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## 29 thoughts on “HOMEWORK H31.A”

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my bending moment diagragm does not end as zero, what might be wrong?

Some common errors would be as follows: incorrect calculation for the equivalent forces leading to incorrect reaction forces, incorrect integration from v to x (ie. forgetting to divide by 2 etc.). I'd recommend going back through each step slowly. This happened to me last homework, but I was able to figure it out once I looked closely specifically at the math that seems easy but can lead to minor errors affecting the entire problem.

Also remember that the professor said to graph the negative moments positive on the graph so you may have had a sign change mix up

Also remember the professor said to graph the negative moments positively so you may have had a sign change mix up too

yeah, thanks guys, I've fixed it

When this usually happens for me, it's usually incorrect reaction forces

It could also be wrong because you might have incorrectly identified the point where shear force diagram goes to 0. Use the equation of the line to find that unknown 'x'. One other mistake could be where you might have considered the distance 'x' for negative area calculation instead of (length BA - x).

How should I determine which dimension in the cross section should be to the third power? Or should I be calculating it both ways and taking the larger result?

what we take to the third power is the height/2 in case you're talking about the area moment

this will be for the stress max calculation right?

For a rectangular cross section the dimension that will be cubed will be the one that is perpendicular to the neutral axis of the cross section.

Kushal is right.

since you are finding sigma (x), sigmax=Moment at x times y divided by the resistance inertia for the y, the way I think about it is that I need to cancel a term out for discrete non-hollow systems while keeping both dimensions in the equation, the only possible way here is having the y cubed

if it was sigma y we are interested about then itll be the other way around having moment at a y location times x distance over total y times x^3 in order to consider both in the equation.

Can we leave our answer in Kip/ft^2

from the examples the professor solved it seems that the unit psi is favored

I think that just to be safe you should convert it to psis as it is more common than kip/ft^2

kip/ft^2 is just ksi because kip = 1000 lbs

I thought that because you have to do the square ft to square inches conversion that they wouldn't be equal.

My bad. kips/in^2 is ksi. So, you will have to convert ft to inches. Thanks for pointing that out, Zach.

I see the above comments, but I am still confused. When calculating the second area moment of inertia, are we looking for Ix or Iy?

i think it wants lx. it's rectangular shape. You can use (1/12)*(bh^3)

I think it wouldn't matter if we calculate Ix or Iy as it gives the same value with the formula (1/12)*(bh^3)

Bowen and Chirag are correct

Iy is what we are looking for.

Why do we have to graph our positive moments as negative?

You don't have to. Stay with the sign from calculation when plotting.

I am having trouble on thinking about the bending moment diagram. Can you explain or clarify on how to build it. If positive area on shear positive slope on bending moment and the integral right?

Yes it's correct. First get the shear diagram, and it tells you the information of the slop of moment diagram. Then think about the reaction force at the boundary, and it tells you how the moment diagram start(at the boundary.)