Bolted, Riveted and Welded Joint

The given load is eccentric lateral load which results in

(i) primary shear due to direct loading
(ii) secondary shear due to eccentricity

(i) Primary shear:
F_p=\frac{F}{n}=\frac{10kN}{2}=5kN
(ii) Secondary shear:
F_s=\frac{m}{r_1^2+r_2^2} \times r =\frac{10 \times 3\sqrt{7}a}{a^2+a^2} \times a=15\sqrt{7}kN \; \; \; \; (\because a=50mm)
Finding resultant: R=\sqrt{F_p^2+F_s^2+2F_pF_s \cos \theta }
Here, \theta \text{ is }90^{\circ}
As secondary load is same on both rivets. Both are critical due to loading.

\therefore \;\;R_{max}=\sqrt{F_p^2+F_s^2}=\sqrt{5^2+(15\sqrt{7})^2}=40kN
Design of Rivet: \begin{aligned} \tau _{max} &=\frac{S_{ys}}{FOS} \\ \frac{R_{max}}{A}&= \frac{50}{1}\\ \frac{40 \times 10^3}{A} &=50 \\ A&= 800 mm^2 \end{aligned}
As FOS is considered as 1, A represents the minimum cross section area required.

\text{Screw efficiency}=\frac{\text{Work done by the applied force/rev}}{\text{Work done in lifting the load/rev}}
Efficiency of screw jack \eta =\frac{\tan \alpha }{\tan(\alpha +\phi )}
Efficiency depends on helix angle and friction angle.

\begin{aligned} \tau_{\max }=\frac{2 \times 10^{3}}{0.707 t(40) \times 2}&=\frac{35.36}{t} \mathrm{MPa} \\ \sigma_{\max }=\frac{M_{\max }}{I_{N A}} \cdot \tau_{\max } &=\frac{2000 \times 150 \times 20}{\frac{0.707 t(40)^{3} \times 2}{12}} \\ \sigma_{\max }&=\frac{795.615}{t} \mathrm{MPa}\\ \text { MSST, } \quad \sqrt{\sigma_{\max }^{2}+4 \tau^{2}} &\leq 2\left(\frac{S_{y s}}{N}\right)\\ \sqrt{\left(\frac{795.615}{t}\right)^{2}+4\left(\frac{35.36}{t}\right)^{2}} & \leq 2 \times 60 \\ \frac{798.752}{t} & \leq 2(60) \\ t &=6.65 \mathrm{~mm} \end{aligned}

\begin{array}{l} \text { If } l \gt 3 d \text { then } \\ \qquad \begin{aligned} \text { Die dia } &=1.5 d \\ &=1.5(12) \\ &=18 \mathrm{mm} \end{aligned} \end{array}

\begin{aligned} F_{A} &=F_{B}=F_{C}=F_{D}=\frac{10 \times 400}{4 \times 50 \sqrt{2}}=14.14 \mathrm{kN} \\ \text{Res}_{\mathrm{B}} &=\sqrt{14.14^{2}+2.5^{2}+2(14.14)(2.5) \cos 45} \\ \text{Res}_{\mathrm{B}} &=16.005 \mathrm{kN} \end{aligned}

Pretension increase stiffness of system..

\begin{aligned} P_{P}&=\frac{P}{n}=\frac{P}{3}=2 \mathrm{kN} \\ r_{A}&=r_{C}=50 ;\left(P_{S}\right)_{A}=\left(P_{S}\right)_{C} \\ r_{B}&=0 \Rightarrow\left(P_{S}\right)_{B}=0 \\ \frac{\left(P_{S}\right)_{C}}{r_{C}}\left[r_{C}^{2}+r_{B}^{2}+r_{A}^{2}\right]&=P \times e \end{aligned}

\left(P_{S}\right)_{C} \times 2 \times r_{C}=P \times e
\left(P_{S}\right)_{C}=\frac{6 \times 250}{2 \times 50}=15 \mathrm{kN}
\theta_{A}=180^{\circ} ; \theta_{C}=0^{\circ}
Resultant shear force \left(R_{C}\right)=P_{P}+\left(P_{S}\right)_{C}
[\therefore \theta_{C}=0^{0}]
\begin{aligned} R_{C}&=2+15=17 \mathrm{kN} \\ R_{A}&=\left(P_{S}\right) A-P_{P}=13 \mathrm{kN} \\ R_{B}&=P_{P}=2 \mathrm{kN} \end{aligned}

\therefore \quad P_{1} \times r+P_{2} r+P_{3} r+P_{4} r-T=0
P_{1}=P_{2}=P_{3}=P_{4} \qquad [By symmetry]
\begin{aligned} 4 P r&=T \\ P=& \frac{T}{4 r}=\frac{200}{4 \times \frac{50}{1000}}=1000 \mathrm{N} \\ \text { Shear stress } &=\frac{T}{\text { Area }}=\frac{1000}{25}=40 \mathrm{MPa} \end{aligned}

Total shear force on P ,
\begin{aligned} F_{s} &=\sqrt{(37500)^{2}+(3000)^{2}} \\ &=37619.808\\ \therefore \qquad \tau_{\max } &=\frac{F_{s}}{A}=\frac{37619.808}{\pi \times 0.25 \times 12^{2}} \\ &=332.6321 \mathrm{MPa} \end{aligned}

Primary force \left(P_{1}\right)=100 \mathrm{N}
\begin{aligned} P_{1}^{\prime \prime}&=P_{2}^{\prime \prime}=P_{3}^{\prime \prime}=P_{4}^{\prime \prime}=\frac{P_{e} \times r}{4 r^{2}} \\ &= \frac{400 \times 500 \times(40 \sqrt{2} / 2)}{4 \times\left(\frac{40 \sqrt{2}}{2}\right)^{2}} \\ P_{1}^{\prime \prime}&= \frac{400 \times 500 \times 20 \sqrt{2}}{4 \times(20 \sqrt{2})^{2}} \\&= \frac{400 \times 500}{4 \times 20 \sqrt{2}} \\ P_{1}^{\prime \prime}&= 1767.766 \mathrm{N} \end{aligned}
(1 and 4) are worst loaded
\therefore P_{\text {net }}^{2}=P_{1}+P_{1}^{\prime \prime 2}+2 P_{1} P_{1}^{\prime \prime} \cos 45^{\circ}
P_{\text {net }}=1839.83 \mathrm{N}