Question

A main cable in a large bridge is designed for a tensile force of 2,600,000 lb. The cable consists of 1470 parallel wires, each 0.16 in. in diameter. The wires are cold-drawn steel with an average ultimate strength of 230,000 psi. What factor of safety was used in the design of the cable

Answer 1

the factor of safety was used in the design of the cable is 2.6146

Step-by-step explanation:

Given the data in the question;

Load on the main capable
`P_{initial` = 2600000 lb

number of parallel wires n = 1470

Diameter d = 0.16 in

average ultimate strength
`S_{ultimate` = 230000 psi

First we calculate the Load acting on each cable;

`P_{initial` = P × n

P =
`P_{initial` / n

we substitute

P = 2600000 lb / 1470

P = 1768.70748 lb

Next we determine the working stress acting in a member;

`S_{working` = P/A

{ Area A =
`(\pi )/(4)`d² }

`S_{working` = P /
`(\pi )/(4)`d²

we substitute

`S_{working` = 1768.70748 /
`(\pi )/(4)`(0.16)²

`S_{working` = 1768.70748 / 0.02010619298

`S_{working` = 87968.29 psi

Now we calculate the factor of safety F.S

F.S =
`S_{ultimate` /
`S_{working`

we substitute

F.S = 230000 psi / 87968.29 psi

F.S = 2.6145785 ≈ 2.6146

Therefore, the factor of safety was used in the design of the cable is 2.6146