AS 4100 Tension Member Design — Clause 7, Equal Angle Grade 300 Worked Example

Complete AS 4100:2020 tension member design walkthrough: gross section yielding (N_ty), net section fracture (N_tn), shear lag factor (k_t), block shear, and the combined tension-plus-bending check for eccentrically connected single angles. The worked example designs a 125x125x10EA Grade 300PLUS equal angle bracing member in a roof truss with one M20 bolt per leg.

The Steel Calculator WASM engine performs tension member checks automatically for AS 4100 (Australia), AISC 360 (USA), EN 1993 (Europe), CSA S16 (Canada), and IS 800 (India).

PRELIMINARY — NOT FOR CONSTRUCTION. All results are for educational and reference use only. Must be independently verified by a Chartered Professional Engineer (CPEng) or RPEQ before use in any project.

Design Problem Definition

A diagonal bracing member in a roof truss is fabricated from a 125x125x10 equal angle (Grade 300PLUS) connected at each end by one M20 Grade 8.8 bolt through each leg. The brace is 3.8 m long between working points and carries a factored tension force from wind uplift on the roof.

Design Data:

• Section: 125x125x10 EA (Equal Angle, AS/NZS 3679.1, Grade 300PLUS)
• Steel: f_y = 300 MPa, f_u = 440 MPa
• Bolt: M20 Grade 8.8, 22 mm hole (standard clearance per AS 4100 Table 9.6.1)
• Bolt position: 55 mm from heel of angle, on the gauge line
• Design tension force (factored): N* = 320 kN (ULS wind combination)
• Connection: One bolt per leg, both legs connected

Section Properties — 125x125x10 EA Grade 300PLUS

Step 1 — Gross Section Yielding (AS 4100 Clause 7.2)

The nominal section capacity for yielding of the gross cross-section:

N_ty = A_g x f_y = 2,300 x 300 = 690.0 x 10^3 N = 690.0 kN

Design capacity (yielding):

phi = 0.90 (AS 4100 Table 3.4, tension)

phi x N_ty = 0.90 x 690.0 = 621.0 kN

Step 2 — Net Section Fracture (AS 4100 Clause 7.3)

The net section fracture capacity accounts for the reduced cross-sectional area at the bolt holes and the non-uniform stress distribution (shear lag) in the connected element.

Net Area Calculation

With one M20 bolt through each leg with 22 mm diameter holes:

For one leg: Net width = leg width - hole diameter = 125 - 22 = 103 mm (but must account for the staggered hole pattern since both legs have holes at the same cross-section).

For an equal angle with holes in both legs on the same cross-section:

• Net area of leg 1: b_net1 = (125 - 22) x 10 = 1,030 mm^2
• Net area of leg 2: b_net2 = (125 - 22) x 10 = 1,030 mm^2
• Total net area: A_n = 1,030 + 1,030 = 2,060 mm^2

But the hole diameter must deduct the hole clearance + 2 mm for damage per AS 4100 Clause 9.1.10. Hole diameter for area deduction = 22 + 2 = 24 mm.

For one leg: A_n_leg = (125 - 24) x 10 = 1,010 mm^2 For both legs: A_n = 2 x 1,010 = 2,020 mm^2

Alternatively, using the gross area minus hole deductions: A_n = A_g - 2 x 24 x 10 = 2,300 - 480 = 1,820 mm^2. The net area is the lesser of the straight net section and the area through the critical path. For a single line of bolts on each leg at the same cross-section, the critical net section = A_n = 1,820 mm^2.

Shear Lag Factor k_t (AS 4100 Clause 7.3)

For an angle section connected by both legs with one bolt per leg, the shear lag factor from AS 4100 Table 7.3.2:

Since both legs are connected with at least one fastener each, k_t = 0.90 for sections where both elements are directly connected.

Note: If only one leg were connected (common for single-angle bracing), k_t would be 0.75 for one bolt and 0.85 for two bolts per AS 4100 Table 7.3.2.

Correction Factor for Eccentric Connection

AS 4100 Clause 7.3.2(b) requires an additional reduction when the centroid of the connected element does not coincide with the centroid of the bolt group. For an equal angle connected through both legs with bolts at identical gauge distances, the bolt group centroid coincides with the angle centroid — no additional eccentricity correction is needed for the tension capacity. However, the eccentricity moment must be checked separately per Clause 7.5 (see Step 5).

