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.
