Find out the deflection at mid-span - cantilever beam:
A double overhanging beam of 12 m length rests on supports symmetrically, 8 m apart. A load of 80 kN acts at one free end & a load of 40 kN acts at other free end. Find out the deflection :
(a) at free ends, and
(b) at mid-span.
Solution
∑ Fy = 0, so that R_{A }+ R_{B } = 80 + 40 = 120 kN -------- (1)
Σ M about A = 0,
⇒ R_{B} × 8 = 40 × 10 - 80 × 2 -----------(2)
R_{B} = 30 kN ( ↑ )
From Eqs. (1) and (2),
R_{A} = 90 kN ( ↑ ) ----------(3)
M = - 80 x + 90 [ x - 2] + 30 [ x - 10] ---------- (4)
EI (d ^{2} y/ dx^{2}) = M = - 80 x + 90 [ x - 2] + 30 [ x - 10] -------- (5)
EI dy/ dx = - 40 x ^{2}+ 45 [ x - 2] ^{2}+ 15 [ x - 10] ^{2}+ C_{1 } --------- (6)
EIy =- (40 /3)x^{2 } + 15 [ x - 2]^{3} + 5 [ x - 10]^{3} + C_{1} x + C_{2} ----------- (7)
Boundary conditions are following :
At A, x = 2 m, y = 0 --------- (8)
At B, x = 10 m, y = 0 -------- (9)
From Eqs. (7) and (8),
0 = - (40/3) × 2^{3} + C1
∴ C_{2} =+ 320 /3 ---------- (10)
From Eqs. (7), (8) and (9),
0 =( - 40 /3 )× 10^{3} + 15 [10 - 2]^{3} + C_{1} × 10 + (320/ 3)
C_{1} =+1664 / 3 ----------- (11)
EIy =- 40 x^{3} + 15 [ x - 2]^{3} + 5 [ x - 10]^{3 } + 1664 x + 320 ---------- (12)
Deflection at free ends:
At D, x = 0
y _{D} =+ 320 / 3 EI ------- (13)
At E, x = 12 m,
EIy =- (40/3) × 12^{3 }+ 15 [12 - 2]^{3} + 5 [12 - 10]^{3} + (1664/3) × 12 + (320/3)
∴ y _{E} =- 3712/3 EI
Deflection at mid-span :
x = 6 m
EIy C =- (40 /3)× 63 + 15 [6 - 2]^{3 } + (1664/3) × 6 + (320/3)
y_{C} =- 4544/ 3 EI