The following is a validation problem for a detailed calculation of the additional clauses for Belgian National Annex. (Belgian NA document: “NBN EN 1993-1-1:2005”)
The Belgian National Annex implementation is a part of the Eurocode 3 design code (EN 1993-1-1:2005) in the Steel design module of STAAD.Pro.
To invoke and use the Belgian National Annex, the user needs to set the value of the “NA” parameter to ‘8’.
A European I section cantilever beam (HD320X127, S275 grade steel) , 5 m has been subjected to a concentrated point Load of 25 KN (FY) and concentrated moments 5 KN-m (MX) and 10 KN-m (MZ) at the free end.
Outputs: MA= 10 KN-m, MB= 10 KN-m, SFA= 0 KN, SFB= 0 KN
The following is the text input information of the STAAD model:
STAAD SPACE
START JOB INFORMATION
ENGINEER DATE 21-Feb-14
END JOB INFORMATION
INPUT WIDTH 79
UNIT METER KNJOINT COORDINATES1 0 0 0; 2 0 5 0;
MEMBER INCIDENCES1 1 2;
MEMBER PROPERTY EUROPEAN1 TABLE ST HD320X127
DEFINE MATERIAL STARTISOTROPIC STEELE 2.05e+008POISSON 0.3DENSITY 76.8195ALPHA 1.2e-005DAMP 0.03END DEFINE MATERIALCONSTANTSMATERIAL STEEL ALL
SUPPORTS1 FIXED
LOAD 1 LOADTYPE None TITLE LOAD CASE 1JOINT LOAD2 FY -25 MX 5 MZ 10PERFORM ANALYSIS
PARAMETER 1CODE EN 1993-1-1:2005NA 8 ALLCMM 2 ALLC2 1.554 ALLFU 295000 ALLPY 275000 ALL*SGR 1 ALLCMN 0.7 ALLTRACK 2 ALLCHECK CODE ALLFINISH
We get: D= 320 mm, B= 300 mm, Tf= 20.5 mm, Tw= 11.5 mm, Iz= 30820 cm4, Iy= 9239 cm4,
Zz= 1926.5 cm3, Zy= 615.93 cm3, J= 225.1 cm4, H= 2.069*10^12 mm6
Moment Capacity, Mckd = Wply*fy/GM0 = 2149*1000*275/ (1*10^6) = 590.975 KN-m
STAAD.Pro value is 591 KN-m [OK]
Here, CMM=2.0, CMN=0.7
So, C1 = 2.578, C2 = 1.554
Π2*E*Iy/ (kL)2 = 7477192.555
(k/kw)2 * (Iw/Iy) = 22394.19851
(kL) 2 *G*It/( Π2*E*Iy) = 23736.5416
C2Zg = 1.554*160 =248.64
So, Mcr = 1540.6 KN-m
STAAD.Pro value is 1541.5 KN-m [OK]
From Belgian NA, λLT, 0 = 0.4, β = 0.75
So, λLT = √ (wy*fy/Mcr) = √ (2149*10^3*275/1540600000) = 0.619355
H/b = 320/300 = 1.067 < 2. So, from Table 6.5 of Eurocode 3, we get αLT = 0.34
From Cl. 6.3.2.3 of Eurocode3, we get:
ФLT = 0.5[1+αLT (λLT- λLT, 0) + β* λLT2]
= 0.5*[1+0.34*(0.619355-0.4) +0.75*0.6193552] = 0.681141
So, χLT = [ФLT+ √ (ФLT2- β* λLT2)]-1 = 0.908294
So, MB = χLT*wy*fy/GM1 = 0.908294*2149000*275/(1*10^6)
= 536.77 KN-m
STAAD.Pro value is 536.8 KN-m [OK]
NEd = 25 KN, NRk = A*fy = 4435.75 KN
MyEd = 5 KN-m, MzEd = 10 KN-m, MyRk = 258.3 KN-m
Determination of χy:
h/b = 1.067 < 2 & tf = 20.5 mm < 100 mm
So, for y-y axis, we will use curve ‘c’. Hence, α = 0.49 [Table 6.2, EC3]
So, ф = 0.5[1+α (λ-0.2) +λ2)] = 0.9363
Determination of χz:
So, for z-z axis, we will use curve ‘c’. Hence, α = 0.34 [Table 6.2, EC3]
So, ф = 0.5[1+α (λ-0.2) +λ2)] = 0.6266
Compression Capacity = χy*NRk = 0.68087*4435.75 =3020.169 KN
Compression Ratio = 25/3020.169 = 0.00828
STAAD.Pro value for compression ratio is 0.008 [OK]
Critical Axial Loads for Flexural and Flexural Torsional Buckling:
From NCCI document SN001a-EN-FU:
Here, y0 = z0 = 0
iy = 75.68 mm, iz = 138.23 mm
So, i02 = 75.682 + 138.232 = 24834.995
Interaction Check:
From Annex A, Table A.1 of EC3, we get the auxiliary terms:
χLT = 0.908294
Wz = wplz/welz = 2149/1926 = 1.1158
Wy = wply/wely = 939/615.933 = 1.525 > 1.5, so, wy = 1.5
CMM = 2.0, ψ=1.0
Determination of λ0:
C1=1, C2=0
Mcr = 1605.96 KN