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Ideal solutions concerning rotating solid and hollow shafts

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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

© BrainMass Inc. brainmass.com March 22, 2019, 3:26 am ad1c9bdddf
https://brainmass.com/engineering/mechanical-engineering/ideal-solutions-concerning-rotating-solid-hollow-shafts-615709

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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 ...

Solution Summary

Various parameters are determined such as power delivery, shear stress for both solid and hollow rotating shafts given input parameters such as rotational speed and radii of the shafts in question

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