chapter 4 (final f) 18 march 07
TRANSCRIPT
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F-1
Welding (chap, 5)
Structural
Connections
Fasteners
Bolts
Non-Structural
(A-307)
Structural
A325
A449
A490
Rivets (obsolete)
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F-2
Bolt
Head
Shank
ThreadedSection
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F-3
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.
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F-4
When slip resistance is required, the pre-
tensioning of bolts should as high as theirProof 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 turn of nut from
the snug-fit position. Otherwise a calibrated
wrench must be used to achieve the proof
load for bolts.
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F-5
(70 % Fu)
(70 % Fu)
(80 % Fu)
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F-6
The most typical types of bolted connections are shown below:-
In shear connection, threads could be:-
A) In the shearing plane (N), or
B) Outside the shearing plane (X).
P
P
Thread outside (X)
Lap Joint
(single shear)
Butt Joint
(double shear)
(a) Shear Connections
(b) Eccentric shear
Connection.
W section
Structural tee
(c) Tension Connections
(d) Combined shear and tension connections
P
P
Thread inside (N)
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F-7
There are several limit-
states or failure modes that
may control the strength
of a bolted connection.
As can be seen from the
above figure, three failuremodes (b, d & g) are relevant
to the connected plates and
not relevant to the fasteners.
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F-8
The nominal tensile strength of a bolt (Rn) can be found as:
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:
297430
7850
ndb
..
Where, 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.
n
b
un AFR
bb
un AFR 750.
Thread cross section area*
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F-9
The nominal-shear strength for one fastener (Rn) will be
Rn = m AbuWhere,
= (1) for single shear
m = number of shear planes
= (2) for double shearAb = 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)
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F-10
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:
pu
den LtR 22
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.
2
1241
d
LdtFR eun .
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F-11
Thus:
Since LRFD recommends spacing of bolts inthe direction of force to be (2.67 d) [see
LRFD J 3.3]. Consider (Le 2.67 d) above:
The lower resistance value guarantees that no tear-outdue 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)
* Rn = 2.4 Fu dt
or
* Rn = 3.0 Fu dt
(When deformation of the hole at service
load is a design consideration). (Eq J3-6a)
(When deformation of the hole at service
load is not a design consideration). (Eq J3-6b)
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F-12
(1) Threads are included in shear plane (N)
(2) Threads are excluded from shear plane (X)
In general:
n
Rn
iQ
i
Where, Rn = Nominal resistance (strength)
n = Strength reduction factor (resistance factor)
iQi = Factored load on structure (ASCE 7)
for a single bolt :
Rn PuWhere,
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- Tension capacity
2- Shear capacity
3- Bearing capacity
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F-13
)75.0(75.075.0 bb
ub
b
un AFAFR
b
b
un AFR ).(. 750750
Reference is made to (LRFD) J 3.6), and page (F-8):
or
Where
= 0.75 as stipulated in AISC J3.6 (page 16.1-108)
b
uF= 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).
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F-14
A No th reads in shear plane (X):-
Rn = 0.75 (Rn) (AISC J3-6)Rn = 0.62m AbFu
b (page F-9)
For long Connections (upto 50 inch long), a 20% reduction is needed
(strength of long connection < 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 Fu
b 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)
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F-15
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 Fu
b) m Ab
(see Table J-3.2)
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F-16
Nominal bearing strength (Rn) was developed earlier (page F-11).
However LRFD-J3.10 further reduces (Rn) in order to prevent boltelongations exceeding 0.25 in.
Hence several categories are presented here below:
1: Deformation Lim it 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.
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F-17
2: Deformation Lim it 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: Streng th Limit State (ho le elongat ion 0.25 in.):
Rn = (3.0 dt Fu). (Equ. J3 6b)See page (F-11) of these notes.
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F-18
A) Minimum Spacing Bol ts along l ine 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)
Normally min. edge distance (sheared edge)
Where d = Nominal diameter of the bolt.
(Reference table J3-4 page 16.1 107)
d4
11
d4
31
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F-19
C) Maximum spacing for Bol ts:
for Painted members:
s 24t but 12 in.for unpainted members:
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.
(AISC page 16.1 108)
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F-20
Example F-1
Determine the required number of 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:
(say 6 bolts)
use 6 in A 490 bolts.
5.837.3
218
R
Tn
n
u
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F-21
Example F-2Compute 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:-Check Plate Capacity:
.ConnectedFullyin.sq..AA
in.sq...A
in.sq...
ne
n
52
52625026
75362506
8
1
8
7
U
Ag
Tension Member Bearing-Type Connection-LFRD
Tn = 0.90 50 3.75 = 169 kips.or
Tn = 0.75 65 2.50 = 122 kips. (controls)
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F-22
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.
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F-23
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
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F-24
Example F-3
Determine the number and spacing of ( 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:
in.sq.1.590.37526A 8143n
Ae
= An
= 1.59 sq. in. (well connected).
max. An = 0.85 Ag = 0.85 6 0.375 = 1.91 sq. in.
Tn = Fy Ag = 0.90 65 6 0.375 = 132 kips.or
Tn = FuAe = 0.75 80 1.59 = 96 kips. (controls)
(check T-16 of these notes)
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F-25
Bolt shear capacity for 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 = =2.4 bolts.
