68402: Structural Design of
Buildings II
Design of Connections
Monther Dwaikat
Assistant Professor
Department of Building Engineering
An-Najah National University
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Slide # 1
Bolted Connections

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
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



Types of Connections
Simple Bolted Shear Connections
Bearing and Slip Critical Connections
Eccentric Bolted Connections
Moment Resisting Bolted Connections
Simple Welded Connections
Eccentric Welded Connections
Moment Resisting Welded Connections
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Slide # 2
Types of Connections
Simple Connections
Bolted Connections
Common Bolts
Eccentric Connections
Welded Connections
High Strength Bolts
Slip Critical
Filet Weld
Groove Weld
Bearing Type
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Slide # 3
Types of Connections
Simple Connections
Eccentric Connections
Bolted Connections
Elastic
Analysis
Ultimate
Analysis
Welded Connections
Moment
Resisting
Elastic
Analysis
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Ultimate
Analysis
Moment
Resisting
Slide # 4
Simple Bolted Connections

There are different types of bolted connections.
They can be categorized based on the type of
loading.
•
•
•
Tension member connection and splice. It subjects the
bolts to forces that tend to shear the shank.
Beam end simple connection. It subjects the bolts to forces
that tend to shear the shank.
Hanger connection. The hanger connection puts the bolts
in tension
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Slide # 5
Simple Bolted Connections
P
P
Tension member
Connection/ splice
P
P
Beam end
Simple shear connection
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Slide # 6
Simple Bolted Connections
P
P
P
Hanger connection
(Tension)
Moment resisting
connection
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Slide # 7
Simple Bolted Connections
The bolts are subjected to shear or tension loading.

•
•
•
In most bolted connection, the bolts are subjected to shear.
Bolts can fail in shear or in tension.
You can calculate the shear strength or the tensile strength of a bolt

Simple connection: If the line of action of the force acting on
the connection passes through the center of gravity of the
connection, then each bolt can be assumed to resist an equal
share of the load.

The strength of the simple connection will be equal to the
sum of the strengths of the individual bolts in the connection.
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Slide # 8
Bolt Types & Materials
A307 - Unfinished (Ordinary or Common) bolts
low carbon steel A36, Fu = 413 MPa,
for light structures under static load
A325 - High strength bolts, heat-treated medium
carbon steel, Fu = 827 MPa,
for structural joints
A490 - High strength bolts, Quenched and
Tempered Alloy steel, Fu = 1033 MPa
for structural joints
A449 - High strength bolts with diameter > 1 ½”,
anchor bolts, lifting hooks, tie-downs
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Slide # 9
Common Bolts


ASTM A307 bolts
Common bolts are no longer common for current structural
design but are still available
Pu   Rn
Rn 
  0.75
f v Abolt
f v  165 MPa
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Slide # 10
High Strength Bolts

High strength bolts (HSB) are
available as ASTM A 325 and ASTM
A490
Bolt
Courtesy of Kao Wang Screw Co., Ltd.
Washer
Slip Critical
Nut

Advantages of HSB over A307 bolts
Bearing Type

Fewer bolts will be used compared to 307  cheaper connection!

Smaller workman force required compared to 307

Higher fatigue strength

Ease of bolt removal  changing connection
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Slide # 11
High Strength Bolts

Snug tight
•
•

Pre-tensioned
•
•
•

All plies of the connection are in firm contact to
each other: No pretension is used.
Easer to install and to inspect
Courtesy of www.halfpricesurplus.com
Bolts are first brought to snug tight status
Bolts are then tensioned to 70% of their tensile
stresses
Bolts are tensioned using direct tension indicator, calibrated wrench or other
methods (see AISC)
Slip critical
•
•
•
Bolts are pre-tensioned but surfaces shall be treated to develop specific friction.
The main difference is in design, not installation. Load must be limited not to
exceed friction capacity of the connection (Strength Vs. Serviceability!)
Necessary when no slip is needed to prevent failure due to fatigue in bridges.
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Slide # 12
HSB – Bearing Type Connections

The shear strength of bolts shall be determined as follows
  0.75
Pu   Rn
Rn 
f v Abolt
AISC Table J3.2
The table bellow shows the values of fv (MPa) for different types of bolts
•
Type
Type N Thread
Type X Thread
A325
330
413
A490
413
517
If the level of threads is not known, it is conservative to
assume that the threads are type N.
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Slide # 13
Bolted Shear Connections

We want to design the bolted shear connections so that
the factored design strength (Rn) is greater than or equal
to the factored load.  Rn  Pu

So, we need to examine the various possible failure
modes and calculate the corresponding design strengths.

Possible failure modes are:
•
•
•
•
•
Shear failure of the bolts
Failure of member being connected due to fracture or yielding or ….
Edge tearing or fracture of the connected plate
Tearing or fracture of the connected plate between two bolt holes
Excessive bearing deformation at the bolt hole
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Slide # 14
Failure Modes of Bolted
Connections

Bolt Shearing

Tension Fracture

Plate Bearing

Block Shear
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Slide # 15
Actions on Bolt
Shear, bearing, bending
P
P
P
Lap Joint
P
Bearing and single plane Shear
P
P
Bending
Butt Joint
P/2
P/2
P
Bearing and double plane Shear
P/2
P
P/2
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Slide # 16
Bolted Shear Connections

Possible failure modes

Failure of bolts: single or double shear
PSingle Shear 
f v Abolt
Single shear
P
P
Double shear
PDouble Shear  2 f v Abolt

