DETERMINING THE DESIGN SHEAR
STRENGTH OFA BEAM WEB
The end of a W12 x 86 beam (A3 6 steel) has been prepared as shown in Fig. 49 for con-
nection to a supporting member. The three holes are 15/16 in (2.38 cm) in diameter for
7/8-in (2.22-cm)-diameter bolts. Determine the design shear strength of the beam web.
Calculation Procedure:- Find the applicable limit states
The applicable limit states are shear yielding, shear fracture, and block shear rupture. For
shear yielding [of gross section (1) in Fig. 49 ]
0^ = 0.90x0.64^ (J5-3)Avg = (</-cope)f = (12.53 in - 2 in) x 0.515 in = 5.42 in^2 (34.96 cm^2 )<j>Rn = 0.9 x 0.6 x 5.42 in^2 x 36 ksi = 105 kips (467 kN)For shear fracture [of net section (1) in Fig. 11-9]
0^ = 0.75x0.64^ (J4-1)Ans = (d-copQ-3>dh)t = (12.53 in - 2 in - 3 x i*/ 16 in) x 0.515 in - 3.97 in
2
(25.6 cm
2
)
(I)Rn = 0.75 x 0.6 x 3.97 in^2 x 58 ksi - 104 kips (462.6 kN)For block shear rupture [of section (2) in Fig. 11-9] 0 = 0.75 and Rn = the greater value of
0.6Av 8 Fy +A J?u (C-J4-1)0.6AnF^A 8 Fy (C-J4-2)FIGURE 49