Answer:
Qcd=0.01507rad
QT= 0.10509rad
Explanation:
The full details of the procedure and answer is attached.
Answer:
d= 4.079m ≈ 4.1m
Explanation:
calculate the shaft diameter from the torque, \frac{τ}{r} = \frac{T}{J} = \frac{C . ∅}{l}
Where, τ = Torsional stress induced at the outer surface of the shaft (Maximum Shear stress).
r = Radius of the shaft.
T = Twisting Moment or Torque.
J = Polar moment of inertia.
C = Modulus of rigidity for the shaft material.
l = Length of the shaft.
θ = Angle of twist in radians on a length.
Maximum Torque, ζ= τ × \frac{ π}{16} × d³
τ= 60 MPa
ζ= 800 N·m
800 = 60 × \frac{ π}{16} × d³
800= 11.78 × d³
d³= 800 ÷ 11.78
d³= 67.9
d= \sqrt[3]{} 67.9
d= 4.079m ≈ 4.1m
Answer:
See explaination and attachment.
Explanation:
Navier-Stokes equation is to momentum what the continuity equation is to conservation of mass. It simply enforces F=ma in an Eulerian frame.
The starting point of the Navier-Stokes equations is the equilibrium equation.
The first key step is to partition the stress in the equations into hydrostatic (pressure) and deviatoric constituents.
The second step is to relate the deviatoric stress to viscosity in the fluid.
The final step is to impose any special cases of interest, usually incompressibility.
Please kindly check attachment for step by step solution.