P/2
P
P/2
Failure of connected elements:

Shear, tension or bending failure of the connected elements (e.g. block shear)

Bearing failure at bolt location
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Slide # 17
Bolted Shear Connections

Shear failure of bolts
•
•
•
•
•
•
Average shearing stress in the bolt = fv = P/A = P/(db2/4)
P is the load acting on an individual bolt
A is the area of the bolt and db is its diameter
Strength of the bolt = P = fv x (db2/4)
stress = 0.6Fy
where fv = shear yield
Bolts can be in single shear or double shear as shown above.
When the bolt is in double shear, two cross-sections are effective
in resisting the load. The bolt in double shear will have the twice
the shear strength of a bolt in single shear.
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Slide # 18
Bolted Shear Connections
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Slide # 19
Bolted Shear Connections

Failure of connected member
•
•

We have covered this in detail in this course on tension members
Member can fail due to tension fracture or yielding.
Bearing failure of connected/connecting part due to
bearing from bolt holes
•
•
•
Hole is slightly larger than the fastener and the fastener is loosely
placed in hole
Contact between the fastener and the connected part over
approximately half the circumference of the fastener
As such the stress will be highest at the radial contact point (A).
However, the average stress can be calculated as the applied force
divided by the projected area of contact
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Slide # 20
Bolted Shear Connections
•
•
•
•
Average bearing stress fp = P/(db t), where P is the force applied
to the fastener.
The bearing stress state can be complicated by the presence of
nearby bolt or edge. The bolt spacing and edge distance will have
an effect on the bearing strength.
Bearing stress effects are independent of the bolt type because
the bearing stress acts on the connected plate not the bolt.
A possible failure mode resulting from excessive bearing close to
the edge of the connected element is shear tear-out as shown
below. This type of shear tear-out can also occur between two
holes in the direction of the bearing load.
Rn = 2 x 0.6 Fu Lc t = 1.2 Fu Lc t
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Slide # 21
Bolted Shear Connections
•
The bearing strength is independent of the bolt material as it is failure in
the connected metal
PBearing 
•
f p d bolt t
The other possible common failure is shear end failure known as “shear
tear-out” at the connection end
Pu   Rn
  0.75
Rn  1.2 Lc t Fu  2.4 d t Fu
Shear limitation
Lc
Lc
Bearing limitation
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Slide # 22
Bolted Shear Connections
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Slide # 23
Bolted Shear Connections
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Slide # 24
Spacing and Edge-distance
requirements

The AISC code gives guidance for edge distance and spacing to
avoid tear out shear
h
AISC Table J3.4
Lc  Le 
2
h is the hole diameter h  dbolt  1.6 mm
Le
Le
S
NOTE: The actual hole diameter is 1.6 mm bigger than the bolt,
we use another 1.6 mm for tolerance when we calculate net area. Here use 1.6 mm only not 3.2

Bolt spacing is a function of the bolt diameter
S  3 d bolt

Common we assume

The AISC minimum spacing is
S  2 23 d bolt
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Slide # 25
Bolt Spacings & Edge Distances

Bolt Spacings
- Painted members or members not subject to corrosion:
2 2/3d ≤ Bolt Spacings ≤ 24t or 305 mm
(LRFD J3.3)
(LRFD J3.5)
- Unpainted members subject to corrosion:
3d ≤ Bolt Spacings ≤ 14t or 178 mm

Edge Distance
Values in Table J3.4M ≤ Edge Distance ≤ 12t or 152 mm
(LRFD J3.4)
(LRFD J3.5)
d - bolt diameter
t - thickness of thinner plate
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Slide # 26
Bolted Shear Connections
•
To prevent excessive deformation of the hole, an upper limit is
placed on the bearing load. This upper limit is proportional to the
fracture stress times the projected bearing area
Rn = C x Fu x bearing area = C Fu db t
yp
•
•
•
•
•
If deformation is not a concern then C = 3, If deformation is a
concern then C = 2.4
C = 2.4 corresponds to a deformation of 6.3 mm.
Finally, the equation for the bearing strength of a single bolts is
Rn
where,  = 0.75 and Rn = 1.2 Lc t Fu < 2.4 db t Fu
Lc is the clear distance in the load direction, from the edge of the
bolt hole to the edge of the adjacent hole or to the edge of the
material
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Slide # 27
Bolted Shear Connections
•
This relationship can be simplified as follows:
The upper limit will become effective when 1.2 Lc t Fu > 2.4 db t
Fu
i.e., the upper limit will become effective when Lc > 2 db
If Lc < 2 db,
Rn = 1.2 Lc t Fu
If Lc > 2 db,
Rn = 2.4 db t Fu
Fu - specified tensile strength of the connected material
Lc - clear distance, in the direction of the force, between the edge
of the hole and the edge of the adjacent hole or edge of the
material.
t - thickness of connected material
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Slide # 28
Important Notes
Lc – Clear distance
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Slide # 29
Design Provisions for Bolted Shear
Connections

T
In a simple connection, all bolts share the load equally.
T/n
T/n
T/n
T/n
T/n
T/n
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T
Slide # 30
Design Provisions for Bolted Shear
Connections

In a bolted shear connection, the bolts are subjected to
shear and the connecting/connected plates are subjected
to bearing stresses.
T
T
Bearing stresses in plate
T
Bolt in shear
T
Bearing stresses in plate
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Slide # 31
Design Provisions for Bolted Shear
Connections

The shear strength of all bolts = shear strength of one bolt
x number of bolts

The bearing strength of the connecting / connected plates
can be calculated using equations given by AISC
specifications.

