Solution to Problem 430 | Relationship Between Load, Shear, and Moment

$20R_1 = 1000(25) + 400(5)(22.5) + 2000(10) + 200(10)(5)$

$R_1 = 5000 \, \text{lb}$
 

$\Sigma M_B = 0$

$20R_2 + 1000(5) + 400(5)(2.5) = 2000(10) + 200(10)(15)$

$R_2 = 2000 \, \text{lb}$
 

To draw the Shear Diagram

1. V_A = -1000 lb
2. V_B = V_A + Area in load diagram; V_B = -1000 - 400(5) = -3000 lb; V_B2 = -3000 + R1 = 2000 lb
3. V_C = V_B2 + Area in load diagram; V_C = 2000 + 0 = 2000 lb; V_C2 = 2000 - 2000 = 0
4. V_D = V_C2 + Area in load diagram; V_D = 0 + 200(10) = 2000 lb

To draw the Moment Diagram

1. M_A = 0
2. M_B = M_A + Area in shear diagram
   M_B = 0 - ½ (1000 + 3000)(5)
   M_B = -10000 lb·ft
3. M_C = M_B + Area in shear diagram
   M_C = -10000 + 2000(10) = 10000 lb·ft
4. M_D = M_C + Area in shear diagram
   M_D = 10000 - ½ (10)(2000) = 0
5. For segment BC, the location of zero moment can be accomplished by symmetry and that is 5 ft from B.
6. The moment curve AB is a downward parabola with vertex at A'. A' is the location of zero shear for segment AB at point outside the beam.