bending forces or beam me up scotty (credit for many illustrations is given to mcgraw hill...
TRANSCRIPT
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Bending Forces
Or Beam Me Up Scotty(Credit for many illustrations is given to McGraw Hill publishers and an array of
internet search results)
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Parallel Reading
Chapter 6Section 6.1 IntroductionSection 6.2 Strain Displacement AnalysisSection 6.3 Flexural Stress in Linear Elastic Beams(Do Reading Assignment Problem Set 6A)
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Consider the Case of a Beam With a Load in the Middle
The members on top are in aSqueeze Play, while themembers on the bottom are inTension.
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Lets Take a Look at That
If we are in tension on one side of theBend and compression on the other,Somewhere there must be a neutral plain
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As We Move Away from the Neutral Axis the Strain Varies Linearly
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Hooke’s Law Now Tells Us About Stress in Beam Bending
Where E is Young’s Modules
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Now We Get A Tip from Statics
The forces (stress * area) above and below the neutral plane haveto be equal.
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Only One Way that is True
That neutral plain has to go right through the centroid of the beam
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So What is a Centroid?(we hope the heck this is a review)
The physical center
The center of massfor the beam
So someone tell me where the centroid isfor the 4 X 6 beam?
Where is it for the 3 X 8 beam?
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So How do We Find the Centroid When its’ not Stupid Obvious
90 mm
20 mm
40 mm
30 mm
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Someone's bound to come up with Integrating Over the Area
A less painful option is usuallyAvailable.
40 mm
30 mm
90 mm
Most of the objects we work with break-down intoSimple parts where the centroid is obvious
90 mm
20 mm
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Lets Peg the Obvious Centroids
30 mm
40 mm
20 mm
90 mm
45 mm 10 mm
20 mm
15 mm
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Next We’ll Weight Each Obvious Centroid by the Area of It’s Object
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Now We’ll Divide Through by Total Area
We just nailed ourselves the Centroid of a T beam
(Of course finding centroids is not a key topicof this course, but if we can’t do it, it will makeour lives miserable for this course).
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Back to Bent Beams
When our beam deflects it bendsalong the arc of a circle of radius ρthrough an angle of θ. The radiusextends from the center point of thearc of the bend to the neutral plane.
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If We Take a Closer Look at Deformation in a Cross-Section
T
The plain of our cross-section remainsA nice flat cross-section – But
On the compression side our nice formerRectangle puffs out increasingly towardThe top (prob not a surprise if weRemember Poisson’s ratio)
And get increasingly skinny on the bottomAs we into higher tension areas awayFrom the neutral plain.
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The Amount of Deflection is Related
To a bunch of terms including theBending moment on the beam,Young’s Modulus (a materialProperty), and something called IThat comes from the geometry ofThe beam.
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Similarly, the amount of thinning or thickening is proportional to
material properties
Any given cross section stays a plain but
The Compression size fattens up
The Tension size skinnies down
Not surprisingly the amount of plumping out orSkinnieing down is proportional to Poisson’sRatio
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Lets Review Our Materials Properties
Young’s Modulus is the slope of the line in a stressStrain plot. It relates change in length from a tensionOr compression load to the stress
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When Things Stretch in One Direction – They Skinny Up in the
other
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The Proportion is Called Poisson’s Ratio
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There is Young’s Modulus(We know what that is)
It makes sense that we might not wantExcessive deflection
So Let Make Sure We Understand the Terms for the amount of
Deflection
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Checking Out More Terms
M is that bending moment couple thatIs deflecting the beam.
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And Then There is I
Right now I’m not seeing what that is
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Let’s Review that Moment Term
Note its just theForce * lever arm y
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Handily a Look at that Last Equation gives us I
Obviously
We call this term the Moment of Inertia(Yes we do hope this is a review for you from Statics)
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Section Modulus is a Closely Related Term
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That Inertia Term is a Measure of the Ability of a Beam to Resist
BendingThere arePrecalculatedTables of theseValues availableFor mostStructural steelshapes
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Lets Try Doing Something With This Stuff
The allowable TensileStress is 12 Ksi
The allowable CompressiveStress is 16 Ksi
What is the largest Moment CoupleI can put on this thing?