The tension strength of the connecting / connected plates
can be calculated as discussed in tension members.
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Slide # 32
AISC Design Provisions

Chapter J of the AISC Specifications focuses on
connections.

Section J3 focuses on bolts and threaded parts

AISC Specification J3.3 indicates that the minimum
distance (s) between the centers of bolt holes is 2.67. A
distance of 3db is preferred.

AISC Specification J3.4 indicates that the minimum edge
distance (Le) from the center of the bolt to the edge of the
connected part is given in Table J3.4. Table J3.4 specifies
minimum edge distances for sheared edges, edges of
rolled shapes, and gas cut edges.
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Slide # 33
AISC Design Provisions

AISC Specification indicates that the maximum edge
distance for bolt holes is 12 times the thickness of the
connected part (but not more than 152 mm). The maximum
spacing for bolt holes is 24 times the thickness of the
thinner part (but not more than 305 mm).

Specification J3.6 indicates that the design tension or
shear strength of bolts is FnAb
•
•
•
•
 = 0.75
Table J3.2, gives the values of Fn
Ab is the unthreaded area of bolt.
In Table J3.2, there are different types of bolts A325 and A490.
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Slide # 34
AISC Design Provisions
•
The shear strength of the bolts depends on whether threads are
included or excluded from the shear planes. If threads are included
in the shear planes then the strength is lower.

We will always assume that threads are included in the
shear plane, therefore less strength to be conservative.

We will look at specifications J3.7 – J3.9 later.
•
•
•
AISC Specification J3.10 indicates the bearing strength of plates at
bolt holes.
The design bearing strength at bolt holes is Rn
Rn = 1.2 Lc t Fu ≤ 2.4 db t Fu
design consideration
- deformation at the bolt holes is a
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Slide # 35
Common bolt terminologies






A325-SC – slip-critical A325 bolts
A325-N – snug-tight or bearing A325 bolts
with
thread included in the shear
planes.
A325-X - snug-tight or bearing A325 bolts
with
thread excluded in the shear
planes.
Gage – center-to-center distance of bolts in p p
direction perpendicular to p
member’s axis
Pitch – ...parallel to member’s axis
Edge Distance – Distance from
center of bolt to adjacent
edge of a member
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g
Edge
distance
p
Slide # 36
Ex. 6.1 - Design Strength

Calculate and check the design strength of the simple
connection shown below. Is the connection adequate for
carrying the factored load of 300 kN.
10 mm
3/8
in.
120x15
5 x ½ mm
A36
301.25
mm
A36
60 2.50
mm
300
65
63kkN
k
301.25
mm
A325-N
¾20
in.mm
bolts
bolts
30 1.25
mm 602.50
mm 1.25
30 mm
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Slide # 37
Ex. 6.1 - Design Strength

Step I. Shear strength of bolts
•
The design shear strength of one bolt in shear = Fn Ab = 0.75 x
330 x  x 202/4000 = 77.8 kN
•
•
 Fn Ab = 77.8 kN per bolt
(See Table J3.2)
Shear strength of connection = 4 x 77.8 = 311.2 kN
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Slide # 38
Ex. 6.1 - Design Strength

Step II. Minimum edge distance and spacing requirements
•
See Table J3.4M, minimum edge distance = 26 mm for rolled edges
of plates
•
•
The given edge distances (30 mm) > 26 mm. Therefore, minimum
edge distance requirements are satisfied.
Minimum spacing = 2.67 db = 2.67 x 20 = 53.4 mm.
(AISC Specifications J3.3)
•
•
Preferred spacing = 3.0 db = 3.0 x 20 = 60 mm.
The given spacing (60 mm) = 60 mm. Therefore, spacing requirements
are satisfied.
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Slide # 39
Ex. 6.1 - Design Strength

