Rayleigh Ritz Method

1

Solved Problems (Rayleigh Ritz method) U

1.Rayleigh Ritz method applied to Beams

#1) Find the deflection at the centre using Rayleigh Ritz method for the simply supported beam subjected to Concentrated load at the centre U

Solution: k

Take y =

a n 1

n

sin

nx . l

According to Rayleigh Ritz method



= U – W and

 C n

 0 where π = Potential

energy U = Strain energy and W is the external work done. 1 d2y 2  EI (  2  d 2 x ) dx  pyc where yc is the deflection at the centre.

dy n  an dx l

k

 Cos n 1

nx l

d2y n 2 2 k nx  2   an 2   sin l l n1  dx  2

d2y n 4 4  2   a 2 n 4 l  dx 



l

We Know that

 sin 0

 sin 0

 sin

2

n 1

4 4 1 2 n    EI a n 4 20 l l

l

k

nx l

K nx n sin dx  P an sin --------(1.a)  l 2 n 1 n 0 k

2

nx mx sin dx  0 when m is not equal to n . l l

nx mx l sin dx  when m = n Hence equation 1.a becomes, l l 2

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2

 an

1 4 l n 4  EI 2an 4  PSin --------------(1b) 2 2 l 2

2 pl 3 1 n sin 4 4 2  EI n 3 k 2 pl 1 n nx There fore y  4 ------------(1c) Sin Sin  4 2 l  EI n1,3,5... n To find out yc substitute x = l/2. in above equation (1.c) From 1 b eqn, a n 

yc 

2 pl 3  4 EI

k

1 4 n 1, 3, 5... n



Hence we have y c 

2 pl 3  1 1 1  4  4  which is the deflection at the centre. 4  EI  3 5 

#2) Find the maximum deflection of a simply supported beam subjected to central concentrated load P at the centre and uniformly distributed load P0/m through out the beam U

P

P0/m

P0/m

L

Let y  a1 sin EI U 2

x l

 a 2 sin

3x l

2

d2y 0  dx 2  dx L

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3

dy  x 3 3x  a1Cos  a2 Cos dx L L L L 2 2 2 d y  x 9 3x   2 a1 Sin  a2 2 Sin 2 L L dx L L EI U 2

U=

2

d2y EI 0  dx 2  dx = 2 L

2

   4 a1 x a 2 2 9 3x    dx which yields Sin  Sin 2 0  L2 L L  L L

EI 4 2 (a1  81a 22 ) 4 L L

Now W =

 P ydx  Py 0

max

0

ymax = a1 – a2 L  x 3x   Hence W =   P0  a1 Sin  a 2 Sin dx   P(a1  a 2 ) L L   0  a  L a1  2  + P(a1 – a2)   3 =U–W

W = 2 P0



2P L a EI 4 2 2  = 4L3 (a1  81a2 )  0 (a1  32 )  P(a1  a2 )    0 and  0 which gives the following equations: a1 a 2

2P L EI 4 a1  0  P  0 3  2L 4 81EI a 2 2 P0 L  P0 3 2 L3 From the above equations we get a1 

2P L 2 L3 a2  ( 0  P) 4 3 81EI ymax = a1-a2 P0 L4 PL3  ymax = 48EI 76.82 EI

C.RameshBabu/AP/Civil/MSEC

2 L3 2 P0 L (  P) and EI 4 3

4

U

2.Rayleigh Ritz Method applied to Columns:

#3) A Uniform Column is fixed at the bottom and free at the top. It carries a compressive load at the free end. Investigate the critical load of the column by assuming a second degree polynomial as y = a0 +a1x + a2 x2. Load P U

Take the origin at the fixed end. At x = 0, y = 0 At x = 0, dy/dx = 0 The above conditions give a0 = a1 = 0 and hence y = a2 x2 When x = L, y = q q =a2L2 and so we have a2 = q/L2 Hence y = qx2/L2



L

L

1 d2y 2 P dy EI ( ) dx   ( ) 2 dx 2  20 2 dx d x 0

x

d2y 2q dy 2qx  2 and ( 2 )  2 dx dx l l L L 1 2q P 2qx   2  EI ( L2 ) 2 dx   2 ( L2 ) 2 dx which gives 0 0

y

2 EIq 2 2 pq 2  3L L3 3EI  4 EIq 4 pq and this equation yields P = 2 0 3  q 3L L L



U

Repeat the above problem by using

y  a0  a1 x  a2 x 2  a3 x 3 and satisfying the bending

moment condition at the top When x = 0 , y = 0 Hence a0 = 0. When x = 0 dy/dx = 0 and we get a1 = 0 d2y  0. When x = l, dx 2 d2y dy  2a 2  6a 3 x  a1  2a 2 x  3a3 x 2 and dx dx 2

