In the figure, shaft AB tr mits power to shaft CD through a set of bevel gears contacting at point The contact force at E on the gear of shaft CD is determined to be (FE)CD = –92.8i – 362.8j + 808.0k lbf.

Question

In the figure, shaft [latex]A B [/latex]tr mits power to shaft [latex]C D [/latex]through a set of bevel gears contacting at point [latex]E [/latex]. The contact force at [latex]E [/latex]on the gear of shaft [latex]C D [/latex]is determined to be [latex]\left(\mathbf{F}_{E}\right)_{C D}=-92.8 \mathbf{i}-362.8 \mathbf{j}+ [/latex][latex]808.0 \mathrm{k} \mathrm{lbf} [/latex]. For shaft [latex]C D:(a) [/latex]draw a free-body diagram and determine the reactions at [latex]C [/latex]and [latex]D [/latex].

assuming simple supports (assume also that bearing [latex]C [/latex]carries the thrust load), [latex](b) [/latex]draw the shearforce and bending-moment diagrams, [latex](c) [/latex]for the critical stress element, determine the torsional shear stress, the bending stress, and the axial stress, and [latex](d) [/latex]for the critical stress element, determine the principal stresses and the maximum shear stress.

Accepted Answer

(a)

[latex]\begin{aligned}&\left(\Sigma M_{D} ight)_{z}=6.13 C_{x}-3.8(92.8)-3.88(362.8)=0 \&C_{x}=287.2 ext { lbf } ext { } \&\left(\Sigma M_{C} ight)_{z}=6.13 D_{x}+2.33(92.8)-3.88(362.8)=0 \&D_{x}=194.4 ext { lbf } ext { } \&\left(\Sigma M_{D} ight)_{x}=0 \Rightarrow C_{z}=\frac{3.8}{6.13}(808)=500.9 \mathrm{lbf} \&\left(\Sigma M_{C} ight)_{x}=0 \Rightarrow D_{z}=\frac{2.33}{6.13}(808)=307.1 \mathrm{lbf}\end{aligned} [/latex]

[latex] ext { (b) For } D Q C ext {, let } x^{\prime}, y^{\prime}, z^{\prime} ext { correspond to the original }-y, x, z ext { axes. } [/latex]

(c) The critical stress element is just to the right of [latex]Q [/latex] where the bending moment in both planes is maximum, and where the torsional and axial loads exist.

[latex]\begin{aligned}&T=808(3.88)=3135 \mathrm{lbf} \cdot \mathrm{in} \&M=\sqrt{669.2^{2}+1167^{2}}=1345 \mathrm{lbf} \cdot \mathrm{in} \& au=\frac{16 T}{\pi d^{3}}=\frac{16(3135)}{\pi\left(1.13^{3} ight)}=11070 ext { psi } \quad ext { } \&\sigma_{b}=\pm \frac{32 M}{\pi d^{3}}=\pm \frac{32(1345)}{\pi\left(1.13^{3} ight)}=\pm 9495 \mathrm{psi} \quad ext { } \&\sigma_{a}=-\frac{F}{A}=-\frac{362.8}{(\pi / 4)\left(1.13^{2} ight)}=-362 \mathrm{psi} \quad ext { }\end{aligned} [/latex]

(d) The critical stress element will be where the bending stress and axial stress are both in compression.

[latex]\begin{aligned}&\sigma_{\max }=-9495-362=-9857 \mathrm{psi} \& au_{\max }=\sqrt{\left(\frac{-9857}{2} ight)^{2}+11070^{2}}=12118 \mathrm{psi}=12.1 \mathrm{kpsi} \quad ext { . } \&\sigma_{1}, \sigma_{2}=\frac{-9857}{2} \pm \sqrt{\left(\frac{-9857}{2} ight)^{2}+11070^{2}} \&\sigma_{1}=7189 \mathrm{psi}=7.19 \mathrm{kpsi} \quad ext { . } \&\sigma_{2}=-17046 \mathrm{psi}=-17.0 \mathrm{kpsi} \quad ext { }\end{aligned} [/latex]

Question 3.74: In the figure, shaft AB tr mits power to shaft CD through a ...

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