Question 3.13: A shaft of length L = 1.8 m is subjected to torques T = 5 kN...
A shaft of length L = 1.8 m is subjected to torques T = 5 kN m at either end (Fig. 3-48). Segment AB (L_{1} = 900 mm) is made of brass (G_{b} = 41 GPa) and has a square cross section (a = 75 mm). Segment BC (L_{2} = 900 mm) is made of steel (G_{s} = 74 GPa) and has a circular cross section (d = a = 75 mm). Ignore stress concentrations near B.
(a) Find the maximum shear stress and angle of twist for each segment of the shaft.
(b) Find a new value for the dimension a of bar AB if the maximum shear stress in AB and BC are to be equal.
(c) Repeat part (b) if the angles of twist of segments AB and BC are to be equal.
(d) If dimension a is reset to a = 75 mm and bar BC is now a hollow pipe with outer diameter d_{2} = a, find the inner diameter d_{1} so that the the angles of twist of segments AB and BC are equal.
(a) Maximum shear stress and angles of twist for each segment. Both seg-ments of the shaft have internal torque equal to applied torque T. For square segment AB, we obtain torsion coefficients k_{1} and k_{2} from
Table 3-1,
Dimensionless Coefficients for Rectangular Bars
| b/t | 1.00 | 1.50 | 1.75 | 2.00 | 2.50 | 3.00 | 4 | 6 | 8 | 10 | ∞ |
| k_{1} | 0.208 | 0.231 | 0.239 | 0.246 | 0.258 | 0.267 | 0.282 | 0.298 | 0.307 | 0.312 | 0.333 |
| k_{2} | 0.141 | 0.196 | 0.214 | 0.229 | 0.249 | 0.263 | 0.281 | 0.298 | 0.307 | 0.312 | 0.333 |
then use Eqs. (3-72) \tau_{\max }=\frac{T}{k_{1} b t^{2}} and (3-73) \phi=\frac{T L}{\left(k_{2} b t^{3}\right) G}=\frac{T L}{G J_{r}} to compute maximum shear stress and angle of twist as
\begin{aligned}\phi_{1}=\frac{T L_{1}}{\left(k_{2} b t^{3}\right) G_{b}}=\frac{T L_{1}}{k_{2} a^{4} G_{b}} &=\frac{(5 \mathrm{kN} \cdot \mathrm{m})(900 \mathrm{~mm})}{0.141(75 \mathrm{~mm})^{4}(41 \mathrm{GPa})} \\&=2.46 \times 10^{-2} \text { radians }\end{aligned} (b)
The maximum shear stress in AB occurs at the midpoint of each side of the square cross section. Segment BC is a solid, circular cross section, so we use Eqs. (3-14) \tau_{\max }=\frac{16 T}{\pi d^{3}} and (3-17) \phi=\frac{T L}{G I_{P}} to compute the maximum shear stress and angle of twist for segment BC:
\tau_{\max 2}=\frac{16 T}{\pi d^{3}}=\frac{16(5 \mathrm{kN} \cdot \mathrm{m})}{\pi(75\mathrm{~mm})^{3}}=60.4\mathrm{MPa} (c)
\phi_{2}=\frac{T L_{2}}{G_{s} l_{p}}=\frac{(5 \mathrm{kN} \cdot \mathrm{m})(900 \mathrm{~mm})}{74 \mathrm{GPa}\left[\frac{\pi}{32}(75 \mathrm{~mm})^{4}\right]}=1.958 \times 10^{-2} (d)
Comparing the shear stress and angle of twist values for square segment AB and circular segment BC, we see that the steel pipe BC has 6% greater maximum shear stress but 20% less twist rotation than the brass bar AB.
(b) New value for dimension a of bar AB so that maximum shear stress in AB and BC are equal. We equate expressions for \tau_{\max 1} and \tau_{\max 2} in Eqs. (a) and (c) and solve for the required new value of dimension a of bar AB:
\tau_{\max 1}=\tau_{\max 2} \text { so } \frac{16}{\pi d^{3}}=\frac{1}{k_{1} a_{\text {new }}^{3}} \text { or } a_{\text {new }}=\left(\frac{\pi d^{3}}{16 k_{1}}\right)^{\frac{1}{3}}=73.6 \mathrm{~mm} (e)
The diameter of bar BC remains unchanged at d = 75 mm, so a slight reduction in dimension a for bar AB leads to the same maximum shear stress of 60.4 MPa [Eq.(c)] in the two bar segments.
(c) New value for dimension a of bar AB so that twist rotations in AB and BC are equal. Now, we equate expressions for \phi_{1} and \phi_{2} in Eqs. (b) and (d) and solve for the required new value of dimension a of bar AB:
\begin{gathered}\phi_{1}=\phi_{2} \\\text { so } \frac{L_{1}}{k_{2} a_{\text {new }}^{4} G_{b}}=\frac{L_{2}}{G_{s} I_{p}}\end{gathered} (f)
or a_{\text {new }}=\left[\frac{L_{1}}{L_{2}}\left(\frac{G_{S} I_{p}}{k_{2} G_{b}}\right)\right]^{\frac{1}{4}}=79.4 \mathrm{~mm}
The diameter of bar BC remains unchanged at d = 75 mm, so a slight increase in dimension a for brass bar AB leads to the same twist rotation of 0.01958 radians, as in Eq. (d) in each of the two bar segments.
(d) Change segment BC to hollow pipe; find inner diameter d_{1} so that twist rotations in AB and BC are equal. Side dimension a of square segment AB is equal to 75 mm, and outer diameter d_{2} = 75 mm (Fig. 3-49). Using Eq. (3-19) I_{P}=\frac{\pi}{2}\left(r_{2}^{4}-r_{1}^{4}\right)=\frac{\pi}{32}\left(d_{2}^{4}-d_{1}^{4}\right) for the polar moment of inertia of segment BC, we get for twist angle \phi_{2} :
\phi_{2}=\frac{T L_{2}}{G_{s}\left[\frac{\pi}{32}\left(d_{2}^{4}-d_{1}^{4}\right)\right]} (g)
Once again, we equate expressions for \phi_{1} and \phi_{2} but now using Eqs. (b) and (g). Solving for d_{1}, we get the expression:
So the square, solid brass pipe AB (a × a, a = 75 mm) and hollow steel pipe BC (d_{2} = 75 mm, d_{1} = 50.4 mm) are each 900 mm in length and have the same twist rotation (0.0246 radians) due to applied torque T. However, additional calculations will show that the maximum shear stress in segment BC is now increased from 60.4 MPa [Eq.(c)] to 75.9 MPa by using a hollow rather than solid bar for BC.
Note that by deriving the formula for inner diameter d_{1} in Eq. (h) (rather than finding a numerical solution alone), we can also investigate other solutions of possible interest using different values of the key vari-ables. For example, if bar AB is increased in length to L_{1} = 1100 mm, inner diameter d_{1} for BC can be increased to 57.6 mm and the angles of twist for AB and BC will be the same.
