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# File - Mechanics of Solids Lecture 6 by Dr Emre Erkmen Lecturer School of Civil and Environmental Engineering The University of Technology Sydney

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Unformatted text preview: Mechanics of Solids Lecture 6 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 Mechanics of Solids - Lecture 6 Outline Outline- Combined bending and axial load- Composite beams under bending moment 2 Mechanics of Solids - Lecture 6 Normal force N causes: uniform normal stress Bending moment M causes: linear longitudinal stress distribution Members under combined bending moment and axial force N A σ = M y I σ = − N σ σ M Resultant stresses by superposition: Use the principal of superposition to determine the resultant normal stress distribution N M y A I σ = − M N Mechanics of Solids - Lecture 6 Eccentric force x e P ide view of the beam x P = M=Pe x e P = + P P y 4 • An eccentric load P can be replaced by a centric load P and a couple M = Pe. • Normal stresses can be found from superposing the stresses due to the centric load and couple. centric bending P M y A I σ σ σ σ = + = + − Tensile force Compression above neutral axis From the formula P P e y A I × = − M P e = × Mechanics of Solids - Lecture 6 Eccentric force x e P x P M=Pe = y 5 • Normal stresses centric bending P M y A I σ σ σ σ = + = − − Compressive force From the formula M P e = × P P e y A I × = − − Mechanics of Solids - Lecture 6 Eccentric force x e P x P M=-Pe = y x P M=Pe = 6 • Normal stresses centric bending P M y A I σ σ σ σ = + = + − Tensile force From the formula M P e = − × P P e y A I × = + + Mechanics of Solids - Lecture 6 Eccentric force x e P Side view of the beam x P = M=-Pe y 7 • Normal stresses can be found from superposing the stresses due to the centric load and couple. centric bending P M y A I σ σ σ σ = + = − − Compressive force Compression above neutral axis From the formula P P e y A I × = − + M P e = − × Mechanics of Solids - Lecture 6 Internal forces under combined axial force and bending moment N=P C 8 M=P.e x N My A I σ = − − y N.A Neutral axis not passing from centroid Mechanics of Solids - Lecture 6 Neutral axis and maximum stress under eccentric force x e P x P M=Pe = y + y 9 • Neutral axis is where the normal stress is zero. P P e y A I σ × = − = N P P e y A I σ × = − N I y A e = × Mechanics of Solids - Lecture 6 Neutral axis and maximum stress under eccentric force x e P x P M=Pe = y + y 10 P P e y A I σ × = − − N I y A e = − × • Neutral axis is where the normal stress is zero. N P P e y A I σ × = − − = Mechanics of Solids - Lecture 6 Neutral axis and maximum stress under eccentric force x e P = y x P M=Pe = y N 11 N P P e y A I σ × = + N I y A e = − × P P e y A I σ × = + = • Neutral axis is where the normal stress is zero. Mechanics of Solids - Lecture 6 Neutral axis and maximum stress under eccentric force x e P x P = M=Pe y 12 N P...
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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 - Mechanics of Solids Lecture 6 by Dr Emre Erkmen Lecturer School of Civil and Environmental Engineering The University of Technology Sydney

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