Bolt Connection Design Using Eurocode 3

Axially loaded bolt connection design is discussed in detail with a worked example and while discussing the technical background related to the design.

In this article, we are concentrating on the design of steel plates with regular bolt arrangements having applied the axial tension.

Firstly, let’s discuss the desing procedure of the steel plate having the bolts.

Desing Procedure 

The following step can be followed during the desing.

• Evaluate the material properties

Initially, you select a plate thickness to proceed with the design. Based on the plate thickness and the steel grade the yield strength (f_y) and ultimate strength (f_u) can be obtained from Table 3.1 of Eurocode 3.

Then obtain the γ_m0 and γ_m2 from Clause 6.1. These values may change as per the national annexures.

• Calculate the cross-section of the plate or the area that applies the axial tension.

Different arrangement of the bolts needs to be considered and we have to find the critical cross-section having the smaller area which leads to increase the stress.

The plate is check for its gross cross-section and critical section (after reducing the hole area). The minimum of the following capacities is taken as plate capacity (N_t,Rd).

1. 1. 1. N_pl,Rd = Af_y / γ_m0  Plate capacity of gross cross section (Plastic Resistance). Here, A – gross crosection
      2. N_u,Rd = 0.9A_netf_y /γ_m2  Plate capacity for gross section (Ultimate Resistance). Here, A_net – net cross section

• Check the plate capacity found above is less than the applied load.

Plate Capacity = (N_t,Rd) = min[N_pl,Rd , N_u,Rd ]

Applied Load = N_Ed

N_Ed/N_t,Rd < 1

Worked Example: Bolt Connection Design; Axially Loaded Plate 

Design Data

• Applied load on plate 300kN
• Thickness of the plate 12mm
• Width of the plate 180mm
• Bolt spacings; s = p = 60mm (refer above figure for s & p)
• Bolt Diameter 16mm
• Steel Grade S275

Steel grade S275 and plate thickness 10mm < 40mm, the following design parameter can be obtained from Table 3.1 of EN 1993-1-1 (EC3).

f_y = 275 N/mm²

f_u =430 N/mm²

From clause 6.1, obtained the partial factors.

γ_m0  = 1.0

γ_m2  = 1.25

Cross sectional area of the plate

A = 180 x 10 = 1800 mm²

Plastic Resistance N_pl,Rd = Af_y / γ_m0

N_pl,Rd = 1800 x 275 / 1.0 = 495000 N = 495 kN

Calculate the Capacity of Plate with reduction of Holes Area

The method of calculating the deduction of bolt hole area is given in Cl 6.2.2.2.

It is quite easy to calculate the deduction as there is an equation for that.

Deduction = t [ nd₀ – Σ (s²/4p) ]

Cross section 1 – 1

Deduction = 10 [ 2 x 18 – 0 ] = 360 mm²

Cross section 2-2

Deduction = 10 [ 2 x 18 –  (60²/4×60) ] = 210 mm²

Cross section 3-3

Deduction = 10 [ 2 x 18 –  {(60²/4×60) + (60²/4×60)} ] = 240 mm2

The maximum deduction was in part 1-1

Therefore,

A_net = 1800 – 360 = 1440 mm²

Calculate ultimate resistance, N_u,Rd

N_u,Rd = 0.9 x 1440 x 430 / 1.25 = 445824 N = 445.8 kN

Plate Capacity = (N_t,Rd) = min[N_pl,Rd , N_u,Rd ]

N_t,Rd = min[495 , 445.8 ] = 445.8 kN

N_Ed/N_t,Rd < 1

300/ 445.8 < 1  Plate can carry the applied load.

Further, reading on steel beam design the article Steel Beam Desing Worked Example may be reffered.