Maximum slope and maximum deflection, Mechanical Engineering

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

 

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

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)

Posted Date: 1/21/2013 5:26:35 AM | Location : United States







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