*3–32. The rubber block is subjected to an elongation of 0.03 in. along the x axis, and its vertical faces are given a tilt so t
hat θ = 89.3 ° . Determine the strains ε x , ε y and γ x y . Take ν r = 0.5.
1 answer:
Rubber block is not shown. I have attached an image of it.
Answer:
A) ε_x = 0.0075
B) ε_y = 0.00375
C) γ_xy = 0.0122 rad
Explanation:
We are given;
δ = 0.03 in
L = 4 in
ν_r = 0.5
θ = 89.3° = 89.3π/180 rad
Let's calculate ε_x in the direction of axis x
Thus, ε_x = δ/L = 0.03/4 = 0.0075
Let's calculate ε_y in the direction of axis y;
ε_y = v•ε_x = 0.5 x 0.0075 = 0.00375
Now, shear strain is angle between π/2 rad surfaces at that point.
Thus,
γ_xy = π/2 - θ = π/2 - 89.3π/180
γ_xy = π(0.003889) = 0.0122 rad
You might be interested in
The exit temperature is 586.18K and compressor input power is 14973.53kW
Data;
- Mass = 50kg/s
- T = 288.2K
- P1 = 1atm
- P2 = 12 atm
The exit temperature of the gas can be calculated isentropically as
Let's substitute the values into the formula
The exit temperature is 586.18K
<h3>The Compressor input power</h3>The compressor input power is calculated as
The compressor input power is 14973.53kW
Learn more on exit temperature and compressor input power here;
Answer:
15,000 psi
Explanation:
The solution / solving is attach below.
Answer:
Yes, the flow is turbulent.
Explanation:
Reynolds number gives the nature of flow. If he Reynolds number is less than 2000 then the flow is laminar else turbulent.
Given:
Diameter of pipe is 10mm.
Velocity of the pipe is 1m/s.
Temperature of water is 200°C.
The kinematic viscosity at temperature 200°C is m2/s.
Calculation:
Step1
Expression for Reynolds number is given as follows:
Here, v is velocity, is kinematic viscosity, d is diameter and Re is Reynolds number.
Substitute the values in the above equation as follows:
Re=64226.07579
Thus, the Reynolds number is 64226.07579. This is greater than 2000.
Hence, the given flow is turbulent flow.