Concept:

Efficiency of Riveted Joints: It is the ratio of the strength of the riveted joint to the strength of the connected plates.

Efficiency of riveted Joint is given by \(= \frac{{p - d}}{p} \times 100\)

Where, p = pitch of rivets

d = Gross diameter of rivets

Concept:

The throat is the part of the weld, which is assumed to be effective in transferring the stresses. It is the weakest plane in the fillet weld. It has thickness called throat thickness. It is the minimum dimension in fillet weld.

Calculation:

For the given fillet weld of triangular shape ABC, draw a perpendicular from point A to meet side BC at D. The length AD represents the throat thickness and it is easily calculated using trigonometry.

In ΔABD

∠B = 30°, ∠D = 90°, ∠A = 60°

AB is Hypotenuse.      

\(\cos 60^\circ = \frac{{AD}}{{AB}}\)  ⇒ AD = AB cos 60° \(= 5 \times \frac{1}{2} = 2.5\;mm\) 

∴ Throat thickness, t = 2.5 mm

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_n = 2.5 × 20 = 50 mm

e_min = 1.5 × d_h = 1.5 × 22 = 33 mm

e = edge distance

P = Pitch

Let n be the no of bolts that can be accommodated in a single row of width 140 mm.

 2 × e + (n - 1) × P = 140

To get maximum value of n, e and P should be minimum i.e. e = e_min and P = P_min

2 × 33 + (n - 1) × 50 = 140

∴ n = 2.48

Concept:

The shear stress distribution in welded joints is complex and non-uniform.  The variation of shear stress in the weld between points C and D and between points A and B is dependent on the length of weld and ratio of the width of plates.

There is stress concentration at points A and B due to which shear stress at these points will be more as compared to shear stress at the center. A typical shear stress distribution at service load for longitudinal fillet weld is shown in the figure below :

For points between C and D shear stress will be maximum at centre because the load passes through it. A typical shear stress distribution for fillet welds loaded transversely to the weld axis is shown in the following figure:

Concept:

Bolt Value, BV = Min (S, P)

Where ,

S is the strength of single bolt in shear.

P is the strength of single bolt in bearing.

Strength of joint or bolt = n × BV

Where n is the number of bolts in a row

For joint to be safe,

Applied load ≤ Strength of joint

Calculation:

Given

Applied load, T = 500 kN

Strength of single bolt in shear, S = 100 kN

Strength of single bolt in bearing = P kN

Case 1:

Let S ≤ P ⇒ BV = S = 100 kN

Now, for safe joint

Applied load ≤ strength of joint

500 ≤ 5 × 100

⇒ 500 ≤ 500 ⇒ hence, ok

This means that P ≥ S or P ≥  100 kN

Case 2:

Let S > P ⇒ BV = P kN

Now, for safe joint

Applied load ≤ strength of joint

500 ≤ 5 × P ⇒ P ≥ 100 kN  

In both cases, we get the same answer. Hence, P ≥ 100 kN is the correct answer. 

Concept

The force resisted by any weld or strength of joint having throat thickness t, length L is given as:

P = Effective area of weld × permissible stress in weld 

The effective area of weld = throat thickness × Effective length of the weld

i.e. A_e = t × L

The throat thickness of the welded connection can be written as:

t = KS

Where,

K is the factor that depends on the angle between fusion faces.

S is the size of the weld.

Calculation:

Angle b/w fusion forces is 75°

⇒ K = 0.7

Throat thickness, t = K × S

⇒ t = 0.7 × 8 = 5.6 mm

Welding is done on both sides as there are two plates,

So length of weld, L = (100 + 100 + 50) × 2 ⇒ L = 500 mm

∴ The strength of Joint, P is given as,

P = 5.6 × 500 × 120 ⇒ P = 336 kN

Concept:

As per IS 800:2007 Codal provisions

For weld in compression Zone,

Maximum pitch or maximum clear spacing between welds = minimum (12t, 200 mm)

For welds in tension zone,

Maximum pitch or maximum clear spacing between welds = minimum (16t, 300 mm)

Where,

t is the thickness of the main plate

In this case, ‘t’ is the flange thickness as it is the main plate.

Due to external loading, the plate girder will bend as shown in the figure, due to which the upper flange plate is subjected to bending compression and lower flange plate is subjected to bending tension.

