# Homework Help: Tensile & Shear Strain

1. Nov 7, 2004

### Tycho

A 1260kg freight elevator is supported by a steel cable of diameter 34.9mm. It is loaded with a crowd of people collectively having a mass of 2850kg and it is descending.

a) What is the strain on the cable while the elevator is descending at the constant speed of 2.30 m/s?

b) What is the strain in the cable when it is brought to a stop in 0.600s?

okay this doesn't seem to be very hard, but as i tried to do it, i just keep getting stuck.
okay firstly: strain.
i only know of 2 types of strain: tensile, and shear. this definitely isn't a shear strain problem since the force applied is not directed parallel to the surface.
that leaves tensile strain. Tensile strain is (the change in length in response to applied stress) / (relaxed length) is it not? But if this is what i am supposed to use, shouldn't the problem have given me an initial legth of the cable? This leads me to believe that i am confused on the concept of strain. Is there a different type of strain that i am missing, or should there be another step?

Thanks physics gurus!

2. Nov 7, 2004

### arildno

The questions make more sense if you regard "strain" as a typo; I think it must be the stress they're after.
Otherwise, they would have needed to give you a stress/strain-relationship+initial length.

However, they just MIGHT be mean to you, and assume you find out the particular relationship in that case.
In order to proceed then, you should derive a relationship between relative longitudinal strain, and relative radial strain.
Hint: This can be done by assuming that the cable's volume must be constant!

3. Nov 7, 2004

### PerennialII

Likewise ... the problem sounds pretty basic in order to focus on strain. Thinking the problem axisymmetric makes it easy to work out in such a case no matter.

4. Nov 7, 2004

### Tycho

Hookes law= Y (Change in length in response to applied stress / relaxed length) where Y= Young's modules
I must have the wrong Hookes law. This just works to take the same problem into a different equation? (i don't know the initial length of the cable.)

It could be a typo.... but i don't want to trust my grade on a typo, y'know? any other suggestions?

5. Nov 7, 2004

### ashfaque

You do not need initial length for this problem.

You know the diameter of the cable ...

Compute Area , Acceleration,Force , Stress ... etc
For constant speed there is no acceleration (or acceleration = 0 )

Tensile Stress = (Axial Force) / (Cross- Sectional Area)
Modulus of Elasticity = Stress / Strain ( this is hooke's law)

For Steel the modulus of elasticity = 200 GPa (approx. or u can use 207 GPa)

Then compute Strain

For constant speed there is no acceleration (or acceleration = 0 )

Last edited: Nov 7, 2004
6. Nov 7, 2004

### arildno

If you assume that the volume of the cable is constant, with unkown length L, KNOWN radius R (you have this one!), then:
$$V=L\pi{R}^{2}=\pi(L+\delta{L})(R-\delta{R})^{2}$$
Or:
$$1=(1+\frac{\delta{L}}{L})(1-\frac{\delta{R}}{R})^{2}$$
We assume:
$$\frac{\delta{L}}{L},\frac{\delta{R}}{R}<<1$$
Multiplying the brackets out, and neglecting products of small terms, yields:
$$1\approx1+\frac{\delta{L}}{L}-2\frac{\delta{R}}{R}$$
Or:
$$\frac{\delta{R}}{R}=\frac{1}{2}\frac{\delta{L}}{L}$$

7. Nov 7, 2004

### Tycho

Okay, I did like you suggested, but nowhere in my work did i factor in the speed at which the elevator was moving?

here's my work:
(1260kg+2850kg)*9.81 = 40319.1 Newtons
R=17.45
Cross-sectional area of the cable = 2piR = 2pi*17.45 = 109.64mm = .10964m
Tensile Stress = (axial force) / (cross-sectional area) = 40319.1 N / .10964m = 367740.788 Pa

Young's Modulus for steel = 20*10^10 = Stress/Strain
20*10^10 = 367740.788 Pa / Strain
Strain = 367740.788 Pa / 20*10^10
Strain = 1.83*10^-6?????????

That number MUST be wrong? did i make a mistake, or is the process incorrect?
Again, where am i supposed to account for the speed at which the elevator is moving? (I.E. the second problem)

8. Nov 7, 2004

### ashfaque

ooooops
cross sectional area = Pi * Radius^2
:)

9. Nov 7, 2004

### Tycho

Wow, this went way over my head! lol. Is this the derivation of the relationship you were talking about? If so, then (since it looks unfamiliar) i would assume this isn't wasn't what my instructor was looking for?

10. Nov 7, 2004

### arildno

You bet that's not what he asked you for if you haven't seen it before.

Stick with what the other poster is doing; the fact that you in a) seems to get a tiny strain is simply because steel hardly lengthens.

11. Nov 7, 2004

### ashfaque

and you should also know:

Summation of Froces(using directions) = Mass * Acceleration

and the v = u - at ( calculate the acceleration, v = 0 , t = 0.6 sec)

12. Nov 7, 2004

### Tycho

oops! haha i knew that!
Strain = 2.11*10^13

That's more like it, but is it right?

And what about the speed of the elevator? If this is the answer to part a, then how do i solve for b?

13. Nov 7, 2004

### ashfaque

part a)
M = 1260 + 2850 = 4110
F = M*g = 4110 * 9.81 = 40319.1 N
A = pi * dia ^2 / 4 = 9.566E-4
stress = F/A = 42.147E6 Pa
Strain = stress / Y = 42.147E6 / 200E9 = 2.1E-4

14. Nov 7, 2004

### Tycho

A= pi * dia ^2 / 4 = 9.566 ^ -4 ??
shouldn't it be:
A= pi * 34.9mm^2 / 4 = 956mm = .956m ?
stress = F/A = 4.21 * 10^4 Pa ?
Young's modulus = stress/strain
strain = stress / young's
strain = 4.21*10^4 / 20*10^10 = 2.105 * 10 ^ 13 ??

or did i screw it up again?

15. Nov 7, 2004

### ashfaque

hmm
34.9mm^2 = (34.9mm)^2 = (0.0349 m) ^ 2 = 1.218E-3 (metre squared)
1.218E-3 *Pi/4 = 9.566E-4 (metre squared)
:)

16. Nov 7, 2004

### ashfaque

hey Tycho .. i joined here today .. u?

17. Nov 7, 2004

### Tycho

I saw what you did in the last one. Area=pi * R^2. you used the diameter instead of the radius. Welcome to the physics forum! this is my second month on.

do you know how to take the speed of the lift into effect?

18. Nov 7, 2004

### ashfaque

ooops on the diameter matter :)

initially moving in a constant speed 'u ' the elevator stops
that means 'v' = 0 ...
noy from v = u - at ( a = acceleration)

we get a = u/t

Mass * a = Summation of forces (vector addition)

19. Nov 7, 2004

### ashfaque

no oooops ......