Determine the deflection under the loads:
Determine the deflection under the loads as shown in Figure
Solution
∑ F_{y} = 0, so that R_{A} + R_{B } = 80 + 20 = 100 kN -------- (1)
Σ M about A = 0.
80 × 1 + 20 × 3 = R_{A} × 2 ∴ R_{B} = 70 kN (↑)
From Eqs. (1) and (2),
R_{A} = 30 kN ( ↑ ) ------------ . (2)
M = 30 x - 80 ( x - 1) + 70 ( x - 2) -------- (3)
EI (d ^{2 }y /dx^{2}) = M = 30 x - 80 [ x - 1] + 70 [ x - 2] ------------- (4)
EI dy/ dx = 15 x - 40 [ x - 1]^{2} + 35 [ x - 2]^{2} + C_{1 } --------- (5)
EIy = 5 x^{3} - 40 [ x - 1]^{3} + 35 [ x - 2]^{3} + C x + C --------- (6)
The boundary conditions are :
At A, x = 0, y = 0 -------- (7)
At B, x = 2 m, y = 0 -------- (8)
From Eq. (6) and (7), C_{2} = 0
From Eqs. (6) and (8)
0 = 5 × 2^{3 } - 40 [2 - 1]^{3} + C_{1 } × 2
C_{1} =- 40 /3
∴ EIy = 5 x^{3} - (40/3) [ x - 1]^{3} + (35/3) [ x - 2]^{3} - (40/3) x . . . (7)
Deflection at C,
x = 1 m
EIyC = 5 × 1^{3} - (40/3) × 1 = - 25/3
∴ y_{C} = - 25 /3 EI --------------- (8)
Deflection at D,
x = 3 m
EIy _{D} = 5 × 3^{3 } - (40 /3)(3 - 1)^{3} + 35 (3 - 2)^{3} - (40 /3)× 3 = 0
Y_{D}= 0