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.
Answer:
Explained
Explanation:
This situation can occur because of various factors such as:
- Gradual deterioration of lubrication and coolant.
- change of environmental condition such as temperature, humidity, moisture, etc.
- Change in the properties of incoming raw material
- An increase or decrease in the temperature of the heat treating operation
- Debris interfering with the manufacturing process.
Answer:
1.2727 stokes
Explanation:
specific gravity of fluid A = 1.65
Dynamic viscosity = 210 centipoise
<u>Calculate the kinematic viscosity of Fluid A </u>
First step : determine the density of fluid A
Pa = Pw * Specific gravity = 1000 * 1.65 = 1650 kg/m^3
next : convert dynamic viscosity to kg/m-s
210 centipoise = 0.21 kg/m-s
Kinetic viscosity of Fluid A = dynamic viscosity / density of fluid A
= 0.21 / 1650 = 1.2727 * 10^-4 m^2/sec
Convert to stokes = 1.2727 stokes
Answer:
0.245 m^3/s
Explanation:
Flow rate through pipe a is 0.4 m3/s Parallel pipes have a diameter D = 30 cm => r = 15 cm = 0.15 m Length of Pipe a = 1000m Length of Pipe b = 2650m Temperature = 15 degrees Va = V / A = (0.4m3/s) / (3.14 (0.15m)^2) = 5.66 m/s h = (f(LV^2)) / D2g (fa(LaVa^2)) / Da2g = (fb(LbVb^2)) / Da2g and Da = Db; fa = fb LaVa^2 = LbVb^2 => La/Lb = Vb^2/Va^2 Vd^2 = Va^2(La/Lb) => Vb = Va(La/Lb)^(1/2) Vb = 5.66 (1000/2650)^(1/2) => 5.66 x 0.6143 = 3.4769 m/s Vb = 3.4769 m/s V = AVb = 3.14(0.15)^2 x 3.4769 m/s = 0.245 m^3/s
