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

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 = 2600000 lb

number of parallel wires n = 1470

Diameter d = 0.16 in

average ultimate strength = 230000 psi

First we calculate the Load acting on each cable;

= P × n

P = / n

we substitute

P = 2600000 lb / 1470

P = 1768.70748 lb

Next we determine the working stress acting in a member;

= P/A

{ Area A = d² }

= P / d²

we substitute

= 1768.70748 / (0.16)²

= 1768.70748 / 0.02010619298

= 87968.29 psi

Now we calculate the factor of safety F.S

F.S = /

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

this is true

explanation:

i looked it up

answer:

a

explanation: