.

Since the algebraical sum of the moments of all the
forces about a point is equal to the moment of the
resultant about the same point (§ 48), and the moment
of the resultant about the point through which it
acts is zero (§ 43), then the algebraical sum of the
moments of all the other forces about that point is
zero.

57. First Case. To find the point of application
of the resultant of two forces when they act in opposite
directions.

Let two forces, A and
B, of which B is the
greater, act in opposite
directions.

Draw a line m n per-
pendicular to their lines of
action, and meeting them at x and y (Fig. 26).

It is now necessary to find a point in mn such that
the algebraical sum of the moments of A and B about
that point will be zero.

An examination of the figure will show that the
point cannot be between x and y because the two forces
would cause rotation in the same direction about any
point in x y.

Take any point 0 outside the smaller force.

If this be the point, then A.o x = B.o y.

But A is less than B, and o a; is less than o y, and the
product of two smaller quantities cannot be equal to
the product of two greater ones.

Hence, the required point cannot be outside the

(n 0

X y p n

Fig. 26.

PAKALLEL FORCES

smaller force. It must therefore be outside the greater
force, and, if p be the point, will be such that ^.py =
A.px. By § 37 the resultant is equal to B—A.

Example. — If two forces equal to 9 lbs. and 4 lbs.
act in opposite directions, find the resultant force and
its point of application when the distance between the
forces is 5 ft.
Adopt a lineal and a force scale.
Draw any line xy per-
pendicular to the lines of
action of the forces (Fig.
27).

Let A be the point where
the smaller force would meet
it, and B the point of inter-
section of the greater, 5 ft. from it.

From A erect a perpendicular A C = 9 lbs., and
from B, BD = 4 lbs.

Join C D and produce it to meet xy in 0.
Then 0 is a point in the line of action of the re-
sultant.

The resultant force is (9— 4) lbs. =5 lbs., and it acts
in the same direction as the 9 lbs.

Proo/.— The triangles C AO and D BO are similar.
Therefore C A : A 0 : : D B : B 0
and 0 ^ xBO = DBxAO.
But O ^ =: 9 lbs., siYid DB = 5 lbs.,
therefore 9 lbs. x BO = 5 lbs. x A 0.
The distance of 0 from either A or B can be obtained
by applying the lineal scale.

58. Second Case. To find the point of application
of the resultant of two forces when they act in the
same direction.

34 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

Let two forces, A and B, of whicli B is the greater,
act in the same direction.

Draw a line m n perpendicular to their lines of action,
and meeting them at x and y (Fig. 28).

Suppose a point to be taken to the left of A .
The forces A and B would cause rotation in the same
direction around this point, hence the algebraical sum
of their moments cannot be zero.

The resultant, therefore, cannot be to the left of A.
Similarly it cannot act to the right of B.

It must therefore act be-
tween A and B.
Q Suppose it acts through

the point o.
Then A. o x — B. o y = o,

m X 0 y n smd A.o x = B.oy.

-p .^g But B was taken greater

than A , therefore o y must
be less than o x, that is, the point o must be nearer
the greater force.

59. Hence, if two forces act in the same direction,
the line of action of the resultant is between them, and
nearer the greater force, and by § 37 the resultant
is equal to the sum of the two.

60. If the two forces be equal, the line of action of
the resultant will be midway between them.

61. If the point of application of one force be joined
to that of another like force, the resultant must pass
through the line joining them.

62. Example. — Two parallel forces equal to 7 lbs.
and 9 lbs. are 8 ft. apart, and act in the same direction.

Find the magnitude of the resultant and where it acts.

PARALLEL FORCES

two scales — say, lineal scale J'' = 1 ft.
and force scale t'' = 1 lb.

Draw any line x y crossing their lines of action at
right angles.

Let A (Fig. 29) be a point in the line of action of the
smaller force, and at this point erect the perpendicular
A C equal to 9 lbs.

Let J5 be a point 8 ft. from it in the line of action of
the larger force, and from B erect the perpendicular
B D equal to 7 lbs.

(Note the perpendiculars are drawn inversely to the
magnitude of the forces.)

Join A D and B C, then the point o where these lines
intersect is a point in the line of action of the resultant.

Through o draw the perpendicular m n.

Then m n represents the resultant, whose magnitude
is equal to the sum of the forces 16 lbs., whose line of
action is through o, and which acts in the same direction
as the otlier two forces.

Proof. — Draw por parallel to x y.

The triangles AoG and D o B are similar,

36 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

therefore A C : o p : : B D : o r
and AC x or = B D x op.
But ^ C = 9 lbs., SindBD = l lbs.
Therefore 9 lbs. x o r = 7 lbs. x o p,
or 9 lbs. x o r — 7 lbs. x o p = o.

Levers.

to' (b) (c)

Fig. 30.

63. Fig. 30 represents the three orders of levers (a)
being the first order, (h) the second, and (c) the third
order. In each case W means the weight or resistance,
P the power, and F the fulcrum.

The distance from the fulcrum (F) to the weight {W)
is called the " weight arm," and the distance from the
fulcrum to the power (P) is called the " power arm."

64. To solve problems on the lever arithmetically
the moments of W and P about F are taken.

The moment of W about F = the moment of P
about F,

therefore W x the weight arm = P x the power arm,
P X power arm

and W =
and P =

weight arm '
W X weight arm

power arm

, weight arm x W
and power arm =

power arm x P

and weight arm =— —

LEVERS

65. It will be noticed that the equations are similar
to those used for graphically working proportions,
therefore problems on the lever may be similarly solved.

Let B C (Fig. 31) be a lever of r
the first order, with ^ ^ as the
weight arm, and A G the power
arm.

(a) To find the weight,
draw the perpendicular
B D, equal to the power,
at the end of the weight
arm, and join A D. From
C draw G E parallel to
A D and meeting the per-
pendicula;r A E 3i>t E. Fig. 31.

Then A E is the required weight drawn to scale.

(6) To find the power, draw the perpendicular A E
equal to the weight at the end of the power arm, and
join G E. From A draw A D parallel to G E until it
cuts the perpendicular from B at D.

Then B D = the power.

(c) To find the power arm, draw the per-
pendiculars B D and A E equal to the power
and weight respectively. Join DA. From
E draw the line E G parallel to D A. The
interception of the line E G with
the lever determines the length
of the power arm.

{d) In order to find the
weight arm, the diagram is
modified a little, as shown
in Fig. 32.

The weight is set up from Fig. 32.

ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

G instead of from A, and the power from A instead
of from B. The line DB being drawn parallel to
A E determines the length of A B. The power or
weight can also be found with this diagram.
Proofs for (a), (6), and (c) —

The triangles AB D and C AE (Fig. 31) are similar,

therefore D B : A B : : A E : A C,

and DBxAC = EAxAB:

i.e. W X weight arm = P x power arm.

Fig. 33.

Fig. 34.

{(l) can be similarly proved.

66. Second and Third Order of Levers.

Suppose A B and A C (Figs. 33 or 34) to represent
the power arm and the weight arm respectively of a
lever of the Second Order (or Third Order).

(a) To find the weight, draw the perpendicular A D
equal to the power. Join C D, and from B draw B E
parallel toG D until it intercepts the perpendicular A E.

Then AE = thG weight.

(6) To find the power, draw the perpendicular A E
equal to the weight. Join B E, and from G draw G D
parallel to B E.

Then A E = the power.

LEVERS

(c) To find the power arm, set up A D equal to the
power, and A E equal to the weight.^^ Join C D and
draw E B parallel to C D. The point where E B inter-
cepts the lever is the end of the power arm.

{d) To find the weight arm, set up ^ D and yl ^ as
before. Join B E. From D draw D C parallel to B E.
The point C is the position of the weight, and ^ O is
the weight arm.

67. The " weight " and " power " in the First Order
of levers form a good illustration of " like parallel
forces " (§ 58), the point of application of the resultant
being at the fulcrum.

Since the lever is in equilibrium, the resultant must
be balanced by an equal and opposite force. This
second force is the reaction of the fulcrum, and is equal
to the sum of the weight and power.

Similarly the second and third orders of levers illus-
trate " unlike parallel forces " (§ 57), the reaction of
the fulcrum again being equal and opposite to the
resultant of the two forces.

Fig. .35.
68. It will be seen by Fig. 35 that a cantilever is

40 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

a lever of the first order, the fulcrum {F) being the
point on which the lever tends to turn. The power
(P) is supplied by the weight of the wall built on the
lever.

69. A beam supported at both ends is a lever of the
second order.

Either end may be considered as the fulcrum, if the
other be treated as the power.

Fig. 36.

70. Levers need not necessarily be straight bars.
They may be curved, as in Fig. 36 {a), cranked as in
Fig. 36 (c), {d) and (e), or of any other form. The
only essential point is that they should be rigid.

71. The " effective leverage " is the perpendicular
distance from the fulcrum to the line of action of the
force. This does not always correspond with the
lengths of the weight and power arms, as will be seen
by referring to Fig. 36 (a), (b), (c) and (d).

In each of these cases the weight and power arms must
be taken as represented by F A and F B respectively.

LEVERS

72. A little consideration will show that the most
economic way to utilize a power is by placing it at
right angles to the power arm, because that is the way
by which the greatest " effective leverage " can be
obtained.

73. If either the power or the weight (or both) be not
perpendicular to their respective arms, the solutions
can be worked in very much the same way as pre-
viously shown. For, let OPA (Fig. 37) represent
a lever with the weight acting perpendicularly at A,
and the power in the direction shown at 0. Then F A

Fig. 37.

will represent the leverage of the weight, and F B the
leverage of the power.

At A erect the perpendicular A C equal to the power,
and join F G. Make the angle B F D equal to the
angle AFC. Let the line F D intercept the direction
of the power at D. Then B D will represent the weight.

To find the power, make BD equal to the weight
and join F D. Make the angle AFC equal to the
angle B F D, and let the line F C meet the perpen-
dicular from A at C. Then A G will represent the
power.

42 ELEMENTARY PKINCUPLES OF GRAPHIC STATICS

To work the above problem mathematically, B F
and FA will still represent the leverages, therefore,
taking the moments about F,

P X FB=W X FA.

The forces shown in Fig. 36 (c), {d) and (e) are not
parallel. In Chapter IV a better method will be shown
for dealing with these cases.

Examples to Chapter II.

1. If the force scale be f' = 100 lbs., what does
^4 B represent ?

A— '^B

Ex. C'h. II.— Quest. 1.

2. A body weighs 46 lbs. To a scale of i/ = 10 lbs.
draw a line showing the force exerted by it.

3. One force is equal to 12-5 lbs., and another, acting
in the same straight line, is equal to 23 lbs.

Graphically show the resultant —
{a) if they act in the same direction.
(6) if they act in opposite directions.

4. A beam, 12 ft. long and weighing 1|- cwts., sup-
ports a load of 4 cwts. at its centre, and another of
3*5 cwts. 4 ft. from one end.

(a) Draw a line showing the total load.

(6) What is the total reactions of the walls ?

5. A ladder leans against a smooth upright wall.
Wliat is the direction of the force exerted by the wall
to support it ?

6. What is meant by the moment of a force ? How
is it found ?

EXAIVIPLES TO CHAPTER II

7. A cantilever 5 ft. long supports a load of 2
cwts. at its outer end.

What is the moment of the load —
{a) at the wall end ?
(6) at the centre ?
(c) at the outer end ?

8. A beam over a 15'. 0'' span carries a certain con-
centrated load.

If the reactions of the supports due to this load be
4 J tons and If tons respectively, what is the amount
of the load, and where is it placed ?

9. A bar 4'. 6'' long works on a pivot which is V.(j''
from one end.

If a weight of 21 cwts. be placed at the end of the
shorter section, what weight must be placed at the
end of the longer section to balance it ? (Neglect the
weight of the bar.)

10. Two walls 6' apart support a beam on which is
placed a load weighing 1200 lbs.

(a) If the load is placed 2*5 ft. from one end, what
portion of the load does each wall support ?

(6) If the beam weighs 150 lbs., what is the total
load on each wall ?

Chapter III
CENTRE OF GRAVITY— BOW'S NOTATION

74. Centre of Gravity. — We speak of the weight
of a body. The weight is simply a downward force
exerted by gravity. The body is made up of innumer-
able particles, on each of which gravity exerts a down-
ward force. These forces, for all practical purposes,
may be considered as parallel, and the weight of the
body is the resultant of all these smaller forces.

If a solid body be freely supported in any position,
the line of action of the resultant force will pass verti-
cally through the body. If it be held from a different
point, the force exerted on each particle, and the resul-
tant of these forces, will again be vertical, and the
second resultant will intersect the first at a certain
point. In whatever position the body is held the
resultant forces wdll cross each other at the same point.

This point is called the Centre of Gravity (e.g.) of the
body, and w^e may assume that its whole weight is
concentrated there.

75. The e.g. of a thin sheet can easily be found experi-
mentally by suspending the sheet in any position and
marking across it in line with the suspension string, as
shown in Fig. 38 (a). Suspend it in another position,
and mark as before (Fig. 38 (b)).

The intersection of these tw^o lines will give the e.g.
of the sheet.

CENTRE OF GRAVITY

FiCx.

76. The e.g. of a thin sheet or lamina in the form of a
parallelogram is given by the intersection of the dia-
gonals (Fig. 39).

Fig. 39

77. To find the e.g. of the triangular lamina ABC
(Fig. 40), bisect the side B C in D, and join A D. Bisect
another side as A B in E, and join E C.

The point F, where E C cuts A D, is the e.g. of the
trianofle.

FD is I of A D, and F E is 1 of G E, therefore we

ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

may find the e.g. of a triangle by joining the middle
point of any side to the opposite angle, and taking a
point on this line J of its length from the bisected line.
Tlie student may adopt either method.

'■•'."

_ . . .

4 •

• .

, »

■■■

' *->•<■ '

Y ■

■.,.ri

. • • eg.- .

"\-

.iA * * - •

Fig. 41.

78. The e.g. of a trapezium or trapezoid. Let A BC D
(Fig. 41) be the trapezium. On each side oi A B mark
o^ A E and B F equal to the base C D, and on each
side oi C D mark ofi C H and D G equal to the top A B.
Join E O and F H. The intersection of these lines gives
the e.g. oiABGD.

79. To find the e.g. of irregular rectilineal figures it is
generally necessary to divide the figure into triangles
or parallelograms.

As an example, we
will proceed to find the
e.g. of the quadrilateral
ABC D (Fig. 42).

Divide the figure into
two triangles by joining
AC.

Find the e.g. of each
as shown in § 77, and
let these be at E and F.

These two triangles

CENTRE OF GRAVITY 47

are portions of a figure which is supposed to be of
uniform density, therefore the weight of each is in
proportion to its area. But the weight of each is a
do\\Tiward force acting at its e.g. (§ 74), hence we have
two like parallel forces, and it is'necessary to find the
resultant force. ^

Through E, the e.g. of the larger triangle, draw the
perpendicular H I equal to the smaller force, and
through F, the e.g. of the smaller triangle, draw the
perpendicular M N equal to the larger force. Join H N
and M I. Through 0, where they intersect, draw the
perpendicular L K.

Then L K is the line of action of the resultant of the
two forces at E and F, that is, of the two triangles.
But the two triangles make up the figure A B C D,
therefore it is the line of action of the resultant of
the whole figure, hence the e.g. of the whole figure lies
in LK.

But it is clear that the e.g. of the whole figure must
lie in a line joining the c.g.s of its two portions,
therefore the e.g. of the whole figure is at the point P,
where the line L K crosses the line E F.

80. To find the e.g. of a mass whose cross section is
uniform in size and shape, it is sufficient to find the e.g.
of a lamina of the same size and shape as the cross
section.

Fig. 43 represents a wall of regular dimensions. If
this wall be considered divided up into an indefinite
number of thin vertical sections parallel with the end
of the wall, then A (the intersection of the diagonals)
gives the e.g. of the first lamina, and B the e.g. of the
last.

The line xi B passes through the e.g. of each lamina,

48 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

and C, the iniddle point of A B, in the e.g. of the whole
wall.

6s,.

-C.

D E

• •

'VA

Fig. 43.

D E F G is n cross section of the same wall, so it can
be seen that the e.g. of the whole wall comes directly
behind the point A, the e.g. of a lamina of cross section.
The student will now understand why P (Fig. 35) is
placed at half the thickness of the wall, this being the
centre of the pressure exerted by the wall.

8i. The e.g. of a door may be found by drawing
the diagonals.

Since the weight of a door acts vertically through its
e.g., the leverage with which the door
acts on its hinges is half the width of
the door. Either
hinge may be con-
sidered as the ful-
crum, then the other
becomes the power
which maintains

equilibrium.

Thus a door is an
example of a lever, the weight being

BOW S NOTATION

represented by the weight of the door, the weight arm
being half the width of the door, the power being the
reaction at one of the hinges, the power arm the distance
between the two hinges, and the fulcrum the other
hinge.

Reference to Figs. 44 and 45 will make this clear.

82. Bow's Notation. — Before proceeding any fur-
ther, it may be advisable to explain the system of
lettering diagrams as devised by R. H. Bow, C.E.,
F.R.S.E. This system has innumerable advantages,
and will amply repay the student for the time spent
in mastering it.

It will be adopted in all the succeeding exercises.
It consists of lettering (or numbering) all the angles or
spaces formed by the external forces, and when nam-
ing a force to do so clockwise.

Fig. 46 is given as an illustration.
Here there are three forces acting upwards, and
two acting downwards, and, if the system be in equi-
librium, the sum of the three forces is equal to the
sum of the two.

There are five forces, consequently five spaces.
Place a letter in each space as A, B, C, D, and E.
Other letters or numbers would do, and they may be
placed in any order, but it is well to be systematic

Ciirts.
5

cwts.
3

cwts.

cwts

(a)

cwts.
t

ib)

■ b

€

Fig. 46.

50 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

and adopt a uniform practice. There is a force divid-
ing the spaces A and B,. so this force will be named
by those two letters, but as in clockwise order A
would come before B, the force is named A B^ not
B A . Similarly the other forces are B C, CD, D E,
and A E.

83. It will be necessary to draw lines representing
the magnitude and direction of those forces, which lines
will have the same names as the above, but small letters
will be used instead of capitals.

84. Now, if a line as ^ Y represents

a force, the force must be considered as acting from
Z to 7 or from Y to X, that is, X or Y must be first in
its course of action.

In Bow's notation the letter which is placed first in
the course of action of a force is the one which comes
first in the clockwise notation.

Draw a line representing the force A B (Fig. 46).

Since A is the first letter in the clockwise order, and
the force acts upwards, a must be placed at the bottom.
The next force is B C, and this acts upwards, so the h of
the first force becomes the first point in the line of action
of the second force, which is again drawn upwards.
Similarly the c of the second force becomes the first
point of the third force C D, but this force acts down-
wards, so cd must be measured off downwards. D E
acts upwards, hence d e must be measured upwards.
E A acts downwards, and e a is measured in that
direction.

(It should be noted that this is simply an application
of addition and subtraction as shown in § 9.)

As this last point corresponds with the first, the answer.

bow's notation 51

or resultant, is zero. If the last point had not fallen on
the first, there would have been a resultant force acting
upwards or downwards according as the last point
would have been above or below a.

85. The diagram showing the beam (or any other
structure) and the position of its loads is called the
" frame diagram," and the diagram representing the
forces drawn to scale is called the " force diagram."

Fig. 46 (a) is the frame diagram, and Fig. 46 (6) is the
force diagram.

86. This method is extremely useful in finding the
resultant of any number of forces. The forces, as
shown on the frame diagram, should be named in clock-
wise order, then the first and last letters will name the
resultant and give its direction as shown on the force
diagram.

Referring to Fig. 46, let it be required to find the
resultant of the three forces shown on the top of the
beam. These forces are A B, B C, and CD. A is the
first letter, and D the last of this series, so a c? on the
force diagram represents the magnitude and direction
of the resultant force. By measuring ad it is found
equal to 2 cwts., and as a cZ also gives the direction, the
force acts from a to d, that is, in an upward direction.

Again, suppose the resultant of the two forces on the
right is required. These are C D and D E. The first
and last letter are C and E, so c e on the force diagram
fully represents the resultant of these forces, and is
equal to 4 cwts., acting in a downward direction.

87. As in the case of the known forces, the letters
have to be placed with due regard to the direction in
which the force is acting, so will the letters indicate the
direction of the unknown ones.

ELEMENTARY PKINCIPI-ES OF GRAPHIC STATICS

Suppose it had been required to find the direction of
the force on the left of Fig. 46 (a). This force is known
a>»E A, and on referring to the force diagram we find that
to proceed from e to a we go downwards, hence the
force E A acts in that direction.

88. If the structure on which the forces act be an
open framed one, in addition to the spaces between the
external forces being lettered, a letter is placed in every
space of the frame.

As an example see Fig. 47.

The external forces are the load of 3 tons and the
two reactions of the supports. The force exerted by

tons.

