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Homework 9 assignment and open thread

November 6th, 2007 · 20 Comments

This is another brief one to get you up to speed for the exam next week. This homework is due on Friday 11/9 at 4 pm.

First, two shear and moment problems: Prob. 7-77, 7-82
(Use the 5 steps…82 is the hardest one yet, so just take your time…)

Then, two cable problems: Prob. 7-94, 7-96

Note that cables as we have studied in Chapter 7 will NOT be on Exam 2, but shear and moment WILL FOR SURE be on the exam.

As usual, use this post as an open thread for discussion.

Tags: MAE230 Statics · Homework (MAE 230)

20 responses so far ↓

  • 1 tpadden // Nov 7, 2007 at 9:52 pm

    On problem 7-82, how do you deal with the segment pins connecting the structure at points B and E? What forces will they add to the structure and what moments, if any do they add?

  • 2 hambut // Nov 7, 2007 at 10:29 pm

    dude i think your looking at the wrong problem…

  • 3 hambut // Nov 7, 2007 at 11:54 pm

    anyone know if the 2000 lb force in 82 is added onto the load or is it just 2000 lb all together?

  • 4 Alessandro // Nov 8, 2007 at 12:12 am

    I believe it is a separate applied load, meaning you have the distributive loading and ontop of that you have 2000lb. If you do 500(9) + (0.5*500*9) you get something way bigger than 2000lb so I would think it is another external applied force. Any other ideas?

  • 5 kls2ycc // Nov 8, 2007 at 12:35 am

    for 82, on the graphing part, did you get a discontinuity at the 2000 point force for both the V and M graph? Also, did you get really large numbers?

  • 6 Alessandro // Nov 8, 2007 at 12:53 am

    Yeap, I’m getting big numbers, in the order of 5 digits.

  • 7 kls2ycc // Nov 8, 2007 at 1:09 am

    actually, i realized i messed up my moment equation, but when i went back it wasn’t discontinuous on the moment graph, however, I wasn’t sure if the moment at B is supposed to equal zero… because I got about 13500 for the moment at the end of the graph

  • 8 Alessandro // Nov 8, 2007 at 1:16 am

    The graph for the force is discontinuous at x= 9, but yea the moment graph is continuous. I did get zero at the end however when I plugged in x=18. I don’t think the moment at the end has a non zero value. (I might be wrong)

  • 9 tpadden // Nov 8, 2007 at 10:12 am

    sorry about that for some reason i was looking at 7-80

  • 10 jmwakey7145 // Nov 8, 2007 at 3:17 pm

    for 7-82 i made two method of section cuts. one before the 2000lb force and one after. Im having trouble with the moment equation for the cut after the 2000lb load. I don’t know how to model the resultant force of the distributed load when its a weird shape. like a trapezoid. anyone have any suggestions?

  • 11 rsa8q // Nov 8, 2007 at 3:45 pm

    alessandro, i agree that there should be no moment at point b in 7-82.

    wakey, i would suggest using an integral to represent the force and moment for the distributed load.

  • 12 kls2ycc // Nov 8, 2007 at 4:30 pm

    for 7-82, i would suggest looking at example 7.8 on page 357. It was very helpful for me, and made a lot of sense, and helped me work the problem out without having to use integrals.

  • 13 kls2ycc // Nov 8, 2007 at 4:32 pm

    for problem 7-96, is the tension from B to the pulley the same as the force hanging from the pulley (15lb) ?

  • 14 katie // Nov 8, 2007 at 4:59 pm

    thats what i used as the tension..and i got 11.6 as x. anyone else? the answer isnt in the back.

  • 15 katie // Nov 8, 2007 at 5:03 pm

    ah and i also got 8.7, if i change one sign.

  • 16 tpadden // Nov 8, 2007 at 5:21 pm

    I am also lost on the second cut for 7-82…I don’t understand how to model a trapezoids centroid for the moment equation

  • 17 rsa8q // Nov 8, 2007 at 9:57 pm

    if anyone would like to check the answer to 7-96, i got xb = 5.65 ft

  • 18 Vfort // Nov 8, 2007 at 10:16 pm

    I got 5.65 ft as well….I cant get the trapezoid’s center of mass either…anyone??

  • 19 tpadden // Nov 9, 2007 at 12:48 am

    ok mclovin and I finally figured out how to do 7-82. The only way we could do this problem was making two cuts, one before the point load and one after the point load, and then working from B towards A so that we cut out having to use a trapezoid. Be sure to sum the moments around B and everything should work.

  • 20 jmwakey7145 // Nov 9, 2007 at 8:50 am

    thanks tpadden, i never thought to do it that way, but its much easier

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