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Mechanics of Solids Lecture 3 by Dr Emre Erkmen Lecturer, School of Civil and Environmental Engineering, The University of Technology Sydney Office: 2.520 Phone:9514 9769 Email: [email protected] Lecture hours: Tuesdays 11:00-14:00, Wednesdays 15:00-18:00 Office hours: Tuesdays 15:00-17:00 1
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Mechanics of Solids - Lecture 3 Outline - Review of Statically Indeterminate Axially loaded members - Axially loaded members with elasto-plastic behaviour -Composite axially loaded members -Thermal expansion 2
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Mechanics of Solids - Lecture 3 1 2 d 1 d 2 Statically indeterminate structures L Assume that for analysis purposes we want the deflections under the load P for the structure 3 RIGID BODY P 3
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Mechanics of Solids - Lecture 3 P 1 P 2 P 3 IGID BODY Statically indeterminate structures d 1 d 2 RIGID BODY P 4 We know that the bars will elongate according to the formula (assuming the length, the material and the area of all bars are the same) 1 1 PL EA δ = , 2 2 P L EA = and 3 3 P L EA =
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Mechanics of Solids - Lecture 3 ANALYSIS OF STATICALLY DETERMINATE STRUCTURES STATICALLY INDETERMINATE STRUCTURES Statically indeterminate structures However, we can not find the internal forces directly c Equilibrium 0 0 0 = = = + + + M F F y x 1 2 3 d 1 d 2 RIGID BODY P 5 L
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Mechanics of Solids - Lecture 3 Statically indeterminate structures Only the last two equations are useful. That means we have two equations for three unknowns We have three equilibrium equations in a plane, i.e. 0 x F + → = (Does not give us any thing because there is no horizontal force acting on the structure) 1 2 3 0 y F P P P P + ↑ = → + + = 1 1 3 2 @2 0 ... 0 M P d P d P + ↵ = → × × − × = 6 P 1 P 2 P 3 RIGID BODY P d 1 d 2
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Mechanics of Solids - Lecture 3 Compatibility Equation: Compatibility equations How is the compatibility equation useful for us to obtain the deflections due to the applied load? ( ) 1 1 PL EA δ = Force deflection relations 1 2 3 d 1 d 2 3 1 2 1 1 1 2 d d d = + 7 ( ) 2 2 P L EA = ( ) 3 3 P L EA = ( ) ( ) ( ) ( ) 2 1 3 1 1 1 2 P L EA PL EA P L EA PL EA d d d = + RIGID BODY δ 1 δ 2 δ 3 P
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