Use 4- in. (A-490-N) bolts. With:End distance must be at least
and spacing of bolts
39.8
96
.23d41 in
.4
1in1
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F-26
Example F-4Design the connection below to develop the full strength of double
angles (back-to-back), using two lines of ( in.) dia. (A-325-N) bolts
in the bearing type connection of 2Ls 6 4 (A-36) material ?8
3
+
+
Pu
(A-36 Steel)
Number of bolts
To be determined.
Gusset Plate
( Thick) (A-36)
2Ls 6 x 4 x 3/8
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F-27
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.
Solution:-
For a single angle:
Ag = 3.61 sq. in.
in.sq.2.95x23.61A83
81
43
n
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F-28
Shearing capacity each bolt = (cotrols)kips.... 831212040750 243
4
Bearing capacity each bolt = 0.75 2.4 0.75 58 = 39.2 kips(Gusset plate more critical)
Required No. of bolts
Use 8 bolts (Four in line). Use spacing as shown below:
bolts6.4531.8
205.4
2
2
2
1
1 2 2
6
4x
Re-Check:
from L data:
used..
..
..
.
.
80860
11
756222
9330
759
9330
4
1
4
1
4
1
L
xU
inL
inx
Higher value of (U) is always on the safe side.
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F-29
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:-
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F-30
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).
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F-31
Example F -5
The connection shown below uses ( inch) A
325 SC bolts. No slippage is permitted. All
connected members are (A-36) steel. Determine
strength capacity of this connection (Tu)
Solution :-
Even the connection is specified as slip-
critical connection, the direct shear and
bearing capacity must be checked (in the
event of accidental slippage).
Tu
1.5
1.5
3
1.51.5
3
(a)
PlateGusset8
3
6 x
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F-32
Tu for four bolts = 15.90 4 = 63.6 kips.
2) Bearing Capacity of (A325N) bolts.
Since the gusset plate is thinner (3/8 < ), It will control bearing :
Rn = (2.4 d t Fu)
58x
8
3x
4
3x2.4x0.75
= 29.36 kips
Bearing for four bolts : Rn = 4x29.36 = 117.45 kits
1) Shear capacity of bolts (A-325 N):
4418.04875.0FR vn bA
= 15.90 kips per bolt.
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F-33
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.
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F-34
4) Tension on the plate :-
in.sq.2.1252.125x1UAA
in.sq.2.125x4.25A
in4.2526Winsq,36xA
ne
21
n
n
21
g
8
1
4
3
Maximum Capacity Tu = 44.4 kips.
0.9 Fy Ag
Rn=
0.75 Fu Ae
(controls)
kips92.42.125x58x0.75R
or
kips97.23x36x0.9R
n
n
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F-35
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-
inch-thick gusset plate, as shown. Use a
single angle for the tension member. Use
A325 SC bolts and A572 Grade 50 steel forboth the tension member and the gusset plate.
D = 8 kips
L = 22 kips
83t
A 325 SC bolts
A572 Grade 50 steel
Gusset plate
Angle
(A 572 Gr 50)
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F-36
The factored load to be resisted is
Pu = 1.2D + 1.6L = 1.2(8) + 1.6(22) = 44.8 kipsBecause 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 inch to 1 inches in 1/8-inch
increments.
Try 5/8 inch bolts. The nominal bolt area is
22
in.3068.04
)8/5(
gA
The shear strength is
Rn = FvAb = 0.75(48)(0.3068)
= 11.04 kips/bolt
(assuming that the threads are in the shear plane)
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).
Rn = Du hsc Tb Ns
= 1 0.35 1.13 1 19 1
= 7.5 kips (controls)
Solution:
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F-37
Say 6 bolts ;This can be achieved in two rows, which requires long leg forthe angle, or select a larger size bolt to have less number of bolts, in order
to fit in one line.
Try ( inch) diameter bolts.
kips15.90x48x0.441875.0R
in.sq.4418.043
4
n
2
bA
(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)
bolts9.5boltsofNumber5.7
8.44
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F-38
bolts4.011.07
44.8bolts.
4
3ofNo.
in
2y
u
1gin.0.996
0.9(50)
44.8
0.9F
PA
Select four bolts of ( in.) diameter (A-325-SC)
Minimum spacing 3 d = 2 in. say (2 in.)
Minimum Edge distance = 1.5 in (assume sheared edge).
Check Tension on Member :
and the required effective net area is
2
u
ue in.0.9190
0.75(65)
44.8
0.75F
PA
U
AA en
112 22
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F-39
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
.in1.15
0.80
0.9190A
2
n
t1.15A161
43
2g
Assume t = in thick, (Ag)2 = 1.35 in.2
in.0.52300
13x12
300
Lr
req.min.
(controls)
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F-40
angle23LTry41
21
21
Ag = 1.44 in2 > 1.35 in2 (OK)
rmin = rz = 0.544 in. > 0.52 in. (OK)
Use an with
Long leg connected. Use A325-SC bolts as shown.
4
1
2
1
2
1
23L
434
3
2
x
2
2
2 2141
161
430101
9209190
92057
607011
inAinA
L
xU
gn .....
..
.
Our selected Ag = 1.44 in2 is greater then (Ag)2 Then OK.