6 in
4 in
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Our Basic Equation
Neutral Axis or Plane
3 in
Limiting stress
Since our beam is symmetric our most limiting stress will be tension
SM *000,12
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Obviously We Need S
=
6 in
4 in
246
*4 62
S
For a rectangle
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Going for the Answer
inlbM *000,28824*000,12
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Assignment 13
Do Problems 6.3-6 and 6.3-10
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What if We Tried a Different Shape
6
4For our rectangle
A 24 section modulus allowed us to putA 288,000 in*lb moment on our beam whileStaying in allowable tensile stress
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Use a Table of Standard Wide Flange Beams
First term after W(in this case 12 inches)Second term is the weight in lbs per foot
If I pick a weight of 22 lbs/ft I will getA Section Modulus of 25.4 > 24
A wide W12X22 wide flange beam will carry slightly more load than our6X4 beam.
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Here’s the Kicker
The area of my wide flange beam is 6.48 in^2Instead of 24 in^2 for my rectangular beam
I get more from only 25% of the material byUsing a wide flange beam!
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Is there a Down Side?
4 in
6 in
S = 16
Maximum Moment = 12,000 * 16 = 192,000 in*lbs
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So Could Anything Go Wrong with Our Wide Flange Beam?
S for a W12X22 beam aboutThe weak axis
S= 2.31 < < 16 for our 4X6
NA
Things really go to crap aroundThe weak axis.
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Assignment 14
Do problems 6.4-15 and 6.4-16
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Controlling Cost With Beams
We might want to consider a less expensivematerial
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The Problem with Concrete Beams
Lk
Like most brittle rock materials – theyHave little tensile strength
We already saw in our last problem thatTensile strength can form our design Limit
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The Practical Trick
Put steel reinforcingRebar near the tensileEdge of the beam
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Theory of Reinforced Concrete
Compression load area
Neutral Plane or axis(Which is not at the centroid)
Tensile load rides entirely onthe steal reinforcing rods
Concrete holdsThe rods out atA distance toMaximize theirInertial value
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Convert the Steel Cross Sectional Area to an Equivalent Concrete
Area
Here is our concreteCompression area
Hear is the equivalent concrete areaTo replace the rebar
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Lets Walk Through This
We have a concrete beam
E for Concrete is 25 GPa
It has steel rebarReinforcement.
E for Steel is 200 GPa
This is the area ofThe steel
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Conversion to an Equivalent Concrete area is Proportional to
Young’s Modulus
So an equivalent concrete area is
*
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We Now Need to Find the Neutral Axis
(Which is not at the Centroid this time)
b
We exploit the fact that the momentOf the top part must be equal to theMoment of the steel equivilent
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Set Up Our Quadratic Equation
Moment of ConcreteAbout neutral axis
Moment of ourEquivalent concrete(steel) section about axis
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Solving Our Quadratic
177.87mm
302.13mm
Let us now assume the bending moment on this beam is 175 KN*m lets check out the resulting stresses
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We Know the Forces Above and Below the Neutral Plain are Equal
and Opposite
Looks like we need the value of I
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Get the Inertia of the Upper Compression Block around Neutral
Axis
Contribution toInertia of BeamFrom CompressionConcrete
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Now the Inertia of Our Equivalent Concrete Area
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Solve I for Our Equivalent Beam System
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Now Let Get the Stress in Our Concrete Compression Area
(from our previous calculation)
Our given Moment Load
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Going for the Stress on the Steel
Stress in equivalent concreteIs the same.
But we remember the ratio ofOur Young’s Modulus
* 8 =