CA Structural II state exam sample problems

[klip]

I downloaded the sample problems for CA structural II state exam. However, the board did not provide answers for these problems. Does anyone by any chance have answers so that we can share?

Especially for the problem D1 concrete problem, how to obtain the strength level drag force at each end of the fourth floor beam?

Thanks a lot.

 

[mjbikes]

I'd also like to know if I did the problems the way the board wants me to. If there is a copy of the solutions around, I'm interested in getting it.

 

[klip]

mjbikes,

Can we compare our answers? The problems are kind of confusing and I am really not sure what I did is what the board wants. I am in the process to finish the concrete problem currently.

 

[BLMedcalf]

mjbikes,
Can we compare our answers? The problems are kind of confusing and I am really not sure what I did is what the board wants. I am in the process to finish the concrete problem currently.

For the first problems, I got:

1. D 2. B 3. B or C (depends on load?)

4. B 5. C (I actually got 5', and C is the closest answer. Not sure if it is right)

Haven't worked the rest, but will compare once I do.

 

[klip]

mjbikes,
Can we compare our answers? The problems are kind of confusing and I am really not sure what I did is what the board wants. I am in the process to finish the concrete problem currently.

For the first problems, I got:

1. D 2. B 3. B or C (depends on load?)

4. B 5. C (I actually got 5', and C is the closest answer. Not sure if it is right)

Haven't worked the rest, but will compare once I do.

The answer for 3, in my opinion, should be B A325-SC because A490 high strength bolts are vulnerable under fatigue loads. I agree with you for 1, 2, and 4.

Honestly, I have no much idea about question 5. How did you figure this out?

I think the essay problem for steel is very straight forward. For concrete sample,

1. will the drag force at left end be (294/2)*(630/(630+350))=94.5 kip and the right end be (294/2)*(360/(630+350))=52.5kip? This is the first time I conduct such type of analysis. Please correct me if I am wrong.

2. shear friction problem?

3. We got moment so that we can degisn this beam by checking its moment capacity. What about the drag force of 42 kips? Do we need to design this as a beam-column? Any thought?

4. development length per chapter 12. Do you aware of any special requirement about development length in high seismic design category in Chapter 21?

5. Check horizontal wall bar requirement per ACI 318-05 Section 21.7? What is your thoughts about vertical bars? Did you calculate?

6. I just treat this a simple beam, but with d=0.8 * wall length.

7. Any thoughts about the location of splice in the horizontal bars?

Please let me know your thoughts. Thanks a lot!

 

[kevo_55]

#3 should be C. You can't galvanize A490 bolts. Also, for fatigue loads you should use SC bolts.

#5 just wants for you to take a look at EQ 18-2 of the CBC. I got my d=4.47'. Therefore, I got C.

As for the concrete problem, #1 is unfair because drag forces come from diaphragm loads. Since we were not given a Sds, we can't check Fp min or max. I simply used the F at the 4th floor. (392k + 294k)

The front wall would see 441 kips and the rear wall would see 245 kips. (simple ratio, the wall stiffness was given to be the same as what was shown for the total base shear)

At end #2, I got a drag force of 212 kips, and at end #3 I got 16 kips.

Part 2 is a shear friction problem. I got #5 bars at 18" on center.

Part 3, I was unsure as well. The drag force would be either tension or compression. I don't have column interaction diagrams with f'c=3 ksi. I simply assumed that the diaphragm was giving bracing and designed for the moment, and then designed for the short column. I just added both of the steel areas together.

Part 4, nothing special. Just Ld=62". This is a drag beam, not the wall itself.

Part 5, check via 21.7.2. You'll find that minimum reinforcement is required as well as two curtains. I used two curtains of #4 at 10" on center.

Part 6, I checked as a beam as well.

Part 7, I selected for the horizontal bars to be in the center 1/2 of the wall. (Away from any yielding??)

 

[MOOK]

#3

I thought that since the section dimensions and weld are given in the problem, I got the shear capacity of the section and I designed the bolts based on this force.

Steel Problem:

- Did you use the bracing to support any of vertical load (dead + live) ?

- To design the beam, the axial force due to seismic PQE = Pt + Pc or PQE = (Pt + Pc)/2. I myself did not divide by 2 because the seismic forces given on one side of the truss.

Concrete problem:

1 - Even thought this part of the problem does not make sense to me, I do not see any need for SDS. I treated it as diaphragm.

2- Dowel bars

What force did you use to design the dowel bars?? and what is the area Acv you used for shear friction?

 

[mjbikes]

MC#3 I believe sould be A. A490 cannot be galvanized. From the commentary in the RCSC bottom of page 16.2-19 it states there is a lower mean slip coeff for HDG faying surfaces.

I agree with 1,2, & 4, and needed to follow kevo's guidance on using eq 18-2 for #5 and got C. (thanks)

I was paying attention to the time it took me to do the B2 steel problem. It is straight forward, but if they all take that long (2 hours) then I won't be finishing. How have you all been fairing on time?

 

[mjbikes]

Steel Problem:- Did you use the bracing to support any of vertical load (dead + live) ?

- To design the beam, the axial force due to seismic PQE = Pt + Pc or PQE = (Pt + Pc)/2. I myself did not divide by 2 because the seismic forces given on one side of the truss.

I used statics to determine the Pd & Pl for the brace calcs. The beam should be designed as simply supported w/o braces for the Md & Ml & Mqb. I guess that is to ensure it doesn't fail due to gravity loads if the braces fail due to a lateral load event.

I divided by 2 for PQE because the girder is a collector (load is "applied" along its length). Also Pt & Pc are at midspan like ex 6.5(?) in 341.

 

[MOOK]

Steel Problem:- Did you use the bracing to support any of vertical load (dead + live) ?