\ E ^

o\f

' J'

•^

" ci> re

Fig. 47.

the left hand support would be known SbsC A, the 3 tons
load as A B, and the force exerted by the right hand
support as B C.

The spaces in the frame are now lettered as D, E, F,
G, H, I and J.

89. An examination of the figure will show that
there is a letter on each side of every bar, and these
letters will name the bar, but which letter comes first
in considering the forces will depend on which end of
the bar is under consideration. Thus, take the bar

LOAD — STEESS 53

dividing the spaces F and C. The bars meeting at the
left of this are named, according to the clockwise order,
G D, D E, E F and F G, and at the other end they are
F G,GH, H G, and G F. Hence, when dealing with
the one end the bar is named F G, and when dealing
with the other it becomes G F.

90. As this part of the work is devoted to Bow's
notation, there are two other things which it may be
advantageous to point out, but which will not be
thoroughly understood until the student is dealing with
the effects of loads upon framed structures (Chap. VII.).

The first of these is Every bar surrounding

a space in the frame diagram meets at the same point
in the stress diagram, and this point is named by the
letter in the space of the frame diagram.

The second is The external forces and bars

meeting at a point in the frame diagram will form the
sides of a polygon in the stress diagram.

91. Load. — By a load on a structure is meant the
sum of all the forces acting upon it, together with the
weight of the structure itself.

92. Stress — tension and compression. — If a force
acts on a body, it produces from that body an equal and
opposite resisting force.

This resistance is known as stress.

If a weight be suspended by a string, the string exerts
an upward force equal to the downward pull of the
weight ; and, if a prop or strut supports a load, it
pushes against it with a force equal to that of the load.

The force exerted by the string in resisting elongation
is called its tensile stress or tension, and the resistance
to crushing set up in the strut is known as compressive
stress or compression.

54 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

But we know that the string in itself could not support
a weight without being attached to some support to
which the string transmits the force. The string exerts
a downward pull at the support, as well as an upward
pull at the weight. Since each end of the string exerts
a force equal to the weight, it might be supposed that
the tension in the string is twice the force exerted by
the weight, but it is not so.

93. The student can easily satisfy himself that the
tension in the string is only equal to the force at one
end if he will fit up an apparatus as shown in Fig. 48.

^ <X^E}^^

Fig. 48.

This consists of two equal weights, two pulleys, and
a spring-balance attached to the two weights.

Although there is a force acting at each end, the
balance will show that the tension is only equal to one
of them.

Similarly with the strut, the load and the support
each exert a force against it at opposite ends, but the
stress set up is only equal to one of them.

When a bar is in tension each end exerts a force
inwards, and these two forces are equal ; and when a
bar is in compression the two ends exert outward
forces which are equal to one another.

By marking the senses at each end of a bar, a glance
will show the kind of stress in that bar.

94. Since a compression bar exerts an outward force

EXAMPLES TO CHAPTER III.

at each end, the arrows will point outwards, thus : —
< > ; and since a tension bar exerts an inward

pull at each end it is marked thus : > < •

95. Strain. — The forces which produce tension or
compression in a bar also cause an alteration in its form.
This change of form may be so slight that, upon the
removal of the forces, the bar will regain its original
shape, or it may be such that the bar is permanently
inj ured.

In either case this change of form is known as strain.

96. When expressing the amount of stress in a bar,
the sign -t- (plus) is often prefixed for compression bars,
and the sign — (minus) for tension bars, instead of
indicating it by means of arrows.

Another method of indicating the kind of stress is to
draw thick or double lines for compression bars and thin
ones for tension bars.

Examples to Chapter III.

1. Find the e.g. of a wall 6 ft. high, 3' 6'' broad at the
base, and 2' at the top, one face being vertical.

2. Fig. 1 shows a beam supporting a body which
weighs 225 lbs.

Ex. Ch. III.— Fig. 1.

Draw a Hne indicating the position and direction of
the force exerted by the body. Scale l''=:100 lbs.

3. Two loads, 3 tons and 4 tons, are placed 5 ft. apart
on a beam. Where is the centre of pressure ?

56 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

4. Draw a vertical line which will pass through the
e.g. of the wall shown in Fig. 2.

r7
'/

. Ch. III.— Fig. 2.

Ex.

////////////////////
Ch. III.— Fig. 3

If the wall shown weighs 135 lbs. per c. ft., what is
its weight per foot run ?

5. Find the e.g. of the triangle Fig. 3.

6. Fig. 4 shows a triangular prism lying on a beam.
If the prism weighs 1,000 lbs., what are the reactions

of each support due to it ?

i.L

■^ 2.0 ^

L'.i'-

Ex. Ch. III.— Fig. 4.

7. In Fig. 5 the directions of a number of forces acting
at a point are given.

Name them according to Bow's Notation.

8. Fig. 6 represents five parallel forces in equilibrium.

. Ch. III.— Fig.

2 6
cults, cures.

/
cwt

cu/ts

Ex. Ch. III.-

-Fig.

cults.

EXAMPLES TO CHAPTER HI

Draw the force diagram. Scale 1"=4 cwts.

9. Five parallel forces in equilibrium are shown in

Ficr. 7.

Give the magnitude
and direction of the
force A B.

10. A bar 5 ft. long
is secured by a pivot
at one end, while a

mt$ curt,

eufts

B I C tOi £

/\ 2 ]cwts.

Ex. Ch. III.— Fig. 7.

14 lb. weight is suspended at the other.

(a) Neglecting the weight of the lever, what power is
required 2 ft. from the pivot to support it ?

(b) What are the magnitude and direction of the force
exerted by the pivot ?

Chapter IV

PARALLELOGRAM, TRIANGLE, AND POLYGON
OF FORCES, AND RETAINING WALLS

97. Up to the present only forces whose Hnes of
action are parallel have been dealt with. It now
becomes necessary to examine other forces.

The student should again take up his spring balances
and arrange them as in Fig. 18. He may dispense
with the pulley, and must remember that the balances
are only used to register the force exerted by the string
attached to each.

Now that the two strings are exerting a force parallel
and opposite to that exerted by the weight, their sum

is equal to the down-
ward force. With the
same weight attached, he
should increase the dis-
tance between the points
of suspension (Fig. 49). A
glance at the balances
may now cause him no
little surprise. He should
try them in three or four
positions, each time in-
creasing the distance between the points of support,
and noting the results.

His observations may be summarized as follows : —

Fig. 49.

PARALLELOGRAM OF FORCES

{a) When the supporting strings are no longer parallel
to the line of action of the force exerted by the weight,
the sum of the forces exerted by them exceeds that
exerted by the weight, (b) The further they are from
being parallel (that is, the greater the angle between
them) the greater is the force they have to exert to
support the weight.

98. Parallelogram of Forces. — Selecting one of
these positions, and adopt-
ing a convenient scale, as
4'' = 1 lb., a line should be
marked behind, and parallel
to, each supporting string,
and the tension measured
on each as indicated by the
balance.

Complete the parallelo-
gram, and draw the diagonal
as shown in Fig. 50.

Measure the diagonal to
the same scale. ^^^- ^^•

Two things will now be noticed : — First, the
diagonal measured to scale will give the same force as
that exerted by the weight. Second, the diagonal
will be in line with the string supporting the weight.

This should be verified by trying it in each of the
former positions.

To change the direction of the pull of the weights,
three pulleys and three weights should be fitted up as
shown in Fig. 51.

It will be seen that similar results are obtained.

Further, any two of the forces can be utilized to find
the third, as indicated by the dotted lines.

60 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

Let US examine these results further.

The force exerted by the weight keeps the other two
forces in position, so it is the equilibrant of them (§ 33).
The two strings da and dc support the 7 lb. weight,
and keep it in equilibrium, but a force which could
be substituted for these two forces is their resultant,
and to support the weight in that position it is evident
that a force is required which is equal to that exerted

Fig. 51.

by the weight (or equilibrant), and which acts in the
opposite direction and in the same straight line. But
the diagonal of the above parallelogram measured to
scale, gives a force equal to the equilibrant, and is in
the same straight line, so if an arrow be placed on it
indicating that it acts in the opposite direction to the
equilibrant, it will represent the resultant of the other
two forces.

99. Hence, if the magnitude and direction of two

TRIANGLE OF FORCES 61

forces be known, by completing the parallelogram and
drawing the diagonal the magnitude and direction of
the resultant force is obtained.

100. It is evident that the resultant of any two
forces (not parallel) must pass through the point of
intersection of the lines of action of the two forces.

10 1. If three non-parallel forces maintain equili-
brium, the lines of direction of these forces, if pro-
duced, will meet at the same point. Any two of the
forces can be replaced by a resultant, and, since the
third force balanced these two, it will balance their
resultant, and this it can only do by acting at the
same point.

102. — Triangle of Forces. — On further examining
the parallelogram ah c d (Fig. 51), it is seen to be made
up of two equal triangles.

The triangle ah d has the side d a parallel and equal
to the 3 lb. force, the side a h parallel and equal to the
5 lb. force, and the side h d parallel and equal to the
7 lb. force.

li h d he considered as representing the equilibrant,
then it will represent a force acting from h to d, and the
three sides proceeding from d to a, a to h, and h to d,
will give the sense of the three forces.

103. Hence, if three forces be in equilibrium, it is
possible to draw a triangle with sides parallel to the
line of action of each force, and representing them in
magnitude each to each, and whose sides, taken in order
round the triangle, will give the sense of each force.

If, therefore, the sense of one of the forces be known,
the sense of the others is known.

104. The converse is equally true : —

If it be possible to form a triangle with sides parallel

62 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

to the lines of action of the forces and equal to them in
magnitude, and whose sides represent the sense of each
taken in order round the triangle, then the three forces
are in equilibrium.

105. Each of these three forces must be the equili-
brant of the other two, hence if the sense of one be
reversed, then that force becomes the resultant of the
others.

106. If, on examining a triangle of forces, it is found
that the sense of one force is opposite to the others,
then the force represented by that line is the resultant
of the others.

107. If three forces (not parallel) maintain equili-
brium, the sum of any two must be greater than the
third.

Since the three are in equilibrium, it must be possible
to form a triangle with sides parallel to and proportional
to the forces, but unless any two are greater than the
third this is impossible.

108. If three forces, of which the two smaller are
equal to the greater, maintain equilibrium, then they
are all parallel, and the two smaller act in the opposite
direction to the greater.

109. Since, when three non-parallel forces are in
equilibrium, it is possible to form a triangle with sides
equal and parallel to the forces represented by them,
and it is impossible to make any triangle with two
parallel sides, there cannot be three such forces, two
of which are parallel, in equilibrium.

no. If, then, in any structure there be three bars
(or two bars and an external force) meeting at a point,
and any two of thembe parallel, the forces exerted by
the two parallel ones are equal and opposite, and the

NON-PARALLEL FORCES

third bar exerts no force. (N.B. — This is not true if
there be more than three bars or more than three bars
and forces together.)

To illustrate this, three portions of different girders
are shown (Fig. 52).

Fig. 52.

In each of the above cases the force exerted hy A B
acts directly on the end of the bar C A, and produces
from it an equal and opposite force. If there were
either a pull or a thrust in 5 C, it is evident that equili-
brium would not be maintained. But we know the
joint is in equilibrium, hence there is neither tension
nor compression in the bar B G, i.e. the force exerted
hy BG = 0.

III. If the magnitude of two forces maintaining
equilibrium with a third force whose magnitude and
direction is known, be given, then their directions can
be ascertained.

Fig. 53.

64 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

Let A B (Fig. 53) be the known force. Draw a h
parallel and equal to this force. With a as centre,
describe an arc with a radius equal to the line repre-
senting one of the other forces, and with h as centre
and a radius equal to the line representing the re-
maining force describe an arc cutting the first at c.
Join c h and c a.

Their directions are given by 6 c and c a, and a refer-
ence to Fig. 53 will show they can be transferred to
either end of A B.

1X2. If three forces whose directions are given, act
at a point in equilibrium, and the magnitude of one be
known, then the magnitude of the others can be found.

Fig. 54 shows a wall and the foot of a roof truss.
Suppose the reaction of the wall to be 30 cwts., then
this is an upward force resisting the action of the rafter
and the tiebeam.

Letter the spaces as shown, and draw a h parallel to
A B and equal to 30 cwts. From h draw h c parallel
to B C, and from a draw a line parallel to C A. Then

Fig. 54.

b c and c a give the magnitude and directions of the
forces exerted hy B C and C A. B C acts towards the
joint, and C A from it.

113. Two or more forces which have a resultant

RESOLUTION OF FORCES

Fig.

force are called the com-
ponents of that force.

The two forces A and
B (Fig. 55) have a re-
sultant force R. R is
then the force which
could be substituted for
them (§31). It is equally
correct to say that the forces A and B could be sub-
stituted for the force R.

A and B are the components of R.

If the direction of the force R were reversed it would
become the equilibrant of the forces A and B (§ 33),
then the three forces would be in equilibrium. But
in § 112 it was shown that if a force be known, the
magnitude of two others producing equilibrium and
acting along given directions could be found. Hence,
if R were considered as acting in the opposite direction,
the magnitude of the forces A and B could be deter-
mined.

Fig. 56 shows the foot of the rafter of a couple roof
along which a force equal to 2J cwts. is acting. It is
necessary to find the vertical and horizontal components

If we consider the
action of this force re-
versed, then it will act
away from the joint.
Draw ah to represent
the 2J cwts. acting in
that direction. From
h draw a vertical line,
and from a a horizontal
one intersecting it at c.

66 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

Fig.

Then h c and c a will represent the two components
in direction and magnitude.

(It should be noticed that the rafter A B causes a
horizontal thrust equal to c a, which tends to overturn
the wall, and a vertical thrust equal to h c. These have
to be resisted by the wall.)

114. If a body A (Fig. 57) be placed on a smooth

inclined plane, it slides in
the direction shown by the
arrow. This sliding must
be caused by some force
acting in that direction.
One of the forces acting on
it must be its own weight,
but this acts vertically, and

could not of itself produce motion down the plane, so
there must be another force acting on it. The other
force is " the reaction of the plane," which always acts
at right angles to the plane. The force which causes
the body to move along the plane is the resultant of
the force caused by the weight of the body and the
reaction of the plane.

But the force acting along the plane, and the reaction
of the plane, are two forces produced by the vertical
force exerted by the weight of the body, hence they are
components of that force.

The magnitude of this vertical force is known and the
direction of the two component forces, so it is possible
to find their magnitude (§ 113).

If it were necessary to keep the body from sliding,
a force equal and opposite to the one acting along the
plane would do it.

115. It is clear that a force applied horizontally, as

RESOLUTION OF FORCES

Fig. 58.

shown in Fig. 58, would also keep the body in equili-
brium, hence the body must have a horizontal thrust
equal to that necessary to
keep it in position when
applied in that direction.
By resolving the vertical
force exerted by the body
into a force at right angles
to the plane, and a hori-
zontal one, the horizontal
thrust of the body is obtained.

1 1 6. Since we know the weight of the body acts
vertically, and the plane exerts a force at right angles
to its surface, being given the weight of the body, and
the inclination of the plane, we can find either the
force the body exerts parallel to the plane, or the force
it exerts horizontally.

117. Fig. 59 shows a cantilever supported by a strut
and loaded with 2 cwts.
Find the kind and
amount of stress set
up in each member.

Letter the spaces on
the frame diagram, and
draw h c equal to 2
cwts. From c draw a
line parallel to C^, and from h draw one parallel to
A B. Let them meet at a.

Then c a and a h will represent the stresses.

Since the force represented by 6 c acts downwards
and the sense must be in the same direction taken
round the triangle, therefore c to a gives the direction
of the force exerted hy C A, and a to 6 gives the direc-

ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

tion of the force exerted hy A B. ca and a b measured
to the same scale as that by which b c was drawn will
give the magnitude of the forces exerted hy C A and
A B respectively.

Since C A acts outwards towards the joint, and A B
inwards from it, the former is in compression and the
latter in tension (§ 92).

1 1 8. Fig. 60 shows a loaded cantilever supported
by a wrought iron rod.

Fig. 60.

The construction of the force diagram needs no
further explanation.

An examination of the force diagram will show that
in this case the beam is in compression and the rod in
tension.

119. The " triangle of forces " is most useful in the
solution of levers when the forces acting on them are
not parallel.

Let AFB (Fig. 61) be a lever with the "power"
and " weight " acting as shown.

In order to maintain equilibrium there must be
another force acting, and this is the " reaction of the
fulcrum." There are then three " non-parallel " forces
maintaining equilibrium, therefore the lines of direction

NON-PARALLEL FORCES

of these three forces must meet at the same point

(§ 101).
Let the Unes of direction of P and W meet at 0, then

Fig. 61.

the hne indicating the direction of the reaction of the
fulcrum must pass through 0. Since the reaction
acts at the fulcrum, it must also pass through F, there-
fore the line F 0 gives the direction of the reaction of
the fulcrum.

At 0 there are now shown the directions of three
forces in equilibrium, and, if one be known, the others
can be determined (§ 112).

Suppose the weight (T^) to be known.

Letter the spaces, and draw o t parallel, and equal to,
the weight. Complete the triangle of forces by drawing
o s parallel to 0 S, and i s parallel to T S.

t s completely represents the reaction of the fulcrum,
and s o completely represents the power.

120. Again, let Fig. 62 represent a door whose hinges
are at A and B, and let it be required to find the hori-
zontal reaction of the hinge A, and the total reaction
of the hinge B.

Since it is tlic horizontal reaction of A that is re-

70 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

quired, its direction must
Q intercept that of the
force exerted by the door
at 0. Join B 0.

Then B 0 gives the
direction of the total
reaction of the hinge at
B.

At 0 the direction of
three forces in equiHbrium
is given, and, since one
of them, the weight of
the door, is known, by applying the triangle of forces
the others can be found.

121. The rafters of lean-to or pent roofs are often
found fixed as shown in Fig. 63.

Fig. 62.

Fig. 63.

RETAINING WALLS

On examining the forces acting on this, it will be
seen that the reaction of the one wall is in a horizontal
direction, and this intercepts the line of action of the
load on the roof at A. The third force is the reaction
of the lower wall, and its direction must pass through
the point of intersection of the other tvv^o forces

(§ 101).

Its direction is therefore given by the line A G.
Resolving G B parallel to B A and A G the triangle
c & a is formed, c a now gives the magnitude and
direction of the thrust of the rafter, and, by finding
the horizontal and vertical components of this thrust,
it will be seen that the lower wall has to support the
whole weight of the roof as well as resist a horizontal
thrust, whose magnitude is given by c d.

By forming the rafter as shown in Fig. 64, the roof
is supported by two
parallel forces. Each
wall then gets one-half
the weight of the roof,
and there is no horizontal
thrust.

122. Retaining
Walls. — Walls built to
sustain water or earth
are called retaining walls,
and it is now intended

to apply the knowledge gained in the preceding pages
to ascertain whether any proposed retaining wall
is sufficiently strong for its purpose.

Before this can be done, we must ascertain the
forces at work.

It is clear that the wall is put to resist the thrust of

72 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

the water or earth behind it. We will first examine
this resistance.

It was shown in Chapter III that the weight of a
body can be considered as concentrated at its centre
of gravity, so the force (or resistance) exerted by the
retaining wall is its weight acting vertically through its
e.g.

By taking a part of the wall 1 ft. in length, and the
area of the cross section, we have the number of cubic
feet in the part of the wall under consideration. Know-
ing the weight of 1 cub. ft. of the particular walling (say
1 cwt. for brickwork, and 140 lbs. for masonry), we can
now ascertain the weight or vertical force exerted by
the wall, and, since it acts through its e.g., we know its
line of action.

123. Next, we will inquire into the force exerted by
water on a retaining wall or dam.

Hydrostatics teaches us that water always exerts a
pressure at right angles to the sides of the vessel con-
taining it or the containing surfaces, and that the
pressure at any point is in proportion to the vertical
distance of this point below the surface of the water.

124. Since the pressure at the bottom of a retaining
wall depends on the vertical height of the surface of
the water above this point, a line equal to the depth
of the water will represent the magnitude of the pressure
at this point. But the pressure is at right angles to
the surface of the wall, so the line representing the
magnitude of the pressure must be drawn in that
direction.

In Figs. 65 and 66, from h, the bottom of the wall,
draw b c perpendicular to ab and equal to the vertical
depth of the water ; b c now represents the magnitude

RETAINING WALLS

and direction of the pressure of the water at this point,
where of course the pressure is greatest. If other
points be taken on the wall, the pressure at these will
be less as the vertical height to the surface decreases,
until the top of the water is reached, where the pressure
is nil.

Hence, if a c be joined, the triangle ah c will graphi-
cally represent the total pressure on a section of the
wall. The ordinates are drawn showing the relative
amount of pressure at different points.

125. The magnitude of this pressure must now be
obtained.

c^'^rg^:<^?^c.^r^>1 ^^ 3^^^ifo2^^^^<-^^'> g °^ "^i

Fig. 65.

Fig. 66.

