Bolts
Non-Structural
(A-307)
A325
Fasteners
Structural
Connections
Rivets (obsolete)
Welding (chap, 5)
Structural
A449
A490
F-1
Shank
Bolt
Head
Threaded
Section
F-2
The AISC specifications recognize two types of high-strength bolted connection:A) Bearing Type Connection:Where connected parts are allowed to
bear directly on the bolt shank. Hence both shear
strength and bearing strength need to be
checked.
B) Slip-Critical Type Connection (formerly
known as friction-type connection):Where connected parts are not allowed to
slip, and shear forces are transmitted by friction
forces rather than direct shear. This connection
requires high pre-tensioning force of bolts upto
their “proof load” which equals  70% of tensile
strength.
F-3
When slip resistance is required, the pre-
tensioning of bolts should as high as their
“Proof Load”, which is equivalent to the
yielding stress of the bolts material as
obtaining by the 0.2% “offset method” or the
0.5% “extension method”. The proof load
stress is a minimum of 70% and 80% of the
minimum tensile strength of A-325 bolts, and
the A-490 bolts respectively.
The proof load can be achieved by using the
turn-of-nut method, which is ½ turn of nut from
the “snug-fit” position. Otherwise a “calibrated
wrench” must be used to achieve the proof
load for bolts.
F-4
(70 % Fu)
(70 % Fu)
(80 % Fu)
F-5
The most typical types of bolted connections are shown below:-
Lap Joint
(single shear)
Butt Joint
(double shear)
(a) Shear Connections
W section
(d) Combined shear and tension connections
P
P
Structural tee
(b) Eccentric shear
Connection.
“Thread outside” (X)
(c) Tension Connections
In shear connection, threads could be:A) In the shearing plane (N), or
B) Outside the shearing plane (X).
P
P
“Thread inside” (N)
F-6
There are several “limitstates” or failure modes that
may control the strength
of a bolted connection.
As can be seen from the
above figure, three failure
modes (b, d & g) are relevant
to the connected plates and
not relevant to the fasteners.
F-7
The nominal tensile strength of a bolt (Rn) can be found as:
Rn  Fub An
Where
Fub = Tensile strength of the bolt material.
An = Bolt net area through threaded zone*.
This net area  0.75 to 0.79 of the bolt gross
area Ab (area of the shank).
Thus:
*
Rn  Fub 0.75 Ab 
0.9743 


