# Bolted, Riveted and Welded Joint

 Question 1
A cantilever beam of rectangular cross-section is welded to a support by means of two fillet welds as shown in figure. A vertical load of 2 kN acts at free end of the beam.

Considering that the allowable shear stress in weld is 60 $N/mm^2$, the minimum size (leg) of the weld required is _______-mm (round off to one decimal place).
 A 6.6 B 2.8 C 4.6 D 8.2
GATE ME 2021 SET-1   Machine Design
Question 1 Explanation:
\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}
 Question 2
A bolt head has to be made at the end of a rod of diameter d = 12 mm by localized forging (upsetting) operation. The length of the unsupported portion of the rod is 40 mm. To avoid buckling of the rod, a closed forging operation has to be performed with a maximum die diameter of ________ mm.
 A 12 B 18 C 40 D 24
GATE ME 2020 SET-2   Machine Design
Question 2 Explanation:
\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}
 Question 3
A rectangular steel bar of length 500 mm, width 100 mm, and thickness 15 mm is cantilevered to a 200 mm steel channel using 4 bolts, as shown.

For an external load of 10 kN applied at the tip of the steel bar, the resultant shear load on the bolt at B, is ___________ kN (round off to one decimal place).
 A 4 B 16 C 24 D 2
GATE ME 2020 SET-1   Machine Design
Question 3 Explanation:

\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}
 Question 4
Pre-tensioning of a bolted joint is used to
 A strain harden the bolt head B decrease stiffness of the bolted joint C increase stiffness of the bolted joint D prevent yielding of the thread root
GATE ME 2018 SET-2   Machine Design
Question 4 Explanation:
Pretension increase stiffness of system..
 Question 5
A steel plate, connected to a fixed channel using three identical bolts A, B and D, carries a load of 6kN as shown in the figure. Considering the effect of direct load of moment, the magnitude of resultant shear force (in kN) on bolt C is.
 A 13 B 15 C 17 D 30
GATE ME 2017 SET-2   Machine Design
Question 5 Explanation:
\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}
 Question 6
A bolted joint has four bolts arranged as shown in figure. The cross sectional area of each bolt is 25$mm^{2}$. A torque T = 200 N-m is acting on the joint. Neglecting friction due to clamping force, maximum shear stress in a bolt is ______ MPa.
 A 20Mpa B 30Mpa C 40MPa D 50Mpa
GATE ME 2016 SET-3   Machine Design
Question 6 Explanation:

$\therefore \quad P_{1} \times r+P_{2} r+P_{3} r+P_{4} r-T=0$
$P_{1}=P_{2}=P_{3}=P_{4} \qquad [By symmetry]$
\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}
 Question 7
A cantilever bracket is bolted to a column using three M12x1.75 bolts P, Q and R. The value of maximum shear stress developed in the bolt P (in MPa) is ________
 A 332.6321MPa B 568.2654MPa C 986.2547MPa D 745.3256MPa
GATE ME 2015 SET-3   Machine Design
Question 7 Explanation:

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}
 Question 8
A horizontal plate has been joined to a vertical post using four rivets arranged as shown in the figure. The magnitude of the load on the worst loaded rivet (in N) is _______
 A 1253.36N B 1839.83N C 1258.36N D 4587.2N
GATE ME 2015 SET-1   Machine Design
Question 8 Explanation:
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
$\therefore P_{\text {net }}^{2}=P_{1}+P_{1}^{\prime \prime 2}+2 P_{1} P_{1}^{\prime \prime} \cos 45^{\circ}$
$P_{\text {net }}=1839.83 \mathrm{N}$
 Question 9
A butt weld joint is developed on steel plates having yield and ultimate tensile strength of 500 MPa and 700 MPa, respectively. The thickness of the plates is 8 mm and width is 20 mm. Improper selection of welding parameters caused an undercut of 3 mm depth along the weld. The maximum transverse tensile load (in kN) carrying capacity of the developed weld joint is _______
 A 60kN B 70kN C 20kN D 30kN
GATE ME 2014 SET-4   Machine Design
Question 9 Explanation:
\begin{aligned} P_{t} &=\sigma_{t}\left(t_{\text {under cut }}\right) L \\ &=700 \times 20 \times(8-3)=700 \times 100 N \\ &=70 \mathrm{kN} \end{aligned}
 Question 10
A bolt of major diameter 12 mm is required to clamp two steel plates. Cross sectional area of the threaded portion of the bolt is 84.3 $mm^{2}$. The length of the threaded portion in grip is 30 mm, while the length of the unthreaded portion in grip is 8 mm. Young's modulus of material is 200 GPa. The effective stiffness (in MN/m) of the bolt in the clamped zone is _______
 A 468.77MN/m B 548.26MN/m C 154.65MN/m D 985.32MN/m
GATE ME 2014 SET-4   Machine Design
Question 10 Explanation:
$k_{t}=\frac{A_{4} E}{4}$ (Stiffness in threaded portion)
\begin{aligned} k_{t} &=\frac{84.3 \times 200 \times 10^{3}}{30 \times 10^{-3}}=562 \times 10^{6} \mathrm{N} / \mathrm{m} \\ &=562 \mathrm{MN} / \mathrm{m} \end{aligned}
$k_{d}=\frac{A_{d} E}{L_{d}}$ (Stiffness in unthreaded region)
$A_{d}$( major diameter c/s area )
$=\frac{\pi}{4} \mathrm{d}^{2}=0.785 \times 144$
$=113.04 \mathrm{mm}^{2}$
$L_{d}$ (length of unthreaded portion) = 8mm
\begin{aligned} \therefore \quad & k_{d}=\frac{113.04 \times 200 \times 10^{3}}{8 \times 10^{-3}} \\ &=2826 \times 10^{6} \mathrm{N} / \mathrm{m}=2826 \mathrm{MN} / \mathrm{m} \\ \frac{1}{k} &=\frac{1}{k_{t}}+\frac{1}{k_{d}} \Rightarrow k=\frac{k_{d} k_{t}}{k_{d}+k_{t}}=\frac{2826 \times 562}{2826+562} \\ &=468.77 \mathrm{MN} / \mathrm{m} \end{aligned}
There are 10 questions to complete.