VAN NOThAND'S SCIENGE SERIES. 
16.
A GRAPHIC IMETHOD FOR SOLVING CERTAIN ALGEBRAIC EQUATIONS. By Prof.
GEORGE L. VOSE. With Illustrations.
17,
WATER AND WATER SUPPLY.         By Prof.
W. H. CORFIELD, M.A., of the University College, London.
18.
SEWERAGE AND SEWAGE UTILIZATION.
By Prof. W. H. CORFIELD, M.A., of the University College, London.
19.
STRENGTH    OF  BEAMS UNDER      TRANSVERSE LOADS. By Prof. W. ALLAN, author
of''Theory of Arches." With Illustrations.
18 1oo, boards 50 cents each.'Se Sent free by mail on receipt of price.








STRENGTH OF BEAMS
UNDER
TRANSVERSE LOADS.
BY
PROF. WV. ALLAN.
FORMERLY OF WASHINGTON AND LEE UNIVERSITY,
LEXINGTON, VA.
NEW YORK:
D. VAN NOSTRAND, PUBLISHER,
23 MURRAY AND 27 WARREN STREET.
1 8 75.








PREFACE.
The usual discussion of stress in beams
loaded transversely, involving as it does the
calculus, is unintelligible to that large class
of builders, and others, whose mathematical
training has not extended beyond the elements
of conic-sections. Yet, the most important
cases may be explained without using the
higher mathematics.
The aim of the following pages is to put into
brief and convenient shape, without resorting
to the higher mathematics, the discussion of
the most important and common cases of
horizontal beams under vertical loads.  In
beams of rectangular cross-section the formulha
given are exact under the assumptions generally made in regard to the laws of elasticity.
In flanged beams the approximations made use
of are those adopted in ordinary practice.
A graphic method of determining the amount
of resistance is added, which is applicable to all
cross sections, and which, if used with ordinary
skill and care, will give results exact enough




4
for every purpose. This method is so simple,
and of such general application, that it commends itself to practical builders. It is easily
understood, and requires no other appliances
but those in the hands of every draughtsman.
In the case of unsymmetrical cross-sections,
(as for instance, iron rails), it affords an easy
means of obtaining sufficiently accurate results,
without resorting to the tedious calculations
otherwise necessary.
This discussion (in which there is no attempt
at originality) was prepared for the use of the
intermediate classes in Engineering, at Washington and Lee University.  Having been
found serviceable to them, both during their
college career and in subsequent practice, I am
induced to hope that it may save time and
labor to those engaged in the same pursuits.
In preparing the manuscript for the press,
and correcting the proofs, I have been much
indebted to one of my former pupils, Mr.
Julius Kruttschnitt, C.E., now Instructor in the
McDcnogh School.
AUGUST, 1875.








ERRATA.
Page 14, in fig. 5, N and N,' should change
places.
" 19 and 53, figs. 8 and 35 should be
turned, so as to make A C vertical.
41, third line from top, add (Fig. 26) after C P.'
60, near bottom, Wx (l-x) should read
Wx
-   (1-x)
"  78, second line from top "Fig. (2)"
should read (Fig. 52).'   78, near bottom, read d = 5c. x, instead
of d = 5cx.'   86, fifth line from top, "proportional"
should be proportioned.
" 102, fourth line from  bottom, omit
"these."
"  111, second line from bottom, "those "
should be that.
" 112, second line from  bottom, "not"
should be cut; and in bottom line,
" beams " should be bases.




STRENGTH OF BEAMS
UNDER TRANSVERSE LOADS.
IN the following discussion the ordinary cases of loaded beams are treated
without resorting to the higher mathematics.  It is an attempt to compile
from various sources the simplest methods of treating such cases as arise most
frequently in practice.
Iransverse stress is produced by a load,
applied to a beam in a direction perpendicular, or inclined, to its length. A D
(Fig. 1) is a beam subjected to such a
stress.
In this kind of stress a compression of
the particles or fibres on one side of the




6
A —---------------------------------— oB                               D
FIG. 1.
beam, and an extension of those on the
other, are produced. In consequence of
this the beam bends. Experiment shows
that the amount of compression on the
the one side, and of extension on the
other, diminishes as we go inwards from
the top or bottom towards the centre,
and at some intermediate plane, 00',
becomes zero. The fibres at this plane
being neither lengthened nor shortened,
it is called the neutral plane, and its intersection by the plane of vertical section is called the neutral line or axis.
Experiment also shows that, from this
plane towards the top and bottom, the
amount of extension and compression
may, for the stresses that occur in ordi



7
nary practice, be considered as varying
directly with the distance from the
neutral plane.
The extreme top and bottom fibres
suffer the greatest compression and extension, and in case of rupture, the rupture begins with them. Some question
exists as to the exact location of the
neutral line or plane, but for slight deflections it passes through the centre of
gravity of the cross section of the beam,
and it is very probable that it never
deviates from this position.
In discussing transverse stress, the assumptions based upon experiment may
be stated as follows:
1. The forces on the fibres are directly
as the amount of extension or compression they produce; (Ut tensio sic vis,)
and since the extension and compression
increase as the distance from the neutral
axis, the forces vary in the same proportion.
2. Within elastic limits the extension
and compression at equal distances from




8
the neutral axis are equal, and the forces
producing them are equal.
3. The neutral axis passes through the
centre of gravity of the cross section.
RECTANGULAR BEAMS.
Let us now discuss the relations existing between the forces in, and on, transversely loaded rectangular beams, the
load being supposed to be vertical in
direction and the beam horizontal. "
Case I.
Let AD  (Fig. 2) be a beam so thin
that it may be considered as composed
of but one layer of fibres or particles.
Let it be fastened in a wall at A B, and
be loaded with W' at the other end.
Neglect for the time the weight of the
thin beam itself, which is small. Imagine it to be cut by a vertical plane E F
at any point, and let us see under what
forces the part ED is held in equilibrium.
The only external force on E D is W'
acting at D downwards, and E D is pre



9
vented from falling under this weight by
the resistance of the fibres at E F. To
analyze these forces, let us take 0 as an
origin of coordinates, and 0 0' as the
axis of x, and 0 E as the axis of y, and
as the forces are all in one plane, find
their components along these axes. The
internal forces, or resistance of the o fibres
at E F, are:
S 
at E F. are:




10
1. The horizontal forces which are
tensile above and compressive below, and
which increase from zero at O0 just in
proportion as we go from that point
towards the upper or lower edge of the
beam. (Fig. 3.), T
FIG. 3.
2. The vertical force. This is called
the shearing force, or transverse shearing
force. It resists the tendency of the
part of the beam E D to slide down on
the surface EF. The existence of this
force may be realized if we conceive the
beam to be divided into two parts by the
vertical plane E F, and those parts to be
united by some very elastic substance,




11
as india-rubber. Then the beam would
take the form shown in Fig. 4,'the part
<'f
F C sliding down on the other. The
force in the beam that resists this sliding
is represented in Fig. 3, by the vertical
arrow at E. Let it be called T. In Fig.
3 are represented all the forces we have
to deal with. Since this system of forces




12
is balanced, the following equations
must be fulfilled:
2X=o 2Y=o 2M=o               (1)
That is: the sum of all the horizontal
forces (2 X), and the sum of all the vertical forces (2Y) must each be equal to
zero, and the sum of the moments about
any point as 0, must also equal zero.
The only horizontal forces in the system are the two triangular groups of
forces E 01 H and F 01 L, representing
the sum of the tensile and compressive
stresses on the fibres.  As the group
E 01 H acts in a direction opposite to
that of the group FO1L, and as the
algebraic sum of the two groups is zero
(2 X=o), the groups must be equal to
each other.  This is indicated in the
figure by the equality of the triangles
E O H and FO, L.
The vertical forces are W' and the
shearing force T at E F, and since
Y=T-W'=O.          We have
T=W'                  (2)




