1 kip = 4.448 KN
1 inch = 25.4 mm
db = 3/4 + 1/16 = 0.813 in. (2.07 cm)
Ant = (3/8)(1.5.813/2) = 0.41 in.2 (2.64 cm2)
Agv = (3/8)(4.5) = 1.69 in.2 (10.90 cm2)
Anv = (3/8)(4.5 1.5(0.813)) = 1.23 in.2 (7.93 cm2)
Agt = (3/8)(1.5) = 0.56 in.2 (3.61 cm2)
φ n = φ0.6FyAgv + FuAnt)
R
(
AISC LRFD Eq. J43a
φ n = 0.75(0.6(36)(1.69) + (58)(0.41)) = 45.21 k (201 KN) >11.2 k (50 KN), OK
R
φ n = φ0.6FuAnv + FyAgt)
R
(
AISC LRFD Eq. J43b
φ n = 0.75(0.6(58)(1.23) + (36)(0.56)) = 47.2 k (210 KN)> 11.2 k (50 KN), OK
R

Check shear yielding of plate
φVn = φ(0.6)(Fy)(A)
AISC LRFD Eq. J53
v
v
φVn = 0.9(0.6)(36)(6 x 3/8) = 43.7k (194 KN) > 11.2 k (50KN), OK
v

φVn = φ(0.6)FuAn
AISC LRFD Eq. J41
v
v
φVn = 0.75(0.6)(58)(62(0.813))3/8 = 42.8 k (190KN) > 11.2 k (50 KN), OK
v

Check bolt shear and plate bearing
Bearing strength of one bolt: φ n = φ .4dtFu
r
2
AISC LRFD Eq. J31a
φ n = (0.75)(2.4)(3/4)(3/8)(58) = 29.4 k (131KN)
r
Shear strength of bolt: φ n = φ nAb
r
F
AISC LRFD Sec. J3
φ n = 0.75(60)(0.44) = 19.8 k (18 KN) (governs)
r
Determine bolt shear demand based on elastic analysis of bolt group:
eccentricity of horizontal component of brace force = 6.11" (15.52 cm)
M = 5.9(6.11) = 36 kip*in (4.07 KNm)
∑d
2
= 2(0) 2 + 2(1.5) 2 = 4.5 in.2 (29.0 cm2)
Mv
36(1.5" )
=
= 12.0k (53.4 KN)
Force to each bolt =
∑d
2
4.5
Each bolt must also resist of the shear and axial forces;
1/2V = 8.32/2 = 4.16k (18.5 KN), Axial = 11.25/2 = 5.63k (25.0 KN)
4.162 + (12 + 5.63) 2 = 18.11k (80.5KN) < 19.8 kip / bolt(88.1 KN), OK
Resultant =
H157