Riveted Connections MCQ Quiz in मल्याळम - Objective Question with Answer for Riveted Connections - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Last updated on Mar 15, 2025

നേടുക Riveted Connections ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Riveted Connections MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

Latest Riveted Connections MCQ Objective Questions

Top Riveted Connections MCQ Objective Questions

Riveted Connections Question 1:

Prying forces are:

  1. Shearing forces on the bolts because of the joints.
  2. Tensile forces due to the flexibility of connected parts.
  3. Bending forces on the bolts because of the joints.
  4. Forces due the friction between connected parts.

Answer (Detailed Solution Below)

Option 2 : Tensile forces due to the flexibility of connected parts.

Riveted Connections Question 1 Detailed Solution

Prying forces:

The flexibility of connected parts can lead to deformations that increases the tension applied to the bolts. The additional tension is called prying force.

The prying forces can be kept small  by using a thick plate or by limiting the distance between the bolt and plate edge.

The corresponding prying force can be calculated as follows

\({\rm{Q}} = \frac{{{{\rm{M}}_{\rm{p}}}}}{{\rm{n}}}\)

Where,

n = Number of bolts

Mp = Plastic moment

\({{\rm{M}}_{\rm{p}}} = \frac{{{{\rm{f}}_{\rm{y}}}}}{{1.15}}\frac{{{\rm{W}}{{\rm{t}}^2}}}{4}\)

t = Thickness of end plate 

Riveted Connections Question 2:

If the diameter of a rivet body is 'D', then the diameter of snap head or pan head will be:

  1. 1.4 D
  2. 1.5 D
  3. 1.6 D
  4. 2 D

Answer (Detailed Solution Below)

Option 3 : 1.6 D

Riveted Connections Question 2 Detailed Solution

Concept:

Riveting and Rivet:

  • Riveting is one of the methods of making permanent joints of two or more pieces (Metal strips). 
  • A Rivet is a short cylindrical rod having a head and a tapered tail. The main body of the rivet is called a shank.

Uses: Rivets are used for joining metal sheets and plates in fabrication work, such as bridges, shipbuilding, games, structural steel work.

Welder 2 10Q - Reviewed images Q1 

Different types of rivet heads are:

  • Snap heads: These are used mainly for structural work and machine riveting.
  • Countersunk heads: These are employed for shipbuilding.
  • Conical heads: These are used where riveting is done by hand hammering.
  • Pan heads: These are required where very high strength is needed since they have the maximum strength.
  • Mushroom heads: These rivets are used to reduce the height of the rivet head above the metal surface.
  • Flatheads: These are most commonly used in sheet metal fabrication where the metal is very thin and the projecting head of the rivet is objectionable. These are also available for heavy fabrication work.

Explanation:

According to IS 2155: 1982, the Diameter of the snap head or pan head is taken as 1.6D, where D is the diameter of a rivet body.

F1 Akhil Madhuri 11.08.2021 D3

Riveted Connections Question 3:

The strength of a joint in a bolted connection is 620 kN. The size of plate is 180 mm width and 20 mm thick. If the yield strength of material of plate as 250 N/mm2 and partial safety factor as 1.1, then the efficiency of the joint is ______.

  1. 90%
  2. 63.67%
  3. 75.75%
  4. 65.75%

Answer (Detailed Solution Below)

Option 3 : 75.75%

Riveted Connections Question 3 Detailed Solution

Concept:

\(Efficiency(η ) = \frac{{\min (Ps,Pb,Pt)}}{P}\)

where, Ps strength of joint in shear

Pb = strength of joint in bearing

Pt = strength of joint in tearing

P = strength of plate in tearing when no deduction has been made for rivet hole = p.t.ft

Calculation:

Given,

strength of joint in bolted connection = 620 kN

size of plate = (180 × 20) mm

yield strength of material = 250 MPa

FOS = 1.1

strength of plate = \(b.t.\frac{{{f_y}}}{{1.1}}\)= 180 × 20 × (250/1.1) × 10-3 = 818.18 kN

efficiency ( η ) = (620/818.18) × 100 = 75.77%

Riveted Connections Question 4:

