**Find out the slope and deflection at the free end:**

Discover the slope and deflection at the free end of a cantilever shown in Figure. Take EI = 200 × 10^{6 }N-m^{2}.

**Solution**

M =- 30 x - 60 [ x - 2] - 24 ( x - 3) (( x - 3)/ 2)

=- 30 x - 60 [ x - 2] - 12 [ x - 3]^{2} -------- (1)

EI d 2 y/ dx2 = M

=- 30 x - 60 [ x - 2] - 12 [ x - 3]^{2} ----------- (2)

EI (dy/ dx) = - 15 x - 30 [ x - 2] - 4 [ x - 3] + C_{1 } --------- (3)

EIy =- 5 x - 10 [ x - 2] - [ x - 3] + C_{1} x + C_{2} ------------- (4)

** Figure**

Boundary conditions are :

At B, x = 5 m, dy/ dx = 0 ---------- (5)

At B, x = 5 m, y = 0 --------(6)

From Eqs. (3) and (5)

0 = - 15 × 5^{2}- 30 (5 - 2)^{2} - 4 (5 - 3)^{3} + C_{1}

∴ C_{1} = 677 ---------- (7)

From Eqs. (4), (6) and (7)

0 =- 5 × 5 ^{3}- 10 (5 - 2)^{3} - (5 - 3) 4+ 677 × 5 + C_{2}

∴ C_{2} =- 2474 -------- (8)

EI dy/ dx = - 15 x^{2 } - 30 [ x - 2]^{2} - 4 [ x - 3]^{3} + 677 --------- (9)

EIy = - 15 x^{3} - 10 [ x - 2]^{3} - [ x - 3]^{4} + 677 x - 2474 --------- (10)

At free end, i.e. at A, x = 0, From Eq. (9)

θ = + 677 /EI =677 × 10^{3}/200´10^{6} = 3385 × 10- 3 radians

From Eq. (10)

y_{A } =- 2474/ EI =- 2474 × 10^{3} × 10^{3}/200 × 10^{6} = - 12.37 mm

= - 12.37 mm ( ↓ )