$\Sigma M_D = 0$
$7R_1 + 40(3) = 5(50) + 10(10)(2) + 20(4)(2)$
$R_1 = 70 \, \text{kN}$
$\Sigma M_A = 0$
$7R_2 = 50(2) + 10(10)(5) + 20(4)(5) + 40(10)$
$R_2 = 200 \, \text{lb}$
To draw the Shear Diagram
- V_{A} = R_{1} = 70 kN
- V_{B} = V_{A} + Area in load diagram
V_{B} = 70 - 10(2) = 50 kN
V_{B2} = 50 - 50 = 0
- V_{C} = V_{B2} + Area in load diagram
V_{C} = 0 - 10(1) = -10 kN
- V_{D} = VC + Area in load diagram
V_{D} = -10 - 30(4) = -130 kN
V_{D2} = -130 + R_{2}
V_{D2} = -130 + 200 = 70 kN
- 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
- M_{A} = 0
- M_{B} = M_{A} + Area in shear diagram
M_{B} = 0 + ½ (70 + 50)(2) = 120 kN·m
- M_{C} = M_{B} + Area in shear diagram
M_{C} = 120 - ½ (1)(10) = 115 kN·m
- M_{D} = M_{C} + Area in shear diagram
M_{D} = 115 - ½ (10 + 130)(4)
M_{D} = -165 kN·m
- M_{E} = M_{D} + Area in shear diagram
M_{E} = -165 + ½ (70 + 40)(3) = 0
- 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.
- 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
30x^{2} + 20x - 230 = 0
3x^{2} + 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.