The rafter axial thrust forces at ends B and C form a triangle where we can

The rafter axial thrust forces at ends b and c form a

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The rafter axial thrust forces at ends B and C form a triangle where we can determine the value of the axial force at the hinge at the base of the ridge haunch a horizontal distance x from end B. To determine distance along rafter equivalent to horizontal distance x = 14.09m from end B: - Thus distance along rafter to ridge hinge = 14.09 x (15.06/15) = 14.15 m This is equivalent to (15.06-14.15)m = 0.96m from end C . Thus axial force at the plastic hinge= 186.3+(0.96/15.06)(209.6-186.3) = 187.7 kN W pl,y = M / f y = 701.3×10 6 /345 = 2,032,618 mm 3 = 2,033 cm 3 Try 610 × 178 UB 82 where W pl,Ry = 2190cm 3 > W pl,y = 2033cm 3 … ok With this I-beam size t f = 12.8mm < 16mm, thus fy = 355N/mm2 n = N E,d / N R,d = N E,d / Af y = 187.7x1000/(10400x355) = 0.051 < a …ok where a = (A-2bt f ) / A = (10400-2x177.9x12.8)/10400 = 0.710 The section is satisfactory. Check that the rafter section at the end of the haunch remains elastic under factored loads. The actions are: - Here the end of the haunch is at 2.992m from end B or 15.06-2.992 = 12.068m from end C. Thus here N Ed = 186.3+(12.086/15.06)(209.6-186.3) Or N Ed = 205.0 kN And the moment at this location is: - M x = 267.9x - 17.86x(x/2) - H(5.725+(1.31/15)x) M x = 267.9x - 8.93x 2 - 187(5.725+0.0873x) M x = 267.9x - 8.93x 2 - 1070.6 - 16.3x M x = 251.6x - 8.93x 2 - 1070.6 Thus at x = 2.992m, M x = 251.6(2.992) - 8.93(2.992) 2 - 1070.6 M Ed = -397.8 kNm Then check that (N Ed /A) + (M/W el,y ) = (205.0/10400) + (397.8x10 6 /1870x1000) Thus (N Ed /A) + (M/W el,y ) = 212.7 N/mm 2 This is less than f y = 355 N/mm 2 . The section remains elastic. Column restraints and stability A stray is provided to restrain the hinge section at the eaves. The distance to the adjacent restraint using the conservative method is: - L m = 38I z / ((1/57.4)(N Ed /A) + (1/756)C 1 2 (W pl,y 2 /AI t )(f y /235) 2 )) 0.5 L m = 38x34.0/((1/57.4)(270.2x10 3 /10400)+(1/756)(1 2 )*(2190x1000 2) /(10400x48.8x10 4 ))(355/235) 2 )) 0.5 Thus L m = 759 mm > 725mm to the first restraint … Ok The column is checked between the second and the third restraints over a length of 1.35 m. At the third restraint, M = H x 3.65m = 187kN x 3.65m = 682.6kNm At the second restraint, M = H x 5m = 187kN x 5m = 935kNm And the axial force at the second restraint, F x = 267.9+(0.725/5.725)(286.1-267.9) F x = 270.2 kN The effective lengths and slenderness ratios are: - k c = (2 + 0.45L r I c /(L c I r )) = (2+0.45x2x15.06x98600/(5.725x55900)) Thus k c = 6.18 Then L cr,y / i y = 6.18 x 5725/232 = 152.4 And L cr,z / i z = 1350 / 34.0 = 39.7 From Table 6.2 of EN1993-1-1, for an S355 rolled H section, t f less than 40 mm, h/b > 1.2, buckling about the major y–y axis, curve a, imperfection factor, α = 0.21, buckling about the minor z–z axis, curve b, imperfection factor, α = 0.34
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From Table 6.2 of EN1993-1-1, for an S355 rolled I section, t f less than 40mm h/b>1.2, buckling about the major y–y axis, curve a, imperfection factor, α = 0.21,
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