1920-21.] Graphical Treatment of Shear Influence Lines. 71 
shear at X is now either m 2 o or — n 2 o according as W 2 is just to the left or 
just to the right of X. Clearly m 2 n x =p, the distance between W 4 and W 2 ; 
and m 2 n 2 = W 2 . 
Similarly, if n 2 a. 2 m 3 is drawn everywhere parallel to the control line 
m A°h^ 2 a 2 m 3 > the horizontal distance from n 2 to n 3 being equal to q— 
the distance between W 2 and W 3 , — the ordinate between the line n 2 a 2 m 3 
and the moment, or base, line at the appropriate point gives the shearing 
force at X. Finally, m 3 n 3 is made equal to W 3 ; n 3 a 3 m x is drawn parallel 
to the control line for a distance r — the spacing between W 3 and W 4 ; 
m 4 ^ 4 is set down equal to W 4 , and is joined to a 4 . 
The line am 1 n 1 . . . m 4 %a 4 is the influence line for the point X, and 
is easily traced on the diagram by the small circles, to the base line aooa 4 . 
Where the influence line falls above the base line the shear at X is 
positive; while, when it falls below the base line, the shear at X is 
negative. 
By an extension of the diagram of fig. 2, a simple means is afforded by 
which to obtain the appropriate influence line for any point in the beam. 
The process is indicated in fig. 3, and is as follows : — 
Place the loads so that W 4 is over the right abutment. Produce the 
load lines through W 4 , W 2 , W 3 , and W 4 vertically downwards (dotted lines), 
and draw vertical (chain-dotted) lines, similarly spaced, from A. 
Draw ab x = L, the span. Set down b x a x = W 1 at A. In the direction aa x 
draw ac and a^b 2 in the spaces p. Set down b 2 a 2 — W 2 . 
In the direction ca 2 draw cd and a 2 b 3 in the spaces q. Set down 
= W 3 . 
In similar manner, draw de and <x 3 fe 4 in the spaces r. Set down 
bpi/y = W 4 , and join e to ct 4 . 
The line acdea 9 is the base line of all the influence lines, and ab x a^b 2 a 2 
. . . 6 4 a 4 is the control line. 
From a v a 2 , a 3 , and <x 4 draw lines a x h v a 2 h 2 . . . aji i everywhere 
parallel to the control line above, to meet the verticals through W 1? W 2 , W 3 , 
and W 4 respectively, and so obtain a series of parallel-sided figures, each 
of horizontal length L. 
To draw the influence line relative to the point X in the beam, distant 
x from B, draw verticals m x n v m 2 n 2 . . . mpi 4 distant x from the left-hand 
ends of the parallel-sided figures. The resulting influence line is indicated 
by the circles, to the base line acdea 4 . 
In the design of structures, only the greatest maximum positive and 
negative shears at each point in the span may be required. An inspection 
of the diagram shows that these occur on the lines m 1 n 1 , m 2 n 2 . . . m 4 r^ 4 , 
