mdp n161- stress analysis fall 2009 1 stress analysis -mdp n161 bending of beams stress and...

MDP N161- Stress Analysis Fall 2009

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Stress Analysis -MDP N161

Bending of BeamsStress and Deformation

Chapter 6

Instructor: Dr. Chahinaz A.R.Saleh


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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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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 axisY = axis of symmetry


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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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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 8.75 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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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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Solution


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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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Bending of BeamsUnsymmetric 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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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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Normal stress due to positive bending moment My

y

yx I

zM


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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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z

zx I

yM

Then for these cases:


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Arbitrarily Applied Moment


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y

y

z

zx I

zM

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


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Example


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mdp n161- stress analysis fall 2009 1 stress analysis -mdp n161 bending of beams stress and...

Documents

mdp n161 stress analysis

stress analysis mdp

maximum stress slide

normal stress slide

beams stress

solution slide

surfaces slide

saleh slide