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Design Guide: Limit State Design of steel members to AS4100's banner
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Design Guide: Limit State Design of steel members to AS4100

The limit state is the point at which a member, assembly, or entire structure no longer fulfills required design criteria such as structural integrity, fitness for use, or durability. The steel limit state requirements in Australia are specified in Australian Standard AS4100 so that the steel members won’t fail and remain fit for their designed use. Limit state design of steel structures method ensures that structures consider serviceability and ultimate load states. In other words, both the functionality, appearance (Serviceability Limit State, SLS), and the safety of a structure and its users (Ultimate Limit State, ULS) are verified through unique design methods and sets of parameters.
Structural steel design calculations require a complex and thorough process. The impact that the execution of these steps has on the final product is huge, so engineers must understand these intricate steps and make use of their knowledge of material behavior and properties to be able to predict, analyze and understand the suitability of their design to ensure that is it robust, safe and fit for purpose. For that reason, we’ve detailed a quick summary of the limit state design of steel members to AS4100.
Table 1: Summary of Steel Failure modes

Some important parameters for steel design
Basic values of steel consider mostly hot-rolled steel, but the Australian Standard can also apply to cold-rolled steel (Clause 1.4 AS4100):
  1. Young’s modulus of elasticity: E = 200 000 MPa
  2. Shear modulus: G = 80 000 MPa
  3. Coefficient of thermal expansion for steel: αT = 11.7×10-6 per degree Celsius
  4. Poisson’s ratio: ν = 0.25
  5. Density: ρ = 7850 kg/m3

To design an appropriate steel member, calculations must satisfy:
S* < 𝞍R
  1. Capacity Factor 𝞍: Also considered a ‘safety factor, this factor decreases the allowable loading capacity of members. It accounts for potential issues such as uncertainties in the condition of materials and allows for the possibility that a structural element might not achieve its predicted strength properties. It is provided in Table 3.4 AS4100 and depends on the type of structure, and steel members included.
  2. Structural analysis S*: To quantify the effect of design actions, this parameter needs to combine with information from the Design Codes for Strength Limit State, AS1170.1, and AS1170.2, to provide appropriate loading conditions. S* refers to the design action effect, referring to the actions, such as design bending moments (M*), shear forces (V*), or axial loads (N*), calculated from design actions or design loads using the appropriate method of analysis.
  3. Design capacity calculation R: The calculated nominal capacity of a member or connection to resist the imposed design action. What follows is a quick summary.

Design for Axial Compression

  1. Find the form factor kf.
  1. The form factor is always equal to or smaller than 1. It’s determined following Clause 6.2.2 and is calculated as:

kf=AeAgk_f = {\frac{A_e}{A_g}}

Where,

Ae=the effective areaAe = \text {the effective area} \\
and

Ag=the gross area of the sectionAg = \text{the gross area of the section}\\

  1. If kf < 1, then the members will reach localized buckling before reaching yield stress.
  2. kf is found in manufacturer tables, for example, OneSteel Manufacturing.

  1. Find the net area of the cross-section (An) and full yield stress (fy) of the section.
  1. The net area of cross-section (An) and full yield stress (fy) can be found in manufacturer tables or calculated for custom sections.
An = Ag - Area of holes or other connections features
  1. The yield stress of the web and flange might be different, use the lower value.

  1. Calculate section capacity (Ns) by the formula:

Ns=kfAnfyN_s = {k_f}{A_n}{f_y}

  1. Find member slenderness reduction factor ⍺c
Find modified slenderness ratio (λn) using AS4100 Clause 6.3.2.

λn=(ler)Kffy250\lambda_n = {(\frac{l_e}{r})}{\sqrt{K_f}}{\sqrt{\frac{f_y}{250}}}

  1. Where le = effective length = ke x length. ke can be found in Figure 4.6.3.2 AS4100
  2. r = the radius of gyration (In the manufacturer tables)
  3. Members could buckle about 2 axes (x and y) so we need to evaluate lex/rx and ley/ry and use the larger value (the more critical one).
  4. Find yield stress (fy) for steel grade in Table 2.1 AS4100‍
  5. Find member section constant (⍺b) according to Table 6.3.3(1) & 6.3.3(2) AS4100‍
  6. Use AS4100 Table 6.3.3(3) to find ⍺c

  1. Calculate the member capacity (Nc) by the formula:

Nc=αcNsN_c = {\alpha_{c}}{N_s}
  1. Check if the capacity meets the design axial compression force: N* < 𝞍N


Design for Axial Tension

There are two possible modes of failure for tensile axial load, including yielding of members along its entire length or rupture of steel where the cross-section has been locally reduced (holes, connections, etc).
  1. Calculate section capacity due to yielding on gross section