The triangle ah c represents the section of a volume
of water whose height is 6 c and whose base is a h, and,
if the volume pressing on 1 ft. of the length of the wall

be taken, its cubical contents is cubic feet.

But 1 cub. ft. of water weighs 62-5 lbs., therefore the
total pressure exerted on 1 ft. of the length of the wall

ah X he

X 62-5 lbs.

126. Having found the magnitude of the pressure,
its direction and point of application must now be
considered. The weight of the triangular volume of
water represented hy ah c must be treated as if con-
centrated at its e.g., and it presses at right angles to

ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

the inner surface of the wall, therefore a line through
its e.g. perpendicular to this surface will give the direc-
tion of this pressure, and its point of application is
where this line meets the wall.

The point of application is always J of a 6 measured
from h.

127. It is intended to build a stone wall 6 ft. high
to dam a stream of water to a depth of 4 ft,, the width
at the base to be 3 ft., the top, 2 ft. 6 in., and the inner
surface is vertical.

Find whether the wall is sufficiently strong.
Set out the wall and depth of water to some con-
venient scale (Fig. 67). From the base h mark off 6 c
perpendicular to a 6 and equal to 4 ft. Join a c. The
triangle ab c represents the pressure on the wall (§125).

Find the e.g. of a 6 c

and of the wall as

shown in Chapter III.

A vertical line from

the e.g. of the wall

will represent the

^'^^^J''^^ direction of the force

exerted by the wall, and a line

perpendicular to the inner face of

the wall will represent the direction

of the force exerted by- the water.

Let these lines intersect at 0.

To find the magnitude of these
forces, a portion of the wall and of
the water, 1 ft. each in length, is taken,

2' 6" + 3'
then the wall weighs ^- x 6' x 140 lbs. =2310 lbs.,

and the water weighs

4'x 4'

62-5 lbs. = 500 lbs.

RETAINING WALLS 75

These represent the magnitude of the forces acting
in the directions shown.

From o scale off o d = 2Z10 lbs., and from d draw
d e = 500 lbs. Join o e ; then o e represents the magni-
tude and direction of the resultant force, and o e cuts
the base of the wall at /.

128. To fulfil all the usual conditions necessary for
the stability of a retaining wall for water, the resultant
force must not intersect the base outside the middle
third, but this rule is not universal in its application.

An examination of Fig. 67 will show that the point /
is within the middle third, hence the proposed wall will
be strong enough.

129. A retaining wall (brickwork) 7 ft. high has a
batter of 1 in 8 on the outer surface. The base is 4 ft.
and the top 1 ft. thick.

Ascertain whether it is safe to allow the water to
rise to a depth of 6 ft.

Set out the wall to scale, and indicate the water line
as before (Fig. 68).

The construction is similar to that of the last exercise.

The only point to be noted is that h c and the centre
of pressure of the water still remain perpendicular
to a b.

The lengtli of ab is obtained by scaling it on the
drawing.

The weight of the wall and the water is obtained as
shown in the previous exercise.

It will be seen that the resultant falls outside the
middle third, hence the wall is probably not strong
enough (§ 128).

Note. — The question of stability depends partly upon
the crushing force and the strength of material at the

76 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

outer edge, and cases may occur where perfect stability
exists although the resultant may pass beyond the
middle third.

To obtain a wall strong enough the thickness should
be slightly increased, and the above test again applied.

130. In order to understand the thrust caused by

tl8J'5 Ik

-17

. A.^--'

/ >

tqbo lis.

Fig. 68.

earth on a retaining wall a little explanation is
necessary.

If a bank of earth be left exposed to the weather,
it will crumble and fall until it forms a certain natural
slope depending upon the nature of the earth of which

RETAINESTG WALLS

it is composed. The angle which this slope forms
with the horizontal plane is called the angle of repose.
If in Fig. 69 A B shows the natural slope, then ABC
is the angle of repose.

If 5 D be drawn perpen- D L. A

dicular to B O, the 77^ 7 y^llllDin)!,

angle A B D i^ the
complement of the
angle A B G.

It has been shown by
several writers that the
portion of earth which
tends to break away
and overturn a wall is that enclosed between the
vertical line B D from the foot of the wall and the line
B E bisecting the complement of the angle of repose,
that is, in Fig. 54, the portion D B E.

B E is called the plane of rupture.

In order that the portion D B E should break away,
it must slide down the plane B E, acting like a wedge
on B D, and forcing it out horizontally.

It was shown in § 114 that if a body be placed on an
inclined plane a certain force is exerted parallel to the
plane. Each particle of the mass D B E is a, body on
the inclined plane B E, and the sum of them may be
treated as if concentrated at the e.g. of D B E, hence
the whole mass exerts a force through its e.g. parallel
to the plane. This intercepts B D at F, so F is the
point of application of the force.

This point is always J the distance up the wall.
D B E is evidently prevented from sliding by the
friction of the plane acting uip B E and the horizontal
reaction of the wall applied at the point F. This

78 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

reaction must be equal to the horizontal thrust of
the mass, or the wall will be forced over.

This horizontal thrust is obtained by resolving the
vertical force exerted by the mass D B E in sl horizontal
direction and one parallel to the plane.

131. A stone wall is built to retain a bank of ordinary

Scales

Lm&al.

1 z s

/ >r 1000

Force.
0

1000

Fig. 70.

earth 8 ft. high and estimated to weigh 120 lbs. per
cub. ft. The base of the wall is 2' 6'', the batter of the
outer face is 1 in 8, and the inner face vertical.

If the angle of repose be 45°, determine the position
of the resultant force.

Set out the wall to scale as shown in Fig. 70. Draw
the angle of repose and bisect the complement ABC

RETAINING WALLS 79

with the line B D. A B D now represents the mass
of earth whose horizontal thrust has to be determined.

Through the c.g.'s of the wall and oi A B D draw
vertical lines. From E, the point of application of
the force exerted hy A B D (which, as shown before,
is J of ^^) draw a horizontal line to intercept these
vertical lines at F and G.

The weight of 1 ft. length of the wall is 2240 lbs.,
and 1 ft. length of the section A B D weighs 1600
lbs. .

From G, on the line passing through the e.g. of
A B D, measure G H equal to the vertical force exerted
by AB D, i.e. = 1600 lbs. From H draw a line
parallel to B D, meeting the line G F Sit I. G I now
represents the magnitude and direction of the thrust
oi A B D, and its point of application is E.

This line meets the line of action of the force exerted
by the wall at F. From F scale oS F J equal to this
force, i.e. equal to 2240 lbs., and from J draw J K
equal and parallel to the thrust G I. Join F K.

F K now represents the magnitude and direction of
the resultant thrust.

This meets the base of the wall at L.

132. If this resultant crosses the base of the wall at
any point between B and 0, the wall is safe from over-
turning ; if it passes through 0, the wall is on the point
of overturning ; and if it passes outside the point 0,
the wall will be overthrown, unless the tensile strength
at the inner edge is sufficient to prevent it.

133. In considering the stability of retaining walls,
there is another point which it may be well to point
out, but the explanation of which is beyond this work.

The removal of the resultant force from the centre

80 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

causes the pressure on the outer edge to be much in-
creased, and the nearer it is to the edge the greater is
this pressure.

The pressure at this point must not be greater than
the material of which the wall is composed can safely
bear, or the wall will fail by crushing.

134. Polygon of Forces. — Let Fig. 71 represent
four forces of which the magnitude and direction of

two are known. It is
required to determine
the other forces.

Draw a h and 6 c to
represent the forces A B
and B G. Join c a.

c a is now the equi-
librant of the two
known forces (§ 105), therefore a c is the resultant
(§ 106), and we may consider this as being substituted
for ^ ^ and 5 C.

We have now three forces (the resultant of the first
two, and the two unknown ones) of which one, a c, is
known, therefore we can find the other two.

From c draw a line parallel to G D, and from a one
parallel to A D. Let them meet at d. Then c d and
da will give the magnitude and direction oi G D and
DA.

It should be noticed that this result could be arrived
at without finding the resultant, by drawing the lines
c d and d a from the ends of the lines representing the
two other forces. This latter method is the more direct,
and is usually adopted.

135. Fig. 72 represents five forces in equilibrium, of
which three are known.

POLYGON OF FORCES

It is necessary to find the magnitude of the other
two, and the direction in which they act.

Draw 6 c to represent
B G in direction and ^^'

magnitude, from c draw
c d representing C D in
direction and magni-
tude, taking care that
the sense of each force
is in the same direction round the figure, and from d draw
^ e to represent D E in direction and magnitude, again
noting that the force acts in the direction of d to e.

From b draw a fine parallel to ^ -B and from e a line
parallel to E A. Let them meet at a. ea and ab
now represent the magnitude oi E A and A B, and, as the
sense of the forces must form a circuit round the polygon,
eto a and a to b are their respective directions, i.e. E A
acts upwards, and A B upwards towards the right.

136. From the polygon of forces the resultant of any
number of forces can readily be obtained.

Let it be required to find the resultant of ^ ^ and
BG (Fig. 72). The first and last letters of the names
of these forces in clockwise order are A and G. On the
force diagram join a and c. Then a c fully represents
the resultant, that is, the line a c gives its magnitude,
and its direction is from a to c.

Had it been required to find the resultant of the
other three forces, the first and last letters would have
been G and A respectively. The resultant force would
in this case be represented by c a, that is, its magnitude
is the same but its direction is from c to a.

137. For the solution of forces acting at a point and
maintaining equilibrium, there must not be more than

82 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

two unknown ones, and of these two, if the direction
be known then the magnitude can be found, and, if
the magnitudes be known, the directions can be found.

138. If a number of forces keep a body in equilibrium,
the polygon representing the forces (i.e. the force
diagram) must close, and the senses be concurrent.

139. If one of the senses of the forces in a closed
force diagram be opposed to the others, the force repre-
sented by it is the resultant of the others.

140. If the force diagram does not close, then the
system which it ^:epresents is not in equilibrium, and the
closing line would represent the magnitude and direction
of the resultant, but its sense would be non-concurrent.

Examples to Chapter IV
1. What does A G (Fig. 1) represent ?

What would it be called if
its direction were reversed ?

2. What is meant by the
" resultant of two forces ? "

3. If two forces, equal to
5 lbs. and 8 lbs. respectively,

act towards a point at an angle with each other of 120°,
what force is required to produce equihbrium ?

4. Fig. 2 represents a piece of
cord attached to opposite sides of a
room, and supporting a weight.

Find the tension in each sec-
tion of the cord.

5. A ladder weighing 150 lbs.
Ex. Ch. IV.— Fig. 2. ^^g^g against a smooth vertical
wall at an angle of 60° with the horizontal plane.

Find the direction and magnitude of the reaction of
the ground.

EXAMPLES TO CHAPTER IV

6. A rafter, inclined at 30° with the horizontal plane,
exerts a force equal to 200 lbs.

Find the vertical and horizontal reactions of the
wall supporting it.

7. A door is T x 3' 6'', and
weighs 250 lbs. The hinges are
T and 12'' from the top and r|
bottom respectively.

Find the horizontal reaction of
the bottom hinge, and the total
reaction of the top one.

8. Fig. 3 illustrates a bracket
supporting a weight of 60 lbs.

{a) Find the amount and kind
of stress in the horizontal and
inclined members.

(h) Also find the reactions A and B.

215

•* 2-0' -»
Ex. Ch. IV.

-Fig. 4.

Ex. Ch. IV.— Fig. 5.

9. Fig. 4 shows a retaining wall supporting a bank
of earth. The earth weighs 120 lbs. per cub. ft., and
its angle of repose is 45°.

If the wall weighs 140 lbs. per cub. ft., where does
the resultant pressure intercept the base of the wall ?

10. Fig. 5 represents five forces in equilibrium.
Find the magnitudes and directions of ^4 ^ and B C,
What is the resultant of the three given forces ?

Chapter V
THE FUNICULAR POLYGON

141. The Funicular Polygon. — If a system of
forces in equilibrium be applied to a body already at
rest, then that body will still remain at rest (§ 32).

Let five forces in equilibrium be applied to a jointed
frame as shown in Fig. 73.

This frame is supposed to be such that each bar (or

Fig. 73.

link as it is called) will stand either tension or compres-
sion, and each joint (or node) is supposed to be hinged
so that the bars will accommodate themselves to the
best position to withstand the forces applied to them.

Such a frame is called a funicular polygon, and must
always close.

Let the force ^ jB be known.

As the whole frame is in equilibrium, each node is
in equilibrium, and the node on which A B in acting

THE FUNICULAE POLYGON

IS maintained in equilibrium by the action of that
force and the stresses set up in B 0 and 0 A . Draw
a b equal and parallel to the force A B, and draw lines
from a and b parallel to B 0 and 0 A . Then b o and o a
will represent the direction and magnitude of the
stresses set up in the links B 0 and 0 A . But at the
other end each link will exert an equal and opposite
force (§ 92). Taking the node where B C acts, we have
three forces, but 0 B has just been found and is repre-
sented in direction and magnitude by o 6. By drawing
parallel to B G and C 0, b c and c o are obtained, and
these represent the force B C and the stress set up in
G 0. Proceeding to the next node, by means of o c, cd
or the force G D and the stress d o in the next bar are
ascertained. By repeating this operation the whole of
the forces and the stresses in the bars are obtained.

When completed it will be seen that the lines repre-
senting the forces form a closed polygon, proving that
the forces represented by these lines are in equilibrium.

142. Further, the lines representing the stresses in
the links all meet at the same point. This point is
called the pole, and the lines radiating from it polar
lines or vectors.

143. If all the forces applied to a funicular polygon,
and the direction of two of the links be known, then the
funicular polygon can be completed, because from the
forces the force diagram can be formed and the inter-
section of the two lines parallel to these two links will
give the pole. The directions of the remaining links
are obtained by drawing the other vector lines.

144. If a system of forces be in equilibrium any
funicular polygon can be found to which they can be
applied.

ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

For, let ah cd e (Fig. 74 (a)) be a reproduction of the
force diagram Fig. 73.

Fig. 74.

Take any pole o and draw the vectors.

Draw AB, BO and 0 A (Fig. 74 (b)) parallel to a b,
b o, and o a. Mark off 5 0 any length, and at its extre-
mity draw B C and C 0 parallel to 6 c and c o. Cut off
C 0 any length, and draw C D and D 0 parallel to c d
and d o. Set off Z) 0 any length, and draw D E and E 0
parallel to d e and e o. Produce ^ 0 and ^ 0 to meet,
and from this point draw a line E A parallel to e a.

Fig. 74 {b) now represents the same five forces as
those in Fig. 73, but they are applied to another funicular
polygon.

Hence, if a system of forces be in equilibrium and a
force diagram drawn, any pole can be taken, and a
funicular polygon found in respect of that pole.

145. Fig. 75 shows a funicular polygon and the forces
applied.

It is necessary to find how equilibrium may be main-
tained in each part if a section be taken at x y.

It is evident that the forces on one side of the section

THE FUNICULAR POLYGON

are kept in position by those on the other, hence the
resultant of the forces on the one side will maintain
equilibrium with the forces on the other.

Draw the force diagram ab cd e.

The forces on the left are E A and A B. Join e 6,
then e & is the magnitude of the resultant, and e to & its
direction. (§ 136.)

It is now necessary to find where this resultant acts.
By substituting the resultant e h for the two forces E A
and A B the force diagram eb cd e is obtained. But o

is a pole from which the vectors ob, o c, o d and o e are
already drawn, hence a funicular polygon may be
obtained whose sides are parallel to these vectors. On
examination it will be seen that 0 B, 0 C, 0 D and 0 E
are already parallel to ob, oc, o d and o e respectively,
but a funicular polygon must close, hence 0 B and 0 E
must be produced to meet.

There must be a force acting at each node of a funi-
cular polygon, and each of the forces B C, C D and D E
are already acting at a node, hence the remaining force,

88 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

which is represented by e h, must act at the node formed
by the production of these Hnes.

Again, the resultant of the forces B C, C D and D E
is h e, that is, the resultant of the forces on the right of
the section is of the same magnitude as the resultant of
the forces on the left, but it acts in the opposite direction,
and by obtaining a new funicular polygon as before, it
will be found to act through the same point.

146. Hence, if a section be taken across a funicular
polygon, the resultant of the forces on either side will
act through the node formed by the production of the
intersected sides, and the magnitude and direction of
each resultant are found by the force diagram ; also, the
resultant of all the forces on one side of the section is
equal to the resultant of all the forces on the other side,
but they act in opposite directions.

147. Similarly the resultant of any of the forces and
a point in its line of action may be obtained. Deter-
mine the resultant of CD, DE and E A (Fig. 75).
C and A are the first and last letters in the clockwise
notation, so c a on the force diagram gives the resultant,
and it acts through the point where the links C 0 and
A 0 would meet if produced.

This is practically the same thing as taking a section
cutting the links G 0 and A 0.

148. Therefore it should be noticed, that the first and
last letters of the forces, when named in clockwise order,
not only give the magnitude and direction of the resul-
tant force on the force diagram, but also name the links
on the funicular polygon whose intersection, when
produced, gives a point in the line of action of tliat
resultant.

149. Before applying parallel forces to a funicular

THE FUNICULAR POLYGON

1b1

polygon, a little explanation of the force diagram may
not be out of place.

If the system be in equilibrium the force diagram
must close. (§ 138.)

Suppose a beam loaded
and supported as shown
in Pig. 76.

Draw a b, b c and c d
to represent the known
forces A B, B C and
CD.

From d draw a line
parallel to D E, and from a one parallel to A E. de
and e a give the magnitude and direction of the force
exerted by the two supports.

Suppose D E and E A to be vertical, as shown in
Fig. 77, then it is evident that the lines d e and e a will
be in a straight line and

Fig. 76.

Ul c I

Fig. 77.

lie upon d a, i.e. d a will
be the closing line of the
force diagram, but the
point of intersection is not
known, hence the reactions
of the two supports are
equal to the total load
a d, but what proportion each bears is not determined.

The point e will, however, lie somewhere between
a and d, and abcda (Fig. 77) will form a closed poly-
gon in quite the same sense asab cd ea (Fig. 76).

Hence, the force diagram of a system of parallel forces
is a straight line.

150. We will now proceed to utilize the funicular
polygon to determine parallel forces.

ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

The most common case is that of a simple beam loaded
at various points.

Two scales are necessary — a lineal scale to set out the
beam and the positions of the loads, and a force scale
for all the measurements on the force diagram. Fig. 78
shows a beam with the position and amount of each load.

Taking a convenient force scale ah =3 cwts., 6 c=
2 cwts., and c d=5 cwts. ; a d is now the sum of the

5 cwts. 2 cwts. Scwts

R-eb

Fig. 78

TO 0

Lineal rTTTilMMI

1 0 Z 1^ 6 S 10 12 Ik

Force, mill I I i I

I II I I I II niLcwfe.

loads, and the sum of the reactions is equal to this, there-
fore dais the closing line of the force diagram.

It is necessary to determine the position of e to ascer-
tain what proportion of the load is borne by each
support.

Take any pole o and draw vectors to a, 6, c and d.

From any point on the support EA draw a line
1-2 parallel to a o until it cuts a perpendicular from the
first load. From the point 2 the line 2-3 is drawn parallel
to 6 o until it intersects the perpendicular from the

THE FUNICULAR POLYGON 91

second load. From 3 the line 3-4 is drawn parallel to
c 0, till it meets the line of action of the third force, and
from this point the line 4-5 is drawn parallel to cZ o as far
as the support. Join 1 and 5.

123451 is now a funicular polygon, and to enable the
student to compare it with the others, the forces are
shown dotted at the nodes.

An examination of the funicular polygon will show
that it has five sides, whereas there are only four vectors.
A vector must now be drawn parallel to the remaining
side of the polygon, and this determines the position of e.

d e now represents the magnitude and direction of
the reaction D E, and e a that oi E A.

151. It will be necessary to know the names of the
links of the funicular polygon. The one parallel to a 0
is A 0, the one parallel to b o is B 0, and so on. It is
not necessary to put the names on the polygon, because a
glance at the force diagram will at once supply them.

152. The name of each link may also be ascertained
by referring to the beam.

As each link is terminated by the lines of action of
some two forces, it has, as its distinguishing letter, the
one which names the space on the beam between those
two forces. Thus the space between the 3 and 2 cwts.
is B. The link terminated by the perpendiculars from
these two forces is known as B 0.

Again, the space E extends from the one support to
the other, and the link which crosses this space is E 0.

153. Find the resultant oi A B and B C (Fig. 78) and
a point in its line of action.

The first and last letters in the clockwise sequence are
A and C, therefore a c on the force diagram gives the
resultant force, which is 5 cwts., and the intersection of

92 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

the links A 0 and C 0 gives the point x through which
it acts.

A perpendicular from x gives the point on the beam
where it acts.

154. Find a point on the beam where the three forces
(Fig. 78) could be accumulated without interfering with
the reactions.

This is practically asking for the resultant of the three
forces and its point of application.

