Analysis of an Axially Compressed Welded I-beam Column

Objective: Check of the Resistance of Sections mode

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:

4.6.sav;
report — Kristall4.6.doc

Initial data:

Ry = 24 kN/cm2	Steel grade C245;
l = 6,5 m	Column height;
N = 5000 kN	Design longitudinal compressive force;
μ = 0,7	The lower restraint is rigid and the upper one is pinned for both principal planes of inertia;
γc = 1	Service factor;
A = 230,4 cm2, Iy = 118243,584 cm4, Iz = 33184,512 cm4 Wy = 4583,085 cm3, Wz = 1382,688 cm3 iy = 22,654 cm, iz = 12,001 cm	Geometric properties for a welded I-section with a web 480×12 mm and flanges 480×18 mm;

 

KRISTALL parameters:

Steel: C245
Group of structures according to the table 50* of SNiP II-23-81* 3
Importance factor 1
Service factor  1
Limit slenderness for members in compression: 180 - 60α
Limit slenderness for members in tension: 250

Section:

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

1. Load-bearing capacity of the element under axial compression/tension:

\[ N=AR_{y} \gamma_{c} =230,4\cdot 24\cdot 1=5529,6 \quad кН. \]

2. Slenderness of the element for both principal planes of inertia:

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

3. Conditional slenderness of the element for both principal planes of inertia:

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

4. Buckling coefficients under axial compression:

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

5. Load-bearing capacity of the element at its buckling:

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

6. Limit slenderness:

\[ \lambda_{uy} =180-60\cdot \frac{N}{\varphi_{y} AR_{y} \gamma_{c} }=180-60\cdot \frac{5000}{0,9622\cdot 230,4\cdot 24\cdot 1}=123,615; \] \[ \lambda_{uz} =180-60\cdot \frac{N}{\varphi_{z} AR_{y} \gamma_{c} }=180-60\cdot \frac{5000}{0,902\cdot 230,4\cdot 24\cdot 1}=119,852. \]

Comparison of solutions:

Factor	Source	Manual calculation	KRISTALL	Deviation %
Strength under combined action of longitudinal force and bending moments, no plasticity	–	5000/5529,6 =  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 = 0,94	0,94	–
Strength under axial compression/tension	0,904	5000/5529,6 =  0,904	0,904	–
Limit slenderness in XoY plane	–	37,9135/119,852 = 0,316	0,316	–
Limit slenderness in XoZ plane	–	20,085/123,615= 0,162	0,162	–