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# File Lecture6_Slides - Mechanics of Solids Lecture 6 by Dr...

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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

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Mechanics of Solids - Lecture 6 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

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Mechanics of Solids - Lecture 6 Eccentric force x e P Side 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 × = −

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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 = − ×

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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 × = 0 = N P P y A I σ = N I y A e = ×

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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 y A I σ × 0 =
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 A I σ = 0 = Neutral axis is where the normal stress is zero.

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Mechanics of Solids - Lecture 6 Neutral axis and maximum stress under eccentric force x e P x P = M=Pe y 12 N P P e y A I σ × = − + Neutral axis is where the normal stress is zero.
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