Unequal Angle Capacity Check to SCI P291

1. Inputs

Section Dimensions (mm)

L_a = mm   Length of leg a

L_b = mm   Length of leg b

t = mm   Thickness

Figure 1: Geometry

L-Section Dimensions

Material Properties (MPa)

f_y = MPa   Yield strength

f_u = MPa   Ultimate tensile strength

G = MPa   Shear modulus

E = MPa   Elastic modulus

Material Factors

γ_m0 =   Material factor for cross-section resistance

γ_m1 =   Material factor for buckling resistance

Applied Loads

N_ed = kN   Axial force (compression is positive)

M_ed,x = kNm   Moment about major axis

M_ed,y = kNm   Moment about minor axis

V_ed,x = kN   Shear force parallel to leg b

V_ed,y = kN   Shear force parallel to leg a

L_buckling = mm   Buckling length

2. Calculate Section Properties

A_g = (L_a + L_b − t) · t = 1900 mm²   Gross area

A_v,x = L_b · t = 1000 mm²   Shear area for V_ed,x

A_v,y = L_a · t = 1000 mm²   Shear area for V_ed,y

Z_pl,x = 10000 mm³   Plastic modulus (major axi)

Z_pl,y = 10000 mm³   Plastic modulus (minor axi)

3. Design Checks

Axial Compression Check

N_c,Rd = f_y · A_g / γ_m0 = 362.73 kN

Axial utilization: Util_axial = 0.634

Shear Check

τ_y = f_y / √3 = 121.24 MPa

V_pl,Rd,x = A_v,x · τ_y / γ_m0 = 110.22 kN

V_pl,Rd,y = A_v,y · τ_y / γ_m0 = 110.22 kN

Shear utilization (x): Util_shear,x = 0.000

Shear utilization (y): Util_shear,y = 0.000

Max shear utilization: Util_shear,max = 0.000

Bending Check

M_c,Rd,x = Z_pl,x · f_y / γ_m0 = 1.91 kNm

M_c,Rd,y = Z_pl,y · f_y / γ_m0 = 1.91 kNm

Bending utilization (x): Util_bend,x = 0.000

Bending utilization (y): Util_bend,y = 0.000

Combined bending utilization: Util_{bend,combined} = 0.000

Combined Forces Check (Simplified)

Combined utilization (axial + bending): Util_combined = 0.634

4. Summary