This homework is a single problem, but I want you to be exceptionally diligent in your approach. To complete this shear and moment problem, you MUST (!!!) follow the 5 steps, and be absolutely explicit about your approach and your solution. Do NOT use the expedited approach presented in Section 7.3. If you do not take the time to do this correctly, not only will your grade on this homework not be very good, but more importantly you will not become skilled at solving for V & M. As such, please take your time and do this problem well!
Assignment: Prob. 7-75
Due date is Friday 11/2, 4 pm.
I expect you to avail yourself of all the resources at your disposal, including V & M links in previous posts. In addition, if you want to know the definition of “explicit”, check out the video below.
22 responses so far ↓
1 Josh // Nov 1, 2007 at 12:49 pm
The answer in the back of the book only gives the equations for one part of the problem, I finished the problem but my moment equation for the last part seems a little strange, so could anyone tell me bassically what theres looks like, mine is parabolic from -367 to 173.9
2 jrh3 // Nov 1, 2007 at 1:21 pm
im working on it now. im trying to figure out how to use the HAMLET to check my work. anyone know how to delete forces from the diagram in the HAMLET program?
3 ryan04583 // Nov 1, 2007 at 2:01 pm
for the portion of the moment graph from x=5 to x=8, the endpoint of your moment graph should be zero. According to my calculations, the starting point for that portion is -75.5, which is also the low point on my graph.
4 Anthony Amadeo Gallo // Nov 1, 2007 at 2:29 pm
hey kids, i hope this isn’t too obnoxious, but where exactly do i find these “5 steps”??? i tried to watch that module but everything was all fuzzy and i couldn’t read it - anyone care to list what they are? i know step 5 is the checks, but what are all of the checks?
5 ryan04583 // Nov 1, 2007 at 2:58 pm
step 1 is FBD of the entire structure with support reactions. step 2 is how many cuts you are making and their locations. step 3 if FBD of all cuts and corresponding equilibrium equations. step 4 is the graphs.
6 kls2ycc // Nov 1, 2007 at 2:59 pm
Is it necessary to make a cut in between the moment and th 8 kN force? Right now I only have two cuts but I wasn’t sure if I needed more.
7 Anthony Amadeo Gallo // Nov 1, 2007 at 3:28 pm
also, ryan, i’m thinking that the endpoint of the moment graph shouldn’t be zero - after all, there’s external forces on the load all the way to the end. in fact, there should be a very large moment at the endpoint!
8 Anthony Amadeo Gallo // Nov 1, 2007 at 3:29 pm
thanks ryan, but what are the actual checks that he wants us to perform?
9 Anthony Amadeo Gallo // Nov 1, 2007 at 3:41 pm
whoops, ignore that post about the endpoint of the moment graph
10 Josh // Nov 1, 2007 at 3:46 pm
The checks are looking at the two graphs and for correlations between them, like the moment graph should be linear when the sheer is constant, and that it should have a maximum when the sheer crosses zero, things like that
11 katie // Nov 1, 2007 at 3:48 pm
im having the same trouble with the hamlet program too.
12 katie // Nov 1, 2007 at 4:14 pm
oh we figured out that you can hide the first distributed load under the other one.
my graphs are the same as the hamlet ones, but im not getting any positive moments at x=2. at x=2 to x=3, my equation is -22.3x+44 for the moment. is there something i am doing wrong?
13 hambut // Nov 1, 2007 at 4:32 pm
i am a little confused with my reaction forces, im getting a negative number for the Yforce @A. just curious if this is somehow correct or what might be wrong with my arithmetic? (Ay = -14.3???)
14 jrh3 // Nov 1, 2007 at 4:43 pm
i got the same thing hambut.
15 jrh3 // Nov 1, 2007 at 4:44 pm
i was also wondering the same thing about the area between teh moment and the 8kN force. is a cut necessary or is just the two cuts in the distributed force area and teh non-distributed force area needed?
and what was figured out about HAMLET? you just hide forces behind one another?
thanks anyone who helps
16 kls2ycc // Nov 1, 2007 at 4:57 pm
on the program the more you lower the magnitude of the point force, you’ll see that the Moment stays negative instead of going positive, so it never really crosses the axis
17 katie // Nov 1, 2007 at 5:03 pm
jrh3, if you make the two distributed loads the same width, you can put them at the end of the beam, on top of one another. then you will get the right shape, no matter the magnitude
18 hambut // Nov 1, 2007 at 5:18 pm
katie, i got the same moment equation from above, but wouldnt that then result in the moment graph not starting at 0? and, if i am using the HAMLET correctly, they have it starting at 0…?
19 dtr5a // Nov 1, 2007 at 10:26 pm
Does anyone want to check their cuts, values/eq’s for shear, and values/eq’s for moment with me? I’m seeing a bunch of familiar number above, but would feel comforted if I got to check with someone since we don’t have answers in the back.
20 McLovin // Nov 2, 2007 at 12:26 am
im also getting -14.3 as Ay…i did the moments around both A and B and still got Ay to be negative…so i have no idea if these are right but….
a-a
V= -22.3
M= -22.3x + 44
b-b
V= 45 - 15(x-5)
M= (75-15x)(x-5) + 7.5(x-5)^2 + 45x - 292.5
yes? no?
21 jrh3 // Nov 2, 2007 at 8:11 am
im gettin the same thing McLovin
22 dtr5a // Nov 2, 2007 at 8:21 am
That’s what I got too, except I made more cuts…I also made cuts before the 20Kn*m moment 2m away from A and I made a cut between that moment and the 8kN force applied to the top of the beam. Why are these two cuts not necessary?
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