Civil Engineering Reference
In-Depth Information
c
d
b
a
a
P
vs
a
a
d
b
c
FIGURE 9.16
Block shear failure in a member at an axial tension connection.
1.5
+
4
t
p
≤
6 in. and, based on bearing stress considerations,
2
d
b
σ
bc
F
u
e
b
≥
.
(9.47)
Block shear failure occurs in members at ultimate shear stress on planes along
the bolt lines (lines a-a in Figure 9.16) and ultimate tensile stress on planes between
bolt lines (line b-b in Figure 9.16). The failure ultimately results in the tear-out of a
section (shaded area of the member in Figure 9.16). If the allowable shear stress is
limited to
F
vU
=
0.60
F
u
/FS (see Equation 9.19) and allowable tensile stress is
F
u
/FS,
the allowable block shear strength,
P
vs
, using FS
=
2.0, is
P
vs
=
0.30
F
U
A
nv
+
0.50
F
u
A
nt
.
(9.48)
However, a combination of yielding on one plane and fracture on the other
plane is likely, depending on the connection configuration. When the net fracture
strength in tension is greater than the net fracture strength in shear,
F
u
A
nt
≥
F
vU
A
nv
≥
0.60
F
u
A
nv
, yielding will occur on the gross shear plane and the allowable block shear
strength,
P
vs
, becomes
P
vs
=
0.35
F
y
A
gv
+
0.50
F
u
A
nt
.
(9.49a)
Conversely, when the net fracture strength in tension is less than the net fracture
strength in shear,
F
u
A
nt
≺
0.60
F
u
A
nv
, yielding will occur on the gross
tension plane and the allowable block shear strength,
P
vs
,is
F
vU
A
nv
≺
P
vs
=
0.30
F
u
A
nv
+
0.55
F
y
A
gt
,
(9.49b)
where
F
u
is the ultimate tensile stress of connection element,
F
y
is the tensile yield
stress of connection element,
A
gv
is the gross area subject to shear stress (thickness
times gross length along lines a-a in Figure 9.16),
A
nv
is the net area subject to shear
