L4 - Change in length due to temperature change No stress...

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Quick review Chapter 2: Axial loading Normal Stress Normal Strain Deformation due to axial loading
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Static Indeterminate Problems Statically indeterminate: • internal forces and reactions cannot be determined from statics alone. • More supports than required to maintain equilibrium. Will learn today: • Establishing compatibility equation • Superposition method • Thermal strain
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Statically indeterminate P A B Draw free body diagram P F A F B 0 A B F F F P = + =
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0 A B F F F P = + = 2 unknowns 1 equation Cannot be solved w/ statics Need another equation
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Can be solved w/ internal forces P F A F B L1 L2 F A F B F B 1 1 A F L AE δ = 2 2 B F L AE = F A
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Compatibility Equation: P A B 1 1 A F L AE δ = 2 2 B F L AE = 1 2 0 = + = 1 2 0 A B F L F L AE AE = 0 A B F F F P = + =
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superposition P A B P A A B = + Deformation due to loading Deformation due to support
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P A 2 1 P 2 =0 P 1 =P P P 1 1 PL AE δ = 2 0 = 1 1 2 loading PL AE = + =
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A B F B F B B reaction F L AE δ = 1 B total loading reaction PL F L AE AE = + =
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Thermal Strain Thermal strain:
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Unformatted text preview: Change in length due to temperature change No stress associated from thermal strain unless the elongation is restrained by supports. Thermal Strain cont. Treat the additional support as redundant and apply the principle of superposition. Thermal elongation: Load elongation: T = ( T ) L = thermal expansion coef. P = PL AE Total elongation: T + P Thermal Strain cont. ( ) = + = + = AE PL L T P T Thermal deformation and the deformation from the redundant support must be compatible. ( ) ( ) T E A P T AE P P T = = = = + = 2.37, 2.47, 2.39, 2.48 Only consider axial deformation. 2.39...
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This note was uploaded on 12/14/2010 for the course ENGR 2530 taught by Professor Lindaschadlerfeist during the Fall '08 term at Rensselaer Polytechnic Institute.

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L4 - Change in length due to temperature change No stress...

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