Question

Two solid steel shafts are fitted with flanges that are then connected by bolts as shown. The bolts are slightly undersized and permit a 1.5-degree rotation of one flange with respect to the other before the flanges begin to rotate as a single unit. Knowing that G=11.2x10^6 psi, determine the maximum shearing stress in each shaft when the torque of T of magnitude 420kip.ft is applied to the flange indicated. Determine the torque T is applied to flange C.

Answer 1

Given that,

\(G=11.2 \times 10^{6} \mathrm{psi}=11.2 \times 10^{6} \mathrm{lb} / \mathrm{in}^{2}\)

\(T=350 \mathrm{lb} \mathrm{ft}=4200 \mathrm{lb} \cdot \mathrm{in}\)

\(\phi=\phi_{B}=\phi_{c}=1.5^{\circ}=0.026179 \mathrm{rad}\)

\(L_{A B}=2 \mathrm{ft}=24 \mathrm{in}\)

\(L_{C D}=3 \mathrm{ft}=36 \mathrm{in}\)

\(D_{A B}=1.25 \mathrm{in}\)

\(D_{C D}=1.5\) in

\(r_{A B}=0.625\) in

$$ r_{c D}=0.75 \text { in } $$

\(T=T_{A B}+T_{C D}\)

$$ \begin{array}{l} J_{A B}=\frac{\pi}{2} \times 0.625^{4}=0.2396 \mathrm{in}^{4} \\ J_{C D}=\frac{\pi}{2} \times 0.75^{4}=0.497 \mathrm{in}^{4} \end{array}  $$

\(T_{\Delta}=\frac{G_{A B} J_{A B}}{L_{A B}} \phi_{B}\)

\(T_{A}=\frac{11.2 \times 10^{6} \times 0.239}{24} \phi=111852.76 \phi\)

\(T_{C D}=\frac{G_{C D} J_{C D}}{L_{C D}} \phi\)

Clearance rotation for flange \(B=\phi=1.5^{\circ}=0.026179 \mathrm{rad}\) T or que to rem ove clearance \(T_{A}=111852.76 \times 0.026179=2928.2981 \mathrm{~b}\).in T or que to remove clearance \(T_{C D}=154622.22 \times 0.026179=4047.851 \mathrm{b.in}\)

Torgue \(T^{\prime}\) to cause additional rotation \(\phi\) :

\(T^{\prime}=T-T_{C D}=4200-4047.85=152.151 \mathrm{~b} . \mathrm{in}\)

\(T^{\prime}=T_{A B}+T_{C D}\)

\(152.15=(111852.76+154622.22) \varphi\)

$$ \begin{array}{l} \varphi=5.709 \times 10^{-4} ! \\ T_{A B}=111852.76 \times 5.109 \times 10^{-4}=63.851 \mathrm{b.in} \end{array} $$

\(T_{C D}=154622.2 \times 5.109 \times 10^{-4}=88.27 \mathrm{lb} \cdot \mathrm{in}\)

Maximum shearing stress in \(A B\)

\(\tau_{A B}=\frac{T_{A B} r_{A B}}{J_{A B}}=\frac{63.86 \times 0.625}{0.2396}=0.166 \mathrm{ksi}\)

Maximum shearing stress in \(C D\)

\(\tau_{\mathrm{CD}}=\frac{T_{C D} r_{C D}}{J_{C D}}=\frac{(4047.85+88.27) \times 0.75}{0.497}=6.24 \mathrm{ksi}\)