Determine the deflection under the loads:
Determine the deflection under the loads as shown in Figure
Solution
∑ Fy = 0, so that RA + RB = 80 + 20 = 100 kN -------- (1)
Σ M about A = 0.
80 × 1 + 20 × 3 = RA × 2 ∴ RB = 70 kN (↑)
From Eqs. (1) and (2),
RA = 30 kN ( ↑ ) ------------ . (2)
M = 30 x - 80 ( x - 1) + 70 ( x - 2) -------- (3)
EI (d 2 y /dx2) = M = 30 x - 80 [ x - 1] + 70 [ x - 2] ------------- (4)
EI dy/ dx = 15 x - 40 [ x - 1]2 + 35 [ x - 2]2 + C1 --------- (5)
EIy = 5 x3 - 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), C2 = 0
From Eqs. (6) and (8)
0 = 5 × 23 - 40 [2 - 1]3 + C1 × 2
C1 =- 40 /3
∴ EIy = 5 x3 - (40/3) [ x - 1]3 + (35/3) [ x - 2]3 - (40/3) x . . . (7)
Deflection at C,
x = 1 m
EIyC = 5 × 13 - (40/3) × 1 = - 25/3
∴ yC = - 25 /3 EI --------------- (8)
Deflection at D,
x = 3 m
EIy D = 5 × 33 - (40 /3)(3 - 1)3 + 35 (3 - 2)3 - (40 /3)× 3 = 0
YD= 0