Evaluate the deflection at the free end of the cantilever:
Discover the slope and deflection at the free end of the cantilever illustrated in Figure . Take EI = 200 × 10^{6} N-m^{2}.
Solution
Apply Udl in both the upward and downward directions in the portion CB.
Figure
Figure
M = - 24 x . (x/2) + 24 ( x - 1) (( x - 1)/2) - 30 [ x - 2]
= - 12 x^{2 } + 12 [ x - 1]^{2} - 30 [ x - 2] --------- (1)
EI (d ^{2 }y /dx^{2} )= M = - 12 x 2 + 12 [ x - 1]^{2} - 30 [ x - 2] ----------- (2)
EI dy/ dx = - 4 x^{3} + 4 [ x - 1]^{3} - 15 [ x - 2] ^{2}+ C_{1} --------- (3)
EIy =- x + [ x - 1] - 5 [ x - 2] + C_{1 }x + C_{2} --------------- (4)
The boundary conditions are :
At B, x = 4 m, dy/ dx = 0 -------- (5)
At B, x = 4 m, y = 0 ----------- (6)
From Eqs. (3) and (5)
0 = - 4 × 4 ^{3}+ 4 (4 - 1)^{ 3}- 15 (4 - 2)_{3} + C_{1}
∴ C_{1} = 208 ----------- (7)
From Eqs. (4), (6) and (7)
0 =- 4 ^{4}+ (4 - 1)^{4} - 5 (4 - 2) ^{3}+ 208 × 4 + C_{2}
∴ C_{2} = - 617 --------- (8)
∴ EI (dy / dx )=- 4 x^{3} + 4 [ x - 1]^{3} - 15 [ x - 2]^{2} + 208 ---------- (9)
EIy =- x^{4} + [ x - 1]^{4} - 5 [ x - 2]^{3 } + 208 x - 617-------- (10)
At free end A, x = 0, From Eq. (9)
θ _{A} = + 208 /EI = + 208 × 10/200 × 106 = + 1.04 × 10^{- 3} radians
From Eq. (10)
y_{A} = 617/EI = - 617 × 10^{3} × 10^{3} / 200 × 10^{6} = - 3.085 mm ( ↓ )