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# File Lecture3 - Lecture 31 Outline-Statically indeterminate...

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Lecture 3 1 Lecture 3 1 Outline: -Statically indeterminate axially loaded members continues… -Composite axially loaded members -Thermal expansion 1. Statically indeterminate axially loaded members continues… When a material is stressed beyond the elastic range, it starts to yield and thereby causes permanent deformation. Among various inelastic behaviour, the common cases exhibit elastoplastic or elastic- perfectly-plastic behaviour. Fig. 1 Stress-strain plot of an elasto-plastic material Y σ is the stress at the elastic limit Y ε is the strain at the elastic limit Example Find the forces for cables BE and CF for a limit allowable stress of σ Y =310MPa . Material is Elasto- plastic. Beam AD is rigid and both BE and CF are elastic with a modulus of E =200GPa and cross- sectional areas of A BE =400mm 2 and A CF =500mm 2 . The horizontal rod is rigid so it does not deform. 1 Mechanics of Solids Dr Emre Erkmen

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Lecture 3 2 Free Body Diagram and the equilibrium equations 0 x F + → = 0 BX N = 0 y F + ↑ = 200 0 A B C F F F + + = @2 0 M + = 200 3.5 2 3 0 B C F F × × × = Compatibility Equations 2 3 C B δ δ = We know that the cables will elongate according to the formula 3 3 4 1 10 200 10 400 8 10 B B B F F δ × × = = × × × 3 3 4 2 10 200 10 500 5 10 C C C F F δ × × = = × × × By using these in the compatibility equation 3 2 B C δ δ = 16 15 C B F F = By substituting the above relation in to the moment equilibrium equation 145.46 B F kN = 136.36 C F kN =
Lecture 3 3 Stresses B B F A σ = C C F A σ = 363.65 B MPa σ = 272.72 C MPa σ = Elasto-plastic material behaviour B Y σ σ > is not acceptable B σ can be as much as B Y σ σ = 123.55 B F kN = Equilibrium should still be satisfied @2 0 M + = 200 3.5 123.55 2 3 0 C F × × × = 150.97 C F kN = 301.93 C MPa σ = acceptable From Force-deflection relations PL EA δ = 3 3 3 1 150.97 10 2 10 150.97 3.02 200 10 500 5 10 C mm δ × × × = = = × × × From Compatibility 2 3 C B δ δ = 2.01 B mm δ =

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Lecture 3 4 Example Bar BF is made of steel ( E =210GPa), and bar CE is made of aluminium alloy ( E =73GPa). The cross- sectional areas are 1200 mm 2 for bar BF and 900mm 2 for bar CE. As a result of a misalignment of the pin holes at A , B and C , a force of P =50kN upward must be applied at D , after pins A and B are in place, to permit insertion of pin C . Determine the normal stress in bar CE when the force P is removed with all pins in place and the vertical component of the displacement of pin D from the initial horizontal position. Free Body Diagram and the equilibrium equations @ 0 80 320 50 0 BF A M T + = × × = Initially T CE = 0 so the system is statically determinate. From the moment equilibrium 200 BF T kN = is found. Under P =50kN point B has an upwards vertical deflection of 3 3 200 10 1000 0.794 1200 210 10 BF B T L mm EA δ × × = = = × × For a vertical deflection of 0.794 mm at point B , vertical deflection of C can be found by using similar triangles because the horizontal bar AD is rigid. And thus from the figure below for any deflection at B which is shown with b , we can say that the deflection at C which is shown with c is always three times, i.e. 3 c b = × .
Lecture 3 5 δ C0 δ CE 0 3 0.794=2.382mm c δ = × Bar

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File Lecture3 - Lecture 31 Outline-Statically indeterminate...

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