$\Sigma M_{R2} = 0$
$5R_1 = 80(3) + 90(2)$
$R_1 = 84 \, \text{kN}$
$\Sigma M_{R1} = 0$
$5R_2 = 80(2) + 90(3)$
$R_2 = 86 \, \text{kN}$
Checking
$R_1 + R_2 = F_1 + F_2$ (okay!)
To draw the Shear Diagram
- V_{A} = R_{1} = 84 kN
- V_{B} = V_{A} + Area in load diagram
V_{B} = 84 - 20(1) = 64 kN
- V_{C} = V_{B} + Area in load diagram
V_{C} = 64 - ½ (20 + 80)(3) = -86 kN
- V_{D} = V_{C} + Area in load diagram
V_{D} = -86 + 0 = -86 kN
V_{D2} = V_{D} + R_{2} = -86 + 86 = 0
- Location of zero shear:
From the load diagram:
y / (x + 1) = 80 / 4
y = 20(x + 1)
V_{E} = V_{B} + Area in load diagram
0 = 64 - ½ (20 + y)x
(20 + y)x = 128
[20 + 20(x + 1)]x = 128
20x^{2} + 40x - 128 = 0
5x^{2} + 10x - 32 = 0
x = 1.72 and -3.72
use x = 1.72 m from B
- By squared property of parabola:
z / (1 + x)^{2} = (z + 86) / 4^{2}
16z = 7.3984z + 636.2624
8.6016z = 254.4224
z = 73.97 kN
To draw the Moment Diagram
- M_{A} = 0
- M_{B} = M_{A} + Area in shear diagram
M_{B} = 0 + ½ (84 + 64)(1) = 74 kN·m
- M_{E} = M_{B} + Area in shear diagram
M_{E} = 74 + A_{1} (see figure for A_{1} and A_{2})
For A_{1}:
A_{1} = 2/3 (1 + 1.72)(73.97) - 64(1) - 2/3 (1)(9.97)
A_{1} = 63.5
M_{E} = 74 + 63.5 = 137.5 kN·m
- M_{C} = M_{E} + Area in shear diagram
M_{C} = M_{E} - A_{2}
For A2:
A_{2} = 1/3 (4)(73.97 + 86) - 1/3 (1 + 1.72)(73.97) - 1.28(73.97)
A_{2} = 51.5
M_{C} = 137.5 - 51.5 = 86 kN·m
- M_{D} = M_{C} + Area in shear diagram
M_{D} = 86 - 86(1) = 0