75 =5714 pounds per square inch,
1.75 x 7-5 x 7-5
which is the maximum stress in the girder, since the
ordinate at C gives the maximum bending moment.
This method is, as stated, approximate. For an
accurate determination of the stresses, the top of the
combustion chamber should be treated as a series of
continuous beams, the end supports of which are the
walls of the chamber, the bolts being the intermediate
supports. The reactions at these intermediate supports
form the loads on the bolts, and, when these loads are
found, the girder may be treated, as before, as a simple
beam.
Referring to the top plan of the combustion chamber
in Fig. 1 8, it will be seen that there are 8 equally spaced
stays, each supporting a transverse section, 7.5x25
inches, of the top, which section is virtually a continuous
beam, supported from above. In this section, there are
4 equal spans, 6\ inches long, the load on each being
6.25 x 7.5 x 1 60 = 7500 pounds. In the transverse
direction, parallel to the stay, the top of the chamber
STATIONARY LOADS: SHEARS AND MOMENTS 59
will sag under the load and between the bolts, thus
producing bending moments in that direction. Let
w = span load = 7500 pounds. Then, according to the
Theorem of Three Moments, the reactions at the 5 sup-
ports are :
Support 12345
Reaction iiw/28 32^/28 267^/28 327^/28 nw/28
It will be seen that the distribution of the load is not
uniform and that the greater loads come on the second
and fourth supports, i.e., on the two outer bolts, aa, of
the stay. Each of these bolts supports one-half of each
of the adjacent spans of the transverse strip, so that,
owing to the bending stress, the area thus supported
may be considered as having a load of ii\ = 32 w/2%
= 8571 pounds. The load on the similar central area
and bolt is w 2 = 26 w/28 = 6964 pounds, and the total
load on the 5 supports is 112^/28 = 30,000 pounds, as
previously computed.
Again, the top of the combustion chamber will sag,
under the load and between the bolts, in the longitudi-
nal direction, at right angles to the stays. It may,
therefore, be considered as divided in that direction into
3 continuous beams, each with 9 equal spans and 10
supports. The reaction at support 10 is the same as
that at 2, etc. Neglecting the load just computed for
transverse bending, we have, for the original span load
w ( Theorem of Three Moments) :
Support i 2 3 4 5
Reaction, 2097^/530, 6017^/530, 5117^/530, 535^/530, 5297^/530
60 GRAPHIC STATICS
The greater loads are therefore those on supports 2
and 9, which supports, for the 3 beams, are the 3 bolts
in each of the two end girders of the series. These
two stays are therefore those which are under the maxi-
mum stress. .
Considering both transverse and longitudinal bending,
the resultant loads on the bolts of the two end girders
are :
On each outer bolt, a, load = (wx 32/28) (601/530) 1.29^ = 9675^3.
On middle bolt, b, load = (w x 26/28) (601/530) = i .05 w= 7875 Ib.
That is, the outer two of the three continuous beams
assumed for longitudinal bending stresses are considered
as having a span load throughout of w l 8571 pounds,
and similarly, the middle beam, a uniform span load of
w^ = 6964 pounds.
Treating the girder as a simple beam and following
the approximate method in all respects, except the
changes as above in the bolt loads, the maximum stress
5 in the girder is found to be 6686 pounds per square
inch, or 17 per cent more than that given by the ap-
proximate method. These calculations apply only to
the two girder stays at the ends of the row ; the inter-
mediate girders are under less stress, as is shown by the
coefficients for the reactions at the corresponding sup-
ports. The values given for these coefficients refer
only to beams of constant cross-section, under a uniform
load, and with all supports at the same level. Strictly
speaking, they are limited also to beams with free ends
(Art. 7), although the difference in this respect is not
material in this case.
STATIONARY LOADS: SHEARS AND MOMENTS 6 1
13. COUNTER SHAFT: TWISTING AND BENDING COM-
BINED. Figure 19 represents a counter shaft AB, sup-
FIG. 19
ported in bearings at A and B, and carrying a 26-inch
driving pulley C and a 24-inch driven pulley D, both of
62 GRAPHIC STATICS
5-inch face. The belt drive at C is assumed to be hori-
zontal; that at D, vertical. Take the vertical load at D
as 250 pounds, this including the weights of pulley and
belting and the pull of the belt tension. At C, let the
vertical load, or weights of pulley and belting, be 100
pounds, and the horizontal load, or belt tension, 200
pounds.
The shaft is bent in a vertical plane by a force
7^ = 250 pounds acting at D and a force P 2 = 100
pounds acting at C, and, in a horizontal plane, by the
force P B = 200 pounds acting at C. It is also twisted,
between C and D, by a driving moment = 200 x %f- =
2600 pound-inches.
The diagram for the vertical bending moments is
marked V. The moments shown by the portion of the
diagram above the zero line are positive, i.e. the lower
side of the shaft is in tension and is bent in a convex
curve; the moments below the line are negative, the
lower side of the shaft being compressed. The reac-
tions are: R' v = 150 pounds and R" v = .200 pounds.
Diagram H gives the horizontal bending moments.
The latter are all positive, and, assuming that the pulley
driving C is in front of it, the rear side of the shaft is
in tension and convex. The reactions are: ^^=33-3
pounds and R" h =2^.^ pounds, the two acting in op-
posite directions.
The linear scale of the original shaft diagram was
i foot= I inch, or 12 to I ; the scale of the force poly-
gon was 200 pounds = I inch ; and the pole distance F
was i^ inches = 250 pounds. Hence (Art. 12), the
STATIONARY LOADS: SHEARS AND MOMENTS 63
moment scale for F"and H was 12 x 250 = 3000 pound-
inches per inch of measured length of any ordinate.
The actual reactions at the bearings are the resultants
of the vertical and horizontal reactions, as above. Thus,
as in (a\ laying off R" v and R" h perpendicular to each
other, their resultant dm = R 2 , inclined at the angle
to the horizontal ; R l can be found similarly, but its
inclination differs and its line of action lies on the other
side of the shaft.
Diagrams V and H are combined to form the result-
ant bending moment diagram R by a method (Art. 14)
similar to that used for the reactions. Thus, as in (b\
let v be the vertical bending moment at the section D
of the shaft and h the similar horizontal moment ; their
resultant r is, on the same moment-scale, the correspond-
ing ordinate of diagram R. The latter diagram gives
simply the magnitudes of the resultant ordinates at all
points, without regard to their sign ; it is not a plane
surface, but is warped, since the inclination of the ordi-
nates changes continually.
The diagram for the twisting moment between C and
D is represented by the rectangle T, whose height is
made equal to this moment, or 2600 pound-inches, on
the same moment scale as before. Diagrams R and T
are combined to form diagram E for the equivalent bend-
ing moments by the method shown in (c). The equation
for the equivalent bending moment (Art. 14) is :
EM b =
in which M b and M t are respectively the bending and twist-
64 GRAPHIC STATICS
ing moments at the section considered and EM b is the
corresponding equivalent bending moment. To apply
this expression graphically for any section, as that at D,
lay off, as in (c\ fg and gk equal respectively to one-half
the bending and twisting moments. Revolve/ to/' and
g to g 1 . Then, for the section at D :
EM b k = ordinate e.
This follows since, for that section,
= I M h an<
This method is used for all sections to the right of D ;
to the left of it, no twisting moment exists and the
original bending moment holds.
Considering strength only, the shaft should be de-
signed for the equivalent bending moment shown by
the maximum ordinate in diagram E. This ordinate is
that at section D and is equal to 2516 pound-inches.
The resisting moment (Art. 10) of a solid cylinder to
bending strain is SI/c, in which 5 is the maximum per-
missible working stress and I/c = 7rd s /$2, d being the
diameter in inches. Equating the equivalent bending
and resisting moments and taking 5 as 16,000 pounds
per square inch :
2516= 16,000 x 7n/ 3 /3 2 >
d= 1.17 inches,
which is less than is customary in practice, since the
stiffness of the shaft and the bearing surface at the
journals must be considered, as well as its strength to
resist bending and shearing. The small diameter of
STATIONARY LOADS: SHEARS AND MOMENTS 65
the shaft in this case is produced mainly by the rela-
tively low value of the bending moment, and this low
value is due to the fact that the shaft is short and the
weight of pulley C acts to balance that of pulley D.