The standard clearance in diameter and width of slots for 20 mm nominal size of fastener is

  1. 1 mm
  2. 2 mm
  3. 3 mm
  4. 4 mm

Answer (Detailed Solution Below)

Option 2 : 2 mm

Riveted Connections Question 4 Detailed Solution

Explanation:

According to IS 800:2007 Clause 10.2.1, page no. 73,

The diameter of the hole should be the nominal diameter of the bolt plus the clearance as given below: 

Standard Clearance for Fastener Holes (IS 800: 2007, Table no. 19) given below:

Nominal size of fastener, d (mm) Size of the hole = Nominal diameter of the fastener + Clearance            
Standard clearance in diameter and width of slot (mm) Oversize clearance in diameter (mm) Clearance in the length of the slot
Short slot (mm) Long slot (mm)
12-14 1.0 3.0 4.0 2.5d
16-24  2.0 4.0 6.0 2.5d
24 2.0 6.0 8.0 2.5d
More than 24 3.0 8.0 10.0 2.5d

Riveted Connections Question 5:

For rivet diameter up to 24 mm, the diameter of the rivet hole is larger than the diameter of the rivet by

  1. 1 mm
  2. 1.5 mm
  3. 2 mm
  4. 2.5 mm

Answer (Detailed Solution Below)

Option 2 : 1.5 mm

Riveted Connections Question 5 Detailed Solution

Concept:

Gross diameter is the diameter of rivet when it is fully inserted in the rivet hole or body. This diameter is more than the diameter of shank (when it is cold or before insertion i.e. nominal diameter)

Important points:

Due to the heating effect, the size of rivets gets expanded which upon cooling gets reduced (called shank diameter).

Rivet hole diameter is calculated as:

Rivet size, d (in mm)

Rivet hole size, do (in mm)

d ≤ 25mm

d + 1.5mm

d > 25mm

d + 2mm

Riveted Connections Question 6:

The minimum edge distance in member with rolled edge (for bolted hand cut connection) is approximately

  1. 1.5 × bolt diameter
  2. 1.5 × hole diameter
  3. 1.7 × bolt diameter
  4. 1.7 × hole diameter

Answer (Detailed Solution Below)

Option 4 : 1.7 × hole diameter

Riveted Connections Question 6 Detailed Solution

Specification for pitch of bolts or rivets

1. Minimum pitch and minimum gauge length

= {2.5 × nominal diameter of the bolt

2. Minimum end and edge distance

\(= \left\{ {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {1.5 \times diameter\;of\;the\;bolt\;hole,for\;machine\;cut\;element}\\ {\;\;\;\;\;\;\;} \end{array}}\\ {1.7 \times diameter\;of\;the\;bolt\;hole,for\;hand\;cut\;element} \end{array}} \right.\)

3. Maximum end and edge distance

\(= \{ 12 \times t \times \epsilon,\)

Where, \(\epsilon = \sqrt {\frac{{250}}{{{f_y}}}}\)

fy = yield stress

t = thickness of thinner plate

Riveted Connections Question 7:

Two plates 160 mm × 16 mm and 160 mm × 12 mm are to be butt welded. If Double V-butt weld is used to connect the plate, then the strength of the welded joint in tension will be

(Permissible tensile stress of steel = 140 N/mm2)

  1. 200.8 kN
  2. 300.6 kN
  3. 330.4 kN
  4. 268.8 kN

Answer (Detailed Solution Below)

Option 4 : 268.8 kN

Riveted Connections Question 7 Detailed Solution

Concept:

In case of double V-butt joint, complete penetration of weld takes place.