Step III. Bearing strength at bolt holes.
•
Bearing strength at bolt holes in connected part (120x15 mm plate)
•
•
•
•
•
•
•
At edges, Lc = 30 – hole diameter/2 = 30 – (20 + 1.6)/2 = 19.2
Rn = 0.75 x (1.2 Lc t Fu) = 0.75 x (1.2 x19.2 x15x400)/1000 = 103.7 kN
But, Rn ≤ 0.75 (2.4 db t Fu) = 0.75 x (2.4 x 20x15x400)/1000 = 216 kN
Therefore, Rn = 103.7 kN at edge holes.
At other holes, s = 60 mm, Lc = 60 – (20 + 1.6) = 38.4 mm.
Rn = 0.75 x (1.2 Lc t Fu) = 0.75x(1.2 x 38.4 x15 x400)/1000 = 207.4 kN
But, Rn ≤ 0.75 (2.4 db t Fu) = 216 kN. Therefore Rn = 207.4 kN
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Slide # 40
Ex. 6.1 - Design Strength
•
•
•
Therefore, Rn = 216 kN at other holes
Therefore, bearing strength at holes = 2 x 103.7 + 2 x 207.4 = 622.2 kN
Bearing strength at bolt holes in gusset plate (10 mm plate)
•
•
At edges, Lc = 30 – hole diameter/2 = 30 – (20 + 1.6)/2 = 19.2 mm.
•
But, Rn ≤ 0.75 (2.4 db t Fu) = 0.75 x (2.4 x 20 x 10 x 400)/1000 = 144
kN.
•
Therefore, Rn = 69.1 kN at edge holes.
Rn = 0.75 x (1.2 Lc t Fu) = 0.75 x (1.2 x 19.2 x 10 x 400)/1000 = 69.1
kN
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Slide # 41
Ex. 6.1 - Design Strength
•
•
•
At other holes, s = 60 mm, Lc = 60 – (20 +1.6) = 38.4 mm.
•
•
•
But, Rn ≤ 0.75 (2.4 db t Fu) = 144 kN
Rn = 0.75 x (1.2 Lc t Fu) = 0.75 x (1.2 x 38.4 x 10x 400)/1000 = 138.2
kN
Therefore, Rn = 138.2 kN at other holes
Therefore, bearing strength at holes = 2 x 69.1 + 2 x 138.2 = 414.6 kN
Bearing strength of the connection is the smaller of the bearing
strengths = 414.6 kN
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Slide # 42
Ex. 6.1 - Design Strength
Connection Strength
Shear strength = 311.2
Bearing strength (plate) = 622.2 kN
Bearing strength (gusset) = 414.6 kN
Connection strength (Rn) > applied factored loads (gQ).
311.2 > 300
Therefore ok.
• Only connections is designed here
Need to design tension member and gusset plate
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Slide # 43
Eccentrically-Loaded Bolted Connections
P
P
P
P
Pe
CG
CG
Pe
e
e
Eccentricity in the plane of
the faying surface
Eccentricity normal to the plane
of the faying surface
Direct Shear + Additional Shear due to
moment Pe
Direct Shear + Tension and Compression
(above and below neutral axis)
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Slide # 44
Forces on Eccentrically-Loaded Bolts
Eccentricity in the plane of the faying surface
LRFD Spec. presents values for computing design
strengths of individual bolt only. To compute
forces on group of bolts that are eccentrically
loaded, there are two common methods:
-
Elastic Method:
Conservative. Connected parts
assumed rigid. Slip resistance between
connected parts neglected.
Ultimate Strength Method (or Instantaneous Center of
Gravity Method):
Most realistic but tedious to apply
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Slide # 45
Forces on Eccentrically-Loaded Bolts
with Eccentricity on the Faying Surface

Elastic Method
e
P
P
Pe
r3
d1
d3
P/3
P/3 CG
CG
r1
d2
r2
P/3
Assume plates are perfectly rigid and bolts perfectly elastic 
rotational displacement at each bolt is proportional to its
distance from the CG  stress is greatest at bolt farthest from
CG
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Slide # 46
Forces on Eccentrically-Loaded Bolts
with Eccentricity on the Faying Surface
MCG = Pe = r1d1 + r2d2 + r3d3
Since the force on each bolt is proportional to its distance
from the CG:
r3
r1 r2


d1 d 2 d3
r1d3
r1d1
r1d 2
r1 
; r2 
; r3 
d1
d1
d1

Substitute into eqn. for MCG:
2
M CG
M CG 
r1 
2
2

r1d 3
r1d1
r1d 2
r1
2
2
2




d1  d 2  d 3
d1
d1
d1
d1

r1
d2
d1
M CG d1
M CG d 2
M CG d 3
;
r

;
r

2
3
2
2
d
d


d2
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Slide # 47
Forces on Eccentrically-Loaded Bolts
with Eccentricity on the Faying Surface
M d y
ry
pmx  r1 sin   1 1  CG 1 1
d1
d1
d2

H1
CG
d1
y1 r1

V1
x1
pmx 
M CG y1
pmy 
M CG x1
Total Forces in Bolt i:

d2



d2
Pex1

d2
Peyi
-Horizontal Component =

-Vertical Component
P

n
=

d
2
Pe y1
d2
Pexi

d2
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Slide # 48
Ex. 6.3 – Eccentric Connections –
Elastic Method
Determine the force in the most stressed bolt of the group
using elastic method
e
125 mm
100
mm
100
mm
100
mm
CG
100
mm
P=140 kN
Eccentricity wrt CG:
e = 125 + 50 = 175 mm
Direct Shear in each bolt:
P/n = 140/8 = 17.5 kN
Note that the upper right-hand and
the lower right-hand bolts are the
most stressed (farthest from CG and
consider direction of forces)
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Slide # 49
Ex. 6.3 – Eccentric Connections –
Elastic Method
Additional Shear in the upper and lower right-hand bolts
due to moment M = Pe = 140x175 = 24500 kN.mm:
d
2
  x 2   y 2 (8)(50) 2  (4)(502  1502 )  120000
pmx 
My
(22500)(150)

 30.6 kN
2
d
120000

pmy 
Mx
(24500)(50)

 10.2 kN
2
d
120000

The forces acting on the upper right-hand bolt are as
follows:
The resultant force on this bolt is:
30.6 kN
10.2 kN
R  (10.2  17.5) 2  (30.6) 2  41.3 kN
17.5 kN
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Slide # 50
Forces on Eccentrically-Loaded Bolts

Eccentricity Normal to Plane of Faying Surface
(a) Neutral Axis at CG
Shear force per bolt due to concentric force Pu
2rut
ruv = Pu/n
n: # of bolts
Bolts above NA are in tension. Bolts below NA
are in compression. Tension force per bolt:
rut = (Pue)/n’dm
n’: # of bolts above NA
dm: moment arm between resultant tensile and
compressive forces
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Slide # 51
Forces on Eccentrically-Loaded Bolts