Net Section Fracture Capacity

N_tn = 0.85 x k_t x A_n x f_u

N_tn = 0.85 x 0.90 x 1,820 x 440 = 613.2 x 10^3 N = 613.2 kN

Design capacity (fracture):

phi = 0.90

phi x N_tn = 0.90 x 613.2 = 551.9 kN

Step 3 — Governing Tension Capacity

The design tension capacity is the lesser of yielding and fracture:

phi x N_t = 551.9 kN

Tension Utilisation

N* / (phi x N_t) = 320 / 551.9 = 0.580 — design is adequate with 42% reserve.

Step 4 — Block Shear (AS 4100 Clause 9.1.8)

Block shear failure involves a block of material tearing out along a perimeter defined by bolt holes. For the 125x125x10EA with one M20 bolt per leg at 55 mm from the heel:

Connected Leg Block Shear

The bolt is at 55 mm from the heel. The critical block shear path:

• Shear plane: from the bolt centre to the end of the angle in the direction of force (longitudinal). End distance = 35 mm (measured from bolt centre to the end of the angle, assuming a 90 mm end plate dimension with bolt at 55 mm).

Minimum end distance per AS 4100 Clause 9.6.2: 1.5 x d_f = 1.5 x 20 = 30 mm. 35 mm > 30 mm — acceptable.

• Shear area (gross): A_gv = 2 x (end_distance + bolt_spacing) x t

For a single bolt, the shear area is the end distance plus any distance to the nearest edge in the direction of shear. Assuming the shear path extends from the bolt to the angle end: L_v = 55 mm (distance from bolt centre to the angle end in the direction of force — measured from bolt to the end of the member).

A_gv = 2 x 55 x 10 = 1,100 mm^2 (both legs) A_nv = 2 x (55 - 12) x 10 = 2 x 43 x 10 = 860 mm^2 (deducting half a hole per shear plane)

• Tension area: Distance from bolt centre to the vertical edge of the connected leg perpendicular to the force. For a bolt at 55 mm from the heel on a 125 mm leg, the edge distance perpendicular to force = 125 - 55 = 70 mm.

A_nt = 2 x (70 - 12) x 10 = 2 x 58 x 10 = 1,160 mm^2

Block shear capacity per leg:

V_bs1 = 0.60 x f_u x A_nv + f_u x A_nt = 0.60 x 440 x 860 + 440 x 1,160 = 227,040 + 510,400 = 737,440 N

V_bs2 = 0.60 x f_y x A_gv + f_u x A_nt = 0.60 x 300 x 1,100 + 440 x 1,160 = 198,000 + 510,400 = 708,400 N

Block shear capacity = min(737.4, 708.4) = 708.4 kN (both legs combined).

phi = 0.75 (AS 4100 Table 3.4, block shear)

phi x V_bs = 0.75 x 708.4 = 531.3 kN

N* = 320 kN < 531.3 kN — block shear does not govern.

Step 5 — Combined Tension and Bending (AS 4100 Clause 7.5 and 8.3)

When a tension member is connected by bolts in only one leg (or eccentrically in both legs), the eccentricity moment must be considered. For our equal angle with both legs connected by a single bolt each, the bolt group centroid coincides with the angle centroid, so the eccentricity e = 0 and M* = 0 for the connection itself.

However, for single-angle members where only ONE leg is connected (the more common case for diagonal bracing), the eccentricity moment is:

e = distance from bolt centre to angle centroid = 55 - 36.5 = 18.5 mm (for a bolt in one leg only).

For completeness, we'll check the single-leg-connection case to illustrate:

M*_ecc = N* x e = 320 x 0.0185 = 5.92 kN.m about the minor axis.

Section Moment Capacity (Minor Axis Bending)

For an equal angle bent about the minor axis (v-v axis), the section is subject to bending about a non-principal axis because the angle section has its principal axes rotated 45 degrees from the geometric axes. The simplified approach per AS 4100 Commentary:

For bending about the axis parallel to the connected leg (geometric axis):

Z_leg = I_leg / (b - c_y) = (t x b^3 / 3) / (125 - 36.5) = (10 x 125^3 / 3) / 88.5

I_leg = 10 x 1,953,125 / 3 = 6.51 x 10^6 mm^4 Z_leg = 6.51 x 10^6 / 88.5 = 73.6 x 10^3 mm^3

M_s = Z_e x f_y. For a Class 2 angle: Z_e = min(S, 1.5 x Z). Approximating S = 1.25 x Z = 92.0 x 10^3 mm^3.

Z_e = min(92.0, 1.5 x 73.6) = min(92.0, 110.4) = 92.0 x 10^3 mm^3

M_s_leg = 92.0 x 10^3 x 300 = 27.6 kN.m phi x M_s_leg = 0.90 x 27.6 = 24.8 kN.m

Interaction Check (AS 4100 Clause 8.3.4)

For the single-leg-connected case:

N* / (phi x N_t) + M* / (phi x M_s) = 320 / 551.9 + 5.92 / 24.8 = 0.580 + 0.239 = 0.819 — still adequate.

