Applies To
	Product(s):	STAAD.Pro
	Version(s):	All
	Environment:	N/A
	Area:	Steel Design
	Subarea:	Verification
	Original Author:	Anisurya Ghosh BSW-Structural Design, Bentley Kolkata

•  Introduction:

The following is a validation problem for a detailed calculation of the axial check for the Russian SP 16 I section design.

• Validation Problem:

A 7.5 m long column formed from a HD320X127 I section (Steel grade: S235), is checked for an applied concentrated load of 3500 KN at the tip.

 

• Input File:

STAAD PLANE
INPUT WIDTH 79
UNIT METER KN
JOINT COORDINATES
1 0 0 0; 2 0 7.5 0;
MEMBER INCIDENCES
1 1 2;
DEFINE MATERIAL START
ISOTROPIC STEEL
E 2.05e+008
POISSON 0.3
DENSITY 76.8195
ALPHA 1.2e-005
DAMP 0.03
TYPE STEEL
STRENGTH FY 253200 FU 407800 RY 1.5 RT 1.2
END DEFINE MATERIAL
MEMBER PROPERTY EUROPEAN
1 TABLE ST HD320X127
CONSTANTS
MATERIAL STEEL ALL
SUPPORTS
1 PINNED
2 FIXED BUT FY MZ
LOAD 1 LOADTYPE None TITLE LOAD CASE 1
JOINT LOAD
2 FY -3500
PERFORM ANALYSIS
PARAMETER 1
CODE RUSSIAN
KY 0.75 ALL
KZ 0.75 ALL
SGR 1 ALL
TRACK 2 ALL
CHECK CODE ALL
FINISH

• Manual Calculation:

Section Properties:

HD320X127

D= 320 mm, B= 300 mm, Tf= 20.5 mm, Tw= 11.5 mm, Ix= 30820 cm⁴, Iy= 9239 cm⁴, Wx= 1926.5 cm³, Sx= 1070 cm³, J= 225.1 cm⁴, H= 2.069*10^12 mm⁶

rx = 13.8 cm, ry = 7.57 cm, An = 161 cm²

Material Properties:

S235

E = 205000 MPa, fy = 235 MPa, γc1 = 1, γc2 = 1

Check for Axial Force:

• As per clause 7.1.1, we need to satisfy eqn. 5 of SP-16.13330.2011:

N/(A_n*R_y*γ_c) < = 1

So, 3500/(161*10^-4*235*1000*1.0) = 0.9251 < 1 [OK]

STAAD.Pro value is 0.92 [Matching]

KY = KZ = 0.75

λx = KL/r = 7.5*0.75/0.138 = 40.76

λy = KL/r = 7.5*0.75/0.0757 = 74.3

λ̅x = λx*√(Ry/E) = 1.38

λ̅y = λy*√(Ry/E) = 2.516

 λ̅ = 2.516

 

• As per clause 7.1.3, we need to satisfy eqn. 5 of SP-16.13330.2011:

N/(ф*A*R_y*γ_c) < = 1 [eqn. 7]

From eqn. 8, we get:

Ф = 0.5*[∂-√(∂²-39.48*λ̅²)]/(λ̅)²

From eqn. 9, ∂ = 9.87*(1-α+β*λ̅) + λ̅²

From table 7, we get α = 0.04, β = 0.09

So, ∂ = 9.87*(1-0.04+0.09*2.516) + 2.516² = 18.040

Now, ф = 0.5*[18.04-√(18.04²-39.48*2.516²)]/2.516² = 0.7385 < 7.6/λ̅², i.e. 1.2006 [OK]

N/(ф*A*R_y*γ_c) = 3500/(0.7385*0.0161*235000*1.0) = 1.2526 > 1.0

STAAD.Pro value is 1.25 [Matching]

 

• As per clause 7.3.2: λ̅w < λ̅uw

h_ef = D-4*t_f = 320-4*20.5 mm = 238 mm

λ̅w = (h_ef/t_w)*√(R_y/E) = 0.693

STAAD.Pro value is 0.699 [Matching]

From table 9, < λ̅uw = 1.2+0.35*λ̅ = 2.0806 (eqn. 24)

STAAD.Pro value is 2.08 [Matching]

Hence, λ̅w < λ̅uw [OK]

• As per clause 7.3.8: λ̅f < λ̅uf

b_ef = 0.5*(B-2*t_w) = 0.5*(300-4*11.5) = 127 mm

λ̅f = (b_ef/t_f)*√(R_y/E) = 0.209

STAAD.Pro value is 0.209 [Matching]

From table 10, < λ̅uf = 0.36+0.1*λ̅ = 0.6116 (eqn. 37)

STAAD.Pro value is 0.611 [Matching]

Hence, λ̅f < λ̅uf [OK]

• Summary Table:

Item	STAAD.Pro Value	Hand Calculated Value	% Deviation
Ratio-Cl 7.1.1	0.92	0.9251	0.5543
Ratio-Cl 7.1.3	1.25	1.2526	0.2080
λ̅uw	2.08	2.0806	0.0288
λ̅uf	0.611	0.6116	0.0982