Bolted Connections MCQ Quiz - Objective Question with Answer for Bolted Connections - Download Free PDF

Concept:

A bolt is a metal pin at one end and shank at the other end which is threaded to receive a nut.

Various types of bolts are:

1) High strength bolt

2) Turned bolts (close Tolerance bolts)

3) Ordinary bolts.

High strength bolts: 

• The bolts that are tightened to develop some initial tension are known as high strength friction grip bolts. The bolts are tightened until they have very high tensile stresses so that the connected parts are tightly clamped together between bolt head and nut.
• Friction develop between the plate surfaces subjected to clamping forces and thus in HSFG bolts (High Strength Friction Grip Bolts) load is primarily transferred through friction.
• Such connections are called no-slip connection or slip-critical connection. The induced initial tension is called the proof load.
• These are used in bridges or where stress reversal may occur.

Explanation:

Table 20: Typical Average Values for Coefficient of Friction (μₜ)

(Clause 10.4.3)

Explanation:

Bolted Joints

Bolted joints are a common method of joining structural elements. Depending on the application, different types of bolted joints are used to ensure the desired performance. These types include:

• Bearing type joint: Primarily used where the bolt bears the load and transfers the shear force directly through the bearing on the bolt.

• Slip-critical joint: Designed to transfer shear forces through friction between the connected elements. This type requires the bolts to be tightened with a great force to generate sufficient clamping pressure.

• Friction grip joint: Similar to slip-critical joints, they transfer loads through friction but may have different applications and specifications.

• Pinned joint: Utilized where rotation is required between the connected elements, typically not designed to transfer significant shear forces.

Concept:

• The Unwin’s formula is used to determine the size of the rivet.
• Unwin's formula gives a relation between the hole diameter of the rivet and the thickness of the connected plates.
• The 'd' in the formula represents the nominal diameter of the rivet in mm and 't' represents the thickness of the plate

According to Unwin’s formula,

The nominal diameter (ɸ) of the rivet is given by:

ɸ = 6.04√t        {t = thickness of plate in “mm”}

ɸ = 1.91√t        {t = thickness of plate in “cm”}

Explanation:

Turned bolts

• Turned bolts are bolts that have been machined or turned to precise dimensions. These bolts typically have a smooth and uniform surface finish. The term "turned" refers to the machining process rather than any specific feature of the bolt, such as ribs or flutes.

Fluted bolts

• Fluted bolts have grooves or flutes running along their length. These flutes provide additional grip and help prevent the bolt from rotating under load. The term "ribbed" and "fluted" can be used interchangeably because both describe the presence of ridges or grooves on the bolt's surface.

Unfinished bolts

• Unfinished bolts refer to bolts that have not been fully processed or treated. They typically lack any special surface finishes or coatings and are usually used in less critical applications. "Unfinished" describes the state of the bolt rather than a specific feature like ribs or flutes.

High strength bolts

• High strength bolts are made from high-grade materials and are designed to withstand higher loads and stress. The term "high strength" refers to the bolt's material properties and mechanical strength, not to the presence of ribs or flutes.

Concept:

Riveted Joint:

A riveted joint is a permanent joint which uses rivets to fasten two materials.

Types of riveted joints:

1. Lap Joint

A lap joint is that in which one plate overlaps the other and the two plates are then riveted together.

Number of shearing planes = 1

2. Butt joint

A butt joint is that in which the main plates are kept in alignment butting each other and a cover plate is placed either on one side or on both sides of the main plates. The cover plate is then riveted together with the main plates.

Number of shearing plane in single cover butt joint = 1

Number of shearing plane in double cover butt joint = 2

Concept of shearing strength:

Strength of joint per pitch length in shearing, P_d = P_s × N × n

Where, P_s = Strength of one rivet in single shear, n = Number of shearing planes, N = Number of rivets

Calculation:

Given, P_s = strength of rivet in single shear, Number of rivets (N) = 2 in double riveted joints and number of shearing planes (n) = 2 for double cover butt joint

∴ \({P_d} = {P_s} \times 2 \times 2 = 4{P_s}\)

Explanation:

M means bolt is of Metric Thread, and 20 refers to its diameter. It's cap diameter is 1.5 D (1.5D is for hexagonal bolt). Strength is determined from grade of bolt. Here grade 4.6

So ultimate strength, fu = 4 x 100 = 400 N/mm2

Concept:

Minimum pitch of bolted connection = 2.5d

Where, d = Nominal diameter of the bolt

Calculation:

For M16,

d = 16 mm

Minimum pitch of M16 = 2.5 × 16 = 40 mm

Explanation:

As per IS 800: 2007, Cl No: 10.2.3.2, the limits on Pitch in case of bolted connection are:

Maximum pitch < minimum (16t, 200 mm) → For Tension Members

Maximum pitch < minimum (12t, 200 mm) → For Compression Members

Where, t is the thickness of thinner plate.

Note:

1. The above values can be increased by 50 % if bolts are staggered and gauge distance does not exceed 75 mm.

2. Minimum Pitch = 2.5 times the nominal diameter of bolt.

Concept:

Minimum pitch of bolt connection = 2.5 d

Where, d = diameter of bolt

Calculation:

Given;

d = 20 mm

Minimum spacing = 2.5 × 20 = 50 mm

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

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

3. Maximum end and edge distance

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

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

P_max = 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

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

Where, t = thickness of the thinner plate

The relation between nominal and bolt hole diameter is given below:

The nominal diameter of bolt, d_n = 20 mm

∴ Diameter of hole, d_h = 20 + 2 = 22 mm

As per IS 800:2007,

P_min = 2.5 × d = 2.5 × 20 = 50 mm

e_min = 1.5 × d_h = 1.5 × 22 = 33 mm

e = edge distance

P = Pitch

Trail 1) Take the number of bolts = 2

Taking the minimum pitch value of 50 mm and minimum edge distance as 33 mm

33 × 2  + 50 = 116 < 140 (section is safe)

Trail 2) Take the number of bolts = 3

2 × 33 + 2× 50 =166 > 140 (section is unsafe)

So, this fails the minimum edge criteria and hence the maximum number of bolts that we can take here is 2

Concept:

The bearing strength of a single bolt is given as

Bearing Strength, F_B = Permissible stress in bearing (f_bp) × bearing area (A_b)

Bearing area, A_b = hole diameter × thickness (t)

Where ,

 t = minimum (Sum thickness of Covering plate / gusset plate thickness, thickness of main plate)

Calculation:

Given: f_bp = 250 MPa, bolt diameter d = 16 mm;

Thickness of gusset plate = 5 mm and Thickness of main plate = 6 mm

∴ t = 5 mm

Bearing Strength, F_B = Permissible stress in bearing (f_bp) × bearing area (A_b)

F_B = 250 × 16 × 5 = 20,000 N

∴ F_B = 20 kN

Explanation:

Bearing strength of all bolts (Pb)

Pb = 2.5 × kb × (d × t) × fu

where ‘t’ is the thickness of thinner main plate and

\({k_b} = min\;\left\{ {\frac{e}{{3{d_h}}},\frac{P}{{3{d_h}}} - 0.25,\frac{{{f_{ub}}}}{{{f_u}}},1.0} \right\}\)

Here we can see Kb depends upon

edge distance,

pitch distance,

bolt hole diameter,

ultimate strength of the bolt and

the ultimate strength of the plate

Additional InformationShearing strength of all bolts (Ps)

\({P_s} = m \times \frac{\pi }{4}{d^2}*{f_S} \times n\)

Where n is number of bolts, d is diameter of bolt

m = 1 for single shear and m = 2 for double shear

Here m = 2 because of double shear in double cover butt joint

\({f_s} = \frac{{{f_u}}}{{\sqrt 3 \times 1.25}}\) (fu is ultimate stress)

Explanation:

As per IS 800:2007, Table 5, Partial safety factors for materials are given as:

(i) When resistance is governed by ultimate stress, γ_m1 = 1.25

(ii) When resistance is governed by yielding, γ_m0 = 1.1

(iii) Resistance of member to buckling,γ_m0 = 1.1

(iv) Resistance of connection:

• Bolts-Friction Type or Rivets = 1.25
• Welds = 1.25 (shop fabrications) and 1.5 (Field fabrications)