Ideal solutions concerning rotating solid and hollow shafts

3a.
A solid shaft of diameter 30mm rotating at 1000rpm delivers a power of 25kW determine (assume Shear Modulus for the shaft, G=80 GPa)
The Shear Stress in the shaft
The angular twist of the shaft as a result of the Shear Stress from part (i)
The important parameters to note here are
Power, P=25kW=25,000W
Diameter of shaft (needs to be in metres), D=30mm=0.03m
Radius of shaft, r=D/2=0.015m
Rotational speed (needs to be in radians per second),
ω=2π×1000rpm=2000π rads/s
Shear Modulus, G=80 GPa=8×〖10〗^10 Pa

(i) First we need to use the Power-Torque equation {1} to determine the Torque in the shaft
P=ωT {1}
So re-arranged to derive the Torque in the shaft
T=P/ω=25,000/2000π=3.978 N.m
We now need to use the equation {2} for the polar second moment of area,
〖 J〗_solid for a solid shaft
J_solid=(πD^4)/32 {2}
J_solid=(π(0.03)^4)/32=7.95×〖10〗^(-8) m^4

One now knows T,J_solid and shaft radius r so we use the Torque-Shear
Stress, equation {3} to determine the maximum Shear stress, τ in the shaft

T/J_solid =τ/r {3}
τ=Tr/J_solid =(3.978×0.015)/(7.95×〖10〗^(-8) )=7.51×〖10〗^5 N.m^(-2)=751 kN.m^(-2)

(ii) For the angle of twist we need to use the Angle (in radians)-Shear stress equation{4} to determine angular twist in the shaft over length, L , for Shear Modulus, G=80 GPa
θ=τL/Gr {4}

θ=(7.51×〖10〗^5×2)/(8×〖10〗^10×0.015)=1.25×〖10〗^(-3) rads

You can convert to degrees if you want noting the conversion factor (2π= 360^0
θ=(360/2π)1.25×〖10〗^(-3)=0.072^0

3b.
Find the diameter of a solid shaft resulting in a transmission Torque of 50 kNm for a maximum allowable Shear Stress in the shaft of 100 MPa
The important parameters to note here are
Transmission Torque, T=50kNm=50,000 Nm
Maximum allowable Shear stress in shaft τ=100 MPa=〖10〗^8 Pa

We need to use the equation {2} for the polar second moment of area,
J_solid for a solid shaft

J_solid=(πD^4)/32 {2}

And equation {3} relating Shear stress to Torque and radius of shaft
T/J_solid =τ/r {3}
Noting that radius r=D/2 and substituting {2} in {3} we get

T/((πD^4)⁄32)=τ/(D⁄2) {5}

Simplifying {5}
32T/(πD^4 )=2τ/D
32T/(πD^3 )=2τ {6}

Rearranging {6} to make D the subject

D^3=16T/πτ {7}

Putting in given values to {7} we get

D^3=(16×50,000)/(π×〖10〗^8 )=2.55×10^(-3) m^3
D=∛(2.55×〖10〗^(-3) )=13.7 cm
3c.
Calculate the power delivered by a rotating hollow shaft with an outer diameter of 150mm, an inner diameter of 100mm, rotating at a speed of 3 rps for maximum Shear Stress of 50 MPa
The important parameters to note here is that the shaft is hollow so we need to use equation {8} defining the polar second order of area for a hollow tubular shaft of outer diameter D and inner diameter d
J_tube=π(D^4-d^4 )/32 {8}

We are told outer diameter, D=150mm=0.15m
Inner diameter, d=100mm=0.1m
Using {8} then
J_tube=π(〖0.15〗^4-〖0.10〗^4 )/32=4×〖10〗^(-5) m^4

We can now use equation {3} to work out the maximum Torque, T on the shaft for maximum Shear stress, τ=50MPa=5×〖10〗^7 Pa

T/J_tube =τ/r {3}
T=(τJ_tube)/r
Putting in values noting that the value radius value, r in the above relates to the radius of the outer diameter which is r=D/2=0.15/2=0.075m
Maximum Torque T=(5×〖10〗^7×4×〖10〗^(-5))/0.075=2.67×〖10〗^4 N.m
We now can use equation {1} to determine the maximum power delivered by this shaft where the given angular velocity of shaft is ω=2πN and where N=3 rps
P=ωT {1}
P=2π×3×2.67×〖10〗^4=503.3kW