The three forces are A B, B C and C D, therefore a d
gives the magnitude and direction of the resultant force,
and y the intersection of the links A 0 and D 0 gives a
point in its line of action. A perpendicular to the beam
from y will give the point where the 10 cwts. would be
placed.

155. Ascertain the force which could be substituted
for A B and the reaction on the left, and the point where
it should be applied.

The sequence is E A and A B, hence e b on the force
diagram gives the direction and magnitude of the force,
and z, where the links E 0 and B 0 meet, is a point
through which it acts.

Of course, the student should note that, having drawn
the lines of action of these resultant forces through the
beam, the distance of these points from either end can
be obtained by applying the lineal scale.

156. It was pointed out in Chapter II that questions
on the three orders of levers could be solved by means of
similar triangles.

The knowledge of the funicular polygon supplies an
easier and more interesting method of solving them, as
the following examples will show :

A lever of the first order supports a weight of 60 lbs.,

LEVERS

r 6'' from the fulcrum. Find the power necessary to
balance this if the power arm be 2 ft.

Set out the lever with the position of the weight, etc.,
to a convenient lineal scale (Fig. 79) and adopt a force
scale.

Draw a 6=60 lbs., and take any pole o. Join a o and
b o. Draw the link A 0 across the space A, and the
link B 0 across the space B, and parallel to a o and b o
respectively. Close the polygon, and draw the vector
c 0 parallel to the closing line. Let o c terminate in a 6

i B I

-•^

Fig. 79.

produced. Then 6 c is the required power, which is
45 lbs.

157. A lever is 6 ft. long. Where is the fulcrum if a
force of 15 lbs. supports a weight of 50 lbs. ?

Draw the lever =6' (Fig. 80) and with a convenient
force scale draw a 6=50 lbs. and b c—l5 lbs. Join a, b,
and c to any pole o. Draw the link B 0 across the space
B and parallel to 6 o. From the ends of this draw the
links A 0 and G 0 parallel to a o and c o and across the
spaces A and C respectively.

A perpendicular from the point of intersection of
these links will give the position of the fulcrum.

94 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

; b

Fig. 80.

158. Fig. 81 shows a lever wdth the relative positions
of the fulcrum, weight, and power.

What weight will a power equal to 24 lbs. sustain ?

Draw c a=:24 lbs. Take a pole and draw the vectors
c o and a 0. Across the space A draw the link A 0
parallel to a 0, and across the space C the link C 0
parallel to c o.

Close the polygon, and draw o h parallel to the link
B 0 thus formed.

a h represents the weight drawn to scale.

Fig. 81.

159. A lever 6 ft. long, weighs 20 lbs. If a weight of
80 lbs. be placed 2 ft. from the fulcrum, which is at the
end, find the power necessary at the other end to support
it. The weight of the lever acts at its centre of gravity.

FUNICULAR POLYGON

Fig. 82 shows the lever and the position of the forces.

Draw the forces a h and h c and the three vectors.

Draw the Hnks as before, and close the polygon. The
closing line indicates how the vector o c? is to be drawn.

c dis the power drawn to scale.

1 60. Reactions of the supports of framed struc-
tures.— It should be noted in the case of a simple beam
that the proportion of a load borne by each support is in
the inverse ratio to the perpendicular distance of its
line of action from the support.

SO Ik. 20 lbs

Fig. 82.

The same rule holds good for framed structures of
every description, hence the reactions on their supports
can be found in a similar manner.

To prove this we will take one simple example and
compare the results arrived at graphically and arith-
metically.

Fig. 83 illustrates a roof truss with a load of 6 cwts. at
the ridge.

The reactions B C and C A are given by 6 c and c a on
the force diagram, which are found to represent 3 J cwts.
and 2\ cwts. respectively.

161. The span shown in the figure is 12 ft. and the

96 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

line of action of the 6 cwts. is 7 ft. from x and 5 ft. from y.

Taking the moments about x, we have the reaction

of 5Cxl2'=6 cwts. X 7', therefore the reaction of

6 cwts. X 7' , , . 1

B G= ^-7 =3J cwts. ; and taking the moments

about y, we have the reaction of C ^ x 12' =6 cwts. x 5',

6 cwts. x 5'

therefore the reaction of C ^

12'

=2J cwts.

Fig. 83.

It will thus be seen that the results obtained graphi-
cally correspond with those found arithmetically.

162. The advantage of the graphic method over the
arithmetical one for finding the reactions of the sup-
ports is quite apparent, when it is pointed out that the
former method is the same for all kinds of structures,
whereas the latter often involves difficult calculations.

In order to show the application of the graphic
method a few typical cases are given. As the method
of procedure is exactly the same as that for a simple
beam, only the points not previously noted will be
commented upon.

REACTION OF SUPPORTS

163. Suppose a truss as shown in Fig. 84 carrying an
evenly distributed load of 3 tons on the top beam.

Find the reactions at the supports.

As the load is evenly distributed along the entire
length of the beam, it may be considered as being
accumulated at its centre, or as being transmitted by
the beam to the joints, one-half being on each.

J \tons. li \tons

nitons

Both cases are worked out, and it should be noticed
that the result is the same in each case.

^m Fjg. 85 shows a Warren girder witli three loads on
^B the bottom boom.

ELEMENTARY PRINCIPLES OP GRAPHIC STATICS

Fig. 86 is a diagram of a short N girder with two loads
on the top boom.

164. Before proceeding with the roof truss it is neces-
sary to understand what the load is composed of, and
how it is transmitted to the truss.

The load consists of the weight of the truss itself, the
weight of the covering, snow, and wind pressure. The
weight of the covering depends on its nature. The
wind pressure is not vertical, but its vertical component
can be found.

n [tons) B ytons.

Fig. 86.

The student may assume that the total vertical load
on a roof is 56 lbs. per square foot of the external sloping
surfaces.

Suppose a space 30' x 24' to be roofed. Fig. 87.

This would necessitate two king-post trusses at 10'
centres. These with the ridge and purlins would divide
the roof into 12 equal spaces.

Taking the rise to be J of the span, the slope would
measure nearly 14 J ft. The area of each slope would
be 14i'x30'=435 sq.ft.

The total weight would then be (2 x 435) half cwts.,
or 435 cwts.

WEIGHT ON ROOF TRUSS

This is spread over 12 spaces, so the weight of each

435

cwts.=:36J cwts.

Taking the spaces A and D, the common rafters
transmit half of the loads to the wall-plates and half
to the purlins. Taking the spaces B and C, half the
weight on each is transmitted to the purlins, and the

■^///////y////////////rrA r////////////777:^.^/. -'^ Vy'.'.v///////^////// ^//.

Fig. 87.

her half to the ridge. Thus the ridge and each purlin
et 36J cwts. and each wall-plate 18J cwts.
But the purlins and ridge are beams with distributed
ads, one end of each being supported by the wall, and
the other by the truss, hence the truss supports half the
load on each. In the same manner it can be shown
that the same truss supports a like amount from the
adjoining spaces.

Therefore, in the above example, the purlins and ridge

100 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

transmit 36 J cwts. each to the truss and 18 J cwts.
comes directly on each wall, and the ridge, purlins, and
walls are the points of application of these loads.

165. If the roof be symmetrically planned, the magni-
tude of the loads at the different points of the truss can
be obtained as follows : —

Divide the total weight of the roof by the number of
spaces into which the trusses divide it, and the weight
thus obtained is again divided, so that each purlin and
ridge gets twice as much as each wall.

166. If a roof truss be symmetrical and symmetri-
cally loaded, the reactions of the supports will be equal,
each equal to half the sum of the loads.

Fig. 88.

167. As an application of the funicular polygon to a
roof truss Fig. 88 is given. It shows a total load of 50
cwts. on one side, and a load of 40 cwts. on the other.

It should be carefully noted how these loads are
applied to the truss.

KE ACTIONS OF SUPPORTS 101

Letter the spaces between the external forces, and
draw the line of loads ah cd ef.

The loads A B and E F come directly on the wall, and
are entirely independent of the truss.

What we have to find out is what proportion of the
loads EC, C D and D E each wall bears, and for this
purpose our line of loads \^h cd e.

Take a pole and join 6, c, d, and e to it. Make a
funicular polygon with the links parallel to these vectors,
and draw o g parallel to the closing link.

e g now represents the proportion of the three loads
borne by the wall F G, but in addition to this, it sup-
ports the load E F which is represented by e /, therefore
the total reaction of the wall F G is shown by the
line / g.

Similarly g a represents the total reaction of the
wall G A.

1 68. Since the resultant of all the forces exerted by
a body passes through its e.g., and since the funicular
polygon proves the most convenient method of obtain-
ing the resultant of a number of parallel forces, it can
be applied to find the e.g. of a body which has to be
divided into a number of segments.

Suppose it is required to find the e.g. of the section
shown in Fig. 89.

Divide the figure into three parts as shown, and find
the e.g. of each portion. The weight of each part may

I now be considered as accumulated at its e.g., and acting
in a vertical direction.
Through the e.g. of each draw vertical lines. D E
now gives the line of action of a force which is equal to
the weight of the bottom portion ; E F gives the line of

102 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

S c:

Fig. 89

EXAMPLES TO CHAPTER V 103

middle portion ; and F 0 the line of action of a force
equal to the weight of the top portion.

Draw a force diagram making d e equal to the weight
of the bottom portion, e f equal to the weight of the
middle portion, and / g equal to that of the top portion.
Take any pole and draw the funicular polygon.

The forces are represented by D E, E F, and F O,
therefore the resultant force is d g, and the intersection
of the links D 0 and G 0 will give a point in its line of
action.

Let them meet at y. Through this point draw the
perpendicular x y, then the e.g. is in this line, and the
resultant force of the whole mass acts along it.

All that is required for present use is the magnitude
of the resultant force and its line of action. The above
method will give it whatever be the number of segments
into which the figure is divided.

Should it be required to ascertain where in the line
X y the e.g. is situated, the whole figure may be con-
sidered as lying on the side B G.

In this case K L, L M, and M N would give the
directions and positions of the forces exerted by the
bottom, middle, and top portions respectively.

By drawing a new force diagram, and proceeding as
before, x'y' is obtained, and the e.g. of the whole figure
is at the point where this intersects x y.

Examples to Chaptepv V

1. A beam rests on two supports, A and B, 10 ft.
apart. If a load of 15 tons be placed 3 ft. from B,
what are the reactions of the supports ?

2. A girder weighing 1 ton, and 15 ft. long, carries a

104 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

load of 1 J tons 4 ft. from one end, and another of 2 tons
5 ft. from the other end.

Find the total load on each support.

3. A beam weighing 75 lbs., and 9 ft. long, is sup-
ported on two props.

If a weight of 25 lbs. be placed 3' 6'' from one end,what
are the thrusts of the props ?

4. A beam rests on two walls 12 ft. apart. If it
weighs 90 lbs., where must a weight of 60 lbs. be placed
so that the one wall will carry twice as much as the
other ?

5. Fig. 1 shows a beam supporting three weights.

cwts

mil

Clfftt

\^'

-*«-

-»«-

J' H

Ex.

Ch.

v.—

Fig.

At what point could they be accumulated so as not
to interfere with the reactions ?

6. A rod 6 ft. long, and weighing 3 lbs., acts as a lever,
the fulcrum being 2 ft. from one end.

If a 6 lb. weight be placed at the end of the shorter
section, what weight must be placed at the other end of
the bar to balance it ?

7. A king-post truss carries a distributed load of
5 tons.

Draw a line diagram of the truss, and indicate the
amount of load at each point of support. Span =25 ft, ;
pitch=30°.

8. Fig. 2 shows a girder loaded at two points.
Find the reactions of A and B,

EXAMPLES TO CHAPTER V

105

\/ho*. V V \/

2 \tGns

3 tons.

Ex. Ch. v.— Fig. 2.

9. Find the reactions due to the three loads shown in
Fig 3.

tons.

tons

Ex. Ch. v.— Fig. 3.

10. Find the total reaction of each support due to the
three loads shown in Fig. 4.

/J I mt%

Jims

Ex. Ch. v.— Fig. 4.

Chapter VI

BENDING MOMENTS, AND SHEARING FORCE

169. In Chapter II it was explained what Bending
Moment (B.M.) means, and how to find it arithmetically.

We will now proceed to find it graphically. We
will take the case of a simple beam loaded at different
points, as shown in Fig. 90.

A oci ^ iC^gi D

Draw the funicular polygon, and find the reactions
as shown in the last chapter.

First, let it be required to find the moment about
X. Draw the perpendicular x y, cutting the funicular
polygon in m and n.

Now the resultant of all the forces on one side of

BENDING MOMENT 107

X is equal to the resultant of all the forces on the other
side (§ 145).

There is only one force on the left of x, 80 E A is
the resultant, and e a on the force diagram represents
this.

Taking the triangles mnp and eao, since mn is
parallel to ea, pm parallel to e o, and p n parallel to
a o, these two triangles are similar in every respect.

Draw the perpendiculars p r and o s from p and o to
m n and e a respectively.

Then mn : pr : : ea : o s

and ea x pr = mn x o s.

But e a is the force E A and pr is its perpendicular
distance from the point x.

Therefore ea x pr = the moment of all the forces

about X.
But ea X pr =z mn x o s,
therefore mn x o s = the moment of all the forces

about X.

mn is the perpendicular distance across the funicular
polygon directly beneath x, and o 5 is the perpendicu-
lar distance of the pole from the line of loads.

Next, let it be required to find the moment about g.

Draw the perpendicular g h, meeting the funicular
polygon in i and j.

The forces on the right of this section are CD and
D E, therefore the resultant is represented by c e, and
it acts where the links C 0 and E 0 meet, that is at k
(§ 145).

Because i k is parallel to eo, jk parallel to c o, and
j i parallel to c e, the two triangles j i k and ceo are
similar.

Draw k I perpendicular to g h.

108 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

Then j i : kl : : c e : o s,
and c e x kl = j i x o s.

Since c e is the resultant of the forces on one side
of g, and k I the perpendicular distance of its point of
application from the section,

therefore c e x A; Z = the moment of the resultant of
C D and D E about the point g,
or, c e X kl =: the moment oi C D and D E
about g (§ 44).
But c e ^= ec, and e c is the resultant of
E A, AB, and BG,
therefore c e x kl = the moment of E A, A B, and
B C about g.
Again, c e x kl = j i x o s,
therefore ji x o s = the moment of CD and Z> ^
about g^
and ^* i x o s = the moment of E A, A B and
J5 O about gr.
But j i is the perpendicular distance across the
funicular polygon directly beneath the point g, and
o 5 is the perpendicular of the pole from the line of
loads.

o 5 is called the " polar distance."
It is now evident that the B.M. at any part of the
beam is given by multiplying the perpendicular across
the funicular polygon beneath that point by the polar
distance.

But for any one funicular polygon the polar dis tance
is constant, therefore the B.M. varies directly as the
perpendiculars (or ordinates) across the polygon.

For this reason the funicular polygon is called the
Bending Moment diagram.

170. The ordinates of the B.M. diagram must be

BENDING MOMENT 109

measured with the lineal scale, but the polar distance,
being on the force diagram, must be measured with the
force scale.

171. Up to the present the pole has been taken at
any point, but it will now be seen that, if the bending
moment is required, it is advisable to place it so that
its perpendicular distance from the line of loads will
represent a definite number of lbs., cwts., or tons.

172. It will also be noticed that it would be much
more convenient if a scale could be found with which
the bending moment could be measured directly off the
ordinates, instead of measuring the ordinates with the
lineal scale and multiplying this by the polar distance.

Suppose in Fig. 89 that o s represents 4 cwts., and
that the lineal scale is J'' = 1 ft., then if an ordinate
measures V it represents 4 ft., but this must be multi-
plied by 4 cwts., so an ordinate of V represents a bend-
ing moment of 4 ft. x 4 cwts, or 16 ft. -cwts.

This gives a new scale of V = 16 ft.-cwts., by which
the moment can be measured directly off the Bending
Moment diagram.

173. This new scale is called the " Bending Moment
scale," and is obtained by multiplying the lineal scale
by the polar distance expressed in lbs., cwts., or tons.

In (§ 172) a bending moment scale of T' = 16 ft.-
cwts. was obtained. This is not a convenient scale
with which to read off the bending moment by applying
the rule to the diagram.

To obtain a B.M. scale such that the bending moment
can be read off directly, the polar distance must be
taken as 1, 5, 10, 50, or 100, etc. (lbs., cwts., or tons).

Let the lineal scale in Fig. 90 be J'' = 1 ft., and o s
(the polar distance) = 5 cwts.

110 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

Then the B.M. scale is Hneal scale x polar dist ince,
that is J'' = 1 ft. X 5 cwts, — 5 ft.-cwts., or B.M.
scale is Y = 1^ ft.-cwts.

Again, suppose the lineal scale to be j'' =: 10', and
the polar distance 10 lbs.

Then the B.M. scale is f = 10 ft. x 10 lbs.,
or B.M. scale is |^' = 100 ft.-lbs. .

By judiciously selecting the polar distance, as shown
above, a decimally divided scale is obtained, with
which the readings can be taken directly off the diagram,
as explained in Chapter I.

174. Example. — A beam is 20 ft. long and loaded with
6 cwts. 4 ft. from one end, and 8 cwts. 5 ft. from the
other.

Find the greatest B.M. and the B.M. at the centre.
Adopt two scales (say J"' = 1 ft. and ^ = 10 cwts.).
Set out the beam with the positions of the loads and
draw the line of loads (Fig. 91).

Place the pole any convenient distance from this
(say 15 cwts.) ; draw the vectors and B.M. diagram.
Then the Bending Moment scale
= Lineal scale x polar distance
= Y = l ft. X 15 cwts. =: 15 ft.-cwts.

=3V" = 10 ft.-cwts.

With the Bending Moment scale measure the ordinates
at the widest part of the B.M. diagram and at the centre.

These represent a bending moment of 36 ft.-cwts. and
32 ft.-cwts. respectively.

The greatest B.M. is at the point where the 8 cwts. is
placed, and it should be noticed that it is always under-
neath one of the loads.

175. Problem. — A beam 16 ft. long is supported at
both ends. A load of 12 cwts. is placed 3 ft. from one

BENDING MOMENT

111

bcufts.

Scwts.

\ ' 1 ' 1

y»

, ~\

--^

■^

^^ 1

Lvneal. mL

s b

I I I I

Scaks: Force.
BM

IIIIIIIH

0 10 30 so 70

mrrTT

FiG. 91.

10 11 n rt

riQ it.

{Uuft%.

end, a load of 10 cwts. is placed 4 ft. from the other,
and a load of 15 cwts. is placed at the centre.
Find : 1. The reactions of the supports,

2. the greatest bending moment,

3. the moment of the 12 cwts. about the centre,

4. the moment of the three loads about the centre.
Taking two scales, such as Y = 1 ft. and |'' = 10 cwts.,

set out the beam (Fig. 92), and draw the force and B.M.
diagrams.

12 cwts 15 cults lOcu/ts
A i B i C i D

Scales: Force. Tmiiiiiii

Fio. 92.

00 p.

112 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

The B.M. scale = Lineal scale x polar distance.

= f ' =1 ft. X 30 cwt. = 30 ft.-cwt.
= -^" = 10 ft.-cwt.

1. The reactions oi D E and E A are represented by
d e and e a on the force diagram and are 17-25 cwts. and
19-75 cwts. respectively.

2. The greatest B.M. is at the centre, and is repre-
sented by I f, which scales 98 ft.-cwts.

3. Note that the first and last letters of the 12 cwt.
load are A and B. The moment of ^ i5 at the centre
will, therefore, be represented by the length of the
ordinate from the centre intercepted between the links
^ 0 and 5 0.

Referring to the diagram, it will be seen that the link
A 0 does not extend as far as the perpendicular. It
must be produced until it does, thus cutting off the
ordinate / g, which scales 60 ft.-cwts.

4. The three loads constitute the three forces A B,
B C, and C D, the first and last letters of which are
A and D. The moment of these about the centre will
be represented by the portion of the perpendicular from
the centre intercepted between the links A 0 and D 0.

The link D 0 must be produced until it meets the
perpendicular at h.

h g gives the required moment, which is 20 ft.-cwts.

176. The Bending Moment diagrams can be applied
to beams supported at one end (cantilevers) as well as
to beams supported at both ends.

Fig. 93 shows a cantilever loaded at the outer end.
It is required to draw the B.M. diagram.

Draw the load line a b, and select a pole.

(In the case of cantilevers it is more convenient to
place the pole so that the closing line of the B.M.

BENDING MOMENT

113

diagram will be horizontal. To obtain this, the pole
is placed opposite the top or bottom of the line of loads.)

Draw the vectors a o and b o. From any point on the
support draw the link A 0 parallel to ao until it inter-
cepts the line of action of the load. From this point
draw the link B 0 parallel to bo. The triangle thus
formed is the B.M. diagram, and from it the moment
at any part of the beam can be obtained by dropping
perpendiculars as previously shown.

In the case of a cantilever with the single load, the
B.M. can be much more conveniently found by multi-
plying the load by its distance from the point selected.