Hence, effective throat thickness = Thickness of thinner plate joined = 12 mm

Strength of the weld = Permissible tensile stress of steel × Leff × effective throat thickness

Calculation:

Effective throat thickness = Thickness of thinner plate joined = 12 mm

Strength of welded joint = 140 × 12 × 160 = 268.8 kN

Riveted Connections Question 8:

If the strength of rivets in bearing and shearing is 30 kN and 45 kN respectively. If 90 kN load is acting at a joint two steel plates, then Number of rivets required at a joint is:

  1. 4
  2. 2
  3. 3
  4. 1

Answer (Detailed Solution Below)

Option 3 : 3

Riveted Connections Question 8 Detailed Solution

Concept:

Rivet value: 

  • Rivet value is defined as the minimum of shearing strength or (capacity ) and bearing strength of rivet.
  • But joint value is minimum of shearing strength of rivet, bearing strength of rivet, and tensile strength of the plate.

Rivet value (Rv) = minimum of (Pand Pb)

Where Ps = Shear strength of the bolt

Pb = Bearing strength of the bolt

Number of rivets (n) = \(\frac{{Force\;at\;the\;joint}}{{{R_v}}}\)

Calculation:

Given, Force acting at a joint = 90 kN

Pb = 30 kN, Ps = 45 kN

Rivet value (Rv) = minimum of (30 and 45) = 30 kN

Number of rivets (n) = \(\frac{{Force\;at\;the\;joint}}{{{R_v}}}\)

\(n = \frac{{90}}{{30}} = 3\)

Riveted Connections Question 9:

if the pitch is 60 mm and rivet value is 40 kN, the number of rivets required for a riveted connection carrying an eccentric load of 150 kN at a distance of 300 mm from the centre line is

  1. 10
  2. 12
  3. 6
  4. 8

Answer (Detailed Solution Below)

Option 4 : 8

Riveted Connections Question 9 Detailed Solution

Concept:

The number of rivets for eccentric connection is given by(provided rivets in two vertical row)

n = \(\sqrt { 6M \over p (2R_v )}\)

where

M = P × e

p = pitch

Rv = Rivet value

Calculation:

Given:

P = 150 kN

e = 300 mm = 0.3 m

Rv = 40 kN

p = 60 mm

The number of rivets for eccentric connection is

n = \(\sqrt { 6M \over p (2R_v )}\)

n = \(\sqrt {{6 \times 150\times 0.3}\over {40\times 2 \times0.06}}\)   = 7.5

Hence, the number of rivet is 8.

Riveted Connections Question 10:

Calculate the value of minimum edge distance for 20 mm diameter bolt using machine flame cut.

  1. 25 mm
  2. 22 mm
  3. 20 mm
  4. 33 mm

Answer (Detailed Solution Below)

Option 4 : 33 mm

Riveted Connections Question 10 Detailed Solution

Concept:

Specification for the pitch of bolts or rivets

1. Minimum pitch and minimum gauge length

  • P = 2.5 × nominal diameter of the bolt

2. Minimum end and edge distance

  • emin = 1.5 × diameter of the bolt hole      ....... (for machine cut element)
  • emin = 1.7 × diameter of the bolt hole      ....... (for hand-cut element)

 

Calculation:

Given,

Diameter of bolt = 20 mm, 

as the bolt diameter is in the range of (16 - 24) mm

Hole Dia = Bolt dia + 2 mm = 20 + 2 = 22 mm

Minimum edge Distance for machine flame cut = 1.5 × 22 = 33 mm

Additional Information 

3. Maximum end and edge distance

  • emax = 12 × t × ϵ 
  • Where, ϵ \(= \sqrt {\frac{{250}}{{{f_y}}}} \), f= yield stress and t = thickness of thinner plate.

4. The maximum pitch of bolts or rivets or welds in the compression zone 

Pmax = minimum { 12 × t or 200 mm }     ....... (For compression zone)

Where, t = thickness of the thinner plate

5. The maximum pitch of bolts or rivets or welds in the tension zone

Pmax = minimum { 16 × t or 200 mm }     ....... (For tension zone)

Where, t = thickness of the thinner plate

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