- 14 -
FIG. 20
14. CENTRE CRANK SHAFT: TWISTING AND BENDING
COMBINED. Figure 20 represents the centre crank
shaft of a vertical, high pressure engine, the shaft being
supported in bearings at A and B and delivering its
66 GRAPHIC STATICS
power as a twisting moment at the left-hand end. For
simplicity, neglect the weight and friction of the shaft
and reciprocating parts. It is required to find the
equivalent bending moments on the shaft and cranks
when the engine is developing its maximum power, i.e.,
when, as in {A), the connecting rod and cranks are at
right angles on the downward stroke.
The data are: cylinder diameter, 15 inches; stroke,
1 8 inches; length of connecting rod, 29 inches; maxi-
mum unbalanced pressure on piston, 100 pounds per
square inch; force on piston at position shown in
(A} = Q TT x 15x15 x J x 100=17,671 pounds; cor-
responding thrust on connecting rod = P = Q/cos =
17,671 x 1.04= 18,400 pounds, about.
In the position shown in (A), the thrust P is resisted
by an equal, opposite, and parallel reaction R t whose
line of action passes through the centre of the shaft.
To simplify the diagrams to be drawn, the forces P and
R, considered as bending forces only, are assumed to be
revolved through 90 degrees, as shown by dotted lines,
and hence to act in the plane of the crank axes an
assumption which evidently does not alter the magni-
tude of the bending action of these forces on the shaft
and crank pin. With regard to the bending of the
crank and the twisting of the latter, of the crank pin,
and of a portion of the shaft, the two forces are con-
sidered as acting in their original planes.
Let abcdef represent the neutral axes of the shaft,
cranks, and crank pin, i.e. a line coinciding with the
axes of the shaft and pin and passing through the
STATIONARY LOADS: SHEARS AND MOMENTS 67
centres of gravity of all transverse sections of the
cranks.
The shaft is subjected to bending from a to b and
from e to /, and to twisting from b to a and onward to
the left; the crank pin is under both bending and
twisting strains throughout its entire length ; and both
cranks are twisted on their neutral axes and are also
bent as cantilevers (Art. 7).
On the load-line gh, lay off P and draw the force
polygon Ogh and the bending moment diagram aDf.
The ray Og' , parallel to the closing line af, determines
the magnitude of the reactions, which are : R l = 11,040
pounds and R 2 = 7360 pounds. With a linear scale of
8 to i, a force scale of 8000 pounds per inch, and a
pole distance of i^ inches, the moment scale (Art. 12)
is 96,000 pound-inches per inch.
Shaft. The right-hand section of the shaft acts simply
as a support and is therefore under bending strain only,
the moments being given by the portion fek of the
diagram.
The similar bending moment diagram for the left-
hand section is abm. This section is also subjected to
twisting by the driving moment, P x r 18,400 x 9 =
165,600 pound-inches. Using the same moment scale
as above, the twisting moment diagram is bnoa. Com-
bining the two diagrams by the method employed in
Example 13, the equivalent bending moment diagram
is bn'd a.
Crank Pin. The crank pin and the parts of the
cranks lying between the two neutral axes of the latter
68 GRAPHIC STATICS
are under bending strain, the moments of which are
given by the portion bmDke of the bending moment
diagram.
The pin is also twisted by a moment,
R 2 x r = 7360 x 9 = 66,240 pound-inches.
This follows since if, in the original plane of the right
reaction R%, there be applied at e two equal and
opposite forces R 2 f and R 2 ", each equal to R%, equilib-
rium will still exist, but the forces R 2 and R 2 will
form a couple of arm ef tending to twist the crank on
its neutral axis, and the force R 2 " = R 2 will act, with
the leverage ed = r, to twist the crank pin. These
forces are shown in the diagram as if they were in the
plane of the paper ; they really act in the plane of R as
in (A).
The twisting moment diagram is cpqd. Combining
this with diagram bmDke y we have the equivalent bend-
ing moment diagram cp'd'q'd. Since the pin is cylin
drical, the maximum ordinate of the latter diagram, that
at the centre, is the only one used in designing the pin.
The equivalent bending moment corresponding with
this ordinate is 159,360 pound-inches.
Crank C 2 . As has been shown, the right-hand crank
is subjected to twisting on its neutral axis ed and to
bending as a cantilever.
The bending is due to the fact that the crank pin is
twisted, through the medium of the crank, by the force
R 2 acting at e. The bending moment at the outer end
d of the crank is therefore R 2 x r= twisting moment
STATIONARY LOADS: SHEARS AND MOMENTS 69
on crank = 66,240 pound-inches; at the inner end e,
the bending moment is zero. Revolving dq to dr, the
bending moment diagram for the crank is edr, the ordi-
nates being horizontal.
As explained previously, the crank is twisted on its
neutral axis by a couple of force R z and of arm ef'
The twisting moment is therefore R% X ef. Considering
the right reaction and not the left as previously, the
bending moment ek at the section e is equal to the
product of the right reaction by the distance from e to
the right support, or R 2 x ef, which is also the twisting
moment as above. Revolving ek to ek 1 , the twisting
moment diagram is the rectangle ek'r'd. Combining
the two diagrams, we have the equivalent bending mo-
ment diagram ek"r"d. The maximum ordinate dr" of
this diagram, measured by the moment scale, gives an
equivalent bending moment of 100,800 pound-inches.
Crank C v The left-hand crank C^ is subjected to
similar bending and twisting actions.
With regard to bending strain, the crank is a cantilever,
fixed at b and carrying the load P at c. The bending
moment at the inner end b is therefore P x r 165,600
pound-inches, which is the twisting moment bn on the
left-hand section of the shaft; the moment at the inner
end c is zero. Revolving bn to bn, the bending mo-
ment diagram is cbu, the ordinates being horizontal.
By the same reasoning as that followed for the right-
hand crank, it can be shown that crank C^ is twisted on
its neutral axis be by the left reaction R^ acting with
the leverage ab. The twisting moment is therefore
70 GRAPHIC STATICS
R 1 x ab, which moment is equal to bm, the bending
moment on the shaft at the section b. Revolving bm
to bm' , the twisting moment diagram is the rectangle
bm'c'c.
The combination of the two diagrams gives the equiv-
alent bending moment diagram bm"c"c. The maxi-
mum ordinate bm" of this diagram corresponds with a
bending moment of 180,000 pound-inches.
Diameter of Crank Pin. The maximum equivalent
bending moment on the crank pin has been found to
be 159,360 pound-inches, which is the value of M in
the formula, M SI/c. If the pin be a solid cylinder
of diameter d lt the rectangular moment of inertia
/= Trd-f/64. and c d-^/2. Taking a suitable value for
the maximum stress S, the value of d l can be deter-
mined as in Example 13.
If the crank pin be of steel and made hollow for
lightness, the method is the same, except that the value
of / is changed, being now the difference between the
moments of inertia of the inner and outer circular sec-
tions. For example, if the external diameter be d^ and
the internal diameter d^, the value of /is Tr(d^ d^}/6^.
and the area of the cross-section of the pin is
A = Tr(df 4z 2 )/4' Dividing :
I = d i + 4 2
A 16 '
Making d% d-^/2 = c and 5 = 10,000 pounds per
square inch,
STATIONARY LOADS: SHEARS AND MOMENTS 71
1 0,000
d l = 5.6 inches, about.
The maximum shearing stress S, on the pin can be as-
certained from the formula R% x r= S 8 J/c (Example
13), in which the twisting moment, R 2 x r= 66,240 pound-
inches and the polar moment of inertia, J=A(d-f + d)/%.
With d l = 6 inches and d^ = c = 3 inches, J = 1 19 and
S 8 is 1670 pounds per square inch, a stress which is
immaterial. The diameter of the shaft can be found
similarly; its twisting moment is greater, and its maxi-
mum equivalent bending moment less, than those of the
crank pin.
With steel having an elastic limit of 50,000 pounds
per square inch, the factor of safety for the crank pin
is 50,000/10,000=5. The diameters as computed, 6
and 3 inches, external and internal, would serve for both
a hollow shaft and a hollow pin, if the shaft, cranks,
and pin were forged in one piece. If the crank shaft
be built up, keys would be required for both pin and
shaft, and the external diameters should be increased
to allow for the keyways.
Dimensions of Cranks. The cranks should be dupli-
cates and should be designed for the maximum moments,
which are given by bm" and dr", at the inner and
outer ends, respectively. As a rule, the cross-section is
rectangular and the thickness t of the web is constant.
For uniform strength throughout, the transverse width
72 GRAPHIC STATICS
w should vary with the maximum moments, as given by
either of the two equivalent bending moment diagrams
a refinement which is possible only with a casting.