Fig. 93.

Thus, if the load be 5 cwts., and the cantilever 8 ft.,
the B.M. at the wall end is 8 ft. x 5 cwts., or 40 ft. -cwts.,
and 2 ft. from the wall it is 6 ft. x 5 cwts., or 30 ft.-cwts.
177. Problem. — A cantilever 8 ft. long supports a
load of 3 cwts. at its centre., 2 cwts. at its outer end,
and 1 cwt. midway between these.
It is required to find —

(a) the greatest bending moment,

(6) the resultant of the 3 cwts. and 1 cwt., and

its point of application,
(c) the resultant of the three loads and its point of
application,

114 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

and (d) what load can be placed at the end of the
beam to produce the same strain at the wall
end as that caused by the three loads.
Having decided upon the scales, set out the beam as

shown in Fig. 94. Draw a line of loads and select a

A Lb J. c i

Lineal

Fore

Scales: Force, ii if t ^

10- 0. 1

isft.mts.

I I I

FfG. 94.

pole. Join a, b, c, and d to o, and draw the links A 0,
B 0, C 0 and D 0 parallel to a o, b o, c o and d o respec-
tively. The figure thus obtained is the Bending
Moment diagram.

(a) A glance will show that the greatest B.M. is at the

wall end, and this measured by the B.M. scale
gives 34 ft.-cwts.

(b) The two loads are A B and B G, therefore a c,

which is equal to 4 cwts., is the resultant, and
this acts where the links A 0 and G 0 meet. A
perpendicular from this point to the beam gives
the point x, 4J ft. from the wall.

the resultant is represented by a d, which equals

BENDING MOMENT

115

6 cwts., and the intersection of the links A O
and D 0 gives a point in its Hne of action. A
Hne through this point parallel to a d gives the
point y on the beam, 5f ft. from the wall.
(d) With the load to be substituted, the B.M., I m, is
to remain the same. Since the same pole can
be used, the link D 0 remains. Join I n. Then
Imn is the new B.M. diagram.
From o draw a vector parallel tol n. Let this meet

the load line at e.
e d is the new load, which is equal to 4 J cwts.
178. Bending Moment with Distributed Loads. —
We have now to consider the strain caused by an evenly
distributed load.

We will first draw a B.M. diagram as if the whole load
X y were concentrated at the centre of the beam (Fig. 95).
The triangle p qr is this diagram.

Fig. 95.

Now divide the load x y into a number of equal parts,
and place them at equal distances apart on the beam.

Draw the vectors and complete the B.M. diagram.

It will be seen that the B.M. is much less at the centre
when the load is so split up, and that the links appear
to form the chords of a curved line. If the load had

116 ELEMENTARY PRINCIPLES OP GRAPHIC STATICS

been divided into a greater number of equal parts
equidistantly placed, this would have been more ap-
parent still.

But the limit to the division of a load is to evenly
distribute it along the entire length of a beam, and in
that case the B.M. at the centre is one-half what it
would be if the load were concentrated at the centre,
and the links would form one continuous curve of a
parabolic form with the vertex at the centre.

179. In order, then, to draw the B.M. diagram for
a beam with a distributed load, we must know how to
draw a parabola.

To show this we will take an example. Fig. 96 shows

Fig. 96.

a beam with an evenly distributed load which is equal to
a 6 on the force diagram.

Draw the B.M. diagram k el as if the whole load
were at the centre.

Bisect d e Sbt f, and through / draw g f h parallel to
k I, and complete the parallelogram k g hi.

Divide k g into any number of equal parts, and join
each point to /. Divide k d into tlie same number of

BENDING MOMENT

117

equal parts, and from each point thus obtained, drop
perpendiculars.

By numbering the points in both directions from k,
as shown, the perpendicular from 1 should meet the
line 1 /, the perpendicular from 2 should meet 2 /, and
so on. A curved line through these intersections forms
the figure kdf, which is half the parabola, and is half
the required B.M. diagram.

The other half can be drawn in the same way, but,
since the diagram will be symmetrical, this is unneces-
sary. If the B.M. be wanted at any point on the second
half of the beam, a point can be taken similarly placed on
the first half and the B.M. at that point ascertained.

Of course, the bending moment at any point is ascer-
tained from the diagram as previously shown, i.e. by
finding the B.M. scale and measuring the ordinate
beneath that point.

i8o. A cantilever with a distributed load gives, like
the beam, a maximum B.M. equal to one-half what it
would be if the load were concentrated at the farthest
point from the support.

Fig. 97 is given as an illustration.

118 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

Set out the cantilever to scale, and draw a h equal
to the total load. Select a pole, draw the vectors, and
the bending moment diagram I mn.

I m gives the B.M. at the wall if the total load were
concentrated at the end. Bisect this in /, and complete
the parallelogram m f g n.

Draw the semi-parabola m f n 3bS explained in § 179,
then m f n is the B.M. diagram.

It should be noticed that in whatever form a can-
tilever is loaded, the B.M. at the unsupported end is
nil, and that it increases as the wall is approached,
reaching its maximum at that end.

i8i. Cantilevers and beams supported at both ends
may have concentrated and distributed loads at the
same time. If the weight of the beams themselves be
considered, then there is always a distributed load.

A B.M. diagram of a cantilever under the two systems
of loading is shown in Fig. 98.

The figure I m n p
concentrated loads.

is the B.M. diagram for the two
and the figure s pn the B.M.

BENDING MOMENT

119

diagram for the uniformly distributed load which
is represented by c c? on the line of loads.

In order to find the B.M. at any point, verticals
must be drawn across the figure slmn and measured
as before on the bending moment scale.

182. Fig. 99 shows the B.M. diagram for two con-
centrated loads and a uniformly distributed load on a
beam supported at both ends.

The B.M. diagram for the concentrated loads is drawn
as already explained. This diagram being below the
closing link, it will be necessary to place the B.M.
diagram for the distributed load above it. On the line
of loads set off cZ e and d /, each equal to one-half of the
distributed load, e / will then be equal to the whole
distributed load.

From the ends of the closing link draw links parallel
to the vectors e o and / o. The perpendicular g h would
give the B.M. at the centre if the whole of the distri-
buted load were concentrated there. Bisect gh in i,

120 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

and with the closing link as the chord make a parabola
pa,ssing through the point i. This parabola, together
with the B.M. diagram for the concentrated loads, gives
the required Bending Moment diagram.

183. Shearing Force. — It has been shown that a
load placed on a beam tends to produce rotation, which
tendency is called " the bending moment." A load
placed on a beam, besides tending to produce rotation,
also tends to cause one portion to slide vertically past
another, as shown in Fig. 100.

This second effect of the
load is like that of the jaws
of a shearing machine, so
the tendency of the weight
to produce vertical move-
ment at any section is
called the Shearing Force
(S.F.) at that section.
To prevent this movement, the end B C oi AG must
exert forces or stresses on the end D E oi D F sufficient
to keep it in position. The amount of the stress is

Fig. 100.

clearly equal to the vertical force W.

Taking

Fig. 101, the
Shearing Force at c c is
equal to W, but, proceed-
ing to the section 6 6,
this force is diminished
by the upward force W\
so the Shearing Force at
& 6 is W —W. Proceeding
again to the section a a
the Shearing Force is augmented by the downward
force W", therefore the S.F. at a a is W -W + W\

Fig. 101.

SHEARING FORCE 121

Since, in Fig. 101, there are four parallel forces in
equilibrium, the " reaction of the wall " is equal to
W - W + W\ Commencing on the left, the S.F. at
a a is equal to the reaction of the wall, i.e., W —
W + W\

To the left of & & the forces are the reaction of the wall
and W" acting in the opposite direction, therefore the
S.F. at 6 6 is (If - IT' + W") - W" = W - W\

To the left of c c the forces are the reaction of the
wall, W", acting downwards, and W\ acting upwards,
therefore the Shearing Force Q>icc \^ (W — W + W")
- W + W = W.

These results correspond with those obtained for the
various sections when considering the forces on the
right of those sections.

Hence the S.F. at any section is obtained by finding
the algebraical -sum of all the forces on either side of
the section.

184. By drawing ordinates from each point on the
force diagram across the space represented on the beam
by the same letter as that which distinguishes
that point on the force diagram, a Shearing Force
diagram can be obtained which will graphically repre-
sent the S.F, at every point of the beam.

An examination of Figs. 102, 103 and 104 will make
this clear.

185. Ordinates across the Shearing Force diagram
perpendicularly under any point of the beam, and
measured on the force scale, will give the Shearing Force
at that point of the beam.

For the S.F. at any part of the space A (Fig. 104) is
given hy ah — h c + cd or da; the S.F. at any part
of the space B is given hy d a — ah orhd; and the S.F.

122 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

A iB ia A IBl

c
b

AiSTCi

Fig. 102.

Fig. 103.

Fig. 104.

at any part of the space C is given hy da — ah + he
or cd.

1 86. If a cantilever carries a uniformly distributed
load, the S.F. at the unsupported end is nil, but it
gradually increases as the wall is approached until at
that end it is equal to the total load.

The S.F. diagram is therefore drawn as shown in
Fig. 105.

iQQQQQQO

■pr

Fig. 105.

Fig. 106.

187. Fig. 106 shows a cantilever with a load dis-
tributed over a portion of its length. It will be noticed
that the S.F. for the whole space A is equal to the
total load, but that under the load it gradually dimi-
nishes towards the outer end.

188. If necessary the Shearing Force due to concen-

SHEARING FORCE

123

i B i

trated and uniformly distributed loads on a cantilever
can easily be shown on one diagram.

Fig. 107 shows such a Shearing Force diagram where
c d represents the uniformly distributed load.

189. The Shearing Force
diagrams for beams sup-
ported at both ends are
obtained in the same way as
those for cantilevers, but as
the S.F. is the algebraical
sum of all the forces on either
side of the section taken, and
the reactions of the supports
are forces acting on the
beam, it will be necessary to
find them. This has been done, as will be seen on
referring to Figs. 108 and 109, by means of the funi-
cular polygon.

If the polar distance be known, then these funicular
polygons also serve as Bending Moment diagrams

(§ 171).

The Shearing Force diagrams are obtained, as pre-
viously explained, from the force diagram. An ex-
amination of the two shearing force diagrams should
make this quite clear.

Figs. 108 and 109 show how to draw the B.M. and
S.F. diagrams for an irregularly loaded beam in one
figure.

190. If a beam carries a uniformly distributed load,
the S.F. at each end is equal to the reactions of the
supports, each of which is equal to one-half of the load.
From each end it gradually diminishes as the centre of
the beam is approached.

124 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

Fio. 108.

A i B i

Fig. 109.

Figs. 110 and 111 show alternative ways of drawing
the Shearing Force diagram under these circumstances.

191. In order to show on one diagram the S.F. due to
tlie two systems of loading of a beam supported at both

EXAMPLES TO CHAPTER VI

125

^ cxxxxxn, ^

P M

^j;;|jiiii^^^[[|

Fig. 110.

Fig. 111.

ends, it is necessary to modify the S.F. diagram of
Fig. 109.

This modification, together with the diagram showing
the Shearing Force of the uniformly distributed load,
is shown in Fig. 112.

; e I

Fig. 112.

Examples to Chapter VI

1. What does " the bending moment " mean ?

2. How is the B.M. scale found ?

126 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

In a certain exercise the lineal scale was Jj^and the
" polar distance " 5 cwts.
Give the B.M. scale.

3. A beam, 15 ft. long and supported at both ends,
carries a load of 2*5 tons 6 ft. from one end.

Find the greatest B.M. and the B.M. at the centre.

4. Draw the B.M. diagram for a beam 20 ft. long with
a distributed load (including its own weight) of 15 tons.

5. A cantilever, 8 ft. long, supports a load of 5 cwts.
at its outer extremity.

Find, geometrically, the moment about the centre.

6. What is meant by " the Shearing Force ? "

7. A cantilever 10 ft. long has a distributed load of
3 cwts. per ft. on the outer half.

Draw the Shearing Force diagram, and give the S.F.
at the wall end.

^ 8. A beam, which is supported at both ends, and
is 20 ft. long, has a load of 6 tons placed 6 ft. from
one end.

Find the " bending moment " and the " shearing
stress " at the centre of the beam.

9. A beam, fixed at one end and IT 6'' long, supports
three loads — 5 cwts. 3' 10" from the wall, 6 cwts.
7' 8"' from the wall, and 2 cwts. at the unsupported
end.

Find the B.M. and S.F. at the centre.
V 10. A girder 20 ft. long supports a load of 5 cwts.
6 ft. from one end and a load of 7 cwts. 4 ft. from the
other.

What load could be placed at the centre of a similar
beam, so that the maximum bending moment may be
the same as that at the centre of the given beam ?

Chapter VII
STRESS OR RECIPROCAL DIAGRAMS

192. It has been shown in Chapter IV that forces can
be apphed along certain directions to resist the action
of some force or forces and so maintain equihbrium.

The usual method of introducing these new forces
is by means of bars of iron or wood. The members thus
introduced have to exert a certain amount of force,
depending on the magnitude of the force or forces they
have to resist, and on the angles at which they are
applied (§ 97). The resistance thus brought forth from
the bar is called " the stress," and by means of the tri-
angle or polygon of forces its magnitude and direction
can be obtained.

193. It was also shown in § 92 that a bar exerting a
force at one end exerts an equal and opposite force at
the other. Hence these new members are only intro-
duced to transmit the force from one point to a more
convenient one, either to the point of support, or to a
point where other members can be introduced to further
transmit it.

As an example of the former see Figs. 58 and 59, where,
to support the weight, two members are introduced.
These are secured to the wall, to which the force exerted
by the weight is conveyed.

To illustrate the latter the cantilever shown in Fig.
113 will be examined.

1«7

128 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

If we only consider the force A B and the bar-s B G
and C A,wQ> have a repetition of Fig. 59, and the magni-
tude and direction of the stresses set up in J^ O and G A
are given by the lines 6 c and c a in the triangle ah c.
The bar 5 C is exerting a force towards the load, there-
fore it exerts an equal force towards the joint at the

Z tons. ^

Fig. 113.

opposite end, and is in compression (§ 92). G A acts
from the load, consequently it acts from the opposite
joint, and is in tension.

We will now examine the joint at the lower end of
B G. We have there three bars, B D, D G, and G B,
but we have just determined the magnitude and direction
of the force exerted hj G B. It is represented by c 6.
By reproducing c h, and drawing lines parallel to B D

RECIPROCAL DIAGRAMS 129

and D G the triangle ch d \^ obtained, and the hnes h d
and d c give the magnitude and direction of the stresses
set up in ^ D and D C. It will be seen that 5 i) is a
compression bar and D C & tension bar.

Proceeding to the opposite end of D C, we have four
bars, AC, C D, DE Sind E A. Two of these, A C and
G D, have already been determined, and, since they are
tension bars, they act away from the joint. Draw a c
and c d to represent these in magnitude and direction,
remembering that the order in which the letters are
placed must indicate the direction of the force repre-
sented by the line.

From a draw a line parallel to E A, and from d a line
parallel to D E. Let these intersect at e. Then d e and
e a will represent the stresses in the bars D E and E A .

194. The student will no doubt have noticed that
instead of reproducing c b, the triangle cb d could have
been made on the c 6 of the first triangle, and that the
figure thus obtained could have been utilized to form
the last figure. The figure ab d e c is formed by com-
bining the three figures in this way, and, since the stress
in each member of the cantilever can be obtained from
it, it is called the " stress diagram."

The combination of the various force diagrams in this
manner saves time and prevents mistakes arising through
inaccurately transferring the measurements.

1 95. The magnitude of the stress in each bar is obtained
by measuring with the force scale.

196. The stress diagram, being a force diagram, must
close.

197. The loads and the reactions of the supports are
called the exterior or external forces, and the stresses are
called the interior or internal forces.

130 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

198. The method of procedure in drawing a stress
diagram is almost the same in every case, and, if the
student thoroughly understands one, he will have less
difficulty in applying his knowledge to new pro-
blems. For this reason it is intended to more fully ex-
plain how the forces act in the cantilever shown in
Fig. 113. The frame and stress diagrams of that figure
are reproduced in Fig. 114, with the joints of the frame
diagram numbered for reference.

199. It was explained in the chapter on Bow's nota-
tion, that, if the known force (or forces) acting at a
point be named in clockwise order, and if the first letter
of such name be placed first in the line of action of the
line representing such force (or forces), then the letters
naming the other forces will, when taken in the same
direction round the figure, give the direction of the un-
known forces.

It was also explained that compression bars exert an
outward force (i.e. towards the joint) at each end, and
the tension bars an inward force (i.e. from the joint)
at each end.

If, then, it is necessary to find the kind of stress in a
bar, all that is required is to select a joint at one of the
ends of a bar, name it in clockwise order, and follow the
direction of the corresponding letters on the stress
diagram.

As an example, we will take the vertical bar (Fig.
114).

If we select the joint marked 2, this bar is DC D
is the first letter, so d on the line dcoi the stress diagram
is the first point in its course of action, hence the force
acts upwards from the joint 2. Proceeding to the joint
3, the bar, when named in clockwise order, becomes

RECIPROCAL DIAGRAMS

131

CD. C is now tlie first letter, and, turning to the stress
diagram, we find that from c to t? is a downward direc-
tion. C D, therefore, acts from the joint 3. Hence the
bar C D or D (7 is in tension. Since both ends of a bar
exert the same kind of force (i.e. inwards or outwards)
only one end need be examined. It should be noticed
that, when considering one end of the bar, the stress
was given by c d, and, when considering the other end,
it was represented hy dc.

3 tons.

Fig. 114.

What has been said about the bar C D and its re-
ciprocal line c d, applies to all the bars and their recipro-
cals.

200. Taking the joint marked 1, the forces are ^ B,
B G and C A, and these are represented in magnitude
and direction hj ah,h c, and c a, that is, to maintain the
load A B, the bar G A must exert a pull equal to
3 tons, and the bar B G must push with a force equal
to 4-2 tons. In order that G A should exert a pull, the
other end must be attached to something to which it
transmits the force as shown by the string (§ 93). It is
secured to the joint 3, and here it exerts a pull equal to
3 tons, as shown by a c on the stress diagram.

132 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

B C pushes against the load, and, by domg so, presses
towards the joint 2 with a force, as shown by c &, equal
to 4-2 tons. To resist this thrust the two bars B D and
D G are introduced. The bar B D, as shown by h d,
pushes towards the joint with a force equal to 3 tons,
and by doing this exerts an equal pressure against the
wall at the joint 4. The bar D C has to exert an upward
pull at the joint 2, which, as we have already seen,
means a downward pull at the joint 3. This pull is
given by c (^ as being equal to 3 tons. This down-
ward pull of C Z> is resisted by the action of the bars
D E and E A. D E pushes towards the joint 3 with a
force equal to 4*2 tons, and consequently exerts an
equal pressure against the wall at the joint 4. The pull
in ^ ^ caused by the action oi C D and D E is shown
by e c to be equal to 3 tons, but in addition to this, it
has to resist the pull of A C, therefore the total
tension in .27 ^ is equal to 6 tons, as shown by e a on
the stress diagram, and this acts away from the joint 5.

201. Having ascertained the kind and amount of
stress of each bar, we will now consider the effect on the
wall.

We have seen that at the joint 5 there is an outward
pull of 6 tons. At the joint 4:, B D gives a direct thrust
of 3 tons, and E D sm oblique thrust of 4-2 tons.
E D must be resolved into its vertical and horizontal
components, each of which is equal to 3 tons, as
shown by e a; and x d. The total horizontal thrust at
the joint 4 is therefore equal to 6 tons.

Hence the two horizontal reactions of the wall are
each equal to 6 tons, but opposite in direction, and the
vertical reaction of the wall is equal to 3 tons.

202. The same principle underlies the construction

RECIPROCAL DIAGRAMS 133

of all framed structures, viz. the transference of a force
from one point to another where it can be more con-
veniently dealt with, and, as the known forces can be
utilized to discover the unknown ones, the stress in
every part of a framed structure can be ascertained by
means of the stress diagram.

203. The stress diagram is a valuable check on results
arrived at arithmetically, and if a structure be badly
designed, the stress diagram at once makes it apparent
by exposing the redundant members, and refusing to
close if necessary members are omitted.

204. Having determined the kind and amount of
stress which the proposed load will produce in each
member of a framed structure, the sectional area of the
members can be determined.

205. It is now intended to find the stresses produced
by given loads in the members of the more common
structures. After what has been said about the action
of forces, and the explanation of the stress diagram in
this chapter, the student will have no difficulty in fol-
lowing them.

The method of procedure is as follows : —

(1) Set out the structure to scale.

(2) Adopt a force scale and draw the line of loads.

(3) Determine the position of the loads, and indicate
them on the structure. The loads are to come on the
joints of the structure. If the true position of a load
be at some intermediate point of a bar, it must be so
divided that each joint at the end of the bar gets its
proper proportion of the load.

