Problem: F6-13 (R.C.Hibbeler.) Fundamental Problems
Draw the Shear and Moment diagram for the simply supported beam.
Solution:
As per the static equilibrium, ∑MA = 0,
RB*12 = 600*9 + 200*6*3, hence RB = 750.0 lbs,
Also ∑Fy = 0, hence RA +RB = 200*6 + 600
Check: Book Solution Manual of Mechanics of Materials by R C Hibbeler
End support reactions are:
Vertical Reactions, RA = 1050.0 lbs, and RB = 750.0 lbs
Moment Reactions are MA = 0 lbsft, and MB = 0 lbsft
In between A and B,
for 0.0ft <= x <= 6.0ft:
Shear, Vx = RA -200.0(x -0.0) = 1050.0 – 200.0(x)
Moment, Mx = VA(x) + -200.0(x)^2/2 = 1050.0* (x) -200.0*(x)^2/2
for 6.0 ft <= x <= 9 ft:
Shear, Vx = RA – 200.0*6= 1050.0 – 1200.0 = -150.0 lbs
Moment, Mx = RA*(x) – 200.0*6*(x-3)= 1050.0* (x) – 1200.0*(x-3) = 3600 – 150*x
for 9 ft <= x <= 12.0 ft:
Vx = -RB = -750.0 lbs
Mx = RB*(L-x) = 750(12-x) = 9000 – 750*x
shear and moment nodes for right point loads are :[-150. -750. 0.]lbs, [2700. 2250. 0.] lbs-ft
Maximum Bending Moment, Mmax = 2756.25 lbs-ft
at x = 5.25 ft from the left end.
