Answer:
Tmax= 46.0 lb-in
Explanation:
Given:
- The diameter of the steel rod BC d1 = 0.25 in
- The diameter of the copper rod AB and CD d2 = 1 in
- Allowable shear stress of steel τ_s = 15ksi
- Allowable shear stress of copper τ_c = 12ksi
Find:
Find the torque T_max
Solution:
- The relation of allowable shear stress is given by:
τ = 16*T / pi*d^3
T = τ*pi*d^3 / 16
- Design Torque T for Copper rod:
T_c = τ_c*pi*d_c^3 / 16
T_c = 12*1000*pi*1^3 / 16
T_c = 2356.2 lb.in
- Design Torque T for Steel rod:
T_s = τ_s*pi*d_s^3 / 16
T_s = 15*1000*pi*0.25^3 / 16
T_s = 46.02 lb.in
- The design torque must conform to the allowable shear stress for both copper and steel. The maximum allowable would be:
T = min ( 2356.2 , 46.02 )
T = 46.02 lb-in
Answer:
1. True
2. True
3. False
Explanation:
The office location is where the soil layer is not uniform. The thickness of the soil varies which could lead to doors being jammed. The engineer needs to estimate the differential in clay soil.
The inclined surface can hold less weight than a vertical surface. The capacity to hold the weight is due to the gravitational force which is exerted to the load.
Answer:
Explanation:
Given conditions
1)The stress on the blade is 100 MPa
2)The yield strength of the blade is 175 MPa
3)The Young’s modulus for the blade is 50 GPa
4)The strain contributed by the primary creep regime (not including the initial elastic strain) was 0.25 % or 0.0025 strain, and this strain was realized in the first 4 hours.
5)The temperature of the blade is 800°C.
6)The formula for the creep rate in the steady-state regime is dε /dt = 1 x 10-5 σ4 exp (-2 eV/kT)
where: dε /dt is in cm/cm-hr σ is in MPa T is in Kelvink = 8.62 x 10-5 eV/K
Young Modulus, E = Stress, 
/Strain, ∈
initial Strain, 
creep rate in the steady state
but Tinitial = 0
solving the above equation,
we get
Tfinal = 2459.82 hr
Explanation:
150 divide by 150 and that how you do the is you what to divide together 15/ 150 you welcome have a good day is you need something else
Answer:
Yes this claim is correct.
Explanation:
The shear stress at any point is proportional to the velocity gradient at any that point. Since the fluid that is in contact with the pipe wall shall have zero velocity due to no flow boundary condition and if we move small distance away from the wall the velocity will have a non zero value thus a maximum gradient will exist at the surface of the pipe hence correspondingly the shear stresses will also be maximum.
