Question

To analyze two built-up members that have the same geometry but are fastened differently, determine the maximum applicable shear force on each cross section, and determine the adjustment in spacing between the weaker member’s fasteners that would allow the member to support the equivalent maximum shear force of the stronger member.

The two cross sections shown below, (a) and (b), are subjected to a vertical shear force as shown. The members are fastened by nails that can support a load of 25.00 kN

each and are spaced perpendicularly to the page in increments of s = 110.0 mm . The geometries of the cross sections are given by a = 235.0 mm , b = 40.00 mm , c = 285.0 mm , d = 155 mm , and e = 325 mm . Assume the cross sections are uniform along the entire lengths of the members.
Learning Goal: To analyze two built-up members that have the same geometry but are fastened differently, determine the maximum applicable shear force on each cross section, and determine the adjustment in spacing between the weaker member's fasteners that would allow the member to support the equivalent maximum shear force of the stronger member. The two cross sections shown below, (a) and (b), are subjected to a vertical shear force as shown. The members are fastened by nails that can support a load of 25.00 kN each and are spaced perpendicularly to the page in increments of s = 110.0 mm. The geometries of the cross sections are given by a = 235.0 mm, b = 40.00 mm , c = 285.0 mm, d = 155 mm, and e = 325 mm Assume the cross sections are uniform along the entire lengths of the members. MTT bk +5k bk bk (a) Part A - Maximum applicable shear force on the member (a) Determine the maximum applicable shear force on the member (a). Express your answer to four significant figures and include the appropriate units. View Available Hint(s) HA ? Value Units Submit

Answer 1

a=235mm 0 b= lomm let the centroid of section from top fibre is 9 = 155 x 40 x 12x (25x40 x 2) 1155x40)+2x (325X40) j 30=155 325mm NA 285 mm ý = 135.06 mm b I 12 & de I - 12 2 length ( ) moment of inertia about NA db² +42 +2x[ded the da] [ da distance of centroidal axis of individual areas from NA ) 155x403 +(155440) x (25:06 -*)* + 2x5 40 x325+ (40x325) I 33.1341222 X 107 mm. 4 X(325-135.00) @ Shear Shear force per unit flow VAY vx (235 x40 ) x (135.06- 33.1341222 X 107 each nail com support 25 kN load spacing of nails (s) = 110 mm =(3.2642 x 103) v » Shear force resisted by a nails per unit length for maximum applicable = 0.45455 kW Shear force >> (3.2642x70%) v = 0.45455 NA. 2X 25 ܘܬܐ V 139.25 kn NA Shear flow = VAY v (155x40) (135.06-20 ) 33.1341222 X 10? shear flow = - 12.153x1003) V = 0.45455 (2-153x2030 maximum, V = 211.122 kn Ans. shear force =)

С. Shear force resisted by stronger section V = 211.122 kW required spacing in the weaker member for equivalent maximum applicable shear force is (5) => shear flow for weaker section @ for equivalent max. Shear force 3.2642x10^(-3) x211.122 =0.68914 kN/mm for required spacing 24 25 -0.68914 = s = 72.55 mm Aus.