Analysis of a Bolted Connection between an Angle and a Gusset Plate with Ordinary Bolts

Objective: Check the mode for calculating bolted connections

Task: Check a bolted connection between two 90х9 mm angles and a 20 mm thick gusset plate with bolts of 8.8 strength class for a shear force of 400 kN.

Source: Moskalev N.S., Pronosin J.A. Steel Structures. Handbook / M.: ASV Publishing House, 2010. p. 102.

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

Initial data file:

2.7.sav;
report — Kristall2.7.doc

Initial data:

N = 400 kN	Shear force;
а = 100 mm	Distance between bolts;
γb = 0,9	Service factor  of the bolted connection; Diameter of bolts 20 mm, diameter of holes 22 mm.

 

KRISTALL parameters:
Steel: C255

Group of structures according to the table 50* of SNiP II-23-81* 3

Importance factor	1
Service factor	1
Service factor  of members to be joined	1
Product of the joint service factor (γb) and the service factor  of members to be joined (γc)	1
Design shear strength of bolts Rbs	320 N/mm2
Design bearing strength of bolt elements Rbp	440.64 N/mm2

 

Type:	Bolts:	Parameters:
Attachment of double angles	Diameter of bolts 20 mm Clearance 2 mm Diameter of holes 22 mm Class of bolts 8.8 Accuracy class B or C	m = 2 a = 100 mm c = 50 mm r = 49 mm t0 = 20 mm
Section - Full assortment of GOST profiles.. Equal angle GOST 8509-93 L90x9

 

Internal forces:

N = 400 kN

Manual calculation (SNiP II-23-81*):

1. Design shear resistance of the bolts was calculated as follows R_bs = 0.40R_bun = 0.40 × 800 = 320 MPa (see table 5*).

2. Design bearing resistance of the bolts was taken as (see table 5*):

– when a 20 mm thick gusset plate is in bearing, R_un= 370 MPa:

\[ R_{bp} =\left( {0,6+340\frac{R_{un} }{E}} \right)R_{un} =\left( {0,6+340\cdot \frac{370}{2,06\cdot 10^{5}}} \right)\cdot 370=447,95 \quad MPa; \]

– when a 9 mm thick angle is in bearing, R_un = 380 MPa:

\[ R_{bp} =\left( {0,6+340\frac{R_{un} }{E}} \right)R_{un} =\left( {0,6+340\cdot \frac{380}{2,06\cdot 10^{5}}} \right)\cdot 380=466,33 \quad MPa. \]

3. Shear strength of the bolts was calculated according to the following formula:

\[ N_{bs} =R_{bs} A_{b} n_{s} \gamma_{b} \gamma_{c} =320\times 10^{3}\times 3,14\times 10^{-4}\times 2\times 1,0\times 0,9=180,864 \quad kN. \]

4. Bearing strength of the bolts was calculated according to the following formula:

– when a 20 mm thick gusset plate is in bearing, R_bp = 447,95 MPa:

\[ N_{bp} =R_{bp} D\left( {\sum\limits_i {t_{i} } } \right)_{\min } \gamma_{b} \gamma_{c} =447,95\times 10^{3}\times 20\times 20\times 10^{-6}\times 1,0\times 0,9=161,262 \quad kN; \]

– when a 9 mm thick angle is in bearing, R_bp = 466,33 MPa:

\[ N_{bp} =R_{bp} D\left( {\sum\limits_i {t_{i} } } \right)_{\min } \gamma_{b} \gamma_{c} =466,33\times 10^{3}\times 20\times 18\times 10^{-6}\times 1,0\times 0,9=151,091 \quad kN. \]

5. Design force per one bolt of the connection calculated taking into account the eccentricity e = 2,35 mm:

\[ N_{red} =\sqrt {\left( {\frac{N}{3}} \right)^{2}+\left( {\frac{eN}{2a}} \right)^{2}} =\sqrt {\left( {\frac{400}{3}} \right)^{2}+\left( {\frac{400\cdot 2,35}{2\cdot 100}} \right)^{2}} =133,416 \quad kN, \]

where  – bolt spacing in the connection.

6. Cross-sectional area of one angle weakened by the holes:

\[ A_{net} =A-td_{0} =15,6-0,9\cdot 2,2=13,62 \quad cm^{2}. \]

Comparison of solutions:

Factor	Shear strength	Bearing strength	Strength of the weakened section
Manual calculation	133,416/180,864 = 0,737	133,416/151,091 = 0,883	400/2/13,62/25 = 0,587
KRISTALL	0,737	0,885	0,587
Deviation from the manual calculation, %	0,0	0,2	0,0
Source	0,737	0,857	–