Question

L152 X 102 x 10 mm QUESTION 1 102 mm 38 mm xx There are 2 failure line to solve. Yes Ο Νο 76 mm 152 mm dll 38 17m Cross section QUESTION 2 r 13@38 mmf 13 mm plate thickness The failure lines are and O OB o Do QUESTION 3 Elle GIVEN: M20 standard holes are used. Neglect the block shear. (a) failure lines Fy = 248 MPa Fu = 400 MPa The angle is L152.102.10 from Profile Table Ag = 2329 mm2, Xg = 23.70 mm and t = 10 mm Calculate the yield strength (Write with the unit) Calculate the rupture strength (Write with the unit) Which one is selected? A or B

There are 2 failure line to solve.

Yes

No

The failure lines are ....... and ...............

M20 standard holes are used. Neglect the block shear.

Fy = 248 MPa

Fu = 400 MPa

The angle is L152.102.10 from Profile Table A_g = 2329 mm2, x_g = 23.70 mm and t = 10 mm

Calculate the yield strength (Write with the unit)

Calculate the rupture strength (Write with the unit)

Which one is selected? A or B

Answer 1

Answer: YES Line Line ABC A-B-D-E Section: L 152 x 102 x 10 Ag = 2322 2329 mm² X = 23.70 mm t = 10 mm s= 38 mm, g = 76 mm. Net with Area Lime A-B-C An Ag - 20x 10 = 2329 – 200 = 2129 mm² Lime A-B-D-E Rue = Ag - Edut + e t = 2329 - 2 x(20x10) + 38X10 = 1976.5 mm2 4x76 Line A-B-D-E te sectect An = Ana = 1976.5 mm² U = 1 - - 2317 = 0.792 3x38 - Effective Area de = Uan = 0'792 x 1976'5 = 1565.40 mm²

Design gield strength Pin = $f Ag = 0.9 x 248 x 2329 = 519.83X103N = 519.83 KN Dirs Design rupture strength & Pn = $ Fute = 0'75 x 400 x 156514 = 469'62 Xion = 469.62 KN Bers Design strength of connection dpn = 469.62 KN secte (select minimum volue) Azeer сhооѕе в.