Cold formed sections BS5950Cold formed sections BS5950COLD FORMED THIN GAUGE SECTION DESIGN (BS5950-

5:1998)

y
+ve My 3

16
200

x x
Mx

0.1

75

Basic section details

Section details
Section name;
Section type; Lipped channel
Depth; D = 200 mm
Breadth of top flange; Bt = 75 mm
Breadth of bottom flange; Bb = 75 mm
O/all breadth; B = Bt = Bb = ;75; mm
Design thickness; t = 3.0 mm
Internal radius of bends; r = 0.1 mm
; ;;
Depth of top stiffening lip; DLt = 16 mm
Depth of bottom stiffening lip; DLb = 16 mm
The section breadth to stiffening lip depth ratio is less than or equal to 5 therefore the flanges are stiffened (cl 4.6)
Number of 90 degree bends; N90 = 4; ;
Design forces and moments
Ultimate axial load; F = 1.0 kN; (Compression);
Ultimate bending moment about x axis; Mx = ;23.20; kNm
; ;;
Ultimate positive bending moment about y axis; Myp = ;0.28; kNm
Ultimate negative bending moment about y axis; Myn = ;-0.51; kNm
Gross section properties
Area of element a; Aa = (DLt - t/2)  t = 44 mm2
Area of element b; Ab = (Bt - t)  t = 216 mm2
Area of element c; Ac = (D - t)  t = 591 mm2
Area of element d; Ad = (Bb - t)  t = 216 mm2
Area of element e; Ae = (DLb - t/2)  t = 44 mm2
Total gross area; Ag = Aa + Ab + Ac + Ad + Ae = ;1110; mm2
Position of x axis from centreline of flange; xbar = [Aa(D-t-(DLt-t/2)/2)+Ab(D-t)+Ac(D-t)/2+Ae(DLb - t/2)/2]/Ag
xbar = 98.5 mm
Second moment of area about x axis
Contribution of element a; Ixa = t  (DLt - t/2)3/12 + Aa  (D - t- (DLt - t/2)/2 - xbar)2 = 36.3 cm4
Contribution of element b; Ixb = Ab  (D-t- xbar)2 = 209.6 cm4
Contribution from element c; Ixc = t  (D - t)3/12 + Ac  ((D-t)/2 - xbar)2 = 191.1 cm4
Contribution from element d; Ixd = Ad  xbar2 = 209.6 cm4
Contribution of element e; Ixe = t  (DLb - t/2)3/12 + Ae  (xbar - (DLb - t/2)/2)2 = 36.3 cm4
Total second moment of area; Ixg = Ixa + Ixb + Ixc + Ixd + Ixe = ;682.9; cm4
Radius of gyration of gross cross section; rxg = (Ixg/Ag) = 78.43 mm
Position of y axis from centreline of web; ybar = [Aa  (Bt - t) + Ab  (Bt - t)/2 + Ad  (Bb - t)/2 + Ae  (Bb - t)]/Ag
ybar = 19.7 mm
Second moment of area about y axis
Contribution from element a; Iya = Aa  (Bt - t - ybar)2 = 11.9 cm4
Contribution from element b; Iyb = t  (Bt - t)3/12 + Ab  ((Bt - t)/2-ybar)2 = 15.1 cm4
Contribution from element c; Iyc = Ac  ybar2 = 22.8 cm4
Contribution from element d; Iyd = t  (Bb - t)3/12 + Ad  ((Bb - t)/2-ybar)2 = 15.1 cm4
Contribution from element e; Iye = Ae  (Bb - t - ybar)2 = 11.9 cm4
Total second moment of area; Iyg = Iya + Iyb + Iyc + Iyd + Iye = 76.9 cm4
Radius of gyration of gross cross section; ryg = (Iyg/Ag) = 26.32 mm

