HW07_Solution - EML 4507 FEA & DESGN 4 Fall 2010 HW07...

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4 Fall 2010 HW07 Solution
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3 4 3 2 1 5 4 3 2 1 5 6 1 2 3 4 1 2 1 x y F 2 3 1 2 3 4 1 2 3 10,000 N L L L x y 7,656 828 6,482 1,000 The properties of the members of the truss in the left side of the figure are given in the table. Calculate the nodal displacement and element forces. Show that force equilibrium is satisfied as Node 3. Elem L (m) A (cm 2 ) E (GPa) α (/ o C) T ( o C) 1 1 1 100 20 × 10 –6 0 2 1 1 100 20 × 10 –6 0 3 1 1 100 20 × 10 –6 0 4 1 1 100 20 × 10 –6 0 5 1 100 20 × 10 –6 –200 Solution: Element stiffness matrices , , For Element 5, l = cos(45) and m = sin(45). Thus, Thermal force in Element 5:
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All other elements do not have thermal force. After assembly, the global stiffness matrix becomes Boundary condition: Reduced equation: Solve the reduced equation: Using the formula , we can obtain the changes in length as , and . Thus, from , we obtain:
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At node 3:
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30. Repeat Problem 29 for the truss on the right side of the figure. Properties of Element 6 are same as that of Element 5, but T = 0 o C. Solution:
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This note was uploaded on 06/07/2011 for the course EML 4507 taught by Professor Kim during the Fall '11 term at University of Florida.

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HW07_Solution - EML 4507 FEA & DESGN 4 Fall 2010 HW07...

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