Hollow shaft power transmission

[Ipodbob]

Hello all this is my first post so i hope i can give you the relevant data in the correct format, also thank you for any assistance :)

I am to find the OD of a hollow shaft when the shafts ID is 0.75 of the OD dimension, the shaft transmits 3MW at 200Rpm. The Shear stress max is 55 MN/m² and the modulus of rigidity is 80 GN/m². The OD can be no larger than 270mm / 0.27m.

Power = 2.Pi.N.T/60 n = 200rpm

Torque = (p.60)/(2.Pi.N)

Torque applied = 143.2 KNm

J (second moment of area) = Pi(D⁴-d⁴)/32

I now know i can use Ta/J = τ/r and transpose that,

Ta/J = 2τ/OD

Subbing J so i can then transpose the equation to find the D(OD), by changing out the d(ID) and r for OD parameters.

Ta/[Pi/32(D⁴-(0.75.D)⁴)] = 2τ/D

I need to transpose this equation for D (the outer diameter) but I'm stuck, i hope this makes sense. The problem is i have calculated all the the answer assuming the max size of 0.27 which is strong enough to transmit the torque but i realize now i need to find the outer diameter with the figures i have given above.

Regards

 

[SteamKing]

Well, you know the applied torque Ta and the max. shear stress τ from your calculations and the OP.

you can substitute these values into your equation:

Ta/[Pi/32(D$^{4}$-(0.75.D)$^{4}$)] = 2τ/D

and solve for D. All it requires is a little algebra.

 

[Ipodbob]

I got this far i am not sure how to move the (0.75)⁴ away so i can leave D⁴ - D⁴ on that side, leaving me with just D on the other side. It seems i need to catch up on some basic algebra.

https://scontent-a-cdg.xx.fbcdn.net/hphotos-prn2/q80/s720x720/1480744_581451648571507_706677602_n.jpg

 

[SteamKing]

Remember, 0.75^4 is a number. You can evaluate numbers. Cross-multiply your expression above to make an algebraic equation in the unknown variable D.

 

[rezeew]

Thank you for sharing your question with us. I would approach this problem by first considering the basic principles of power transmission and stress analysis. The power transmitted by a shaft is directly proportional to its torque and rotational speed, as shown in the equation you provided. However, in order to calculate the outer diameter of the hollow shaft, we also need to consider the stress and strength of the material.

Based on the given information, we can calculate the maximum torque applied to the shaft as 143.2 kNm. We can then use this value to calculate the maximum shear stress using the equation τ = Tc/J, where Tc is the torque and J is the second moment of area. This gives us a maximum shear stress of 55 MN/m2.

Next, we need to consider the strength of the material, which is given by the modulus of rigidity. We can use this value to calculate the maximum allowable shear stress for the material using the equation τ_max = G/2, where G is the modulus of rigidity. In this case, the maximum allowable shear stress is 40 MN/m2.

Now, we can use the maximum allowable shear stress to calculate the maximum allowable outer diameter of the hollow shaft. We can rearrange the equation for shear stress to solve for the outer diameter, which gives us D = (2Tc/τ_max)^1/4. Substituting in the values for Tc and τ_max, we get D = (2*143.2 kNm/40 MN/m2)^1/4 = 0.38 m or 380 mm.

Therefore, the outer diameter of the hollow shaft should not exceed 380 mm in order to meet the given specifications and ensure that the material does not exceed its maximum allowable shear stress. I hope this helps to clarify the problem and provide a solution. Please let me know if you have any further questions or need any additional assistance. Thank you.