Element flat widths

Element a; ba = DLt - t - r = 12.9 mm
Element b; bb = Bt - 2  (t + r) = 68.8 mm
Element c; bc = D - 2  (t + r) = 193.8 mm
Element d; bd = Bb - 2  (t + r) = 68.8 mm
Element e; be = DLb - t - r = 12.9 mm
Steel details (Table 4)
;
Nominal yield strength; Ys = 450 N/mm2
Nominal ultimate tensile strength; Us = 480 N/mm2
Modified tensile yield strength due to cold forming (cl 3.4)
Average yield strength; Ysa = min(Ys + 5  N90  t2  (Us-Ys)/Ag, 1.25Ys, Us) = 454.9 N/mm2
Design strength in tension; pyt = min(Ysa, 0.84  Us) = 403.2 N/mm2
Modified compressive yield and design strengths due to cold forming (cl 3.4)
Yield strength element a; Ysaca = Ysa = 454.9 N/mm2;
Design strength element a; pyca = Ysaca = 454.9 N/mm2
Yield strength element b; Ysacb = Ys + (Ysa - Ys)  (48(280/Ys)0.5 - bb/t)/(24(280/Ys)0.5) = 453.8 N/mm2;
Design strength element b; pycb = Ysacb = 453.8 N/mm2
Yield strength element c; Ysacc = Ys = 450.0 N/mm2;
Design strength element c; pycc = Ysacc = 450.0 N/mm2
Yield strength element d; Ysacd = Ys + (Ysa - Ys)  (48(280/Ys)0.5 - bd/t)/(24(280/Ys)0.5) = 453.8 N/mm2;
Design strength element d; pycd = Ysacd = 453.8 N/mm2
Yield strength element e; Ysace = Ysa = 454.9 N/mm2
Design strength element e; pyce = Ysace = 454.9 N/mm2
Minimum modified compressive yield strength; Ysac = min(Ysaca, Ysacb, Ysacc, Ysacd, Ysace) = 450.0 N/mm2
Minimum design strength in compression; pyc = min(Ysac, 0.84  Us) = 403.2 N/mm2

Axial compression (Section 6)

Effective length for compression
Effective length about x axis; LEx = 9000 mm
Effective length about y axis; LEy = 9000 mm

Local buckling coefficients (fig. B.1)

Dimension b1; b1 = D - t = 197.0 mm
Dimension b2; b2 = B - t = 72.0 mm
Ratio of b2 to b1; h = b2 / b1 = 0.365
Element a; Ka_f = 0.425
Element b; Kb_f = max(4.0 , [7 - (1.8  h) / (0.15 + h) - 1.43  h3]  h2) = 4.000
Element c; Kc_f = max(4.0, 7 - (1.8  h) / (0.15 + h) - 1.43  h3) = 5.654
Element d; Kd_f = max(4.0 , [7 - (1.8  h) / (0.15 + h) - 1.43  h3]  h2) = 4.000
Element e; Ke_f = 0.425
Effective element widths
Element a (cl. 4.5.1)
Local buckling stress; pcra_f = 0.904  ES5950  Ka_f  (t / ba)2 = 4259.7 N/mm2
Basic effective width; ba_f_bas = ba[1+14((max(0.123,pyca/pcra_f))1/2-0.35)4]-0.2 = 12.9 mm
Actual effective width; ba_f = 0.89  ba_f_bas + 0.11  ba = 12.9 mm
Effective area; Aa_f = t  (ba_f + r + t / 2) = 43 mm2

Element b (cl. 4.4.1)

Local buckling stress; pcrb_f = 0.904  ES5950  Kb_f  (t / bb)2 = 1409.4 N/mm2
Effective width; bb_f = bb[1+14((max(0.123,pycb/pcrb_f))1/2-0.35)4]-0.2 = 68.4 mm
Effective area; Ab_f = t  (bb_f + 2  r + t) = 215 mm2

Element c (cl. 4.4.1)

Local buckling stress; pcrc_f = 0.904  ES5950  Kc_f  (t / bc)2 = 251.1 N/mm2
Effective width; bc_f = bc[1+14((max(0.123,pycc/pcrc_f))1/2-0.35)4]-0.2 = 113.7 mm
Effective area; Ac_f = t  (bc_f + 2  r + t) = 351 mm2

