For the beam and loading shown, ldentify the maximum absolute values of the shear force and bendlng moment. tw max on AB and CD max 4w at the Center max 32 at the center imax 32 on AB and CD Solution Given data It is UDL acting at the centre. We know that RA + RB = w(l/2) RB(l) = w(l/2)(l/2) RB = wl/4 ==> RA = wl/4 Now calculating Shear force and Bending moment Fx = wl/4 - w(x - l/4) At FAB = x = l/4 FAB = wl/4 At centre x = l/2 Fc = wl/4 - w(l/2 - l/4) 0 Now for bending moment Mx = RA.x - w(x - l/4)(x -l/4)/2 At x = l/4 M = wl/4(l/4) = wl2/16 At x = l/2 Mc = wl/4(l/2) - w(l/4)(wl/4)/2 = 3wl2/32.

1. For the beam and loading shown, ldentify the maximum absolute values of the shear force and
bendlng moment. tw max on AB and CD max 4w at the Center max 32 at the center imax 32 on
AB and CD
Solution
Given data
It is UDL acting at the centre.
We know that
RA + RB = w(l/2)
RB(l) = w(l/2)(l/2)
RB = wl/4 ==> RA = wl/4
Now calculating Shear force and Bending moment
Fx = wl/4 - w(x - l/4)
At FAB = x = l/4
FAB = wl/4
At centre x = l/2
Fc = wl/4 - w(l/2 - l/4)
0
Now for bending moment
Mx = RA.x - w(x - l/4)(x -l/4)/2
At x = l/4
M = wl/4(l/4) = wl2/16
At x = l/2
Mc = wl/4(l/2) - w(l/4)(wl/4)/2
= 3wl2/32