The hubs on crank pin and shaft should be long
enough for adequate keying and to prevent the crank
from jarring loose. In built up shafts, this factor usu-
ally determines the thickness /, which is made uniform
throughout and equal to the length of the hub, while
the width w is also uniform and equal to the maximum
external diameter required for the larger of the two
hubs. In any event, the maximum stress at any section
of a crank of rectangular cross-section can be deter-
mined by proper substitution in the general formula,
M SI/c. M is here the maximum bending moment
at that section, as given by the equivalent bending
moment diagram ; 6" is the corresponding maximum
stress which occurs at either of the outer ends of the
width w\ /=/2^/i2; and c=w/2. The values for
/ and w are, in each case, taken from the dimensions- of
the section under consideration.
CHAPTER IV
LIVE LOADS: SHEARS AND MOMENTS
The shear and bending moment diagrams which have
been considered thus far represent these functions for
all sections of a beam carrying stationary loads only.
This limitation applies also when these diagrams have
been constructed for the moving parts of machines,
since the members in motion have been assumed to be
at rest for an instant and in momentary equilibrium
under the action of the driving, resisting, and support-
ing forces ; and, further, the shears and moments which
have been thus determined, refer only to the given
position of the moving parts.
While the general effects of a live or travelling load
as a train passing over a bridge are, at any instant,
precisely the same as those of an equivalent stationary
or dead load in the same position, there is, so far as the
maximum stresses in the members are concerned, an
important difference between the two cases, in that,
with every change in the location of the moving load
there are corresponding changes in the shears and mo-
ments which that load produces. Hence, in each mem-
ber of a structure traversed by a moving load, there is a
continual variation of stress, the maximum range of
73
74
GRAPHIC STATICS
which must be determined in order that the member
shall be proportioned for adequate strength under all
conditions of service. These variations in the shears,
moments, and their corresponding stresses will now be
examined.
15. Variation of Live Load Shear at Sections to the
Left of the Load, (a) Concentrated Loads. Let Fig.
21 represent a simple beam of span s, traversed by a
FIG. 21
single concentrated load W. The vertical shear V
(Art. 8) is the resultant or algebraic sum of all forces
to the left of the section considered. With a single
moving load, as in this case, Fis therefore equal to the
left reaction /? r Assume that the load is moving from
right to left, and let x be the distance from the right
support. Taking moments about the right support :
W y. x = R x s
which is the vertical shear for all sections between the
points A and B, and to the left of the load.
LIVE LOADS: SHEARS AND MOMENTS 75
This expression is the equation of a straight line db
inclined to the horizontal by the angle whose tangent
is W/s. When x = s, R l =V=W; when ;r = o,
R = V o. Therefore, the horizontal line ac is the
zero line, and, at any position of the load, the ordinate^
just below the load or, strictly speaking, immediately
to the left of its centre of gravity and included in
the diagram abc will give Ffor that section, and similarly
for all sections to the left of W. It is evident that this is
true, whether the load be moving to the right or left,
since, in this case, we have but one concentrated load
to consider. The diagram abc, therefore, represents the
variation of the vertical shear at all sections to the left
of the load.
For sections to the right of the load, V R 1 W, an
expression which is the same as the preceding equation,
except that there is a constant deduction W. It is,
therefore, the equation of the straight line cd, which
is parallel to ba t and starts at the point c of the zero
line. Hence, the diagram cad represents the variation
of the shear, now negative, for the sections to the right
of the load.
(b) Uniform Load. Now, consider a moving uniform
load, long enough to cover the span when located be-
tween the two supports of the beam. In practice, the
cars of a railroad train, although not the locomotive and
its tender, are considered as such a load in passing over
a bridge. This case differs from the one just discussed,
in that the total load on the beam varies with the length
of the segment of the span covered by the load.
7 6
GRAPHIC STATICS
Let Fig. 22 represent a simple beam of span s, carry-
ing a partial uniform load W t composed of unit loads
spaced at a unit distance apart. If x be the number of
unit distances covered by the load at a given time and
ft __ _ 5 _ w De the weight
of the unit load,
then the total
load at that time
will.be wx = W,
which is there-
fore variable.
This total load,
or resultant of
all of the unit
loads, will be
taken as concen-
trated at the
centre of gravity of the uniform load. It is required to
determine the variation in the vertical shear for all sec-
tions of the beam to the left of the head of this moving
uniform load.
Assume the load to be travelling from right to left.
For all sections to the left of the head H of the load,
V= R^ Taking moments about the right support :
^ x s = W x x/2 = w* 2 /2,
FIG. 22
which is the equation of a parabola. If x = o, V= o ;
if x s, V= ws/2. Plotting ab as a parabolic curve in
LIVE LOADS: SHEARS AND MOMENTS 77
accordance with the equation for V, the diagram abc
represents the variation of V, for all sections to the left
of H t while the load of final length s moves from right to
left. Wherever H may be located, the ordinate y im-
mediately below it will give the value of V from H to
the left support, for that position of the load.
Again, assume the load to be moving from left to
right and measure x from the left support. Taking
moments about the right support:
R^ X S = W(S X/2) WX(S X/2\
R l = wx zvx 2 /2s,
V= R l W wx wx^/2 s wx wx^/2 s,
which is the same equation as before, except that V is
now negative. If x o, VQ\ if x=s, V= ws/2.
Hence, the diagram cda represents the variation of V
for all sections between H and the left support, the
ordinate y under H giving the required magnitude of
F, which is, in all cases, negative.
16. Influence Diagrams : Influence Lines. The line
ab and the diagram abc, Fig. 21, are termed an influence
line and an influence diagram, respectively. In this
case, since VR^, they represent the variation of both
the vertical shear and the left reaction. Influence
diagrams are usually constructed on the basis of the
unit load a pound, kilogram, or ton. The ordinates
of the diagram then refer to the effect of that unit load
only, and, to obtain the actual reaction, shear, etc., for
the given load, the length of each ordinate is multiplied
by the number of pounds, tons, etc., in that load.
GRAPHIC STATICS
17. Variation of Live Load Shear at any Given
Section of a Beam, (a) Concentrated Load. Let Fig.
23 represent a single concentrated load W> moving
FIG. 23
from right to left over a simple beam of span s. It
is required to find the variation in the shear at the
section //, distant m units from the left support, while
the load traverses the beam.
Assume the load to be to the right of H and distant x
units from the right support. Taking moments about
that support :
W x x R x s,
which is the equation of the straight line ab, inclined to
the horizontal by the angle whose tangent is W/s. If
x = o, V=o\ iixs, V= W. Hence, ac is the zero
line, and the diagram abc gives the magnitude of V
for all sections between W and the left support ; and,
for the section at H, the ordinate y, immediately under
LIVE LOADS: SHEARS AND MOMENTS 79
the load and included within the partial diagram ade,
represents V, while W is to the right of H.
When W is between H and the left support, R l is
still equal to Wx/s t but V= R 1 W and is negative, so
that the line cf is parallel to ba and begins at the point
c, where xs and V=o. Hence, the ordinate immedi-
ately below the load and included within the partial
diagram gee
gives the shear "*
i """ ~ ~ ~ """
between the 1
? 1 H A
P, f*-- n >j
m ---- i
load and the
left support,
which shear is ^--j
always negative, w j
The complete
diagram, repre-
senting the vari-
ation of Fat the FIG. 24
section H for all
positions of the load, is adegce. This diagram serves
whether the load moves toward the right or the left.
(b) Uniform Load. Let Fig. 24 represent a simple
beam of span s, traversed from right to left by a uni-
form load of w pounds per unit of length of the span,
the load being long enough to cover the span. It is
required to find the variation in shear at the section H,
distant m units from the left support, while the head of
the load crosses the span.
Draw the zero line ab = s ; from b, erect be, and, from
a y let fall ad, both equal to w ; draw ac and bd. Let
8o GRAPHIC STATICS
x= length of the load, as measured from the right support.
Then, as shown previously, while the head of the load is
to the right of section H, the shear at that section is
positive and is :
V= R l = wx*/2 s.
With the head of the load at H, this expression is the
area of the triangle aef, forming the part of the diagram
below the load, since :
ef : x : : w : s,
ef wx/2,
area aef= ef x xJ2 = zux^/2 s = V.
When the lead moves to the section If', distant n
units from H, the shear at section H is the resultant of
the positive shear produced by the partial load of length
x n (x being now the distance from the right support
to the point H') and the negative shear due to the load
of length n, both in the positions shown. This resultant
shear is the difference between the areas of the positive
and negative sections of the diagram below the load,
or area aef minus fghk. As before, R l = wx^/2 s, but
the shear at section His now equal to R 1 minus the load
nw to the left of H, or
F= R l nw = wx*/2 s nw.