Element d (cl. 4.4.1)

Local buckling stress; pcrd_f = 0.904  ES5950  Kd_f  (t / bd)2 = 1409.4 N/mm2
Effective width; bd_f = bd[1+14((max(0.123,pycd/pcrd_f))1/2-0.35)4]-0.2 = 68.4 mm
Effective area; Ad_f = t  (bd_f + 2  r + t) = 215 mm2

Element e (cl. 4.5.1)

Local buckling stress; pcre_f = 0.904  ES5950  Ke_f  (t / be)2 = 4259.7 N/mm2
Basic effective width; be_f_bas = be[1+14((max(0.123,pyce/pcre_f))1/2-0.35)4]-0.2 = 12.9 mm
Actual effective width; be_f = 0.89  be_f_bas + 0.11  be = 12.9 mm
Effective area; Ae_f = t  (be_f + r + t / 2) = 43 mm2

Total effective area

Total effective area; Af = Aa_f + Ab_f + Ac_f + Ad_f + Ae_f = 867 mm2
Position of y axis of effective area from web; yf = [Aa_f(B-t) + Ab_f(B-t)/2 + Ad_f(B-t)/2 + Ae_f(B-t)] / Af = 25.1 mm
Dist. between gross and effective area y axes; es = yf - ybar = 5.4 mm; (Generating a +ve moment);
Torsional flexural buckling capacity (cl 6.3 & annex D)
Dimension b (Table D.1); b = B - t = 72.0 mm
Dimension d (Table D.1); d = D - t = 197.0 mm
Dimension bL (Table D.1); bL = DLt - t / 2 = 14.5 mm
Position of shear centre (Table D.1); e = d2  b  bL  t  [1/2 + b / (4bL) - 2bL2 / (3d2)] / Ixg = 30.9 mm
Warping constant (Table D.1); Cw = b2  t  [4  bL3 + 3  d2  bL - 6  d  bL2 + b  d2] / 6 - Ixg  e2
Cw = 4472.3 cm6
St. Venant torsion constant; J = [(bat3) + (bbt3) + (bct3) + (bdt3) + (bet3)]/3 = 3214.800 mm4
Distance from shear ctr to centroid (along x axis); xo = ybar + e = 50.6 mm
Polar radius of gyration; ro = (rxg2 + ryg2 + xo2)1/2 = 97.0 mm
Beta constant;  = 1 - (xo/ro)2 = 0.728
Short strut capacity; Pcs = Af  pyc = 349.7 kN
Torsional buckling load; PT = min(Pcs, 1/ ro2  (GS5950  J + 2  2  ES5950  Cw/max(LEx,LEy)2))
PT = 50.7 kN
Elastic flexural buckling load (x axis); PEx = min(Pcs, 2  ES5950  Ixg / LEx2) = 170.6 kN
Torsional flexural buckling load; PTF = min(Pcs, 1/(2)  [(PEx+PT) - ((PEx+PT)2 - 4PExPT)1/2])
PTF = 46.1 kN
Elastic flexural buckling load (y axis); PEy = min(Pcs, 2  ES5950  Iyg / LEy2) = 19.2 kN
Slenderness ratio factor (x axis); x = max(1.0, (PEx / PTF)1/2) = 1.92
Slenderness ratio factor (y axis); y = max(1.0, (PEy / PTF)1/2) = 1.00

Flexural buckling capacity (cl 6.2)

