Analysis of an Erection Joint in the Beam Web with High Strength Bolts

Objective: Check the mode for calculating friction bolted connections

Task: Check the erection joint of the compound I-beam web with high strength bolts.

Source: Steel Structures: Student Handbook / [Kudishin U.I., Belenya E.I., Ignatieva V.S and others] - 13-th ed. rev. - M.: Publishing Center "Academy", 2011. p. 216.

Compliance with the codes: SNiP II-23-81*, SP 16.13330, DBN B.2.6-163:2010.

Initial data file:

2.4.sav; report — Kristall2.4.doc

Initial data from the source:

М = 1216 kNm	Bending moment acting in the web plane;
Ry = 240 MPa,	Steel C245; Thickness of the beam web 8 mm;
Rbun = 110 kN/cm2,	High strength bolts from 40H “select” steel; Diameter of bolts 24 mm, diameter of holes 27 mm;
γc = 1	Service factor;
γb = 1	Service factor  of the friction connection; Method of cleaning the surfaces – flame treatment, without preservation;
μ = 0,42	Coefficient of friction;
γh = 1,12	Tightening control – by the torque.

 

KRISTALL parameters:
Steel: C245

Importance factor	1
Service factor	1
Service factor  of members to be joined	1

Diameter of bolts 24 mm
Steel: 40H "select"
Clearance 3 mm
Method of cleaning the surfaces to be joined: Flame treatment of two surfaces, without preservation

Type	Parameters
	m = 1 n = 9 a = 70 mm b = 170 mm c = 50 mm t = 16 mm t0 = 16 mm

 

Internal forces and moments:

N = 0 kN
M = 1216 kNm
Q = 0 kN

Manual calculation:

1. Design tension resistance of high strength bolts was calculated according to the following formula:

\[ R_{bh} =0,7R_{bun} =0,7\times 1100=770 \quad Н/мм^{2} \quad =77,0 \quad kN/cm^{2}. \]

2. Design force which can be resisted by each plane of friction:

\[ Q_{bh} =\frac {kR_{bh} A_{bn} \mu} {\gamma_{h} }=\frac {2 \times 77,0\times 3,53\times0,42}{1,12}=203,8575 \quad kN, \]

3. Force on the end bolt:

\[ N_{\max } =\frac{M\cdot y_{\max } }{2\sum\limits_i {y_{i}^{2} } }=\frac{1216\cdot 1,53}{2\left( {\left( {1\cdot 0,17} \right)^{2}+\left( {3\cdot 0,17} \right)^{2}+\left( {5\cdot 0,17} \right)^{2}+\left( {7\cdot 0,17} \right)^{2}+\left( {9\cdot 0,17} \right)^{2}} \right)}=195,080 \quad kN. \]

Comparison of solutions:

Factor	Friction force limit
Manual calculation	195,080/203,8575 = 0,957
KRISTALL	0,956
Deviation from the manual calculation, %	0,1
Source	0,96

 

Comments:

In the considered example the vertical distance between bolts is taken as 170 mm, which exceeds the limiting distance 12t = 12*8 mm = 96 mm, calculated according to the codes. To perform the computer-aided calculation in KRISTALL the thickness values of the web and the gusset plates were specified as t = t₀ = 16 mm, which does not affect the result of the calculation.