Thread cross section area  0.785   d b 
n 

Where,
2
db = dia. of bolt (shank)
n = number of threads per inch.
Note: Tensile strength values Fub are given in LRFD
table J3.2 page (16.1)-104.
F-8
The nominal-shear strength for one fastener (Rn) will be
Rn = m A b  u
Where,
= (1) for single shear
m = number of shear planes
= (2) for double shear
Ab = Bolt gross area.
u = Ultimate shear strength for the bolt material,
found experimentally to be around (0.62 Fub) of the
ultimate tensile strength (Fub).
Thus:
Rn = m Ab (0.62 Fub)
F-9
The bearing failure near the edge of the plate is related to the
shear failure (tear-out) as shown in figure 4.6.4 b & d (page F-7).
Assuming angle () 0 as shown in figure 4.6.5 below:
Rn  2t Le  d2  up
Where
up = Shear strength of plate material = 0.62 Fu
Fu = Ultimate tensile strength of plate material.
Le = Distance from center of last bolt to the edge
of plate along direction of force.
d = Nominal Bolt diameter.
 Le 1 
Rn  1.24Fu dt   
 d 2
F-10
Thus:
Since LRFD recommends spacing of bolts in
the direction of force to be (2.67 d) – [see
LRFD J – 3.3]. Consider (Le  2.67 d) above:
* Rn = 2.4 Fu dt
(When deformation of the hole at service
load is a design consideration). (Eq J3-6a)
or
* Rn = 3.0 Fu dt
(When deformation of the hole at service
load is not a design consideration). (Eq J3-6b)
The lower resistance value guarantees that no tear-out
due to bearing shall occur below this (Rn) value, and no
Deformation at the hole shall occur at service load level.
(AISC J3-10 Page16.1-111)
F-11
In general:
n Rn  Σ i Qi
Where,
Rn = Nominal resistance (strength)
n = Strength reduction factor (resistance factor)
iQi = Factored load on structure (ASCE – 7)
for a single bolt :
Rn  Pu
Where,
u = 0.75 for fracture in tension, shear or bearing.
Rn = Nominal strength of one fastener
Pu = factored load on one fastener
The strength of a fastener is based on:
(1) Threads are included in shear plane (N)
1- Tension capacity
2- Shear capacity
(2) Threads are excluded from shear plane (X)
3- Bearing capacity
F-12
Reference is made to (LRFD) – J 3.6), and page (F-8):
Rn  Fub 0.75 Ab   0.75Fub (0.75 Ab )
or
Rn  0.75(0.75Fub )Ab
Where
 = 0.75 as stipulated in AISC J3.6 (page 16.1-108)
Fub = Tensile strength of bolt material
(120 ksi for A – 325 Bolts & 150 ksi for A – 490 Bolts).
Ab = Gross area of the bolt (Shank section).
F-13
A – No threads in shear plane (X): Rn = 0.75 (Rn)
(AISC J3-6)
Rn = 0.62m AbFub
(page F-9)
For long Connections (upto 50 inch long), a 20% reduction is needed
(strength of long connection < Σ of strength of individual bolts).
Thus:Rn = (Reduction due connection length)(0.62 Fub)m Ab
= 0.8 x 0.62 Fub . m Ab = 0.5 Fub m Ab
Rn = (0.75)(0.5) Fub m Ab
Where,
Fub = Ultimate strength of bolt material:
120 ksi for A – 325 ; 150 ksi for A – 490
m = 1 for single shear , 2 for double shear
Ab = Area of the shank portion of bolt
(See LRFD J 3.6 & table J 3.2)
F-14
B - Threads in Shear Plane (N):The area of the threaded section of the bolt is about 75%
of the gross area at the shank section,
Thus:
Rn = [Reduction for length] (0.62 Fub) m (0.75 Ab)
= (0.8)(0.62 Fub) m (0.75 Ab)
= 0.37 Fub m Ab
 (0.40 Fub) m Ab
Hence:
Rn = 0.75 (0.4 Fub) m Ab
(see Table J-3.2)
F-15
Nominal bearing strength (Rn) was developed earlier (page F-11).
However LRFD-J3.10 further reduces (Rn) in order to prevent bolt
elongations exceeding 0.25 in.
Hence several categories are presented here below:
1: Deformation Limit State (Standard Holes):Where Le  1.5 d and c/c spacing  3d, and there are two or more bolts:
 Rn =  (2.4 dt Fu) …………..(Equ.. J3-6a)
 = 0.75
d = Nominal diameter of bolt at shank section.
t = Thickness of the least connected plate.
Fu = Ultimate tensile strength of the connected plate.
F-16
2: Deformation Limit State (Long-Slotted Holes):
Same as previous conditions.
 Rn =  (2.0 dt Fu)… (Equ. J2 6c)
Where  = 0.75 ......... (AISC - J3 – 10 Page 16-1-111)
3: Strength Limit State (hole elongation  0.25 in.):
 Rn =  (3.0 dt Fu) ……. (Equ. J3 – 6b)
See page (F-11) of these notes.
F-17
A) Minimum Spacing Bolts along line of Force:
2.67 d  s  3d
(AISC J3.3)
where,
s = spacing of bolt along force line.
d = diameter of bolts-shank section
B) Minimum Edge Distance along Line of force:
Normally min. edge distance (rolled edge) 
1 14 d
Normally min. edge distance (sheared edge)  1 3 d
4
Where d = Nominal diameter of the bolt.
(Reference table J3-4 page 16.1 – 107)
F-18
C) Maximum spacing for Bolts:
for Painted members:
s  24t but  12 in.
for unpainted members:
(AISC page 16.1 – 108)
s  14t but  7 in.
t = thickness of the thinner connected plates.
D) Maximum Edge Distance:
max. edge distance = 12 t but ≤ 6“ ...
(AISC J 3.5 Page 16.1 – 106)
t = thickness of the thinner connected plates.
Reference:
AISC J 3.5.
F-19
Example F-1
Determine the required number of ¾ in. dia.
A490 bolts for the connection below.
Assuming DL = 14 kips, LL = 126 kips?
Solution:a) Calculate the factored load:
Tu = 1.2 DL + 1.6 LL
= 1.2  14 + 1.6  126 = 218 kips.
b) Design strength Rn per bolt:
Rn = 0.75 Fub (0.75 Ab)
= 0.75  113  0.4418 (AISC Table J 3.2)
= 37.3 kips.
c) Number of bolts required:
n
Tu
218

