Find out Deflection under the load:
A beam of span 4 m is subject to a point load of 20 kN at 1 m from the left support and a Udl of 10 kN/m over a length of 2 m from the right support.
Find out :
1. Slope at the ends.
2. Slope at the centre.
3. Deflection under the load.
4. Deflection at the centre.
5. Maximum deflection.
Take EI = 20 × 10^{6} N-m^{2}.
Solution
∑ Fy = 0, so that R_{A} + R_{B} = 20 + 10 × 2 = 40 kN --------- (1)
Figure
Taking moments around A,
R_{B} × 4 = 20 × 1 + 10 × 2 × 3 = 80
R_{B} = 20 kN -------- (2)
R_{A} = 20 kN
M = 20 x - 20 [ x - 1] - 10 [ x - 2] ([ x - 2]/2)
= 20 x - 20 [ x - 1] - 5 [ x - 2]^{2}
EI (d ^{2} y/ dx^{2}) = M
= 20 x - 20 [ x - 1] - 5 [ x - 2]^{2} ---------- (4)
EI (dy / dx )= 10 x^{2} /3 - (10/3) [ x - 1]^{2} - (5/3) [ x - 2]^{3} + C_{1} ------ (5)
EIy = 10 x^{3} /3- (10/3) [ x - 1]^{3} - ( 5/12) [ x - 2]^{4} + C _{1} x + C_{2 } ---------6
at A, x = 0, y = 0, C_{2} = 0
at B, x = 4 m, y = 0
0 =(10 × 4^{3} )/3- 10 (4 - 1)^{3} - (5/12) (4 - 2)^{4} + C_{ 1} × 4
C_{1} =- 29.17
EI dy/ dx = 10 x^{2} - 10 [ x - 1]^{2} - (5 /3 )[ x - 2]^{3} - 29.17
(a) Slope at A, (x = 0),
θ _{A} = - 29.17 / EI = - 29.17 × 10/(20 × 10^{6})
= - 1.46 × 10^{- 3} radians
(b) Slope at B, (x = 4 m),
EI θ_{B} = 10 × 4^{2} - 10 (4 - 1)^{2} - 5 (4 - 2)^{3} - 29.17 = + 27.5
θ_{B} = + 1.38 × 10^{- 3} radians
(c) Slope at centre, (x = 2 m),
EI θ_{C} = 10 × 2^{2} - 10 (2 - 1)^{2 } - 29.17
θ_{C} = + 0.04 × 10^{- 3} radians
Deflection under the load :
EIy = 10 x^{3} /3- 10 [ x - 1]^{3} - (5/12) [ x - 2]^{4} - 29.17 x
At x = 1 m,
EIy _{D} = (10/3) - 29.17
EIy_{D} = - 25.84 × 10^{3} × 10^{3}/20 × 10^{6}
= - 1.29 mm
(d) Deflection at the centre :
x = 2 m
EIy =10 × 2^{3} - (10/3) (2 - 1)^{3} - 29.17 × 2
y_{C } = - 1.75 mm
(e) Maximum deflection : Let the maximum deflection b/w D and C (x < 2 m).
dy/ dx = 0
10 x^{2} - 10 ( x - 1)^{2 } - 29.17 = 0
10 x^{2} - 10 x^{2} - 10 + 20 x - 29.17 = 0
x = 1.96 m < 2 m
EIy_{ max} = (10/3) (1.96)^{3} - 10 (1.96)^{3} - 29.17 × 1.96 = - 35
∴ y_{max} = - 1.7501 mm