IIA-2 EXAMPLE II.A-1 ALL-BOLTED DOUBLE-ANGLE CONNECTION Given

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II A-2

EXAMPLE II.A-1 ALL-BOLTED DOUBLE-ANGLE CONNECTION

Given:

Select an all-bolted double-angle connection between an ASTM A992 W36231 beam and an ASTM A992
W1490 column flange to support the following beam end reactions:

R = 37.5 kips
D
R = 113 kips
L

Use w-in.-diameter ASTM A325-N or F1852-N bolts in standard holes and ASTM A36 angles.

Solution:

From AISC Manual Table 2-4, the material properties are as follows:

Beam
ASTM A992
F = 50 ksi
y
F = 65 ksi
u

Column
ASTM A992
F = 50 ksi
y
F = 65 ksi
u

Angles
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II A-3

ASTM A36
F = 36 ksi
y
F = 58 ksi
u

From AISC Manual Table 1-1, the geometric properties are as follows:

Beam
W36231
t = 0.760 in.
w

Column
W1490
t = 0.710 in.
f

From Chapter 2 of ASCE/SEI 7, the required strength is:

LRFD ASD
R = 1.2(37.5 kips) + 1.6(113 kips) R = 37.5 kips + 113 kips
u a
= 226 kips = 151 kips

Connection Design

AISC Manual Table 10-1 includes checks for the limit states of bearing, shear yielding, shear rupture, and block
shear rupture on the angles, and shear on the bolts.

Try 8 rows of bolts and 2L532c (SLBB).

LRFD ASD
R = 247 kips > 226 kips o.k. R
n n 165 kips151 kips o.k.


Beam web strength from AISC Manual Table 10-1: Beam web strength from AISC Manual Table 10-1:

Uncoped, L = 1w in. Uncoped, L = 1w in.
eh eh

R = 702 kips/in.(0.760 in.) R
n n 468 kips/in.(0.760 in.)
= 534 kips > 226 kips o.k.

= 356 kips > 151 kips o.k.

Bolt bearing on column flange from AISC Manual Bolt bearing on column flange from AISC Manual
Table 10-1: Table 10-1:

R = 1,400 kips/in.(0.710 in.) R
n n 936 kips/in.(0.710 in.)
= 994 kips > 226 kips o.k.

= 665 kips > 151 kips o.k.
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II A-4

EXAMPLE II.A-2 BOLTED/WELDED DOUBLE-ANGLE CONNECTION

Given:

Repeat Example II.A-1 using AISC Manual Table 10-2 to substitute welds for bolts in the support legs of the
double-angle connection (welds B). Use 70-ksi electrodes.

Note: Bottom flange coped for erection.

Solution:

From AISC Manual Table 2-4, the material properties are as follows:

Beam
ASTM A992
F = 50 ksi
y
F = 65 ksi
u

Column
ASTM A992
F = 50 ksi
y
F = 65 ksi
u

Angles
ASTM A36
F = 36 ksi
y
F = 58 ksi
u

From AISC Manual Table 1-1, the geometric properties are as follows:

Beam
W36231
t = 0.760 in.
w

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II A-5

Column
W1490
t = 0.710 in.
f

From Chapter 2 of ASCE/SEI 7, the required strength is:

LRFD ASD
R = 1.2(37.5 kips) + 1.6(113 kips) R = 37.5 kips + 113 kips
u a
= 226 kips = 151 kips

Weld Design using AISC Manual Table 10-2 (welds B)

Try c-in. weld size, L = 23 2 in.

t = 0.238 in. < 0.710 in. o.k.
f min

LRFD ASD
R = 279 kips > 226 kips o.k. R
n n 186 kips > 151 kips o.k.


Angle Thickness

The minimum angle thickness for a fillet weld from AISC Specification Section J2.2b is:

t = w + z in.
min
= c in. + z in.
= a in.

Try 2L432a (SLBB).

Angle and Bolt Design

AISC Manual Table 10-1 includes checks for the limit states of bearing, shear yielding, shear rupture, and block
shear rupture on the angles, and shear on the bolts.

Check 8 rows of bolts and a-in. angle thickness.

LRFD ASD
R = 286 kips > 226 kips o.k. R
n n 191kips > 151 kips o.k.

Beam web strength: Beam web strength:

Uncoped, L = 1w in. Uncoped, L = 1w in.
eh eh

R = 702 kips/in.(0.760 in.) R
n n 468 kips/in.(0.760 in.)
= 534 kips > 226 kips o.k.

= 356 kips > 151 kips o.k.

Note: In this example, because of the relative size of the cope to the overall beam size, the coped section does not
control. When this cannot be determined by inspection, see AISC Manual Part 9 for the design of the coped
section.
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II A-6

EXAMPLE II.A-3 ALL-WELDED DOUBLE-ANGLE CONNECTION

Given:

Repeat Example II.A-1 using AISC Manual Table 10-3 to design an all-welded double-angle connection between
an ASTM A992 W36231 beam and an ASTM A992 W1490 column flange. Use 70-ksi electrodes and ASTM
A36 angles.

Solution:

From AISC Manual Table 2-4, the material properties are as follows:

Beam
ASTM A992
F = 50 ksi
y
F = 65 ksi
u

Column
ASTM A992
F = 50 ksi
y
F = 65 ksi
u

Angles
ASTM A36
F = 36 ksi
y
F = 58 ksi
u

From AISC Manual Table 1-1, the geometric properties are as follows:

Beam
W36231
t = 0.760 in.
w
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II A-7

Column
W1490
t = 0.710 in.
f

From Chapter 2 of ASCE/SEI 7, the required strength is:

LRFD ASD
R = 1.2(37.5 kips) + 1.6(113 kips) R = 37.5 kips + 113 kips
u a
= 226 kips = 151 kips

Design of Weld Between Beam Web and Angle (welds A)

Try x-in. weld size, L = 24 in.

t = 0.286 in. < 0.760 in. o.k.
w min

From AISC Manual Table 10-3:

LRFD ASD
R = 257 kips > 226 kips o.k. R
n n 171 kips > 151 kips o.k.


