D4.E.6.1 Members Subject to Axial Forces

Axial Tension

The criteria governing the capacity of tension members are based on two limit states: resistance due to yielding and resistance due to rupture. The resistance due to rupture depends on effective net section area. You may specify the net section area through the NSF design parameter. Additionally, the shear lag factor, U, may be entered using the SLF parameter. STAAD.Pro calculates the tension capacity of a member based on these two limits states per Cl.13.2 of CAN/CSA-S16-09. Design parameters FYLD, FU, NSF, and SLF (Refer to D4.E.7 Design Parameters) are applicable for these calculations

Yielding, per Cl. 13.2(a)
$T_{r} = ϕ A_{g} F_{y}$
Rupture, per Cl. 13.2 (b)
$T_{r} = ϕ_{u} A_{n e} F_{u}$

Note: Pin connection equations in S16-14 and 19 are not checked by the program.

Axial Compression

The compressive resistance of columns is determined based on Clause 13.3 of the code. The equations presented in this section of the code assume that the compressive resistance is a function of the compressive strength of the gross section (Gross section Area times the Yield Strength) as well as the slenderness factor (KL/r ratios). The effective length for the calculation of compression resistance may be provided through the use of the parameters KT, KY, KZ, LT, LY, and LZ (Refer to D4.E.7 Design Parameters). Some of the aspects of the axial compression capacity calculations are :

For doubly symmetric sections meeting the requirement of Table 1, resistance is:

Resistance due to Major axis buckling per Cl. 13.3.1.

Resistance due to Minor axis buckling per Cl. 13.3.1

C_{r} = ϕ A F_{y} {(1 + λ^{2 n})}^{- 1 / n}

where

n	=	1.34 for hot-rolled, fabricated structural sections and hollow structural sections manufactured in accordance with CSA G40.20, Class C (cold-formed non-stress-relieved) 2.24 for doubly symmetric welded three-plate members with flange edges oxy-flame-cut and hollow structural sections manufactured in accordance with CSA G40.20, Class H (hot-formed or cold-formed stress-relieved) Design parameters `NCR` and `STP` are used to evaluate the value of n for a member.
λ	=	$\sqrt{F_{y} / F_{e}}$
F_e	=	$\frac{π^{2} E}{{(\frac{k L}{r})}^{2}}$

Note: For CSA S16-19, the value of F_e is used per 13.3.1.1 a) for double-symmetric sections.

For any other section not covered under Cl. 13.3.1, the factored compressive resistance, C_r , is computed using the expression given in Cl. 13.3.1 with a value of n = 1.34 and the value of F_e taken as follows:

For doubly symmetric sections and axisymmetric sections, the least of F_ex , F_ey , and F_ez .

For singly symmetric sections with the Y axis taken as the axis of symmetry, the lesser of F_ex and F_eyz where

F_eyz	=	$\frac{F_{e y} + F_{e z}}{2 Ω} [1 - \sqrt{1 - \frac{4 F_{e y} F_{e z} Ω}{{(F_{e y} + F_{e z})}^{2}}}]$
F_ex	=	$\frac{π^{2} E}{{(\frac{k_{x} L_{x}}{r_{x}})}^{2}}$
F_ey	=	$\frac{π^{2} E}{{(\frac{k_{y} L_{y}}{r_{y}})}^{2}}$
F_ez	=	$[\frac{π^{2} E C_{w}}{{(K_{z} L_{z})}^{2}} + G J] \frac{1}{A {\bar{r}}_{0}^{2}}$
x₀,y₀	=	the principal coordinates of the shear center with respect to the centroid of the cross section
${\overset{⎯}{r}}_{0}^{2}$	=	$x_{0}^{2} + y_{0}^{2} + r_{x}^{2} + r_{y}^{2}$
Ω	=	$\frac{r_{x}^{2} + r_{y}^{2}}{{\overset{⎯}{r}}_{0}^{2}}$

For asymmetric sections the smallest root of:
$(F_{e} - F_{e x}) (F_{e} - F_{e y}) (F_{e} - F_{e z}) - F_{e}^{2} (F_{e} - F_{e y}) {(\frac{x_{0}}{{\overset{⎯}{r}}_{0}})}^{2} - F_{e}^{2} (F_{e} - F_{e x}) {(\frac{y_{0}}{{\overset{⎯}{r}}_{0}})}^{2} = 0$

For Class 4 member subjected to axial compression, the factored compressive resistance is:
$C_{r} = ϕ A_{e} F_{y} {(1 - λ^{2 n})}^{- 1 / n}$

A_e is calculated using reduced element widths meeting the maximum width to thickness ratio specified in Table 1.
Effective width required for the calculation of effective area A_e , for different section shapes are as follows.
- For flanges of I-section, T-section and channel section and legs of angle section
  $b_{e} = 200 t / \sqrt{F_{y}}$
- For stem of T-section
  $b_{e} = 340 t / \sqrt{F_{y}}$
- For flanges of HSS rectangular or Tube sections
  $b_{e} = 670 t / \sqrt{F_{y}}$
- For circular HSS or Pipe section
  D = 23,000t/(F_y