Eccentricity Normal to Plane of Faying Surface
(b) Neutral Axis Not at CG
Bolts above NA resist tension
tf
y
Depth
d=Depth/6
CG
(tension
group)
X
X
beff
Bearing stress below NA resist compression
2rut Shear per bolt due to concentric
NA
force Pu:
ruv= Pu/n
Select first trial location of NA as 1/6
of the total bracket depth.
Effective width of the compression
block:
beff = 8tf ≤ bf (for W-shapes, Sshapes, welded plates and
angles)
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Slide # 52
Forces on Eccentrically-Loaded Bolts
Check location of NA by equating the moment of the bolt area
above the NA with the moment of the compression block area
below the NA:
Ab x y = beff x d x d/2
Ab = sum of areas of bolts above the NA
y = distance from X-X to the CG of bolts above NA
d = depth of compression block (adjust until satisfy)
Once the NA has been located, the tensile force per bolt:
rut = (PuecAb)/Ix
c = distance from NA to most remote bolt in group
Ix = combined moment of inertia of bolt group and compression block
about NA
68402/61420
Slide # 53
Bolts Subjected to Shear and Tension
• Nominal Tension Stress Ft of a bolt subjected to combined
factored shear stress (fv =Vu/NbAb) and factored tension stress (ft
= Tu/NbAb) can be computed as functions of fv as:
•
•
•
•
•
Fnt
  1.3Fnt 
Fnt
f v  Fnt
 Fnv
 = 0.75
F’nt = nominal tensile strength modified to include the effect of shear
Fnt = nominal tensile strength from Table J3.2 in (AISC Spec.)
Fnv = nominal shear strength from Table J3.2 in (AISC Spec.)
fv = the required shear stress
Bolt Type
Fnt (MPa)
A325
A490
620
780
68402/61420
Slide # 54
Ex. 6.5 – Combined Tension & shear
Is the bearing-type connection below satisfactory for the
combined tension and shear loads shown?
Shear stress per bolt: fv = Vu/NbAb=537000/(8x380)= 176.6 MPa
Fnv=(0.75)(413)=310 MPa> fv = 176.6 MPa (OK)
1200 kN 537 kN
1073 kN
Tension stress per bolt:
ft = Tu/NbAb=1073000/(8x380)= 353 MPa
1
2
Eight 22 mm
A325X bolts
Nominal Tension Strength Ft (Table J3.5)
Ft = 0.75[(1.3x620 – (620/310)x176.6) ≤ 620]
= 496 MPa ≤ 620]
= 496 MPa > ft = 353 MPa (OK)
68402/61420
Slide # 55
Simple Welded Connections
Structural welding is a process by which the parts that
are to be connected are heated and fused, with
supplementary molten metal at the joint.
A relatively small depth of material will become molten,
and upon cooling, the structural steel and weld metal will
act as one continuous part where they are joined.


P
P
P
Fillet weld
Fillet weld
P
68402/61420
Slide # 56
Introductory Concepts
Welding Process – Fillet Weld
68402/61420
Slide # 57
Introductory Concepts

The additional metal is deposited from a special
electrode, which is part of the electric circuit that includes
the connected part.
•
•
•
In the shielded metal arc welding (SMAW) process, current arcs
across a gap between the electrode and the base metal, heating
the connected parts and depositing part of the electrode into the
molten base metal.
A special coating on the electrode vaporizes and forms a
protective gaseous shield, preventing the molten weld metal
from oxidizing before it solidifies.
The electrode is moved across the joint, and a weld bead is
deposited, its size depending on the rate of travel of the
electrode.
68402/61420
Slide # 58
Introductory Concepts
•
•
As the weld cools, impurities rise to the surface, forming a
coating called slag that must be removed before the member
is painted or another pass is made with the electrode.
Shielded metal arc welding is usually done manually and is
the process universally used for field welds.

For shop welding, an automatic or semi automatic
process is usually used. Foremost among these is the
submerged arc welding (SAW),

In this process, the end of the electrode and the arc are
submerged in a granular flux that melts and forms a
gaseous shield. There is more penetration into the base
metal than with shielded metal arc welding, and higher
strength results.
68402/61420
Slide # 59
Introductory Concepts

Other commonly used processes for shop welding are
gas shielded metal arc, flux cored arc, and electro-slag
welding.

Quality control of welded connections is particularly
difficult, because defects below the surface, or even minor
flaws at the surface, will escape visual detection. Welders
must be properly certified, and for critical work, special
inspection techniques such as radiography or ultrasonic
testing must be used.
68402/61420
Slide # 60
Introductory Concepts

The two most common types of welds are the fillet weld
and the groove weld. Fillet weld examples: lap joint – fillet
welds placed in the corner formed by two plates
Tee joint – fillet welds placed at the intersection of two
plates.

Groove welds – deposited in a gap or groove between two
parts to be connected
e.g., butt, tee, and corner joints with beveled (prepared)
edges

Partial penetration groove welds can be made from one or
both sides with or without edge preparation.
68402/61420
Slide # 61
Welded Connections

Classification of welds
•
According to type of weld
Groove weld
Fillet weld
•
According to weld position
Flat, Horizontal, vertical or overhead weld
•
According to type of joint
• Butt, lap, tee, edge or corner
• According to the weld process
• SMAW, SAW
68402/61420
Slide # 62
Introductory Concepts
68402/61420
Slide # 63
Weld Limit States
The only limit state of the weld metal in a
connection is that of fracture
Yielding is not a factor since any deformation that
might take place will occur over such a short
distance that it will not influence the performance of
the structure
68402/61420
Slide # 64
Design of Welded Connections

Fillet welds are most common and used in all structures.