(4) Determine the reactions of the supports. If the
structure be symmetrical and symmetrically loaded,
each reaction wiU be equal to half the load ; if not, the

134 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

reactions may be determined by means of the funicular
polygon.

(5) Place letters (or numbers) between all the exterior
forces, and then in all the spaces of the frame.

(6) An exterior force (usually one of the reactions,
though not always), is now chosen where not more than
two members act, and resolved along their directions.

(7) At the next joint one of these is combined with the
exterior force (if there be one), and resolved along the
direction of the other bars. This process is continued
until all the bars are resolved.

2o6. Fig. 115 shows a span roof with a load concen-
trated at the ridge.

Fig. 115.
At the ridge there are three forces, the load A B and
the two rafters resisting this. Since there are three
forces in equilibrium, and A B is known, the triangle of
forces can be applied to determine the others, ah c is
this triangle, and, as the names of the rafters in clock-
wise order are B C and C A,b c and c a give the directions

RECIPROCAL DIAGRAMS 135

as well as the magnitudes of the forces exerted by them.
These act towards the joint, and are therefore in com-
pression. Taking the rafter C ^ , since it exerts a pres-
sure represented by c a at the ridge, a c will represent the
pressure it exerts at the foot. This force is resisted by
the vertical and horizontal reactions of the wall C D, and
by resolving the force a c along those directions the tri-
angle a cd is obtained, c d and d a represent those
reactions, therefore d c shows the proportion of the load
A B borne by the wall C D, and a d represents the force
which tends to overturn the wall.

The rafter B C presses with a force represented by
h c towards the ridge, and consequently with a force c h
towards the foot. This, being resolved in vertical
and horizontal directions, gives c e the vertical thrust
and e h the horizontal thrust of the rafter.

It should be noticed that the two vertical reactions
of the walls are together equal to the total load. This
is shown by c c and cd (or ed) being equal io ah.

The two horizontal reactions are also equal, as shown
hy da and h e.

207. If one of the rafters be more inclined than the
other, as in Fig. 116, we find that the stress in this is
greater than in the less inclined one, and that this in
turn produces a greater vertical reaction.

This is what we should expect, because the line of
action of the load ^ ^ is nearer to E C than to G D.

Again, the sum of the two vertical reactions is equal
to the load, and the horizontal reactions are equal to
each other.

208. With six spring balances, a few pieces of string,
and a weight, fitted up as shown in Fig. 117, the student
may perform a very interesting experiment for himself.

136 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

The only difference between this and the two previous
examples is that the directions of the forces are reversed,
but this will not affect their magnitudes.

=>o

Fig. 117.

RECIPROCAL DIAGRAMS 137

The weight can be sUpped to different positions, and
the effect on each balance noted.
Briefly, they are as follows : —

(a) The more inclined any section of the string becomes,
the greater is the stress produced.

(b) The nearer the weight is to one side, the greater is
the proportion of the weight supported by the vertical
balance on that side.

{d) And the forces registered by the horizontal bal-
ances are always equal to one another and opposite in
direction.

As an exercise he should select one of the positions,
graphically determine the forces, and compare the
results with those shown on the balances.

209. In the two previous exercises we have assumed
that the total load is on the ridge. Each rafter is in
reality a beam with a distributed load, and half this load
is supported at each end. Thus the load on the ridge
is only one half the total load, the other half being sup-
ported directly by the walls. Fig. 118 shows the load
apportioned in this manner. A new member is also
introduced.

Draw ab, be and c d to represent the three loads.
a d now represents the total load, and d a the total
reactions. The frame and the loading being symmetrical
the two reactions will be equal. Bisect d a. The re-
actions oi D E and E A are represented hy d e and e a
respectively. Since e a represents the total reaction of
the wall E A, and the load represented by a 6 is sup-
ported directly by the wall, the remainder e 6 is the
reaction caused by the load on the frame.

138 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

By resolving eh in directions parallel to B F and F E
the triangle eb fis obtained. The sides of this represent
the forces in magnitude and direction when taken in
order round the triangle. If we follow them we find
that j5 i^ is in compression and F E in tension.

We can proceed in a similar manner with the other
reaction, and obtain the triangle c e /, which determines
the forces exerted hjEF and F G. It should be noticed

(PC

Fig. 118.

that the new member is introduced to resist the outward
thrust of the rafter, which it does by exerting an inward
force at each end.

(We may, if we like, treat the foot of the rafter as if the
four forces, E A, A B, B F and F E were acting. E A
and A G are known, and by resolving them along the
directions of the others, we get a polygon of forces
whose sides are ea, ab, b f and / e. But a h Hes on a

RECIPROCAL DIAGRAMS

139

portion of e a, and the polygon appears as shown by
eah f e.)

210. We will now take a roof, as shown in Fig. 119, and

8^/

l^ '

1 f

^ - X - -^

- y ^

^-^ ~~

-^ .

Fig. 119.

suppose the load on each rafter to be 6 cwts. Appor-
tioning the load, we get 6 cwts. on the ridge and 3 cwts,
on each wall.

Since the line of action of 5 C is nearer the one wall
than the other, they will not support equal shares, so
we must fall back on one of the methods of determining
the reactions.

The usual way is by means of the funicular polygon.
Draw the line of loads. Since it is the proportion of
B C which each wall supports that has to be determined,
join b and c to a pole. Draw the funicular polygon and
o e parallel to the closing link, d e and e a now represent
the total reactions oiD E and E A respectively, (c e and
e h could have been found by means of similar triangles
(§ 51), or, if X and y be known, they could have been
ascertained by taking the moments about either sup-
port (§ 50).)

140 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

Having ascertained the position of e, the stress diagram
can be drawn as in the previous exercise.

211. Taking Fig. 118, and adding a vertical member
(a king-rod) we obtain Fig. 120.

Fig. 120.

The stress diagram shows / and g at the same point.
The distance between them being nil, shows that the
load produces no stress m. F G.

This is what we should expect, since at the bottom of
F G there are three bars, two of which are parallel, meet-
ing at a point (§ 110).

212. If we camber the tie-rod, we obtain Fig. 121.

Fig. 121.

RECIPROCAL DIAGRAMS

141

Resolve e 6, the resultant oi E A and A B, in directions
parallel to B F and F E, thus obtaining h f and / c. h f
represents the force exerted by B F Sit the foot, therefore
/ b represents the force it exerts at the ridge. At this
latter point there are four forces, but i^ J5 and B C are
known, and are represented by / 6 and h c, therefore by
drawing lines from c and / parallel to C G and G F, eg
and g f are obtained, and these give the stresses in
those members. By combining the force g c with c e,
the resultant oi C D and D E, and resolving parallel to
E G, eg ia obtained, and this gives the tension in E G.

It will be seen that by cambering the tie-rod, a tensile
stress is produced in the king-rod.

213. Fig. 122 shows another kind of roof truss. It is
formed by the addition of two members to Fig. 120.

/)i^Z<^

Fig. 122.

Resolve g 6 as before, and so obtain h h and h g. Combine
h h with the load h c, and resolve parallel to C I and / H,
thus obtaining c i and i h. By combining i c and c d, and
resolving parallel to D J and J I, dj and j i are deter-
mined, and by combining j d with d e and drawing lines
parallel to E K and K J, e k and k j are obtained. Since
k and h are at the same point, g kis equal to h g, therefore
the stresses in all the members are found.

142 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

The lines h i and k j coincide with a portion of the
lines h e and k b respectively, but this should present
no difficulty, if it be remembered that each is a side
of a polygon of forces.

214. By cambering the tie-rod of the last figure we
obtain Fig. 123.

Fig. 123.

h and k do not now come together, neither do h i and
k j coincide with h e and A; 6 as in the last exercise.
215. In Fig. 124 we have a different arrangement.

CX^D

B^l/J

k .

^^^f^

^^^^^

t ^

e
f

Fig. 124.

By means of the reactions, B H and H G can be deter-
mined. The stress in HB and the load CD can be
utilized to obtain the stresses in O / and / H. If we
proceed to the ridge, we find that of the five forces acting

RECIPROCAL DIAGRAMS

143

Fig. 127.

144 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

there, we only know / G and G D, so we cannot i)roceed
at that point. If we go to the joint at the bottom of
H I, by means of the stresses oi G H and H I, we can
find those oi I J and J G. Now we know three of the
five forces acting at the ridge, so that we can determine
the others. This problem presents no further difficulty.

216. Figs. 125, 126, and 127 show trusses which are
loaded at a greater number of points.

The construction of the stress diagrams for these
should present no difficulty, if it be remembered that the
line representing the stress in one bar may lie partly or
wholly on the line representing the stress in another bar.

The bars J K and R 8 (Fig. 127) may at first prove a
little disconcerting, but an examination of them will
show that at the bottom joint of each there are three
bars, two of which are parallel, therefore the stress in
J K and RS due to the loading is nil (§ 110). They
are introduced to prevent the tie-rod sagging.

217. Fig. 128 shows a braced cantilever with a con-
centrated load at its outer end.

a f

Fig. 128.

The stresses in B C and C A are obtained by resolving
A B along those directions and so obtaining the triangle

RECIPROCAL DIAGRAMS

145

ah c. ca represents the force exerted by ^ O at its
lower end, therefore a c is that which it exerts at the
upper end. By resolving this parallel to G D and D A,
we get c d and d a, which give the stresses in those mem-
bers. By combining the stresses of D C and C B and
resolving parallel to B E and E D, a polygon whose sides
are dc,cb,b e, and e d, is obtained, b e and e d give the
stresses inB E and E D respectively. Similarly by com-
bining the stresses oi A D and D E, those oi E F and
F A are obtained.

The reactions of the wall are determined as shown in
§201.

2x8. Fig. 129 shows a braced cantilever with a dis-
tributed load.

Fig. 129.

The first thing we have to consider is how to divide
the load. It will be seen that it is supported by two
bars of equal length. Each supports one half of the
total load, therefore one quarter of the total load is
supported at each end of the two bars.

Thus A B equals one quarter of the load, B C equals
one half of the load, and C I) equals one quarter of the
load.

146 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

By means of the load G D the stresses oi D E and E G
are obtained. The stress oi E D ia utilized to find those
ol D F and F E. By combining the load B G with the
stresses oi G E and E F, the stresses of F G and G B are
determined.

219. Fig. 130 shows another form of braced cantilever
with a distributed load.

In allocating the load to its various points of support,
we find that the bar B G is one half the length of the
bar G E, and consequently receives only one third of
the load. Half of this third, or one sixth of the total
load, is supported at each end of the bar B G. Half

}b\

Yr^

Fig. 130.
of the remainder, or one third of the total load, is sup-
ported at each end of G E. The total load, then, is so
divided that A B equals one sixth of the load, B G equals
6 + J or i of the load, and G D equals J of the load.

The stress diagram is drawn in exactly the same
manner as that of the last exercise.

220. When dealing with the roof trusses, we obtained
the magnitude of the stress in each bar by measuring
with the force scale its corresponding line on the stress
diagram.

RECIPROCAL DIAGRAMS

147

Looking at the stress diagrams of Figs. 128 and 129,
we see that the former is made up of equilateral triangles
and half such triangles, and that the latter is made up of
right-angled isosceles triangles. As the sides of these
triangles always bear a certain relationship to each other,
if we know the length of one side we can easily obtain that
of tlie others.

We will first examine the stress diagram of Fig. 128.
If we know the lengths of the sides of the triangle ah c,
we know the lengths of all the other lines in the diagram.
The relationship between the sides of this triangle is as
follows : —

h c='&ll xab or ^ a c,
ab=-S66xa c,
and a c= 1-155 xab or 2 h c.

Proceeding to the stress diagram of Fig. 129, and taking
the triangle c d e, we have —

c d=c e='707 xd e,
and d 6=1-414 xc d.

To show how this knowledge is applied, a load of 3
tons is taken in Fig. 128. a b now represents 3 tons,
therefore a 0=1-155 x a 6=3-46 tons, and 6c=l-73
tons.

ad, c d, d e and e f are each =a c=3-46 tons.

a /=2 a c=6'93 tons.

6e=6cxcc=(l-73 + 3-46) tons =5-19 tons.

In this manner the stress in the bars of the foregoing
cantilevers and the following girders can be obtained
with mathematical accuracy without making use of the
force scale. The magnitude and kind of stress in each
bar should be indicated on the frame diagram as shown
in Fig. 128.

221. The remaining figures show a few short Warren

148 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

and N girders under different kinds of loads, with their
respective stress diagrams.

The difficulty which will be found in drawing the
stress diagrams for these is caused by some of the
bars not being called upon to resist the action of the
load, i.e. the stress in them due to the load = 0.

Having ascertained these and determined the ex-
terior forces, the drawing of the stress diagrams be-
comes comparatively easy.

Fig. 131. The load A B being midway between the
supports, the two reactions B C and C A are equal.

Fig. 131

At each end of the top flange there are only two
bars, and these are at right angles to each other. It
is evident two forces acting at right angles to each
other cannot maintain equilibrium, therefore, since
these corners are in equilibrium, we know that these
bars exert no force.

To indicate that the stress in A D =^ 0, d must be
placed at the same point as a on the stress diagram.
Similarly, j must be placed at the same point as h.

The stress diagram can now be drawn in the ordinary
way.

RECIPROCAL DIAGRAMS

149

222. Fig. 132 is the same girder as the last, with a
distributed load on the top flange.

Fig. 132.

The chief difficulty in this is in apportioning the load
to the several points of support.

The bars B H and E N being half as long at the bars
C J and D L, only receive half as much of the load, so
J of the load comes on each of the two former, and J on
each of the two latter. Half of each of these loads is
supported at each end of the bars, therefore A B = 3^,
BC = -^ + iorlCD = i + iori, DE = i + '-^
or i, and E F = ^.

At either end of the top flange there are apparently
three forces, two of which are parallel, hence we know
that the stresses in ^ ^ and N F are equal to ^ ^ and
E F respectively, and that the stresses in B H and
EN = 0 {^ 110).

To indicate this on the stress diagram, place h at the
same point as h, and n at the same point as e.

Of course, the bars B H and E N are necessary to
support the distributed load. They are subject to a
cross strain which an ordinary stress diagram is unable
to take account of.

150 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

223. Fig. 133 shows a distributed load on the bottom
flange of a girder.

/\m

L \

D E]

Fig. 133.

Here the load is distributed over three beams of
equal length. Each receives J the load, and transmits
^ of the total load to each end. The two middle
joints, therefore, receive twice as much as the outer
ones.

Another difficulty presents itself — the two outer
loads are in line with the reactions. How can these
be indicated on the frame diagram ? The plan usually
adopted is to open out the line as shown, and place a
letter in the space. The two outer lines must repre-
sent the reactions.

The point / coincides with the point c. At the ends
of the top flange there is a repetition of what was
found in Fig. 131, therefore g and m must be placed at
the point /.

It will be seen that the letters i, j, and k are at the
same point. This indicates that the stress in both I J
md J K = 0.

RECIPROCAL DIAGRAMS

151

224. Fig. 134 : The load is shown in the middle of
a bar. To draw the stress diagram this load must be
divided between the two joints at the end of the bar.

At each end of the bottom flange we have the re-
action and two bars, but in each case two of these are
parallel. We therefore know that the stresses in D C
and G J = 0, and those oi A D and J B = the re-
actions (§ 110).

This is shown on the stress diagram by placing d
and j at the point c.

The load does not affect F G and G H.

225. Fig. 135 illustrates an N girder with a distributed
load on the top flange.

i B I C i D ..

Fig. 135.

152 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

An examination of the bars H G and 0 G will show
that the stress in each of them = 0.

By placing h and o at the point g, we have h a and / o
equal to the reactions, and these represent the stresses
oi H A and F 0.

226. In Fig. 136 we have, as in Figs. 131 and 133,
at each top corner, two theoretically useless bars. To

FiCx. 136.

indicate their values on the stress diagram d must be
placed at the point a, and k at the point h.

G H is another bar which exerts no force, as can
easily be seen by examining the bottom joint. Its
work is to prevent sagging in G C and H G.

227. In Fig. 137 the reactions are not equal. They

I ^ 6%W6't

Fig. 137,

\ k

kr\iK

— -^

y\/

RECIPROCAL DIAGRAMS

153

have been determined by means of the funicular poly-
gon, and, of course, if the polar distance be known,
the bending moment at any point can be obtained.

The stress in the bars G A, S E and L M = 0.

The two former make the girder more rigid, and
L M strengthens the top flange by resisting any bending
that is likely to occur in F L and F M through being in
compression.

228. In Fig. 138 we have a lattice girder supporting
a number of loads. In this a new difficulty presents
itself, because at no point are there less than three
unknown forces. In order to overcome this difficulty,
we may consider the girder as being made up of the two
girders shown in Fig. 138. The loads retain the same
positions as on the original girder.

1 2 S U i

i 1 i i i

EXXZ

T T

S I

(a) ^^ ib) ^

Fig. 138.

Fig. 139.

On examining Fig 139 (a) it will readily be seen that
the compression of each of the vertical bars is equal to
the reaction of the wall that supports it. These re-
actions are due to the three loads 1, 3 and 5, and can
easily be determined whatever the loads may be.
Turning to Fig. 139 (6) we see that the two loads 2
and 4 do not in any way affect the vertical bars referred
to. Therefore, to ascertain the stresses of the two
verticals or pillars, it is only necessary to find the
reactions due to the loads 1, 3 and 5.

154 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

Having found the force exerted by each pillar, we
have only two unknown forces at the points of support
of the girder, and can proceed with the solution.

229. Fig. 140 shows the stress diagram for a lattice
girder supporting a uniformly distributed load.

From the preceding paragraph it will be seen that,
since the load is uniformly distributed, the stresses of
B H and T G are equal to one another, each being
equal to one quarter of the load. These are shown by
b h and t g on the stress diagram.

230. By adding vertical bars to Fig. 140 we obtain
Fig. 141. This is termed a lattice girder with verticals.

I c

> '

F i

G not

Fig. 140.

We may imagine this girder as being a combination
of the two single girders {a) and (b) (Fig. 142).

The load on the original girder has now to be divided
between (a) and (b) (Fig. 142). There is on each of
these a point of support corresponding to every point
of support on the combined girder, therefore half of
each load on the original girder must be placed on the
corresponding points of the single girders.

This is plainly shown in Figs. 141 and 142.

RECIPROCAL DIAGRAMS

155

Having divided the combined girder into two single
ones and apportioned the loads as described, the stresses
in the bars of the former may be obtained by finding
those in the bars of the single ones and combining
them.

In considering Fig. 141 as being made up of {a) and
(b) Fig. 142, it should be noted that the two booms
and all the verticals are duplicated. Where members
are duplicated, the stresses must be added or sub-
tracted according as they are alike or unlike. When
added, the sum will represent a stress of the same kind
as those added, and when subtracted, the difference

6 20 36 JO 10 2i 10 Hi IS 6 2i W Hi 15 6

. ^

i i i J

XXX

(a)

I (i) 1

Fig. 141.

Fig. 142.

will represent a stress of the same kind as that of the
greater.

In order to more clearly explain the method of pro-
cedure, the stress diagrams of (a) and {b) (Fig. 142) are
drawn, and the values indicated on the girders (Figs.
143 and 144).

These results must now be combined and figured on
the original girder, as already explained (see Fig. 145).
Thus, the stress of the vertical B H (Fig. 143) is + 2-5
and that oi B I (Fig. 144), which coincides with it
when superposed, is + 22-5. Being stresses of the same
kind, they must be added. The sum is + 25, which
gives the total stress oi B I (Fig. 145). Taking the

156 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

vertical / J (Fig. 143), the stress is — 10, while that of
the corresponding vertical H K (Fig. 143) is + 20.
These, being unlike, must be subtracted and the differ-
ence given the sign of the greater. Therefore, the

E F

22-i

Cv
■f ,

, /

Fig. 143.

i ^ i ^ 1 M M

Fig. 144.
stress oiKL (Fig. 145) is + 10. Again, K A (Fig. 143)
coincides with K A (Fig. 144). The stresses of these

RECIPROCAL DIAGRAMS

157

Fig. 145.
are — 30 and — 20 respectively, therefore the stress
oi A N (Fig. 145) is — 50. Proceeding in this manner,
the stresses of all the members can be obtained. None
of the diagonals are duplicated, therefore the stresses
of these remain as found in Figs. 143 and 144.

231. Fig. 146 shows the same girder as that given in

i C i D U i F A ' ^

'iTi

Fig. 146.

158 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

the last paragraph, but in this case all tlie stresses are
obtained by means of one diagram. The loads are
equal to those of the last exercise.

The first difficulty met with is that at the points of
support there are four forces, only one of which is
known. In order to find the stresses oi B I and W G
we must suppose the girder and loads divided as shown
in Fig. 144, and find the reactions of the supports.
The stresses in B I and W G due to this half of the load
are respectively equal to these reactions, but to find
the total compression in them, we must add to each
the other liaK of the load which comes directly upon it.
The compressive stress oi B I due to the half of the total
load is 22-5, and half the load which comes directly
upon it is 2-5, therefore the total stress in 5 / is + 25.
Similarly, W G equals + 32-5.

