Applications of Torsion Or You Will Learn This And Like It (Credit for many illustrations is given to McGraw Hill publishers and an array of internet search results)

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 inches We are allowed 8 ksi of shear stress We need to make the shaft as thin and light as Possible.

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. This We 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 of Our shaft thickness. We will try J for shafts of different standard thicknessesses Until 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 = lb*in We know c2 (the radius of our Shaft) is 1.75 inches We don’t know c1 (the radius of the Hollow center) Do a little algebra to isolate what it Is we do not know Plug and Chug

The Final Solution inches 1.69 inches inner radius inches outer radius That shaft is pretty thin. We may want to ask whether 3.5 inch diameter is Really the best choice

The Thin Walled Torsion Member If C2 and C1 are about the same size There will be very little shear difference Over the thickness. Simplifies to (pg. 4 of FE exam book)

Lets Illustrate How it Works If I put 24,000 lb*in or torque On this what be the stress in Each wall? For thin walled members its not How far from the center that controls Stress. There is a uniform flow of Shear that affects the entire Material surface 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 area Enclosed by the member And we get The 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 be Accomodated.

Thin Walled Shear Members Drive shafts usually aren’t thin walled Members - but Airplane wings are thin And have very high Shear loads.

Now Lets Try Doing a Statically Indeterminate Problem What are the reactions at the wall For A and C From Statics we know A+ C = 1.4 KN*m Like most problems that have Duplicate support points this Problem will be statically indeterminate (we cannot break down how much of That balancing reaction comes from A And 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 a Little data as a bonus

Apply the Principle But since the angle of twist has to be the same For 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 for T AB

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 to Solve for the angle of twist

Now Its’ Trivial We use our equation for T AB And our equation for T BC

We’re Done

Assignment 9 Do problems and Do problem 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 whether The answer shown is correct or not.