Strength Of Materials Miscellaneous
- Two shafts AB and BC, of equal length and diameters d and 2d, are made of the same material. They are joined at B through a shaft coupling, while the ends A and C are built-in (cantilevered). A twisting moment T is applied to the coupling. If TA and TC represent the twisting moments at the ends A and C, respectively, then
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θA = θC
TC = 16TA.Correct Option: C
θA = θC
TC = 16TA.
- A torque of 1.0 Nm is transmitted through a stepped shaft as shown in figure. The torsional stiffness's of individual sections of lengths M N, NO and OP are 20 Nm/rad, 30 Nm/rad and 60 Nm/rad respectively. The angular deflection between the ends M and P of the shaft is
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T = 20 Nm/rad θMN
T = 1 N.mθMN = 1 rad 20 θNO = 1 rad 30 θOP = 1 rad 60
The part MN, NO & OP are connected in series
θNP = θMN + θNO + θOP= 1 + 1 + 1 = 0.1 rad. 20 30 60
θ = 0.1 radCorrect Option: B
T = 20 Nm/rad θMN
T = 1 N.mθMN = 1 rad 20 θNO = 1 rad 30 θOP = 1 rad 60
The part MN, NO & OP are connected in series
θNP = θMN + θNO + θOP= 1 + 1 + 1 = 0.1 rad. 20 30 60
θ = 0.1 rad
- Two shafts A and B are made of the same material. The diameter of shaft B is twice that of shaft A. The ratio of power which can be transmitted by shaft A to that of shaft B is (If maximum shear stress remains the same)
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dB = 2dA.
τB = τATr = Tr 2 A 2 B TA = JA rB TB JB rA 2 = 1 = 1 24 23 8 Correct Option: C
dB = 2dA.
τB = τATr = Tr 2 A 2 B TA = JA rB TB JB rA 2 = 1 = 1 24 23 8
- The compound shaft shown is built-in at the two ends. It is subjected to a twisting moment T at the middle. What is the ratio of the reaction torques T1, and T2 at the ends?
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θ 1 = θ 2T1I = T2I J1G J2G T1 = 1 T2 16 Correct Option: A
θ 1 = θ 2T1I = T2I J1G J2G T1 = 1 T2 16
- A hollow circular shaft of inner radius 10 mm outer radius 20 mm and length 1 m is to be used as a torsional spring. If the shear modulus of the material of the shaft is 150 GPa, the torsional stiffness of the shaft (in kN-m/rad) is ______ (correct to two decimal places).
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35.34 KN- m/rad
G = 1504 PaT = GJ θ I = I = 150 × 103 × π × (0.044 - 0.024) θ 1 32 I = 35.34 θ Correct Option: A
35.34 KN- m/rad
G = 1504 PaT = GJ θ I = I = 150 × 103 × π × (0.044 - 0.024) θ 1 32 I = 35.34 θ