## Determine the deflection at free end, Mechanical Engineering

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Determine the deflection at free end:

For the beam illustrated in Figure, determine the deflection at free end and the maximum deflection. Figure

Solution

RA + RB  = 48 × 6 + 60 = 348 kN         ----------- (1)

Σ M about A = 0.

48 × 6 × (2 + 3) + 60 × 16 = RB  × 12

∴         RB  = 200 kN (↑)                        ---------- (2)

From Eqs. (1) and (2),

RA  = 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/ dx2      = M = 148 x - 24 [ x - 2]2  + 24 [ x - 8]2  + 200 [ x - 12]          -------- (5)

EI (dy/ dx )= 74 x2 - 8 [ x - 2]3 + 8 [ x - 8]2 + 100 [ x - 12] 2+ C1          ------- (6)

EIy (74/3) x3  - 2 [ x - 2]4  + 2 [ x - 8]4  + 100 [ x - 12]3 + C1 x + C2   ---------- (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),

C2 = 0   ------------- (10)

From Eqs. (7) and (9),

0 = (74 /3)× 123 - 2 [12 - 2]4  + 2 [12 - 8]4  + C1 × 12

∴          C1 = - 1928                              ------------ (11)

∴          EIy = 74 x3  - 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)× 163  - 2 [16 - 2]4  + 2 [16 - 8]4  + 100 [16 - 12]3  - 1928 × 16 = 3680

∴          yC  =+  3680 / EI

Maximum Deflection : It occurs between D and E.

dy / dx = 0

0 = 74 x2  - 8 [ x - 2]3  - 1928

= 74 x2  - 8 x3  + 64 + 48 x2  - 96 x - 1928

= 8 x3  - 122 x2  + 96 x + 1928

By trial and error, x = 5.7 m

EIy max     = 74 × 5.73  - 2 [5.7 - 2]4  - 1928 × 5.7 = - 6796.3

∴          ymax = - 6796.3 / EI

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