THE
ELEMENTARY PRINCIPLES
OF
GRAPHIC STATICS
Th.
Elementary Principles
of
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
BY
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
7
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
10
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.
|
A: |
^SC :- |
^D |
|||||
|
■ ,. |
Mil Ml! |
\\\\\\\\\ |
h |
||||
|
30 20 |
n |
27 |
2 |
/ |
0 |
||
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0 |
1 |
/4 |
|||||
|
^ |
01 |
0 |
J |
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|
0 |
1 |
It |
\zz |
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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
11
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.
B
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
13
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
D
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
EF
(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
15
Take a scale such as V = 1 unit. Draw B G = I unit, and BD = 23 units (Fig. 9).
r
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)
BD
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
17
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.
18
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
19
^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
E
20
30
UO
SO
bo
10
|
nil |
1 |
1 |
1 |
1 |
1 |
1 |
\ / |
|
W 0 |
i |
2 |
J |
U |
5 |
(9 |
7 |
|
lllllllllll |
1 |
1 |
1 |
1 |
I |
1 |
h |
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.
A
Examples to Chapter I.
Ex. Ch. I.— Fig. 1.
^B
1. What does A B (Fig. 1) show :—
(a) With a scale of J'' = 1 unit.
(b) With a scale of f '' = 1 yard.
(c) With a scale of f = 1 ton. {d) With a scale of -// = 1000 lbs.
2. Taking a convenient scale, graphically determine the following : —
(a) The diagonal of a square whose sides are 36 ft.
EXAMPLES TO CHAPTER I
21
(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
B
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
A
C
Fig. 17. spring balances the student can prove for
EEACTION — MOMENTS
25
u
f
Fig. 1!
I
KS
5
I
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 ^
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
29
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.
I
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
30
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
31
) 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
|
A |
6 |
|
|
(n 0 |
X y p n |
|
|
Fig. 26. |
PAKALLEL FORCES
33
I
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
35
two scales — say, lineal scale J'' = 1 ft. and force scale t'' = 1 lb.
D
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 =— —
W
i
LEVERS
37
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.
38
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
39
(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
41
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
0,
V
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 ?
(c) What is the direction of the reactions ?
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
43
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
45
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.
h
FD is I of A D, and F E is 1 of G E, therefore we
46
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.
B
F
|
'■•'." |
1 |
_ . . . |
4 • |
|
|
• . |
\ |
, » |
||
|
■■■ |
' *->•<■ ' |
Y ■ |
||
|
■.,.ri |
. • • eg.- . |
"\- |
.iA * * - • |
|
|
H |
c |
Fig. 41. |
D |
Q |
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 |
|
|
\y |
F |
C |
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
49
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
B
cwts. 3
cwts.
b
D
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.
52
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
|
A |
tons. |
J |
B |
||||||
|
\ E ^ |
A |
G |
/ |
/\ |
/ |
/ |
|||
|
o\f |
\ |
/ |
H |
V |
' J' |
I |
|||
|
•^ |
" ci> re |
m. |
Cf |
c |
^c |
f |
n |
m |
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.
55
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.
.W
|
t |
||||
|
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. |
t3 5. |
2 6 cults, cures. |
/ cwt |
|
|
cu/ts Ex. Ch. III.- |
-Fig. |
cults.
EXAMPLES TO CHAPTER HI
57
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,
(0
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
69
{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
63
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).
c
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
c
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
65
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
67
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-
68
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.
<N
m
B
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
w
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
71
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
73
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
IS
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
74
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
2
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
a
tl8J'5 Ik
|
1 -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
77
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
I
/ >r 1000
Force. 0
1000
Uh
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
I
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
81
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
83
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
85
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.
86
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
87
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
X
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
89
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
D
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.
90
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
93
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
t
-•^
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
i
B
i
t
; 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.
W
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
95
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
h
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
97
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
A
nitons
Both cases are worked out, and it should be noticed that the result is the same in each case.
A
I
^m Fjg. 85 shows a Warren girder witli three loads on ^B the bottom boom.
L
98
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
99
This is spread over 12 spaces, so the weight of each
435
cwts.=:36J cwts.
12
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///////^////// ^//.
1
D
B
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
ft
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.
I
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:
t,
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 I |
mil \ |
Clfftt i |
|||
|
\^' |
C |
-*«- |
J' |
-»«- |
J' H |
|
Ex. |
Ch. |
v.— |
Fig. |
1. |
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
|
A |
\/ho*. V V \/ |
B |
|
|
P |
2 \tGns |
3 tons. |
M |
Ex. Ch. v.— Fig. 2.