13
Next obtain the moments of all the
forces about 0, and place the sum of
these moments = zero.    Replace the
tensile and compressive forces by their
resultants. The resultant or sum of all
the tensile forces represented by the
triangle E 01 H (Fig. 3) may evidently
be represented by the area of the triangle
of which the base E 0O is the distance
over which the forces are distributed,
and the altitude E H is the stress in the
outside fibre. Let
S = this stress = EH
and   d~=E F= depth of beam
Then area E 01 H=  S d=N = resultant
of tensile forces. Similarly
Area F O, L=I S d=N'= resultant of
compressive forces.
These resultants will pass through
the centres of gravity of the triangles
E 01 H and 01 F L, since the little forces
of which they are composed are represented by these triangles.  Hence the
direction of Nwill intersect 01 E at a
point G (Fig. 5), whose distance from 0,




14
li
X d
IS =-  E O —. 2-d. This is the lever
arm of N about 0,. That of Kf is G' 01
d
also= -. Hence the sum of the moments
3
of these two forces about 01 (since they
both tend to produce left-handed rotation) is
d     Id    1
-N — N'          Sd2
3     3     6




15
The force T since its direction passes
through 01 has no lever-arm, and hence
its moment is zero.
If the distance from O' to 01 be called
x, the moment of the weight W' is
= + W'x (since it tends to produce
right-handed rotation).W'x — S d'=2     M=O.W' x=- S d          (3)
So far we have considered a beam
whose breadth is that of only one row
of fibres, but a beam of any breadth
may be made up of a number of such
slices placed side by side, and if b = the
number of slices, or breadth of the
beam, and W = the weight hung at the
end of it, then eq. (3) becomes
1
W x=-S b d           (4)
6
This discussion is general and will
apply to any section as well as to E F. S
and x are the variables in eq. (4), and
these quantities will have different values
at the different sections, which values in



16
crease as we go towards A B (Fig. 2),
but the form of the equation will evidently be unchanged. If A C (Fig. 2)
be = 1, we have for the section at A B
(Fig. 2)
Wl= -Sbd              (5)
A B is the section of greatest stress, and
the beam if overloaded will break there.
The quantity - S b d is called the moment of resistance of the fibres, or moment of the internal forces, and is often
written M for brevity.  Wx is called
the moment of the weight, or moment
of the external forces. Let the maximum value of - S bd2 (eq. 5) be called
6
Mo.
We may illustrate geometrically the
variation of the moments M=W x, and
consequently of the stresses produced on
the outside fibres from A to C.
In Fig. 6 let A C be the beam. Take
a line on some scale to represent the




17
FIG. 6.
value of Mo=W 1, and lay it off from A
perpendicular to A C. Let A L be this
line. Draw LC. Then the dotted perpendiculars in the triangle L A C will represent the moments of resistance in the
beam  at the several points at which
they are drawn.
From eq. (2) it is seen that the shearing force is constant at every section of
the beam. This force we may assume
with sufficient accuracy, for our present
purpose, to be uniformly distributed over
the cross section of the beam on which
it acts. Hence if A - area of cross sec



18
tion, and t = shearing force on a unit of
the surface,
T=W=At              (6)
Lay off AC (Fig. 7) =  and CP=W.
point of the beam.
Corollary. When several weights as
the beam at different points, the moment
of resist e tane  y point is equal to the
joint moments of the weights at that
point. Thus, calling distances measured
from C, G, and E towards A, x, x and x2




19
respectively, we have for the equation of
moments for points between C and G
W x- S bd'      Between G and E
6      1
Wx+W1 x1=      S b d  Whileat K,
for instance, it is




20
Wx+W1 x+W2 x-        Sbd2   (7)
The shearing force at K is
T=W+W1+W,              (8)
Geometrically. Let A C (Fig. 8) = I
AG=l and AE=l,. Layoff AL=W,
LH=W1 1, and A I=W      1,. Draw the
triangles as in Fig. 8. Then NP=total
moment at K, for instance.
The shearing force is represented by
the rectangles AP, N 0, and ST (Fig.
9), and at any point in the beam is equal
to the sum of the weights between that
point and C.
EXAMPLES.
(1.) Suppose the safe stress per square
inch to be 1,000 lbs. (= S), and I =10
ft., b=3 inches, and d=12 inches, what
weight will the beam support?
(2.) Suppose W=- ton, 1=12 ft., b=2
inches, what must be the depth (d) of a
rectangular cast iron beam, so that S shall
not exceed 4 tons?




21
gI        J
Case II.
Let the beam be as in the last ease,
but with the load distributed uniformly
over it (Fig 10). Let w-weight on a
unit of length, W-total weight on A C,
-- A C= length, d=AB==depth, x=E C
as before. Then the forces to be considered are represented in Fig 11, the




22
little arrows along E C representing the
weight distributed along the beam.
jr
0Q            ^oi —--..QO...._..FIG. 10.
Replace the weights along E C by their
resultant, which is = wx, and which
should be applied at the middle point of
E C, since the little weights on the beam
are uniformly distributed. Then putting
the resultants N and N' in place of the
tensile and compressive force, and proceeding as before, we have
T-wx=0
T=wx                    (9)




23
x  1
Also        x. -- Sb cP  =o
2 6
2 --   Sbd' =M    (10)
2   6
At A B (Fig. 10) these equations become
T0=w_ W                )
-2    2   6S      M       (11)
By comparing the last equation with
eq. (5), we see that if the weight and
beam be the same the stress on the fibres
in this last case is only one-half what it
was in the former, -or, what amounts to
the same, the beam will bear twice as
much distributed over it, as it will when
the weight is concentrated at the extremity.
From eq. (9) we see that the shearing
force is not constant as in the last case,
but varies as x. It is greatest at A.
Geometrically. The equation M = X
wx2 corresponds with that of a parabola
with vertex at C and axis vertical. Lay
off AL (Fig. 12) =  w 1, and through
L and C draw a parabola. The ordinates




-7>
4 — >  14mmf..10;____FI.llvj H-   -G
AW  ~- 




25
of this parabola (dotted in the figure)
will represent the moments at the several
points.
-  t  i,
fi I i i,-'
L...........-..........
L.....................................i
FIG. 12.
The equation T=w z is represented by
the triangle A PC (Fig. 13), which
therefore gives the shearing stress at
every point of the beam when A P is
taken = w 1.
Corollary 1. When the load is distributed over only a part of the beam
as in Fig. (14), let R C=m=the loaded
part, and take the other letters as before.
Then the equations for any section in
the loaded part are evidently the same
as those just obtained, viz.:




26
FIG. 13.
~ad-  --- - ----   -    - 
~ p
a'*'*19 
53_^  1t 
~   A;
I g~




27
~w x-      d26
wx21=    Sbdq-M           (12)
At R        i wm2=M
And          T=wx
But at any section E F between A and
R  the moment of the load is = w m
(x —i m), the latter factor being the distance from the centre of gravity of the
load to the section E F. The moment of
resistance having the same form as before, we have for the equation of moments for any section in R A
wm (x-im) =6 Sbd =M          (13)
At A this becomes
w m (I — m,) =M0 the greatest moment.
The shearing force at E F being equal
in amount and opposite in direction to
the whole load between E F and C will
be
T==w m              (14)
Geometrically. For the moments: lay
off AC    I andCR  — m (Fig. 15). At
R erect D R =   2, and at A make AL




28
I            r
thy
M..  C a   D d    
-=Mo. Through C and D draw a parabola as in the last case, and (since eq.
[13] is of the first degree) through D
and L draw a straight line. Then from
C to R the ordinates of the parabola
represent the moments, and from R to
A they are represented by the ordinates
of the trapezoid R L. For the shearing
stress (Fig. 16), lay off from R, RN=wm.




