Page 125 - Structural Steel Designers Handbook AISC, AASHTO, AISI, ASTM, and ASCE-07 Design Standards

Brockenbrough_Ch03.qxd 9/29/05 5:05 PM Page 3.57

CONNECTIONS

CONNECTIONS 3.57

This results in a conservative connection capacity of

φR n = 2.06 × 19.6 = 40.4 kips < 51 kips No good

Alternatively, the direction of the force on each bolt can be found using the instantaneous center
method. The bolt tear-out can then be found along the direction of the force. Though this proce-
dure (Muir and Thornton, 2004) is too complex to show here, it results in an effective bolt value
of 24.8 kips. Thus,

φR n = 2.06 × 24.8 = 51.0 kips ≥ 51 kips OK

Determine maximum plate thickness to ensure connection ductility at bolts: The maximum nom-
inal strength of a 1-in-diameter A325X bolt (with the 1.25 factor discussed in Art. 3.1.5) is

1
×
.
.
R = 125 60   2 π = 58 9 kips/bolt
 
n
2
For the “moment only” condition, C′= 26.0 from AISC Table 7-9. This results in a moment M of
M = 58.9 × 26.0 = 1530 in⋅kips
The maximum plate thickness of

t pmax = 6 M = ( 6 1530 ) = . 128 in
FL 2 50 12 2
()
y
1
5
A plate up to 1 / 4-in-thick is allowed for ductility. Thus, proceed with checks for / 8-in-thick plate.
Design strength for interaction of shear and bending in plate:
51 0
• Shear stress on gross section of plate = f v = 0 625 . ×12 = 6.80 ksi
.
• Plastic section modulus of gross section of plate = Z gross = . 0 625 4 ×12 2 = 22 . in 3
5
.
• Bending stress on gross section of plate = f b = 51 0 . 22 5 . ×10 5 . = 23 8 ksi 2 2
≤
23.8
6.80
×
×
×
• Shear and bending interaction on gross section of plate = ( 0.9 0.6 50 ) +( 0.9 50 ) = 0.343 1
• Shear stress on net section in plate = f v = 0 625 ( 51 0 . − ×1 125) = 10 9 ksi
.
.
× 12 4
.
• Plastic section modulus of net section of plate = Z net = 22 1 − ( . 1 125 ) = 14 . in 3
1
.5
3
• Bending stress on net section of plate = f b = 51 0 . 14 1 . ×10 5 . = 38 0 ksi
.
10 9 . 2 38 0 . 2 ≤
×
×
.
.
• Shear and bending interaction on net section of plate = ( 075 06 65 ) +( 075 65 ) = 0 746 1.
× .
Plate buckling:
LF y 12 50
λ= = = 0 468
.
+
t 47 500 +112 000, (/ 2 0 625 47 500 112 000 12 2 10 5)] 2
,
[/
(
.
L e 2 )
.
,
0
,
p
Since λ= 0.472 ≤ 0.7, buckling does not govern.
5
Size weld: For ductility, the weld size must be equal to / 8 t p = 0.625(0.625) = 0.391 in. Therefore,
7
use / 16-in fillet welds. Checks show that the welds meet design strength requirements for shear
and bending.
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