Question 1 |

A bracket is attached to a vertical column by
means of two identical rivets U and V separated
by a distance of 2a = 100 mm, as shown in the
figure. The permissible shear stress of the rivet
material is 50 MPa. If a load P = 10 kN is applied
at an eccentricity e=3\sqrt{7}a, the minimum crosssectional area of each of the rivets to avoid failure
is ___________ mm^2

.

.

800 | |

25 | |

100 \sqrt{7} | |

200 |

Question 1 Explanation:

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.

(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.

Question 2 |

A square threaded screw is used to lift a load W
by applying a force F. Efficiency of square threaded
screw is expressed as

The ratio of work done by W per revolution to work done by F per revolution | |

W/F | |

F/W | |

The ratio of work done by F per revolution to work done by W per revolution |

Question 2 Explanation:

\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.

Efficiency of screw jack \eta =\frac{\tan \alpha }{\tan(\alpha +\phi )}

Efficiency depends on helix angle and friction angle.

Question 3 |

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 3 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 4 |

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 4 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 5 |

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 5 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}

There are 5 questions to complete.

The answer in the ques 20 is wrong it should be A not D as gate does not rely upon emperical formulas!