29
\r
Draw C N and N P. Then the triangle
C RN (corresponding to the equation
T=w x) gives the shearing force at each
point in C R while the rectangle R P
(corresponding to the equation T=w m)
gives the force in the remaining segment
of the beam.
Corollary 2. When there is a load W
at the extremity C, in addition to the




30
load uniformly distributed over the beam
(Fig. 17), we have a combination of
A   -     _1j 9     AC
B ~
4,
FIG. 17.
Cases I. and II. and the moment of the
external forces at any section, E F, is
w?2 + w x
Hence the equation of moments is
MI= 6 Sb d2=  wx +Wx       (15)
This is greatest at A, or
M= — Sbd -2=  I+Wl         (16)
The shearing force
T=wx+W                     (17)
At A
T,=wo +W                   (18)




31
Geometrically. The simplest way of
representing the moments is to construct
those due to each kind of weight, and
t0  2
then combine them. Thus, let MI'= -
2
and M" =Wx. Construct M' as in Case
I., it being represented by a parabola
with vertex at C and axis vertical, and
M"as in Case I., it being represented by
a triangle (placed under AC for convenience). Since M == M+ M' from eq.
(15) we have the total moment at any
point E (Fig. 18) represented by the
sum of the ordinates of these two figures=N P Fig. (18.)
The moments may also be represented
by the parabola corresponding to eq.
(15) as in Fig. (19.) This parabola has
its vertex atC' and not at C. Of course,
only that part of the curve between C
and A is applicable to our purpose.
The shearing force is represented by
adding the triangle N PP' Fig. (20)
representing the variable part w x of T
to the rectangle C P which represents
the constant part W of T.




32
EXA__ PLE.
Discuss the forces when the load is
distributed as shown in Fig. (21) assuming various values for L N and N C
as well as for w and W.
Case III.
Let the beam whose length is I rest




upon supports at B and D (Fig. 22) and
let it be loaded at some point G with a
single weight W. Let m and n be the
segments into which the beam is divided
at the point G of the application of the
weight.
First find the proportions of the




34
weight supported at B and D, or in
other words, the reactions of the supports. By the principle of the lever the
respective portions of the weight supported at B and D are inversely proportional to the distances of these points
from G. Thus, let W'==reaction at D
W"1 + W' ~ W'::m?+n: M




C)~~~ I ic
I. _-  -k
0
^~ ~ ~~ ~~              g      c^i
1  
C1~~~~~3<       <<3.  
0  Volt  1^
14              c
0                  IAc




36
But W'+W"=W       and n+ n=l..W'=-W           )
1 "              (19)
And so      W"=- W
1
Now apply the conditions of equilibrium to any part A E of the beam counting from A. Fig. (23).
4PFZI~      c
}:




37
1st, Between A and G. 2 X==O merely
indicates the equality of N and N', as
these are the only horizontal forces.
2YY=O shows that the shearing force
at the section EF is downwards
and-W".. T-=W'=-W             (20)
1
2 M=O. The joint moment of N and
N' is as before=1 S b d. That of T is
zero. The only other force acting on
A E is W", the reaction of the abutment.
Let 0 0= x. Then the moment of W'"
is
W" x -- Wx
U        1
Hence    -- Wx —Sbd2 = 
IV      6
1       6.'tWx=-Sb — M (21)
2d, Between G and C (Fig. 24). Here,
between A and E are the two external
forces W' at B and W  at G. Hence the
shearing force at E F is upwards and




3.8
T==V"-W-=- -W'= ~ W.  (22)
For M —0 we have
-- S b d + W" x-W (x —n)=O
6
n-        1
IWx-W(x-6n)=Sd2==M (23)
^BB~~~o__^




39
The greatest value of eq. (21) is at G,
where it becomes identical with eq. (23)
for the same point. This value is
m. n 
m  W=     S b d2==M    (24)
6
At A and C the moments are zero.
-.. --. —- --
Geometrically. From eq. (21), which
is of the first degree, it is seen that the




40
moments vary in AG as they did in
Case I. Hence they may be represented
by the ordinates of the triangle A L G.
(Fig. 25).
Eq. (23) is also that of a straight line
cutting the axis of X at C. Hence the
moments in G C are represented by the
triangle GLC (Fig. 25). The shearing
_ 
cc...




41
force in A G is represented by the rectangle A P and that in G C by the rectangle CP'.
Note, That the maximum moment (at
G) corresponds to the point where the
shearing force passes through zero.
Corollary 1. When the weight W   is
at the middle of the beam we have
n -y m  -7I1
W' —W"-== W
Then eq. (21) becomes
}Wx=      Sbd'=M 
and (23) is                      (25)
tW(I-x)6SbG=M           }
W   (l- x)=_ S d==M
At the centre  M,==i W. I< W I (26)
The triangles A L G and G L C (Fig,
27) represent the moments in this case.
G L having been laid off=j; W 1.
The shearing force throughout the
beam is then T=- W    (27) as is shown
in the rectangles (Fig. 28).
Comparing the value of MIo given in
2




42?L —......
(26) we see that the load and the length
being the same, a beam will bear four
times as much with both ends supported,
and the load placed in the middle as it
will do with one end fixed and the other
loaded.
Corollary 2. When there are several
weights, as in Fig (29), the moment of




43
^ ___        W
the external forces at any section is that
due to the action of all the weights.
Let the segments into which the weights
W, WI and W2 divide the beam be m
and n for W, m    and n, for W1 and
mQ and n2 for W2.
Let R1 reaction of abutment at B and
R= reaction of abutment at D and
l= length and let x be counted from
A as before.




44
i
The reaction of the abutment at B
due to these weights is
R1=-W+LW +12W 
-1      1 w+T W,+ 12  (28)
So at D R2=  WM    +TlW   W +
For any section between A and G, R,




45
is the only external force and hence the
equation of moments is
R1 x=SS b d2= A       (29)
The shearing force is  T=R1    (30)
For every section between G and G'
the weight W is to be taken into consideration and the equations are:
R x-W(x-mo))=-6 S6d'=zM      (31)
T=RI-W              (32)
For any section between G' and G")
we have:
(33)
Rx-W(x- m)- Wl(x —m,)=-    S bd'-=M
T=R1-W -W1
For any section between G"' and C
R, x-W(x-mV)-W,(x-nm,)-W, )
(x-m-)=- Sbd2=IM           (34)
T=R,-W-W,-W2             J
The location of the greatest moment
is most readily determined by geometrical
construction.




46
Geometrically. The moments are represented in Fig. (30) by constructing....V...
separately those due to each weight and
then combining them. Thus, the moments produced at every point in the
beam by the weight W are represented




47
by the triangle AL C (Fig. 30), in which
G L equals the greatest moment due to
W=-   N Wm. Similarly A L' 0 represents
those due to W1, G' L' being=-'W1 mr
and A L" C gives those due to W,, G" L'
being= —2 W.n2
Now, if at every point we add together the ordinates of these three triangles for that point, and lay them off
above A C we shall get a polygon A H
H' HIt C, which represents eqs. (29) (31)
(33) and (34) and gives the total moment
at any section. The greatest ordinate of
this polygon will, of course, show the
location of the maximum moment. This
will be at G or G' or G, according to
the relative amounts and positions of
the weights W, W, and W. In the
fig. it is at G'. Hence from eq. (31)
R1, M — W(m1-m)== S b d==M, (35)
The shearing force may be represented




48
as in Fig. (31). It is greatest in that
__ _________
rt   e:
one of the two end segments which corresponds to the greater of the two quantities R1 and R2.
Note, That in this case the simplest
way of finding the point of maximum




