Question

The maximum stress in a section of a circular tube subject to a torque is τmax = 22 MPa . If the inner diameter is Di = 3.75 cm and the outer diameter is Do = 5.5 cm , what is the torque on the section? Express your answer with appropriate units to three significant figures. View Available Hint(s)

Answer 1

Answer:

The torque on the section is 563.373 Nm

Explanation:

The torque in the section can be calculated through torsion formula. The torsion formula is given as follows:

Torque = T = τJ/ r

where,

r = maximum radius = 5.5 cm/2 = 2.75 cm = 2.75 x 10^-2 m

τ = maximum shear stress = 22 MPa = 22 x 10^6 Pa

J = Polar moment of inertial = π/2 (ro^4 - ri^4)

J = π/2 (0.0275^4 - 0.01875^4) m^4 = 7.042 x 10^-7 m^4

Therefore:

T = (7.042 x 10^-7 m^4)(22 x 10^6 Pa)/(2.75 x 10^-2 m)

T = 563.373 Nm