Question

A solid titanium alloy [G 114 GPa] shaft that is 720 mm long will be subjected to a pure torque of T 155 N m. Determine the minimum diameter required if the shear stress must not exceed 150 MPa and the angle of twist must not exceed 7?. Report both the maximum shear stress ? and the angle of twist ? at this minimum diameter. ?Part 1 where d is the shaft diameter. The polar moment of inertia is also a function of d. Find Incorrect. Consider the elastic tors on formula. The maximum shear stress occurs at the radial location ? = (d 2 the value of d for which the maximum shear stress in the shaft equals 150 MPa. Based only on the requirement that the shear stress must not exceed 150 MPa, what is the minimum diameter of the shaft? dr 16.1763 the tolerance is +/-2% Click if you would like to Show Work for this question: Open Show Work Attempts: 1 of 3 used SAVE FOR LATER SUBMIT ANSWER Part 2 Based only on the requirement that the angle of twist must not exceed 7°, what is the minimum diameter of the shaft?

Answer 1

Answer:

Part 1: The diameter of the shaft so that the shear stress is not more than 150 MPa is 17.3 mm.

Part 2: The diameter of the shaft so that the twist angle  is not more than 7° is 16.9 mm.

Explanation:

Part 1

The formula is given as

$\dfrac{T}{J}=\dfrac{\tau}{R}$

Here T is the torque which is given as 155 Nm

J is the rotational inertia which is given as $\dfrac{\pi d^4}{32}$

τ is the shear stress which is given as 150 MPa

R is the radius which is given as d/2 so the equation becomes

$\dfrac{T}{J}=\dfrac{\tau}{R}\\\dfrac{155}{\pi d^4/32}=\dfrac{150 \times 10^6}{d/2}\\\dfrac{155 \times 32}{\pi d^4}=\dfrac{300 \times 10^6}{d}\\\dfrac{1578.82}{d^4}=\dfrac{300 \times 10^6}{d}\\d^3=\dfrac{1578.82}{300 \times 10^6}\\d^3=5.26 \times 10^{-6}\\d=0.0173 m \approx 17.3 mm$

So the diameter of the shaft so that the shear stress is not more than 150 MPa is 17.3 mm.

Part 2

The formula is given as

$\dfrac{T}{J}=\dfrac{G\theta}{L}$

Here T is the torque which is given as 155 Nm

J is the rotational inertia which is given as $\dfrac{\pi d^4}{32}$

G is the torsional modulus  which is given as 114 GPa

L  is the length which is given as 720 mm=0.720m

θ is the twist angle which is given as 7° this is converted to radian as

$\theta=\dfrac{7*\pi}{180}\\\theta=0.122 rad$

so the equation becomes

$\dfrac{T}{J}=\dfrac{G\theta}{L}\\\dfrac{155}{\pi d^4/32}=\dfrac{114 \times 10^9\times 0.122}{0.720}\\\dfrac{1578.81}{d^4}=1.93\times 10^{10}\\d^4=\dfrac{1578.81}{1.93\times 10^{10}}\\d=(\dfrac{1578.81}{1.93\times 10^{10}})^{1/4}\\d=0.0169 m \approx 16.9mm$

So the diameter of the shaft so that the twist angle  is not more than 7° is 16.9 mm.