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,
R_{A } = R_{B} = 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)
The equation for deflection is :
EI = d ^{2} y/dx^{2} = M = W x - W[x -(l/3) ]- W[x -(2/3)] --------- (3)
Integrating the Equation (3),
EI (dy/ dx) = W x^{2}/2 - (W /2)[x- (l/3)]^{2}- (w/2 ) [x - (2l/3) ]^{2} + C_{1} ------- (4)
EI y= W x^{2}/6 - (W /6)[x- (l/3)]^{3}- (w/6 ) [x - (2l/3) ]^{3} + C_{1}x + C_{2} -------- (5)
The boundary conditions :
at A, x = 0, y = 0 ∴ C_{2 } = 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} + C_{1} l
C_{1} =- W l^{3} / 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} - Wl^{2}/9
Actually since the problem is symmetric the maximum deflection takes place in the centre.
y_{1C } + y_{2C} = y_{3C}
θ_{1A} + θ_{2 A }= θ_{3 A }= θ_{3B}
Deflection under the load, (x = l/3) ,
EIyD = W/6(1/3)^{3} - (W l ^{2}/9 l )×(l/3)
= Wl^{3}/27 (1/6 - 1) = - 5 W l ^{3} / (27 × 6)
y_{D } = - 5 W l ^{3} / 162 EI --------- (7)
At A, (x = 0),
θA = - W l 2 / 9 EI ---------- (8)
At B (x = l),
EI θ_{B} = 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] = +Wl^{2}/9
∴ θ = + W l ^{2}/9 EI ---------- (9)
For maximum deflection, slope is zero.
0 = W x^{2} /2 -(w/2) [ x-(l/3)]^{2 }- Wl^{2}/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 x^{2} - 9 (x - l/3)^{2} - 2l ^{2}
= 9 x^{2} - 9 ( x^{2 } + 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) x^{3} - W (x - (l /3))^{3} - (Wl^{2}/9 )x
Now put x = l /2
EIy _{max } = (W /6 )(l/2)^{3} -(w/6)((l/2)-(l/3))^{3} -Wl^{3}/18
= (w/6)((l/2)-(2l/3))^{3}-(wl^{2}/9)(l/2)= (wl^{3}/6)(1/216)+(1/3)-(1/8))
= - Wl ^{3}/6 [(1/ 8 )-(1/ 36) -(1/3) ]= - wl^{3}/6 ((72+1-27)/216)
= (Wl ^{3} /(36 × 8 × 6)) [36 - 8 - 96] = - Wl ^{3} (23/648)
∴ y _{max} = 23 Wl ^{3}/ 648 ------ (11)