With the head of the load at If', wx?/2 s is the area
of the triangle akp, while nw is the area of the parallelo-
gram eghp. Hence
area akp area nw = area aef zxv&fghk.
18. Maximum Live Load Shear. From Fig. 24, it will
be seen that, when the given section, as that at If', divides
LIVE LOADS: SHEARS AND MOMENTS 81
the span into two unequal segments, the maximum numer-
ical value of the live load shear due to a uniform load will
occur when the load covers fully and only the greater
segment of the span, and that this value will grow with
increase in the length of the segment, as compared with
that of the span. If this greater and loaded segment lies
to the right of the section considered, the maximum live
load shear will be positive ; if to the left, negative.
With a series of concentrated loads differing in mag-
nitude and unequally spaced, the principle as to the
loading of the greater segment of the span to obtain the
maximum live load shear still holds, except that the maxi-
mum shear may occur when one or more of the loads
have passed beyond the section considered and are sup-
ported by the smaller segment of the beam. This fol-
lows from the fact that the value of the shear depends
primarily on that of the left reaction, and that the value
of the latter grows with every added load borne by the
beam. Hence, with some of the loads on the lesser
segment, the shear, or algebraic sum of the increased
reaction and these loads, may be greater than the reac-
tion only with the smaller number of loads which the
length of the larger segment will admit. The final
shear at any section is, in any case, the algebraic sum
of those due to the dead and live loads.
19. Counterbracing. The upper diagram, Fig. 25, is
an outline of a bridge truss* of the Pratt type, in which
* Bridge trusses are built either as deck or through spans. In the former,
the roadway rests on the upper chord; in the latter, on the lower chord.
82
GRAPHIC STATICS
V(-)
(a)
FIG. 25
the diagonal members are built to withstand tension
only. As shown in Fig. 24, a uniform load, like a train
of cars crossing a bridge truss, produces positive shear
in moving from the right abutment to the left and nega-
tive shear in passing in
the other direction. The
dead load shear is pos-
itive from the left abut-
ment to the middle of
the truss and negative
from the middle to the
right abutment. With
the train coming from
either direction and pass-
ing over a 5-panel truss, like that shown in Fig. 25,
the resultant of the shears due to dead and live loads
is always positive in the first and second panels
(counting from the left), and always negative in the
fourth and fifth panels; but, in the middle panel, this
resultant shear is positive when the train comes from
the right and negative when it enters from the
left.
The vertical shear which exists in a panel is a force
which acts upward if the shear be positive, and down-
ward if it be negative. In the Pratt truss, under any
system of loading, the vertical members are always in
compression, and the diagonals are built to take tension
only. Hence, when the shear in the middle panel
changes from positive to negative, or the reverse, there
must be a corresponding shifting, from one diagonal to
LIVE LOADS: SHEARS AND MOMENTS 83
the other, of the tensile stress whose vertical component
is to resist this shear, i.e. the main tie (diagonal) ceases
to act and the counter-tie takes the load, or vice versa,
the two ties being oppositely inclined.
Thus, in (a), Fig. 25, the shear in the middle panel is
positive ; the stress in the vertical 5-4 is compressive,
and therefore acts toward panel-point 4 ; and the stress
in the main diagonal 4-7 is tensile, and acts from point
4. The panel is subjected to the upward force V, and
the only vertical forces at point 4 are the compressive
stress 3-4 and the vertical component of the tensile
stress 4-7. For equilibrium, these vertical forces must
be equal and opposite. Hence, Fis equal in magnitude
to stress 4-5, and also to the vertical component of the
stress 4-7, and is the same in direction as stress 3-4
with regard to panel-point 4. When, as in (b\ the
shear in the panel becomes negative, V is equal and
opposed to the vertical component of the stress in the
counter-tie 5-6 1 and is equal and like the stress 4-3 in
the vertical, both with regard to panel-point 5.
When the main diagonal is acting, the counter-tie
buckles and is not under stress ; owing to their opposite
inclination, both ties cannot be strained simultaneously.
This is the fundamental principle of counterbracing,
which is applied whenever a member is subject to stress
reversal and is fitted to take tension only. In the War-
ren truss, the members are built for both stresses, and
counterbraces are not required. In roof trusses the
wind loads may cause a reversal of stress in some mem-
bers, thus necessitating counterbracing.
04 GRAPHIC STATICS
20. Variation of Bending Moment at any Given Sec-
tion of a Beam, (a) Concentrated Loads. Let Fig. 26
represent a simple beam of span s, traversed from right
to left by a single concentrated load W. It is required to
find the variation in the bending moment at the section
FIG. 26
A, distant s 1 from the left support and s z from that at
the right, while the load crosses the beam.
The bending moment (Art. 9) at any section is the
moment of the left reaction about any point in that sec-
tion, minus the similar moments of the loads, if any, to
the left of the section. First, let the load W be
between the section A and the right support, at the dis-
tance^ from the latter. The left reaction is R 1 =
and the bending moment at section A is :
J/o
Ws l x
LIVE LOADS: SHEARS AND MOMENTS 85
which is the equation of the straight line ab, inclined to
the horizontal by the angle whose tangent is s-^/s, the
weight W being taken, for convenience, as a unit load.
Now, assume the load Wto be between the section A
and the left support, at the distance x l from the latter.
Then, R l = W(s^ l )/s, and the bending moment at
section A is :
Taking ^as the unit load, this expression is the equa-
tion of the straight line cd, inclined to the horizontal by
the angle whose tangent is s%/s.
The triangle aec is the influence diagram for the
bending moments at section A, when W is equal to the
unitload, and the ordinate y 1 or j/ 2 immediately below
the load gives the bending moment at A for that posi-
tion of the load. This follows, since :
and
j 2 = s 1 x .r 2 /s = Mr
The diagram has been constructed for the unit load.
To obtain the bending moment for a load IV, greater
than unity, each ordinate should be multiplied by the
number of units in W.
(b) Uniform Load. The influence diagram in Fig^
26 is drawn for the unit load. With a total uniform
load W covering a segment or all of the span, there is
a series of unit loads, each equal to w t spaced at unit
86 GRAPHIC STATICS
distance apart, over the segment covered. The bending
moment due to each unit load is given by the ordinate
immediately below that load. Hence, the total bending
moment produced by the total load W is the sum of
these unit moments, or the area of the portion of the
influence diagram below the total load W.
Thus, in Fig. 26, let the uniform load be moving
from right to left and let the head of the load be at the
distance x z from the right support. Taking moments
about that support :
R^ x s 2 = wx 2 x ^ 2 /2 ;
and the left reaction is :
R^ = wx^/2 s.
The bending moment at the section A due to the total
load W= ivx, is :
R 1 xs 1 = ws^xf/2 s = M.
But:
and the area of the triangle below the load is :
j 2 x ^2/ 2 = V 2 2 /2 s=M t
since w t as the unit load, is equal to unity.
21. Maximum Bending Moments due to Live Loads.
The maximum bending moment which is possible at
any given section of a beam under any system of load-
ing, depends on two conditions : first, the value of the
moment rises with increase in the magnitude of the left
and, second, this value is lessened by increase
LIVE LOADS: SHEARS AND MOMENTS 87
in the moments of any loads to the left of the section.
Hence :
(a) With uniform live loads, the maximum moment
at any section of a beam occurs when the load covers
the entire span.
(b) With concentrated live loads, unequal in magni-
tude and unequally spaced, the maximum moment at
any section occurs when the span carries the greatest
possible load, consistent with having one of the largest
loads at or near the given section.
22. Live Load Stresses in Trusses and Plate Girders.
Either a truss or a girder, under the action of live loads,
is essentially but a beam, and as such is subjected to
bending moments and vertical shear.
In a truss such as that shown in Fig. 25, the primary
function of the web members as the vertical 4-5 and
the diagonal 4-7 is to resist the vertical shear, since,
if the chords be parallel, these web members are the
only ones having either vertical stresses or stresses
which can be resolved vertically. Hence, the vertical
component of the stress in any diagonal of this truss is
equal in magnitude to the vertical shear in the panel in
which the diagonal is located, and the stress in the
vertical meeting a diagonal at a panel-point in the
unloaded chord is equal in magnitude and opposite in
direction to the vertical component of the stress in the
diagonal.