Weld sizes are specified in 1 mm increments

A fillet weld can be loaded in any direction in shear,
compression, or tension. However, it always fails in
shear.

The shear failure of the fillet weld occurs along a plane
through the throat of the weld, as shown in the Figure
below.
68402/61420
Slide # 65
Design of Welded Connections
hypotenuse
L
Throat = a x cos45o
= 0.707 a
a
a
root
Failure Plane
L – length of the weld
a – size of the weld
68402/61420
Slide # 66
Design of Welded Connections

Shear stress in fillet weld of length L subjected to load P
= fv =
P
0.707 a L w
If the ultimate shear strength of the weld = fw

Rn = f w  0.707  a  L w

Rn = 0.75  f w  0.707  a  L w

i.e.,  factor = 0.75
fw = shear strength of the weld metal is a function of the
electrode used in the SMAW process.
•
•
The tensile strength of the weld electrode can be 413, 482, 551, 620,
688, 758, or 827 MPa.
The corresponding electrodes are specified using the nomenclature
E60XX, E70XX, E80XX, and so on. This is the standard terminology
for weld electrodes.
68402/61420
Slide # 67
Design of Welded Connections
•


The two digits "XX" denote the type of coating.
The strength of the electrode should match the strength of
the base metal.
•
If yield stress (y) of the base metal is  413 - 448 MPa, use
E70XX electrode.
•
If yield stress (y) of the base metal is  413 - 448 MPa, use
E80XX electrode.
E70XX is the most popular electrode used for fillet welds
made by the SMAW method.
E – electrode
70 – tensile strength of electrode (ksi) = 482 MPa
XX – type of coating
68402/61420
Slide # 68
Fillet Weld

Stronger in tension and compression than in shear
Concave
Surface
Convex
Surface
Leg
Throat

Unequal leg
fillet weld
Leg
Leg
Throat
Leg
Fillet weld designations:
12 mm SMAW E70XX: fillet weld with equal leg size of 12 mm,
formed using Shielded Metal Arc Welding Process, with filler metal
electrodes having a minimum weld tensile strength of 70 ksi.
9 mm-by-12 mm SAW E110XX: fillet weld with unequal leg sizes,
formed by using Submerged Arc Metal process, with filler metal
electrodes having a minimum weld tensile strength of 758 MPa.
68402/61420
Slide # 69
Fillet Weld Strength
Stress in fillet weld = factored load/eff. throat area
Limit state of Fillet Weld is shear fracture through
the throat, regardless of how it is loaded
  0.75
f w  0.6 FEXX
Design Strength:
Vn  f wte Lw
For equal leg fillet weld:
Vn  f w ( 0.707a )Lw
68402/61420
Slide # 70
Design of Welded Connections

Table J2.5 in the AISC Specifications gives the weld design
strength
•
•

fw = 0.60 FEXX
For E70XX, fw = 0.75 x 0.60 x 482 = 217 MPa
Additionally, the shear strength of the base metal must also
be considered:
•
•
Rn = 0.9 x 0.6 Fy x area of base metal subjected to shear
where, Fy is the yield strength of the base metal.
68402/61420
Slide # 71
Design of Welded Connections

For example
T
Plan
Elevation
Strength of weld in shear = 0.75 x 0.707 x a x Lw x fw

In weld design problems it is advantageous to work with
strength per unit length of the weld or base metal.
68402/61420
Slide # 72
Limitations on Weld Dimensions



•
•
•
•
•
•
•
•
Minimum size (amin)
Function of the thickness of the thinnest connected plate
Given in Table J2.4 in the AISC specifications
Maximum size (amax)
function of the thickness of the thinnest connected plate:
for plates with thickness  6 mm, amax = 6 mm.
for plates with thickness  6 mm, amax = t – 2 mm.
Minimum length (Lw)
Length (Lw)  4 a
otherwise, aeff = Lw / 4 a = weld size
Read J2.2 b page 16.1-95
Intermittent fillet welds:
Lw-min = 4 a and 38 mm.
68402/61420
Slide # 73
Limitations on Weld Size – AISC
Specifications J2.2b Page 16.1-95




The minimum length of fillet weld may not be less than 4 x
the weld leg size. If it is, the effective weld size must be
reduced to ¼ of the weld length
The maximum size of a fillet weld along edges of material
less than 6 mm thick equals the material thickness. For
material thicker than 6 mm, the maximum size may not
exceed the material thickness less 2 mm. (to prevent
melting of base material)
The minimum weld size of fillet welds and minimum effective
throat thickness for partial-penetration groove welds are
given in LRFD Tables J2.4 and J2.3 based on the thickness
of the base materials (to ensure fusion and minimize
distortion)
Minimum end return of fillet weld  2 x weld size
68402/61420
Slide # 74
Limitations on Weld Dimensions


Maximum effective length - read AISC J2.2b
•
•
•
If weld length Lw < 100 a, then effective weld length (Lw-eff) = Lw
If Lw < 300 a, then effective weld length (Lw-eff) = Lw (1.2 – 0.002 Lw/a)
If Lw > 300 a, the effective weld length (Lw-eff) = 0.6 Lw
Weld Terminations - read AISC J2.2b
•
•
Lap joint – fillet welds terminate at a distance > a from edge.
Weld returns around corners must be > 2 a
68402/61420
Slide # 75
Guidelines for Fillet Weld design

