Question

A bolt is subjected to an axial pull of 10 kN together with a transverse shear force of 5 kN.

Solve the diameter of the bolt by using

(i) Maximum principal stress theory

(ii) maximum strain theory

(iii) Octahedral shear stress theory​

Answer 1

Answer:

sorry I didn't you the answer

Answer 2

Answer:

Maximum principal stress theory=$12.4\ mm$

Maximum shear stress theory=$13.42 \ mm$

Maximum principal strain theory=$12.7 \ mm$

Explanation:

Given $: P_{t 1}=10 \mathrm{kN} ; P_{s}=5 \mathrm{kN} ; \sigma_{t(e l)}=100 \mathrm{MPa}=100 \mathrm{~N} / \mathrm{mm}^{2} ; 1 / m=0.3$

Let$d=$Diameter of the bolt in $\mathrm{mm} .$

$\therefore$ Cross-sectional area of the bolt,

$A=\frac{\pi}{4} \times d^{2}=0.7854 d^{2} \mathrm{~mm}^{2}$

We know that axial tensile stress,

$\sigma_{1}=\frac{P_{t 1}}{A}=\frac{10}{0.7854 d^{2}}=\frac{12.73}{d^{2}} \mathrm{kN} / \mathrm{mm}^{2}$

and transverse shear stress,

$\tau=\frac{P_{s}}{A}=\frac{5}{0.7854 d^{2}}=\frac{6.365}{d^{2}} \mathrm{kN} / \mathrm{mm}^{2}$

1. According to the maximum principal stress theory

We know that maximum principal stress,

$\sigma_{t 1} &=\frac{\sigma_{1}+\sigma_{2}}{2}+\frac{1}{2}\left[\sqrt{\left(\sigma_{1}-\sigma_{2}\right)^{2}+4 \tau^{2}}\right]$

$&=\frac{\sigma_{1}}{2}+\frac{1}{2}\left[\sqrt{\left(\sigma_{1}\right)^{2}+4 \tau^{2}}\right] \\&=\frac{12.73}{2 d^{2}}+\frac{1}{2}\left[\sqrt{\left(\frac{12.73}{d^{2}}\right)^{2}+4\left(\frac{6.365}{d^{2}}\right)^{2}}\right] \quad \ldots\left(\because \sigma_{2}=0\right)$

$&=\frac{6.365}{d^{2}}+\frac{1}{2} \times \frac{6.365}{d^{2}}[\sqrt{4+4}] \\\\&=\frac{6.365}{d^{2}}\left[1+\frac{1}{2} \sqrt{4+4}\right]\\=\frac{15.365}{d^{2}} \mathrm{kN} / \mathrm{mm}^{2}\\=\frac{15365}{d^{2}} \mathrm{~N} / \mathrm{mm}^{2}$

According to the maximum principal stress theory,

$\begin{aligned}&\sigma_{t 1}=\sigma_{t(e l)} \text { or } \frac{15365}{d^{2}}=100 \\&d^{2}=15365 / 100=153.65 \text { or } d=12.4 \mathrm{~mm} \text { Ans. }\end{aligned}$

According to the maximum shear stress theory

We know that maximum shear stress,

$\tau_{\max } &=\frac{1}{2}\left[\sqrt{\left(\sigma_{1}-\sigma_{2}\right)^{2}+4 \tau^{2}}\right]=\frac{1}{2}\left[\sqrt{\left(\sigma_{1}\right)^{2}+4 \tau^{2}}\right] \quad \ldots\left(\because \sigma_{2}=0\right)$

$&=\frac{1}{2}\left[\sqrt{\left(\frac{12.73}{d^{2}}\right)^{2}+4\left(\frac{6.365}{d^{2}}\right)^{2}}\right]=\frac{1}{2} \times \frac{6.365}{d^{2}}[\sqrt{4+4}] \\&=\frac{9}{d^{2}} \mathrm{kN} / \mathrm{mm}^{2}=\frac{9000}{d^{2}} \mathrm{~N} / \mathrm{mm}^{2}$