Design of Weld Between Column Flange and Angle (welds B)

Try 4-in. weld size, L = 24 in.

t = 0.190 in. < 0.710 in. o.k.
f min

From AISC Manual Table 10-3:

LRFD ASD
R = 229 kips > 226 kips o.k. R
n n 153 kips > 151 kips o.k.


Angle Thickness

Minimum angle thickness for weld from AISC Specification Section J2.2b:

t = w + z in.
min
= 4 in. + z in.
= c in.

Try 2L43c (SLBB).

Shear Yielding of Angles (AISC Specification Section J4.2)

A 224.0 in.c in.
gv
= 15.0 in.2

R 0.60F A (Spec. Eq. J4-3)
n y gv
=0.6036 ksi15.0 in.2
= 324 kips

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II A-8

LRFD ASD
 = 1.50
R = 1.00(324 kips) R 324 kips
n n 
= 324 kips > 226 kips o.k.
 1.50
216 kips151 kips o.k.
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II A-9

EXAMPLE II.A-4 ALL-BOLTED DOUBLE-ANGLE CONNECTION IN A COPED BEAM

Given:

Use AISC Manual Table 10-1 to select an all-bolted double-angle connection between an ASTM A992 W1850
beam and an ASTM A992 W2162 girder web to support the following beam end reactions:

R = 10 kips
D
R = 30 kips
L

The beam top flange is coped 2 in. deep by 4 in. long, L = 14 in., L = 1w in. Use w-in.-diameter ASTM A325-
ev eh
N or F1852-N bolts in standard holes and ASTM A36 angles.

Solution:

From AISC Manual Table 2-4, the material properties are as follows:

Beam
W1850
ASTM A992
F = 50 ksi
y
F = 65 ksi
u

Girder
W2162
ASTM A992
F = 50 ksi
y
F = 65 ksi
u

Angles
ASTM A36
F = 36 ksi
y
F = 58 ksi
u
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IIA -10

From AISC Manual Tables 1-1 and 9-2 and AISC Manual Figure 9-2, the geometric properties are as follows:

Beam
W1850
d = 18.0 in.
t = 0.355 in.
w
S = 23.4 in.3
net
c = 4.00 in.
d = 2.00 in.
c
e = 4.00 in. + 0.500 in.
= 4.50 in.
h = 16.0 in.
o

Girder
W2162
t = 0.400 in.
w

From Chapter 2 of ASCE/SEI 7, the required strength is:

LRFD ASD
R = 1.2(10 kips) + 1.6(30 kips) R = 10 kips + 30 kips
u a
= 60.0 kips = 40.0 kips

Connection Design

AISC Manual Table 10-1 includes checks for the limit states of bearing, shear yielding, shear rupture, and block
shear rupture on the angles, and shear on the bolts.

Try 3 rows of bolts and 2L4324 (SLBB).

LRFD ASD
R = 76.4 kips > 60.0 kips o.k. R
n n 50.9 kips > 40.0 kips o.k.


Beam web strength from AISC Manual Table 10-1: Beam web strength from AISC Manual Table 10-1:

Top flange coped, L = 14 in., L = 1w in. Top flange coped, L = 14 in., L = 1w in.
ev eh ev eh

R = 200 kips/in.(0.355 in.) R
n n 133 kips/in.(0.355 in.)
= 71.0 kips > 60.0 kips o.k.

= 47.2 kips > 40.0 kips o.k.

Bolt bearing on girder web from AISC Manual Table Bolt bearing on girder web from AISC Manual Table
10-1: 10-1:

Rn 526 kips/in.(0.400 in.) Rn 351 kips/in.(0.400 in.)
= 210 kips > 60.0 kips o.k. 
= 140 kips > 40.0 kips o.k.

Note: The middle portion of AISC Manual Table 10-1 includes checks of the limit-state of bolt bearing on the
beam web and the limit-state of block shear rupture on coped beams. AISC Manual Tables 9-3a, 9-3b and 9-3c
may be used to determine the available block shear strength for values of L and L beyond the limits of AISC
ev eh
Manual Table 10-1. For coped members, the limit states of flexural yielding and local buckling must be checked
independently per AISC Manual Part 9.
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IIA -11

Coped Beam Strength (AISC Manual Part 9)

Flexural Local Web Buckling
d
Verify c2d and d  .
c
2

c4.00 in.218.0 in.36.0 in. o.k.
18.0 in.
d 2.00 in.  9.00 in. o.k.
c
2

c 4.00 in.

d 18.0 in.
= 0.222

c 4.00 in.

h 16.0 in.
o
= 0.250

c
Since 1.0,
d
2c
f  (Manual Eq. 9-8)
d
= 2(0.222)
= 0.444

c
Since 1.0,
h
o
1.65
h 
k 2.2 o  (Manual Eq. 9-10)
 c 
1.65
16.0 in.
=2.2 
4.00 in.
= 21.7

2
t 
Fcr 26,210 w  fk (Manual Eq. 9-7)
ho 
2
0.355 in.
=26,210  0.44421.7
 16.0 in. 
= 124 ksi ≤ 50 ksi

Use F = 50 ksi.
cr

F S
R  cr net from AISC Manual Equation 9-6
n
e
50 ksi23.4 in.3
=
4.50 in.
= 260 kips

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