When a weight is over the point 1 foot from the centre, another is just over the left point of support, and we may say that either 11 or 10 weights are on the span; that is, the two loci intersect at x=1. The maximum for the 10 weights is the same as the ordinate at x = 1 to the locus for 11 weights, and its height is less than the above calculated value.

109. A beam 39 feet span is subject to the transit of the same load as in No. 108. Find the maximum bending moment during transit.

wM9x
9 × 19.5 −2(4+8+12+16) = 95·50 ft.-tons.

Now

=

38 × 18.5 × 18.5 − 2 (4 + 8 + 12 +16)=95·51 ft.-tons; and this being the greater is the maximum for transit.

Only 10 weights can be on the span at one time; and if we suppose these spread uniformly over their actual length, that is over 9 intervals, we have the approximate locus fig. 41,

Mx = g(1952-x2)(2-3);

and its maximum value is M 105 ft.-tons.

SHEARING FORCES AND SHEARING FORCE DIAGRAMS
FOR FIXED LOADS.

DEFINITION. The Shearing Force at any cross section of a beam, or, as we may more conveniently say, at any point of the span is,-The algebraic sum of the external forces acting upon the portion of the beam to either side of the point.

We will denote this shearing force by F, and calculate its amount systematically from the forces acting on the portion of the beam to the left of the point; thus for a beam

Similarly for a cantilever—

Fx-W1-W2- &c.....

=

1

2

(2)

{positive} according as P is

negative

On the beam then, Fx is

than the sum of the weights to the left of the

{ greater}

section at x. The beam may be divided at a certain point

left

{right}

into two segments, such that for every point in the right negative }; at this point the

segment, the shearing force is { positive }

shearing force changes sign. The shearing force on either segment may be considered as positive; for convenience we have chosen as just stated. Having fixed the sign thus, on a cantilever as shown on the diagrams, that is with its fixed end to the right, the shearing force is everywhere negative; and in considerations regarding beams and cantilevers, it is necessary to take this into account; when, however, the cantilever alone is considered, it will be better to reckon the shearing forces as positive.

It may be observed that the symbolical expression for Fx does not depend for its form upon the position of the origin; in this respect it is unlike Mx.

A Shearing Force Diagram is a figure having a horizontal base representing the span on a convenient scale, and an outline or locus. For a cantilever this locus lies wholly under the base; for a beam, it lies above the base for the segment to the left, under the base for the segment to the right, and crosses the base at an intermediate point. The height of any point on the locus, measured on a scale for forces, say tons or lbs., gives the shearing force at the point of the span over which it stands. Both these scales should accompany the drawing.

Maximum Shearing Force.-On a cantilever it is evident that the greatest value is at the fixed end. On a beam there are two values each greater than the others near it, one at the left end and positive, one at the right and negative; the value of the greater of these is the greatest for the whole span.

This locus, in the diagrams for all the cases of fixed loads which we consider, consists of straight lines. For a portion of the span between any two adjacent loads, the straight line is evidently horizontal; for portions uniformly loaded it slopes at a rate given by the number which indicates the intensity of the load; thus if w lbs. per foot be the intensity of the uniform load, then w vertical to one horizontal is the slope. Where a weight is concentrated at a point, the line is vertical; that is, at such a point the locus makes a sudden change of level, the change being equal to the weight.

On a cantilever, the shearing force at the fixed end is equal to the load, and at the free end it is zero; we can readily draw the straight lines as above described to suit the nature of the load, and so complete the diagram.

On a beam the shearing force at the left end is P, at the right end it is -Q, and at some intermediate point the locus intersects the base and changes sign; the manner of fixing the position of this point will be explained immediately. The whole locus is then readily completed by drawing the lines as above to suit the nature of the load.

Theorem. The Shearing Force at any point of a beam or cantilever is the rate of variation of the bending moment at that point; and on the beam the shearing force changes sign at the point of maximum bending moment.

As before,

Fx = P - Σ(W),

where (W) means the sum of the loads to the left of x; and MxP(c-x) - Σ(W.x),

х

where (W.x) means the sum of the products got by multiplying each load to the left of x by its leverage about x. Hence if we suppose Fx to be positive, and take the bending moment at any interval d further to the right, the second bending moment will exceed the first by F. d, if there be no load on the portion d; and by Fx. d, minus the load over the portion d into its leverage about x, if there be a load on the portion d.

Since the leverage of the load which is on the portion d is less than d, we can by taking d small enough make this

product as small as we please; that is, F.d is the change of the bending moment in passing from a through a small interval d. Now the rate of change of My means the change in passing from a through an unit interval, say one foot, if the change continued uniform throughout that interval, and at the same rate as at x; in other words, the change in M, for d reckoned equal to unity, without taking into consideration any additional loads which may be over that interval; hence the

.........

Rate of change of MxFx ;......... ...(1) also, as we pass from the left to the right end of the span, S positive

if Fx be

negative }, M, is { decreasing}, and at the point

where Fx changes sign M, is a maximum.

COROLLARY. For any system of fixed loads the shearing force only once changes sign, since there is only one maximum bending moment.

These general remarks will suffice for the analysis of shearing forces due to fixed loads, and we will now give the graphical solutions.

Beam under unequal weights at irregular intervals.-Fig. 68. P and Q are determined graphically by finding the point g, fig. 19. Through g draw a horizontal line and produce the lines of action of P, W1, W,,... Q to cut it; join these lines of action in pairs by horizontals through a, b, c, d, e, f respectively. The scale used for forces is also the scale for shearing forces.

The construction is cyclical; through each of the points a, b, c, d, e, f, g a horizontal line being drawn to join the lines of action of the forces in pairs, viz:-From a joining P and W1, from b joining W, and W, from c joining W, and W,, from d joining W, and W, from e joining W, and W,, from ƒ joining W, and Q, and from g joining Q and P.

5

3

4

2

Cantilever under unequal weights at irregular intervals.-Fig. 69. From a, b, c, d respectively, fig. 21, draw horizontals joining the lines of action of

(SeeFig.21)

the forces in pairs, the vertical
through the fixed end being W
reckoned as a line of action.
Thus draw horizontals, from a
joining the vertical through the
fixed end and W1, from b join-
ing W, and W,, from c joining
W, and W,, and from d joining
W and the vertical through
the fixed end. The scale for
forces is also the scale for
shearing forces.

2

3

Fig.69.

COROLLARIES. For a beam with W at the centre, the shearing force diagram consists of two rectangles of height W, one standing above the left half and the other below the right half of span.

For a beam loaded with W at a point dividing the span into any two segments, the shearing force diagram consists of two rectangles, one standing above the left segment, the other below the right segment. The height of each is inversely proportional to the length of the segment on which it stands, and the sum of their heights is W.