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Unformatted text preview: 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: emre.erkmen@uts.edu.au Lecture hours: Tuesdays 11:00-14:00, Wednesdays 15:00-18:00 Office hours: Tuesdays 15:00-17:00 1 Mechanics of Solids - Lecture 4 Outline Outline Members subjected to loads that are perpendicular to their axis- Bending moment and shear force diagrams- Bending stress- Bending properties of sections 2 Mechanics of Solids - Lecture 4 Classification of beams according to how they are supported Statically determinate beams 3 Types of boundary conditions for beams Mechanics of Solids - Lecture 4 Positive sign convention for internal shear force and bending moment 4 Mechanics of Solids - Lecture 4 Example Draw the shear and bending moment diagrams for the cantilever below 5 Mechanics of Solids - Lecture 4 Example Note that directions of the internal shear and the bending 6 From Equilibrium y F = V P = @ B x M = = M P x = moment are according to the positive sign convention Shear force diagram Bending moment diagram 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 Mechanics of Solids - Lecture 4 Example positive sign convention 8 From Equilibrium Shear force diagram Bending moment diagram 1 x < Y F = 100 V = 100 M x = @ X M = 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 Y F = 50 V = ( ) 100 50 1 M x x = + @ X M = Mechanics of Solids - Lecture 4 Example 1m 1m 1m 100kN 50kN 40kN x 10 From Equilibrium positive sign convention Shear force diagram Bending moment diagram 2 3 x < Y F = 90 V = @ X M = ( ) ( ) 100 50 1 40 2 M x x x = + Mechanics of Solids - Lecture 4 Example Draw the shear and bending moment diagrams for the cantilever below 100kN 1m 2m x 11 Mechanics of Solids - Lecture 4 Example Find the support reactions 12 From Equilibrium X F = Ax R = Y F = 100 0 Ay By R R + = @ X A M = 100 1 3 0 By R = 66.67 Ay R kN = 33.33 By R kN = Mechanics of Solids - Lecture 4 Example R Ay R Ax 100kN 1m 2m R By Ax R = 66.67 Ay R kN = 33.33 By R kN = 13 From Equilibrium positive sign convention Shear force diagram Bending moment diagram 1 x < Y F = 66.67 V kN = 66.67 M x = @ X M = 1m 2m B A 66.67kN-33.33kN Mechanics of Solids - Lecture 4 Example Ax R = 66.67 Ay R kN = 33.33 By R kN = 14 From Equilibrium positive sign convention Shear force diagram Bending moment diagram 1 3 x < 1m 2m B A 66.67kN-33.33kN Y F = 33.33 V kN = ( ) 66.67 100 1 M x x = @ X M = Mechanics of Solids - Lecture 4 Example Draw the shear and bending moment diagrams for the cantilever below 15 Mechanics of Solids - Lecture 4...
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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 Lecture4_Slides.pdf - Mechanics of Solids Lecture 4 by...

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