# NCEES sample exam 802(a)



## steelhead (Feb 23, 2013)

Did anybody else work through this problem?

I noticed there is eratta on this problem, but the errata sheet doesnt correct anything, it just lists part of the answer exactly as the book lists? I'm thinking they need an eratta for the eratta?

When the polar moment of inertia is calculated, they dont account for the eccentricity- I (think) it should read:

J= (80-8.1)^2 + (80 +8.1^2) + (60-26.1)^2 + (60+26.1)^2=21492

V2= 75 + 3915*(60-26.1)/21492 = 81.2kips

Anyone else get an different answer, or am I looking at it wrong?

thanks


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## ajk244 (Feb 25, 2013)

The printing of the sample test you have was probably printed after the errata came out, so the errata matches what you have.

The load eccentricity doesn't matter when determining your polar moment of inertia (just like your applied force doesn't matter when figuring out the moment of inertia of a beam). The polar moment is just based on the location of your resisting elements relative to your center of rigidity. It's all just section properties.

The accidental eccentricity only comes into play when determing the torsion applied about the center of rigidity.


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## McEngr (Feb 25, 2013)

good answer akladiva - think of it like the center of rigidity is one side of the equation while the mass (applied torsion) is taken about on the other side.

(reaction sum)x(torsional distance) = force x center of mass distance

center of mass distance changes, but the other side does not...

hopefully, that has simplified it to where you understand the concept...


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