1. Stress Analysis -MDP N161 Bending of Beams Stress and Deformation
Chapter 6
Instructor: Dr. Chahinaz A.R.Saleh
MDP N161- Stress Analysis Fall 2009

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Lesson Overview
Studying the effect of applied transverse force and moment on the beams.
Calculation of stresses and deformations in beams.
Normal stresses due to unsymmetric bending moment
Equation of neutral axis
Deflection of beam
MDP N161- Stress Analysis Fall 2009

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5. Deformation of Straight Beam under Pure Bending
Assumptions
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For this case, top fibers are in compression and bottom in tension…..i.e. bending produces normal strains and hence normal stress
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Neutral surface – does not undergo a change in length
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X = longitudinal axis
Y = axis of symmetry
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Note that:
Normal strain varies linearly with y
Maximum and minimum strains at surfaces
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But x = x/E
 x = -Ey/
The stress varies linearly with y
Remember
E is the modulus of elasticity
 is the radius of curvature
MDP N161- Stress Analysis Fall 2009

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Stress due to Bending
Equilibrium
Fx =0
x dA=0
-Ey/ dA =0
 y dA =0
i.e. z is a centroidal axis
MDP N161- Stress Analysis Fall 2009

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Mz =0
M +x y dA=0
M =- x y dA=  Ey2/ dA
=(E/)  y2 dA
=E Iz /
But x = -Ey/
M = -x Iz/y
x =- M y/Iz
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Flexure Formula
x =- M y/Iz
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Example 6-1
Determine the maximum stress for the shown loaded beam of rectangular cross section having width of in and depth of 27 in.
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Solution
1- External reactions
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2- Internal Reactions
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3- cross section
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4- Maximum stress
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Example 6-2
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Solution
1- External reactions and maximum internal moment
Mmax = 22.5 kN.m
Acting at the middle of the beam
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Example 6-3
Determine the maximum stress for the shown loaded beam.
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Solution
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Deformation (bending angle )
For pure bending
L=
= L/
But 1/ = -x/Ey
=-Lx/Ey
=ML/EI   L
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36. Bending of Beams Unsymmetric Bending
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In all previous cases which have been studied:
Y was an axis of summetry..i.e. Y is a principal axis
M acts about z
Z is a central neutral axis
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38. Normal stress due to positive bending moment Mz
According to right hand role; positive Moment (Mz )could be represented by an arrow pointed to the positive z dn.
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39. Normal stress due to positive bending moment My
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40. Arbitrarily Applied Moment
If M is applied about a centroidal axis which is not a principal
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Then decompose M to Mz and MY, where Y and Z are the centroidal principal axis
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42. Normal Stress due to arbitrary applied moment
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For this shape; axis 2 is not an axis of symmetry.
If M is applied about axis 1; could we apply the flexure formula and say that
 =- M1 y/I1 ?
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What about this section? How can we get the stresses due to bending about axis 1?
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Conditions of application of flexure formula;
z and y axes are centroidal principal axes
M is about z which is a centroidal principal axis
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Then for these cases:
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47. Arbitrarily Applied Moment
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49. Conditions of application the flexure formula
z and y axes are centroidal principal axes
Moments are about these centroidal principal axes
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50. Equation of neutral axis
Equation of neutral axis is :
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Example
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