Expert: Arshad Khan - 9/19/2011

Question
QUESTION: Hi,
Thanks in advance for reading my question. I need to find out how much weight a piece of wire will support. The wire has a diameter of 0.18mm. I found a formula online that stated  - "The  formula  for  computing  the  SWL  of  a  wire rope is the diameter of the rope squared, multiplied by 8 (D x D x 8 = SWL in tons)"
This formula seems to be for wire rope, and I need to calculate the safe working load for steel wire. Can I still use this formula?
I did use the formula to make thw following calculation
0.18 x 0.18 =  0.0324
0.0324 x 8 =  0.2592 tons
Does this look correct to you? I know nothing about engineering, but it seems like a wire that is 0.18 mm in diameter would not be able to take 0.2592 tons of weight. Is this possible, and if not, can you tell me how I can calculate the safe working load of the 0.18mm diameter wire?
Thank you kindly,
Eddie Palmer

ANSWER: Hi Eddie,

Wire rope capacities are difficult to work out and are usually stated in a table by the manufacturer.

However, if you are talking about a single strand of steel wire of only 0.18mm diameter, then the ultimate capacity can be easily worked out.

If it is mild steel, the ultimate stress would be limited to 240 N/mm2; if it is High Tensile steel, it would be 500 N/mm2. For safe working loads you need to divide these by a factor of safety which can be between 1.5 to 2.5, depending on application.

So, X-section area is Pi*d*d/4 = 3.14 * 0.18 * 0.18/4 = 0.025 mm2

Therefore ultimate capacity = 0.025 * 240 = 6.107 N = 0.623 kg

This is, as you suspected, a lot less than 0.2592 tons!

---------- FOLLOW-UP ----------

QUESTION: Hi Mr Khan,

A while ago you gave me some help calculating the safe working load of a steel wire 0.18 mm in diameter. That was a big help to me for which I am very grateful.
I am now trying to calculate the safe working load for a wire 0.08 mm in diameter and I would like to ask for your help again.
You gave me a formula -
X-section area is Pi*d*d/4 = 3.14 * 0.18 * 0.18/4 = 0.025 mm2
Therefore ultimate capacity = 0.025 * 240 = 6.107 N = 0.623 kg

I am trying to apply that to my current problem but I have only got as far as this
3.14 * 0.08 * 0.08/4 = 0.005024 mm2
I am not sure how to calculate the  N value and then the kg value in the second step you showed me.

Could you tell me what the safe working load for a steel wire 0.08 mm in diameter would be?
Also, would the length of the wire make any difference?
This information would be a great help to me. Thank you very much.
Best Regards,
Eddie

Answer
Hi Eddie,

The length of wire makes no difference to tensile capacity but does effect the overall elongation due to strain.

The ultimate capacity of the 0.08 mm wire (mild steel) would be: 0.005024 x 240 = 1.206 N

1 kg is equivalent to approx 9.8N, so that would make it 0.123 kg.