Who's right? Hiner or Mansour? (Seismic)

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ptatohed

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In Hiner's SDR Workbook, he states that Total Shear = Direct Shear + Torsional Shear (Chapt 8)

In Mansour's SD book, he states that Total Shear = Direct Shear +/- Torsional Shear (Chapt 9)

Which is correct?

In particular, I was working Mansour's sample problem 9.5 tonight. I calc'ed 60.0 kips for Direct Shear and 0.85 kips for Torsional Shear. Per Hiner, I'd add the two to get 60.85 kips. That's what I did but I was wrong. Per Mansour's solution, the answer was 60.0 - 0.85 = 59.15 kips.

Why are the formulas different and how would I have known to subtract the Torsional Shear and not add?

Thanks.

 
In Hiner's SDR Workbook, he states that Total Shear = Direct Shear + Torsional Shear (Chapt 8)
In Mansour's SD book, he states that Total Shear = Direct Shear +/- Torsional Shear (Chapt 9)

Which is correct?

In particular, I was working Mansour's sample problem 9.5 tonight. I calc'ed 60.0 kips for Direct Shear and 0.85 kips for Torsional Shear. Per Hiner, I'd add the two to get 60.85 kips. That's what I did but I was wrong. Per Mansour's solution, the answer was 60.0 - 0.85 = 59.15 kips.

Why are the formulas different and how would I have known to subtract the Torsional Shear and not add?

Thanks.
When the effective resultant of the lateral load is to the right of the center of rigidity, the torsional shear stress on the right wall is to be added to the direct shear and on the left wall, it is to be subtracted. Since the lateral load can be considered reversible, the situation can reverse. Therefore, the proper DESIGN shear load for either wall should be considered 'Direct Shear + Torsional Shear'

 
In Hiner's SDR Workbook, he states that Total Shear = Direct Shear + Torsional Shear (Chapt 8)
In Mansour's SD book, he states that Total Shear = Direct Shear +/- Torsional Shear (Chapt 9)

Which is correct?

In particular, I was working Mansour's sample problem 9.5 tonight. I calc'ed 60.0 kips for Direct Shear and 0.85 kips for Torsional Shear. Per Hiner, I'd add the two to get 60.85 kips. That's what I did but I was wrong. Per Mansour's solution, the answer was 60.0 - 0.85 = 59.15 kips.

Why are the formulas different and how would I have known to subtract the Torsional Shear and not add?

Thanks.
In the SDR Workbook, Direct shear is always assumed to be "positive" while the Torsional Shear can be a "positive" value (i.e., in the direction of the Direct Shear) or a "negative" value (i.e., in the opposite direction of the Direct Shear) ... so sometimes the Torsional shear adds (+) and sometimes it subtracts (-).

So you might want to think of it this way: Total Shear = Direct Shear + (+/- Torsional Shear)

I'm not familiar with Mansour's sample problem 9.5, but I suspect it would be solved using the SDR Workbook approach as 60.0 kips + (- 0.85 kips) = 59.15 kips

Ultimately you need to know when to add and when to subtract. Hopefully that helps.

 
ptatohed,

Two opposite answers, and I am still looking for a definitive solution... have you found one? Or which do you believe, and why?

After Googling, I found yet another answer, which says just ignore it;

"Negative Torsional Shear – The base shear causes a shear stress that acts in

the same direction in all vertical base members. The torsional shear stresses,

however, have different signs on either side of the center of rigidity. On one side

the torsion increases the stress from the base shear; on the other side the stress

is decreased. The amount of decrease is known as negative torsional shear.

Negative torsional shear should be neglected; that is, it should not be considered

to decrease the design capacity of a lateral resisting element."

/>http://www.prepareinc.us/PREPARE%20Advanced%20Torsion%20Calcs.pdf

Then there's this page, with discussion going both ways:
/>http://www.euken.com/group/seaint/mailarchive/2000c/msg00345.html

And this one says neglect it:
/>http://books.google.com/books?id=uvUvkANZ8QwC&pg=PA575&lpg=PA575&dq=negative+torsional+shear&source=bl&ots=vpRmhBJdmh&sig=ma-j2fq3IHE_Bez3Qu4lcguIvZI&hl=en&ei=QXSlToquOuWwiQLF1ZH7BQ&sa=X&oi=book_result&ct=result&resnum=7&ved=0CEsQ6AEwBg#v=onepage&q=negative%20torsional%20shear&f=false

So, yeah, as if this test weren't difficult enough, it seems the best we can do is hope the possible answers on the exam will point us in the right direction, and hope some day the pain will be over.

Thanks

 
Last edited by a moderator:
ptatohed,

Two opposite answers, and I am still looking for a definitive solution... have you found one? Or which do you believe, and why?

After Googling, I found yet another answer, which says just ignore it;

"Negative Torsional Shear – The base shear causes a shear stress that acts in

the same direction in all vertical base members. The torsional shear stresses,

however, have different signs on either side of the center of rigidity. On one side

the torsion increases the stress from the base shear; on the other side the stress

is decreased. The amount of decrease is known as negative torsional shear.

Negative torsional shear should be neglected; that is, it should not be considered

to decrease the design capacity of a lateral resisting element."

http://www.preparein...ion%20Calcs.pdf

Then there's this page, with discussion going both ways:

http://www.euken.com...c/msg00345.html

And this one says neglect it:

http://books.google....20shear&f=false

So, yeah, as if this test weren't difficult enough, it seems the best we can do is hope the possible answers on the exam will point us in the right direction, and hope some day the pain will be over.

Thanks
Civil and CA - thanks for the answers

Bole, good research. I have been struggling with this still too. In my opinion the formula will always be F = Direct Shear + Torsional Shear. With that said, Torsional Shear (VT) can be negative. But, I believe the only thing that makes it negative is when your d (distance from CR to wall being analyzed is negative). This would be the south wall for V in the W-E direction and the west wall for V in the S-N direction. The tricky part is to know when to use MT1 or MT2 (which either adds or subtracts the Accidental Eccentricity). Typically you use MT1 to calc F in the north wall and MT2 for the south wall (V in W-E direction); and MT1 for the west wall and MT2 for the east wall (V in the S-N direction).

I think there is more to it (like if the direction of the torsional moment is CW vs. CCW) but, the test is tomorrow and my brain is fried and I have looked at this as much as I am going to - lol. Good luck everyone!

 
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