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Chapter 4 Pure Bending

# Chapter 4 Pure Bending - Chapter 4 Pure Bending Ch 2 Axial...

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Chapter 4 Pure Bending Ch 2 – Axial Loading Ch 3 – Torsion Ch 4 – Bending -- for the designing of beams and girders

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4.1 Introduction

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4.2 Symmetric Member in Pure Bending M = Bending Moment Sign Conventions for M: -- concave upward -- concave downward
Force Analysis – Equations of Equilibrium 0 x dA σ = 0 x z dA σ = ( ) x y dA M σ - = Σ F x = 0 Σ M y-axis = 0 Σ M z-axis = 0 τ xz = τ xy = 0 (4.1) (4.2) (4.3)

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4.3 Deformation in a Symmetric Member in Pure Bending Assumptions of Beam Theory: 1. Any cross section to the beam axis remains plane 2. The plane of the section passes through the center of curvature (Point C). Plane CAB is the Plane of Symmetry
The Assumptions Result in the Following Facts: 1. τ xy = τ xz = 0 γ xy = γ xz = 0 2. σ y = σ z = τ yz = 0 The only non-zero stress: σ x 0 Uniaxial Stress The Neutral Axis (surface) : σ x = 0 & ε x = 0

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L ρθ = ' ( ) L y ρ θ = - ' = - L L δ Where ρ = radius of curvature θ = the central angle Line JK (4.5) Before deformation: DE = JK Therefore, Line DE (4.4) (4.6) ( ) y y δ ρ θ ρθ θ = - - = -
The Longitudinal Strain ε x = l l o - = = = - x x y L y δ θ ε ρθ ε ρ ε x varies linearly with the distance y from the neutral surface (4.9) The max value of ε x occurs at the top or the bottom fiber: = m c ε ρ (4.8)

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Combining Eqs (4.8) & (4.9) yields x m y c ε ε = - (4.10)
4.4 Stresses and Deformation is in the Elastic Range For elastic response – Hooke’s Law x x E σ ε = ( ) x m y E E c ε ε = - x m y c ε ε = - max x m y y c c σ σ σ = - = - Therefore, (4.11) (4.10) (4.12)

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Based on Eq. (4.1) 0 x dA σ = max x m y y c c σ σ σ = - = - 0 ( ) m x m y dA dA ydA c c σ σ σ = - = - = (4.1) (4.12) (4.13) Hence, 0 ydA first moment of area = =
Therefore, Within elastic range, the neutral axis passes through the centroid of the section.

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Chapter 4 Pure Bending - Chapter 4 Pure Bending Ch 2 Axial...

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