Calculating Strain in a Steel Cable Supporting a Freight Elevator

In summary: Tensile stress is the maximum stress a material can withstand before it breaks. In summary, a 1260kg freight elevator is supported by a steel cable of diameter 34.9mm. The cable is loaded with a crowd of people collectively having a mass of 2850kg and it is descending at a constant speed of 2.30 m/s. The strain on the cable is computed to be 367740.788 Pa.
  • #1
Tycho
23
0
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!
 
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  • #2
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!

Rewrite the appropriate Hooke's law using the relative radial strain instead.
 
  • #3
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
relative radial strain?
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
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:
  • #6
If you assume that the volume of the cable is constant, with unkown length L, KNOWN radius R (you have this one!), then:
[tex]V=L\pi{R}^{2}=\pi(L+\delta{L})(R-\delta{R})^{2}[/tex]
Or:
[tex]1=(1+\frac{\delta{L}}{L})(1-\frac{\delta{R}}{R})^{2}[/tex]
We assume:
[tex]\frac{\delta{L}}{L},\frac{\delta{R}}{R}<<1[/tex]
Multiplying the brackets out, and neglecting products of small terms, yields:
[tex]1\approx1+\frac{\delta{L}}{L}-2\frac{\delta{R}}{R}[/tex]
Or:
[tex]\frac{\delta{R}}{R}=\frac{1}{2}\frac{\delta{L}}{L}[/tex]
 
  • #7
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
ooooops
cross sectional area = Pi * Radius^2
:)
 
  • #9
arildno said:
If you assume that the volume of the cable is constant, with unkown length L, KNOWN radius R (you have this one!), then:
[tex]V=L\pi{R}^{2}=\pi(L+\delta{L})(R-\delta{R})^{2}[/tex]
Or:
[tex]1=(1+\frac{\delta{L}}{L})(1-\frac{\delta{R}}{R})^{2}[/tex]
We assume:
[tex]\frac{\delta{L}}{L},\frac{\delta{R}}{R}<<1[/tex]
Multiplying the brackets out, and neglecting products of small terms, yields:
[tex]1\approx1+\frac{\delta{L}}{L}-2\frac{\delta{R}}{R}[/tex]
Or:
[tex]\frac{\delta{R}}{R}=\frac{1}{2}\frac{\delta{L}}{L}[/tex]



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
Tycho said:
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?
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
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
ashfaque said:
ooooops
cross sectional area = Pi * Radius^2
:)


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
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
ashfaque said:
part a)
M = 1260 + 2850 = 4110
F = M*g = 4110 * 9.81 = 40319.1 N
A = pi * dia ^2 / 4 = 9.566 ^ -4
stress = F/A = 42.147 ^ 6 Pa
Strain = stress / Y = 42.147 ^ 6 / 200 ^ 9 = 2.1 ^-4

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
Tycho said:
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?
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
hey Tycho .. i joined here today .. u?
 
  • #17
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
Tycho said:
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?

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
Tycho said:
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?
no oooops ...

Radius^2 = (Diameter)^2 / 4
(R)^2 = (D/2)^2 = D^2 / 4
 
  • #20
ashfaque said:
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)


you're going to have to put these two things together for me. how would i put this summation of forces into the equaition for strain?? therein is my biggest problem!
 
  • #21
Tesnsion+Weight (with direction) = Summetion of forces
u know weight
and also Mass*a ...
figure out Tension , that's the force to calculate stress ... then do like as (a) ..

i have to get my semseter results today ... got to sleeep (i live in another part of the world)
pray .. okay
:)
 

What is tensile strain?

Tensile strain is the deformation or elongation of a material when a tensile force is applied to it. It is a measure of how much a material stretches or lengthens under tension.

What is shear strain?

Shear strain is the deformation or distortion of a material when a shear force is applied to it. It is a measure of how much a material twists or changes shape when subjected to a shear stress.

What is the difference between tensile and shear strain?

Tensile strain is the elongation of a material under a tensile force, while shear strain is the distortion of a material under a shear force. Tensile strain occurs in the direction of the applied force, while shear strain occurs perpendicular to the applied force.

How is tensile strain calculated?

Tensile strain is calculated by dividing the change in length of a material by its original length. It is usually expressed as a decimal or a percentage.

What factors affect tensile and shear strain?

The material properties, such as elasticity and strength, as well as the magnitude and direction of the applied force, can affect tensile and shear strain. The shape and size of the material can also play a role in determining the amount of strain experienced.

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