Beams I -- Bending Stresses (1)
-
Upload
patrick-lee -
Category
Documents
-
view
235 -
download
12
Transcript of 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