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5

d2y d2y . Hence  0  2a 2  6a3 x =0 and a2 = -3a3l dx 2 dx 2 So y =  3a3lx 2  a3 x 3  a3 ( x 3  3lx 2 ) At x = l,y = q q There fore q= a3 (l 3  3l 3 ) which gives a3   3 2l d2y q q dy q Hence y =  3 ( x 3  3lx 2 ) and  3 (6l  6 x)  3 (6lx  3x 2 ) ; 2 dx 2l dx 2l 2l At x = l,



l

l

1 d2y 2 P dy EI ( ) dx   ( ) 2 dx 2  20 2 dx d x 0

EI  2

l

l

q2 P q2 2 ( 6 l  6 x ) dx  (6lx  3x 2 ) 2 dx On simplifying 0 4l 6 2 0 4l 6

3EIq 2 3 pq 2 .  5l 2l 3 15EI EI  3EIq 6 pq  3   0 From this equation we get P= 2  2.5 2 q 5l 6l l l



#4) A uniform column hinged at both ends is subjected to a compressive load P at the two ends. Find the critical load using Rayleigh Ritz method if the trial function is 4hx(l  x) i) y  l2 x ii) y  aSin l Solving (i) L L 1 d2y 2 P dy   2  EI ( d 2 x ) dx   2 ( dx ) 2 dx 0 0 U

U

y

d2y 8h 4hx(l  x) dy 4h and so we have ;  ( l  2 x )  2 2 2 2 dx l dx l l l

l

EI 64h 2 dx P 16h 2 2    4 (l  4 x 2  4lx )dx 4  2 0 l 20 l  64 EIh 16 ph 12 EI    0 . This equation gives the critical load.Pcr= 2 3 h l 3l l

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6

U

Solving (ii) d 2 y  a 2 x x dy a x and  Sin y  aSin . This provides  Cos 2 2 l l dx l l dx l L L 2 1 d y P dy   2  EI ( d 2 x ) 2 dx   2 ( dx ) 2 dx 0 0

 l

 Sin

2

0 l

 Cos o

a 2 4 P a 2 2 x 2 x Sin dx  Cos 2 dx ------------------------------------------------A 2 0 l 4  l 20 l l l

EI 2

x l

l

dx  

2 x

l

l

2x l dx  l 2 2

1  Cos

0 l

dx   0

2x l dx  l 2 2

1  Cos

Substituting the above results in the equation A and finding The above expression yields critical Load Pcr=

 2 EI 4 a l P 2a 2 l  .  . 2 . 0 a 2 2 l 2 2l 4

 2 EI l2

#5) A uniform column is fixed at one end and is kept on rollers at the other end. In other words one end is fixed and other end is hinged. The length is 1m. Find the critical load. U

x At x = 0, y=0; -------(1) At x = 1, y = 0 ; -----(2) dy At x = 1,  0 -----(3) dx Let y = A0  A1 x  A2 x 2  A3 x 3  A4 x 4 . On applying (1), A0 = 0 When x = 1, y = 0. This implies that A1+A2+A3+A4 = 0-----(4) dy When x = 1, 0 dx dy Hence  A1  2 A2  3 A3  4 A4  0 ---------------------------(5) dx (4) – (5) Implies that –A2 – 2A3 – 3A4 =0 And so A2= - (2A3 +3A4).-----(6) Substitute (6) in (4) so that we get A1=A3 + 2A4 ------------------------------(7) Hence y = (A3 + 2A4)x - (2A3 +3A4)x2 +A3x3 + A4x4 y = A3 ( x  2 x 2  x 3 )  A4 (2 x  3x 2  x 4 ) Put A3 = a and A4 = b y= a( x  2 x 2  x 3 )  b(2 x  3x 2  x 4 )

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7

d2y dy  a(4  6 x )  b(6  12 x 2 ) -------(8)  a(1  4 x  3x 2 )  (b(2  6 x  4 x 3 ) and 2 dx dx L L 2 1 d y P dy   2  EI ( d 2 x ) 2 dx   2 ( dx ) 2 dx ---------------------------------------------------------(9) 0 0 Substituting (8) in (9)



 



  1680a 2  7056b 2  6720ab  P 56a 2  288b 2  252ab ----------------------------(10)    0and  0 ----------------------------------------------------------------------------(11) a b Equation 11 gives us the following two results: a[3360  112 P]  b[6720  252 P]  0

----------------------------------------------------(12) a[6720  252 P]  b[14112  576 P]  0 ----------------------------------------------------(13) These are linear homogeneous equations. We know a and b cannot be equal to zero. If they are zero they will not be buckled form. Therefore to get trivial solution the following must be true (3360 – 112P )

(6720 – 252P ) =0

(6720 – 252P ) (14112 – 576P) Upon expanding and simplifying, we get 1008 P2 – 129024 P + 2257920 = 0 which gives P= (128 + 86.166)/2 P= 20.92 is the lower value. During simplification EI was omitted. The same can be reintroduced. The length 1m can also represented by “l”. 20.92 EI Hence P = l2

References: “Theory of Elasticity “ by Sadhu Singh “Finite Element Primer” by V.K.Manickaselvam. U

C.RameshBabu/AP/Civil/MSEC