Shear Lag
Phenomenon (e.g. Angle with connection to one leg)
Member has elements not in common plane and connectors are not provided in all
elements
Number of fasteners per gage line may not be sufficient to develop full section strength
Tensile force is not uniformly distributed over net area
o
Non-Uniform Tensile Stress
→
concentration of shear stress → shear distortion →
shear lag in vicinity of connection
AISC D3 (pg. 16.1-27): Effective net area
A
e
= UA
n
where
U
depends on
member shape, and
number of fasteners per gage line.
An empirical measure for the shear lag reduction coefficient
𝑈 = 1 −
?̅
𝑐??
𝐿
𝑐??
?̅
???
: distance measured from the plane of connection to the centroid of cross sectional
area (tabulated in the section tables -AISC Part I)
L
con
: length of connection
Note: For Splice plates,
A
e
= A
n
≤ 0.85A
g
(AISC J4.1). Members such as single angles,
double angles and WT sections shall have connections proportioned such that
U
≥ 0.6.

35

36
Examples

37
Example
: Determine the strength of the tension member in a truss
Member: W8×24 and Grade 50 steel
Two lines of ¾-in diameter bolts
Three bolts per line
Solution
: Properties of W8×24:
A
g
= 7.08 in;
d
=7.93 in;
b
f
=6.5 in;
t
f
=0.40 in
Based on gross area:
P
u
=
φ
t
F
y
A
g
= 0.9(50)(7.08) = 319 kips
Based on effective net area:
P
u
= φ
t
F
u
A
e
where
A
e
= UA
n
𝐴
?
= 𝐴
?
− 4 (
7
8
) 0.40 = 7.08 − 1.40 = 5.68
in
2
Since
𝑏
?
= 6.5
in
>
2
3
𝑑 (= 0.67 × 7.93 = 5.31
in),
𝑈 = 0.90
1
A
e
= UA
n
= 0.9(5.68) = 5.11 in
2
φ
t
F
u
A
e
=0.75(65)(5.11) = 249 kips
The design strength of this member is 249 kips as controlled by rupture failure based on least net
area.
1
Suppose we assume bolt-to-bolt separations are 3 in, then
𝑈 = 1 −
?
̅
𝐿
= 1 −
0.695
6
= 0.88.
(see
WT4×12 page 1-66 for
?
̅
).

38
Example
: Design a 9 feet single angle subject to 30 kips dead load and 40 kips live load.
Member: A36 steel
7/8-in diameter bolts
At least 4 bolts per line
Solution
:
Load combinations:
P
u
=1.4(30) = 42 kips
P
u
=1.2(30) + 1.6(40) = 100 kips
Based on gross area:
min
𝐴
?
=
𝑃
?
𝜑
?
𝐹
?
=
100
0.9(36)
= 3.09
in
2
Based on effective net area
2
:
U
=0.80 since
o
bolts connected to one leg only
o
at least 4 bolts per gage line
o
only one bolt at any one cross section
min
𝐴
?
=
𝑃
?
𝜑
?
𝐹
?
𝑈
=
100
0.75(58)(0.80)
= 2.87
in
2
min
A
g
= min
A
n
+Δ
A
due to one bolt where Δ
A
=(
d
bolt
+ 1/8)
t
=
t
.
Slenderness ratio:
min
𝑟 =
𝐿
300
=
12(9)
300
= 0.36
in
Section selection:
Angle
t
(in)
Hole
area
ΔA
Gross Area Required
max
(
𝑃
?
𝜑
?
𝐹
?
;
𝑃
?
𝜑
?
𝐹
?
𝑈
+ ∆𝐴 )
Lightest Angles (
A
(in
2
),
r
(in))
5/16
0.312
max(3.09, 2.87+0.312) = 3.18
6x6x
5
/
16
(
A
=3.67,
r
=1.19)
3/8
0.375
max(3.09, 2.87+0.375) = 3.25
6x3
1
/
2
x
3
/
8
(
A
=3.44,
r
=0.762)
7/16
0.438
max(3.09, 2.87+0.438) = 3.30
5x3x
7
/
16
(
A
=3.31,
r
=0.644)
4x4x
7
/
16
(
A
=3.30,
r
=0.777)
1/2
0.500
max(3.09, 2.87+0.500) = 3.37
4x3x
1
/
2
(
A
=3.25,
r
=0.633)
NG
3
1
/
2
x3
1
/
2
x
1
/
2
(
A
=3.25,
r
=0.679)
NG
5/8
0.625
max(3.09, 2.87+0.625) = 3.50
4x3x
5
/
8
(
A
=3.99,
r
=0.631)
Use L5x3x
7
/
16
or L4x4x
7
/
16
(with minimum area (weight))
2
Assume L6
x
6
x
5
/
16
(see page 1-42) and 3 in. bolt-to-bolt separations,
𝑈 = 1 −
?̅
𝐿
= 1 −
1.6
9
=
0.82.

39
Block Shear
: Tearing out (fracture) failure (AISC J4.3)

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- Spring '14
- GregoryG.Deierlein
- Strength of materials, Tensile strength, in2, net area