When both legs are connected (our bracing member), M* = 0 and the interaction reduces to the pure tension check.

Step 6 — Slenderness Check

While slenderness is not a strength limit state for tension members, AS 4100 Clause 7.4 recommends a maximum slenderness ratio L/r of 300 for main tension members to prevent excessive sag and vibration during handling and erection.

L = 3,800 mm (brace length) r_v = 24.5 mm (minimum radius of gyration for an equal angle)

L / r_v = 3,800 / 24.5 = 155

155 < 300 — slenderness is acceptable.

AS 4100 vs AISC 360 Tension Member Comparison

Key difference: AS 4100 uses a single phi = 0.90 for both tension limit states, while AISC 360 uses phi_t = 0.75 for fracture, making AS 4100 less conservative for net section fracture. AS 4100 also separates the eccentricity moment check (Clause 7.5) from the shear lag factor, whereas AISC 360 incorporates eccentricity effects into the shear lag factor U.

Frequently Asked Questions

How does AS 4100 calculate the tension capacity of a steel member?

AS 4100 Clause 7.2 specifies two limit states for tension members: gross section yielding (N_t = A_g x f_y) and net section fracture (N_t = 0.85 x k_t x A_n x f_u). The design tension capacity phi-N_t is the lesser of the two, with phi = 0.90 for both limit states. For a 125x125x10EA Grade 300PLUS angle with one M20 bolt hole in each leg, gross yielding gives 621 kN and net fracture with shear lag factor k_t = 0.75 gives 493 kN — net section fracture governs.

What is the shear lag factor k_t in AS 4100 tension member design?

The shear lag factor k_t per AS 4100 Clause 7.3 accounts for non-uniform stress distribution when a tension member is connected by only part of its cross-section. For angle sections connected by one leg with one bolt: k_t = 0.75. For two bolts in line: k_t = 0.85. For welded connections extending at least the outstand length: k_t = 0.90. For sections where all elements are connected (e.g. universal beams with flange connections): k_t = 1.0. The k_t factor directly reduces the net section fracture capacity and is one of the most frequently overlooked parameters in Australian steel tension design.

How does block shear failure differ from net section fracture in AS 4100?

Block shear failure per AS 4100 Clause 9.1.8 is a bolt-group failure mode where a block of material tears out along a perimeter defined by bolt holes. It combines shear on one plane and tension on a perpendicular plane. The nominal block shear capacity V_bs = 0.60 x f_u x A_nv + f_u x A_nt, capped at 0.60 x f_y x A_gv + f_u x A_nt. For an equal angle with a single bolt in each leg, the block shear capacity is checked at both the connected leg bolt group and the outstanding leg, but block shear rarely governs for single-bolt angle connections where the bolt spacing is small.

When does an equal angle tension member need to be checked for combined tension and bending?

Single-angle tension members connected by one leg carry the tension force eccentrically because the centroid of the bolt group lies outside the centroid of the angle cross-section. AS 4100 Clause 7.5 requires this eccentricity moment M* = N* x e (where e is the distance from the bolt centre to the angle centroid) to be considered even if the connection is nominally pinned. For a 125x125x10EA connected through one leg with one M20 bolt at 55 mm from the heel, the eccentricity e = 36.5 mm and the moment M* = 0.75 x 493 x 0.0365 = 13.5 kN.m. The combined tension plus bending interaction per Clause 8.3 must be checked, and this often governs for single-bolt connections despite the primary load being pure tension.

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Related Pages

• AS 4100 Beam Design — 310UB40.4 Worked Example — Flexure, shear, LTB
• AS 4100 Column Design — 200UC46 Worked Example — Axial and buckling
• AS 4100 Bolt Design Guide — Clause 9.3 — Bolted connections
• AS 4100 Fillet Weld Design — Clause 9.7 — Weld capacity
• AS 4100 Load Combinations — ULS/SLS/STB
• Tension Member Design — Multi-Code Reference — AISC, EN 1993, AS 4100
• Block Shear — AISC Design Guide — Block shear across codes
• Bolt Grades — AS 1252 Reference — Australian bolt specification

This page is for educational reference. All resistance formulae are per AS 4100:2020 with AS/NZS 3679.1 section properties. Verify the applicable edition of the National Construction Code for your project jurisdiction. Results are PRELIMINARY — NOT FOR CONSTRUCTION without independent review by a registered structural engineer (CPEng/RPEQ).