Two types of fillet welds can be used
•
Shielded Metal Arc Welding (SMAW)
0.707 a
teff  0.707 a
a
•
Automatic Submerged Arc Welding (SAW)
teff  a
AISC – Section J2.2
68402/61420
Slide # 76
Weld Symbols
(American Welding Society AWS)
10
12
75@125
6
10
200
200
50@150
Fillet weld on arrow side. Weld’s leg size is 10 mm.
Weld size is given to the left of the weld symbol.
Weld length (200 mm) is given to the right of the
symbol
Fillet weld, 12 mm size and 75 mm long intermitten
welds 125 on center, on the far side
Field fillet welds, 6 mm in size and 200 mm long, both
sides.
Fillet welds on both sides, staggered intermitten 10
mm in size, 50 mm long and 150 mm on center
Weld all around joint
Tail used to reference certain specification or process
68402/61420
Slide # 77
Guidelines for Fillet Weld design
0.707 a
a

Fillet weld design can be governed by the smaller value of
•
Weld material strength
Pu _Weld   ( 0.707 a Lweld f w )
  0.75
•
&
f w  0.6 FExx
Electrode
FEXX (MPa)
E70XX
482
E80XX
551
Base Metal Strength
Pu _ BM   ( tbase Lweld 0.6FY )
AISC Table J2.5
  0.9
Yield Limit State
68402/61420
Slide # 78
Guidelines for Fillet Weld design

The weld strength will increase if the
force is not parallel to the weld

Pu _Weld   ( 0.707 a Lweld f w )


f w 0.6 FExx 1  0.5 sin1.5  &   0.75

Maximum weld size

if tbase  6 mm
6 mm
tweld _ max  
tbase  2 mm if tbasemetal  6 mm
Minimum weld size
tweld _ min  tthinner part
AISC Table J2.4
68402/61420
Slide # 79
Capacity of Fillet Weld
The weld strength is a function of the angle 

Pu _Weld  weld 0.707wLweld 0.6 FExx 1  0.5 sin1.5 
Strength



Weld governs
w = weld size
P
u _ BM
Base metal governs
  y (tbaseLweld 0.6 FY )
Angle ()
68402/61420
Slide # 80
Ex. 7.6 – Design Strength of Welded
Connection

Determine the design strength of the tension member and connection
system shown below. The tension member is a 100 mm x 10 mm
thick rectangular bar. It is welded to a 15 mm thick gusset plate using
E70XX electrode. Consider the yielding and fracture of the tension
member. Consider the shear strength of the weld metal and the
t = 15 mm
surrounding base metal.
a = 6 mm
100 mm x 10 mm
125 mm
12 mm
12 mm
125 mm
68402/61420
Slide # 81
Ex. 7.6 – Design Strength of Welded
Connection

Step I. Check for the limitations on the weld geometry
•
tmin = 10 mm (member)
tmax = 15 mm (gusset)
•
Therefore, amin = 5 mm
- AISC Table J2.4
amax = 10 mm – 2 mm = 8 mm
- AISC J2.2b page 16.1-95
Fillet weld size = a = 6 mm
- Therefore, OK!
Lw-min = 4 x 6 = 24 mm
38 mm
•
•
and
- OK.
Lw-min for each length of the weld = 100 mm (transverse distance
between welds, see J2.2b)
Given length = 125 mm, which is > Lmin. Therefore, OK!
68402/61420
Slide # 82
Ex. 7.6 – Design Strength of Welded
Connection
68402/61420
Slide # 83
Ex. 7.6 – Design Strength of Welded
Connection
•
•

Length/weld size = 125/6 = 20.8 - Therefore, maximum effective
length J2.2 b satisfied.
End returns at the edge corner size - minimum = 2 a = 12 mm
-Therefore, OK!
Step II. Design strength of the weld
•
Weld strength = x 0.707 x a x 0.60 x FEXX x Lw
= 0.75 x 0.707 x 6 x 0.60 x 482 x 250/1000
= 230 kN

Step III. Tension strength of the member
•
Rn = 0.9 x 344 x 100 x 10/1000 = 310 kN
68402/61420
- tension yield
Slide # 84
Ex. 7.6 – Design Strength of Welded
Connection
•
Rn = 0.75 x Ae x 448
•
•
•
•
- tension fracture
Ae = U A
Ae = Ag = 100 x 10 = 1000 mm
Therefore, Rn = 336 kN
The design strength of the member-connection system = 230 kN.
Weld strength governs. The end returns at the corners were not
included in the calculations.
68402/61420
Slide # 85
Elastic Analysis of Eccentric Welded
Connections

It is assumed here that the rotation of the weld at failure occur around the
elastic centre (EC) of the weld. The only difference from bolts is we are
dealing with unit length of weld instead of a bolt
F
M
d
e

The shear stress in weld due to torsion
moment M is
M d
f2 

AISC Manual Part 8
J
M is the moment, d is the distance from the centroid of the
weld to the weld point where we evaluate the stress, J is
the polar moment of inertia of the weld
68402/61420
Slide # 86
Elastic Analysis of Eccentric Welded
Connections – Shear & Torsion

stresses due to torsional moment “M” is
M
Fe
J  Ix  I y
f2 
- Or for teff = 1 mm
M d
J
f2x 
- Calculation shall be done for teff
teff  0.707 w
&
M y
J
f2 y 
68402/61420
M x
J
Slide # 87
Elastic Analysis of Eccentric Welded
Connections – Shear & Torsion

