Question

An 18-in.-long titanium alloy rod is subjected to a tensile load of 24,000 lb. If the allowable tensile stress is 60 ksi and the allowable total elongation is not to exceed 0.05 in., compute the required rod diameter. Assume the proportional limit to be 120 ksi.

Answer 1

Answer:

Required Diameter = 302.65 inches

Explanation:

We are given;

Allowable tensile stress = 60 ksi

Weight of tensile load = 24,000 lb

Elongation = 0.05 in

Original length = 18 in

We'll need to check the diameters under stress and strain.

Now, we know that the formula for stress is;

Stress = Force/Area

Thus,

Area = Force/stress

So for this stress, area required is;

A_req = 24000/60 = 4000 in²

So let's find the required diameter here.

Area = πd²/4

So, 4000 = πd²/4

(4000 x 4)/π = d²

d² = 5092.96

Required diameter here is;

d = √5092.96

d = 71.36 in

For Strain;

Formula for strain is;

Strain = stress/E

We are given E = 120 ksi

stress = P/A = 24,000/A

strain = elongation/original length = 0.05/18 = 0.00278

Thus;

0.00278 = P/(A•E)

0.00278 = 24000/(120 x A)

Making A the subject to obtain;

A = 24000/(120 x 0.00278)

A_required = 71942 in²

Area = πd²/4

So, 71942 = πd²/4

(71942 x 4)/π = d²

d² = 91599.4

Required diameter here is;

d = √91599.4

d = 302.65 in

The larger diameter is 302.65 inchesand it's therefore the required one.