Determine the deflection at free end:
For the beam illustrated in Figure, determine the deflection at free end and the maximum deflection.
Figure
Solution
R_{A} + R_{B} = 48 × 6 + 60 = 348 kN ----------- (1)
Σ M about A = 0.
48 × 6 × (2 + 3) + 60 × 16 = R_{B} × 12
∴ R_{B } = 200 kN (↑) ---------- (2)
From Eqs. (1) and (2),
R_{A} = 148 kN (↑) ------ (3)
Apply the Udl in downward and upward directions in the portion EC
Figure
M = 148 x - (48/2) [ x - 2] [ x - 2] + (48/2) [ x - 8] [ x - 8] + 200 [ x - 12]
= 148 x - 24 [ x - 2]^{2} + 24 [ x - 8]^{2 } + 200 [ x - 12] ----------. (4)
EI d^{ 2} y/ dx^{2} = M = 148 x - 24 [ x - 2]^{2} + 24 [ x - 8]^{2} + 200 [ x - 12] -------- (5)
EI (dy/ dx )= 74 x^{2 }- 8 [ x - 2]^{3} + 8 [ x - 8]^{2} + 100 [ x - 12] ^{2}+ C_{1 } ------- (6)
EIy (74/3) x^{3} - 2 [ x - 2]^{4} + 2 [ x - 8]^{4} + 100 [ x - 12]^{3} + C1 x + C_{2} ---------- (7)
The boundary conditions are :
At A, x = 0, y = 0 ------- (8)
At B, x = 12 m, y = 0 ------- (9)
From Eqs. (7) and (8),
C_{2} = 0 ------------- (10)
From Eqs. (7) and (9),
0 = (74 /3)× 12^{3 }- 2 [12 - 2]^{4} + 2 [12 - 8]^{4} + C_{1} × 12
∴ C_{1} = - 1928 ------------ (11)
∴ EIy = 74 x^{3} - 2 [ x - 2]^{4} + 2 [ x - 8]^{4} + 100 [ x - 12]^{3} - 1928 x
Deflection at free end, x = 16 m
EIy C = (74 /3)× 16^{3} - 2 [16 - 2]^{4} + 2 [16 - 8]^{4} + 100 [16 - 12]^{3} - 1928 × 16 = 3680
∴ y_{C} =+ 3680 / EI
Maximum Deflection : It occurs between D and E.
dy / dx = 0
0 = 74 x^{2} - 8 [ x - 2]^{3} - 1928
= 74 x^{2} - 8 x^{3 } + 64 + 48 x^{2} - 96 x - 1928
= 8 x^{3} - 122 x^{2 } + 96 x + 1928
By trial and error, x = 5.7 m
EIy _{max} = 74 × 5.7^{3} - 2 [5.7 - 2]^{4} - 1928 × 5.7 = - 6796.3
∴ y_{max }= - 6796.3 / EI