Slenderness ratio (x axis); x = x  LEx/rxg = 220.8
Slenderness ratio (y axis); y = y  LEy/ryg = 342.0
The section slenderness is adequate for compression loads due to wind reversal only;
Short strut capacity; Pcs = Af  pyc = 349.7 kN
Perry coefficient (x axis); x = 0.002  (x  LEx/rxg - 20) = 0.402;
Constant phi (x axis); x = [Pcs + (1+x)  PEx]/2 = 294.4 kN
Flexural buckling load (x axis); Pcx = PEx  Pcs/[x + (x2 - PExPcs)] = 130.0 kN
Perry coefficient (y axis); y = 0.002  (y  LEy/ryg - 20) = 0.644;
Constant phi (y axis); y = [Pcs + (1+y)  PEy]/2 = 190.6 kN
Flexural buckling load (y axis); Pcy = PEy  Pcs/[y + (y2 - PEyPcs)] = 18.5 kN
Minimum flexural buckling load; Pc = min(Pcx, Pcy) = 18.5 kN
PASS - Pc >= F - Axial load capacity is adequate (UF = 0.054)
Applied moment due to shift in neutral axis; Mys = F  es = 0.0 kNm
Modified positive design moment; Myp = Myp + Mys = ;0.3; kNm
 Myp>0 kNm - Therefore there is a resultant positive moment
Modified negative design moment; Myn = Myn + Mys = ;-0.5; kNm
Myn<0 kNm - Therefore there is a resultant negative moment

Major axis bending

;
Limiting web stress (cl. 5.2.2.2)
Depth of compression zone; Dc = (D - t) / 2 = 98.5 mm
Depth Dw; Dw = max(D, 2  Dc) = 200.0 mm
Limiting web stress; p0 = min[(1.13 - 0.0019  Dw (Ysac / 280 N/mm2)0.5/ t)  pyc ,pyc]
p0 = 390.9 N/mm2
Element effective widths
Element a (cl. 4.5.2)
Stress at supported edge; fcs = p0 = 390.9 N/mm2
Stress at free edge; fcf = p0  2  (D / 2 - DL) / (D - t) = 333.3 N/mm2
Ratio of fcs to fcf; R = fcs / fcf = 1.173
Local buckling coefficient; Ka_mx = max(0.425, 1.7 / (3 + R)) = 0.425
Local buckling stress; pcra_mx = 0.904  ES5950  Ka_mx  (t / ba)2 = 4259.7 N/mm2
Compressive stress; fca_mx = fcf = 333.3 N/mm2
Basic effective width; ba_mx_bas = ba[1+14((max(0.123,fca_mx/pcra_mx))1/2-0.35)4]-0.2 = 12.9 mm
Actual effective width; ba_mx = 0.89  ba_mx_bas + 0.11  ba = 12.9 mm

Element b (cl. 4.4.1)

Dimension b1; b1 = B - t = 72.0 mm
Dimension b2; b2 = D - t = 197.0 mm
Ratio of b2 to b1; h = b2 / b1 = 2.736
Local buckling coefficient (fig. B.2); Kb_mx = max(4.0, 5.4 - (1.4  h) / (0.6 + h) - 0.02  h3) = 4.000
Local buckling stress; pcrb_mx = 0.904  ES5950  Kb_mx  (t / bb)2 = 1409.4 N/mm2
Compressive stress; fcb_mx = p0 = 390.9 N/mm2
Effective width; bb_mx = bb[1+14((max(0.123,fcb_mx/pcrb_mx))1/2-0.35)4]-0.2 = 68.6 mm

