Solution to Problem 431 | Relationship Between Load, Shear, and Moment

$7R_1 + 40(3) = 5(50) + 10(10)(2) + 20(4)(2)$

$R_1 = 70 \, \text{kN}$
 

$\Sigma M_A = 0$

$7R_2 = 50(2) + 10(10)(5) + 20(4)(5) + 40(10)$

$R_2 = 200 \, \text{lb}$
 

To draw the Shear Diagram

1. V_A = R₁ = 70 kN
2. V_B = V_A + Area in load diagram
   V_B = 70 - 10(2) = 50 kN
   V_B2 = 50 - 50 = 0
3. V_C = V_B2 + Area in load diagram
   V_C = 0 - 10(1) = -10 kN
4. V_D = VC + Area in load diagram
   V_D = -10 - 30(4) = -130 kN
   V_D2 = -130 + R₂
   V_D2 = -130 + 200 = 70 kN
5. V_E = V_D2 + Area in load diagram; V_E = 70 - 10(3) = 40 kN
   V_E2 = 40 - 40 = 0

To draw the Moment Diagram

1. M_A = 0
2. M_B = M_A + Area in shear diagram
   M_B = 0 + ½ (70 + 50)(2) = 120 kN·m
3. M_C = M_B + Area in shear diagram
   M_C = 120 - ½ (1)(10) = 115 kN·m
4. M_D = M_C + Area in shear diagram
   M_D = 115 - ½ (10 + 130)(4)
   M_D = -165 kN·m
5. M_E = M_D + Area in shear diagram
   M_E = -165 + ½ (70 + 40)(3) = 0
6. Moment curves AB, CD and DE are downward parabolas with vertices at A', B' and C', respectively. A', B' and C' are corresponding zero shear points of segments AB, CD and DE.
7. Locating the point of zero moment:
   a / 10 = (a + 4) / 130
   130a = 10a + 40
   a = 1/3 m
   
   y / (x + a) = 130 / (4 + a)
   y = 130(x + 1/3) / (4 + 1/3)
   y = 30x + 10
   
   M_C = 115 kN·m
   M_zero = M_C + Area in shear
   0 = 115 - ½ (10 + y)x
   (10 + y)x = 230
   (10 + 30x + 10)x = 230
   30x² + 20x - 230 = 0
   3x² + 2x - 23 = 0
   x = 2.46 m
   
   Zero moment is at 2.46 m from C

Another way to solve the location of zero moment is by the squared property of parabola (see Problem 434). The point of zero moment is an ideal location for the construction joint.