Analysis of an Axially Compressed Welded I-beam Column

Objective: Check the mode for calculating columns of solid cross-section

Task: Check the design section of a welded I-beam for the axially compressed column with a height of 6,5 m.

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. 256.

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

Initial data file:

5.1.sav;
report — Kristall-5.1.doc

Initial data:

l = 6,5 m	Column height;
μ = 0,7	The lower restraint is rigid and the upper one is pinned;
N = 5000 kN	Design compressive force;
γ_c = 1	Service factor;
R_y = 24 kN/cm²	Steel grade C245;
A = 230,4 cm²	Geometric properties of
I_x = 118243,584 cm⁴, I_y = 33184,512 cm⁴ i_x = 22,654 cm, i_y = 12,001 cm.	the selected section;

KRISTALL parameters:

Steel: C245
Group of structures according to the table 50* of SNiP II-23-81* 3
Importance factor 1
Service factor 1
Member length – 6.5 m

Length between out-of-plane restraints – 6.5 m
Limit slenderness for members in compression: 180 - 60α
Limit slenderness for members in tension: 250

Section:

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

1. Strength check of the selected column section:

\[ \frac{N}{AR_{y} \gamma_{c} }=\frac{5000}{230,4\cdot 24\cdot 1}=0,904. \]

2. Slenderness of the column:

\[ {\lambda}_{x} =\frac{l_{ef,x} }{i_{x} }=\frac{0,7\cdot 6,5\cdot 100}{22,654}=20,08475; \] \[ {\lambda}_{y} =\frac{l_{ef,y} }{i_{y} }=\frac{0,7\cdot 6,5\cdot 100}{12,001}=37,9135. \]

3. Conditional slenderness of the column:

\[ \bar{{\lambda }}_{x} =\frac{l_{ef,x} }{i_{x} }\sqrt {\frac{R_{y} }{E}} =\frac{0,7\cdot 6,5\cdot 100}{22,654}\sqrt {\frac{240}{2,06\cdot 10^{5}}} =0,68555; \] \[ \bar{{\lambda }}_{y} =\frac{l_{ef,y} }{i_{y} }\sqrt {\frac{R_{y} }{E}} =\frac{0,7\cdot 6,5\cdot 100}{12,001}\sqrt {\frac{240}{2,06\cdot 10^{5}}} =1,2941. \]

4. Buckling coefficients:

\[ \varphi_{y} =1-\left( {0,073-5,53\frac{R_{y} }{E}} \right)\bar{{\lambda }}_{y} \sqrt {\bar{{\lambda }}_{y} } =1-\left( {0,073-\frac{5,53\cdot 240}{2,06\cdot 10^{5}}} \right)\cdot 0,68555\sqrt {0,68555} =0,9622; \] \[ \varphi_{y} =1-\left( {0,073-5,53\frac{R_{y} }{E}} \right)\bar{{\lambda }}_{y} \sqrt {\bar{{\lambda }}_{y} } =1-\left( {0,073-\frac{5,53\cdot 240}{2,06\cdot 10^{5}}} \right)\cdot 1,2941\sqrt {1,2941} =0,902. \]

5. Strength of the column from the condition of providing the general stability under axial compression:

\[ N_{b,x} =\varphi_{x} AR_{y} \gamma_{c} =0,9622\cdot 230,4\cdot 24\cdot 1=5320,58 \quad кН; \] \[ N_{b,y} =\varphi_{y} AR_{y} \gamma_{c} =0,902\cdot 230,4\cdot 24\cdot 1=4987,7 \quad кН. \]

6. Limit slenderness of the column:

\[ \left[ \lambda \right]_{x} =180-60\alpha_{x} =180-60\cdot \frac{N}{\varphi _{x} AR_{y} \gamma_{c} }=180-60\cdot \frac{5000}{5320,58}=123,615; \] \[ \left[ \lambda \right]_{y} =180-60\alpha_{y} =180-60\cdot \frac{N}{\varphi _{y} AR_{y} \gamma_{c} }=180-60\cdot 1=120. \]

Comparison of solutions:

Factor	Source	Manual calculation	KRISTALL	Deviation, %
Strength under combined action of longitudinal force and bending moments, no plasticity	–	0,904	0,904	–
Stability under compression in XoY (XoU) plane	23,69/24=0,987	5000/4987,7 = 1,002	1,002	–
Stability under compression in XoZ (XoV) plane	–	5000/5320,58 = 0,940	0,94	–
Strength under axial compression/tension	5000/230,4/24= 0,904	0,904	0,904	–
Limit slenderness in XoY plane	–	37,9135/120 = 0,316	0,316	–
Limit slenderness in XoZ plane	–	20,08475/123,615 = 0,1625	0,1625	–