Beams I -- Bending Stresses (1)

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 1/57

Bending Stresses in Beams

Hide Text1

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 2/57

OverView

Hide Text2

In this stack, our goal is to developa means for determining thestresses in a beam.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 3/57

OverView

Hide Text3

We will proceed by first determining thestrains due to bending…

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 4/57

OverView

Hide Text4

…and then use Hooke's law to determinethe stresses.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 5/57

Beam

Hide Text5

  To begin our detailed look

at the deformations of a bent beam, consider a beam with asymmetric cross section.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 6/57

Centroid of Section

Hide Text6

  We can calculate thelocation of the centroid bysetting the first moment ofarea to zero. The centroidwill always lie on the axis ofsymmetry.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 7/57

Neutral Axis!

Hide Text7

  We define the neutral axis 

of the beam as a line whichexperiences no strain as the beam is bent. As we willdemonstrate at the end of thisstack, the neutral axis passesthrough the centroid of thesection at any point along the beam.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 8/57

Plane Section

Hide Text8

  Let's now look at the material which lies in aplane passed through the beam. This particularplane is normal to the neutral axis.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 9/57

Apply Load

Hide Text9

  We now load the beamand allow it to deflect.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 10/57

Deformed Beam

Hide Text10

  After deformations, we observe that the plane section we were viewing remainsplane, and further, it remains normal to the neutral axis.

This observation is the fundamental assumption in the derivation of the beam

 bending equations. Note that we will not observe this same behavior for very largedeformations.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 11/57

Two Adjacent Sections

Hide Text11

  Let's take another look at the deformation ofthe loaded beam, only this time we will look at thematerial which lies between two adjacent planes.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 12/57

After Loading

Hide Text12

  As we have observed before, the sectionsremain plane and normal to the neutral axisafter deformation.

  For clarity, we will continue by looking at aside view of the deformed beam.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 13/57

Hide Text13

  As we have assumed, the sections remainplane and normal to the neutral axis afterdeformation.

  For clarity, we will continue by looking at aside view of the deformed beam.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 14/57

Hide Text14

  As we have assumed, the sections remainplane and normal to the neutral axis afterdeformation.

  For clarity, we will continue by looking at aside view of the deformed beam.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 15/57

Hide Text15

  As we have assumed, the sections remainplane and normal to the neutral axis afterdeformation.

  For clarity, we will continue by looking at aside view of the deformed beam.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 16/57

Hide Text16

  What we are really interested in is the displacedshape of the element lying between the two planes. Wefocus in on this particular element.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 17/57

Hide Text17

  This is the displaced shape of theelement after loading. Note that thetwo planes defining the element arenormal to the neutral axis.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 18/57

Hide Text18

  Here the element is shown before and after deflection.

How can we relate theposition of a point in thematerial before deformation to

the position of the same pointafter deformation?

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 19/57

Hide Text19

  In order to rigorously

define how a point in thematerial moves when the beam is loaded, we must firstdefine a coordinate system.  Here we define the x axisto run along the neutral axis.We also assume that thelength of the beam element issome value dx.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 20/57

Hide Text20

  The distance of the pointabove or below the x axis(neutral axis) we will defineas "y". y is positive whenthe point lies above theneutral axis.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 21/57

Hide Text21

  The displacement of theneutral axis from the

undeformed to the deformedconfiguration is described bythe function !  (x).

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 22/57

Hide Text22

  If ! " x) is thedisplacement of the beam at any point, x,then the firstderivative of thedisplacement,! ' (x) ,is the slope of the beam at the point x.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 23/57

Hide Text23

  Let's compare ! ' (x) (the slope) of one side

of the deformed elementto the other. To do thiswe extend theundeformed orientationof the two planes downonto the deformedelement.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 24/57

Hide Text24

  At this point we need toassume that the left face ofthe element is located a

distance x along the beam.From this assumption wecalculate that the right faceof the element is located adistance x + dx along the beam.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 25/57

Hide Text25

  The angle between theplane defining the left faceof the deformed element

and its undeformedorientation is simply theslope of the beam at thatpoint, !  '(x).

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 26/57

Hide Text26

  Similarly, the angle between the plane definingthe right face of thedeformed element and itsundeformed orientation isthe slope of the beam at thatpoint, ! ' (x + dx)  .

