A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
A metal beam with Y as the young's modulus is supported at the two ends and loaded at the center. The depression is directly proportional to:
aY
bY^2
c1/Y
anonymous
 5 years ago
A metal beam with Y as the young's modulus is supported at the two ends and loaded at the center. The depression is directly proportional to: aY bY^2 c1/Y

This Question is Closed

JamesJ
 5 years ago
Best ResponseYou've already chosen the best response.0If the mass were on the end of a wire of this metal, what would say the answer is, out of a,b,c?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0c? I am lost at this question

JamesJ
 5 years ago
Best ResponseYou've already chosen the best response.0What's the definition of Young's Modulus?

JamesJ
 5 years ago
Best ResponseYou've already chosen the best response.0Go back and understand what it means. Then answer my question about the wire. Then think again about the beam. ==== This lecture might be helpful, if a bit advanced: http://ocw.mit.edu/courses/physics/801physicsiclassicalmechanicsfall1999/videolectures/lecture26/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I studied the topic again, I never get this for some reason. But I got the answer for your question If the mass were on the end of a wire of this metal, what would say the answer is, out of a,b,c? it would then be Y

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0\[ \frac{F}{A} = Y \frac{\Delta \ell}{\ell} \] So for a given force and length, \( Y \Delta \ell = constant \). Hence \( \Delta \ell \) is inversely proportional to \( Y \). Therefore with the wire alone, the answer will be c) 1/Y. The same principle applies with the beam. Here the cross sectional area is a bit more complicated and related to the length of the beam, but the basic physical principles are these same.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0Intuitively, this also makes sense. Y is a measure of 'stiffness'. Y of iron is higher than that of copper say. You'd expect the copper beam to bend more, consistent with the fact that 1/Y for copper > 1/Y for iron as \[ Y_{iron} > Y_{copper} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and what about this formula, how do we get this e=12MgL^3/3bd^3Y how do we get this?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0I don't know what that is. What is this formula and what are all the variables?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0e=depression M=mass of load g=gravitational acceleration l=length of beam b is the breadth of the beam d is the thickness of the beam Y=modulus

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0Oh, you have a formula for it. Why didn't you tell me?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0Anyway, you can see immediately that E is proportional to 1/Y. So the answer remains the same.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah, but how do we get this formula?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0anyways, I like you answer better, sweet and simple.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0I'm not sure, but it'll come from manipulation of the basic formula for Young's modulus one way or another.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0or experiment, I found this: http://amrita.vlab.co.in/?sub=1&brch=74&sim=550&cnt=1
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.