# Question(s) of the Week #1



## phecke (Apr 2, 2015)

We're gonna give this a go.

See the attached for the diagram.

For the attached diagram, give the location of the Center of Rigidity, the value of the Torsional Rigidity, and determine the shear force in all walls shown. Please note the CM is at the geometric center of the "structure". Also, assume the CENTER of the walls at the perimeter are collinear with the perimeter dimensions (meaning the center of walls A and B are 100'-0" apart and walls C and D are 40'-0" apart).

torsional analysis problem.pdf


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## phecke (Apr 2, 2015)

For the attached diagram, how many zero force members are there?

truss problem.pdf


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## phecke (Apr 2, 2015)

For the attached diagram, determine if a 5-1/2" x 36" 20F-V12 Glulam timber can support the loads shown. The compression edge is fully laterally supported and temperature conditions are normal. Moisture content is &lt; 16. Be sure to check deflection vs IBC 2012 criteria.

glulam problem.pdf


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## BoilerEng (Apr 2, 2015)

For the first problem - are you assuming the shears in the x and y direction act simultaneously? Or would you look at it as direct shear plus torsional shear from either the x-direction or y-direction but not together?


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## phecke (Apr 2, 2015)

BoilerEng said:


> For the first problem - are you assuming the shears in the x and y direction act simultaneously? Or would you look at it as direct shear plus torsional shear from either the x-direction or y-direction but not together?






Direct shear and torsional moment shear in the x and y directions independently.


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## BoilerEng (Apr 2, 2015)

phecke said:


> BoilerEng said:
> 
> 
> > For the first problem - are you assuming the shears in the x and y direction act simultaneously? Or would you look at it as direct shear plus torsional shear from either the x-direction or y-direction but not together?
> ...


Center of Rigidity from bottom left.... X = 50 ft Y = 40 ft

I got max shears of:

A = B = 54.5 k

C = 59.4 k

D = 143 k

When checking the torsion from Vx, I got close to having a torsional irregularity, but still stayed under.


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## darius (Apr 2, 2015)

phecke said:


> For the attached diagram, how many zero force members are there?


Are all the interior members 0? These are my thoughts.


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## Antnyt23 (Apr 3, 2015)

BoilerEng said:


> phecke said:
> 
> 
> > BoilerEng said:
> ...




I got similar answers

Center of Rigidity

dx=50' from left dy=40' from bottom

Torsional Rigidity

J=112,000

Wall Shears

D = 140.6 k

C = 59.4 k

Amax=Bmax=54.5 k


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## BoilerEng (Apr 3, 2015)

Antnyt23 said:


> BoilerEng said:
> 
> 
> > phecke said:
> ...




For the shear in D, when you look at torsional moment, you correctly had it reduced from the direct shear. However, you can also look at the (-) sign of the accidental eccentricity. This leads to less torsional moment and less reduction. When you move the accidental eccentricity in the other direction, you'll get the torsional shear in D as approx -7 kips. So 150-7 = 143


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## Antnyt23 (Apr 3, 2015)

Thank you boiler! makes perfect sense now. Appreciate it.


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## bassplayer45 (Apr 8, 2015)

Quick explanation on why we reduce shear in D from torsion? Are we only assuming the load is going to the right? I came up with 157 clearly because i assumed the max eccentricity movement to produce the largest moment, and i just want to be sure i understand the theory (bridge guy).


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## bassplayer45 (Apr 8, 2015)

Yep, brain fart. Wall D is above center of rigidity and center of mass, so torsion is reducing shear, must reduce moment as much as possible scooting accidental eccentricities toward center of mass to maximize. I think studying has fried my brain


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## Jake47 (Apr 9, 2015)

Per the attached calculations, the glulam beam is acceptable as designed. Eng Boards Week 1--Glulam Problem.pdf


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## BoilerEng (Apr 9, 2015)

bassplayer45 said:


> Yep, brain fart. Wall D is above center of rigidity and center of mass, so torsion is reducing shear, must reduce moment as much as possible scooting accidental eccentricities toward center of mass to maximize. I think studying has fried my brain


Exactly right. This is also the case regardless of which direction the load is coming from.

And I know how you feel - I'm pretty fried at this point. Worked a rigidity problem last night and tried putting the load through the CR instead of CM so I was getting the opposite effect from torsional shear.


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## Antnyt23 (Apr 9, 2015)

Jake47 said:


> Per the attached calculations, the glulam beam is acceptable as designed.
> 
> 
> 
> ...


Jake I believe in your fb calculation you need to convert your Mu from lb-ft to lb-in.... in my answer i got 1704.5psi as the actual fb &lt; Fb = 1761psi .... Can someone confirm or did i do mine incorrectly?


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## Jake47 (Apr 9, 2015)

Atnyt23,

You are absolutely correct. I did not convert units....that's what I get for not double checking when I saw how low the demand was. Fail.


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## Antnyt23 (Apr 10, 2015)

Trust me I've been having the same problem doing practice problems. So worried about my speed i forget to double check little things and make minor errors that cause incorrect answers.


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