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/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.

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)

=          Wl³/27 (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] = +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)