49
moment is to construct the figure representing the shearing force, and the
point when the shearing force passes
through zero (G' Fig. 31) is the point
sought.
EXAMPLES.
1. Let ==20 ft. m=5 ft. ml=-10 ft. m215 ft.W==l ton W1,=2 tons W2,3 tons.
Find the maximum moment.
2. Find the size of a rectangular wooden
beam where
1=15 ft. m==3 ft. m,=l-6 ft. mn,=14 ft.
W==1 ton W=-I ton W.2,2 tons
S=1,000 lbs. and d=4 b.
Case IV.
Let the load be equally distributed
over the beam (Fig. 32). In this case the
reaction of each abutment=I the load, or
R=-2     2          (36)
Take any section E F whose distance
from A=x.   Then the external forces




50
a        /
n A __         art   
acting between A and E F are, R1 and
the resultant of all the little weights
from A to E (==wx). This last force
acts at its centre of gravity (Fig. 33),
which is half way from A to E. Its
lever arm is therefore = -.  Hence the
eqation of moments will be
equation of moments will be




51
co
Co
w    w_ 1. I nv
x-      S b d-=S =M 
2    2  6         (37)
or   - (I-x) =-Sbd 2M )
2       6
This is a maximum at the centre where
M0= 21       (38)
The shearing force at E F is




52
T=-2-w x              (39)
2
This is greatest at the abutments where
X=l, or 0. To                (40).
At the centre    T= 
Geometrically. The values of M    in
eq. (37) may be represented by a para-...........
/.........




53
bola with vertex at L, the ordinates G L
(Fig. 34) being taken =Mo. The shearing force is represented by the two triangles APG and G Q C (Fig. 35).
The maximum moment exists at the
point (G) when the shearing force is
zero.




54
Corollary. When the uniform load
extends only over a certain distance
from one of the supports, as in (Fig. 36),
let A D=loadecl segment —to.  The reactions of the abutments are:
At A, Ri      (  17)       nS  (41
At C, R-2w m-, 




55
Then for any section in AD the moments of the external forces will be as
in the case just discussed,
Il-4m\      w c2
=-~(   I —Win ")     X22
=wmN      --- )a-I  2
And the equation of moments in AD
will be:'m (-         — m)  w  =- Sbd'=M  (42)
x    2 =6
For any section in D C the whole load
(w m) is to be considered as acting
through its centre of gravity (G), and
the equation of moments is:
zwm( -)x-win(x —m)=6 S bd2=M
Reducing
W m2         1
2  (I-x)=   Sbd12M         (43)
The shearing force in A D is
T.( m)           (44)
In DC T — m.      ^     -WM 




56
T is a maximum at A, or
To=w (- -m)
Geometrically. Eq. (42) corresponds
to the parabola AL K (Fig. 37), which...........,..
I...................
s..........C  a A a  K (w  
2 in
from A=-j- (I — m) ) and whose axis is




57
vertical. Eq. (43) corresponds to the
straight line H C. We only use the part
ALH of the parabola, the moments in
D C being represented by the triangle
D H C. The maximum moment is at N
corresponding to the vertex L of the
parabola. The value of this moment is:
Mw 2n2 n         2l-,
~MO=) 2.I~m'       (45)
which is obtained from eq. (42) by substituting for x the value A N (=- A K) -
T-(1 — m). This M0 is always less than
the maximum moment that exists when
the load extends all over the beam as
will appear by making m to vary in eq.
(45) and applying the tests for a maximum to it.
The shearing stress for the loaded
segment is represented by the triangles
A P K and KP'D (Fig. 38), and for the
other segment by the rectangle D H.
The point K, where the stress is zero, is
found by making in eq. (44)




58
T= —m(        ) - w- -0
and finding the value of x.
This point co
This point corresponds to the maximum
moment. It may also be found graphically, by constructing Fig. 38, and, as
before, affords the easiest method of de



59
termining the point of the beam where
the maximum moment exists and where
consequently there is greatest danger of
rupture.
EXAMPLES.
1. Let 1=20 ft. w=500 lbs. per ft.
n7= 15 ft. and let there be a weight in
A                              c
Fig. 39.
addition. W=5 tons at a point 18 ft.
distant from  A. Required the maximum moment.
G.. 40
FIG. 40.




60
2. Let one-half of the above beam be
loaded with a uniform load, w=l- ton
per foot, and the other half with a uniform load of wq'=-  ton per foot. Required the moments.
Case V.
A single moving load. When a single
moving load passes over a beam, as in
(Fig. 41), the maximum moment at each
instant (as appears from  Case III.)
takes effect at the section just under the
weight. To determine the law of variation of these maxima as the weight
travels over the beam: Let x=the distance at any instant from A, and then
the reaction of A at that instant (=the
I-x
part of W  transmitted to it) =W  I
Multiplying this by the lever arm x we
have for the moment under the weight:
W r                        (4) 1
(l-x)-     Sbd= c M     (46)
This is a maximum at the centre, where




61
M,=i W              (47)
^\ —--------------
I                   2_
Eq. (46) corresponds to a parabola
(Fig. 41) with vertex at L, the ordinate
G L being= W I. The shearing force
for each segment into which W  (Fig.
42) at any instant divides the beam is
equal to the reaction of the abutment
equal to the reaction of the abutment




62
corresponding to that segment. Thus,
if W is at a distance x from A the rel-x
action of A is=W -    and of C it is=
wI.
If W  has the position marked 2 in
(Fig. 42), then the shearing stress in the




63
left segment is shown by the rectangle
AN N' V and in the right segment by
the rectangle W N" N"' C. The diagram shows in a similar way the stress
at other points. If the third position of
W in the figure is at the centre of the
beam then evidently the greatest shearing stress to be provided for in the left
half of the beam will be represented by
the locus of the points like L, N', P', and
for the right half it will be the locus of
the points P", Q", L', etc.
These loci are given by the equations:
T=     (1-x)=eq. of L C
W T                   (48)
T=-.x     =eq. of AL'
Turn the triangle A L' C down for convenience, as in (Fig. 43), and then the
shearing stress to be provided for is
given by the figure A L P L' C. In this
figure A L  C L' W andD P=W.
2




64
v     Z
5........................................................
moving lod gradually covers  beam,
Q:. —-" —-~ - --  a -- ^
Case VLI
A distributed moving load. When a
moving load gradually covers a beam,
(Fig. 44), moving on from one end as a
long train of cars, the maximum moments produced is that due to the load
when it covers the entire length of the




65
beam, and consequently this case is provided for in Case IVT
But with the shearing stress it is different.!Here, as in Case V, we need
the locus of the greatest shearing stresses
that can be brought upon the beam.
This maximum at any section D occurs
when the longer segment into which D




66
divides the beam is loaded, and the
other is not. In that case the shearing
force at D (=the reaction of the abutment C) is
T= 2-1            (49)
This equation gives the parabola
A N P' (Fig. 45) with vertex at A, where','P...'"
I,




67
C P' =      When the load comes from
2
the other end of the beam we get the
parabola C N P.    Hence   the figure
A P N P' C gives the maximum shearing
stress to be provided for.
It is easy to see that the shearing
stresses thus obtained are greater than
those which exist when the load covers
the entire beam. In the latter case the
forces are represented by the triangles
A P G and GP' C (Fig. 45), the shearing
stress at D being given by eq. (39)
T=w x — =D H
2
In-the case of the passing load we have
just seen that
T= 1=D K
21
The value of D H is always less than that
of D K when x>; for if 2x be a cer-2
tain quantity, then the product of
the halves of that quantity (=x2) is




