Applications of Torsion

Or You Will Learn ThisAnd Like It

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Parallel Reading

Chapter 4 Section 4.6 Section 4.8 (Do Reading Assignment Problems 4B)

Lets Design a Drive Shaft

Our Drive Shaft must transmit 240 HP at 1800 rpm.It is to have a diameter of 3.5 inchesWe are allowed 8 ksi of shear stressWe need to make the shaft as thin and light asPossible.

Strategy

We are going to go for a hollow shaft to save weight

We need to get the Moment or Torque to which our shaft will be subjected. ThisWe will do with the formula

With the Torque in hand we will use the formula

We know the maximum shear, the torque, and the shaft radius c. J is a function ofOur shaft thickness. We will try J for shafts of different standard thicknessessesUntil we find one that works.

We Need Some Conversions

Now Lets Solve For Torque

Now Lets Set Up to Find a Solution

We know T = 8304.4 lb*inWe know c2 (the radius of ourShaft) is 1.75 inchesWe don’t know c1 (the radius of theHollow center)

Do a little algebra to isolate what itIs we do not know

8000*

75.1*4.8304*221.8 75.1

4

Plug and Chug

The Final Solution

6927.121.84 inches

1.69 inches inner radius - 1.75 inches outer radiusThat shaft is pretty thin. We may want to ask whether 3.5 inch diameter isReally the best choice

The Thin Walled Torsion Member

If C2 and C1 are about the same sizeThere will be very little shear differenceOver the thickness.

Simplifies to

(pg. 4 of FE exam book)

Lets Illustrate How it Works

If I put 24,000 lb*in or torqueOn this what be the stress inEach wall?

For thin walled members its notHow far from the center that controlsStress. There is a uniform flow ofShear that affects the entireMaterial surface

tconstq tan* Its shear flow

So How Do I Get This Shear Flow that Must Move Through the Skin of the Member?

Where A is the total areaEnclosed by the member

And we getThe Shear Flow

Now Lets See How Much Shear that Puts in the Skin

Lets Try a Twist Where the Skin Thickness Changes

The same shear flow must beAccomodated.

Thin Walled Shear Members

Drive shafts usually aren’t thin walledMembers - but

Airplane wings are thinAnd have very highShear loads.

Now Lets Try Doing a Statically Indeterminate Problem

What are the reactions at the wallFor A and C

From Statics we knowA+ C = 1.4 KN*m

Like most problems that haveDuplicate support points thisProblem will be statically indeterminate(we cannot break down how much ofThat balancing reaction comes from AAnd how much from C)

Enter Strength of Materials

• We know that the angle of twist has to be the same for shaft AB and shaft BC

We’ll through in aLittle data as abonus

Apply the Principle

But since the angle of twist has to be the sameFor shaft BC we also know

And from Statics we know

Lets Start Filling In Some Numbers

Now We’ll Crunch J

Now We’ll Plug In to Find the Reaction at the Wall A

Rearrange the equation to solve forTAB

More of the Same Gets Us an Expression for the Torque in Shaft

BC

Now We’ll Take Our Statics Equation

Now if that does not look like an invitation toSolve for the angle of twist

Now Its’ Trivial

We use our equation for TAB

And our equation for TBC

We’re Done

Assignment 9

Do problems 4.8-8 and 4.8-10Do problem 4.6-2

Warning – you must show your work and explain step by step what you are doing.Simply showing work and an answer will be marked wrong regardless of whetherThe answer shown is correct or not.