Question 3.13: A shaft with a length L = 1.8 m is subjected to torques T = ......

Use a four-step problem-solving approach. Combine steps as needed for an efficient solution.
Part (a): Maximum shear stress and angles of twist for each segment.
1,2. Conceptualize, Categorize: Both segments of the shaft have internal torque equal to the applied torque T. For square segment AB, obtain torsion coefficients k_1 and k_2 from Table 3-1.
3. Analyze: Use Eqs. (3-72)

\tau_{max}=\frac{T}{k_1bt^2}                               (3-72)

and (3-73)

\phi =\frac{TL}{(k_2bt^3)G}=\frac{TL}{GJ_r}               (3-73)

to compute the maximum shear stress and angle of twist as

\tau_{\max 1}=\frac{T}{k_1 b t^2}=\frac{T}{k_1 a^3}=\frac{(5 \ kN \cdot m )}{0.208(75 \ mm )^3}=57 \ MPa                          \hookleftarrow (a)

\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 \ kN \cdot m )(900 \ mm )}{0.141(75 \ mm )^4(41 \ GPa )}  \\ =2.46 \times 10^{-2}   radians                                       \hookleftarrow (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 use Eqs. (3-14)

\tau_{max}=\frac{16T}{\pi d^3}                           (3-14)

and (3-17)

\phi =\frac{TL}{GI_p}                                              (3-17)

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 \ kN \cdot m )}{\pi(75 \ mm )^3}=60.4 \ MPa                                           \hookleftarrow (c)

\phi_2=\frac{T L_2}{G_s I_p}=\frac{(5 \ kN \cdot m )(900 \ mm )}{74 \ GPa \left[\frac{\pi}{32}(75 \ mm )^4\right]}=1.958 \times 10^{-2}                              \hookleftarrow (d)

4. Finalize: Compare the shear stress and angle of twist values for square segment AB and circular segment BC. Steel pipe BC has 6% greater maximum shear stress but 20% less twist rotation than the brass bar AB.
Part (b): New value for dimension a of bar AB so that maximum shear stresses in AB and BC are equal.
1,2. Conceptualize, Categorize: 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.
3. Analyze:

\tau_{\max 1} = \tau_{\max 2}     so    \frac{16}{\pi d^3}=\frac{1}{k_1a^3_{new}}      or       a_{new} =\left\lgroup\frac{\pi d^3}{16\ k_1}\right\rgroup ^{\frac{1}{3} }=73.6 \ mm    \hookleftarrow (e)

4. Finalize: 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.
Part (c): New value for dimension a of bar AB so that twist rotations in AB and BC are equal.
1, 2. Conceptualize, Categorize: Now, 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.
3. Analyze:

so

or                                                                                                              (f)

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 \ mm       \hookleftarrow

4. Finalize: 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.
Part (d): Change segment BC to hollow pipe; find inner diameter d _{1} so that twist rotations in AB and BC are equal.
1,2. Conceptualize, Categorize: 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} (r^4_2-r^4_1) =\frac{\pi }{32} (d^4_2-d^4_1)                                            (3-19)

for the polar moment of inertia of segment BC, twist angle \phi _{2} is

\phi_2=\frac{T L_2}{G_s\left[\frac{\pi}{32}\left(d_2^4-d_1^4\right)\right]}                                                (g)

3. Analyze: Once again, equate expressions for \phi _{1} and \phi _{2} but now use Eqs. (b) and (g). Solve for d _{1} to get

4. Finalize: 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), you can also investigate other solutions of possible interest using different values of the key variables. 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.