The chord members as 4-6 and 5-7, Fig. 25
take the tensile and compressive stresses, respectively,
88
GRAPHIC STATICS
produced by the bending moments. The stress in either
chord, at any given section of such a truss, is equal to
the quotient of the bending moment at that section,
divided by the depth of the truss. Thus, Fig. 27 repre-
sents a segment of a Pratt truss and a corresponding
section of the equilibrium
polygon constructed for
z the loads and reactions, as
for a simple beam. Ap-
plying Rankine's Method
of Sections, let the truss
be cut on the line ab, im-
mediately to the right of
the vertical 2-3 1 so that
the bending moments at
ab and 2-3 will be virtu-
ally the same. To main-
tain equilibrium, apply to the severed members, the forces
F, F v and F^ each of which is equal to and like the stress
in the member to which it is applied. The segment of
the truss is then in equilibrium under the action of these
applied forces, the load P at panel-point j, and the left
reaction R v The two latter can be replaced (Art. 11)
by their resultant r, acting vertically through the point
E, the intersection of the prolonged sides BC and AD of
the equilibrium polygon. For equilibrium to exist, the
algebraic sum of the moments of r and the three applied
forces, about an axis perpendicular to their plane of action,
must be zero. Let d be the depth of the truss. Then,
taking moments about panel-point 2, where the lines of
LIVE LOADS: SHEARS AND MOMENTS 89
action of F l and F% meet and their moments are zero,
we have :
Fx d r X /
F = rl/d = Mjd.
since, by Art. 1 1, r x / is equal to the bending moment
at the section ab or 2-3. It will be observed that these
principles relating to the web- and chord-stresses, apply
only to trusses with parallel chords. When the upper
chord is ' broken ' and thus inclined to the lower, there is a
vertical component of its stress which must be considered.
In a plate girder, the web plate is usually assumed to
take only the same stresses as the web members of a
truss with parallel chords, and the flanges composed
of angles and cover-plates, if any to act as the chord
members. The customary method of design regards
the web plate as resisting vertical shear only and the
flanges as taking the full bending stress ; in more accu-
rate work, the resistance of the web plate to bending is
considered.
EXAMPLES
15. PLATE GIRDER BRIDGE WITH LOCOMOTIVE WHEEL
LOADS : MAXIMUM MOMENTS AND SHEARS. The weight
of a locomotive, as distributed among its driving and
pilot wheels, forms a system of vertical loads, more or
less unequal and unequally spaced. In designing a
bridge, the maximum moments and shears must be de-
termined for the greatest live load which will pass over
it. For simplicity in the following example, the loco-
motive only will be considered ; it is usual to design the
9 o
GRAPHIC STATICS
bridge for two coupled locomotives with their tenders,
followed by a uniform train load.
Let AB, Fig. 28, represent a deck, plate girder bridge
for a single track railroad, the effective span and depth
of the girders being 60 feet and 6 feet, respectively.
The load carried by the bridge is one Northern Pacific
tandem compound locomotive (1901), whose total weight
is 198,000 pounds, 175,000 pounds being on the driving
wheels and 23,000 pounds on the engine truck. These
weights, when distributed as shown in the figure, are, in
LIVE LOADS: SHEARS AND MOMENTS 91
tons of 2000 pounds, 10.94 tons on each driver and 5.75
tons on each pilot wheel. The weights on the wheels at
one side of the locomotive constitute the system of live
loads borne by each of the two girders forming the
bridge. The spacing of these loads is marked in inches
below the line A'B' indicating the span.
(a) Maximum Moments. The system of loads may
be located in any position on the span, although prefer-
ably the first driving wheel should be near the middle of
the girder, as in the figure. The moments and shears
should be found at sections reasonably close together
throughout the span. Therefore, divide the girder into
any number of segments, say 10, at intervals of 6 feet.
On the load line win, lay off the loads and draw the
force polygon Omn, with pole .distance equal to 5
times the depth, or 30 feet, since, by using thus a multi-
ple of the depth, the moment diagram is, by Art. 22,
available for the measurement of flange stresses. Con-
struct the corresponding equilibrium polygon acdefgb.
Then, the bending moment at any one of the numbered
sections of the girder will be given by the length of the
ordinate immediately below it and between the closing
line ab and the perimeter a e b of the equilibrium
polygon, this length being measured by the proper
moment scale (Art. 12).
Now, shift the span one interval to the right, as at
a $\\ prolong the side gb\ project a^ and b l to a^ and
'&i-, respectively, and draw the new closing line a^b^.
This operation is, in effect, moving the system of loads
a distance equal to one interval to the left on the span
92 GRAPHIC STATICS
as originally placed, and the equilibrium polygon for
this new location is a^ e b^ . The bending moment
at any section of the girder is then given by the length
of the corresponding ordinate above that section (as
marked on the line a-fi^ and included between the
perimeter and the closing line a^b^.
The span is then shifted successively to several new
positions to the left, as a 2 l? 2 , etc., each position differing
from the one immediately preceding it by the space of
one interval ; and the corresponding equilibrium poly-
gon is constructed in each case. This process is con-
tinued until every section, from the middle of the girder
to the left abutment, has been subjected to its maximum
bending moment. Each numbered section from I to 5,
in the various locations of the span, is then projected
upon the corresponding closing line of the equilibrium
polygon, and a free curve, known as the stress curve,
is drawn through the points thus determined. Thus,
CD and EF are the stress curves for sections i and 2,
respectively. The maximum bending moment at section
I is evidently given by the length of the greatest ordi-
nate included between the stress curve CD and the per-
imeter a e - b. The method for all other sections is
similar.
The maximum moments thus obtained for the live
load are plotted upward from the axis GK, for sections
i to 5, giving points on the semi-curve of maximum
moments GL ; the full curve GLK\s symmetrical about
its centre line LM, and is cusped at L, as shown.
The dead load, i.e., the total weight of one girder, of
LIVE LOADS: SHEARS AND MOMENTS 93
one-half the lateral bracing and cross-framing, and of
the floor system, may be taken, in this case, roughly, as
about 600 pounds per linear foot of the span, or 18
tons for each girder. This load is considered to be uni-
formly distributed- over the span, and the moments due
to it are represented by the curve GNK, which is
drawn, for convenience, below the axis G K, to the same
moment scale as before. The maximum moment at any
section, due to the dead and live loads, will then be
given by the length of the corresponding ordinate
included between the two curves. The curve GNK is
a parabola whose equation (Art. n) is :
M= WSX/2 ZVX^/2,
in which M is the bending moment in ton-feet at the
section distant x feet from the left support, w is the
uniform load in tons per lineal foot of the span, and s is
the span in feet.
(b) Maximum Shears. It can be shown by compu-
tation or by measurement from the diagrams that the
maximum live load shear at any section occurs when the
first driving wheel is immediately above that section, the
shear being then equal to the left reaction, less the load
of 5.75 tons on the pilot wheel. The maximum shear
can hence be found from a diagram similar to that in
Fig. 28, by successively locating each section of the
span below the first driver, constructing the correspond-
ing equilibrium polygon, and then (Art. n) drawing a
ray in the force polygon parallel to the 'closing line of
the equilibrium polygon. The distance between the in-
94
GRAPHIC STATICS
tersection of this ray with the load line mn and the point
m will then be the left reaction for that position of the
loads, and, deducting from this the first load of 5.75 tons
FIG. 29
on the pilot wheel, we have the maximum live load shear
at the section immediately under the first driving wheel.
A more convenient method is shown in Fig 29. A
new force polygon O'm'n' is constructed, having the pole
O' at a horizontal distance from m 1 equal to the span.
LIVE LOADS: SHEARS AND MOMENTS 95
Then, assuming the loads to be located as on the span
A'B' , Fig. 28, the corresponding equilibrium polygon will
be a'c'd'e'fg'b 1 . In the force polygon, draw the ray
O'a" parallel to the closing line b'd. Since the tri-
angles O'm'a" and b"a'b' are equal in all respects, b'b n is
equal to a"m' , and hence to the left reaction for this
position of the loads. This principle is general, applying
to all other values of the left reaction, when found
similarly.
Prolong the sides g'b' and a 'c', and draw the indefinite
line/^ parallel to d c\ and at a vertical distance from it
equal to the pilot load, 5.75 tons. Lay out the span
with section 5 immediately below the first driving
wheel ; the ordinary y above the right end of the span is
the left reaction for this position of the loads, and the
ordinate j/ 5 is the maximum live load shear at section 5
since the pilot load must be deducted. Similarly, j/ 3 is
the similar shear at section 3, and so on. When the
first driver reaches the left support, its line of action co-
incides with that of the left reaction, and the pilot load
is no longer on the girder. Hence, the side c'd' is pro-
longed to d", and the live load shear at section o is
equal to the left reaction, and is represented by the
ordinate j . In this method, it is assumed that the given
sections as 5, j, and o are successively located im-
mediately below the first driving wheel, which is only
approximately the position shown in Fig. 29, the latter
position being, for comparison, the same as that in Fig.
