In these two cases it is evident, and it can easily be proved for a general case fig. 68, that the area of the part of the shearing force diagram above the base is equal to the area of the part below the base.

W

W

For a cantilever with W at its free end, the shearing force diagram is evidently a rectangle of height W, standing below the span.

Beam uniformly loaded.-Fig. 71. Draw verticals, one upwards from the left end, another downwards from the right

end of span, each equal to W; join their extremities with a straight line, which will cut the base at O the centre.

The shearing force diagram consists of two right-angled triangles, whose common height

is W; one is above the left half, the other is below the right half, of span, and the right angle of each is at the end of span.

Cantilever uniformly loaded.-Fig. 72. Draw a vertical downwards from the fixed end of

the base equal to W, and join its extremity to the free end of the base.

The shearing force diagram is a right-angled triangle of height W, standing below the span, and with the right angle at the fixed end.

W

Fig.72.

F. = W

Beam under an uniform load together with weights fixed at intervals.—Fig. 73. The two diagrams figs. 68 and 71 are to be directly superimposed by shifting the portions of the triangle till they stand upon the steps of the other figure. Another method is,-Draw DE to represent the shearing force for the uniform load alone, and on this line as a sloping base make the construction for fig. 68.

Cantilever under an uniform load together with weights fixed at

intervals. Fig. 74. The diagram fig. 72 is to be superimposed upon the diagram fig. 69, by shifting the portions of the triangles till they stand upon the steps of the other figure.

W

Cantilever uniformly loaded on a portion of its span.—Fig. 75. The left end of the load is the free end of the cantilever. Draw downwards from the fixed end a vertical equal to the total load, and through its extremity draw a horizontal to meet the vertical through the nearest end of the load; join the point of intersection to the free end with a straight line.

Fig.75.

W

Beam uniformly loaded on a portion of its span.-Figs. 76 and 77. Fix the point S at which the maximum bending moment occurs as on fig. 32; draw verticals through S and through the two extremities of the load; lay off MN W, the total load, upon the vertical through one of the extremities of the load; from N draw a horizontal meeting

the vertical through the other end in K, and draw MK cutting the vertical through S in s. Draw for base a horizontal through s and extending the whole length of span, a horizontal through M and extending to the vertical through the right end of span, and a horizontal from N to the vertical through the left end of span.

Note that this construction determines P and Q the supporting forces.

Beam loaded uniformly on two segments with different inten

sities of load.— Fig. 78. Fix the point T, at which the maximum bending moment occurs, as on fig. 36; draw verticals through T1, the junction of the loads, and the extremities of the span; lay off MN upon the vertical through the junction, equal to W, the load on that segment

1

in which T, lies; project N at L and K on the verticals through the extremities, join MK cutting the vertical from

T at t. Through t draw the base; make Lh equal to the total load, and draw Mh.

This construction also determines P and Q.

SHEARING FORCES DUE TO MOVING LOADS.

In the same way as was explained (see page 102) when speaking of bending moments, the Shearing Strain produced by a moving load is greater than that produced by the same load when fixed. In the cases which follow, it is to be understood that the loads as given are dead loads, or the equivalent reduced dead loads. When the load is partly fixed and partly moving, the equivalent dead load is the sum of the actual dead load and the dead load equivalent to the actual moving load.

DEFINITION. For any point x, the Range of Shearing Force due to a moving load is its extent; and the limits of this extent are the maximum positive and maximum negative values which Fx assumes during the transit of the moving load.

Beam under a rolling load.-Fig. 79. At any point of the span, Fx the shearing force is positive and equal to P, so

long as R is to the right of the point; since P increases as R moves towards the left support, Fx is evidently a positive maximum when R is indefinitely close to, and on the right side of, the point. When R passes to the left of the point, FP-R-Q; since Q increases as R comes closer to the right support, it is again evident that Fx is a negative maximum when R is indefinitely close to, and on the left side of, the point. When R is indefinitely close to the point, P and Q have sensibly the same values as for R exactly at the point, and we have the following,—

To find the maximum shearing force at any point x:Place R over the point, and calculate P and Q for that position of the load; these are respectively the max. positive and max. negative values of Fx.

During the passage of the load, the shearing force assumes all values lying between these maxima; and it is important to observe that the shearing force not only changes sign at any point as R passes over the point, but that it changes from its greatest positive to its greatest negative value, or vice versa, according as R is moving to

left or right, and does so even although the load be moving slowly.

The positive maximum at each point is the value of P as R comes to the point; and since P is proportional to the remote segment, it follows that the positive maximum at each point is proportional to the distance of the point from the right end of span; it is zero for the right end, and increases uniformly till it is R for the left end; for, when R is