a , 
BEAMS AND BENDING 95 
; ab When the load is in the position shown at (a), the positive shearing 
force at D is - = 
j 7 F=2,-5(- a) “Cc 2 
{ which shows that F in- A 
 ereases as x, decreases, b DG 
_ Whenthe loadisinthe |r 
_ position shown at (2), the | 
; apg shearing force at 4 
is F=R, — w(x --2,) D 
Ry 
(a) 
mm 
Ry 
Sain 
don't : 
= GP ti) — tex, which A vs Cc B 
shows that F decreases porta pegens be) ta LD. 
1 R 
j 
im 
<t 
Pare 
as x, decreases. 
t is therefore evi- 
dent that the positive 
‘shearing force at D is 
greatest when 2, = 2, 
that is, when the left- 
hand end of the load 
is at D. The shearing 
force at D is then 
Pa=5l-2), which is 
the equation to a parabola 
having for its axis the 
vertical through B, the 
right-hand endof thespan, 
the vertex being on the 
base line XX, as shown 
at (d). The maximum 
_ positive shearing force is 
sl at A where «= 0. Fig. 125. 
In like manner it can be shown that the negative shearing force at D 
__ is greatest when the right-hand end of the load is at D. The shearing 
_ force at D is then F--< 
for its axis the vertical through A, the left-hand end of the span, the 
vertex being on the base line XX, as shown at (d). The maximum negative 
shearing force is 7 at B where x=1. 
The positive and negative shearing force diagrams are shown plotted 
on the same side of the base line XX at (e). 
Exampte III.—Two loads W, and W, at a fixed distance ¢ 
nied moving along a girder AB (Fig. 125) supported at its 
c 
' 
we 
SES BR OR Ce 
| ht a tnaaaeas <ael Riaeas | 
, Which is the equation to a parabola having 
Let W, the resultant of W, and W,, act at a distance a from W, and 
