Mechanics and Materials Change in Length and Stress/Strain

JimCrown

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A 35mm diameter steel cable carries a mine skip to a depth of 625m. The mass of the skip is equal to the mass of the 625m length cable. The unit mass of the steel cable is 7.85 x 10^3 kg/m^3 . The cable unwinds from the surface. Calculate the extension in:

a) The full length of the cable

b) When the cable is extended for half itslength, (present your answers in mm).

Assume E for the steel cable = 210000N/mm^2 and use g = 10m/s^2

When I try to do this :

7.85 * 10^3 = 7850

1kg = 9.81N

7850 * 9.81 = 77008.5 N

Stress = Force/Area

Area = π * (35)^2/4 = 962.1127502

77008.5/962.1127502 = 80.04103468

E = Stress/Strain

Strain = Stress/E

80.04103468/210000 = 3.811477842^-04 or 0.0003811477842

625 * 1000 = 625000

Strain = Change in Length/Length = 0.0003811477842 * 625000 = 238.2173651

The answer is 292mm but, I keep getting 238mm. I do not knowwhere I am going wrong. Also, I have no idea how to answer part B. The answer to this is 109mm. Please can anyone help?
 
Last edited:
A 35mm diameter steel cable carries a mine skip to a depth of 625m. The mass of the skip is equal to the mass of the 625m length cable. The unit mass of the steel cable is 7.85 x 10^3 kg/ms^3 . The cable unwinds from the surface. Calculate the extension in:

a) The full length of the cable

b) When the cable is extended for half itslength, (present your answers in mm).

Assume E for the steel cable = 210000N/mm^2 and use g = 10m/s^2

When I try to do this :

7.85 * 10^3 = 7850

1kg = 9.81N

7850 * 9.81 = 77008.5 N

Stress = Force/Area

Area = π * (35)^2/4 = 962.1127502

77008.5/962.1127502 = 80.04103468

E = Stress/Strain

Strain = Stress/E

80.04103468/210000 = 3.811477842^-04 or 0.0003811477842

625 * 1000 = 625000

Strain = Change in Length/Length = 0.0003811477842 * 625000 = 238.2173651

The answer is 292mm but, I keep getting 238mm. I do not knowwhere I am going wrong. Also, I have no idea how to answer part B. The answer to this is 109mm. Please can anyone help?
The question instructs you to use g=10m/s^2. You've used 9.81. Maybe that's all it is.
 
A 35mm diameter steel cable carries a mine skip to a depth of 625m. The mass of the skip is equal to the mass of the 625m length cable. The unit mass of the steel cable is 7.85 x 10^3 kg/ms^3 . The cable unwinds from the surface. Calculate the extension in:

a) The full length of the cable

b) When the cable is extended for half its length, (present your answers in mm).

Also, I have no idea how to answer part B. The answer to this is 109 mm. Please can anyone help?
Check the unit carefully.

The unit for "unit mass" should be kg/(m^3) or kg/(mm^3). I know what it should be - but you need to read and present the problem carefully.
 
Check the unit carefully.

The unit for "unit mass" should be kg/(m^3) or kg/(mm^3). I know what it should be - but you need to read and present the problem carefully.

Yes sorry I meant kg/m^3 not ms. I made a typo. I still have no idea how to reach the answer. I know I am missing something. Please, can you help me?
 
A 35mm diameter steel cable carries a mine skip to a depth of 625m. The mass of the skip is equal to the mass of the 625m length cable. The unit mass of the steel cable is 7.85 x 10^3 kg/m^3 . The cable unwinds from the surface. Calculate the extension in:

a) The full length of the cable

b) When the cable is extended for half itslength, (present your answers in mm).

Assume E for the steel cable = 210000N/mm^2 and use g = 10m/s^2

When I try to do this :

7.85 * 10^3 = 7850

1kg = 9.81N

7850 * 10 = 78500 N/m^3 = 0.785 * 10^-4 N/mm^3

Weight of the chain =
0.785 * 10^-4 * 625000 * 962.128 = 47203.6568 N

Force = Weight of the (chain + skip) = 2 * 47203.6568 N

Continue....

Stress = Force/Area

Area = π * (35)^2/4 = 962.1127502

77008.5/962.1127502 = 80.04103468

E = Stress/Strain

Strain = Stress/E

80.04103468/210000 = 3.811477842^-04 or 0.0003811477842

625 * 1000 = 625000

Strain = Change in Length/Length = 0.0003811477842 * 625000 = 238.2173651

The answer is 292mm but, I keep getting 238mm. I do not knowwhere I am going wrong. Also, I have no idea how to answer part B. The answer to this is 109mm. Please can anyone help?
.
 
Last edited by a moderator:
Why does the
78500 N/m^3 = 0.785 * 10^-4 N/mm^3

As both are not the same/equal to each other. I could understand 10^4 but why 10^-4?


Also, for the half length do, I just halve the 625m to 312.5m?
1 m^3 = 10^(9) mm^3 ... that means .... 1 /m^3 = 10^(-9) /mm^3


Also, for the half length do, - Are you defining "do" to be your total stretch at full length of cable?
 
Also, for the half length do, I just halve the 625m to 312.5m?
At half length, the stress will be reduced - so length will not be half (for half stretch).

Another point, for the first problem, you are assuming constant load at all the cross-sections of the hanging chain. That can be very wrong assumption for a "hanging" cable under gravitational force.
 
At half length, the stress will be reduced - so length will not be half (for half stretch).

Another point, for the first problem, you are assuming constant load at all the cross-sections of the hanging chain. That can be very wrong assumption for a "hanging" cable under gravitational force.

So do I halve the weight instead?
 
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