TITLE: PROJECT: JOB NUMBER:

user13/09/20158:29:27 AM

----- [ DESIGN SUMMARY ] ------------------------------------------------------------------------------------

SECTION: 200 UB 22.3TYPE: OneSteel 300PLUSSOURCE: OSMM 2014 [14]DESIGN RATIO: 1.37 (PASS)

Mx*: 23 kNm (COMPRESSION ZONE: Bottom Flange)SUBSEGMENT LENGTH: 3500 mmSUBSEGMENT RESTRAINT: LLLOAD POSITION: AT SUBSEGMENT ENDSLOAD HEIGHT: SHEAR CENTRE

My*: 5 kNm

----- [ SECTION PROPERTIES ] --------------------------------------------------------------------------------

Class: HR Grade: 300 Mass: 22.5 kg/m d: 201.6 mm b: 133.0 mm t,f: 7.0 mm t,w: 5.0 mm r1: 8.9 mm r2: 0.0 mm fy,f: 320 MPa fy,w: 320 MPa fu: 440 MPa Ag: 2870 mm Ix: 2.10E+7 mm Iy: 2.75E+6 mm rx: 85.5 mm ry: 31.0 mm Zx: 2.08E+5 mm Zy: 4.13E+4 mm Sx: 2.31E+5 mm Sy: 6.34E+4 mm J: 4.50E+4 mm Iw: 2.60E+10 mm sx: NON-COMPACT sy: NON-COMPACT Zex: 2.27E+5 mm Zey: 6.03E+4 mm kf: 1.0 b: 0

----- [ BENDING ] -------------------------------------------------------------------------------------------

No holes specified in the flanges, such that the full effective section modulus will be used (5.2.6). Upgrade to CHECKSTEEL Plus to add section penetrations.

Msx = fy Zex = 0.90 320 2.27E+5 = 65.38 kNm [2.84]

Any element assumed to provide full, partial, or lateral restraint must be designed to transfer 2.5% of the critical flange force (0.025 23 kNm / 0.1946 m) = 2.95 kN (5.4.3.1).

Subsegment assumed to be bent in single curvature and not subject to transverse loads, such that m = -1.0 (5.3.2.4 (a)). This assumption tends to be conservative, and the result can be more accurately calculated by specifying the appropriate moment modification factor (m) using the wizard.

Lmax = ry (80 + 50 m) (250/fy) = 31.0 (80 + 50 -1.00) (250/320) = 822.01 mm

Full lateral restraint cannot be assumed as Lmax < Subsegment Length, lateral buckling check required for this subsegment (5.3.2.4).

Leb = kt kl kr Subsegment Length = 1.0000 1.0 1.00 3500 = 3500.00 mm

Mo = [( E Iy / Leb) (G J + E Iw / Leb)] = [( 200000 2.75E+6 / 3500.00) (80000 4.50E+4 + 200000 2.60E+10 / 3500.00)] = 58.75 kNm

Moment modification factor (m) calculated in accordance with Table 5.6.1.

m = 1.0

s = 0.6 {[(Msx/Mo) + 3] - Msx/Mo} 1.0 = 0.6 {[(72.64/58.75) + 3] - 72.64/58.75} = 0.5350

Mbx = m s Msx Msx = 1.0 0.5350 65.38 65.38 kNm

CHECKSTEEL v1.3.1 Page 1

TITLE: PROJECT: JOB NUMBER:

user13/09/20158:29:27 AM

= 34.98 kNm [1.52]

Msy = fy Zey = 0.90 320 6.03E+4 = 17.37 kNm [3.47]

Biaxial = (Mx*/Msx) + (My*/Msy) 1.0 = (23/65.38) + (5/17.37) = 0.6397 [1.56]

Biaxial = (Mx*/Mbx) ^ 1.4 + (My*/Msy) ^ 1.4 1.0 = (23/34.98) ^ 1.4 + (5/17.37) ^ 1.4 = 0.7310 [1.37]

CHECKSTEEL v1.3.1 Page 2