9. Find the reactions due to the three loads shown in Fig 3.
|
5 |
tons. |
If |
tons. |
L |
tons |
||
|
A |
\ |
\ |
\ |
/ |
B |
||
|
M |
Wa |
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
i
BENDING MOMENT
111
bcufts.
Scwts.
|
A |
i |
5, |
I |
C |
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|||||
|
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i 1 |
D |
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||
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, ~\ |
--^ |
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^^ 1 |
Lvneal. mL
s b
I I I I
Scaks: Force. BM
w
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0 10 30 so 70
mrrTT
XI
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
BM
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.
(c) The three loads are A B, B G and G D, therefore
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.
y^
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.
it
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
F
B
D
F
Fig. 100.
clearly equal to the vertical force W.
Taking
i:
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\
a
I
w
Fig. 101.
iW
I
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
i
A iB ia A IBl
I
a
c b
d
AiSTCi
D
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.
P
iQQQQQQO
a
b
B
A
B
■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^^^[[|
h
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
I
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.
I
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.
i
B
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.
(c) The two vertical balances together always register a force equal to that of the weight.
{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^/ |
\C |
|
l^ ' |
\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.
a
Fig. 120.
(i
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.
a
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.
i
i
CX^D
B^l/J
i
F
|
k . |
b |
|
^^^f^ |
c |
|
^^^^^ |
|
|
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\ |
c |
1 |
|
i |
Yr^ |
E v |
/o |
|
i |
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.
a
|
F |
||||
|
'A |
1 |
A |
K |
/\m |
|
/" |
V |
J |
V |
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.
K
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.
a
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
r
I ^ 6%W6't
k-
Fig. 137,
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1 |
|
|
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\ k |
|
f |
\ |
kr\iK |
|
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/ |
y\/ |
|
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/ |
/" |
|
^ |
/. |
[n
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
I
(a) ^^ ib) ^
Fig. 138.
Fig. 139.
I
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.
B
|
I c |
> ' |
D |
i |
£ |
i |
F i |
|
^ |
t |
N |
t |
Q |
t |
G not
A
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 |
|
X |
XXX |
|
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(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.
fi
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 ' ^
B
|
'iTi |
Mi |
^A |
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.
Butler & Tanner, The Selwood Printing Works, Frome, and Londoa
i
j[ List of Standard Books
RELATING TO
BUILDING ARCHI- TECTURE, SANITATION ^ DECORATION
PUBLISHED AND SOLD BY
B. T. BATSFORD,
94, HIGH HOLBORN, LONDON,
I
A List of Standard Books
PUBLISHED BY
B. T. BATSFORD, 94, high holborn, London.
HOW TO ESTIMATE: being the Analysis of Builders'
Prices. A complete Guide to the Practice of Estimating,
and a Reference Book of Building Prices. By John T. Rea,
F.S.I., Surveyor. With typical examples in each trade, and
a large amount of useful information for the guidance of
Estimators, including thousands of prices. Second Edition,
revised and enlarged. Large 8vo, cloth, 7s. 6c/. net.
This work is the outcome of many years' experience in the personal
supervision of large contracts. It is applicable for pricing in any part of
the country, and is adaptable to every class of building and circumstance.
" The book is excellent in plan, thorough in execution, clear in exposition, and will be a boon alike to the raw student and to the experienced estimator. For the former it will be an invaluable instructor; for the latter a trustworthy remembrancer and an indispensable work of reference."— T/ie Building World.
BUILDING SPECIFICATIONS, for the use of Archi- tects, Surveyors, Builders, &,c. Comprising the Com- plete Specification of a Large House, also numerous clauses relating to special Classes of Buildings, as Warehouses, Shop- Fronts, Public Baths, Schools, Churches, Public Houses, &c., and Practical Notes on all Trades and Sections. By John Leaning, F.S.I. 650 pages, with 150 Illustrations. Large 8vo, cloth, 18s. net.
" Cannot but prove to be of the greatest assistance to the specification writer, whether architect or quantity surveyor, and we congratulate the author on the admirable manner in which he has dealt with fhesuhiect.'"— The Builder^ s Journal.