According to maximum shear stress theory,

We know that maximum shear stress,

$\tau_{\max } &=\frac{1}{2}\left[\sqrt{\left(\sigma_{1}-\sigma_{2}\right)^{2}+4 \tau^{2}}\right]=\frac{1}{2}\left[\sqrt{\left(\sigma_{1}\right)^{2}+4 \tau^{2}}\right] \quad \cdots\left(\because \sigma_{2}=0\right) \\&=\frac{1}{2}\left[\sqrt{\left(\frac{12.73}{d^{2}}\right)^{2}+4\left(\frac{6.365}{d^{2}}\right)^{2}}\right]=\frac{1}{2} \times \frac{6.365}{d^{2}}[\sqrt{4+4}] \\&=\frac{9}{d^{2}} \mathrm{kN} / \mathrm{mm}^{2}=\frac{9000}{d^{2}} \mathrm{~N} / \mathrm{mm}^{2}$

According to maximum shear stress theory,

$\tau_{\max } &=\frac{\sigma_{t(e l)}}{2} \text { or } \frac{9000}{d^{2}}=\frac{100}{2}=50 \\\therefore \quad d^{2} &=9000 / 50=180 \text { or } d=13.42 \mathrm{~mm} \text { Ans. }$

3. According to the maximum principal strain theory

We know that maximum principal stress,

$\sigma_{t 1}=\frac{\sigma_{1}}{2}+\frac{1}{2}\left[\sqrt{\left(\sigma_{1}\right)^{2}+4 \tau^{2}}\right]=\frac{15365}{d^{2}}$

and minimum principal stress,

$\sigma_{22} &=\frac{\sigma_{1}}{2}-\frac{1}{2}\left[\sqrt{\left(\sigma_{1}\right)^{2}+4 \tau^{2}}\right] \\&=\frac{12.73}{2 d^{2}}-\frac{1}{2}\left[\sqrt{\left(\frac{12.73}{d^{2}}\right)^{2}+4\left(\frac{6.365}{d^{2}}\right)^{2}}\right] \\&=\frac{6.365}{d^{2}}-\frac{1}{2} \times \frac{6.365}{d^{2}}[\sqrt{4+4}] \\&=\frac{6.365}{d^{2}}[1-\sqrt{2}]=\frac{-2.635}{d^{2}} \mathrm{kN} / \mathrm{mm}^{2} \\&=\frac{-2635}{d^{2}} \mathrm{~N} / \mathrm{mm}^{2}$

We know that according to maximum principal strain theory,

$\frac{\sigma_{t 1}}{E}-\frac{\sigma_{t 2}}{m E}=\frac{\sigma_{t(e l)}}{E} \text { or } \sigma_{t 1}-\frac{\sigma_{t 2}}{m}=\sigma_{t(e l)} \\\therefore \quad \frac{15365}{d^{2}}+\frac{2635 \times 0.3}{d^{2}}=100 \text { or } \frac{16156}{d^{2}}=100 \\d^{2}=16156 / 100=161.56 \text { or } d=12.7 \mathrm{~mm} \text { Ans. }$

4. According to maximum strain energy theory

We know that according to maximum strain energy theory,

$\left(\sigma_{t 1}\right)^{2}+\left(\sigma_{t 2}\right)^{2}-\frac{2 \sigma_{t 1} \times \sigma_{t 2}}{m}=\left[\sigma_{t(e t)}\right]^{2} \left[\frac{15365}{d^{2}}\right]^{2}+\left[\frac{-2635}{d^{2}}\right]^{2}-2 \times \frac{15365}{d^{2}} \times \frac{-2635}{d^{2}} \times 0.3=(100)^{2} \frac{236 \times 10^{6}}{d^{4}}+\frac{6.94 \times 10^{6}}{d^{4}}+\frac{24.3 \times 10^{6}}{d^{4}}=10 \times 10^{3}$

$\frac{23600}{d^{4}}+\frac{694}{d^{4}}+\frac{2430}{d^{4}}=1 or \frac{26724}{d^{4}}=1 \therefore \quad d^{4}=26724 or d=12.78 \mathrm{~mm} Ans.$