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







Related Discussions:- Maximum slope and maximum deflection, Assignment Help, Ask Question on Maximum slope and maximum deflection, Get Answer, Expert's Help, Maximum slope and maximum deflection Discussions

Write discussion on Maximum slope and maximum deflection
Your posts are moderated
Related Questions
Mounting Systems For The  Welding Head-Boom mounted type These are the most versatile type of welding systems. Normally the boom is supported by means of a column. The welding

Pattern Practices - Manufacturing processes: Pattern Practices : A pattern may be defined as a replica or fascimile model of the desired casting which, when packed or embedd

FLUX FEEDING & RECOVERY SYSTEM The flux is fed by gravity from the flux delivery hopper mounted on the welding head through a tube into the welding zone. Flux feeding is normally

By introducing their model of neural excitability in 1952, Alan Lloyd Hodgkin and Andrew Huxley not only shed light on the mechanism underlying a very important phenomenon in Physi

Crankcase: It is connected to engine cylinder. It splits into two parts, named as right crankcase and left crank case. It supports the kick-starter spindle, crankshaft, main shaf

Ratio of applied stress and yield stress: What should be the ratio of applied stress, s, to yield stress s Y in a large thin plate so that plastic spread along lint of crack

MECHANISM OF LASER WELDING   Laser welding of metals is performed with beams focused to irradiate levels in the range of 106 to 107 W / cm 2 . At such power densities, melti

Expression for strain energy: Derive an expression for strain energy because of torsion.                                          Sol.: The work done in straining shaf

A String LM long is tied to the ends of a uniform rod that weight 60 N and is 1.6m long. The string passes over a nail so that the rod hangs horizontally. Determine the tension in

Develop an EES Code for a Rankine cycle with reheat and with 2 open feedwater heaters.  The cycle is shown below:  Use the following conditions: 1.  700 psi, 850F 2.  500