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 27/57

Hide Text27

  We can expand thisexpression as shown below.We will neglect the higher

order terms since we aredealing with smalldisplacements, and thereforesmall changes in slopes.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 28/57

Hide Text28

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 29/57

Hide Text29

  The previous few steps mayhave been a bit confusing, solet's go through them againwith a bigger picture.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 30/57

Hide Text30

  We begin by projectingthe deformed element ontothe undeformed element.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 31/57

Hide Text31

  We were able tocalculate the angle between theundeformed anddeformed planes asshown above.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 32/57

Hide Text32

  Recall that the length of an arc

may be calculated as the angledefining the arc times the radiusof the arc.  Using this knowledge we areable to calculate the horizontaldisplacement of the upper-leftcorner of the element as:  y! ' (x)  .

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 33/57

Hide Text33

  We can calculate thedisplacement of theupper-right corner of theelement in a similar fashion.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 34/57

Hide Text34

  The change in length of thetop chord of the element can

now be calculated as thedifference between thedisplacement of the upper-leftand upper-right corners.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 35/57

Hide Text35

  Eliminating like terms, wefind that the change in lengthof the top chord is given by:

 – y! '' (x)  dx  

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 36/57

Hide Text36

  Recall that strain is

calculated as change in lengthdivided by original length.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 37/57

Hide Text37

The strain of the material at the topof the element is then calculated asthe change in length of the topchord divided by the originallength of the top chord.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 38/57

Hide Text38

  Note that we have actually kept the location where we calculate the strain in terms

of the y coordinate. Our equation tells us that when y is zero there is no strain. Wecan confirm this by noting that the element does not change length at the neutral axis.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 39/57

Hide Text39

  Now that we have anexpression for the strain at anypoint in the beam, how do wecalculate the stress?

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 40/57

Hide Text40

  Hooke's Law tells us thatstress is linearly related to strain by the material constant, E.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 41/57

Hide Text41

  Combining the two equationswe can relate the stress at a pointin the beam to the displacedshape of the beam and Young'sModulus.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 42/57

Hide Text42

  Plotting the stress on the

element, we see that it varieslinearly with y , the distancefrom the neutral axis. Also, ifthe function! '' (x)  (thecurvature of the beam) is notconstant, then the stressvaries along the length of the beam as well.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 43/57

Hide Text43

  Since it is not common thatwe begin with an expressionfor the displaced shape of the beam, let's try to manipulateour equation for stress due to bending until it is more useful.  Here we introduce thenotion that internal moment,M, is actually the sum of manysmaller moments caused by

stress acting away from theneutral axis.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 44/57

Hide Text44

  Next, we substitute ourexpression for stress.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 45/57

Hide Text45

  Rearranging terms...

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 46/57

Hide Text46

  Recall the expression for

moment of inertia.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 47/57

Hide Text47

  Performing thesubstitution we find thatinternal moment is related to

the curvature of the beam.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 48/57

Hide Text48

  How can we use this resultto simplify our expression forstress in the beam?

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 49/57

Hide Text49

  We use the secondexpression to eliminate ! '' (x) from the stress equation and wearrive at the famousrelationship:

  " = –M y/ I

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 50/57

Hide Text50

  Here is a summary of the importantrelationships we have derived usingthe displacement assumption that"plane sections remain plane, andnormal to the neutral axis."

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 51/57

Hide Text51

So far we have worked on the assumptionthat the neutral axis and the centroidal axiscoincide. We will now demonstrate thevalidity of this assumption. To this end,

consider the net horizontal force, P.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 52/57

Hide Text52

The net force can be calculated by integrating thestresses over the cross section. Since no horizontalloads have been applied to the beam, the net force

must be zero.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 53/57

Hide Text53

Using our previous results, we canexpress the stresses in terms of thedisplacement and material

properties as shown.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 54/57

Hide Text54

We can pull E and ! '' out of theintegral, since they do not vary

over the cross-section.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 55/57

Location of Neutral Axis

Hide Text55

Since E and ! '' are notzero in general, the

 boxed result must hold.This is simply thedefinition of the centroid,and so the proper placefrom which y ismeasured is the centroid.Thus, the neutral axis andthe centroidal axis arecoincident.

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 56/57

Summary

Hide Text56

  The relation between bending moment and the resulting

stresses is extremely important, and you are likely to encounterit again and again. You should store " = My/I somewhere inyour brain near F = ma.  Remember, the linear distribution of stress predicted bythis equation is based on the assumed "plane sections remainplane and normal to the neutral axis" assumption, which is anapproximation (but a darn good one as long as the beam's

length is more than about 3 to 4 times its depth).

8/12/2019 Beams I -- Bending Stresses (1)

http://slidepdf.com/reader/full/beams-i-bending-stresses-1 57/57

End

Hide Text57