Effective section properties

Effective area of element a; Aa_mx = (ba_mx + r +t / 2)  t = 43 mm2
Effective area of element b; Ab_mx = (bb_mx + 2  r + t)  t = 215 mm2
Total effective area; Amx = Aa_mx+Ab_mx + Ac + Ad + Ae = ;1109; mm2
Position of neutral axis from tension flange ctrline; xmx = [Aa_mx(D-t-(t/2+r+ba_mx)/2)+Ab_mx(D-t)+Ac(D-t)/2+Ae(DL-
t/2)/2]/Amx =
xmx = 98.5 mm
Second moment of area about neutral axis
Contribution from element a; Ia_mx = t  (t/2 + r + ba_mx)3 / 12 + Aa_mx  (D - t - (t/2 + r + ba_mx)/2 - xmx)2
Ia_mx = 36.3 cm4
Contribution from element b; Ib_mx = Ab_mx  (D - t - xmx)2 = 209.2 cm4
Contribution from element c; Ic_mx = t  (D - t)3/12 + Ac  ((D - t)/2 - xmx)2 = 191.1 cm4
Contribution from element d; Id_mx = Ad  xmx2 = 209.4 cm4
Contribution from element e; Ie_mx = t  (DL - t/2)3/12 + Ae  ((DL - t/2)/2 - xmx)2 = 36.3 cm4
Total second moment of area; Imx = Ia_mx + Ib_mx + Ic_mx + Id_mx + Ie_mx = ;682.3; cm4
Section modulus (compression edge); Zxc = Imx / (D - t - xmx) = 69.2 cm3
Section modulus (tension edge); Zxt = Imx / xmx = 69.3 cm3
Moment capacity
Stress at tension flange; p0t = p0  xmx / (D - t - xmx) = 390.5 N/mm2
Moment capacity at tension face; Mcxt = p0t  Zxt = 27.06 kNm
Moment capacity at compression face; Mcxc = p0  Zxc = 27.06 kNm
Moment capacity; Mcx = min(Mcxt, Mcxc) = 27.06 kNm
PASS - The moment capacity exceeds the applied moment
Shear in web (cl. 5.4)
Applied shear force; Fvy = 0.33 kN
Maximum applied shear stress; vmax_y = Fvy(B-t/2)(D-t)(1+(D-t)/(4(B-t/2)))/(2Ixg) = ;0.58; N/mm2
Maximum allowable shear stress; pv_max = 0.7  py = 282.24 N/mm2
Average shear stress; vy = Fvy / (t  D) = ;0.55; N/mm2
Shear yield strength; pv = 0.6  py = 241.92 N/mm2
Shear buckling strength; qcry = (1000  t / D)2  1.0 N/mm2 = 225.00 N/mm2
Minimum shear strength; pvy_min = min(pv , qcry) = 225.00 N/mm2
PASS - The shear capacity is not exceeded
Combined bending and shear (cl. 5.5.2)
Shear/shear buckling resistance; Pvy = min(pv , qcry)  D  t = ;135.00; kN
Bending moment at position of max shear; Mvx = 0.51 kNm
Section utilisation; UFvy1 = (Fvy / Pvy)2 + (Mvx / Mcx)2 = 0.000
PASS - The section utilisation is less than 1.0
Shear force at position of max moment; Fvy_m = 0.3 kN
Section utilisation; UFvy2 = (Fvy_m / Pvy)2 + (Mx / Mcx)2 = 0.735
PASS - The section utilisation is less than 1.0

Minor axis positive bending capacity

Limiting web stress (cl. 5.2.2.2)
Depth of compression zone; Dc = ybar = 19.7 mm
Depth Dw; Dw = max(B , 2  Dc) = 75.0 mm
Limiting web stress; p0 = min[(1.13 - 0.0019  Dw (Ysac / 280 N/mm2)0.5/ t)  pyc ,pyc]
p0 = 403.2 N/mm2
Element effective widths
Element c (cl. 4.4.1)
Dimension b1; b1 = D - t = 197.0 mm
Dimension b2; b2 = B - t = 72.0 mm
Ratio of b2 to b1; h = b2 / b1 = 0.365
Local buckling coefficient (fig. B.2); Kc_my = max(4.0, 7 - (1.8  h) / (0.15 + h) - 0.091  h3) = 5.719
Local buckling stress; pcrc_my = 0.904  ES5950  Kc_my  (t / bc)2 = 254.0 N/mm2
Compressive stress; fcc_my = p0 = 403.2 N/mm2
Effective width; bc_my = bc[1+14((max(0.123,fcc_my/pcrc_my))1/2-0.35)4]-0.2 = 120.9 mm

