Question 4.3: A hollow LEHI cylinder has an inner radius a=15 mm, an outer...

From Eqs. (4.37) and (4.40),

                          \sigma _{1}=+\frac{Tr}{J}    and \sigma _{2}=-\frac{Tr}{J},                                            (4.37)

                          \sigma ^{\prime }_{z\theta })_{\max /\min }=\sqrt{\left(\frac{\sigma _{\theta \theta }-\sigma _{zz}}{2} \right)^{2}+\sigma ^{2}_{z\theta } }\rightarrow \sigma ^{\prime }_{z\theta })_{\max /\min }=\pm \frac{Tr}{J},                 (4.40)

                     \sigma ^{\prime }_{z\theta })_{\max }=\frac{Tr}{J}           and           \sigma ^{\prime }_{zz })_{\max }=\frac{Tr}{J},

where

Given

a=15 mm=0.015 m

c=20 mm=0.02 m

L=0.5 m

T=600 Nm

\Delta \theta =3.57^{\circ }=0.0623 rad

first calculate \sigma ^{\prime }_{z\theta })_{\max } and \sigma ^{\prime }_{zz })_{\max }:

     \sigma ^{\prime }_{z\theta })_{\max } and  \sigma ^{\prime }_{zz })_{\max }=\frac{Tc}{\pi (c^{4}-a^{4})/2}=\frac{2(600Nm)(0.02m)}{\pi \left[(0.02 m)^{4}-(0.015 m)^{4}\right] }

Second, calculate G. Assuming \gamma_{c}\ll 1, we have

Hence, \sigma _{z\theta }(r=c)=2G\varepsilon _{z\theta }(r=c)=G\gamma _{c} implies that

Third, calculate \varepsilon ^{\prime }_{z\theta })_{\max } and \varepsilon _{zz} using Hooke’s law:

and, finally,

Thus, the shaft does not extend and the maximum   shear strain is indeed small, consistent with our small-strain assumption in the derivation of the governing  equations and our use of a LEHI descriptor of the  behavior. Also note that in reference to Table A2.1, a shear modulus G \sim 28 GPa suggests that the material is a 2024-T4 aluminum. The yield strength of this  erial is \sim 170 MPa in shear; hence, we would not expect that yield would have occurred.