mdp n161- stress analysis fall 2009 1 stress analysis -mdp n161 bending of beams stress and...
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MDP N161- Stress Analysis Fall 2009
1
Stress Analysis -MDP N161
Bending of BeamsStress and Deformation
Chapter 6
Instructor: Dr. Chahinaz A.R.Saleh
MDP N161- Stress Analysis Fall 2009
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Lesson Overview
1. Studying the effect of applied transverse force and moment on the beams.
2. Calculation of stresses and deformations in beams.
3. Normal stresses due to unsymmetric bending moment
4. Equation of neutral axis5. Deflection of beam
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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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y
yx
xx
s
sx
)(lim
'lim
0
0
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
y
x
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Stress due to Bending
EquilibriumFx =0
x dA=0
-Ey/ dA =0 y dA =0i.e. z is a centroidal
axis
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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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Example 6-1
Determine the maximum stress for the shown loaded beam of rectangular cross section having width of 8.75 in and depth of 27 in.
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Solution1- 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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Deformation (bending angle )
L
For pure bending
L=
= L/
But 1/ = -x/Ey
=-Lx/Ey
=ML/EI
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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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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.
z
zx I
yM
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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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Normal Stress due to arbitrary applied moment
y
y
z
zx I
zM
I
yM
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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
z
zx I
yM
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Conditions of application the flexure formula
z and y axes are centroidal principal axes Moments are about these centroidal
principal axes
y
y
z
zx I
zM
I
yM
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Equation of neutral axis
0y
y
z
z
I
zM
I
yM
Equation of neutral axis is :
y
y
z
zx I
zM
I
yM