28.
As stated previously, the live load shear is positive
96 GRAPHIC STATICS
when the load moves from right to left. Laying off
these shears from the axis PQ upward, PQRS becomes
the diagram of live load shears from the left support to
the middle of the girder. The total dead load on each
girder is 18 tons. As this load is uniformly distributed,
the dead load shear is one-half of it, or 9 tons, at the left
support, and decreases uniformly to zero at the middle
(Art. 8). This shear is also positive, but, for conven-
ience, is laid off below the axis in the diagram TPQ.
The final shear at any section is then represented by
the corresponding ordinate in the combined diagram
STQR. The diagram for the right half of the girder,
and with the load moving from left to right, will be
similar but reversed, corresponding sections having
shears of the same magnitude, but both those due to
dead and live loads being negative.
(c) Flange Stresses: Impact. By Art. n, the bend-
ing moment at any section of a beam is equal to the
product of the corresponding ordinate in the equilibrium
polygon by the pole distance H of the force polygon.
By Art 22, the flange stress, which corresponds with the
chord stress in a truss, is equal to the bending moment,
divided by the depth of the truss or girder. In this
case, the pole distance in the first force polygon was
taken as 5 times this depth. Hence :
M=y x H,
77=54
A x S = M/d= Hy/d= 57,
in which M is the bending moment at the given section,
y is the corresponding ordinate of the bending moment
diagram, p is the depth of the girder, A is the total area
LIVE LOADS: SHEARS AND MOMENTS
97
of either flange, and 5 is the flange stress, per square
inch, at the given section. Hence, the moment diagram
GNKL, Fig. 28, will serve as a total (A x 5) flange
stress diagram, if its ordinates be measured by a force
scale 5 times greater than that used for the force poly-
gon, i.e., if the scale of Omn be 20 tons to the inch, the
flange stress scale will be 100 tons to the inch. The
unit stress at any given section of the web, due to the
vertical shear, can be found by dividing the total shear,
as given by the corresponding ordinate of the diagram
STQR, Fig. 29, by the cross-sectional area of the web.
In bridges, the stresses, as determined above, are in-
creased by an allowance for the impact resulting from
the sudden application of the live load. Formulas for
this allowance are given in works on this subject; the
increase is usually about 70 per cent.
(d) Summary. Tabulating the results, as measured
from the diagrams by the moment scale, we have, neg-
lecting the allowance for impact :
SECTION
I
1 2
3 4
5
Bending moment,
i
dead load, ton-feet o I 48.6 ! 86.4 i 113.4 \ 129.6 135.
Maximum moment,
live load, ton-feet . o
198.3 1 354- 1 498.5
558.
588.5
Total bending moment o 246.9
440.4 ! 6II.9
687.6
723.5
Vertical shear, dead
load, tons ... 9
7.2
5'4
3.6 1.8
Maximum live load
shear, tons .
37-5
33.2
28.5
23-5
18.6
13-7
Vertical shear, total,
tons 46.5
40.4
33-9
27.1
20.4
13-7
i ! '
9 GRAPHIC STATICS
1 6. WARREN GIRDER FOR OVERHEAD CRANE : MAXI-
MUM STRESSES. Girders for overhead cranes are of sev-
eral types. The rolled beam, the plate girder, the box
girder, and the truss in different forms have each its place
for this service. The selection depends mainly on the
span and the maximum live load to be transported,
although economy in weight and construction are pri-
mary requisites.
The upper diagram, Fig. 30, is the elevation in out-
FIG. 30
line of a Warren girder or truss with parallel chords,
as employed in such cranes. In this girder, the web
members are usually all diagonals and they are built to
withstand both tension and compression ; verticals may
be inserted as struts between each joint of the lower
chord and the middle of the panel above it, when the
length of a panel is considerable. The crab, which car-
ries the live load, has four wheels, tracking on the upper
LIVE LOADS: SHEARS AND MOMENTS 99
chords of the two parallel girders. The conditions are
thus, in general, the same as those of Example 15, ex-
cept that the plate girder has been replaced, in this
case, by the Warren truss. Let the girder, Fig. 30, be
a 6-panel, deck truss of 72 feet span and 6 feet in depth.
Consider first :
(a) Dead Load Stresses. The dead load on each
girder consists of the weight of one girder, its cross-
shafting, and platform. Take it as 6 tons and consider
it as uniformly distributed over the girder. Then, as in
the roof truss, Example 7, there will be a weight W
= i ton, acting at each of the inner joints of the upper
chord, and a weight W ' /2 = % ton, on each of the end
joints. Since the load is uniformly distributed, the re-
actions R 1 and R 2 are equal and each is 3 tons.
Lay off the loads ab, be, etc., on the load-line aa f .
Bisect this line at m. Then, ma R 1 and a'm = R 2 .
The closed force polygon for the loads and reactions is
then abcdd' c' b' a 1 ma. From m, draw the indefinite hori-
zontal line ml.
Joint I is in equilibrium under the action of the reac-
tion R l and the load W/ 2t . which are known, and the
stresses in the members BE and EM, which stresses
are to be determined. From b and m lay off be and
me, parallel respectively to BE and EM and meeting
at e. Then, taking the loads in clockwise order, the
closed force polygon for the joint is abema. The stress
be thus acts toward the joint and is therefore compres-
sive, while em acts from it and is tensile.
At joint 2 there are acting the known stress me and
100
GRAPHIC STATICS
the stresses in EFand FM, which are to be determined.
From e and m, lay off ef and mf t parallel to their cor-
responding members and meeting at/. The force poly-
gon is mefm, efis compressive, and/;;/ is tensile.
In a similar way, the force polygon for joints j to 8,
inclusive, are drawn. Thus, for joint j, the polygon is
bcgfeb\ for joint 4, fghmf\ etc. Since the load is uni-
formly distributed, the stress diagram is symmetrical
about its axis Im. It will be seen that the diagonals.
when under dead load, are alternately struts and ties,
the order changing at the middle of the truss.
(b) Live Load Stresses. The maximum weight to be
transported may be taken as 15 tons and the weight of
the crab as 5 tons, which will bring a load of 20/4 = 5
tons on each of the four wheels or runners of the crab.
The distance between these runners, measured parallel
to the girder, is 6 feet.
The graphic analysis of a truss is based on the as-
sumption (Art. 5) that all loads are transferred to, and
LIVE LOADS: SHEARS AND MOMENTS 101
concentrated at, the joints of the truss. Hence, if the
centre of the crab be above joint j, as in the partial
diagram, Fig. 31, each runner will be 3 feet from that
joint and 9 feet from the joint on the other side of the
runner. The load of 5 tons on each runner will then
be divided in the proportion of 3 to I between the two
adjacent joints. In the position shown, there will thus
be a load P = 3.75 x 2 = 7.5 tons on joint 3, a load P l =
1.25 tons on joint i, and a load P 2 = 1.25 tons on joint 5.
The crab and the weight transported thus form a sym-
metrical system of three loads spaced 12 feet apart.
The maximum stresses produced in the members of
the truss by the travel of the crab and its load can be
found by locating the system of loads centrally above
panel-points j, 5, and 7, in succession, and taking, for
each member, the greatest of the three stresses developed
in it by these three locations of the system of loads.
Since the crab may approach the middle of the girder
from either end, it will be necessary to determine the
stresses in one-half only of the members.
There are several ways of rinding the changes in the
stresses due to these changes in the location of the loads :
first, the method of the equilibrium polygon can be em-
ployed, as in Example 1 5 ; second, the system can be
located at each of the three panel-points in succession,
and a separate stress diagram drawn for each case ; and,
finally, a single stress diagram, for one load only at one
panel-point, can be constructed, and the stresses due
to the system at each of the three panel-points can be
found by proportion from this stress diagram. This
IO2
GRAPHIC STATICS
method of multiples is essentially the use of the influence
diagram in modified form. While possibly not the most
convenient process in this case, it is given below for in-
formation.
Assume the central load P= 7.5 tons to be the only
one on the girder and to be located, as in Fig. 31, at
panel-point j. The reactions are : R 1 = 6.2$ tons and
R = 6, the radius of the friction circle :
r l sin < = 0.625 x 0.1045 = 0.065 inch.
