Applications of Beams and Bending

(Credit for many illustrations is given to McGraw Hill publishers and an array ofinternet search results)

Parallel Reading

Section 6.4 Design of Beams for StrengthSection 6.5 Flexural Stress in nonhomogeneous beams(Do Reading Assignment Problem Set 6B)

Chapter 6

Lets Try Using Our Radius of Curvature Formula

Old McDonald had a beam

The beam had a Young’s ModulusOf 165 GPa

If was acted on by a 3 KN*mBending couple

What are the maximum tensile and compressive stresses and radius of curvature?

Our Key Formulas

Obviously we are going nowhere with thisUntil we get the Inertia of our beam

Getting I

We previously got our centroid using a weighted average of area but nowwe are after the second moment - I

Getting the inertia of each of the componentrectangles about their own axis would notbe a trick. There is a formula for that.

20 mm90 mm

40 mm

30 mm Of course we want the moment about theCentroid axis

Fortunately We Have the Parallel Axis Theorem

000,6012

*90 203

200,259*1800 122

The moment of inertia of an element aboutA different axis is the moment of inertia ofThe element around its own axis

90

20

+

180012

Plus the area of the element time theDistance between the axis squared.

000,6012

*90 203

So Work Through and Get the Inertia

1800

90

2012

From component #1

1200

30

4018

From component #2

1800

1200

Stresses Go Down Easy Now

Our basic formula

Compression side on bottom σ max = -131.3 MPa (Negative is for compression)

Tension side on top σ max = 76 MPa (Positive is for tension)

Now for the Radius of Curvature

Lets Try Bending a Beam

How much of a bending moment can I puton this puppy before the stress goes outof tolerance?

Our basic formula

1.8

(Interesting note Wood is usually weaker incompression than tension)

Getting a Hold of I

86412

*6 123

Our Solution Formula

We know the critical stress is going to be tension(Compression is the high strength measure foralmost all materials)

We know the neutral plain passesThrough the centroid

6 in So the distance y from the centroid toThe point of maximum compression mustbe 6 inches

Set Up Our Equation

lbsinMy

I*000,288

6

864*2000*

Lets Do a Bit of Reinforcing

We’ll put a reinforcing plate on the bottom ofthe beam and see how much it can take now.

1.8

For Composite Materials Life Get Tougher

It is no longer true that theNeutral axis or plane passesthrough the centroid.

Finding the Neutral Axis

Start by getting the ratio of the Young’s Modulus of theTwo materials(Steel will be much more resistant to strain than wood)

Now Weight Each Component Area According to Young’s Modulus

Remember steel hasa Young’s Modulus of16 times that of wood.

Now Use That Weighted Area to Apply a Weight to the Centroid of

Each Component Area

6 in

0.25 in

Now Divide Through By Weighted Area to Find the Neutral Plane

3.758 in

CompressionArea

TensionSide

Avoid Confusion

What we just found was the neutral axis –The axis of the first moment.

This is NOT the second moment or momentof inertia. Do not confuse the calculations!

Now We Will Take the Inertia Moment of Each Component

Around the Neutral Axis

Wood area (I1) aroundNeutral Axis

Steel Area (I2 ) aroundNeutral Axis

Obviously I’m using theParallel Axis Theorem

Now We Will Use Our Young’s Modulus Weighted Stress Formula

Lets see what kind of stress our wood compressionArea will take.

8.242 in

lbsinMyn

I*2.688,454

242.8*1

77.1873*2000

*

*

So Lets See What the Tension Side Will Take

9.600911258.4*111.16

77.1873*000,22M

4.258

Comparison

• Will Our New Beam be limited by compression or tension– Wood Compression limit 454,688 in*lbs– Tension Limit for Steel 600,912 in*lbs– Obviously the wood at the top will crush out first –

limit is 454,688 in*lbs

• But without the plate the limit was 288,000 in*lbs

A little steel goes a long ways

Now Lets Try a Few FE Style Problems

We know we get this answer by taking the area weighted average of theCentroids of each component area.

6 cm2

10 cm2

5.5 cm

Of course picking off centroidsand areas of rectangles iseasy prey.

2.5 cm

Take That Area Weighted Average

3.625 cm Pick C

Take That Area Weighted Average

3.625 cm Pick C

Lets Do One With a Composite Beam

We will weight the two componentAreas by their Young Modulus.Then we will weight the centroidsof the two component areas bytheir weighted areas.

Lets Get the Weighted Area

Area = 4 cm2

2 cm

2 cm

0.125 cm

Steel Area Equivalent = 2*0.125*20 = 5 cm2

Identify the Centroids of the Component Areas With Respect to

the Beam Bottom

0.125

2 cm

1.125 cm

0.0625 cm

Now weight them by the weighted areas

1.125* 4 + 0.0625* 5 = 4.8125

Now divide through by the totalArea

4.8125/(4 + 5) = 0.535 cm

Pick B

Assignment 15

Do problems 6.5-8 and 6.5-14(note that these are the same problem, but you are being ask to solve it byTwo different methods).

For each step in your work show the formulas and explain the logic andApproach you are using to work toward a solution. Circle your answer atThe end. Warning – explaining your work is a key part of the answer.Failure to include step by step explanations and your work will get youMarked wrong – even if the answer is correct.