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anonymous
 3 years ago
can anyone help with mechanics of materials?
anonymous
 3 years ago
can anyone help with mechanics of materials?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I can't tell what the 4 shaded shear planes are. Can you try taking a clearer picture of the two pictorials?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the planes are parallel to the front plane

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is that understandable? uploading pics is taking forever right now

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I think I know the orientation it's supposed to be, but the bottom picture is difficult to decipher

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1340762720808:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you can also try the engineering study group

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thx i have, no one's ever there. my ? was posted there first

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0haha i see.. with the level of the course you're taking, you're bound to get a lot less help here in openstudy..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I took this course a couple semesters ago, I just have to go through my old book to refresh my memory >.<

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh okay haha. my sister's about to take it next semester or so. she's taking statics and particle dynamics right now.. how's mechanics of materials compared to that so far?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Mechanics of materials does build off the very basics of statics, not so much dynamics. I personally liked MME far more than statics. Except I had to take statics and dynamics crammed in the SAME course...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0haha same with my sister. i guess it's the standard course material across universities..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0nonsense, plenty of smart people here :) @abstracted if you figure it out, pls feel free to help. i have to sleep. what i did was avg shear stress=shear force/area. i used 500kPa to set as avg shear stress and for shear force i set V=P/2 since i used just one side of connection and divide by 2 since everything else is assumed to be equal so load is half if only one bolt is considered. used the plane for area which is 0.1m*0.1m. my answer for P is 10kN but book says 9.05

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i prefer statics much more but this is summer course so materials are very condensed.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0haha yep. my sister's taking summer school too, but with statics and particle dynamics lol. by the way I can't help on this one, I'm a computer engineer lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it's ok, thx for looking. particle dynamics sounds nasty 0_o

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0mme gets better, the first 6 or so chapters were very boring to me. It gets interesting when you get to bending moments/beam deflections

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0bending moments were covered in the first chapter

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0typo, bending/flexural stresses*

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0o yea, haven't heard of it yet. anyway thx. gonna try again tomorrow.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I got 9.05 kn, see below: \[\tau = \frac{V}{A}\rightarrow V=\tau A\] In the case for the bolts:\[V=\frac{P}{4}\]\[A= \pi(0.003m)^2=2.8274(10^{5})m^2\]\[\tau = 80MPa=80(10^6)Pa\] The shear force is 1/4 of the load because you must take into consideration both ends of both bolts; just like they take into consideration both sides of both bolts (the four shaded planes). See the attached image, all four shaded areas represent an individual shear. Plugging everything in and solving for P; \[P_{bolts}=4(80MPa)(2.8274(10^{5})m^2)=9.048kN\] And I got \[P_{planes}=20kN\] So the answer is the smaller one: 9.048kN

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0There are some tools that may prove helpful for your class (they did for me). The authors of my text book created a website full of animated examples for every chapter (does not require log in) http://web.mst.edu/~mecmovie/ The authors of a different mme book created a very simple software that you can use to calculate nearly everything you come across in this class. http://www.mdsolids.com/

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thx so much and thx for the references!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No problem, feel free to ask me any mme questions in the future, I loved this class!
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