In Fig. 57, draw the upper sheave, the friction circle
and the dotted tangent at G. Let fall the tangent at B
until it meets at o 1 the indefinite horizontal line HK.
Draw the line of action cd parallel, to the left, and at
a distance / from the tangent at G\ this line meets HK
at H t thus locating the latter point. From o l set off to
the left the distance o^ = 2(^ sin + /) = 0.866 inch ;
from and K-6, and dividing their sum by
H-7, we have 3.7, which is the value of the ratio W/P.
Hence W 3700 pounds. If stiffness and friction
were not considered, the value of this ratio would be 6.
The efficiency of the tackle is thus 61.66 per cent, which
is a fairly accurate result. This method assumes that
the section 7\ of the rope is vertical and thus parallel
to the other sections.
The value of the ratio in Equation (3) is 0.865.
This value is constant throughout the system, and gives
the relation between the tensions in any two consecutive
sections of the rope.
FRICTION
173
26. SPUR GEARS : RELATION OF LOAD AND POWER.
Figure 58 represents the pitch circles of a train of
spur gears, A, B, and C, with involute teeth. The power
applied to the driving gear A is equivalent to a force P
of arm L ; the resistance acting
on the gear C is equal to a force
Wot arm L 1 . Assuming fric-
tion at the wheel journals and
between the engaged teeth, it
is required to determine the
magnitude of the force P for
a known resistance W.
The driving gear A is in
equilibrium under the action
of the force P, the reaction
7*! from the teeth of gear B,
and the virtual reaction A\ from
the bearing. By Art. 38, the
reaction 7\ is parallel to the
line aa' passing through the
point of tangency of the pitch
circles, the distance of 7\ from
the centre of A being greater
than that of aa 1 by the amount
given in the article cited. The
lines of action of P and 7^ intersect at c, and the reac-
tion R l drawn from c is, by Arts. 34 and 35, tangent to
the lower side of the friction circle.
The intermediate gear B is in equilibrium under the
FIG. 58.
174 GRAPHIC STATICS
action of the force T v the reaction T 2 from the teeth of
gear C, and the virtual reaction R^ from the bearing.
Since the gear B is the driver for gear C and its motion
is opposite to that of gear A, the reaction T 2 has, by
Art. 38, the direction and location shown in the figure,
being parallel to the line bb' passing through the point
of tangency of the pitch circles. The lines of action of
TI and T 2 intersect at d t from which point the reaction
R 2 is drawn tangent to the lower side of the friction
circle.
The driven gear C is in equilibrium under the action
of the resistance J/F, the force 7" 2 , and the reaction R%
from the bearing. The lines of action of T 2 and W
meet at ^, from which point the reaction R z is drawn
tangent to the upper side of the friction circle.
The directions of all the forces acting on the train and
the magnitude of one, the resistance W, are known.
Starting with W, the force triangle for each gear and
the force polygon for the train can be drawn by the
general method, as shown by the diagram below.
For these gears, the angle of obliquity (Art. 38) is
usually about 15 degrees, that is, the line bb' makes an
angle of 15 degrees with the horizontal.
PROBLEMS
1. What weight can be drawn up an inclined plane rising I in 5
by a pull of 200 pounds : (a) when the pull is parallel with the
plane ; (b} when it is horizontal ? Disregard friction.
2. A wheel of 5 feet diameter weighs 2 tons, including its load.
What is the least horizontal force necessary to pull it over a stone
4 inches high ?
3. A boiler weighing 5000 pounds is supported by tackles from
the fore and main yards. If the tackles make angles of 30 and 35
degrees, respectively, with the vertical, what is the tension in each
of the two ?
4. Sheer legs 50 feet long are spread 18 feet at the base. The
back stay is 80 feet long. Find the stresses acting in each member
when lifting a load of 25 tons at a distance of 20 feet from the foot
of the sheer legs, the weight of the latter being disregarded.
5. A crane post is 12 feet high ; its jib, 32 feet long, and stay, 23
feet long, meet at the peak A. Two backstays, making angles of 45
degrees with the horizontal, are in planes due north and due west
from the post. A weight of 6 tons is suspended from A. Find the
forces in the jib and stays : (a) when A is southeast of the post ;
(b) when A is due east; (<:) when A is due south.
6. A rod weighing 10 pounds rests in a smooth hemispherical
bowl which is fixed with its rim horizontal. The rod is 15 feet long,
and a length of 3 feet is outside of the bowl. If the inclination of
the rod to the horizontal be 30 degrees, find the reactions of the
bowl.
7. A simple triangular truss, 24 feet span and 3 feet deep, is
supported at the ends and carries a load of 3 tons concentrated at
the middle. Find the stresses in each member. . .
175
I7 6 GRAPHIC STATICS
8. The span of a roof is 16 feet ; length of rafters, 9 and 12 feet,
respectively. The rafters are spaced 2 feet apart and the roofing
material weighs 16 pounds per square foot. Find the thrust on each
rafter and the stress in the tie bar.
9. Let a roof truss, whose general plan is that shown in Fig. n,
be unsym metrical, the rafters to left and right having inclinations of
50 and 45 degrees, respectively, to the horizontal. The span is 45
feet ; the trusses are spaced 10 feet apart ; and the weight of the roof
covering and snow is 40 pounds per square foot of roof surface. Find
the stresses in the members.
10. Find the stresses in the members of this truss due to wind
load, assuming the right end of the truss to be free. Take the hori-
zontal wind pressure as 40 pounds per square foot.
11. A king post truss has a span of 16 feet and a rise of 8 feet.
Find the stresses in the members due to a load of 18,000 pounds at
the middle.
12. A floor beam, 18 feet long and carrying a uniform load of 180
pounds per linear foot, is trussed by rods which are i| feet below
the middle of the beam. Consider the rods as jointed in the middle
of the beam, and find the stress in each rod. Disregard the weight
of the beam.
13. A simple beam, 18 feet long, carries a load of four tons.
Draw the bending moment and shearing force diagrams : (#) when
the load acts at the middle of the beam ; (b) when it acts 5 feet
from one of the ends. Disregard the weight of the beam.
14. A simple beam is 18 feet long and loaded with one ton evenly
distributed along its span. Find the bending moment and the shear-
ing force at a distance of 6 feet from one end and also at a distance
of i foot from the middle. Disregard the weight of the beam.
15. On a common steelyard, a load of 1600 pounds is balanced
by 360 pounds. Draw the diagrams of bending moment and shear-
ing force in the steelyard bar, when thus loaded.
PROBLEMS 177
16. A countershaft, 6 feet long between centres of bearings, car-
ries a driving pulley A, 20 inches diameter and distant i foot from
the centre of the left-hand bearing, and a driven pulley J3, 18 inches
diameter and distant 3! feet from A. In both cases, the belt-drive
is horizontal, being toward the front at A and to the rear at B.
The tension of the driving side at A is 200 pounds; at J3, 170
pounds. The weight at A is 80 pounds ; at B, 60 pounds. Find
the maximum equivalent bending moment on the shaft.
17. In Example 14, let the cranks be offset, the neutral axis of
each making an angle of 120 degrees with the centre line of the
shaft. With this modification, draw the moment diagrams through-
out, as in Fig. 20.
18. In a 4-cylinder, triple expansion engine, the high-pressure
and intermediate-pressure cylinders are 32 and 52 inches, respec-
tively, in diameter ; the stroke is 48 inches ; length of connecting
rod, 96 inches ; the H.P. and I. P. cranks are 90 degrees apart, the
H.P. leading; the initial, unbalanced pressure on the H.P. piston
is 200 pounds per square inch ; on the I. P. piston, 50 pounds; the
length of each crank pin is 19 inches; and the average length of
the three bearings one at each end of the section of the shaft and
one between the two pairs of cranks is 24 inches. Find the
maximum bending moment on the H.P. and I. P. section of the
shaft, due to the loads on the two pistons and the corresponding
reactions at the bearings.
19. Assume three concentrated loads, P\, P^ and Ps, spaced at
the constant distance a between the first two loads and b between
the second and third. Let this series of loads cross a beam of span
s from left to right. Find the influence line for the left reaction.
20. Find the influence line for maximum shear for the series of
loads, as above.
21. In Fig. 28, let there be two coupled locomotives crossing the
plate girder bridge from left to right, the weights and spacing being
those given in Example 15 and the locomotives being 9 feet apart.
Find the maximum moments and shears.
178 GRAPHIC STATICS
22. Let the 5-panel truss, Fig. 25, have a span of 125 feet and a
depth of 26 feet. Take the live load as the two coupled locomotives
in Problem 21, followed at a distance of 5 feet by a train weighing
5000 pounds per lineal foot. Determine the maximum stresses in
the members and find in which panel or panels counterbracing is
necessary.