68
greater than the product of any other
two parts (such as I and [2 x-l] ) into
which it can be separated.
That is      x2>1 (2x-) 
wx2 wl (2 x-1)
*'* T^> — 27~ \       (50)
21'      21i         (50)
Wc X2      w l
or             >   X-       J
In the expressions for the moment of
resistance M-I  Sb d the quantity denoted by S (= the stress on the outside
fibres) varies directly as M. Hence, all
the geometrical illustrations we have
given of the moments may apply equally
well to the values of S. The maximum
moments give the maximum stress on
the fibres, and indicate the points of rupture when the beam is loaded with its
breaking weight.
ULTIMATE VALUES OF S.
If beams are loaded transversely
until fracture takes place, the value
of S or the stress on the outside fibre




69
which exists at the moment of fracture, gives us a value for the tensile or
compressive strength of the material according to the manner of rupture. If
the beam yields by tearing, S gives us
the tensile strength, if by crushing S
gives the compressive strength.   We
readily obtain the value of S answering
to the ultimate strength from any of the
formulas under "Transverse Stress," by
substituting given values for 1, b, and d
and the actual breaking weight for W.
But the tensile and'compressive
strengths of materials are also obtained
by direct tension and compression, the
force being applied in the direction of
the length of the bars until rupture takes
place.
If our theory were perfect the values
of tensile and compressive strength thus
deduced would agree with the ultimate
values of S found in transverse stress;
but they do not.
The difference is very wide sometimes.
Thus in cast-iron, S (in this case it represents the tensile strength) derived from




70
breaking rectangular beams by a transverse load is nearly 20 tons per square
inch, while the tensile strength obtained
directly is only about 8 tons. This discrepancy has been accounted for in two
ways.
1. That the neutral axis moves towards
the compressed side, and that therefore
a larger portion of the beam is subjected to tension than the formula supposes.
2. That the neutral axis always remaining at the centre of gravity of the beam,
the additional strength is due to the
adhesion of the fibres which is developed
by the unequal lengthening, and shortening of them as we go from the neutral
axis towards the surfaces. In favor of
this view is the fact that we know such
adhesion to be an element of strength;
for the compression or extension due to
a given force is not so great in a transversely loaded beam as in one directly
compressed or extended.
The action of this adhesive force may
be illustrated as follows:




71
A                 A _ S _ _
F        4w6. 9W
FIG. 46.
In the beam AB (Fig. 46), strained by
the weight W, all the fibres are equally
elongated, and they only resist by their
direct tenacity. But in the beam A' C
to the one-half of which is appended the
weight, while the other half, E C, is less
strained or altogether prevented from
extending, evidently W  will have to
overcome not merely the tenacity of the
fibres in A' B' but the adhesive force of




72
the fibres along the plane E F, where the
two parts of the beam   join; for this
force will tend to prevent the stretching
of the fibres in A' B', and consequently
increases the strength of A'B'. This
kind of force exists between every two
layers of horizontal fibres in a beam
under transverse loading, and is called
the longitudinal shearing stress.  It is
neglected in the formulhe we have given.
From the variation between the ultimate valves of S (called moduli of rupturse) and the values for strength obtainedl by direct tension and compression, it
results that the values should be determined in b( th ways, and that the values
gotten by one method should not be used
in calculations involving the other kind
of stress.
BEAMS OF UNIFORM STRENGTH.
As already stated, in solid rectangular
beams, S has different values for the various points in the length of the beam.
There is always a point of maximum




73
stress where the beam, if loaded sufficiently, will break. Now at all other
points there is an excess of material
which is useless and injurious from its
own weight. To secure the requisite
strength with the least material is an object usually desirable, and this can be
readily accomplished in certain materials
(as cast-iron), by giving the beam such a
shape as will make S, the stress on the
outside fibre, constant throughout its
length. In wood the injury resulting
from the cross cutting of the fibres frequently prevents the putting of the
theory into practice.
The application of the theory of uniform strength to beams of rectangular
cross section may be most simply explained by taking up the cases we have discussed in detail.
In Case I. from eq. (5) the maximum
stress in the outside fibre is,
6W/
S=   W — (51)
This stress only occurs at A, where the




74
beam will ultimately break, and it is
evidently possible to take away some
of the material between that point and
C
w
FIG. 47.
C without diminishing the strength. If
this be so done that at every point between A and C there shall exist on the
outside fibre a stress equal to that at A,
the beam will be one of uniform strength,
and we shall have attained the greatest
economy of material. Let us suppose,
the use we have for the beam requires
the depth to be uniform. What must be
its plan in order that S shall be constant
in value, or the beam be as liable to
break at any other point as at A5.
6 ( W x
In eq. (4) S-= 6d,if we assume S to
In eci. (4)b d2 )




l75
be constant, the other side of the equation must be constant also, and since
6 W is constant, and we have made d
constant by assuming the depth to be
uniform, the whole expression can be
constant only when b varies as x.
*.  b x  whence b=cx     (where
c-some constant factor). This equation
which is that of a straight line shows
that the breadth must vary directly as
the length. Hence the plan should be a
triangle with vertex at C (Fig. 48.)
FIG. 48.
On the other hand, if we suppose the
breadth to be uniform and wish to have
6 Wx
S constant, in the eq. S= 6 d,  x must
vary as d', or
=c cx




76
This corresponds to a parabola, and
the beam, if the top be straight will
have the elevation shown in (Fig. 49).
Suppose that b varies as d, then d-n b
(n being a constant) and
6Wx
n2 b3
\k             oC




77
To render S. constant we must have,
b' xx.'. b —cx  and  d' —' cx.
These are the equations of a cubic parabola. Hence the horizontal section (Fig.
- __ —
FIG. 50.
50), and the vertical section (Fig. 51),
should be curves of that kind.  The
FIG. 51.




78
cross section is rectangular as in (Fig.
2).
In acse II we have
S=   b 3           (55)
b d2
FIG. 52.
Hence, if we make suppositions sililar to those above we shall have, when
the depth is uniform (or d= a constant)
x2 varying as 7, or
b=CX2
Hence the iplan (Fig. 53) should consist of parabolas with vertices at C. If
b be constant then
dc2=cx   or d=v/c.x
This is the equation of a straight line,




79
A      -C
FIG. 53.
and gives for the elevation the triangle
(Fig. 54).
IQ                        C
Bj
FIG. 54.
In Case III. the analysis gives results
similar to those in Case 1.
Thus from equation (21) S= -Wn
x       6 Wn              I.
2 Here  I   is constant, and if d
b d'    I
be constant also, b must vary as x..'.   b=cx.




80
This gives the triangle A GH (Fig.
55. We obtain similarly the triangle
G C H for the other end of the beam.
If the breadth be constant we have,
d2-c
d -ex
which gives a parabola AK (Fig. 56) with
vertex at A. So, for the right hand end
of the beam the proper elevation is the




81
parabola C K (Fig. 56). The elevation
(Fig. 56) assumes that the top of the
beam needs to be horizontal.
In  C   I   e   (  
In Case IV., equation (37) gives
3 wI x  — xc)
S= 3        -: HIhere, if d be constant
b (I2
in order to render S constant we have
b=cx ( —x).




82
This may be represented by parabolas
with vertices at G and H (Fig. 57) opposite the middle of the beam. If b is
constant, then,
d2-cx ( —x)
FIG. 57.
which is the equation of an ellipse, and
the beam (if it is required to be horizontal on top) may be made as in (Fig.
58.)
In these cases of beams of uniform
strength, we have so far only considered
the moments of the weights or the bending moments as they are called. But the
results are to be modified by the transverse shearing stress. In ordinary rec



83
FIG. 58.
tangular beams this shearing stress is so
small compared with the bending moment, that it may be left out of consideration.  But in beams of uniform
strength the ends must not taper to a
point, but must always be left large
enough to bear the shearing stress. In
the case represented in (Fig. 57) the beam
should have, near the ends, the shape
shown in (Fig. 59).
I     _Ai. 9
FIG. 59.




