# NCEES Structural Problem 804



## STRENG62 (Mar 22, 2011)

I am working on p213 and trying to determine hold-down requirements. It seems they applied the 0.6 reduction to the dead loads and also applied the 0.6 factor to the resisting moment MR. Does this effectively "double dip" the dead load reduction for computing the hold-down force? Thank you in advance for your help.


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## mjbikes (Mar 26, 2011)

STRENG62 said:


> I am working on p213 and trying to determine hold-down requirements. It seems they applied the 0.6 reduction to the dead loads and also applied the 0.6 factor to the resisting moment MR. Does this effectively "double dip" the dead load reduction for computing the hold-down force? Thank you in advance for your help.


Looks like you're right. Their calc: 0.6 x 0.6 x w x L x L/2 has an extra 0.6.

This practice exam is much better than the previous one, but there are still quite a few errors...


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## mjbikes (Mar 26, 2011)

mjbikes said:


> STRENG62 said:
> 
> 
> > I am working on p213 and trying to determine hold-down requirements. It seems they applied the 0.6 reduction to the dead loads and also applied the 0.6 factor to the resisting moment MR. Does this effectively "double dip" the dead load reduction for computing the hold-down force? Thank you in advance for your help.
> ...


Also, the reactions for the wind loading on the beam over the opening are wrong.


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## lhpriest (Mar 27, 2011)

mjbikes said:


> mjbikes said:
> 
> 
> > STRENG62 said:
> ...




It also appears as though the solution has the points "C" and "D" backwards in the summing of moments. i.e. the moment summation for point D' is actually for point C', and the moment summation for point C is actually for point D. Is this correct or did I miss something?

Also, it seems that you would have to reverse the wind load direction to calculate the maximum holddown force required at points C' and C. Is this what should be done in an exam situation or would it suffice to simply make a statement about the load reversal and move on? Any opinions welcomed.


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## mjbikes (Mar 28, 2011)

lhpriest said:


> It also appears as though the solution has the points "C" and "D" backwards in the summing of moments. i.e. the moment summation for point D' is actually for point C', and the moment summation for point C is actually for point D. Is this correct or did I miss something?


I think they have it correct. I did take me a while to go thru their answer term by term to figure it out with their incorrect loads.



lhpriest said:


> Also, it seems that you would have to reverse the wind load direction to calculate the maximum holddown force required at points C' and C. Is this what should be done in an exam situation or would it suffice to simply make a statement about the load reversal and move on? Any opinions welcomed.


In calculating C and C' under my test simulation I reversed the wind loads to find the maximum uplift as you would in real life. Their solution states the wind vectors are right to left only, so the overturning moments were calculated without reversing the lateral loading. On the real exam, I guess it would benefit us to state our intentions in these situations.


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## lhpriest (Mar 28, 2011)

mjbikes said:


> lhpriest said:
> 
> 
> > It also appears as though the solution has the points "C" and "D" backwards in the summing of moments. i.e. the moment summation for point D' is actually for point C', and the moment summation for point C is actually for point D. Is this correct or did I miss something?
> ...


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## mjbikes (Mar 28, 2011)

lhpriest said:


> I must have missed something, the right side reactions of the vertical loads over the opening pass through points C and C', so there would be no moment from these loads about points C and C'. The only moments about C and C' would come from the horizontal load and the vertical loads on the floor/roof. Is that incorrect?
> For the reaction of the loads on each side of the opening, did you use a simply supported member and solve for the statics reaction at each side rather than use the tributary width (like the solution)?
> 
> Thanks for your help!


The moments for overturning and resisting are about the opposite wall end. Therefore, for reactions at D' and D the force from the beam spanning the opening has an arm of zero. For reactions at C' and C the force from the beam spanning the opening has an arm of 10'. Their solution for D &amp; D' does not have the middle beam load in the calc, the C &amp; C' calcs do. So what you're saying is correct.

Technically, when finding the uplift/compression they should be dividing Mo-Mr by ~9.4 for tension holdowns shown in their elevation sketch.

The beam is a simply supported span. Proper statics analysis results in 2050lbs @C and 1750lbs @B for wind loads.


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## lhpriest (Mar 28, 2011)

mjbikes said:


> lhpriest said:
> 
> 
> > I must have missed something, the right side reactions of the vertical loads over the opening pass through points C and C', so there would be no moment from these loads about points C and C'. The only moments about C and C' would come from the horizontal load and the vertical loads on the floor/roof. Is that incorrect?
> ...


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