The verticals K L, OP, and S T will next prove
troublesome, but the rule is to assume that the stress
in each of these is equal to one half of the load bearing
directly upon it, which agrees with the results obtained
in the last exercise.

The exercise presents no further difficulty.

If the student remembers how to find the stresses
of the outside verticals, and the rule relating to the
inner ones, he should be able to find the stresses in a
lattice girder with verticals under any system of
loading.

232. In actual practice the framed structures usually
consist of a much greater number of parts, so that the
alphabet is insufficient to name them all. The prin-
ciple of Bow's Notation is retained, but numerals are
used instead of letters. They have not the advantage
of capitals and small letters respectively for the frame

EXAMPLES TO CHAPTER VII

159

and force diagrams, but there is no difficulty in using
them when once the principle is fully understood.

233. In conclusion, it must not be supposed that
complex structures can now be readily analyzed.
There are some which may take years of study to eluci-
date, and others in which the result depends upon the
nature of the workmanship, which cannot be foreseen.
Students who have carefully worked through the
preceding pages should be able to prepare stress dia-
grams for all ordinary cases, and, if they are ambitious,
may try their hands at a collar-beam truss where the
walls are not rigid, a hammer-beam truss, an arched
roof truss, etc.

Examples to CiiArTER VII
1. Fig. 1 shows a couple close roof witli a rise one
quarter the span, carrying a distributed load of 16 cwts.

Ex. Ch. VII.— Fig. 1.
What are the stresses produced in each member ?

2. The roof truss given in Fig. 122 supports a dis-
tributed load of 5 tons.

Draw the truss and figure the stresses on it.

3. Fig. 2 shows a
roof truss carrying a
load of 4 tons. Pitch

Give the stresses of
all the members. 1 Ex. Ch. VII.— Fig

160 ELEMENTARY PRINCIPLES OF GRAPHIC STATICS

4. Taking the load on the Queen-post truss shown
in Fig. 126 as 5 tons, determine the various stresses.

5. Find the amount and kind of stress produced in
each member shown in Fig. 128, with a load of 2-5 tons
at its extremity.

6. Find the stresses in each member, and the re-
actions of the wall, due to the load given in Fig. 3.

Ex. Ch. VII.— Fig. 3.

7. Fig. 4 is an elevation of a trussed girder with a
concentrated load.

Ex. Ch. VII.— Fig. 4.

Draw to twice the scale, showing the members in
compression by double lines, and those in tension by
single lines, omitting any bars not affected by the load.

8. Determine graphically to a scale of f' to a ton
the reactions at the points of support, and the stresses
set up in the different members of the loaded girder
shown in Fig. 5,

EXAMPLES TO CHAPTER VH

161

' $ tor?^ ^ tons '

Ex. Ch. VII.— Fig. 5.

9. A girder 21 ft. long and 2 ft. deep is trussed as
shown in Fig. 6.

E
B

Ex. Ch. VII.— Fig. 6.

Determine tlie stresses when it carries a distributed
load of 1 cwt. per foot.

10. Draw the stress diagram of Fig. 7 ; mark on the

Ex. Oh. VII.— Fig. 7

truss the amounts of the stresses in cwts., distinguishing
between compression and tension bars.
What is the bending moment at a: z/ ?

ANSWERS TO EXAMPLES

Examples to Chapter 1 (pp. 20, 21)

1. (a) 9-4 units; {h) 6-26 yards; (c) 3-13 tons; {d)
5371 lbs.

2. {a) 50-91 ft. ; (b) 12-99 inches ; (c) 22-79 ft ; {d)
Tiebeam, 25'.0 ; Principal, 14-43 ft. ; King-post, 7-21 ft. ;
Struts, 7-21 ft.

3. 1-11 inches.

4. 4-05 units.

5. 1-809 units.

6. 4-53 units.

Examples to Chapter II (pp. 42, 43)

1. 273 lbs.

2. A vertical line 2-3 inches long,

3. {a) 35-5 lbs. ; (6) 10-5 lbs.

4. {a) total load = 8-75 cwts. ; (b) 8-75 cwts, ; (c)
vertically upwards.

5. Horizontal.

6. See pars. 40-44.

7. (a) 10 ft. cwts. ; (b) 5 ft. cwts. ; (c) nil.

8. SJ tons, at 3-75 ft. from one end.

9. 10| cwfcs.

10. (a) 503 lbs. and 700 lbs. ; {b) 575 lbs. and 775 lbs.

Examples to Chapter III (pp. 55-57)

1. See par. 78.

2. A vertical line 2-25 inches long.

3. Between the loads and 2-85 ft. from the smaller.

Answers to examples 163

4. 1,485 lbs. ; the e.g. is -93 ft. from the vertical face.

5. See par. 77.

6. 400 lbs. and 600 lbs.

7. (!) OX; (2) XB; (3) BF; (4) F A ; (5) A 0.
6. A vertical line 2'' long.

9. 3 cwt. acting vertically downwards.

10. (a) P = 35 lbs ; (b) 21 lbs. vertically downwards.

Examples to Chapter IV (pp. 82, 83)

1. The resultant oi A B and A D ; equilibrant.

2. See par. 31.

3. 7 lbs.

4. Tension in longer cord, 10-8 lbs. ; in shorter cord,
14-4 lbs.

5. The direction of the reaction of the ground is
found by joining the foot of the ladder to the point of
intersection of the vertical through the e.g. of the ladder
and the direction of the reaction of the wall ; magnitude
of force = 156 lbs.

6. Vertical reaction = 100 lbs. ; horizontal =
17320 lbs.

7. Horizontal reaction = 80-7 lbs. ; total reaction of
top hinge = 262-7 lbs.

8. {a) - 34-64 lbs. and + 69-28 lbs. ; (6) both
34-64 lbs.

9. 1-28 ft. from inner face of wall.

10. AB = 1-25 ; BC =z -75, both acting towards
the point.

Examples to Chapter V (pp. 103-105)

1. Reaction of ^ = 4-5 tons ; of J5 = 10-5 tons.

2. 2-23 tons and 2*26 tons.

3. 47-2 and 52-7.

164 ELEMENTARY PRINCIPLES OP GRAPHIC STATICS

4. 1 ft. from the end.

5. 6-13 ft. from that ead near which the force of
4 cwts. acts.

6. 2-25 lbs. •

8. ^ = 3-625 tons ; B = 1-375 tons.

9. ^ = 3-92 tons ; B = 4-58 tons.
10. 14-25 tons and 11-75 tons.

Examples to Chapter VI (pp. 125, 126). '^

1. See par. 42.

2. By multiplying the lineal scale by the polar dis-
tance; V= 40 ft.-cwts. or 1"= 10 ft.-cwts.

3. 9 ft.-cwts. ; 7-5 ft.-cwts.

5. 20 ft.-cwts.

6. See par. 183.

7. S.F. = 15 cwts.

8. B.M. = 18 ft.-tons ; S.F. = 1-8 tons.

9. B.M. = 23 ft.-cwts, ; S.F. = 8 cwts.
10. 2-9 cwts.

Examples to Chapter VII (pp. 159-161)

1. B F a>nd G F = 8-96 cwts. ; E F = 8-01 cwts.

2. BHsindE K = 75 cwts. ; 0 / and D J = 50 cwts. ;
IJ, H I siiid J K = 25 cwts. ; G H and G K = 64-95
cwts.

3. BH and EK = 90-44 cwts. ; C I and D J =
60 cwts. ; / / = 40 cwts. ; H I and K J = 26-39 cwts. ;
GH and 6^ Z = 79-79 cwts.

4. BlandF M = 80 cwts. ; C J and ^ L = 60 cwts. ;
KHandKD = 51-96 cwts. ; J / and it/ L = 20 cwts. ;
J K and KL = 10 cwts. ; H I and H M = 69-28 cwts.

5. BG =+ 1-44 ; A G, A D, D E = - 2-88 ; D G
and EF=+ 2-88: A F = - 5-76 : B E = + 4:32

ANSWERS 165

tons. The horizontal reaction = 5-76 tons ; vertical
reaction =2-5 tons.

Q.BE, BD, DG=+M2 tons; E D = - 4:62
tons ; AE = - 2-31 tons ; AG = - 6-93 tons. The
horizontal reactions = 6-93 tons ; vertical reaction =
4 tons.

8. A J and I L = 0 ; AM = + 3-625 ; B M z= +
209 ; DM = + 3-97 ; F M = + 2-38 ; ^ Jf = + -79 ;
1 M = + 1375 ; CK = - 303 ; E L = - SIS ; G L,
DE, F G'a.nd HI = - 1-59 ; E F smd G H = + 1-59 ;
AB = - 418 ; BG = + 1-87 ; G D = - 1-87 tons.

9.0/, Z)iH^, EG=+2i'5; A I and ^ 6^ = -
25-4 ; AH = - 24-5 ; I H s^nd H G = + 70 cwts.

10. ^ J, i? 7 = 0 ; ^ L = - 1045 ;GN = - 1760 ;
Z)P=- 214-5; ES= + 197-3; #t/ = - 148-6; (? If
= -82-7; /J= + 104-375; / iT = + 104-5 ; / ilf
= + 1760 , 10= + 214-5 ; / 0 = + 221-0 ; I R= +
221-0; / I^ zz: + 197-3 ; / F = + 148-6 ; IX= +
82-7; 7 7= +82-625; J iT = - 148-0 ; L if = -
101-0 ; N0= - 55-0 ;Pg=- 9-5 ;i2/8'=- 33-75 ;
TU = - 68-7 ; F If = - 93-0 ; Z 7 = - 117-0 ; iT L
= + 71-4 • MN = + 38-4 ;OP=+6-4;(gi2=:0;
/S 7^ = + 24-0 ; C7 F = + 49-0 \ W X= + 65-63 cwts.
B.M. at X 7 = 779-75 ft.-cwts.

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Stands and Stages, Coffer Dams, Foundations, Bridges, Gates,
Tunnels, Excavations, Wood and Half-Timber Houses, and
various Structural Details; together with new and simple
methods of Finding the Bevels in Roofs, Setting Out Domes,
Steeples, tfec. ; the Uses of the Steel Square ; Notes on the
Woods used; and a Glossary of Terms. 450 pages, with
1,100 clear and practical Illustrations. Large 8vo, cloth,
12s. 6c?. net.

" A handsome and substantial volume. The project has been well earned out. It
excels nearly all in its completeness." — The Carpenter and Builder.

" The book is full of sound practical matter. It is profusely illustrated with the
clearest of line drawings and photographs, not mere sketches, but working drawings of
the highest possible value. Anyone confronted with an unusual difficulty would almost
surely find its solution somewhere in the volume." — The Building News.

MODERN PRACTICAL JOINERY. A Guide to the
Preparation of all kinds of House Joinery, Bank, Office,
Church, Museum and Shop-fittings, Air-tight Cases, and
Shaped Work, with a full description of Hand-tools and their
uses. Workshop Practice, Fittings and Appliances, also
Directions for Fixing, the Setting-out of Rods, Reading of
Plans, and Preparation of Working Drawings, Notes on
Timber, and a Glossary of Terms, &c. By George Ellis,
Author of " Modern Practical Carpentry." Containing
380 pages, with 1,000 practical Illustrations. Large 8vo,
cloth, 125. 6c?. net.

" In this excellent work the mature fruits of the first-hand practical experience of an
exceptionally skilful and intelligent craftsman are given. It is a credit to the author'*
talent and industry, and is likely to remain an enduring monument to British craftsman-
ehi]^."— Building World.

PLASTERING — PLAIN AND DECORATIVE. A

Practical Treatise on the Art and Craft of Plastering and
Modelling. Including full descriptions of the various Tools,
Materials, Processes, and Appliances employed, and chapters
on Concrete Work, both plain and reinforced. By William
Millar. Containing 600 pp. Illustrated by 53 full-page
Plates, and about 500 smaller Diagrams in the Text. 3rd
Edition, Revised and Enlarged. Thick 4to, cloth, 18s. net.

*' ' Millar on Plastering ' may be expected to be the standard authority on the subject
for many years to come, and we congratulate the author on having left such a legacy to
his craft as will connect his name with it as intimately and as durably as that of
Tredgold with Carpentry A truly monumental v^ovk."— 'The Builder.

PKOFESSOR BANISTER FLETCHER'S VALUABLE TEXT-BOOKS FOR
ARCHITECTS AND SURVEYORS.

Arranged in Tabulated Form and fully indexed for ready reference.
QUANTITIES. A Text-Book explanatory of the best methods
adopted in the measurement of builders' work. Seventh
Edition, revised and enlarged by H. Phillips Fletcher,
F.R.LB.A., F.S.I. With special chapters on Cubing, Priced
Schedules, Grouping, the Law, &c., and a typical example
of the complete Taking-off, Abstracting, and Billing in all
Trades. Containing 460 pages; with 82 Illustrations. Crown
8vo, cloth, 7s. 6cZ.
The most Complete, Concise, and Handy Work on the Subject.

" It is no doubt the best work on the subject extant."— ITie Builder.

*' A good treatise by a competent master of his subject. . . . Indispensable to
every architectural or surveying student."— JTie Building News.

" Those who remember the earlier editions of this work will thoroughly appreciate the
increase in size and the great improvement in quality of this last edition, which certainly
makes it one of the most complete works upon the subject."— TTie Builder's Journal.

"We compliment Mr. Fletcher on his revision and on the accuracy of the book
generally."- TTie Surveyor.

DILAPIDATIONS. A text-book on the Law and Practice of.

With the various Acts relating thereto, and special chapters
on Ecclesiastical Dilapidations and on Fixtures. Sixth Edition,
revised and remodelled, with recent Legal Cases and Acts, by
Banister F. Fletcher, F.Pv.I.B.A., F.S.I., and H. Phillips
Fletcher, F.R.LB.A., F.S.L, Barrister-at-Law. Crown 8vo,
cloth, 6s. 6(i.

LIGHT AND AIR. With Methods of Estimating Injuries, &c.
Fourth Edition, revised and enlarged by Banister F.
Fletcher and H. Phillips Fletcher. With full Reports and
Digests of Ruling Cases, and 27 Coloured Diagrams, kc.
Crown 8vo, cloth, Qs. 6o?.

"By far the most complete and practical text-book we have seen. In it will be found
the cream of all the legal definitions and decisions."- BuiZding News.

VALUATIONS AND COMPENSATIONS. A Text-Book
on the Practice of Valuing Property, and the Law of Com-
pensations in relation thereto. Third Edition, rewritten
and enlarged by Banister F. Fletcher and H. Phillips
Fletcher, with Appendices of Forms, ifec, and many new
Valuation Tables. Crown 8vo, cloth, 6s. M.

" Very useful to students preparing for the examination of the Surveyors' Institution."

— The Sui-veyor,

ARBITRATIONS. Third Edition, revised and largely re-
written, by Banister F. Fletcher and H. Phillips Fletcher.
With references to the chief governing cases, and an Appendix
of Forms, Statutes, Rules, &c. Crown 8vo, cloth, gilt, 5s. M.

'• It is as well written and revised as can be, and we doubt if it would be possible to
find a more satisfactory handbook." — The Builder.

PROFESSOR SANISTER FLETCHER'S VALUABLE TEXT-BOOKS FOB
ARCHITECTS AND SVaY-EYORS— continued.

THE LONDON BUILDING ACTS, 1894-1905. A

Text-Book on the Law relating to Building in the Metropolis.
Containing the Acts, in extenso, together with the unrepealed
Sections of all other Acts affecting building, the latest Bye-
Laws and Regulations, Notes on the Acts and abstracts of
the latest decisions and cases. Fourth Edition, thoroughly
revised by Banister F. Fletcher, F.R.I.B.A., F.S.L, and
H, Phillips Fletcher, F.R.I.B.A., F.S.L, Barrister-at-Law.
With 23 Coloured Plates, showing the thickness of walls,
plans of chimneys, &c. Crown 8vo, cloth, 6s. 6d.

" It is the Law of Building for London in one volume." — Architect.
" Illustrated by a series of invaluable coloured plates, showing clearly the meaning of
the various clauses as regards construction."— 2'Ae Surveyor.

CONDITIONS OF CONTRACT relating to Building
Works. By Frank W. Macey, Architect. Revised, as to
the strictly legal matter, by B. J. Leverson, Barrister-at-
Law. Royal 8vo, cloth, 15s. net.

ESTIMATING : A Method op Pricing Builders' Quantitibs
FOR Competitive Work. By George Stephenson. Showing
how to prepare, without the use of a Price Book, the Estimates
of the work to be done in the various Trades throughout
a large Villa Residence. Sixth Edition, the Prices carefully
revised to date. Crown 8vo, cloth, 4s. 6d. net.

" The author, evidently a man who has had experience, enables everyone to enter, aa
It were, into a builder's office and see how schedules are made out. The novice will find a
good many ' wrinkles ' in the hook."— Architect.

REPAIRS : How to Measure and Value them. A Hand-
book for the use of Builders, Decorators, &c. By George
Stephenson, Author of " Estimating." Fourth Edition, the
prices carefully revised. Crown 8vo, cloth, 3s. net.

'• ' Repairs ' is a very serviceable handbook on the subject. A good specification for
repairs is given by the author, and then he proceeds, from the top floor downwards, to
show how to value the items, by a method of framing the estimate in the measuring book.
The modus operandi is simple and soon learnt." — TTie Building News.

FACTS ON FIRE PREVENTION. An enquiry into the
Fire-Resisting Qualities of the chief Materials and Systems of
Construction, conducted by the British Fire Prevention Com-
mittee. Edited by Edwin 0. Sachs, Author of *' Modern
Theatres." Containing Accounts of Tests of Floors, Ceilings,
Partitions, Doors, Curtains, &c., with 100 Full-page Plates.
2 vols. Large 8vo, cloth, 25s. net.

STRESSES AND THRUSTS. A Text-Book on their
Determination in Constructional Work, with Examples
OF the Design of Girders and Roofs, for the use of
Students. By G. A. T. Middleton, A.R.I.B.A. Third Edi-
tion, revised and much enlarged. With 170 Illustrative
Diagrams and Folding Plates. 8vo, cloth, 4s. 6d. net.

" The student of building construction will find all he ought to know as to the relation
of stresses and thrusts to the work he may be engaged in. The varying degrees of streea
are calculated in a simple way, so that the merest tyro in mathematics will be able to
appreciate and apply the principles laid down." — The Surveyor.

THE ELEMENTARY PRINCIPLES OF GRAPHIC

Statics. Specially prepared for the use of Students enter-
ing for the Examinations in Building Construction of the Board
of Education. By Edward Hardy, Teacher of Building Con-
struction. With 150 Illustrations. Crown 8vo, cloth, 3«. net.

Prof. Henry Adams, writing to the Author, says : — " You have treated the subject
in a rery clear and logical manner, and I shall certainly recommend the book to my
elementary students as the best of its kind."

TREATISE ON SHORING AND UNDERPINNING
and generally dealing with Dangerous Structures.
By C. H. Stock, Architect and Surveyor. Third Edition,
thoroughly revised by F. R. Farrow, F.R.I.B.A., fully
illustrated. Large 8vo, cloth, 4s. 6c?.

"The treatise is a valuable addition to the practical library of the architect and
builder, and we heartily recommend it to all readers." — Building News.

DANGEROUS STRUCTURES and How to Deal with
them. A Handbook for Practical Men. By G. H. Blagrove,
Certified Surveyor under the London Building Act, 1894.
Second Edition, re-written and much enlarged, With 35
Illustrations. Crown 8vo, 4s. 6d. net.

This volume deals with some of those awkward problems in
building which demand prompt solution. It describes ready
means for getting over difiiculties which frequently occur in
practice, and supplies data from which efficient, and at the same
time economical, remedies may be designed to counteract evils
arising from structural defects.

SCAFFOLDING : A Treatise on the Design and Erec-
tion of Scaffolds, Gantries, and Stagings, with an
account of the Appliances used in connection therewith,
and a Chapter on the Legal Aspect of the Question. By
A. G. H. Thatcher, Building Surveyor. Illustrated by 146
Diagrams and 6 Full-page Plates. Large 8vo, cloth, 6s. net.

" A really valuable little treatise."— TAe Builder.

CONCRETE: ITS USE IN BUILDING. By Thomas
Potter. Second Edition, greatly enlarged. 500 pp. of Text,
and 100 Illustrations. 2 vols., crown 8vo, cloth, 7s. 6d.

THE DRAINAGE OF TOWN AND COUNTRY

Houses. A Practical Account of Modern Sanitary Ar-
rangements and Fittings for the Use of Architects, Builders,
Sanitary Inspectors, and those preparing for examinations
in Sanitary Science. By G. A. T. Middleton, A.R.I.B.A.
With a special chapter on the Disposal of Sewage on a small
scale, including a description of the Bacterial Method. With
93 Illustrations. Large 8vo, cloth, 4s. Qd. net.