84
DOUBLE FLANGED BEAMS,
So far we have considered beams with
rectangular cross sections, and beams of
uniform strength deduced from these.
We will now consider beams of - shape.
It is evident from the investigation already given of the condition of stress
in transversely loaded beams, that those
portions of the beam nearest the centre
bear but a small proportion of the stress,
while the contrary is the case with the
outside fibres. Hence we would gain
strength by moving a considerable portion of that about the neutral axis and
placing it on the top and bottom.
The first form in which the idea was
applied was in the T or J. cast iron beam.
The fact that rectangular cast-iron
beam always broke by the tearing of the
fibres on the side subjected to tension,
suggested the idea of reinforcing that
side of the beam with a flange. The result of this is, that the neutral axis still
passing through the centre of gravity of
the cross section, the extreme fibres sub



85
jected to compression are farther off
than those subjected to tension, and consequently are strained more nearly to
their full strength before fracture. This
form of beam gives a large increase of
strength for the same amount of iron.
It was still plain that the fibres in that
part of the web about the neutral axis
were but little strained as compared with
the fibres on the outside, and it was proposed to leave as little material there as
possible, and to place the mass of it in
two flanges (I), one above and the
other below, giving to these flanges sizes
inversely proportional to the tensile and
compressive strength of the material.
The question then was, how much of the
material should be left in the web, for
plainly all could not be taken.  The
amount to be left is determined by experiment. If the web is left too thin,
the beam will twist and break under the
shearing force, and in some cases, from
the want of stiffness in the compressed
flange.
To simplify the calculations, the web




86
is considered as bearing all the shearing
stress, and no other, and the flanges as
bearing all the extension and compression due to the bending moment; and
these parts should be proportional accordingly with due reference to the practical difficulties that sometimes occur.
The ordinary formulas for the strength.of such beams are gotten by the following approximation:   We first neglect
the compressive and tensile forces of the
web, which are small compared with
those of the flanges, and consider it as
bearing only the shearing stress. Then
as the depth of the flanges is generally
small as compared with the depth of the
beam, we consider all the fibres in
each flange as strained alike, and as bearing the average stress that is brought on
that flange. (Fig. 60.)
The resultant of the force on each
flange, then, is equal to the stress on a
unit of surface (S) multiplied by the
flange area (A): that is = S A.
The point of application of the force
will be at the middle of the depth of the




87
FIcT. 60.
flanges (at 0 and O', Fig. 61). Fig. 61
0o.           XFIG. 61.
FIG. 61.
shows the forces we have to deal with
in the Case corresponding to Case I.
under rectangular beams.




88
Let O' 0=d.
S'   stress on upper flange per
unit of surface.
S" = stress on lower flange per
unit of surface.
A". area of lower flange.
A'   area of upper flange.
Then if we take O (Fig. 61) as a centre of moments we have:
-N d-:'.o. + W      =  (But N=S'A').. S'A'd=Wx                   (57)
If we take O' as the centre of moments we will get
S" A" cd  W x        (58)
The formula for shearing force is identical with that under Case I. of rectangular beams; that is:
T=Wx              (59)
If A'=A", then plainly S'=S" (from
equations 57 and 58), or, the forces of
tension and compression are equal (as in
rectangular beams); but if A' and A"
are not equal, we have:




89
S': S"Wx        "'       A''A' d  A" d
That is, the unit stresses in the flanges
are inversely as the areas. Now, to have
the material distributed  between the
flanges most efficiently for strength the
unit stress should be in proportion to
the ultimate strength of the material
against tension and compression, and
hence the areas of the cross sections of
the flanges should be inversely as the
ultimate strength.
Thus, if A D (Fig. 62) be of cast-iron,
which is six times as strong against colmpression as against tension, the unit
stress in the lower flange should be made
six times as great as in the upper, and to
effect this the area of the lower flange
should be one-sixth that of the upper.
The Cases under = beams are similar
to those under rectangular beams.
Cass I.-Beams fixed at one end and
loaded at the other.
S' A' d=W  x, and S" A" d=W  x (60)




c4
Catse II.-Beams fixed at one end and
loaded uniformly.
S' A' d(g   w x, and S" Al" cd =   wx2 (61)
Cacse III —Beams supported at both
ends and loaded at some intermediate point.




(62)
WM. x-W (x-m)=S' A'd, or, S"A" d
Case IV.-Beams supported at both
ends and loaded uniformly.
S' A' d= - w x (-x) -=S" A" d  (63)
Case Z.-A single moving load over a
beam supported at both ends.
W x
S'A' d=      (I-x) = S A"d   (64)
Case VI.-A   distributed moving load
may be considered as included in
Case IV.
The formulae for shearing stress are
identical with those in rectangular
beams.
The principles of the uniform strength
of beams may be applied to flanged
beams as they were to rectangular beams.
The discussion is analogous to that already given.




92
MOMENT OF RESISTANCE OF BEAMS DETERMINED GEOMETRICALLY.
The following method of obtaining
the moment of resistance of beams is of
easy application, and in many cases of
unsymmetrical cross section is the simplest that can be used:
I. For illustration, take a beam of
rectangular cross section. Let GP (Fig.
63) be the cross section at some point of
this beam. The stresses on the fibres,
as we have already seen, increase just in
proportion as we go from the neutral
axis towards the upper or lower surface
of the beam, and may for any vertical
slice (as that at EF) be represented by
the ordinates of two triangles, as shown
in Fig. 64, where E I (=F J) represents
the stress on the outside fibre. For the
cross section GP (Fig. 63) the stresses
will be represented by two wedges, the
bases of which are GM   and MR, and
the elevations of which are the triangles
shown in Fig. 64. The volumes of these
wedges give the amount of compressive




93
\             /
\ /
1 A- LX
I
TQ
Ao b       1; 
and tensile force exerted at the cross section in question, and the points in G P
under the centre of gravity of the
wedges give the " centres of resistance,"
or the points of application of the resultants of these forces.




94
As a geometrical representation of the
stresses on the fibres, these wedges are
perfect, for the perpendicular ordinate
of the wedge gives in every case the
stress which exists in the fibre over
which it stands. Thus the line T'V'
(Fig. 64) represents the stress on each
fibre in the row TV (Fig. 63).
But it is often difficult to find the
centre of gravity of these wedges in the
case of curved and irregular cross sections, and yet this must be cone before
we can know    the lever-arms of the
stresses. To render this easier to do we
may   represent the  stresses, not by
wedges, but by prisms, the centres of
gravity of which are over the centres of
gravity of their bases.
Thus, if in (Fig. 65) we draw the two
shaded triangles, and conceive prisms of
a height = E I (the stress on the outside
fibre, Fig. 64), to be constructed on them
as bases, we shall have a geometrical
representation of the stress on the section GP, less perfect in some respects




95.. —---           v
R                P
FIG. 65.
than that given by the wedges but better suited to our purpose.
For, note that,
1. The volume of the prism GOH
(Fig. 65) is equal to that of the wedge
G M (Fig. 63), and the volume of any
part of the prism cut off by a plane
parallel to the neutral axis, as that whose
base t'O v' is equal in volume to the
corresponding part of the wedge T M,
or since the height of the prism is constant, the stress on the surface GM as
we go out from the neutral axis varies




96
as the area of the triangle which forms
the base of the prism.
2. The vertical slice of the prism
standing on any line t'v' represents in
amount the stress on the line of fibres
t v, for this stress is equal to the corresponding one in the wedge, the slice of
the prism being as much higher than
that of the wedge as tv exceeds t'v'.
Of course (except in the case of the outside fibres in the row G H) each ordinate
in the slice of the prism no longer represents the stress on the fibre over which
it stands, as was the case in the wedge.
3. The moment of the tensile forces,
for instance, will equal the area of the
prism GOHI multiplied by its height,
(E I = stress on outside fibre = S). The
centre of resistance of these forces, or
the centre of gravity of the prism is at
C (Fig. 65), the centre of gravity of the
base GOH.     The triangle GOH     is
sometimes called the " effective area " of
the surface G M, because a uniform
stress on it of an intensity = the unit
stress at G H gives the same amount of




