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).
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).
6.6 | |
2.8 | |
4.6 | |
8.2 |
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.
12 | |
18 | |
40 | |
24 |
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).
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).
4 | |
16 | |
24 | |
2 |
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
strain harden the bolt head | |
decrease stiffness of the bolted joint | |
increase stiffness of the bolted joint | |
prevent yielding of the thread root |
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.
13 | |
15 | |
17 | |
30 |
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}
\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 25mm^{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.
20Mpa | |
30Mpa | |
40MPa | |
50Mpa |
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 ________
332.6321MPa | |
568.2654MPa | |
986.2547MPa | |
745.3256MPa |
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 _______
1253.36N | |
1839.83N | |
1258.36N | |
4587.2N |
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}
\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 _______
60kN | |
70kN | |
20kN | |
30kN |
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 _______
468.77MN/m | |
548.26MN/m | |
154.65MN/m | |
985.32MN/m |
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}
\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.