 5.8 (say 6 bolts)
 Rn 37.3
use 6 – ¾ in A 490 bolts.
F-20
Example F-2
Compute the tensile factored load capacity for the bearing type connection
shown below, Assume (A-572 Grade 50) steel plates, with (7/8”) diameter
A-325 bolts in standard holes:
A) Bolt threads are excluded from shear plane (X).
B) Bolt threads are included from shear plane (N).
Solution:-
Tension Member Bearing-Type Connection-LFRD
Check Plate Capacity:
Ag  6  0.625  3.75 sq. in.
A n  6  2 78  18  0.625  2.5 sq. in.
A e  U A n  2.5 sq. in. Fully Connected .
or
Tn = 0.90  50  3.75 = 169 kips.
Tn = 0.75  65  2.50 = 122 kips. (controls)
F-21
A) Strength of A-325-X bolts:
Shear:
Rn =  (0.5 Fub) m Ab
= 0.75  0.5  120  1  0.6013
= 27.1 kips per bolt.
Bearing:
Since Le ≥ 1.5d and spacing S > 3d, the bearing strength of a bolt is:
Rn =  (2.4 Fu dt)
= 0.75  2.4  65  0.875  0.625
= 64 kips per bolt.
Thus shear controls, the capacity of the connection:
Tn = 4  27.1 = 108 kips < 122 kips
So, Connection capacity = 108 kips.
F-22
B) Strength of A-325-N bolts:
Shear:
Rn =  (0.4 Fub) m Ab
= 0.75  0.4  120  1  0.6013
= 21.6 kips per bolt.
No need to check bearing again, Since
Shear will control:
Tn = 4  21.6 = 86.4 kips
 Total Connection capacity = 86.4 kips
F-23
Example F-3
Determine the number and spacing of (¾ in.) dia. (A-490-N) bolts
required to develop the full strength of the A-572 Grade 65 plates shown
below. Assume double rows of bolts with standard holes?
Solution:Consider middle plate:
An  6  23 4  18  0.375  1.59 sq. in.
Ae = An = 1.59 sq. in. (well connected).
max. An = 0.85 Ag = 0.85  6  0.375 = 1.91 sq. in.
or
(check T-16 of these notes)
Tn =  Fy Ag = 0.90  65  6  0.375 = 132 kips.
Tn =  Fu Ae = 0.75  80  1.59 = 96 kips. (controls)
F-24
Bolt shear capacity for ¾ in. A-490-N bolt:
Rn =  (0.4 Fub) m Ab
= 0.75  0.4  150  2  0.4418
= 39.8 kips per bolt.
(controls)
Bearing Strength on the 3/8 in. plate:
Rn =  (2.4 Fu dt)
= 0.75  2.4  80  0.75  0.375
= 40.5 kips per bolt.
Number of bolts = 96 =2.4 bolts.
39.8
Use 4-¾ in. (A-490-N) bolts. With:
End distance must be at least 1 1 in.
4
and spacing of bolts  3d  2 in.
1
4
F-25
Example F-4
Design the connection below to develop the full strength of double
angles (back-to-back), using two lines of (¾ in.) dia. (A-325-N) bolts
in the bearing type connection of 2Ls 6  4  3 (A-36) material ?
8
2Ls 6 x 4 x 3/8
+
Pu
(A-36 Steel)
+
Number of bolts
To be determined.
½“
Gusset Plate
( ½“ Thick) – (A-36)
F-26
Solution:For a single angle:
Ag = 3.61 sq. in.
A n  3.61  2   3

