A transmission shaft carries a pulley midway between the two bearings. The bending moment at the pulley varies from 200 N-m to 600 N-m, as the torsional moment in the shaft varies from 70 N-m to 200 N-m. The frequencies of variation of bending and torsional moments are equal to the shaft speed.

Question

A transmission shaft carries a pulley midway between the two bearings. The bending moment at the pulley varies from 200 N-m to 600 N-m, as the torsional moment in the shaft varies from 70 N-m to 200 N-m. The frequencies of variation of bending and torsional moments are equal to the shaft speed. The shaft is made of steel FeE 400 [latex] \left(S_{u t}=540 N / mm ^{2} ext { and } S_{y t}= ight. 400 N / mm ^{2} ext { ). } [/latex]. The corrected endurance limit of the shaft is 200 N/mm2. Determine the diameter of the shaft using a factor of safety of 2.

Accepted Answer

[latex] ext { Given }\left(M_{t} ight)_{\max }=200 N - m [/latex].

[latex] \left(M_{t} ight)_{\min }=70 N - m \quad\left(M_{b} ight)_{\max }=600 N - m [/latex].

[latex] \left(M_{b} ight)_{\min .}=200 N - m \quad S_{u t}=540 N / mm ^{2} [/latex].

[latex] S_{y t}=400 N / mm ^{2} \quad S_{e}=200 N / mm ^{2}(f s)=2 [/latex].

Step I Mean and amplitude stresses

[latex] \left(M_{b} ight)_{m}=\frac{1}{2}\left[\left(M_{b} ight)_{\max }+\left(M_{b} ight)_{\min } ight] [/latex].

[latex] =\frac{1}{2}[600+200]=400 N - m [/latex].

[latex] \left(M_{b} ight)_{a}=\frac{1}{2}\left[\left(M_{b} ight)_{\max .}-\left(M_{b} ight)_{\min .} ight] [/latex].

[latex] =\frac{1}{2}[600-200]=200 N - m [/latex].

[latex] \left(M_{t} ight)_{m}=\frac{1}{2}\left[\left(M_{t} ight)_{\max }+\left(M_{t} ight)_{\min } ight] [/latex].

[latex] =\frac{1}{2}[200+70]=135 N - m [/latex].

[latex] \left(M_{t} ight)_{a}=\frac{1}{2}\left[\left(M_{t} ight)_{\max }-\left(M_{t} ight)_{\min } ight] [/latex].

[latex] =\frac{1}{2}[200-70]=65 N - m [/latex].

[latex] \sigma_{x m}=\frac{32\left(M_{b} ight)_{m}}{\pi d^{3}}=\frac{32\left(400 imes 10^{3} ight)}{\pi d^{3}} [/latex].

[latex] =\left(\frac{4074.37 imes 10^{3}}{d^{3}} ight) N / mm ^{2} [/latex].

[latex] \sigma_{x a}=\frac{32\left(M_{b} ight)_{a}}{\pi d^{3}}=\frac{32\left(200 imes 10^{3} ight)}{\pi d^{3}} [/latex].

[latex] au_{x y m}=\frac{16\left(M_{t} ight)_{m}}{\pi d^{3}}=\frac{16\left(135 imes 10^{3} ight)}{\pi d^{3}} [/latex].

[latex] =\left(\frac{687.55 imes 10^{3}}{d^{3}} ight) N / mm ^{2} [/latex].

[latex] au_{x y a}=\frac{16\left(M_{t} ight)_{a}}{\pi d^{3}}=\frac{16\left(65 imes 10^{3} ight)}{\pi d^{3}} [/latex].

[latex] =\left(\frac{331.04 imes 10^{3}}{d^{3}} ight) N / mm ^{2} [/latex].

[latex] \sigma_{m}=\sqrt{\sigma_{x m}^{2}+3 au_{x y m}^{2}} [/latex].

[latex] =\sqrt{\left(\frac{4074.37 imes 10^{3}}{d^{3}} ight)^{2}+3\left(\frac{687.55 imes 10^{3}}{d^{3}} ight)^{2}} [/latex].

[latex] =\left(\frac{4244.84 imes 10^{3}}{d^{3}} ight) N / mm ^{2} [/latex].

[latex] \sigma_{a}=\sqrt{\sigma_{x a}^{2}+3 au_{x y a}^{2}} [/latex].

[latex] =\sqrt{\left(\frac{2037.18 imes 10^{3}}{d^{3}} ight)^{2}+3\left(\frac{331.04 imes 10^{3}}{d^{3}} ight)^{2}} [/latex].

[latex] =\left(\frac{2116.33 imes 10^{3}}{d^{3}} ight) N / mm ^{2} [/latex].

Step II Construction of modified Goodman diagram

[latex] an heta=\frac{\sigma_{a}}{\sigma_{m}}=\frac{2116.33}{4244.84}=0.4986 ext { or } heta=26.5^{\circ} [/latex].

The modified Goodman diagram for this example is shown in Fig. 5.56.

Step III Permissible stress amplitude
Refer to Fig. 5.56. The co-ordinates of the point X are obtained by solving the following two equations simultaneously:

[latex] \frac{S_{a}}{200}+\frac{S_{m}}{540}=1 [/latex] (a).

[latex] \frac{S_{a}}{S_{m}}= an heta=0.4986 [/latex] (b).

∴ [latex] S_{a}=114.76 N / mm ^{2} [/latex].

[latex] S_{m}=230.16 N / mm ^{2} [/latex].

Step IV Diameter of shaft

[latex] ext { Since } \sigma_{a}=\frac{S_{a}}{(f s)} \quad herefore \quad \frac{2116.33 imes 10^{3}}{d^{3}}=\frac{114.76}{2} [/latex].

d = 33.29 mm.

Question 5.20: A transmission shaft carries a pulley midway between the two...

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