- To design the beam, the axial force due to seismic PQE = Pt + Pc or PQE = (Pt + Pc)/2. I myself did not divide by 2 because the seismic forces given on one side of the truss.

I used statics to determine the Pd & Pl for the brace calcs. The beam should be designed as simply supported w/o braces for the Md & Ml & Mqb. I guess that is to ensure it doesn't fail due to gravity loads if the braces fail due to a lateral load event.

I divided by 2 for PQE because the girder is a collector (load is "applied" along its length). Also Pt & Pc are at midspan like ex 6.5(?) in 341.

When I designed # 3 using the shear capacity, I got answer "A" but it does not make sense to me because we should use SC connections for fatigue loading.

Steel Problem

How did you deal with the concentrated load on the 1/4 and 3/4 length of the beam of the beam? how did you distribute this load to the bracing using statics?? did you assume half of it goes to the mid bracing and the other half goes to the column?

The problem in AISC 341 is different from this one, this problem the seismic loads are on ONE side of the truss while the one in AISC are distributed equally on both sides of the truss, so I think we should not divide by 2.

 

[klip]

For the steel brace, I would calculate the brace axial due to dead and live by applying half of the gravity loads to the brace-beam intersection point.

For the beam axial loads due to seismic loads, it should be 1/2 of the seismic load above this story plus the total seismic load at this story. It should be (121+107+87+64+43)/2+20. This is because the brace above point B can only deliver half the loads to beam segment BC.

Mook, Can you elaborate how did you get the drag force for the concrete problem #1 by using diaphragm?

 

[klip]

In addition, for the steel column in SCBF, did any one pay attention to the requirement of AISC 341 Scetion 8.3 regarding the use of overstrength factor? Thanks,

 

[mjbikes]

I revise my answer for #3 should be a SC joint. Whether or not A490, I don't know. I did hear A490 should be avoided for fatigue, but could not find anything in the code to verify.

For the steel girder design, it seems to be exactly like ex 3.9. I used (Pt+Pc)/2 for Pu

For the gravity loads to the braces, I used table 3-22c.

Section 8.3 of 341 says to use the amplified seismic load when axial loading is over 0.4 of capacity. Does that mean the overstrength factor (omega)?

Did the Masonry problem. Straight forward.

 

[MOOK]

For the steel brace, I would calculate the brace axial due to dead and live by applying half of the gravity loads to the brace-beam intersection point.
For the beam axial loads due to seismic loads, it should be 1/2 of the seismic load above this story plus the total seismic load at this story. It should be (121+107+87+64+43)/2+20. This is because the brace above point B can only deliver half the loads to beam segment BC.

Mook, Can you elaborate how did you get the drag force for the concrete problem #1 by using diaphragm?

Klip

Here is what I did

New diaphragm = (686/60)

New walls = (686/40)

Drag force at line 2 = New diaphragm * 20

4Drage force at line 3 = (New diaphragm *40) - (New walls * 20)

Use the larger force

Hope this helps

 

[klip]

Klip

Here is what I did

New diaphragm = (686/60)

New walls = (686/40)

Drag force at line 2 = New diaphragm * 20

4Drage force at line 3 = (New diaphragm *40) - (New walls * 20)

Use the larger force

Hope this helps

 

[whype]

For the Problem #2 (multiple choice questions), my solution does not match any of the answers given.

This is my solution:

Apparently 1.2D + 1.6L governs. D = 18 psf, L = 20 psf (Partition) + 50 psf (office occupancy) = 70 psf

w = (1.2*18psf + 1.6*70psf)*15 in/(12 in/ft) = 167 plf

M = (1/8)(w)(Span length^2) = (1/8)(167 plf) (14^2) = 4091.5 lb-ft

Did I miss anything? Any help will be appreciated. Thanks.

 

[kevo_55]

^^ Wood is in ASD in CA.

I would have to say that this is my only comment.

 

[whype]

^^ Wood is in ASD in CA.
I would have to say that this is my only comment.

Kevo,

That makes a lot of sense. Using ASD, you will get w =110 plf. Thanks very much.

The only problem is that "Wood" was not mentioned in the problem statement.

 

[kevo_55]

^^ I see your issue.

Typically, when I see a 1 story office building I think of wood joists. That's what lead me to use an ASD design.

I guess that more should have been given for this problem.

 

[mjbikes]

KlipHere is what I did

New diaphragm = (686/60)

New walls = (686/40)

Drag force at line 2 = New diaphragm * 20

4Drage force at line 3 = (New diaphragm *40) - (New walls * 20)

Use the larger force

Hope this helps

Mook,

I doubt that your answer is right. First of all, you did not use the asssumption that the solid concrete wall and the concrete wall with opeinings have the same relative stiffness at each floor level. I think that the answer to this problem relates to the base shearsgiven in figure CS1. on the other and, based on your answer, the drag force is quite big. If you use the same method for 2nd level point A, the drag force would be even higher. However, in problem 3, the drag force assumedis only 42 kips (even though this is an assumption). I guess we did not find the correct approach. what do you think?

Klip, I think you may have misinterpreted the slightly ambiguous wording of the problem. I think the stiffness of the wall at each level is equal to those above and below. The relative stiffness of the solid and perforated walls can then be calc'ed via the base shear of each wall.

That being said, the diaphragm loads at each level is 'sorted out' at that level (i.e. the roof lateral load is distributed to its supporting walls. Therefore, the roof load will not come into play for the drag strut in question. Next, since the 294kips is a diaphragm load, it is uniformly distributed to the walls and drag strut. each 20' portion receives 294/3=98k. relative stiffness requires the walls to receive 189k & 105k. So at 2: 189-98=91k tension (compression) and at 3: 105-98=7k compression (tension).

Hope that is correct!