I- ____
FIG. 66.
resistance, as that on the whole area G MI
acted on as the latter is by a varying
stress.
Considering the stresses represented
by the two prisms whose bases are
GOH and R OP (Fig. 65), as concentrated at the centres of gravity C and C'
(Fig. 67) of these bases, and taking one
of these points (as C', Fig. 67) as the
centre of moments, we have in the case
represented in the figure:
(Vol. of prism GO H) X CC' — Wx
or if b = breadth and d = depth of beam




98
FI                L__P
FIG. 67.
S    ( b). 2  d=  S b  I =W    (65)
as before.
Corollary. If the beam   be square,
b-=d and
M=1   S b3         (66)
II. As a second example, take a square
beam so placed that its diagonal will be
vertical.  Fig. 68 is the cross section.
Here we find the base of the prism of
stress by points. To find the line in the
base of the prism corresponding to the




99
FIG. 68.
stress in any row of fibres, such as A B
whose distance'from the neutral axis is
OX, proceed as follows:
We see that if the cross section were
the square of which HI-LMK     is the
half, then a'' b' would be the line required, since this is the breadth at that point
of the triangle I 0 K, which would in




100
that case represent the base of the prism
of stress. Project the points A and B
upon I-K. Then the actual row (AB)
of fibres is as much shorter than the corresponding row in the supposed section
H M, as R T is less than H K, and consequently to obtain the proper line in the
base of the true prism  of stress, t' b'
must be shortened in this proportion.
Draw lines from R and T to O. These
lines intersect the row of fibres at ca and
b. Then
1HK: RrT (:AB)::ctb': ab (67)
Hence a b is the line required, and a and
b are two points in the outline of the
base of the prism of stress. Any number of lines as n v, &c., may be gotten
similarly and the curve drawn through
the points a. - n - b. v, &c., will give
the form of the base of the prism of
stress. This base is shaded in the diagram.
For any ordinate of the curve O n G,
as a X, we have
OX: a X:: OG: R G-AX




101
d'
But AX=X G and making GO=-2
and putting O X=x and a X=y, we
have
d'   C _ 
x: y: --: (2 - x
2     2
2   ( /I    \ )    (68).'"I.-    2    J   )
This is the equation of a parabola
with vertex at n, half way between H K
and the neutral axis. Hence the base of
each prism is composed of parts of two
symmetrical parabolas.
Areas of the bases. Since the area of
a parabola is two-thicrds of the circumscribing rectangle, the area of each base
=2 (GOXnv)
But n v=  R'T' and R' T'=   H K.. nv — H.K
d' d    1
Area =    -     =- dI
-'2 4     12
The centres of gravity of these bases
(and consequently of the prisms) are at
C and C', and the distance
c c=   d'




i02
JIence the moment of resistance of the
fibres about C or C' is
1
M=-Sc'3            (69)
24
(S=height of prism or stress on external
fibres at G and P.)
Corollary. To compare the resistance
of the beam in this position with its resistance when lying flat:
Let cl=side of the square as L G.
Then cd'=d A/ and eq. (69) becomes
M=12S cVI2 =      S d8    (70)
Comparing this with eq. (66), we see
that the beam offers greater resistance
when flat in the proportion of
1    1
6  6V
In solving these problems with diagonally placed beams, place the above value
of M equal to the moment of the weight
as before.




103
III. Let us apply this nethod to a T
beam. Take for example the cast-iron.J beam, calculated in part on p. 257
Rankine's Civil Engineering, in which
the area of the flange = - that of. the
web. Assume the flange to be 6 inches
by 1 inch, and the web to be 5 inches by.8 of an inch, and draw a figure of the
cross section to scale (Fig. 69).
1. As the top and bottom of this section is not symmetrical, it is necessary to
find the position of the neutral axis,
which is no longer at the half-depth.
This may be done by calculation, or by
a simple mechanical process as follows:
The centre of gravity of the cross
section, since this last is symmetrical
with regard to the vertical line through
the middle of the web, must lie on this
line. Cut accurately the figure of the
cross section out of card board, or tin,
or good paper, and suspend it freely by
one end of the flange, suspending also
from the same point a plummet. Mark
the line of the plummet on the' card
board, and the centre of gravity being




104;A  A/"
i...........
/ / \ "*
/  -** — - ----- --.1
SCG, I'
FIG. 69.
on this line and also on the middle line
of the web, will be at their intersection
0.   By measurement this point was
found to be distant from A B four and
three-tenths inches (4. 3"), which is also
the value by calculations. The line L 
drawn through this point is the neutral
axis.




105
2. To determine the bases of the
prisms of stress. On the upper side the
base is the triangle O A B, if the altitude be taken equal to the stress on the
fibres along A B. For if L' M' H G were
the upper half section, O G H would be
the base of the prism and O A B is less
than O G H, in the same proportion that
L M AB, the real half section, is less
than L' M' H G. Hence (if the upper be
the compressed side) the total compressive force is equal to the prism  erected
on AO B, with the height equal to the
unit stress at A B.
Below the axis L M.-For convenience
we should have the height of the tension
prism equal to that of the compression
one, and the base must be determined
under this condition.   Complete the
large rectangle G H Q Y, making the
distance of Q Y below 0==4. 3 inches.
Draw   0 Q and 0 Y   and the shaded
trapezoid J N R Z cut out on the flange
by these lines, will evidently be the portion of the base due to the flange. Having prolonged the lines of the web to T




106
and V, draw O T and O V, and then the
shaded triangle O K I will be that part
of the base due to the portion (L Z) of
the web below the neutral axis.
The total tensile and compressive
forces being always equal, and the height
of the prisms having been assumed, each
-S'- stress on fibres at the distance of
A B from 0, the bases of these prisms
must be equal also.   This necessary
equality between the area of 0 A B and
that of 0 K I+ J N R Z, affords a means
of testing the accuracy of our work in
finding the position of O.
3. Area of base 0 A B. This is,
=-  (L A  AB) =    (4.3X.8) = 1.72 sq.
inches.
4. To determine the distance C C'
(Fig. 69) between the centres of gravity
of the prisms, which distance is the leverarm to be used when one of these points
is taken as the centre of moments.
These centres of gravity (C and C') can
be readily determined by means similar
to those employed in finding the centre




of gravity of the cross-section itself.
Thus, cut the shaded areas (Fig. 69) out
of card board or paper, and suspending
each of them from two points in succession, draw vertical lines through the
points of suspension. The intersection
of these two lines gives the centre of
gravity. In the present case they may
be so simply obtained by calculation,
that we adopt that method. The centre
of gravity of A 0 B is -= the distance
from O to AB or O C=2 (4.3) =2.87
inches. As to the shaded part below 0,
by using the ordinary formula for the
centre of gravity and taking moments
around 0, we find the distance
OC'
ON RX (1.7)-(0 J K + 0   Z)-(.7)
O NR-2(O J K)
2.2764-.01388
- 1.318 inch.
2.0145-.2975
Hence the distance
CC' —  C +O C'-2.87 + 1.318=4.18 ins.
Hence, since S' is the height of the




