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Mechanics of Solids Lecture 4 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 4 Outline Members subjected to loads that are perpendicular to their axis - Bending moment and shear force diagrams - Bending stress - Bending properties of sections 2
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Mechanics of Solids - Lecture 4 Classification of beams according to how they are supported Statically determinate beams 3 Types of boundary conditions for beams
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Mechanics of Solids - Lecture 4 Positive sign convention for internal shear force and bending moment 4
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Mechanics of Solids - Lecture 4 Example Draw the shear and bending moment diagrams for the cantilever below 5
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Mechanics of Solids - Lecture 4 Example Note that directions of the internal shear and the bending 6 From Equilibrium 0 y F = V P = − @ 0 B x M = = M P x = − × moment are according to the positive sign convention Shear force diagram Bending moment diagram
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Mechanics of Solids - Lecture 4 Example Draw the shear and bending moment diagrams for the cantilever below 1m 1m 1m 100kN 50kN 40kN x 7
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Mechanics of Solids - Lecture 4 Example positive sign convention 8 From Equilibrium Shear force diagram Bending moment diagram 0 1 x < 0 Y F = 100 V = − 100 M x = − × @ 0 X M =
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Mechanics of Solids - Lecture 4 Example positive sign convention 1m 1m 1m 100kN 50kN 40kN x x V 100kN 9 From Equilibrium Shear force diagram Bending moment diagram 1 2 x < M 1m 50kN 0 Y F = 50 V = − ( ) 100 50 1 M x x =− × + × @ 0 X M =
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Mechanics of Solids - Lecture 4 Example positive sign convention 1m 1m 1m 100kN 50kN 40kN x 10 From Equilibrium Shear force diagram Bending moment diagram 2 3 x < 0 Y F = 90 V = − @ 0 X M = ( ) ( ) 100 50 1 40 2 M x x x =− × + × − − ×
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Mechanics of Solids - Lecture 4 Example Draw the shear and bending moment diagrams for the cantilever below 100kN 1m 2m x 11
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Mechanics of Solids - Lecture 4 Example Find the support reactions 12 From Equilibrium 0 X F = 0 Ax R = 0 Y F = 100 0 Ay By R R + = @ 0 X A M = 100 1 3 0 By R × − × = 66.67 Ay R kN = 33.33 By R kN =
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