• Introduction:

The following is a validation problem for a detailed calculation of the NBR 8800 HSS Round section design.

The NBR 8800 implementation was an integral part of STAAD(X) SS3 release.

• Validation Problem:

A 6 m high cantilever column formed from a 12.75 in diameter circular hollow section (HSS12.750X0.500, A500 Gr. B steel), is checked for an applied a concentrated axial compression load of 2340 KN (FY) at the top (free end).

 Sample Model:NBR-8800_Pipe.STDX

Section Properties [HSS Round]:

HSS12.750X0.500

OD = 323.85 mm, TDES = 11.81 mm, AX = 11548.4 mm2, I_xx = I_yy = 141100950 mm4,

J = 282201910 mm4,

r_x = r_y = 110.5 mm, Z_ex = Z_ey = 1150371.89 mm3, S_x = S_y = 871406.22 mm3

Material Properties:

A500 Gr. B

E = 199948 MPa, G = 76903.1 MPa, fy = 300 MPa, fu = 400 MPa, µ = 0.3

• Manual Calculation:

Section Classification:

1. Axial Compression

a. Flange Classification:

D/t = 323.85/11.81 = 27.42, 1.4*  = 36.14

So, D/t < 1.4* . Hence, section is Non-slender.

b. Web Classification:

D/t = 323.85/11.81 = 27.42, 1.4*  = 36.14

So, D/t < 1.4* . Hence, section is Non-slender.

2. Flexure

a. Flange Classification:

λ = D/t = 27.42

From Annex G.2.7, λp = 0.07E/fy = 46.655

λr = 0.31E/fy = 206.613

λ < λp, hence section is compact.

b. Web Classification:

λ = D/t = 27.42

From Annex G.2.7, λp = 0.07E/fy = 46.655

λr = 0.31E/fy = 206.613

λ < λp, hence section is compact.

Axial Tension:

For normal load combination, γa1 = 1.1 and γa2 = 1.35 [Table 3, NBR-8800]

Tensile Yielding:

N_tRdy = Ag*fy/γa1 = 11548.4*300/ (1.1*1000) KN = 3149.56 KN

Tensile Rupture:

N_tRdr = Ae*fe/γa2 = 11548.4*400/ (1.35*1000) KN = 3421.748 KN

Axial Compression:

According to clause E.1.1:

Flexural Buckling about major axis:

Effective Length Factor Kx = 1.0

Elastic flexural buckling force Nex = π²*E*Ix/(Kx*Lx²) = 3.1416^2*199948*141100950/(1.0*6000^2*1000)

= 7734.75 KN

0.11*E/fy = 73.314

So, D/t < 0.11*E/fy. Hence, from clause F.4, Net stress reduction factor Q = 1.0

For λ0 <= 1.5, Reduction factor associated with resistance to compression χ = 0.658 ^{λ0^2} = 0.829 (Clause 5.3.3.1)

Now, Axial Compression Capacity N_c,Rd = χ*Q*Ag*fy/ γa1 = 0.829*1.0*11548.4*300/(1.1*1000) = 2610.99 KN

Ratio = 2340/2610.99 = 0.896

Flexural Buckling about minor axis:

Effective Length Factor Ky = 1.0

Elastic flexural buckling force Ney = π²*E*Iy/(Ky*Ly²) = 3.1416^2*199948*141100950/(1.0*6000^2*1000)

= 7734.75 KN

0.11*E/fy = 73.314

So, D/t < 0.11*E/fy. Hence, from clause F.4, Net stress reduction factor Q = 1.0

For λ0 <= 1.5, Reduction factor associated with resistance to compression χ = 0.658 ^{λ0^2} = 0.829 (Clause 5.3.3.1)

Now, Axial Compression Capacity N_c,Rd = χ*Q*Ag*fy/ γa1 = 0.829*1.0*11548.4*300/(1.1*1000) = 2610.99 KN

Ratio = 2340/2610.99 = 0.896

Flexure:

Bending about Major Axis:

Yielding:

Yielding Capacity M_Rd = 1.5*W*fy/ γa1 = 1.5*871406.22*300/(1.1*1000000) KN-m = 356.484 KN-m

Ratio = 0

Local Buckling:

Plastic Bending Moment M_pl = Zx*fy = 1150371.89*300/1000000 KN-m = 345.112 KN-m

λ < λp, hence from clause G.2.7, Nominal flexural strength M_{Rd, LB} = M_pl/ γ_a1 = 313.738 KN-m

Bending about Minor Axis:

Yielding:

Yielding Capacity M_Rd = 1.5*W*fy/ γa1 = 1.5*871406.22*300/(1.1*1000000) KN-m = 356.484 KN-m

Ratio = 0

Local Buckling:

Plastic Bending Moment M_pl = Zx*fy = 1150371.89*300/1000000 KN-m = 345.112 KN-m

λ < λp, hence from clause G.2.7, Nominal flexural strength M_{Rd, LB} = M_pl/ γ_a1 = 313.738 KN-m

Shear:

Torsion:

Torsion Shear Modulus W_t = π*(D-t) ²*t/2 = 3.1416*(323.85-11.81)²*11.81/2 = 1806306.01 mm³

We will take minimum of these two.

0.60*W_t*fy/γ_a1 = 295.58 KN-m

T_Rd 1 = [1.23*1806306.01*199948]/[1.1*62.7505*4.30431*10⁶] = 1495.202 KN-m > 295.58 KN-m

T_Rd 2 = [0.6*1806306.01*199948]/[1.1*143.5956*10⁶] = 1371.911 KN-m > 295.58 KN-m

So, T_Rd = 295.58 KN-m

• Summary Table: