L8X6X7/8 Steel Angle — Section Properties

Dimensions

Property Value Unit
Leg Length (leg) 8 20.32 cm
Leg Thickness (t) 0.875 2.22 cm
Area (A) 11.5 in² 74.2 cm²
Weight 39.1 lb/ft 58.2 kg/m

Elastic Section Properties

Property Strong Axis (X-X) Weak Axis (Y-Y) Unit
Moment of Inertia (I) 72.4 34.9 in⁴
Elastic Section Modulus (S) 13.4 7.94 in³
Plastic Section Modulus (Z) 24.3 14.4 in³
Radius of Gyration (r) 2.5 1.74 in

Torsional Properties

Property Value Unit
Torsional Constant (J) 2.96 in⁴
Warping Constant (Cw) 11.3 in⁶

Section Profile Summary

Angles are asymmetric sections — the shear center is at the intersection of the two legs, not at the centroid. Any load applied through the centroid still produces torsion. For single-angle beams, lateral-torsional buckling must be checked per AISC 360 Eq. F10-4.

Key Design Checks (AISC 360)

Check Formula This Section
Plastic moment Mp = Zx × Fy 1,215 kip-in
Deflection Δ = 5wL⁴/(384EIx) Use Ix = 72.4 in⁴
Torsion St. Venant = GJ/L J = 2.96 in⁴
Column buckling KL/r → Fcr r_x = 2.5 in

Design Notes

Verification (AISC 360): All designs using this section must be verified by a licensed Professional Engineer. Before finalizing member selection, check beam-column interaction (P-M), lateral-torsional buckling, serviceability deflections, and all connection limit states. See Engineering Disclaimer.

Worked Example: Tension Member Check — l8x6x7-8

Scenario: A single angle used as a diagonal tension brace in a braced frame bay. The brace is connected to a gusset plate by welding one leg. Because the connection is eccentric (load applied to one leg only), shear lag reduces the effective net area.

Given:

Step 1 — Gross section yield (AISC 360 Eq. D2-1):

φTn_yield = φ × Ag × Fy = 0.9 × 11.5 × 50 = 518 kips

Step 2 — Net section fracture with shear lag (AISC 360 Eq. D2-2):

For a single angle welded to one leg, the shear lag factor U depends on the connection length and eccentricity. Per AISC 360 Table D3.1, for a single angle with 4+ fasteners per line or equivalent weld length, U = 0.80.

Ae = U × An where An = Ag (no bolt holes for welded connection)

φTn_fracture = φ × Ae × Fu = 0.75 × (0.80 × 11.5) × 65 ≈ 375 kips

Step 3 — Design check:

The smaller of the two limit states governs: φTn = min(yield, fracture) kips. Ensure the factored axial demand (from wind or seismic load combinations) does not exceed this value. Also verify the gusset plate weld length develops the required force.

Related Resources

Design Resources


Educational reference only. Verify all section properties against the current AISC 360 Manual and mill certificates before design. Results are PRELIMINARY — NOT FOR CONSTRUCTION.