Solution to Problem 446 | Relationship Between Load, Shear, and Moment

$4w_o + 2 \,[ \, \frac{1}{2}w_o(1) \, ] = 20(4) + 2(50)$

$5w_o = 180$

$w_o = 36 \, \text{kN/m}$
 

To draw the Shear Diagram

1. V_A = 0
2. V_B = V_A + Area in load diagram
   V_B = 0 + ½ (36)(1) = 18 kN
   V_B2 = V_B - 50 = 18 - 50
   V_B2 = -32 kN
3. The net uniformly distributed load in segment BC is 36 - 20 = 16 kN/m upward.
   V_C = V_B2 + Area in load diagram
   V_C = -32 + 16(4) = 32 kN
   V_C2 = V_C - 50 = 32 - 50
   V_C2 = -18 kN
4. V_D = V_C2 + Area in load diagram
   V_D = -18 + ½ (36)(1) = 0
5. The shape of shear at AB and CD are parabolic spandrel with vertex at A and D, respectively.
6. The location of zero shear is obviously at the midspan or 2 m from B.

To draw the Moment Diagram

1. M_A = 0
2. M_B = M_A + Area in shear diagram
   M_B = 0 + 1/3 (1)(18)
   M_B = 6 kN·m
3. M_midspan = M_B + Area in shear diagram
   M_midspan = 6 - ½ (32)(2)
   M_midspan = -26 kN·m
4. M_C = Mmidspan + Area in shear diagram
   M_C = -26 + ½ (32)(2)
   M_C = 6 kN·m
5. M_D = M_C + Area in shear diagram
   M_D = 6 - 1/3 (1)(18) = 0
6. The moment diagram at AB and CD are 3rd degree curve while at BC is 2nd degree curve.