section at which the bending moment is required; let c denote the length of the cantilever. Let M denote the bending moment, and S the shearing

out the length of the beam. Hence the bending moments are represented graphically in Fig. 26, where the maximum bending moment We is drawn to scale, and the bending moment at any point may be found by measurement. In a similar manner Fig. 27 represents the constant shearing stress. If W = 10 lbs., and c 8 inches, then We 80 inch-pounds, so that the bending moments are expressed as inch-pounds or foot-tons, according to the units of load and distance used. The shearing force is 10 lbs. Case II., Figs. 28, 29, and 30.-A cantilever loaded with a uniformly distributed

load of W per unit of length over its whole length. Adopting the same notation as before, and remembering that the uniform load to the right of the section may be considered as concentrated at its centre of gravity, and acting with a leverage equal to the distance

of the centre of gravity

w per unit of length

FIG. 28.

FIG. 29.

FIG. 30.

x

from the section under consideration, viz.

we have

2'

which is a maximum when x = c, viz. M =

wc2

fixing. Again, the shearing stress is equal to the load on the right of the section, which increases uniformly to the point of fixing; hence Fig. 29 represents the bending moments, and Fig. 30 the shearing stresses on the same principles as explained in connection with Case I. It may be observed that in Case I. the equation of bending moments is that of a straight line, while w.x2 2

in Case II. the equation M = is that of a parabola, the origin

being at the extremity; hence we may determine points in the curve by solving the equation for M after substituting for x its value in feet or inches, and for w its value in tons or pounds; or the curve may be drawn by any of the geometrical methods, We2 having first calculated the maximum ordinate 2

Let w =

a ton per foot run, and let c = 6 feet; then—
if x = 1; M = 0.25 foot-tons.

bending moments and shearing stresses are shown in Figs.

32 and 33.

Case IV., Figs. 34, 35, and 36.—A cantilever loaded with any number of concentrated loads.

Let W1, W2, W3, etc., denote the loads, and X1, X2, X3, etc., the distances from the fixed end respectively. Then the moment at b is MW3(x3x2); in a similar manner Ma W3(X3-x1)+W2(X2−X1). The moment about the

=

fixed point is

=

MW3x3+W2x2 + W11 +...

FIG. 34.

FIG. 35.

FIG. 36.

The shearing stress is zero from the extremity to c; from c to b it is W3; from b to a it is W3 + W2; from a to the point of fixing it is W3+W2+ W1+...

Hence the diagrams Figs. 35 and 36 represent the bending moments and shearing stresses respectively.

Case V., Figs. 37, 38, and 39.-A cantilever partially loaded

with a uniformly distributed load of w per unit of length (Fig. 37). Let x denote the distance of the section

of the beam at which wa

FIG. 37.

wb'

FIG. 38.

the bending moment

is required from the

wb

point of fixing; then,

FIG. 39.

considering the loads to the right of the section, we have: Load

The bending moment at a section situated at a distance a from the point of fixing is found by making xa in the foregoing equation, thus

To find the bending moment between this last section and the point of fixing, we observe that a is greater than x; thus

The maximum bending moment occurs at the point of fixing, where x = 0; then

Mwb
cv(" + a)

We observe that the diagram of bending moments, Fig. 38, is a parabola immediately under the load, and a straight line (viz. the tangent at the extremity of the parabola) from the left extremity of the load to the point of fixing. The diagram of shearing stresses, Fig. 39, needs no comment.

2

Case VI., Figs. 40, 41, and 42.-A beam supported at both

on the points of support will be equal to one another, and to half the load W, and there will be a corresponding reaction acting upwards of the same amount. We may consider the beam fixed at the point where the bending moment is required, and consider the forces on one side of it as acting on a beam fixed at one end. Considering the forces to the right of the section, and denoting forces acting upwards by the sign —, and those acting downwards by the sign +, we have—

W
(c

which is a maximum, and =

We
2

when x = o, i.e. at the centre

of the beam.

If we had taken the moments of forces to the left of the section, we should have found—

The shearing stress is constant from the supports to the

centre, and equals

W
2

; at the centre the shearing stress is zero.

The diagrams of bending moments and shearing stresses are shown in Figs. 41 and 42 respectively.

Case VII., Figs. 43, 44, and 45.-A beam supported at both ends, and loaded at

a point between the centre and one of the supports (Fig. 43). Let m and n denote the segments into which the load W divides the beam, and let x denote the dis

Wm

tance of any section

FIG. 45.

from the right sup

port. The reaction at the left support is, by the law of the lever,

Wm

which equals a maximum when xm, i.e. at the point of application of the load

Wmn
M =
l

E