A rotator has a weight of 100lb with a centroidal radius of gyration of 9 in. Determine the moment of inertia about the center o
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Answer: Attached below is the missing detail and Mohr's circle.
i) б1 = 9.6 Ksi
б2 = -10.7 ksi
ii) 10.2 Ksi
iii) -0.51Ksi
Explanation:
First step :
direct compressive stress on shaft
бd = P / π/4 * d^2
= -20 / 0.785 * 5^2 = -1.09 Ksi
shear stress at the outer surface due to torsion
ζ = 16*T / πd^3
= (16 * 250 ) / π * 5^3 = 010.19 Ksi
<u>Calculate the Principal stress, maximum in-plane shear stress and average normal stress</u>
Using Mohr's circle ( attached below )
<u>i) principal stresses:</u>
б1 = 4.8 cm * 2 = 9.6 Ksi
б2 = -5.35 cm * 2 = -10.7 ksi
<u>ii) maximum in-plane shear stress</u>
ζ = radius of Mohr's circle
= 5.1 cm = 10.2 Ksi ( Given that ; 1 cm = 2Ksi )
<u>iii) average normal stress</u>
= 9.6 + ( - 10.7 ) / 2
= -0.51Ksi
The question is incomplete. The complete question is :
The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .
When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?
Solution :
Given data :
Diameter of the rod : 46 mm
Torque, T = 85 Nm
The polar moment of inertia of the shaft is given by :
J = 207.6
So the shear stress at point A is :
Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.
Complete question:
Natural gas at 70 ⁰F and standard atmospheric pressure of 14.7 psi (abs) is compressed isentropically to a new absolute pressure of 70 psi. Determine the final density and temperature of the gas.
Answer:
The final density of the natural gas is 0.004243 slugs/ft³ and
The final temperature of the natural gas is 306.6 ⁰F
Explanation:
For Ideal gas: P = ρRT
R is ideal gas constant = 3.099 x 10³ ft lb / slug⁰R
T₁ is initial temperature = 70 ⁰F = (70+460)⁰R = 530⁰R
P₁ is intial pressure of the gas = 14.7 psi = (14.7 lb/in² X 144 in²/ft²) = 2116.8 lb/ft²
From the ideal gas equation, we calculate initial density of the natural gas:
ρ₁ = P/RT ⇒ 2116.8/(3.099 x 10³ X 530) = 0.001289 slugs/ft³
For isentropic process:
where K is the ratio of specific heats for natural gas; K = 1.31, Therefore
0.004243 slugs/ft³
Final density; ρ₂ = 4.243 X10⁻³ slugs/ft³
From ideal gas equation; P = ρRT
P₂ = ρ₂RT₂
T₂ = P₂/ρ₂R
P₂ (lb/ft²) = (70 lb/in²)( 144 in²/ft²) = 10080 lb/ft²
T₂ = 10080/(0.004243 X 3099)
T₂ = 766.6⁰R
Final Temperature; T₂ = (766.6-460)⁰F = 306.6⁰F