Respuesta :
Answer:
Step-by-step explanation:
Given:
Force, f = 9000 lb
Young modulus, E = 10,000,000 psi
Stress, S = 30000 psi
Elongation, x = 0.50 in
Length, L = 16 ft
Converting ft to in,
12 in = 1 ft
= 192 in
Young modulus, E = stress/strain
Stress = force/area, A
Strain = elongation, x/Length, L
Maximum stress = 30,000 psi
Stress = force/area
Area = 9000/30000
= 0.3 in^2
Using minimum area of 0.3 in^2,
A = (pi × d^2)/4
0.3 = (pi × d^2)/4
= sqrt(1.2/pi)
diameter, d = 0.62 inches
The minimum diameter that may be used for the rod is 0.61 inches.
Given that,
A 9,000 lb load is suspended from the roof in a shopping mall with a 16 ft long solid aluminum rod.
The modulus of elasticity of aluminum is 10,000,000 psi.
If the maximum rod elongation must be limited to 0.50 in. and the maximum stress must be limited to 30,000 psi,
We have to determine,
The minimum diameter that may be used for the rod.
According to the question,
Force = 9000lb
Young's modulus Elasticity = 10,000,000 psi
Maximum rod elongation is = 0.50 inches
Maximum stress = 30,000 psi
The elasticity can be expressed as the stress upon the strain.
[tex]\rm Elasticity = \dfrac{stess}{strain}[/tex]
The stress can be determined by,
[tex]\rm Stress = \dfrac{force}{area}\\\\Area = \dfrac{force}{stress}\\\\Area = \dfrac{9000}{30000}\\\\ Area = 0.3 \ inches^2[/tex]
The minimum diameter that may be used for the rod is,
[tex]\rm Area = \dfrac{\pi d^2}{4}\\\\\dfrac{0.3 \times 4}{3.14} = d^2\\\\d^2 = 0.38\\\\d = 0.61 \ inches[/tex]
Hence, The minimum diameter that may be used for the rod is 0.61 inches.
For more details refer to the link given below.
https://brainly.com/question/10603012