108
prisms, the moment of resistance of the
fibres is
M-4.18X1.72. S'=7.19 S'
If it be desired to have M, not in
terms of S', the stress along A B, but of
S" the stress on the lowermost fibres (at
F P), we have since the stresses increase
directly with the distance from 0,
S': S"::L' G: L' F: 4.3": 1.7".S'-2.53. S" and M=18.19. S"
IV. As an illustration of the great
saving of labor sometimes effected by
this process, take the steel rail now
widely used in England, the cross section of which is given to scale in
(Fig. 70). The determination of the
moment here by calculation would be
long and tedious. The dimensions of the
cross section are given on the figure.
1. The centre of gravity O of the cross
section is found by making a template
as in the last case, and suspending it
freely by a corner. The vertical through
the point of suspension intersects AX




109
---—... —---—..-. —--. —^- --- --
~~   -
4,1
at 0, which is 2.55 inches below-A.
Through this point draw L M, the neutral axis.
2. The bases of the prisms of stress




110
are determined by points as in example
11.
Lay off   X —2.55 inches. Draw the
rectangle G H Q Y. Assume the height
of the prisms to be the stress in the fibre
at A. Then proceeding as in example
II., the line of the base corresponding
to any row of fibres, as B D, is b d.
Obtain any number of points in the
same way as b and dc, and through tlese
points draw a curve bounding the shaded
figure A b 0 d.  Similarly below  the
neutral axis, tv is the line in the base of
the stress prism corresponding to T V,
and the shaded figure, O TRP, is that
base where the height is taken equal to
the unit stress at A. The equality of
the bases in area is the test of accuracy.
3. To determine these areas.   The
simplest plan in the present case is first
to find the area of the cross section itself. This is done as follows: The rail
in question weighed 84 lbs. per yard,
and the steel, of which it was made
weighed,.277 lb. per cubic inch. Hence




111
if A = area of cross section in square
inches
(36. A).277=-84.. A=8.4 sq. inches.
Now cut out of the same card board,
or paper, templates of the two shaded
parts in the diagram, and also of the
cross section itself, and weigh them. The
ratio of the weights will equal that of
the areas.
The comparison of weights may be
readily made by means of a suspended
wire, which may serve as a temporary balance, the templates to be compared being
stuck on the opposite ends, and one or
both moved until the wire is evenly balanced. The weights of the templates
being inversely as their distances from
the point of suspension of the wire,
their areas will be in the same proportion. The areas of the prisms in the
case before us were found to be equal
each to 2.49 square inches.
4. The centres of gravity of these
bases are found in the same way as those
of the cross section itself. The poi-,,t




112
C was thus found to be 1.84 inches above.
0 and C' to be 1.66 inches below it.
Hence the distance
CC'=3.5 inches.
Therefore, finally, if S'=unit stress at
A, the moment of resistance is,
A: 2.49 X 3.5 X S'- 8.715. S'
If we desire M in terms of S" (stress
at F P), we have
S': S"  1.84  1.66
S'. S l-.1 S/'.. M=9.67 S"
V. A circular cross section (Fig. 71).
Here the neutral axis of course=L M,
passing through the centre. Draw the
circumscribing rectangle G Y and obtain
the points t, b, v, c, &c., in the curve
bounding the base,as heretofore. Through
the points so found draw the curves.
Determine the areas of the bases of
the stress prisms by comparing them
with the half-square G L M H. Thus, if
a template is not to the surface of one
of these beams it will just equal in




113,,'     ~'~~A.....  _'   l
0                     i
third of the rectangle G M, or ~.- =.
FIG. 7 1.
weight the template cut to the surface
AGLO    t b A, or, in other words, the
shaded surface AbO c  A is just oned2  1
third of the rectangle GM, or  2  6 42.
2The centres of gravity C and C' are
The centres of gravity 0 and C' are




114
found as before by means of templates.
In this way it was found that
0 C=O C'-.58 (O A) =. 587.d
C..   C'=.587 d
The height of the prisms being S
(= stress at A or F) the moment of resistance is,
M=S.6 (.587 d) =.0978 ds S
(The accurate value by calculation is
M=.0982 d3 S.)
Note.-The curve of the base of the
prisms is a lemniscate. To find its equation we have in the triangles O b X' and
OB'A,
bX': AB' (=BX')      OX': OA
And taking the vertical axis as that of
X, and the horizontal one, as that of Y,
the origin being at 0, and calling the coordinates of the circle x' and y', and
those of the lemniscate x and y we have:
y: y'::: R, or y: VR2_-x2::x: R.*. R2 y2=x2 (R2 x2).




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PLATTNER. Manual of Qualitative and Quantitative
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Caswell.  Illustrated with 87 wood cuts, and one
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pages, Svo, cloth................................  7  50
PLYMPTON. The Blow Pipe. A Guide to its Use.
in the Determination of Salts and Minerals. Compiled from various sources, by George W. Plympton,
C. E. A. M., Professor of Physical Science in the
Polytechnic Institute, Brooklyn, New York, J2mo,
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PYNCHON. Introduction to Chemical Physics, designed for the use of Academies, Colleges and High
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9




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ELIOT AND STORER.         A compendious Manual of
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WATT'S Dictionary of Chemistry.   New and Revised
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RANDALL. Quartz Operators Hand-Book. By P. M.
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SILVERSMITH. A Practical Hand-Book for Miners,
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THE USEFUL METALS AND THEIR ALLOYS,
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JOYNSON. The Metals used in construction, Iron,
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GREENE. Graphical Method for the Analysis of Bridge
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BELL. Chemical Phenomena of Iron Smelting. An
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ROGERS. The Geology of Pennsylvania. A Government survey, with a general view of the Geology of
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BO URNE. Treatise on the Steam Engine in its various
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STUART., The Naval Dry Docks of the United
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ATKINSON. Practical Treatises on the Gases met
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CLARK. Theoretical Navigation and Nautical Astronomy. By Lieut. Lewis Clark, U. S. N. Illustrated
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BECKWITH. Observations on the Materials and
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MORFIT. A Practical Treatise on Pure Fertilizers, and
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BARNARD.       Tne Metric System of Weights and
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14




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MYER. Manual of Signals, for the use of Signal officers
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CLEVENGER. A Treatise on the Method of Government Surveying, as prescribed by the U. S. Congress
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complete Mathematical, Astronomical and Practical
Instructions for the Use of the United States Surveyors in the Field.  By S. R. Clevenger, Pocket
Book Form, Morocco............................. 2 50
PICKERT'AND METCALF. The Art of Graining.
How Acquired and How Produced, with description
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HUNT. Designs for the Gateways of the Southern Entrances to the Central Park. By Richard M. Hunt.
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LAZELLE. One Law in Nature. By Capt. H. M.
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BOYNTON. History of West Point, its Military Importance during the American Revolution, and the
Origin and History of the U. S. Military Academy.
By Bvt. Major C. E. Boynton, A.M., Adjutant of the
Military Academy. Second edition, 416 pp. 8vo,
printed on tinted paper, beautifully illustrated with
36 maps and fine engravings, chiefly from photographs taken on the spot by the author.  Extra
cloth............................................  $3  50
WOOD. West Point Scrap Book, being a collection of
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Beautifully printed on tinted paper. 8vo, cloth.... 5 oo
WEST POINT LIFE. A Poem read before the Dialectic Society of the United States Military Academy.
Illustrated with Pen-and-Ink Sketches. By a Cadet.
To which is added the song, " Benny Havens, oh!"
oblong 8vo, 21 full page illustrations, cloth..........  2 50
GUIDE TO WEST POINT and the U. S. Military
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HENRY. Military Record of Civilian Appointments in
the United States Army. By Guy V. Henry, Brevet
Colonel and Captain First United States Artillery,
Late Colonel and Brevet Brigadier General, United
States Volunteers. Vol. I now ready. Vol. 2 in
press. 8vo, per volume, cloth...................  5 oo
HAMERSLY. Records of Living Officers of the U.
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sources.  By Lewis B. Hamersly, late Lieutenant
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MOORE. Portrait Gallery of the War. Civil, Military
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16