# NCEES Sample Questions #523 Machine Design



## M.E. Nebraska (Apr 5, 2008)

Can anyone explain to me how they arrived to the solution on NCEES #523 on the machine design portion. You need the picture to work the problem, that is why I haven't included the problem statement. I don't understand how they got the secondary shear force calculations.


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## M.E. Nebraska (Apr 5, 2008)

Another question on NCEES #537.

ASME Code for Transmission Shafting Equation. For rotating shaft that has a gradually applied load, the values of Cm and Ct, are?

Answer is 1.5 and 1.0.

NCEES Sample Soloution says that this cand be found on page 691 of Shigley and Mischke 4th Edition. I have the 6th edition and I can't find this. Has anybody else found this?


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## Matt-NM (Apr 5, 2008)

ME Nebraska

NCEES #523 MD

I arrived at the solution a little different than they did. I used MERM (pages 51-16 to 51-17) as the template. Here is what I did...

1. Calculate the torsional shear stress on each fastener using Tau_t = Fer/J where F = 500 lbf, e = 19 in, r = 6 in, and J = approx. 10.22 (J is calculated using the parallel axis theorem for the two offset fasteners and then the basic pie*r^4/2 for the fastener located at the centroid. The fastener at the centroid hardly contributes to the total). (See example 51.6 in MERM).

-This comes out to 5,577 psi, which is per fastener. (This is directed completely vertically. See Figure 51.13 in MERM for an explanation on the direction)

2. Calculate the "direct vertical downward shear" using Tau_v = F/nA where F = 500 lbf , n = 3 , and A = .1419 in^2 (tensile stress area for a 1/2-13 UNC fastener - this can be found in table 51.5 in MERM)

- This comes out to 1,174 psi, which is also per fastener.

3. Sum the two downward shear stresses, 5,577 psi + 1,174 psi = 6,751 psi. Multiply 6,751 psi * .1419 in^2. This equates to 958 lb, thus option (D) 960 is correct.

NCEES # 537 MD - I missed this problem also. If I cannot find anything in MERM pertaining to it, I am not going to worry about it, as it seems like a special case that would probably not appear on the exam (hope i'm right about this!)

Hope this helps some.


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## mackintosh (Apr 5, 2008)

M.E. Nebraska said:


> Another question on NCEES #537.
> ASME Code for Transmission Shafting Equation. For rotating shaft that has a gradually applied load, the values of Cm and Ct, are?
> 
> Answer is 1.5 and 1.0.
> ...


This appeared on the online sample Machine Design PM Module as well. I've got the new Shigley's 8th Edition, couldn't find it in there. That was the only one I totally had to guess on. The only code questions I figure I'm likely to get are the Pressure Vessel stuff in the MERM.


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## nickwusz (Apr 7, 2008)

mackintosh said:


> This appeared on the online sample Machine Design PM Module as well. I've got the new Shigley's 8th Edition, couldn't find it in there. That was the only one I totally had to guess on. The only code questions I figure I'm likely to get are the Pressure Vessel stuff in the MERM.


Re: NCEES MD Question #537

I have Shigley's 5th sedition and this equation is on p.704, equation 18-12. (Section 18-4, Static Loading-Bending and Torsion)


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## mackintosh (Apr 7, 2008)

nickwusz said:


> Re: NCEES MD Question #537I have Shigley's 5th sedition and this equation is on p.704, equation 18-12. (Section 18-4, Static Loading-Bending and Torsion)


I pulled out my old Shigley 4th Edition, and note that it says: "This code has been obsolete for many years now..." I figure that's why it's been dropped from the newer editions, and is probably not a likely exam question.


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## GT ME (Jul 25, 2008)

The solution from MD #523 afternoon is in disagreement with MERM.

NCEES has 1 mistake in the solution (and I contacted them on the mistake) -- as mechanical engineers, we should understand (more appropriately, visualize) that the force rotates clockwise around the centroid (in this problem).

This means that Fa should rotate clockwise &amp; Fc should rotate clockwise, and Fb = 0 since it's the centroid. The direct downward shear force is 167.

Fc should rotate clockwise! the solution states CCW, which is wrong!!! We must be able to visualize this as MEs.

The solution neglects the polar moment of inertia J in the solution unlike the MERM solution in chapter 51 -- this is probably to simplify the solution (I hate that assumption).

hope this helps.



M.E. Nebraska said:


> Can anyone explain to me how they arrived to the solution on NCEES #523 on the machine design portion. You need the picture to work the problem, that is why I haven't included the problem statement. I don't understand how they got the secondary shear force calculations.


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## GT ME (Jul 25, 2008)

OMG -- one of the few I skipped -- never could find the damn equation...lmao.



M.E. Nebraska said:


> Another question on NCEES #537.
> ASME Code for Transmission Shafting Equation. For rotating shaft that has a gradually applied load, the values of Cm and Ct, are?
> 
> Answer is 1.5 and 1.0.
> ...


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