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# File Lecture12.pdf - Lecture 121 Outline Energy methods...

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Lecture 12 2 Example The three-bar truss is subjected to a horizontal force of 20kN. If cross-sectional area of each member is 100mm 2 , determine the horizontal displacement at point B , ( ) B h Δ . The modulus of elasticity E = 200GPa. B ) h Since only a single external force is acting on the truss and required displacement is at the same point and in the same direction as the force, we can use the conservation of energy principle. Also note that the support reactions do not produce work because they are not displaced. By using the method of joints, force in each member can be determined as shown on free-body diagrams of pins at B and C . Summation of the strain energies for all members of the truss can be determined as ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 3 1 2 2 2 3 3 3 3 1 2 2 20 10 N 1.732 m 11.547 10 N 1 m 23.094 10 N 2 m 1 20 10 N 2 2 2 2 i i B h i B h N L P AE AE AE AE = Δ = × × × Δ = + + From the above equation, the horizontal deflection at point B can be determined as ( ) 94640.0 N m B h AE Δ = Substituting in numerical values for A and E , we obtain ( ) ( )( ) ( ) 2 2 9 2 3 94640.0 N m 100 mm 1 m /1000 mm 200 10 N/mm 4.73 10 m 4.73 mm B h Δ = = × =
Lecture 12 3 Example

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## This note was uploaded on 08/22/2011 for the course ENG 48331 taught by Professor Brown during the Three '11 term at University of Technology, Sydney.

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File Lecture12.pdf - Lecture 121 Outline Energy methods...

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