Question

A plate 7x3/4” using A36 material is shown below, where the holes are for ¾” diameter bolts. Determine the available tension capacity including checking block shear strength by (a) LRFD and (b) ASD

7" 1.5" 4" 1.5"

Answer 1

GIVEN DATA Member Details Grade of Steel Plate dimension A 36 Connection Details Diameter of bolt, 3 in Number of bolts across a section, Number of bolts in line Pitch of the bolts Gage of the bolts End Distance Edge distance 3 in 4 in 1.25 in 1.5 in CALCULATIONS For grade of steel, A 36 Yield stress, 36 ksi 58 ksi 5.25 n Ultimate stress, Gross area of section, Thickness of member, Diamter of bolt hole 0.75 in 8 0.875 in DESIGN STRENGTH elding in the gross section Design strength of gross section is obtained as: LRFD ASD 0.9 x 36 x 5.250 170 kips 36 x 5.250 /1.67 -113.2 kips Fracture in the net section A-ndht) Net area 5.25 - 2 x 0.875 0.75 3.938 in Length of connection in the direction of loading 6 in Shear lag Factor Effective area, UAn 1 x 3.938 3.938 in2

Design strength of net section is obtained as LRFD ASD 0.75x 58 x 3.938 171 kips -58 x 3.938 /2.0 114.2 kips Block Shear Gross and net sections of shear Ag2 x [Length (a-b)lxt - 2 x 7.25 x 0.750 10.9 in2 2 xLength (a-b) minus[(m-1)+0.5]d hxt An - 27.25 -2.5 x 0.875x 0.750 - 7.6 in2 ons of tension, Atength (b c)l xt 4 x 0.75 Gross and net sections of tension, Ant llength (b-c)] minus [(n-1)xd»])x -41.0 0.8750.750 2.3 in Comparing 0.6F, A and FA pt 0.6FuA 0.6x 58 x 7.6- 264.48 kips 133.40 kips FuAn58 x 2.3 Therefore Hence, shear fracture-tension yielding controls Design rupture strength, LRFD ASD = 0.75 x[ 264.48 - 264.48 + +36 x 3.0 1 36 x 3.01/2 186.2 kips 279 kips Thus, design strength of the member is the smaller of the three values, that is LRFD ASD o,Tm- 170 kips T113.2 kips