Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

Tip our Team

Our Website is free to use.
To help us grow, you can support our team with a Small Tip.

Holooly Tables

All the data tables that you may search for.

Holooly Help Desk

Need Help? We got you covered.

Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Products

Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Holooly Help Desk

Need Help? We got you covered.

Q. 9.19

In order to measure torque applied on a shaft, we often use strain gauge to measure strain and then calculate the torque applied. Figure 9.52 shows a circular solid shaft of 96 mm diameter on which a single strain gauge is cemented at an angle of 20° with a line parallel to the axis of the shaft. If G = 27 GPa for the shaft material, calculate torque T for the gauge reading = 400μ.

Verified Solution

Let us draw the free-body diagram of the shaft as shown in Figure 9.53:

In the figure, we isolate a differential element on the shaft outer surface aligned along the axis of the shaft [refer to Figure 9.53(a)]. Clearly due torque T, the element will be in a state of pure shear and if the element is rotated along 20°, it will be subjected to the stress state as shown in Figure 9.53(b).
From the Mohr’s circle of stresses, we can say that

$\sigma=\tau \sin 40^{\circ} \quad \text { where } \quad \tau=\frac{16 T}{\pi d^3}$

Obviously, $∈_{x^{\prime} x^{\prime}}=∈_1$ is the gauge reading and it is given by

$∈_{x^{\prime} x^{\prime}}=∈_1=\frac{1}{E}\left[\sigma_{x^{\prime} x^{\prime}}-ν\left(\sigma_{z^{\prime} z^{\prime}}+\sigma_{y^{\prime} y^{\prime}}\right)\right]$

For the plane-stress condition, $\sigma_{z^{\prime} z^{\prime}}$ = out of plane stress = 0. Therefore,

$∈_{x^{\prime} x^{\prime}}=\frac{1}{E}(\sigma+ν \sigma) \quad\left(\text { as } \sigma_{y^{\prime} y^{\prime}}=-\sigma\right)$

or          $∈_{x^{\prime} x^{\prime}}=\frac{1+ν}{E} \sigma=\frac{\sigma}{2 G} \quad\left[\text { as } G=\frac{E}{2(1+ν)}\right]$

hence,

$\sigma=2 G \in_{x^{\prime} x^{\prime}}=2 G ∈_1$

or        $\left\lgroup \frac{16 T}{\pi d^3}\right\rgroup \sin 40^{\circ}=2 G \in_1 \quad\left[\text { as } \sigma=\tau \sin 45^{\circ} \text { and } \tau=16 T / \pi d^3\right]$

Therefore,

$T=\left\lgroup \frac{\pi d^3}{8} \right\rgroup \frac{G\in_1}{\sin 40^{\circ}}$

$=\frac{\pi(0.096)^3}{8} \frac{(27)\left(10^9\right)(400)\left(10^{-6}\right)}{\sin 40^{\circ}} N m$

or      T = 5.8375 kNm

Therefore, the required torque applied on the shaft is 5.84 kN m.