23. Let the Warren girder, Fig. 31, have a ' broken' lower
chord, the depth at joints 6 and 8 being 6 feet ; at joints 4 and 10,
5 feet ; and at joints 2 and 12, 3 feet, 4 inches. Find the maximum
live load stresses in the members, due to the load given in Example
1 6.
24. Find the centre of gravity of the deck beam whose dimen-
sions are given in Example 20.
25. Find the centre of gravity of an angle with unequal legs,
7x3! inches by T \ inch thick.
26. Find the rectangular moment of inertia of the bulb angle
whose dimensions are given in Example 19. Take the neutral axis
as perpendicular to the web and use the method of the area of the
equilibrium polygon.
27. The cross-section of a channel has the following dimensions :
depth, 15 inches; area of section, 9.9 square inches; thickness of
web, 0.4 inch ; width of flange, 3.4 inches ; thickness of flange, 0.9,
tapering to 0.4 inch ; distance of centre of gravity from outside of
web, 0.794 inch.
Find the moment of inertia : (a} with neutral axis perpendicular
to web at centre ; () with neutral axis parallel with centre line of
web.
28. A cast iron frame weighing 300 pounds is moved across a
smooth iron floor plate by a force P acting at an angle of 45 degrees
with the horizontal. Take the coefficient of friction as 0.4 and find
the magnitude of P.
29. The joint between the two sections of a vertical pump rod is
made by a through key having one horizontal side and one side
PROBLEMS 179
tapered to an angle of 3 1 degrees with the horizontal. In making
the joint, the total stress to be produced in the rod is 1000 pounds.
Using a coefficient of friction of 0.25, find the force which will be
required to drive the key home.
30. The cover of a steam engine cylinder, 34 1 inches in diameter
is secured to the cylinder by thirty-four ij-inch steel studs. Con-
sidering friction, what power, applied to the nut by a wrench 2 feet
long, will produce a total stress in each nut of 4500 pounds ?
31. Find the efficiency, at the mid-point of the forward stroke,
of a double-acting, oscillating engine, disregarding the friction of
the piston and piston rod. Data: Cylinder diameter, 15 inches;
stroke, 18 inches; diameter of crank pin, 2| inches; of trunnion, 6
inches ; unbalanced pressure on piston, 100 pounds per square inch ;
coefficient of friction, 0.185.
32. The shaft of a marine engine is transmitting 4000 horse
power when revolving at a speed of 125 revolutions per minute, with
the ship making 15 knots per hour. The thrust shaft hits an ex-
ternal diameter of 13! inches. There are 11 thrust collars, each 2
inches wide, working in a cast iron bearing lined with white metal.
Assume as was formerly the practice that the collar bearing is
continuous throughout its circumference, and, with a coefficient of
friction of o.i, find the power expended in overcoming the friction
of the thrust bearing.
33. Following the general methods given in Example 25, ana-
lyze the action of the Weston Differential Pulley Block or chain
hoist. Take the coefficient of journal friction as o.i, that of chain
friction as 0.2, and the ratio between the diameters of the two
sheaves in the upper block as ^.
Show the relation between the power and the load and the
reasons for the large loss by friction. Prove that the tackle is self-
locking with regard to backward motion.
34. In Example 26, let the pitch diameters of gear wheels A, B,
and C be 6, 8, and 12 inches, respectively. Interpose a lo-inch
180 GRAPHIC STATICS
gear wheel between wheels B and C. Taking the force P as 75
pounds, the arm L as 8 inches, the arm L' as 15 inches, the coeffi-
cient of journal friction as o.i, and that of tooth friction as 0.12, find
the magnitude of the resistance W.
35. In Fig. 52, interpose a belt-tightening pulley to the left of
the slack side (7\) of the belt and near the driving pulley A, thus
altering the line of action of the slack side and increasing its arcs
of contact. With this modification, construct the force polygons
for the entire gear by the methods used previously.
INDEX
Angle of friction, 136
Area, moment of inertia of, 121, 130
partial, centre of gravity of, 1 27
Axes of inertia, parallel, 1 20
Axis, friction, 149
neutral, 41
Beam, uniform load, 50
Beams, 41
Bellcrank, 10
Belt gearing, 155
Bending moment, 43, 84
maximum, 86
Bending and twisting moments, 49,
61, 65
Bending of ropes, 152
Bridge trusses, 82
Bulb angle, centre of gravity of, 125
Cantilever beams, 41
Centre of gravity, no
of bulb angle, 1 25
of partial area, 127
Centrifugal force, load due to, 52
Centroid of parallel forces, in, 113,
124
Chain friction, 152
Character of stress, 25
Chord stresses, 87
Circle, friction, 145
Coefficient of friction, 135
Collar, friction of, 142
Cone, friction, 137
Continuous beams, 41, 56
Counterbracing, 81
Counter shaft, 61
Crane truss, 35
pillar, 1 6
Crank shaft, 65
Dead load, Pratt truss, 105
roof truss, 26
Warren girder, 99
Deck beam, moment of inertia of,
128
Diagonals, stresses in, 87
Diagrams, influence, 77
shear and moment, 44
stress, 24
Elastic curve, 41
Engine, stationary, 14, 160
Equilibrium, conditions of, 9
polygon, 4
Equivalent moments, 49
Force polygon, 3
triangle, I
Framed structures, 23
Friction, 134
angle of, 136
axis of, 149
circle, 145
coefficient of, 135
cone, 137
force of, 135
of chains, 152
of collars, 142
of gear teeth, 158
of journals, 145
181
182
INDEX
Friction
of motion, 135
of pivots, 142
of plane surfaces, 137
of rest, 135
of screw jack, 164
of screw threads, 139
of stationary engine, 160
radius of, 145
rolling, 135
sliding, 135
Gear teeth, friction of, 158
Girder, stay, 56
Warren, 98
Gravity, centre of, no, 127
Gyration, radius of, 115
Inertia, moment of, 115, 118, 121,
128, 130
Influence diagrams, 77
lines, 77
Journals, friction of, 146
Lines, influence, 77
Link connections, friction of, 149
Live load, 73
shear, 74, 78
shear maximum, 8l
stresses, 87
Warren girder, 99
Load, dead, 27
snow, 27
uniform, 50, 52, 56
wind, 30
Loads on trusses, 24
Locomotive side rod, 52
wheel loads, 89, 124
Maximum bending moment, 86, 89, 98
shear, 81, 89, 98
Moment, bending, 43, 84, 86, 89, 98
diagrams, 44
equivalent, 49
of inertia, 115
of inertia of area, 121, 130
of inertia of deck beam, 128
of inertia of parallel forces,
118
resisting, 43
scale, 47
twisting, 48
Motion, friction of, 135
Neutral axis, 41
Notation, truss, 27
Parallel axes of inertia, 120
forces, centroid of, in, 113
forces, moment of inertia of,
118
Pawl and ratchet, 1 1
Pillar crane, 1 6
Pivots, friction of, 142
Plane surfaces, friction of, 137
Plate girder bridge, 89
girder stresses, 87
Polygon, equilibrium, 4
force, 3
Pratt truss, 105
Pulley blocks, load vs. power, 167
Radius of friction, 145
of gyration, 115
Rachet, 11
Ratchet-rack, 12
Resisting moment, 43
Rest, friction of, 135
Rolling friction, 135
Roof truss, dead load, 26
wind load, 30
Ropes, bending of, 152
INDEX
183
Scale, moment, 47
Screw jack, friction of, 164
threads, friction of, 139
Shaft, centre crank, 65
counter, 61
Shear diagrams, 44
live load, 74, 78
maximum, 81, 89, 98
vertical, 42
Sheer legs, 20
Simple beam, 41
Sliding friction, 135
Snow load, 27
Spur gear train, load vs. power, 173
Stationary engine, 14, 160
Stress, character of, 25
diagrams, 24
Supports, free, 41
restrained, 41
Triangle, force, I
Truss, bridge, 82
crane, 35
Pratt, 105
roof, dead load, 26
roof, wind load, 30
Warren, 98
Trusses, live load stresses, 87
Twisting and bending, 49, 61, 65
moments, 48
Uniform load, 50, 52, 56, 105
Vertical shear, 42
Verticals, stresses in, 107
Warren girder, 98
Wind load, 30
D. VAN NOSTRAND COMPANY
23 MURRAY AND 27 WARREN STREETS
NEW YORK
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