Bending Forces Or Beam Me Up Scotty (Credit for many illustrations is given to McGraw Hill...

Bending Forces

Or Beam Me Up Scotty(Credit for many illustrations is given to McGraw Hill publishers and an array of

internet search results)

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)

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.

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

As We Move Away from the Neutral Axis the Strain Varies Linearly

Hooke’s Law Now Tells Us About Stress in Beam Bending

Where E is Young’s Modules

Now We Get A Tip from Statics

The forces (stress * area) above and below the neutral plane haveto be equal.

Only One Way that is True

That neutral plain has to go right through the centroid of the beam

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?

So How do We Find the Centroid When its’ not Stupid Obvious

90 mm

20 mm

40 mm

30 mm

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

Lets Peg the Obvious Centroids

30 mm

40 mm

20 mm

90 mm

45 mm 10 mm

20 mm

15 mm

Next We’ll Weight Each Obvious Centroid by the Area of It’s Object

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).

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.

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.

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.

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

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

When Things Stretch in One Direction – They Skinny Up in the

other

The Proportion is Called Poisson’s Ratio

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

Checking Out More Terms

M is that bending moment couple thatIs deflecting the beam.

And Then There is I

Right now I’m not seeing what that is

Let’s Review that Moment Term

Note its just theForce * lever arm y

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)

Section Modulus is a Closely Related Term

That Inertia Term is a Measure of the Ability of a Beam to Resist

BendingThere arePrecalculatedTables of theseValues availableFor mostStructural steelshapes

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

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

Obviously We Need S

=

6 in

4 in

246

*4 62

S

For a rectangle

Going for the Answer

inlbM *000,28824*000,12

Assignment 13

Do Problems 6.3-6 and 6.3-10

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

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.

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!

Is there a Down Side?

4 in

6 in

S = 16

Maximum Moment = 12,000 * 16 = 192,000 in*lbs

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.

Assignment 14

Do problems 6.4-15 and 6.4-16

Controlling Cost With Beams

We might want to consider a less expensivematerial

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

The Practical Trick

Put steel reinforcingRebar near the tensileEdge of the beam

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

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

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

Conversion to an Equivalent Concrete area is Proportional to

Young’s Modulus

So an equivalent concrete area is

*

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

Set Up Our Quadratic Equation

Moment of ConcreteAbout neutral axis

Moment of ourEquivalent concrete(steel) section about axis

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

We Know the Forces Above and Below the Neutral Plain are Equal

and Opposite

Looks like we need the value of I

Get the Inertia of the Upper Compression Block around Neutral

Axis

Contribution toInertia of BeamFrom CompressionConcrete

Now the Inertia of Our Equivalent Concrete Area

Solve I for Our Equivalent Beam System

Now Let Get the Stress in Our Concrete Compression Area

(from our previous calculation)

Our given Moment Load

Going for the Stress on the Steel

Stress in equivalent concreteIs the same.

But we remember the ratio ofOur Young’s Modulus

* 8 =