Effective section properties

Effective area of element c; Ac_my = (bc_my + 2  r +t)  t = 372 mm2
Total effective area; Amy = Aa + Ab + Ac_my + Ad + Ae = 891 mm2
Position of neutral axis from comp flange ctrline; ymy=[Aa(B-t) + Ab(B-t)/2 + Ad(B-t)/2 + Ae(B-t)]/Amy = 24.5 mm
Stress at tension flange; p0t = p0  (B - t - ymy) / ymy = 782.7 N/mm2
The elastic value of p0t exceeds pyt therefore, in accordance with clause 5.2.2.1, allow for plastic redistribution of tensile
stresses with a limiting value of pyt.
Let the distance from the compression flange to the neutral axis equal yp, then:-
Compression force; C = p0  Ac_my + p0  t  yp
Tension force; T = pyt(Aa + Ae) + 2pytt[B - t - yp  (1 + pyt/p0)] + (pyt2/p0)  t  yp
Equating compression and tension force and solving for yp:-
Distance from compression flange to N.A.; yp = 12.2 mm
Compression force; C = p0  Ac_my + p0  t  yp = 164.9 kN
Tension force; T = pyt(Aa + Ae) + 2pytt[B - t - yp  (1 + pyt/p0)] + (pyt2/p0)  t  yp
T = 164.9 kN
Taking moments about neutral axis:-
Moment due to compression force; Mcf = p0  Ac_my  yp + 2  p0  t  yp2 / 3 = 1.96 kNm
Mt due to tension force (elements a and e); Mtf1 = pyt(Aa + Ae)  (B - t - yp) = 2.10 kNm
Mt due to tension force (elmts b and d, stress pyt); Mtf2 = 2pytt[B-t-yp(1 + pyt/p0)]  [B-t-yp-(B-t-yp(1 + pyt/p0))/2]
Mtf2 = 4.14 kNm
Mt due to ten force (elmts b and d, varying stress); Mtf3 = 2  (pyt2/p0)  t  yp2 (pyt/p0) / 3 = 0.12 kNm
Positive moment capacity; Mcyp = Mcf + Mtf1 + Mtf2 + Mtf3 = 8.31 kNm
Positive bending section utilisation; UFMyp = Myp/Mcyp = 0.034
Pass - Mcyp >= Myp - Positive y axis bending capacity is adequate (UF = 0.034)

Minor axis negative bending capacity

Limiting web stress (cl. 5.2.2.2)
Depth of compression zone; Dc = B - t - ybar = 52.3 mm
Depth Dw; Dw = max(B , 2  Dc) = 104.7 mm
Limiting web stress; p0 = min[(1.13 - 0.0019  Dw (Ysac / 280 N/mm2)0.5/ t)  pyc ,pyc]
p0 = 403.2 N/mm2
Element effective widths
Element a (cl. 4.5.1)
Local buckling coefficient; Ka_my = 0.425
Local buckling stress; pcra_my = 0.904  ES5950  Ka_my  (t / ba)2 = 4259.7 N/mm2
Compressive stress; fca_my = p0 = 403.2 N/mm2
Basic effective width; ba_my_bas = ba[1+14((max(0.123,fca_my/pcra_my))1/2-0.35)4]-0.2 = 12.9 mm
Actual effective width; ba_my = 0.89  ba_my_bas + 0.11  ba = 12.9 mm

Element e (cl. 4.5.1)

Local buckling coefficient; Ke_my = 0.425
Local buckling stress; pcre_my = 0.904  ES5950  Ke_my  (t / be)2 = 4259.7 N/mm2
Compressive stress; fce_my = p0 = 403.2 N/mm2
Basic effective width; be_my_bas = be[1+14((max(0.123,fce_my/pcre_my))1/2-0.35)4]-0.2 = 12.9 mm
Actual effective width; be_my = 0.89  be_my_bas + 0.11  be = 12.9 mm

