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# L4 - • Change in length due to temperature change • No...

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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
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
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 δ = F A

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Compatibility Equation: P A B 1 1 A F L AE δ = 2 2 B F L AE δ = 1 2 0 δ δ δ = + = 1 2 0 A B F L F L AE AE = 0 A B F F F P = + =
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 δ = 1 1 2 loading PL AE δ δ δ = + =
A B F B F B B reaction F L AE δ = 1 B total loading reaction PL F L AE AE δ δ δ = + =

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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: δ T = α ( Δ T ) L α = thermal expansion coef. P = PL AE Total elongation: T + P Thermal Strain cont. ( ) = + Δ = + = AE PL L T P T α δ Thermal deformation and the deformation from the redundant support must be compatible. ( ) ( ) T E A P T AE P P T Δ − = = Δ − = = + = σ 2.37, 2.47, 2.39, 2.48 Only consider axial deformation. 2.39...
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L4 - • Change in length due to temperature change • No...

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