Nt=AgfyN_t = {A_g}{f_y}
  1. Find gross cross-section area (Ag) using manufacturer tables or by manual calculation.
  2. Find yield stress (fy) for steel grade in Table 2.1 AS4100
  1. Calculate section capacity due to rupture on a net section per Clause 7.2 AS4100

Nt=0.85ktAnfuN_t = {0.85}{k_t}{A_n}{f_u}

  1. Find the correction Factor for the distribution of forces (kt) based on Clause 7.3 AS4100.
  2. Find cross-sections where there are holes (due to connections, bolts, etc.) where rupture is likely to happen and calculate the net area of cross-section (An)
  3. An be calculated as per Clause 9.1.10 AS4100
  4. Use Table 2.1 AS4100 to find the tensile strength used in design (fu)

  1. Compare section capacity between gross and net sections, and choose the smaller one as the main tensile capacity Nt
  2. Check if the capacity meets the design axial tensile force: N* < 𝞍Nt

Design for Bending/Flexure strength

Bending strength over the steel beams or columns might lead to two failure modes: localized plate buckling and lateral buckling of the compression flange.
For localized plate buckling, the buckling will appear at one specific point (usually at the maximum bending moment). The capacity at which the beam reaches the localized plate buckling is called Ms, and it is because the beam is fully restrained against lateral buckling so another failure mode is likely not to happen.
  1. Check section moment capacity

Ms=fyZeM_s = {f_y}{Z_e}
  1. Ze = Effective section modulus, which can be found in the appropriate tables in manufacturer guides that specify material properties of different steel members or specified in Clauses 5.2.3, 5.2.4, or 5.2.5 AS4100 as appropriate.
  2. fy = yield stress of the steel grade
  1. Find member moment capacity:

Mb=αmαsMsM_b = {\alpha_m}{\alpha_s}{M_s}

  1. Calculate the moment modification factor (αm) according to Tables 5.6.1. and 5.6.2 AS4100 OR Equation in Clause 5.6.1.1(a)iii AS4100:

αm=1.7Mm∗(M2∗)2+(M3∗)2+(M4∗)2≤2.5\alpha_m = {\frac{1.7M^*_m}{\sqrt{(M^*_2)^2 + (M^*_3)^2 + (M^*_4)^2}}\leq 2.5}

Where

Mm∗=maximum design bending moment in the segmentM2∗,M4∗=design bending moments at the quarter points of the segmentM3∗=design bending moment at the midpoint of the segmentM^*_m=\text{maximum design bending moment in the segment}\\ M^*_2, M^*_4 = \text{design bending moments at the quarter points of the segment}\\ M^*_3 = \text{design bending moment at the midpoint of the segment}\\

  1. If the beam has different bending moment profiles along its length, then we segment the beam and assess (αm) and (αs) in each segment.
  2. Calculate slenderness reduction factor (αs) according to the formula (Clause 5.6.1.1(2) AS4100):

αs=0.6[MsMoa2+3−MsMoa]\alpha_s = 0.6[\sqrt{\frac{M_s}{M_{oa}}^2 + 3}-\frac{M_s}{M_{oa}}]
Where

Moa=MoM_{oa}=M_o
where Mo is the reference buckling moment

Find the member’s effective length le with the formula:

le=ktklkrll_e = k_tk_lk_rl
kt is found in Clause 5.6.3 (1), and it is only relevant for a partially restrained section
kl is found in Clause 5.6.3 (2) and it is only relevant if the load applied at the top flange of the beam can move sideways with buckling flange
kr is found in Clause 5.6.3 (3) and it allows for restraint provided at the segment ends against lateral rotation.
l is the segment length.

  1. Using le above, find the reference buckling moment (Mo) with the formula:

Mo=(π2EIyle2)[GJ+(π2EIwle2)]M_o = \sqrt{(\frac{\pi^2EI_y}{le^2})[GJ + (\frac{\pi^2EI_w}{le^2})]}
E, G = the elastic moduli (see Clause 1.4 AS4100)
Iy, J, and Iw = section constants (see Clause 1.4 AS4100)
le = the effective length determined in accordance with Clause 5.6.3 AS4100

💡NOTE: Values of E and G and expressions for J and Iw are given in Paragraph H4 of Appendix H AS4100.
  1. Find αs as the formula above
  2. Calculate (αm . αs) for each segment
  3. Choose the smallest (αm . αs)
  4. If αm. αs > 1 then use αm. αs = 1
  5. Find Mb

  1. Use Mb as the final member capacity and perform design checks: M* < 𝞍Mb

Design for Shear Capacity

There are two possible modes of shear failure; yielding and buckling. That is why there are two shear capacities for two modes of failure.
Compare the ratio dp/tw to:

82fy250\frac{82}{\sqrt{\frac{f_y}{250}}}
  1. dp = total depth of the steel section
  2. tw = thickness of the web
  3. fy = yield stress for steel grade in Table 2.1 AS4100

2. If

dptw≤82fy250\frac{d_p}{t_w}\leq \frac{82}{\sqrt{\frac{f_y}{250}}}
then the capacity of the beam is limited by yielding, so we use Clause 5.11.4 AS4100 to find shear yield capacity (Vv) according to the formula:

Vv=0.6fyAwV_v = 0.6f_yA_w
  1. Calculate the area of the web:

Aw=d1∗twA_w = d_1 * t_w
Design for Shear Capacity diagram

  1. fy = yield stress in the manufacturer tables

3. If

dptw≥82fy250\frac{d_p}{t_w}\geq \frac{82}{\sqrt{\frac{f_y}{250}}}
then the capacity of the beam is limited by buckling, so we use Clause 5.11.5 AS4100 to find shear buckling capacity (Vb) which is dependent on whether the web is stiffened or unstiffened.
4. Check if the capacity (buckling or yielding) meets the design shear force:
V* < 𝞍Vw

Design for Bearing Capacity

Bearing capacity failure usually happens near the supports where highly concentrated reaction forces propagate. For I-sections, this can result in the crushing (yielding) of the web directly above the flange or buckling close to where the force is applied. That is why there are two bearing capacities for the webs of I-sections; bearing yield capacity and bearing buckling capacity. Engineers need to check both.
  1. Calculate the bearing yield capacity (Rby) according to Clause 5.13.3 AS4100, with the formula:

Rby=1.25bbftwfyR_{by} = 1.25b_{bf}t_wf_y
  1. Use the cross-section dimension of the web as An = tw x bb
Where,
tw = thickness of the web
bb can be calculated from Figure 5.13.1.1 AS4100
  1. Use ⍺b = 0.5 & kf = 1 to find ⍺c
  2. The slenderness ratio for an I or C section can be taken as le/r = 2.5 d1/tw
  1. Choose the lesser between Rby and Rbb to be used as the overall bearing capacity Rb
  2. Check if the capacity meets the design shear force: R* < 𝞍Rb

Design for Combined Axial load and Bending

Typically, steel members will be subjected to combined axial load and bending. So as well as checking the axial load capacity and the bending capacity in isolation, engineers must also consider combined axial and bending actions in certain load cases. In many cases, this will be the dominant condition for steel members. This can lead to one of two failure mechanisms: in-plane buckling and out-of-plane buckling.
Combined Axial load and Bending diagram

  1. Assess the secondary effects by calculating the moment amplification factor (𝛿b) according to Clause 4.4.2.2. AS4100, with the formula:

δb=cm1−N∗Nomb≥1\delta_b = \frac{c_m}{1-\frac{N*}{N_{omb}}}\geq 1
  1. cm = 0.6 - 0.4ꞵm

Where
Ꞵm ratio of smaller to the larger bending moment at the ends of the member
Some typical values of Ꞵm could be found in figure 4.4.2.2 (A) AS4100
  1. N* = design axial compressive force
  2. Nomb = elastic buckling load, determined by Clause 4.6.2. AS4100

  1. Calculate the Design bending moment

M∗=δbMm∗M^* = \delta_bM_m^*
  1. Mm* is the maximum bending moment along the member

  1. Check in-plane capacity for compression members according to Clause 8.4.2 AS4100 with the formula:

Mx∗ϕMsx+N∗ϕNcx≤1\frac{M^*_x}{\phi{M_{sx}}} + \frac{N^*}{\phi{N_{cx}}}\leq 1

The tension members won’t cause buckling so there is no need to check combined action
  1. Find ultimate design bending moment (Mx*) and axial compression (N*) (use the moment amplification factor)
  2. Find the section capacity in bending (Msx) and axial compression (Ncx) in the x-axis direction (in-plane direction)

  1. Check out-of-plane capacity for compression members according to Clause 8.4.4 AS4100 with the formula:

Mx∗ϕMbx+N∗ϕNcy≤1\frac{M^*_x}{\phi{M_{bx}}} + \frac{N^*}{\phi{N_{cy}}}\leq 1
  1. Find ultimate design bending moment (Mx*) and axial compression (N*) (use the moment amplification factor)
  2. Find the section capacity in bending (Mbx) as per clause 5.6 and axial tension (Nt) as per Clause 7.2

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📚 References

  1. [1] Gorenc., B.E. Tinyou., R. Syam., A. (2012). The Steel Designer's Handbook (8th Ed.). UNSW Press.