Civil PE Sample Exam by Lindeburg Question 138

Professional Engineer & PE Exam Forum

Help Support Professional Engineer & PE Exam Forum:

This site may earn a commission from merchant affiliate links, including eBay, Amazon, and others.

AlexPE

Member
Joined
Feb 27, 2016
Messages
20
Reaction score
6
For the solution for question 138 in the Lindeburg PE sample exam: why is the equation for the stiffness 12EI/L^3 for each column?

The problem asks you to idealize the structure as a cantilever, so shouldn't the stiffness be 3EI/L^3 for each column (like a cantilevered beam)?

They account for the fact that there are 4 columns later on in the solution, so I must be missing something with the original stiffness equation. There's nothing in the errata for this problem. Could someone help me out?

Thanks a bunch!

 
Hi Alex,

I think 12EI♢/L^3 is the shear force on the column. It is derived from the slope deflection equation Mab = 2EI/L (2○a + ○b -3♢/L) + FEMab. Due to side-sway the moment equation simplifies to Mab = -6EI♢/L^2 because ○a =○b = 0 as well as FEMab. If you draw a FBD and calculate the end shears using the moments you should get 12EI♢/L^3. Hope this helps!

 
Thanks for your help choyos! much appreciated!

Another question for the PPI sample exam by lindeburg.... The solution to problem 154 sets the weld strength equal to shear rupture in the plate. Doesn't shear yielding need to be checked for this as well? I found that the plate actually needs to be 1/2" thick to accommodate shear yielding, which is larger than 0.4" (the "solution" to the problem). Perhaps there is a reason why shear yielding does not need to be checked in this case?

Thanks again for your help!

 
Another question for the PPI sample exam by lindeburg.... The solution to problem 154 sets the weld strength equal to shear rupture in the plate. Doesn't shear yielding need to be checked for this as well? I found that the plate actually needs to be 1/2" thick to accommodate shear yielding, which is larger than 0.4" (the "solution" to the problem). Perhaps there is a reason why shear yielding does not need to be checked in this case?
0.4" is larger than 1/2", so it seems like it would not control. In the "Practice Problems for the CE PE Exam" from Lindeburg I found several questions like that where they didn't show checking all potential failures for the type of question they ask. The trick on the PE exam is to read the question carefully and only check what it asks you to. Otherwise you will be burning time for no reason. Wording is very important, pay close attention to "most nearly" versus " lightest/smallest acceptable member" and choose the answer accordingly.

Now when you take the SE Exam, you had better show ALL your work on the essay questions if you expect to get full credit on the afternoon sessions.

 
^^^^ that guy is an idiot! 0.4" is not larger than 1/2". Where are the moderators? He should have read the question more carefully before making a dumb statement.

Ha ha. Oh well there goes my credibility...

 
I haven't actually read the sample exam question, but it sounds like you might be mixing failure modes.. "shear rupture" (AISC J.4) is a weld failure mode whereas "shear yielding" (AISC G.2) is a member failure mode. 

 
Back
Top