Problems on Torsion
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The document contains 8 questions related to determining the diameter of solid and hollow shafts based on transmitted power, torque, maximum shear stress, and angle of twist specifications. The questions involve calculating shaft diameters, transmitted torque values, shear stresses, and comparing weights of solid versus hollow shaft designs.
![Q2. A 500 mm long solid steel [G = 80 GPa] shaft is being designed to transmit a torque T =
20 Nm. The maximum shear stress in the shaft must not exceed 70 MPa, and the angle of
twist must not exceed 3° in the 500 mm length. Determine the minimum diameter d
required for the shaft.
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Problems on Torsion
- 2. Q1. Determine the diameter of a solid shaft which will transmit 300 kW at 250 r.p.m; The maximum shear stress should not exceed 30 N/mm2 angle of twist should not be more than 1o in a shaft length of 2 m. Take modulus of rigidity = 1 x 105 N/mm2.
- 5. So, D= 107.5 mm is not recommended
- 6. Q2. A 500 mm long solid steel [G = 80 GPa] shaft is being designed to transmit a torque T = 20 Nm. The maximum shear stress in the shaft must not exceed 70 MPa, and the angle of twist must not exceed 3° in the 500 mm length. Determine the minimum diameter d required for the shaft. 𝑻 𝑱 = 𝝉 𝑹
- 8. Q3. Determine the diameter of solid steel shaft which will transmit 90 kW at 160 r.p.m; Also determine the length of the shaft if the twist must not exceed 1o over the entire length. The maximum shear stress is limited to 60 N/mm2. Take the value of modulus of rigidity = 8 x 104 N/mm2
- 9. Q4. A hollow cylindrical steel shaft is 1.5 m long and has inner and outer diameters respectively equal to 40 and 60 mm (a) What is the largest torque that can be applied to the shaft if the shearing stress is not to exceed 120 MPa? (b) What is the corresponding minimum value of the shearing stress in the shaft? A hollow cylindrical steel shaft: length = 1.5 m inner diameter Di= 40 mm outer diameters Do= 60 mm (a)What is the largest torque= ? shearing stress is not to exceed 120 MPa
- 10. (a) What is the largest torque= ? shearing stress is not to exceed 120 MPa Simple Torsion Equation => 𝑻 𝑱 = 𝝉 𝑹 = 𝑮𝜽 𝑳 𝑻 𝑱 = 𝝉 𝑹 => T= Zp 𝝉 T= 𝜋 16 𝐷0 4−𝐷ⅈ4 𝐷0 𝝉 = 𝜋 16 604 −404 60 ∗ 120 = 4084070.45 N-mm =4.08 kN-m (b) What is the corresponding minimum value of the shearing stress in the shaft? 𝒒 𝒓 = 𝝉 𝑹 q= 𝝉 𝑹 . r q= 𝟏𝟐𝟎 𝟑𝟎 ∗ 𝟐𝟎 q = 80 MPa
- 11. Q5. Two shafts of the same material and of same lengths are subjected to the same torque, if the first shaft is of a solid circular section and the second shaft is of hollow circular section, whose internal diameter is 2/3 of the outside diameter and the maximum shear stress developed in each shaft is the same, compare the weights of the shafts.
- 15. Q6. A solid circular shaft and a hollow circular shaft whose inside diameter is ( 𝟑 𝟒 ) of the outside diameter, are of the same material, of equal lengths and are required to transmit a given torque. Compare the weights of these two shafts if the maximum shear stress developed in the two shafts are equal.
- 18. Q7. A solid cylindrical shaft is to transmit 300 kW power at 100 r.p.m. (a) If the shear stress is not to exceed 80 N/mm2, find its diameter. (b) (b) What percent saving in weight would be obtained if this shaft is replaced by a hollow one whose internal diameter equals to 0.6 of the external diameter, the length, the material and maximum shear stress being the same
- 21. Q8. The final drive to a conveyor that feeds coal to a railroad car is a solid shaft loaded in pure torsion and carrying 800 Nm of torque i) if the acceptable shear stress is 39MPa, find its diameter ii) What percentage saving in weight would be obtained if this shaft is replaced by a hollow shaft whose outside diameter of 60 mm is available in the same material as specified for the solid shaft. Compute what maximum inside diameter the hollow shaft can have that would result in the same stress in the solid shaft. The length , material are being the same.
- 22. Torque T= 800 Nm = 8*105 N-mm Acceptable shear stress 𝜏= 39MPa i) If Solid shaft finding diameter: 𝑻 𝑱 = 𝝉 𝑹 T= Zp 𝝉 Zp = 𝜋 d3 16 d3 = ( 16𝑇 𝜋 𝜏 ) d3 = ( 16 ∗8 ∗105 𝜋 ∗39 ) => d= 47.09 mm
Editor's Notes
- #3: The elastic torsion formula and the angle-of-twist equation will be algebraically manipulated to solve for the minimum diameter required to satisfy each consideration. The larger of the two diameters will dictate the minimum diameter d that can be used for the shaft.
- #4: The elastic torsion formula and the angle-of-twist equation will be algebraically manipulated to solve for the minimum diameter required to satisfy each consideration. The larger of the two diameters will dictate the minimum diameter d that can be used for the shaft.
- #7: The elastic torsion formula and the angle-of-twist equation will be algebraically manipulated to solve for the minimum diameter required to satisfy each consideration. The larger of the two diameters will dictate the minimum diameter d that can be used for the shaft.
- #8: The elastic torsion formula and the angle-of-twist equation will be algebraically manipulated to solve for the minimum diameter required to satisfy each consideration. The larger of the two diameters will dictate the minimum diameter d that can be used for the shaft.
- #9: D=76.8 mm , L=893.6mm
- #22: In many design situations, economy of material usage is a major criterion of performance for a product. In aerospace applications, every reduction in the mass of the aircraft or space vehicle allows increased payload. Automobiles achieve higher fuel economy when they are lighter. Also, since raw materials are purchased on a price per unit mass basis, a lighter part generally costs less. Providing economy of material usage for load-carrying members requires that all the material in the member be stressed to a level approaching the safe design stress. Then every portion is carrying its share of the load.