# Calculating tensile stress? help please?

## Homework Statement

A cable containing 37 strands of 0.0250" diameter steel wire successfully supports 1000 lb load in tension. Calculate tensile stress? Do you think this is a reasonable and safe design stress level?

## Homework Equations

tensile stress = F/A

## The Attempt at a Solution

A = [ pi(0.0250^2)/4 ] x 37 (?)
tensile = 1000/A = 55,059 psi
This is not a safe design stress level because it is too high? is this correct?

Last edited:

PhanthomJay
Homework Helper
Gold Member

## Homework Statement

A cable containing 37 strands of 0.0250" diameter steel wire successfully supports 1000 lb load in tension. Calculate tensile stress? Do you think this is a reasonable and safe design stress level?

## Homework Equations

tensile stress = F/A

## The Attempt at a Solution

A = [ pi(0.0250^2)/4 ] x 37 (?)
F = 1000/A = 55,059 lbf
This is not a safe design stress level because it is too high? is this correct?
Watch your equations and units. The area you calculate is on square inches. Then the tensile stress ( not the force as you have written) is 55,000 psi if the math is correct. Compare that to the ultimate tensile stress of steel stranded wire. Do you have an idea of what that value is?

Watch your equations and units. The area you calculate is on square inches. Then the tensile stress ( not the force as you have written) is 55,000 psi if the math is correct. Compare that to the ultimate tensile stress of steel stranded wire. Do you have an idea of what that value is?

oh yea my mistake. i meant tensile stress = 1000/A = 55,059 psi. i looked it up and i found 40,000 psi for ultimate tensile stress of steel stranded wire. is that correct?

1) The 1000 lb is not the Force acting on the strands, it is simply the mass, to get the Force, you would have to get its weight.

2) Also when calculating you have to convert to SI units first to get a unified answer.

3) Also the area of a circle is pi r^2 but the 0.0250' given is the diameter so you would have to divide it in half first.

4) Yes You do multiply by 37 since this is the part of the cross sectional area of the cable

5) Tensile stress is measured in Pascals

6) Ultimate Yield strength of steel is around 250 Mpa = 250 x 10 (power 6) Pascals, if the tensile stress you got is more than this number (after you correct the calculation errors I pointed out) Then it is not a safe design

7) 250 Mpa / The Tensile stress you calculated, will give you a safety ratio, the higher the safer.

SteamKing
Staff Emeritus
Homework Helper
1) The 1000 lb is not the Force acting on the strands, it is simply the mass, to get the Force, you would have to get its weight.

In the Imperial system, unless stated otherwise, when loads are given in pounds, it is understood that the loads are pounds force, not pounds mass.

2) Also when calculating you have to convert to SI units first to get a unified answer.
It's not clear what is meant by 'unified answer'.

You can certainly calculate stress using Imperial units without having to convert everything to SI. The units for stress are pounds/sq.in. (psi) most commonly.

3) Also the area of a circle is pi r^2 but the 0.0250' given is the diameter so you would have to divide it in half first.

The diameter of each wire is given as 0.025" or 0.025 inches (0.025' = 0.025 feet). Using diameter, A = πD$^{2}$/4

4) Yes You do multiply by 37 since this is the part of the cross sectional area of the cable

5) Tensile stress is measured in Pascals

Only if you are using SI units.

6) Ultimate Yield strength of steel is around 250 Mpa = 250 x 10 (power 6) Pascals, if the tensile stress you got is more than this number (after you correct the calculation errors I pointed out) Then it is not a safe design

It's not clear from the OP what the tensile stress limit for the wire is. When dealing with wire, one must be careful to use proof loads for a particular wire size and construction. These values can vary from the results of tensile testing a steel piece which is not wire.

7) 250 Mpa / The Tensile stress you calculated, will give you a safety ratio, the higher the safer.

1 person
PhanthomJay
Homework Helper
Gold Member
oh yea my mistake. i meant tensile stress = 1000/A = 55,059 psi. i looked it up and i found 40,000 psi for ultimate tensile stress of steel stranded wire. is that correct?
Steel strand comes in varying strengths, the common grade steel might be around 40,000 psi but the higher strength steel strands could be 3 times that or more. You really can't answer the question without knowing the type of steel or the desired safety factor. Assuming you have ordinary steel, your calculated stress exceeds the allowable ultimate stress. That is not a good thing.

1 person
In the Imperial system, unless stated otherwise, when loads are given in pounds, it is understood that the loads are pounds force, not pounds mass.
I usually work with metric Units so I am used to convert from mass to Weight by multiplying the mass by the gravitational acceleration constant (g).

It's not clear what is meant by 'unified answer'.

You can certainly calculate stress using Imperial units without having to convert everything to SI. The units for stress are pounds/sq.in. (psi) most commonly.
I meant by unified answer: answer which you can compare with known values online, for example if you search the Tensile Yield Strength of steel, You would almost always find the units given in Mpa so instead of converting from the answer's units to pascals to compare it with the value present why not just work in SI units so that it is easier to compare??

The diameter of each wire is given as 0.025" or 0.025 inches (0.025' = 0.025 feet). Using diameter, A = πD$^{2}$/4
My apology I thought he used the Diameter with the πr^2 formula. Nothing to say here

Only if you are using SI units.
Like I said previously, much easier to work in SI units to get an answer which you can compare to known values.

It's not clear from the OP what the tensile stress limit for the wire is. When dealing with wire, one must be careful to use proof loads for a particular wire size and construction. These values can vary from the results of tensile testing a steel piece which is not wire.

I took the liberty just to find the Ultimate tensile stress limit for structural ASTM A36 steel which is used in structural steel in the US. I assume this is what OP is talking about. I also quoted the value 250 Mpa just to explain to him how to find the safety ratio. Although the value of UTS should be mentioned in the Question for him to compare.