4
1
8
x 38   2.95 sq. in.
U = 0.80 (assuming 4 bolts or more in one row)
Ae = U An = 0.80  2.95 = 2.36 sq. in.
= 0.9  3.61  36 = 117 kips.
Angle Capacity (Pu)
= 0.75  2.36  58 = 102.7 kips. (controls)
(Pu) for two angles = 205.4 kips.
F-27
Shearing capacity each bolt = 0.75  4  34   0.4  120  2  31.8 kips. (cotrols)
2
Bearing capacity each bolt = 0.75  2.4  0.75  ½  58 = 39.2 kips
(Gusset plate more critical)
205.4
Required No. of bolts 
 6.45 bolts
31.8
Use 8 bolts (Four in line). Use spacing as shown below:
4”
Re-Check:
from L data:
x  0.933 in.
2¼
L  2  2  2  6.75 in.
1
4
U  1
1
4
x
L
x
1
4
 1  09..933
75
 0.86  0.8 used
6”
2½
1¼
1½ 2 ¼ 2 ¼ 2 ¼
Higher value of (U) is always on the safe side.
F-28
In slip-critical type connection (page F-3), plates are not allowed to
slip against each other, hence there is no practical shearing or
bearing contact between the bolts and the joining plates (or shapes).
This necessities the need for high quality control over the bolt
tension (proof load).
The main performance criteria of the slip-critical type connection is
that no slippage of the connected parts should occur at the service
load level at the structure.
However, to make the design of the slip-critical type connection
compatible with the design of the bearing type connection, AISC
stipulates the design of slip-critical connection at the factored load
level of the connected elements, as follows:F-29
 Rn =  Du hsc Tb Ns
(AISC J 3.8) (Page 16.1 – 109)
where:

= 1.0 for connections which resist slip at the service load level.
Rn = Nominal slip resistance of a (sc) bolt in kips.

= Friction coefficient for class A & B surfaces:
 = 0.35 for class (A) surface (clean mill scale).
 = 0.50 for class (B) surface (blast clean surface).
Du = 1.13 a statistical factor related to bolt pretension.
Hsc = Hole factor
hsc = 1.0 for standard holes.
hsc = 0.85 for Oversized holes.
hsc = 0.70 for long-slotted holes.
Ns = Number of slip planes (1 or 2) as (m) before.
Tb = Minimum fasteners tension in kips. (Table J3.1).
F-30
Example F -5
The connection shown below uses (¾ inch) A
325 SC bolts. No slippage is permitted. All
3
connected members are (A-36) steel. Determine

