C.RameshBabu/AP/Civil/MSEC 1 Solved Problems (Rayleigh Ritz method) U1.Rayleigh Ritz method applied to Beams U#1) Find the deflection at the centre using Rayleigh Ritz method for the simply supported beam subjected to Concentrated load at the centre Solution: Take y = =knnlx na1sin t. According to Rayleigh Ritz method [= U W and 0 =cc[nC where = Potential energy U = Strain energy and W is the external work done. cpy dxx dy dEI = }[222) (21 where yc is the deflection at the centre. == knnlx nCoslnadxdy1t t = =||.|

\| knnlx nlnadxy d122 222sin t t ==||.|

\| knnlx nlnadxy d1244 42222sin t t }[ = = = Knnknnlna P dxlx nlna EI0 1244 4202sin sin21 t t t --------(1.a) We Know that 0 sin sin0=} dxlx mlx nlt twhen m is not equal to n . 2sin sin0ldxlx mlx nl=} t t when m = n Hence equation 1.a becomes, C.RameshBabu/AP/Civil/MSEC 2 2 2221444 t tt nPSinlla EIa nn =cc[ --------------(1b) From 1 b eqn, 2sin1 24 43tt nn EIplan = There fore == kn lx nSinnSinn EIply... 5 , 3 , 14 4321 2 t tt------------(1c) To find out yc substitute x = l/2. in above equation (1.c) == kncn EIply... 5 , 3 , 14 431 2t Hence we have |.|

\| + + =4 4 43513112EIplyct which is the deflection at the centre. U#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 P0/m Let lxalxa y t t 3sin sin2 1 + = dxdxy d EIUL} ||.|

\|=02222 P P0/m L C.RameshBabu/AP/Civil/MSEC 3 LxCosLaLxCos aL dxdy t t t t 3 32 1 + = LxSinLaLxSin aL dxy d t t t t 3 9222 12222 = dxdxy d EIUL} ||.|

\|=02222= dxLxSinLaLxSinLa EI L} ||.|

\|022222143 92t t t t which yields U = ) 81 (222144a aLEI+t Now W = max00 Py ydx PL+} ymax = a1 a2 Hence W = ) (32 102 1 0 a a P dxLxSin aLxSin a PL +((

|.|

\| +} t t W = (((

+3221 0aaLPt+ P(a1 a2) [= U W [= ) ( )3(2) 81 (42 1210222134a a PaaL Pa aLEI + +tt 01=c[ caand 02=c[ ca which gives the following equations: From the above equations we get )32(20431 PL PEILa + =t t and )32(8120432 PL PEILa =t t ymax = a1-a2 ymax = EIL PEIPL82 . 76 48403+ 03228102203240134= + = PL PLa EIPL PaLEItt tt C.RameshBabu/AP/Civil/MSEC 4 U2.Rayleigh Ritz Method applied to Columns: U#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. 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 dxdxdy Pdxx dy dEIL L202220) (2) (21} }[ = 22lqxdxdy= and 2 222) (lqdxy d= dxLqx PdxLqEIL L220220)2(2)2(21} }[ = which gives LpqLEIq32 2232 = H LpqLEIqq 34 403 = =c[ c and this equation yields P =23LEI URepeat the above problem by using 3322 1 0 x a x a x a a y + + + = 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 When x = l, 022=dxy d. 23 2 13 2 x a x a adxdy+ + = and x a adxy d3 2226 2 + = x y Load P C.RameshBabu/AP/Civil/MSEC 5 At x = l, 022=dxy d. Hence x a adxy d3 2226 2 + = =0 and a2 = -3a3l So y = ) 3 ( 32 333323 lx x a x a lx a = + At x = l,y = q There fore q= ) 3 (3 33 l l a which gives 332lqa = Hence y = ) 3 (22 33 lx xlq and ) 3 6 (223 x lxlqdxdy = ; ) 6 6 (23 22x llqdxy d = dxdxdy Pdxx dy dEIl l202220) (2) (21} }[ = dx x lxlq Pdx x llq EI l l2 20 062262) 3 6 (4 2) 6 6 (4 2 = H } } On simplifying lpqlEIq5323232 = H . 056 33 = =cH clpqlEIqqFrom this equation we get P=2 25 . 2615lEIlEI= U#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 i) 2) ( 4lx l hxy = ii) lxaSin y t= USolving (i) dxdxdy Pdxx dy dEIL L202220) (2) (21} }[ = 2) ( 4lx l hxy = and so we have ) 2 (42 x llhdxdy = ; 2 228lhdxy d = } } + = H l ldx lx x llh Pldx h EI0 02 24242) 4 4 (162642 0316 643 = =cH clphlEIhh. This equation gives the critical load.Pcr=212lEI C.RameshBabu/AP/Civil/MSEC 6 USolving (ii) lxaSin y t= . This provides lxCosladxdy t t= and lxSinladxy d t t2222= dxdxdy Pdxx dy dEIL L202220) (2) (21} }[ = } } = H l ldxlxCosla PdxlxSinla EI0 0222 2244 22 2t t t t------------------------------------------------A 2 22102 ldxlxCosdxlxSinl l==} }0tt } } =+= l loldxlxCosdxlxCos022 221 tt Substituting the above results in the equation A and finding 02.2.2 2.222244= =cH c lla P lla EIat t The above expression yields critical Load Pcr=22lEI t U#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. At x = 0, y=0; -------(1) At x = 1, y = 0 ; -----(2) At x = 1, 0 =dxdy-----(3) Let y = 443322 1 0 x A x A x A x A A + + + + . On applying (1), A0 = 0 When x = 1, y = 0. This implies that A1+A2+A3+A4 = 0-----(4) When x = 1, 0 =dxdy Hence 0 4 3 24 3 2 1 = + + + = A A A Adxdy---------------------------(5) (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 = ) 3 2 ( ) 2 (4 243 23 x x x A x x x A + + + Put A3 = a and A4 = b y= ) 3 2 ( ) 2 (4 2 3 2x x x b x x x a + + + x C.RameshBabu/AP/Civil/MSEC 7 ) 4 6 2 ( ( ) 3 4 1 (3 2x x b x x adxdy+ + + = and ) 12 6 ( ) 6 4 (222x b x adxy d+ + + = -------(8) dxdxdy Pdxx dy dEIL L202220) (2) (21} }[ = ---------------------------------------------------------(9) Substituting (8) in (9) = H | | | | ab b a P ab b a 252 288 56 6720 7056 16802 2 2 2+ + + + ----------------------------(10) 0 0 =cH c=cH cbanda----------------------------------------------------------------------------(11) Equation 11 gives us the following two results: ----------------------------------------------------(12) ----------------------------------------------------(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. Hence P = 292 . 20lEI UReferences: Theory of Elasticity by Sadhu Singh Finite Element Primer by V.K.Manickaselvam. 0 ] 576 14112 [ ] 252 6720 [0 ] 252 6720 [ ] 112 3360 [= + = + P b P aP b P a