# Shaft Shear Stress with strange shape

1. Oct 31, 2013

### Woopydalan

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
Hello,

I am having difficulty solving for the Area and wondering if this is correct.

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2. Oct 31, 2013

### Staff: Mentor

Are you applying some rough-and-ready "near enough is good enough" approximation involving the centreline of the cross-section?

Why don't you work it out more exactly? It doesn't seem all that difficult. (I'm finding trying to follow your rough approximation procedure to be more difficult.)

Each semicircular portion is the area of one circle minus the area of the smaller one, then halved. The rest is a handful of rectangular shapes.

mathematicians can point out a tiny error, but you are in an engineering subforum and here it won't be noticed http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon6.gif [Broken]

Last edited by a moderator: May 6, 2017
3. Oct 31, 2013

### SteamKing

Staff Emeritus
The area enclosed by the midline can be calculated by taking the average of the outer and inner cross sectional areas.

4. Oct 31, 2013

### Woopydalan

Okay, how about now? The radius of the circles is the same for both parts, since you move over 0.2 inches, but the thickness of that part is 0.2 inches, so they negate each other

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5. Nov 1, 2013

### SteamKing

Staff Emeritus
Looks good.

6. Nov 1, 2013

### Staff: Mentor

No it isn't. Nor are they both semicircles.

But if you are just doing this roughly it may be okay to make that rough approximation. (This being engineering, not physics.)

7. Nov 1, 2013

### SteamKing

Staff Emeritus
Even physics doesn't operate with infinite precision.

As a check, I drafted the cross section in CAD and calculated the area to the outer edge and the area to the inner edge.

The CAD program got: Aouter = 22.9314 in^2; Ainner = 15.2777 in^2

Amidline = (22.9314+15.2777)/2 = 19.1046 in^2

Hand calculations: Aouter = 22.931 in^2; Ainner = 15.931 in^2; Amidline = 19.431 in^2

The problem is good to only 2 sig. figs. (The max. all. stress = 12 ksi)

%error in Amidline = (19.431 - 19.105)/19.105 = +1.7%

which is pretty good for a 5 minute calculation.

8. Nov 1, 2013

### Staff: Mentor

Unless you also work it out accurately, you can't know how good your rough method will approximate it. That's why I suggested that it be done just the once and accurately. (I won't say exactly, because I can see it would be difficult to work it out exactly.)

It is too easy for a rough approximation method to conceal small mistakes in principle.

9. Nov 1, 2013

### SteamKing

Staff Emeritus
I don't know what your standards are for 'rough approximation', but I think I have demonstrated the accuracy of the calculation.

This is only an approximate method for determining shear stress due to torsion anyway. There are several assumptions made to obtain the Bredt formula which are probably worth more than 1% or 2% of the stress values obtained. The big sign in the problem statement says 'neglecting the effect of stress concentrations'.

10. Nov 1, 2013

### nvn

Agreed. It is relatively easy to quickly compute almost the exact area enclosed by the midline. E.g., 5.5*4.8 - 2(0.5*pi*1.6^2) + 2[0.1(2*1.6)] = 18.9975. Since the exact area is 18.9971, the error in this quick calculation is +0.0021 %. Close enough. There is no need to start out with a +2.3 % error, just to compute this area.

Agreed.

Last edited: Nov 1, 2013
11. Nov 1, 2013

### Woopydalan

Who knew I could stir up so much controversy with this problem...does that mean my attempt #2 is wrong?

12. Nov 2, 2013

### Staff: Mentor

Not completely wrong, no. But basing inner area calculations on a radius of 1.5" when the radius is actually 1.7" is, for starters, going to put some marks in jeopardy.

unless your professor has given a clever justification for doing so, but which you have not alluded to.

13. Nov 2, 2013

### nvn

Woopydalan: There is a chance your answer could be marked wrong, because its accuracy is less than +/-0.2 %.

First, in most of these mechanics text books, when a given question provides values such as 15 N, then unless otherwise stated, they expect you to assume the given values are accurate. E.g., 15 N should be read as 15.00 N. Similarly, 1 mm should be read as 1.000 mm, not 1 mm +/-0.5 mm.

Secondly, most of the mechanics text books are expecting students to provide final answers accurate to +/-0.2 %. The way to do this is explained in item 1 in post 4551418.

Your attempt 2 is accurate to 2.1 %, which is 10.5 times less accuracy than is typically required to match the answer in the back of the book, or to match the answer in the lecturer's solution manual.

Last edited: Nov 2, 2013