GATE Past Year Questions: Bolted, Riveted & Welded Joint | Design of Machine Elements - Mechanical Engineering PDF Download

Q1: A bracket is attached to a vertical column by means of two identical rivets U and V separated by a distance of 2 a = 100 mm , as shown in the figure. The permissible shear stress of the rivet material is 50MPa. If a load P = 10 kN  is applied at an eccentricity e = 3 √7 a , the minimum crosssectional area of each of the rivets to avoid failure is ___________ mm² [GATE ME 2022, SET-1]
(a) 800
(b) 25
(c) 100√7
(d) 200
Ans: (a)
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:

(ii) Secondary shear:

Here, θ is 90°
As secondary load is same on both rivets. Both are critical due to loading.

As FOS is considered as 1, A represents the minimum cross section area required.

Q2: A square threaded screw is used to lift a load W by applying a force F. Efficiency of square threaded screw is expressed as   [GATE ME 2022, SET-1]
(a) The ratio of work done by W per revolution to work done by F per revolution
(b) W/F
(c) F/W
(d) The ratio of work done by F per revolution to work done by W per revolution
Ans: (a) 
Screw efficiency= Work done in lifting the load/rev/Work done by the applied force/rev
Efficiency of screw jack η = tan α/tan ⁡( α + ϕ )

Q1: 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.  [GATE ME 2021

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).
Ans: (6.4 to 6.9)

Q1: 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.  [GATE ME 2020, SET-2]
Ans: (18 to 19.5)
If l >3d then
Die dia  =1.5d 
=1.5 (12) 
=18mm  

Q2: 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.  [GATE ME 2020, SET-1]

Ans: (15.9 to 16.1)

Question for GATE Past Year Questions: Bolted, Riveted & Welded Joint

Try yourself:A single riveted lap Joint of two similar plates as shown in the figure below has the following geometrical and material details.

width of the plate ω = 200 mm, thickness of the plate t = 5 mm, number of rivets n = 3, diameter of the rivet d_r = 10 mm, diameter of the rivet hole dr = 11 M mm, allowable tensile stress of the plate σ_p = 200 MPa, allowable shear stress of the rivet σ_s = 100 MPa and allowable bearing stress of the rivet σ_c = 150 MPa.

If the plates are to be designed to avoid tearing failure, the maximum permissible load P in kN is

[2013]

• A.

  83
• B.

  125
• C.

  167
• D.

  501

Explanation

For plates, σ = P/A

P = 167kN

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Question for GATE Past Year Questions: Bolted, Riveted & Welded Joint

Try yourself:A single riveted lap Joint of two similar plates as shown in the figure below has the following geometrical and material details.

width of the plate ω = 200 mm, thickness of the plate t = 5 mm, number of rivets n = 3, diameter of the rivet d_r = 10 mm, diameter of the rivet hole dr = 11 M mm, allowable tensile stress of the plate σ_p = 200 MPa, allowable shear stress of the rivet σ_s = 100 MPa and allowable bearing stress of the rivet σ_c = 150 MPa.

If the rivets are to be designed to avoid crushing failure, the maximum permissible load P in kN is

[2013]

• A.

  7.50
• B.

  15.00
• C.

  22.50
• D.

  30.00

Explanation

For rivet,

∴ P_s = 100 × d × t × 3 = 15 kN
and P_c = 150 × d × t × 3 = 22.5 kN
∴ p = max (15, 22.5) = 22.5 k N

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Question for GATE Past Year Questions: Bolted, Riveted & Welded Joint

Try yourself:A bracket (shown in figure) is rigidly mounted on wall using four rivets. Each rivet is 6 mm in diameter and has an effective length of 12 mm.

Direct shear stress (in MPa) in the most heavily loaded rivet is

• A.

  4.4
• B.

  8.8
• C.

  17.6
• D.

  35.2

Explanation

Given : Number of rivets = 4,
Diameter of each rivet , d = 6 × 10^-3 m Direct load on the bracket = 1000 N since there is no eccentricity
∴ Load on each rivet = 1000/4 = 250 N

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Question for GATE Past Year Questions: Bolted, Riveted & Welded Joint

Try yourself:A steel plate, connected to a fixed Channel using three identical bolts A, B and C, carries a load of 6 kN as shown in the figure.

Considering the effect of direct load and moment, the magnitude of resultant shear force (in kN) on bolt C is

[2017, Set-2]

• A.

  13
• B.

  15
• C.

  17
• D.

  30

Explanation

P_P = P/n = P/3 = 2kN
r_A = r_C = 50, (P_S)_A = (P_S)_C
r_B = 0 ⇒ (P_S)_B = 0

Q_A = 180°, Q_C = 0°
Resultant shear force
(R_C) = P_P + (P_S)_C.
R_C = 2 + 15 = 17 KN

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Question for GATE Past Year Questions: Bolted, Riveted & Welded Joint

Try yourself:A steel bar of 10 × 50 mm is cantilevered with two M12 bolts (P and Q) to support a static load of 4 kN as shown in the figure.

The resultant shear stress on bolt P is closest to:

[2008]

• A.

  132 MPa
• B.

  159 MPa
• C.

  178 MPa
• D.

  195 MPa

Explanation

Resultant load on P = F_sec – F_prim
= 20 – 2 = 18 kN

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Question for GATE Past Year Questions: Bolted, Riveted & Welded Joint

Try yourself:A steel bar of 10 × 50 mm is cantilevered with two M12 bolts (P and Q) to support a static load of 4 kN as shown in the figure.

The primary and secondary shear loads on bolt P, respectively, are

• A.

  2 kN, 20 kN
• B.

  20 kN, 2 kN
• C.

  20 kN, 0 kN
• D.

  0 kN, 20 kN

Explanation

Let F₁ and F₂ resist the 4 kN eccentric oad. The primary shear load is due to a direct shear stress of 4kN, which is equally divided amongst the two bolts, i.e. 2 kN,
∴ F_primary on P = 2 kN

The secondary load is to resist the moment produced by the eccentric load 4 kN about the C.
G. of the joint
Moment M = 4 k N × 2000 mm
= 8000
Nm Now this moment must be balanced by a moment produced by equal and opposite forces i.e. secondary loads on the bolts i .e. F × 400 = 800 N m
⇒ F = 20 k N

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Question for GATE Past Year Questions: Bolted, Riveted & Welded Joint

Try yourself:A bolted Joint is shown below. The maximum shear stress, in MPa, in the bolts at A and B, respectively are

[2007]

• A.

  242.6, 42.5
• B.

  42.5, 242.6
• C.

  42.5,42.5
• D.

  60.15,343.64

Explanation

Shear load at A and C


(τ_max)_A,C = 242.5MPa
(τ_max)B = 42.44MPa

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