" A very complete exposition of the principles and details of modem practice in this
branch of design and work. ... It will well repay consultation by every one called
upon to deal with the problem of domestic sanitation from the constructional side." —
The Surveyor.

" Very reliable and practical." — The Plumber and Decorator.

THE PLUMBER AND SANITARY HOUSES. A Prac-
tical Treatise on the Principles of Internal Plumbing Work.
By S. Stevens Hbllybr. Sixth Edition, revised and enlarged.
Containing 30 lithographic Plates and 262 woodcut Illustra-
tions. Thick royal 8vo, cloth, 12s. 6d.

PRINCIPLES AND PRACTICE OF PLUMBING. By

S. Stevens Hellyer. Fifth Edition. Containing 294 pp. of
text and 180 practical Illustrations. Crown 8vo, cloth, 5s.

TECHNICAL PLUMBING. A Handbook for Students and
Practical Men. By S. Barlow Bennett, Lecturer on Sanitary
Engineering to the Durham County Council. Second Edition,
revised, with about 500 Illustrations. Large 8vo, cloth,
3s. 6d. net.
Entirely New and Improved Edition, superseding all previoics issues.

CLARKE'S TABLES AND MEMORANDAT^FOR

Plumbers, Builders, Sanitary Engineers, &c. By J.

Wright Clarke, M.S.I. With a new section of Electrical
Memoranda. 312 pages, small pocket size, leather. Is. 6d. net.

"It is obviously one of those things a tradesman should carry in his pocket as
religiously as he does a foot rule." — Plumber and Decorator.

"The amount of information this excellent little work contains is marvellous."—
Sanitary Record.

PRACTICAL SCIENCE FOR PLUMBERS AND

Engineering Students. By J. Wright Clarke. Treat-
ing of Physics, Metals, Hydraulics, Heat, Temperature, <kc.,
and their application to the problems of practical work.
With about 200 Illustrations. Large 8vo, cloth, 5s. net.

PUMPS: Their Principles and Construction. By J.

Wright Clarke. With 73 Illustrations. Second Edition,
thoroughly revised. 8vo, cloth, 3s. 6(i. net.

HYDRAULIC RAMS: Their Principles and Construc-
tion. By J. Wright Clarke. With 36 Diagrams. 8vo, 2«.

A thoroughly comprehensive and practical Treatise.
SANITARY ENGINEERING. A Practical Treatise on the
Collection, Removal and Final Disposal of Sewage, and the
Design and Construction of Works of Drainage and Sewerage,
with special chapters on the Disposal of House Refuse and
Sewage Sludge, and numerous Hydraulic Tables, Formulae
and Memoranda, including an extensive Series of Tables of
Velocity and Discharge of Pipes and Sewers. By Colonel
E. C. S. Moore, R.E., M.R.S.I. Second Edition, thoroughly
revised and greatly enlarged. Containing 830 pp. of Text,
140 Tables, 860 Illustrations, including 92 large Folding
Plates. Large thick 8vo, cloth, 32s. net.

" . . . The book is indeed a full and complete epitome of the latest practice in
ganitary engineering, and as a book of reference it is simply indispensable."— T^e Public
Health Engineer.

WATERWORKS DISTRIBUTION. A Practical Guide to
the Laying Out of Systems of distributing Mains for the
Supply of Water to Cities and Towns. By J. A. McPherson,
A.M. Inst. C.E. Fully illustrated by 19 Diagrams and 103
other Illustrations, together with a Large (^hart (29" x 20")
of an Example District. Second Edition, revised and en-
larged with further Diagrams. Large crown 8vo, cloth, 6s. net.

GAS FITTING. A Practical Handbook relating to the
Distribution of Gas in Service Pipes, the Use of Coal Gas,
and the best Means of Economizing Gas from Main to
Burner. By Walter Grafton, F.C.S., Chemist at the
Beckton Works of the Gas Light and Coke Co. With 143
Illustrations. Large crown 8vo, cloth, 5s. net.

" The author is a recognised authority upon the subject of gas lighting, and gasfittera
and others who intend to study gasfitting in practical detail will find the book most
serviceable."— r^e Builder.

STABLE BUILDING AND STABLE FITTING. A

Handbook for the Use of Architects, Builders, and Horse
Owners. By Byng Giraud, Architect. With 56 Plates and
72 Illustrations in the Text. Crown 8vo, cloth, 7s. 6c?.

Adopted as the Text-book by the Surveyors' Institution.
FARM BUILDINGS: Their Construction and Arrangement.
By A. Dudley Clarke, F.S.I. With chapters on Cottages,
Homesteads for Small Holdings, Iron and Wood Roofs, Repairs
and Materials, Notes on Sanitary Matters, &c. Third
Edition, revised and enlarged. With 52 full-page and other
Illustrations of plans, elevations, sections, details of construc-
tion, ifec. Crown 8vo, cloth, 6s. net.

" To architects and surveyors, whose lot it may be to plan or modify buildings of the
kind, the volume will be of singular service." — Builder' s Journal.

RESIDENTIAL FLATS OF ALL CLASSES, including

Artisans' Dwellings. A Practical Treatise on their
Planning and Arrangement, together with chapters on their
History, Financial Matters, (fee. With numerous Illustra-
tions. By Sydney Perks, F.R.I.B.A., P.A.S.I. vVith a large
number of plans of important Examples by leading architects
in England, the Continent, and America; also numerous
Views from Special Photographs. Containing 300 pages,
with 226 Illustrations. Imperial 8vo, cloth, 21s. net.

"A standard -work of considerable importance." — The Building News.
" Altogether it is a book which is not only unique in architectural literature, but is
one of which every page has a practical tendency." — The Architect.

MODERN SCHOOL BUILDINGS, Elementary and
Secondary. A Treatise on the Planning, Arrangement and
Fitting of Day and Boarding Schools. With special chapters
on the Treatment of Class-Rooms, Lighting, Warming, Ventila-
tion and Sanitation. By Felix Clay, B.A., Architect. Second
Edition, thoroughly revised and enlarged. Containing 556
pp. with 450 Illustrations of Plans, Perspective Views, Con-
structive Details and Fittings. Imperial 8vo, cloth, 25s. net.

"Mr. Clay has produced a work of real and lasting value. It reflects great credit on
his industry, ability, and judgment, and is likely to remain for some time the leading
work on the architectural requirements of secondary education."— TAe Builder.

PUBLIC BATHS AND WASH-HOUSES. A Treatise
on their Planning, Design, Arrangement and Fitting ; with
chapters on Turkish, Russian, and other special Baths, Public
Laundries, Engineering, Heating, Water Supply, &c. By
A. W. S. Ckoss, M.A., F.R.I.B.A. 284 pages, with 274 illus-
trations of modern examples. Imperial 8vo, cloth, 21s. net.

PUBLIC LIBRARIES. A Treatise on their Design, Con-
struction, and Fittings, with a Chapter on the Principles of
Planning, and a Summary of the Law. By Amian L.
Champneys, B.A., Architect. Containing about 200 pages,
with over 100 Illustrations of Modern Examples and Fittings
from Photographs and Drawings. Imperial 8vo, 12s. 6d. net.

THE PRINCIPLES OF PLANNING. An Analytical
Treatise for the Use of Architects and others. By Percy
L. Marks, Architect. With Notes on the Requirements of
Different Classes of Buildings. Illustrated by 150 Plans,
mainly of important modern Buildings. Second edition,
revised and enlarged. Large 8vo, cloth, 8s. Qd. net.

" For a single-handed attempt to grapple with such a widely extending subject, the
author has really done very well. Labour he has clearly not stinted, and his success in
arranging his large amount of material is worthy of much praise." — The Builder.

ARCHITECTURAL SKETCHING AND DRAWING
IN PERSPECTIVE. A progressive series of 36 Plates,
illustrating the Drawing of Architectural Details and Sketch-
ing to Scale ; including chapters on the Plan and Measuring
Point Methods, the Simplification of Perspective by R's
method, and on Figures, Foliage, &c. By H. W. Roberts,
Author of "R's Method." Large imperial 8vo, cloth,
7s. 6d. net.

This book provides a progressive course of perspective drawing,
founded to some extent upon the well-known R's Method, showing
its application to various problems of practical work. Its aim is to
present perspective drawing in a simple form, and to aid the draughts-
man by placing at his disposal various practical expedients to simplify
the details of his work.

THE PRINCIPLES OF ARCHITECTURAL PER-
SPECTIVE, prepared for the use of Students, &c. with
chapters on Isometric Drawing and the Preparation of
Finished Perspectives. By G. A. T. Middleton, A.R.LB.A.
Illustrated with 51 Diagrams, and 9 full-page and folding
Plates, including a series of finished perspective views of build-
ings by various Architects. Demy 8vo, cloth, 2s. 6d. net.

ARCHITECTURAL DRAWING. A Text-Book with special
reference to artistic design. By R. Phene Spiers, F.S.A.
With 28 full-page and folding Plates. 4to, cloth, 8s. 6d. net.

ALPHABETS OLD AND NEW. Containing 200 com-
plete Alphabets, 30 series of Numerals, and numerous fac-
similes of Ancient Dates, &c., with an Essay on Art in the
Alphabet. By Lewis F. Day. Second Edition, revised, with
many new examples. Crown 8vo, cloth, 3s. 6d. net.

" Everyone who employs practical lettering will be grateful for ' Alphabets, Old and
New,' Mr. Day has written a scholarly and pithy introduction, and contributes some
beautiful alphabets of his own design." — The Art Journal.

A HANDBOOK OF ORNAMENT. With 300 Plates,
containing about 3,000 Illustrations of the Elements and the
application of Decoration to Objects. By F. S. Meyer,
Third Edition, revised. Thick 8vo, cloth, 12s. 6d.

" A Library, a Museum, an Encyclopaedia, and an Art School in one. The work is
practically an epitome of a hundred Works on Design."— The Studio.

A HANDBOOK OF ART SMITHING. For the use of

Practical Smiths, Designers, Architects, (fee. By F. S.
Meyer. With an Introduction by J. Starkie Gardner.
Containing 214 Illustrations. Demy 8vo, cloth, 65.

" An excellent, clear, intelligent, and, so far as its size permits, complete account
of the craft of working in iron for decorative purposes."— 77i« Athen(Tum.

HOMES FOR THE COUNTRY. A Collection of Designs
and Examples of recently executed works. By R, A. Briggs,
Architect, F.R.I.B.A, Soane Medallist, Author of " Bunga-
lows." Containing 48 full-page Plates of Exterior and Interior
Views and Plans. With descriptive notes. Demy 4to, cloth,
10s. 6d. net.

" The arrangement of the plans generally reveala a masterhand at this class of archi-
tecture."—jTAe Fall Mall Gazette.

BUNGALOWS AND COUNTRY RESIDENCES. A

Series of Designs and Examples of recently executed works.
By B. A. Briggs, F.R.I.B.A. Fifth and Enlarged Edition,
containing 47 Plates, with descriptions, and notes of cost of
each house. Demy 4to, cloth, 12s. Qd.

" Those who desire grace and originality in their suburban dwellings might take
many a valuable hint from this book." — The Times.

IB A BOOK OF COUNTRY HOUSES. Containing 62 Plates
^^M reproduced from Photographs and Drawings of Perspective
^^K Views and Plans of a variety of executed examples, ranging
^^H in size from a moderate-sized Suburban House to a fairly
^^H large Mansion. By Ernest Newton, Architect. Imperial
^^B 4to, cloth, 21s. net.

^^^K The houses illustrated in this volume may be taken as representa-

^^Hf tive of the English Country House of the present day. They offer

V^Kl much variety in their size, their sites, the character of the materials

I^^P in which they are constructed, and their types of plan.

THE COUNTRY HOUSE. A Practical Manual of the
Planning and Construction of Country Homes and their
Surroundings. By Charles E. Hooper. Containing 350 pp.,
with about 400 Illustrations, comprising photographic views^
plans, details, &g. Crown 4to, cloth, 15s. net.

This volume afifords hints and pi-actical advice on the selection of the site, the
planning, the practical details of construction and sanitation, the artistic treatment of the
interior, and the laj'ing-out of the grounds. Although written by an American for
Americans, there is a great deal which is particularly applicable to English homes, and
much of the architecture illustrated is strongly reminiscent of the work of some of our
best Enghsh architects.

MODERN COTTAGE ARCHITECTURE, Illustrated
from Works of well-known Architects. Edited,
with an Essay on Cottage Building, and descriptive notes on
the subjects, by Maurice B. Adams, F.R.I.B.A. Containing
50 plates of Perspective Views and Plans of the best types
of English Country Cottages. Royal 4to, cloth, lOs. 6d. net.

■ "The cottages which Mr. Adams has selected would do credit to any estate ir»
England."— r/ie Architect.
"It should meet with a large sale. The author has been wise enough to get together
a varied style of design by various architects who have shown marked ability in this
direotion."— TAt JBritifth Architect.

MODERN SUBURBAN HOUSES. A Series of Examples
erected at Hampstead, Bickley, and in Surrey, from designs
by C. H. B. QuENNELL, Architect. Containing 44 Plates of
Exterior and Interior Views, reproduced from special photo-
graphs, and large scale plans from the author's drawings.
Large 4to, cloth, 16s. net.

Cleverly planned, of quiet refined design, and financially successful,
Mr. Quennell's examples clearly demonstrate that it is not necessary
to rely on characterless designs and stock patterns for our suburban
houses, as is often the case with the speculative builder.

MODERN HOUSING IN TOWN AND COUNTRY.

Illustrated by examples of mmiicipal and other schemes of
Block Dwellings, Tenement Houses, Model Cottages and
Villages, and the Garden City, together with the Plans and
other illustrations of the Cottages designed for the Cheap
Cottages Exhibition. By James Cornes. With many Plans
and Views from Drawings and Photographs, accompanied by
descriptive text. Royal 4to, cloth, 7s. 6c/. net.

"Its value is great. Its size enables the illustrations to be satisfactory in scale; its
price, for a book so copiously illustrated, is surprisingly low ; it will, doubtless, be accepted
for some time to come as a standard book of reference on the subject."'— T^e Times.

HOUSES FOR THE WORKING CLASSES. Com-
prising 52 typical and improved Plans, arranged in groups,
with elevations for each group, block plans, and details. By
S. W. Cranfield, A.R.I.B.A , and H. I. Potter, A.R.I.B.A.
With descriptive text, including Notes on the Treatment and
Planning of Small Houses, Tables of Sizes of Rooms, Cubic
Contents, Cost, &c. Second Edition, revised and enlarged,
with many additional plans. Imperial 4to, cloth, 21s. net.

This book deals with Cottages suitable for the Working Classes in
Suburban and Eural Districts. The majority of the examples illustrated
consist of two and three-storey dwellings, adapted to be built in pairs,
groups or terraces, and vary in cost from about £1.50 to £400.

" As a book of types of the best examples of houses of this kind, the work is the most
complete we have seen." — The Building News.

" The book meets a distinct want. The subject is not written round, but thoroughly
threshed out ; and what with good illustrations to scale, clear letterpress, and abundant
tables of areas, &c., there is no lack of information for those in search of it. We con-
gratulate the authors on their enterprise." — The Sun-eyor.

THE MODEL VILLAGE AND ITS COTTAGES:
BOURNVILLE. Illustrated by 57 Plates, consisting of
38 views from specially taken photographs, and 19 plans and
details of the Village and its Cottages, with a descriptive
account, and some notes on economic Cottage-building and
the laying-out of Model Villages. By W. Alexander Harvey,
Architect. Large 8vo, cloth, 8s. 6c/. net.

A HISTORY OF ARCHITECTURE ON THE COM-
PARATIVE METHOD for the Student, Craftsman,
and Amateur. By Banister Fletcher, F.R.LB.A., late
Professor of Architecture in King's College, London, and
Banister F. Fletcher, F.R.I.B.A. Containing 800 pp.,
with 300 full-page Illustrations, reproduced from photo-
graphs of Buildings and from specially prepared drawings of
constructive detail and ornament, comprising over 2,000
Illustrations. Fifth Edition, thoroughly revised and greatly
enlarged. Demy 8vo, cloth, 2l5. net.

" Par excellence The Student's Manual of the History of Architecture." — The
Architect.

". . . It is concisely written and profusely illustrated by plates of all the typical build-
ings of each country and period. . . . Will FILL A VOID IN OUR Literature."— ^uiYdm^'jVews.

"... As COMPLETE AS it WELL CAN BE." — The Times.

THE ORDERS OF ARCHITECTURE. Greek, Roman

iand Italian. A selection of typical examples from Nor-
mand's Parallels and other Authorities, with notes on the
Origin and Development of the Classic Orders and descrip-
tions of the plates, by B. Phene Spiers, F.S.A., Master of
the Architectural School of the Royal Academy. Fourth
Edition, revised and enlarged, containing 27 full-page Plates,
seven of which have been specially prepared for the work.
Imperial 4to, cloth, IO5. M.
"An indispensable possession to all students of architecture." — The Architect.

THE ARCHITECTURE OF GREECE AND ROME.

I A Sketch of its Historic Development. By W. J.
Anderson, Author of " The Architecture of the Renaissance
in Italy," and R. Phene Spiers, F.S.A. Containing 300
pages of text, and 185 Illustrations from photographs and
drawings, including 43 full-page Plates, of which 27 are
finely printed in collotype. Large 8vo, cloth, 18s. net.
" It is such a work as many students of architecture and the classics have vainly
earned for, and lost precious years in supplying its place." — The Architect.

" The whole conveys a vivid and scholarly picture of classic ai-t." — The British Architect.

THE ARCHITECTURE OF THE RENAISSANCE IN
ITALY. A General View for the Use of Students and
Others. By William J. Anderson, A.R.I.B.A. Third Edition,
containing 64 full-page Plates, mostly reproduced from
Photographs, and 98 Illustrations in the Text. Large 8vo,
cloth, 12s. 6d net.

"A delightful and scholarly book, which should prove a boon to architects and
idents. "—.^ourwaZ R.I.B.A.

'Should rank amongst the best architectural writings of the day."— r/^e Edinburgh
ieview.

A NEW AND EPOOH-MAKINQ BOOK.

GOTHIC ARCHITECTURE IN ENGLAND. An

Analysis of the origin and development of English Church
Architecture, from the Norman Conquest to the Dissolution
of the Monasteries. By Francis Bond, M.A., Hon. A.R.I.B.A.,
Containing 800 pp., with 1,254 Illustrations, comprising 785
photographs, sketches, and measured drawings, and 469 plans,
sections, diagrams, and moldings. Imperial 8vo. 31s. ^d. net.

"The fullest and most complete illustrated treatise on the subject which. has yet
appeared. ... It is a book which every student of architecture, professional or amateur,
ought to have." — The Builder.

" Perfectly orderly, and most complete and thorough, this g^reat book leaves nothing
to be desired," — The Building News.

"It brings the study of architecture up to the standard of modem ideals, and should,
we expect, long remain the best book of its kind in the language."— TA^ British Architect.

EARLY RENAISSANCE ARCHITECTURE IN ENG-
LAND. An Historical and Descriptive Account of the
Tudor, Elizabethan and Jacobean Periods, 1500 — 1625. By
J. Alfred Gotch, F.S.A. With 88 photographic and other
Plates and 230 Illustrations in the Text from Drawings and
Photographs. Large 8vo, cloth, 21s. net.

•' A more delightful book for the architect it would be hard to find. The author's
well-chosen illustrations and careful, well-written descriptions hold one's interest over the
whole 266 pages of the book. Mr. Gotch shows how architecture developed from the pure
Gothic through Tudor, Elizabethan, and Jacobean phases, imtil the full Renaissance, when
classical features obtained the mastery over our English work. The book is quite a store-
house of reference and illustration, and should be quite indispensable to the architect's
library."— :/7u5 British Architect.

CLASSIC ARCHITECTURE. A Series of Ten Plates (size
20 in. X 15 in.) of examples of the Greek and Roman Orders,
with full details and a Selection of Classic Ornament. By
Charles F. Mitchell and George A. Mitchell, Lecturers
on Architecture, Regent Street Polytechnic, W. With de-
scriptive letterpress, in portfolio, price 6s. net, or the Set of
10 plates without text or portfolio, price 5s. net.

The Examples illustrated are as follows : — Plate I. — Doric Order
from the Parthenon, Athens. II.— Ionic Order from the Erechtheion,
Athens. III. Corinthian Order from the Monument of Lysicrates,
Athens. IV. — Tuscan Order, with Portion of Arcade, based upon the
design of Barozzi of Vignola. V. — Doric Order from the Theatre of
Marcellus, Rome. VI.— Ionic Order from the Temple of Fortuna
Virilis. VII.— Corinthian Order from the Temple of Castor and Pollux
(Jupiter Stator), Rome. VIII. — Composite Order from the Arch of
Septimius Severus, Rome. IX. — Examples of Greek Ornament. X. —
Typical Roman Ornament from Buildings in Rome.

B. T. BATSFORD, 94, high holborn, London.

TG
270

Hardy, Edward

of li! ^}^^^^tary principles
of graphic statics

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