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# File Lecture12_Slides.pdf - Mechanics of Solids Lecture 12...

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Mechanics of Solids Lecture 12 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 12 Outline - Energy methods Conservation of Energy Castigliano’s theorem 2 Principle of Virtual Work
Mechanics of Solids - Lecture 12 Conservation of energy The external loads are applied slowly to a body, so that kinetic energy can be neglected. The loads deform the body and as they are displaced they do external work U e . This external work is transformed into internal work or strain energy U , which is 3 i stored in the body. Assuming material’s elastic limit is not exceeded, conservation of energy for body is stated as e i U U =

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Mechanics of Solids - Lecture 12 Conservation of energy Consider a truss subjected to load P applied gradually, thus U e = 0.5P Δ , is the vertical displacement of truss at the point where P is applied. 4 Assume that P develops an axial force N j in a particular member, and strain energy stored is Summing strain energies for all members of the truss, we obtain the strain energy of the structure as 2 2 j j j j j N L U A E = 2 1 1 2 2 n j j j j j N L P A E = Δ =
Mechanics of Solids - Lecture 12 Conservation of energy Consider a beam subjected to load P . External work is U e = 0.5P Δ . Strain energy in beam due to shear can be neglected (lecture 11). 5 Beam’s strain energy determined only by the internal moment M , can be written as 2 0 1 2 2 L M P dx EI Δ =

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Mechanics of Solids - Lecture 12 Conservation of energy Consider a beam loaded by a moment M 0 . A rotational displacement θ is caused. External work done is U e = 0.5M 0 . 6 2 0 0 1 2 2 L M M dx EI =
Mechanics of Solids - Lecture 12 The three-bar truss is subjected to a horizontal force of 20 kN. If cross-sectional area of each member is 100 mm 2 , determine the horizontal displacement at point B . E =200 GPa. Example - Deflections by Conservation of energy 7

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Mechanics of Solids - Lecture 12 Since only a single external force acts on the truss and required displacement is in the same direction as the force, we can use conservation of energy to find the displacement. The reactive force on the truss do no produce work since they are not displaced. sing ethod ints, rce ch ember etermined own n ee Example - Deflections by Conservation of energy 8 Using method of joints, force in each member is determined as shown on free- body diagrams of pins at B and C .
Mechanics of Solids - Lecture 12 Example - Deflections by Conservation of energy ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 3 3 2 2 3 3 1 2 2 11.547 10 N 1 m 1 20 10 N 2 2 20 10 N 1.732 m 23.094 10 N 2 m 2 B h N L P AE AE E AE Δ =

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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 Lecture12_Slides.pdf - Mechanics of Solids Lecture 12...

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