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Variational method- principles U is solution to continum problem F,E are differential operators
Now, u is exact solutions if for any arbitary i.e if variational integral is made “stationary” now the approximate solution can be found by substituting trial function expansion . since above holds true for any parameter
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Total DOF =O2 (one 2t each node)
u=λ1+λ2x in matrix form Where [p] =parametric matrix II. Displacement function in-terms of nodal displacement :express displacement function in terms of nodal displacement using the coordinates of nodes Where [A] =connectivity matrix obtained Where [N] =shape functions [N] =[p] III. shape functions = Inverse is obtained by using method of adjoin Sum of shape functions is always unity At node 1 x= At node 2, x=
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The relationship between a global coordinates X & a local coordinates x. Linear shape functions in local ordinates 2. Notice |
Interpolation function | In term of global coordinates (x). | In terms of local coordinates (x) |
Linear | ||
Quadratic | ||
Cubic |




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The relationship between local coordinates x& the natural coordinates linear share function in natural coordinates
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Step 1 :- see the order of the equation given :- if ‘n’ then assume polynomial of order/degree “n+1” Now, step 2 – apply boundary conditions one by one i.e y(0) =1 Rewrite the approximately solution is
Step 3- find the reside given by the differential equations i.e ,
The residue becomes
Step 4 – the weighted integral from Wi -> coefficient of Ci in y i.e from equation C Now, putting i=1 , we get Similarly putting i=2 , we get On solving Solving 1 & 2 Therefore,
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Step 1 – formulate differential equation of beam Where v= disp. In ‘y’ dim
Step -2 Residual Weighting function Step -3 Use galerkin’s formulation Step – 4 find the unknown displacement But The maximum disp. In beam at x=l By using 4th boundary con This is a admissible trial solution . Step – 5 trial balance function & boundary conditions Boundary conditions By 1st boundary conditions By 3rd conditions
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At node 1 , At node 2 , 1= 1= At node 3 , At node 7 , At node 8, at node 4, At node 5 = At node 6,
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