I(y)=
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L[y(x)]=
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S[z(x,y)] =
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S(q) = Where, q is the function to be found: q:[a,b] t such that q is differentiable, q(a)=
The Euler–Lagrange equation, then, is given by Here
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We assume that F is twice continuously differentiable A weaker assumption can be used but the proof becomes more difficult. Let Then define, We now wish to calculate the total derivative of It follows from the total derivation that.
So, Where & Using the boundary conditions Applying the fundamental lemma of calculus of variations now yields the Euler- Lagrange eq.
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δWi= δWe Internal virtual work External virtual work
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Our ODF is sepretable , thus we, seprate Now, integrable both sides we get, Thus , So, Applying y© =1 we obtained Finally, |
We are seprating our differential eq. Applying y(x)=4 4=0-0+C C=4 Put this value in eq. 1
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We are seperating our differential equation Integrating both sides Apply boundary value conditions Y(0)=3 Take exponential function both sides C=3 Put in eq. 1
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Element | K(N/m) | Nodes | Displacements m | boundary conditions |
1 | 500 | 1-2 | ||
2 | 100 | 2-3 | ----- |
Step -2 Element stiffness matrices. Step -3 global stiffness matrix Assemble the element stiffness matrices to get global stiffness matrix Step -4 reduced stiffness matrix imposing boundary conditions i.e Therefore, reduced stiffness matrix is. Step-5 determine unknown joint displacement Applying equation of equilibrium Step -6 calculation of spring force. Spring -1 Spring 2 –
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Element | K(N/m) | Nodes | Displacement | Boundary conditions | |
1 | 1000 | 1-2 | |||
2 | 2000 | 2-3 | - | ||
3 | 3000 | 3-4 | |||
Step -2 element stiffness matrices Step -3 global stiffness matrix Assemble the element stiffness matrices to get the global stiffness matrix Step -4 reduced stiffness matrix Imposing boundary conditions Eliminate first raw, first column & fourth raw & fourth column . Therefore reduced stiffness matrix is Step -5 determine unknown joint displacement Applying equation of equilibrium Step -6 calculate of spring force Spring 1 -
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