Effective section properties

Effective area of element a; Aa_my = (ba_my + r +t / 2)  t = 43 mm2
Effective area of element e; Ae_my = (be_my + r +t / 2)  t = 43 mm2
Total effective area; Amy = Aa_my + Ab + Ac + Ad + Ae_my = 1110 mm2
Position of neutral axis from tension flange ctrline; ymy = [Aa_my(B-t) + Ab(B-t)/2 + Ad(B-t)/2 + Ae_my(B-t)] / Amy
ymy = 19.7 mm
Stress at tension flange; p0t = p0  ymy / (B - t - ymy) = 151.4 N/mm2
Second moment of area about neutral axis
Contribution from element a; Ia_my = Aa_my  (B - t - ymy)2 = 11.9 cm4
Contribution from element b; Ib_my = t  (B - t)3/12 + Ab  ((B - t)/2 - ymy)2 = 15.1 cm4
Contribution from element c; Ic_my = Ac  ymy2 = 22.8 cm4
Contribution from element d; Id_my = t  (B - t)3/12 + Ad  ((B - t)/2 - ymy)2 = 15.1 cm4
Contribution from element e; Ie_my = Ae_my  (B - t - ymy)2 = 11.9 cm4
Total second moment of area; Imy = Ia_my + Ib_my + Ic_my + Id_my + Ie_my = 76.9 cm4
Section modulus (compression edge); Zyc = Imy / (B - t - ymy) = 14.7 cm3
Section modulus (tension edge); Zyt = Imy / ymy = 39.1 cm3
Moment capacity
Moment capacity at tension face; Mcynt = p0t  Zyt = 5.92 kNm
Moment capacity at compression face; Mcync = p0  Zyc = 5.92 kNm
Negative moment capacity; Mcyn = min(Mcynt, Mcync) = 5.92 kNm
Negative bending section utilisation; UFMyn = abs(Myn)/Mcyn = 0.085
Pass - Mcyn >= Myn - Negative y axis bending capacity is adequate (UF = 0.085)
Shear in web (cl. 5.4)
Applied shear force; Fvx = 4.00 kN
Maximum applied shear stress; vmax_x = Fvx  ((B - t/2) - ybar)2 / Iyg = 15.09 N/mm2
Maximum allowable shear stress; pv_max = 0.7  py = 282.24 N/mm2
Average shear stress; vx = Fvx / (2  t  B) = 8.89 N/mm2
Shear yield strength; pv = 0.6  py = 241.92 N/mm2
Shear buckling strength; qcrx = (1000  t / B)2  1.0 N/mm2 = 1600.00 N/mm2
Minimum shear strength; pvx_min = min(pv , qcrx) = 241.92 N/mm2
PASS - The shear capacity is not exceeded
Combined bending and shear (cl. 5.5.2)
Shear/shear buckling resistance; Pvx = min(pv , qcrx)  2  B  t = 108.86 kN
Bending moment at position of max shear; Mvy = 0.00 kNm
Section utilisation; UFvx1 = (Fvx / Pvx)2 + (Mvy / Mcyp)2 = ;0.001
PASS - The section utilisation is less than 1.0
Shear force at position of max positive moment; Fvx_mp = 0.0 kN
Section utilisation; UFvx2 = (Fvx_mp / Pvx)2 + (Myp / Mcyp)2 = 0.001
PASS - The section utilisation is less than 1.0
Shear force at position of max negative moment; Fvx_mn = 0.0 kN
Section utilisation; UFvx3 = (Fvx_mn / Pvx)2 + (Myn / Mcyn)2 = 0.007
PASS - The section utilisation is less than 1.0

Combined bending and compression

Local capacity check (cl. 6.4.2)
Section utilisation (-ve My critical); UFlocal = F / Pcs + Mx / Mcx + abs(Myn) / Mcyn = 0.945
Pass - Local compression and bending capacity is adequate (UF = 0.945)
Overall buckling capacity check (cl. 6.4.3)
Section utilisation (-ve My critical); UFo_all1 = F/Pc + Mx/[CbxMcx(1-F/PEx)] + abs(Myn)/[CbyMcyn(1-F/PEy)]
UFo_all1 = 0.968
Pass - Overall buckling capacity is adequate (UF = 0.968)