Question

25m- JL 127 6.4 20 EN 10025-2 Grade S355 M20-8.8 bolts (standard holes) Compute the design strength фР, of this connection considering applicable tension member limit states. Bolt limit states will be covered later in this course.

Answer 1

GIVEN DATA Member Details Grade of Steel Member designation Gross area of section, Thickness of member Gr S355 2L 127x76x6.4 1250 mm 6.4 mm Connection Details Member connected to one/two legs Diameter of bolt, Number of bolts across a section Number of bolts in line Pitch of the bolts, End distance Distance of centroid from edge one leg 20 mm 75 mm 25 mm (in the direction of loading) 20.42 mm CALCULATIONS For grade of steel S355 355 Mpa 470 Mpa 20+ 22 mm Yield stress, Ultimate stress, Diamter of bolt hole, DESIGN STRENGTH Yielding in the gross section Design strength of gross section is obtained as: 0.9 x 355 x 1250 x 2 798.8 kN Fracture in the net section Net area A-n (d, t =2x 1250 1x 22x 6.4 2359.2 in2 NOTE Shear lag Factor, 1.0 when connected to two legs when connected to one leg where is distance of centroid from edge and L is the length of connection Length of connection in the direction of loading, 150 in Since the member is connected to one leg, Shear lag Factor 0.864 Effective area UA

0.864 x 2359.2 2038.3 in2 Design strength of net section is obtained as 0.75x 470x 2038.3 718.5 kN Thus, design strength of the member is the smallest of the three values, that is IS 9,7,- 718.5 kN