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Unformatted text preview: 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 is the stress at the elastic limit Y is the strain at the elastic limit Example Find the forces for cables BE and CF for a limit allowable stress of 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 Lecture 3 2 Free Body Diagram and the equilibrium equations x F + = BX N = y F + = 200 0 A B C F F F + + = @2 M + = 200 3.5 2 3 0 B C F F = Compatibility Equations 2 3 C B = 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 = = 3 3 4 2 10 200 10 500 5 10 C C C F F = = By using these in the compatibility equation 3 2 B C = 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 = C C F A = 363.65 B MPa = 272.72 C MPa = Elasto-plastic material behaviour B Y > is not acceptable B can be as much as B Y = 123.55 B F kN = Equilibrium should still be satisfied @2 M + = 200 3.5 123.55 2 3 0 C F = 150.97 C F kN = 301.93 C MPa = acceptable From Force-deflection relations PL EA = 3 3 3 1 150.97 10 2 10 150.97 3.02 200 10 500 5 10 C mm = = = From Compatibility 2 3 C B = 2.01 B mm = 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 @ 80 320 50 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 = = = For a vertical deflection of 0.794 mm at point B , vertical deflection of C can be found by using similar...
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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 Lecture3.pdf - Lecture 3 1 Lecture 3 1 Outline:...

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