# NCEES Sample Exam

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### #1 mjbikes

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Posted 04 April 2011 - 06:57 PM

The retaining wall is loaded in axial compression and flexure. For the ASD analysis of the wall in part a, shouldn't we use the more complex calculations for 'k' and the moment capacities that takes into accoount the axial loading? In this case the way they used (ignoring the axial load) was ok since the loads were not close to capacity. If the flexure compression stress was closer to the masonry capacity, then the addition of an axial load may make it inadequate, but the method they used would not show that.

### #2 lhpriest

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Posted 04 April 2011 - 07:22 PM

For combined stresses, MSJC requires that fa (axial compression stress) + fb (flexural compression stress) be less than or equal to Fb= 1/3 f'm. It also requires that fa be less than Fa.

What equations for k are you referring to?

### #3 hansel

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Posted 04 April 2011 - 09:30 PM

QUOTE (mjbikes @ Apr 4 2011, 04:57 PM) <{POST_SNAPBACK}>
The retaining wall is loaded in axial compression and flexure. For the ASD analysis of the wall in part a, shouldn't we use the more complex calculations for 'k' and the moment capacities that takes into accoount the axial loading? In this case the way they used (ignoring the axial load) was ok since the loads were not close to capacity. If the flexure compression stress was closer to the masonry capacity, then the addition of an axial load may make it inadequate, but the method they used would not show that.

Yes you are right. They bypassed the interaction diagram. This is because the actual bending (applied) is less than the pure flexural (M, P=0) capacity of the wall. Unless you have a pretty good idea about when to consider the axial load small, I will always calculate the three points of the interaction diagram: (P, M=0); (P_bal, M_bal); (M, P=0) .

### #4 mjbikes

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Posted 05 April 2011 - 06:28 AM

Their solution is valid if there is no tension in the section (i.e. uncracked). But since the bending stress (292psi) is greater than the compression stress (22psi) there is tension in the masonry and the superposition, fa + fb < fm, no longer applies; the section is cracked. According to SERM, to determine the stresses on the cracked section, the strain distribution over the section is estimated, and forces are determined and compared iteratively until balanced.

The 2006 Design of Reinforced Masonry Structures derives a direct solution that is a bit involved and is similar to the solution NCEES provided for this problem. A reinforcing tensile stress and a masonry compression stress, M/bd^2, are found compared and then solved for M. That moment capacity is then compared to the applied moment. It is very tedious...

### #5 lhpriest

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Posted 05 April 2011 - 01:38 PM

Another thing that appears to be missing from the solution is the consideration of accidental eccentricity. (0.1 x the wall dimension) I understand that this is in the Column section of the masonry code, but wouldn't it also be applicable to a wall element with axial load?

### #6 Andrewstructure

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Posted 13 April 2011 - 08:49 PM

A little off topic...

Is the new NCEES Structural Sample Questions & Solutions (NCPSE5) any different from the old one?

http://ppi2pass.com/...U...0&pr=NCPSE5

Does it have all the horrible erratta errors the last one did?

### #7 mjbikes

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Posted 13 April 2011 - 10:08 PM

QUOTE (Andrewstructure @ Apr 13 2011, 01:49 PM) <{POST_SNAPBACK}>
A little off topic...

Is the new NCEES Structural Sample Questions & Solutions (NCPSE5) any different from the old one?

http://ppi2pass.com/...U...0&pr=NCPSE5

Does it have all the horrible erratta errors the last one did?

The new book is a complete exam, instead of half of an exam. It mirrors the change in the exam format with 40 multiple choice morning sections and written afternoon sections. There are still errors, but fewer and (somewhat) less blatant. It is worth while to purchase.

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