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

Answer:

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

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 = $\frac{\pi }{4}$d² }

$S_{working$ = P / $\frac{\pi }{4}$d²

we substitute

$S_{working$ = 1768.70748 / $\frac{\pi }{4}$(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