BUILDING MATERIALS: their Nature, Properties, and Manufacture. With chapters on Geology, Chemistry, and Physics. By G. A. T. Middleton, A.R.I.B.A., Author of " Stresses and Thrusts," &c. With 200 Illustrations and 12 full-page Photographic Plates. Large 8vo, cloth, 10s. net.
" The author has collected his materials with rare diligence, and has handled them with workmanlike skill and judgment; and it would be by no means surprising to find Middleton on Materials ' becoming as popular and as authoritative as ' Leaning on Quantities.' "—The Building World.
THE CONDUCT OF BUILDING WORK AND
the Duties of a Clerk of Works. A Handy Guide TO THE Superintendence of Building Operations. By J. Leaning, Author of " Quantity Surveying," <fec. Second Edition, revised. Small 8vo, cloth, 2s. M. net.
''This most admirable little volume should be read by all those who have charge of buildmg operations .... It deals in a concise form with many of the important pomts arismg during the erection of a building. "— The British Architect. B. 6. 07.
Seventh Edition, Thououghly Revised and greatly Enlarged. BUILDING CONSTRUCTION AND DRAWING.
A Text-Book on the Principles and Details of Modern Con- struction, for the use of Students and Practical Men. By Charles F. Mitchell, Lecturer on Building Construction at the Polytechnic Institute, London, assisted by George A. Mitchell. Part 1. — First Stage, or Elementary Course. 470 pp. of Text, with nearly 1,100 Illustrations, fully dimensioned. Crown 8vo, cloth, 3s.
"An excellent and trustworthy little treatise, prepared and illustrated
IK A VERY THOROUGH AND PRACTICAL SPIRIT."— TAe Builder.
" It seems to have most of the advantages of "Vols. 1 and ti of Rivington's ' Building Construction,' with the additional ones of cheapness and conciseness, and appears to be thoroughly practical."— J/r. J. T. Hurst, Author of the ^'■Surveyor's Handbook."
"A model of clearness and compression, well written and admirably illiistrated, and ought to be in the hands of every student of building construction." — I%e Builder.
Fifth Edition, Thoroughly Revised and greatly Enlarged. BUILDING CONSTRUCTION. A Text-Book on the Principles and Details of Modern Construction, for the use of Students and Practical Men. By Charles F. Mitchell, assisted by George A. Mitchell. Part 2. — Advanced and Honours Courses. Containing 800 pp. of Text, with 750 Illustrations, fully dimensioned, many being full-page or double plates, with constructional details. Crown 8vo, cloth, 5s. 6(/.
"Mr. Mitchell's two books form unquestionably the best guide to all the mechanical part of architecture which any student can obtain at the present moment. In fact, so far as it is possible for any one to compile a satisfactory treatise on building construction, Mr. Mitchell has performed the task as well as it can be performed." — The Builder.
BRICKWORK AND MASONRY. A Practical Text-book for Students and those engaged in the Design and Execution of Structures in Brick and Stone. By Charles F. Mitchell and George A. Mitchell. Being a thoroughly revised and remodelled edition of the chapters on these subjects from the Authors' "Elementary" and "Advanced Building Construc- tion," with special additional chapters and new Illustrations. 400 pp., with about 600 illustrations. Crown 8vo, cloth, 5^.
" Regarded in its entirety, this is a most valuable work. It is not a treatise, as the term is generally understood, but a compendium of useful information admirably collated and well ilhxstrated, and as such has a distinct sphere of usefulness."— T/ie Builder.
FORTY PLATES ON BUILDING CONSTRUCTION.
— Including Brickwork, Masonry, Carpentry, Joinery, Plumb- ing, Constructional Ironwork, &c., &c. By C. F. Mitchell. Revised by Technical Teachers at the Polytechnic Institute. The size of each Plate is 20 in. by 12 in. Price, in sheets, 55. Od. Or bound in cloth, 10s. Gc?.
DRY ROT IN TIMBER. By W. H. Bidlake, A.R.I.B.A. With numerous Diagrams. 8vo, cloth, Is. 6o?.
A71 entirely Hew and Up-fo-Date Treatise, coMaiaing the results of a Unique Practical Experience of Tiventy-five Years,
MODERN PRACTICAL CARPENTRY. By George Ellis, Author of "Modern Practical Joinery," &c. Containing a full description of the methods of Constructing and Erecting Roofs, Floors, Partitions, Scaffolding, Shoring, Centering, 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.
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" 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.
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" 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.
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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.
I
L
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.
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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.
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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.
I
I
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.
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PRACTICAL SCIENCE FOR PLUMBERS AND
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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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