8
Gusset Plate
6x½
strength capacity of this connection (Tu)
1.5”
Solution :-
Even the connection is specified as slip-
3”
Tu
1.5”
critical connection, the direct shear and
bearing capacity must be checked (in the
event of accidental slippage).
1.5” 3” 1.5”
(a)
F-31
1) Shear capacity of bolts (A-325 N):
 R n   Fv Ab  0.75  48  0.4418
= 15.90 kips per bolt.
Tu for four bolts = 15.90  4 = 63.6 kips.
2) Bearing Capacity of (A325N) bolts.
Since the gusset plate is thinner (3/8” < ½”), It will control bearing :
Rn = (2.4 d t Fu)
 0.75 x 2.4 x 3 x 3 x 58
4
8
= 29.36 kips
 Bearing for four bolts : Rn = 4x29.36 = 117.45 kits
F-32
3) Slip-Critical Connection: Rn =   Du hsc Tb Ns
where,
 = 1.0
 = 0.35 assume class (A) surface.
Du = 1.13
hsc = 1.0 standard holes
Tb = 28 kips
(Table J 3.1)
(Table 16.1 – 103)
Ns 1
slip plane.
 Rn = 1  0.35  1.13  1  28  1 = 11.1 kips per bolt
For four bolts Tu = 4  11.1 = 44.4 kips.
F-33
4) Tension on the plate :-
A g  6x 21  3 sq, in
 R n=
0.9 Fy Ag
W n  6  2 34  18   4.25 in
0.75 Fu Ae
A n  4.25 x
1
2
 2.125 sq. in.
A e  UA n  1 x 2.125  2.125 sq. in.
 Rn  0.9 x 36 x 3  97.2 kips
or
 Rn  0.75 x 58 x 2.125  92.4 kips
(controls)
Maximum Capacity Tu = 44.4 kips.
F-34
Example F-6:
A 13-foot-long tension member and its
connection must be designed for a service
dead load of 8 kips and a service live load of
22 kips. No slip of the connection
is permitted. The connection will be to a 3/8-
t3
8
Gusset plate
Angle
(A 572 – Gr 50)
D = 8 kips
L = 22 kips
inch-thick gusset plate, as shown. Use a
single angle for the tension member. Use
A325 SC bolts and A572 Grade 50 steel for
A 325 SC bolts
A572 Grade 50 steel
both the tension member and the gusset plate.
F-35
Solution:
The factored load to be resisted is
Pu = 1.2D + 1.6L = 1.2(8) + 1.6(22) = 44.8 kips
Because the bolt size and layout will affect the net area of the tension member, we will
begin with selection of the bolts. The strategy will be to select a bolt size for trial,
determine the number required, and then try a different size if the number is too large
or too small. Bolt diameter typically range from ½ inch to 1½ inches in 1/8-inch
increments.
Try 5/8 inch bolts. The nominal bolt area is
Ag 
 (5 / 8) 2
4
 0.3068 in. 2
The shear strength is
 Rn = FvAb = 0.75(48)(0.3068)
= 11.04 kips/bolt
(assuming that the threads are in the shear plane)
Rn =   Du hsc Tb Ns
= 1  0.35  1.13  1  19  1
= 7.5 kips (controls)
No slip is permitted, so this connection is slip-critical. We will assume class (A)
surfaces, and for a 5/8-inch-diameter A325 SC bolt, the minimum tension is Tb = 19
kips (from AISC Table J3.1).
F-36
Number of bolts 
44.8
7.5
 5.9 bolts
Say 6 bolts ;This can be achieved in two rows, which requires long leg for
the angle, or select a larger size bolt to have less number of bolts, in order
to fit in one line.
Try (¾ inch) diameter bolts.
Ab 
 3
2
   0.4418 in.sq.
4 4
 R n  0.75x48x0.4418  15.90 kips
(assuming thread in shear plane after slip have occurred).
The Slip-Critical bolt capacity :
Rn = 1  0.35  1.13  1  28  1
= 11.07 kips. (controls)
F-37
44.8
3 
No. of  in.  bolts 
 4.0 bolts
11.07
4 
Select four bolts of (¾ in.) diameter (A-325-SC)
Minimum spacing  3 d = 2¼ in. say (2½ in.)
Minimum Edge distance = 1.5 in (assume sheared edge).
Check Tension on Member :
A  
g 1
Pu
44.8

 0.996 in.2
0.9Fy 0.9(50)
1½” 2½” 2½” 2½” 1½”
and the required effective net area is
Ae 
Pu
44.8

 0.9190 in.2
0.75Fu 0.75(65)
An 
Ae
U
F-38
For the layout shown in Figure above, with four bolts in the line of force,
the value of U from the AISC table – D3.1 (page 16.1-29) is 0.80. (Once
a member has been selected, U can be computed with AISC.
Therefore
An 
A 
g 2
0.9190
 1.15 in2 .
0.80
 1.15   34  161 t
Assume t = ¼ in thick, (Ag)2 = 1.35 in.2
(controls)
rmin. req.
L
13x12


 0.52 in.
300
300
F-39
Try
L 3 21  2 21  41 angle
Ag = 1.44 in2 > 1.35 in2
x
2½”
(OK)
3½”
rmin = rz = 0.544 in. > 0.52 in. (OK)
Use an L 3 21  2 21  41 with
Long leg connected. Use
4
3
4
A325-SC bolts as shown.
x
0.607
1
 0.92
L
7.5
0.919
1
An 
 1.0 in 2 Ag 2  1.0  3  1   1.2 in 2
4
16 4
0.92
U  1


Our selected Ag = 1.44 in2 is greater then (Ag)2 Then OK.
F-40