Maximum slope and maximum deflection:
A simply supported beam of span l is subjected to two concentrated loads at one-third span through two supports. Discover the maximum slope & maximum deflection EI is constant.
Solution
By symmetry,
RA = RB = W ---------- . (1)
Let a section X-X at a distance x from A,
M = W . x - W ?[x - l/3] - W [x - 2l /3] -------- (2)
![202_Maximum slope and maximum deflection.png](https://www.expertsmind.com/CMSImages/202_Maximum%20slope%20and%20maximum%20deflection.png)
The equation for deflection is :
EI = d 2 y/dx2 = M = W x - W[x -(l/3) ]- W[x -(2/3)] --------- (3)
Integrating the Equation (3),
EI (dy/ dx) = W x2/2 - (W /2)[x- (l/3)]2- (w/2 ) [x - (2l/3) ]2 + C1 ------- (4)
EI y= W x2/6 - (W /6)[x- (l/3)]3- (w/6 ) [x - (2l/3) ]3 + C1x + C2 -------- (5)
The boundary conditions :
at A, x = 0, y = 0 ∴ C2 = 0
It must be understood that the Equation (3), (4) & (5) pertain to the region x > 2l /3
Therefore second & third terms vanish while BC at x = 0 is used.
at B, x = l, y = 0
0 = W l 3/6- W /6(2l /3)3- (W/6)(l/3)3 + C1 l
C1 =- W l3 / 6 [1 - 8/27 - 1/27] = W l 2/9 ----------- (6)
∴ EI (dy/dx) = W x2/2 [x-(l/3)] 2 - (W/2) [x-(2l/3)] 2 - Wl2/9
Actually since the problem is symmetric the maximum deflection takes place in the centre.
y1C + y2C = y3C
θ1A + θ2 A = θ3 A = θ3B
Deflection under the load, (x = l/3) ,
EIyD = W/6(1/3)3 - (W l 2/9 l )×(l/3)
= Wl3/27 (1/6 - 1) = - 5 W l 3 / (27 × 6)
yD = - 5 W l 3 / 162 EI --------- (7)
At A, (x = 0),
θA = - W l 2 / 9 EI ---------- (8)
At B (x = l),
EI θB = W l 2 /2- (W/2) (4l 2/9) -( W/2)( l 2/9) - (W l 2/9)
= W l 2/18 [9 - 4 - 1 - 2] = +Wl2/9
∴ θ = + W l 2/9 EI ---------- (9)
For maximum deflection, slope is zero.
0 = W x2 /2 -(w/2) [ x-(l/3)]2 - Wl2/9
Again note down that maximum deflection shall occur between the loads which is easily ascertained from symmetry. Though, to prove this Equation (5) is utilized and since x < 2l/3 among the loads, the third term vanishes.
⇒ 0 = 9 x2 - 9 (x - l/3)2 - 2l 2
= 9 x2 - 9 ( x2 + l 2 /9 - 2l x /3) - 2l 2
=- l 2 + 6 l x - 2l 2
6lx = 3l 2
x = l / 2 -------- (10)
EIy max = (W/6) x3 - W (x - (l /3))3 - (Wl2/9 )x
Now put x = l /2
EIy max = (W /6 )(l/2)3 -(w/6)((l/2)-(l/3))3 -Wl3/18
= (w/6)((l/2)-(2l/3))3-(wl2/9)(l/2)= (wl3/6)(1/216)+(1/3)-(1/8))
= - Wl 3/6 [(1/ 8 )-(1/ 36) -(1/3) ]= - wl3/6 ((72+1-27)/216)
= (Wl 3 /(36 × 8 × 6)) [36 - 8 - 96] = - Wl 3 (23/648)
∴ y max = 23 Wl 3/ 648 ------ (11)