Calculation:

Upper Flange plate is in compression, therefore the maximum pitch S₁ is given as

S₁ = minimum (12t, 200 mm)

⇒ S₁ = minimum (12 × 20, 200) ⇒ S₁ = 200 mm 

Lower Flange plate is in Tension, therefore the maximum pitch S₂ is given as

S₂ = minimum (16t, 300 mm)

⇒S₂ = minimum (16 × 30, 300) ⇒ S₂ = 300 mm  

The connection is lap connection with axial pull so there will not be tearing failure of bolt. So the strength of connection is = 50 KN

Efficiency, \({\rm{\eta \;}} = {\rm{\;}}\frac{{{\rm{strength\;of\;connection}}}}{{{\rm{Strength\;of\;soild\;plate}}}} \times 100\% \)

Strength of solid plate \(= \;\frac{{{A_g}.{f_y}}}{{{\gamma _{mo}}}}\)

\(\begin{array}{l} = \;\frac{{700 \times 250}}{{1.1}}{\rm{\;}} = {\rm{\;}}159.09{\rm{\;KN}}\\ \therefore \eta \; = \;\frac{{50}}{{159.09}} \times 100\% \; = \;31.42\% \end{array}\)

The bolt connection is symmetric, and its centroid lines will be on the symmetric lines. The distance of each bolt from centroid of the bolt group will be.

\({r_x} = \sqrt {{{\left( {30} \right)}^2} + {{\left( {40} \right)}^2}} \)

\(\Rightarrow {r_x} = \sqrt {2500}\)

⇒ r_x = 50 mm

\({\rm{\Sigma }}{r^2} = 4 \times r_x^2\)

= 4 × (50 mm)²

= 10000 mm²

Therefore, the shear force induced in the bolt X is given by

\({F_s} = \frac{{M \times {r_x}}}{{{\rm{\Sigma }}{r^2}}}\)

\({F_s} = \frac{{165 \times {{10}^3} \times 50}}{{10000}}\)

For Fe 410 grade steel: f_u = 410 MPa, f_y = 250 MPa

Partial safety factor for material: governed by yielding (γ_m0) = 1.1

A_vg = (2 × 100) × 10 = 2000 mm², A_vn = (2 × 100) × 10 = 2000 mm², A_tg = 225 × 10 = 2250 mm², and A_tn = 225 × 10 = 2250 mm²

The block shear strength will be minimum of T_db1 and T_db2 as calculated below:

\({{\rm{T}}_{{\rm{db}}1}} = \frac{{{{\rm{A}}_{{\rm{vg}}}}{{\rm{f}}_{\rm{y}}}}}{{\sqrt 3 {\rm{\;}}{{\rm{\gamma }}_{{\rm{m}}0}}}} + \frac{{0.9{\rm{\;}}{{\rm{A}}_{{\rm{tn}}}}{{\rm{f}}_{\rm{u}}}}}{{{{\rm{\gamma }}_{{\rm{ml}}}}}}\)

\({{\rm{T}}_{{\rm{db}}1}} = \left[ {\frac{{2000 \times 250}}{{\sqrt 3 \times 11}} + \frac{{0.9 \times 2250 \times 410}}{{1.25}}} \right] \times {10^{ - 3}}\)

∴ T_db(!) = 926.631 kN

\({{\rm{T}}_{{\rm{db}}2}} = \frac{{0.9{\rm{\;}}{{\rm{A}}_{{\rm{vn}}}}{{\rm{f}}_{\rm{u}}}}}{{\sqrt 3 {\rm{\;}}{{\rm{\gamma }}_{{\rm{m}}1}}}} + \frac{{{{\rm{A}}_{{\rm{tg}}}}{{\rm{f}}_{\rm{y}}}}}{{{{\rm{\gamma }}_{{\rm{m}}0}}}}\)

\(\therefore {{\rm{T}}_{{\rm{db}}2}} = \left[ {\frac{{0.9 \times 2000 \times 410}}{{\sqrt 3 \times 1.25}} + \frac{{2250 \times 250}}{{1.1}}} \right] \times {10^{ - 3}}\)

∴ T_db(2)= 852.23 kN

∴ The block shear strength of the tension member is 852.23 kN.