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 × 106 N-m2.
Solution
∑ Fy = 0, so that RA + RB = 20 + 10 × 2 = 40 kN --------- (1)
Figure
Taking moments around A,
RB × 4 = 20 × 1 + 10 × 2 × 3 = 80
RB = 20 kN -------- (2)
RA = 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/ dx2) = M
= 20 x - 20 [ x - 1] - 5 [ x - 2]2 ---------- (4)
EI (dy / dx )= 10 x2 /3 - (10/3) [ x - 1]2 - (5/3) [ x - 2]3 + C1 ------ (5)
EIy = 10 x3 /3- (10/3) [ x - 1]3 - ( 5/12) [ x - 2]4 + C 1 x + C2 ---------6
at A, x = 0, y = 0, C2 = 0
at B, x = 4 m, y = 0
0 =(10 × 43 )/3- 10 (4 - 1)3 - (5/12) (4 - 2)4 + C 1 × 4
C1 =- 29.17
EI dy/ dx = 10 x2 - 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 × 106)
= - 1.46 × 10- 3 radians
(b) Slope at B, (x = 4 m),
EI θB = 10 × 42 - 10 (4 - 1)2 - 5 (4 - 2)3 - 29.17 = + 27.5
θB = + 1.38 × 10- 3 radians
(c) Slope at centre, (x = 2 m),
EI θC = 10 × 22 - 10 (2 - 1)2 - 29.17
θC = + 0.04 × 10- 3 radians
Deflection under the load :
EIy = 10 x3 /3- 10 [ x - 1]3 - (5/12) [ x - 2]4 - 29.17 x
At x = 1 m,
EIy D = (10/3) - 29.17
EIyD = - 25.84 × 103 × 103/20 × 106
= - 1.29 mm
(d) Deflection at the centre :
x = 2 m
EIy =10 × 23 - (10/3) (2 - 1)3 - 29.17 × 2
yC = - 1.75 mm
(e) Maximum deflection : Let the maximum deflection b/w D and C (x < 2 m).
dy/ dx = 0
10 x2 - 10 ( x - 1)2 - 29.17 = 0
10 x2 - 10 x2 - 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
∴ ymax = - 1.7501 mm