Check of the Load-bearing Capacity of an Axially Compressed Column

Objective: Check of the calculation of the resistance of sections.

Task: Verify the correctness of the stability analysis of normal sections.

References: Nasonov S.B. Manual on design and analysis of building structures. – M: ASV Publishing House, 2013. – p. 66-67.

Initial data file:
Example 4.SAV;
report – Decor 4.doc.

Software version: DECOR 21.1.1.1, 17.02.2016.

Compliance with the codes: SNiP ІІ-25-80, SP 64.13330.2011.

Initial data from the source:

b×h = 10×15 cm	Section sizes of the element
l = 2,5 m	Column height
μx = μy = 1	Effective length factors
N = 60 kN	Compressive force
Material of the element: pine.
Grade of wood: 2.
Operating conditions class: 1 (А2 according to SNiP ІІ-25-80).

 

DECOR initial data:

 

Importance factor γ_n = 1

Service factors
Service factor  for temperature and humidity operating conditions mВ	1
Allowance for the temperature conditions of operation mТ	1
Allowance for the duration of loading md	1
Service factor under short-term loads mn	1
Factor that allows for the effect of impregnation with protective substances mа	1

Wood species - Pine
Grade of wood - 2
Limit slenderness of members in tension - 200
Limit slenderness of members in compression - 120
Member length 2,5 m

Effective length factor in the XoY plane - 1

Effective length factor in the XoZ plane – 1

Section:

b = 100 mm h = 150 mm Non-glued timber section

Forces:

N = -60 kN
M_y = 0 kN*m
Q_z = 0 kN
M_z = 0 kN*m
Q_y = 0 kN

Comparison of solutions:

File	Example 4.SAV
Report file	Decor 4.doc
Check	Stability in the XOZ plane under a longitudinal force	Stability in the XOY plane under a longitudinal force
Theory	0,55/1,3 = 0,423	1/1,3 = 0,769
DECOR	0,42	0,769
Deviation, %	0,83 %	–

 

Comments:

1. Service factor for 1 (А2) class m_в = 1 (table 5 of SNiP ІІ-25-80, table 7 of SP 64.13330.2011).
2. Limit slenderness of the compressed column is equal to λ_max = 120 (table 14 of SNiP ІІ-25-80, table 17 of SP 64.13330.2011).