Forces due to direct applied force is
f1x 

Fx
Aweld
f1 y 
Fy
Aweld
Total stress in the weld is
fx 
f1x  f 2 x
fv 
fy 
&
f1 y  f 2 y
f x  f y   Rn _ weld
2
2
68402/61420
Slide # 88
Ex. 7.7 – Design Strength of Welded
Connection – Shear and Torsion

Determine the size of weld
required for the bracket connection
in the figure. The service dead load
is 50 kN, and the service live load
is 120 kN. A36 steel is used for the
bracket, and A992 steel is used for
the column.
250 mm
D = 50 kN
L = 120 kN
300
mm
15 mm PL
200
mm
Calculations are done
for teff = 25 mm
68402/61420
Slide # 89
Ex. 7.7 – Design Strength of Welded
Connection – Shear and Torsion

Step I: Calculate the ultimate load:
Pu = 1.2D + 1.6L = 1.2(50)+1.6(120) = 252 kN

Step II: Calculate the direct shear stress:
f1 y 

252 1000
 360 N/mm
200  300  200
Step III: Compute the location of the centroid:
x (700)  200(100)(2) or x  57.1mm

Step IV: Compute the torsional moment:
e = 250+ 200 – 57.1 = 392.9  M = Pe = 252(392.9)=99011 kN-mm.
68402/61420
Slide # 90
Ex. 7.7 – Design Strength of Welded
Connection – Shear and Torsion

Step V: Compute the moments of inertia of the total weld
area:
Ix = 1(300)3 (1/12)+2(200)(150)2=11.25×106 mm4
Iy = 2 {(200)3 (1/12)+(200)(100-57.1)2 }+ 300(57.1)2=3.05×106 mm4
J = Ix + Iy = (11.25 + 3.05)×106 = 14.3×106 mm4

Step VI: Compute stresses at critical location:
99011(150) 1000
 1039N/mm
6
J
14.3 10
M x 99011(200 57.1) 1000


 989 N/mm
6
J
14.3 10
f2x 
f2 y
fv 
M y

f 22x   f1 y  f 2 y   (1039) 2  (989 360) 2  1703N/mm
2
68402/61420
Slide # 91
Ex. 7.7 – Design Strength of Welded
Connection – Shear and Torsion

Step VII: Check the shear strength of the base metal
The shear yield strength of the angle leg is:
ΦRn = (0.9)0.6Fyt = 0.9(0.6)(248)(15) = 2009 N/mm
The base metal shear strength is therefore:
2009 N/mm > 1703 N/mm (OK).

Step VIII: Calculate the weld size, assuming Fw = 0.6FEXX
a
Rn
 (0.707) FW

1703
 11.1mm
0.75(0.707)(0.6  482)
 Use 12 mm
Answer: Use a 12-mm fillet weld, E70 electrode.
68402/61420
Slide # 92
Elastic Analysis of Eccentric Welded
Connections – Shear & Tension
68402/61420
Slide # 93
Elastic Analysis of Eccentric Welded
Connections – Shear & Tension

stresses due to torsion moment “M” is
F
fv 
A
M Fe
ft 
Mx c
Ix
- Calculation shall be done for teff
- Or for teff = 1 mm
teff  0.707 a

F = applied force

e = eccentricity of load

Ix = moment of inertia around x-axis

c = distance from neutral axis of weld to the farthest weld point
68402/61420
Slide # 94
Ex. 7.8 – Design Strength of Welded
Connection – Shear & Tension

An L6x4x1/2 is used in a seated beam connection, as shown in the
figure. It must support a service load reaction of 25 kN dead load and a
50 kN live load. The angles are A36 and the columns in A992. E70XX
electrodes are to be used. What size fillet weld are required for the
connection to the column flange?
20
mm
152
mm
20 mm
68402/61420
82 mm
Slide # 95
Ex. 7.8 – Design Strength of Welded
Connection – Shear & Tension

Step I: calculate the eccentricity of the reaction with respect to
the weld is:
e = 20 + 82/2 = 61 mm

Step II: Calculate the moment of inertia for the weld
configuration:
I = 2(1)(152)3 / 12 = 585300 mm4 c = 152/2 = 76 mm

Step III: Calculate the factored-load reaction is:
Pu = 1.2D + 1.6L = 1.2(25)+1.6(50) = 110 kN
Mu = Pue = 110(61) = 6710 kN-mm
68402/61420
Slide # 96
Ex. 7.8 – Design Strength of Welded
Connection – Shear & Tension
•
Step III: Calculate the factored-load reaction is:
M u c 6710(76) 1000
ft 

 871N/mm
I
585300
Pu
110000
fv 

 362 N/mm
A 2(1)(152)
fr 

f t 2  f v2  (871) 2  (362) 2  943 N/mm
Step IV: The required weld size a
•
a = 943/(0.9x0.707x0.6x482) = 6.2 mm
68402/61420
Slide # 97
Ex. 7.8 – Design Strength of Welded
Connection – Shear & Tension

The required size is therefore: a = 7 mm

Step V: Check minimum and maximum weld size

•
•
•
From AISC Table J2.4  Minimum weld size = 5 mm
From AISC Table J2.2b  Maximum weld size = 13 - 2 = 11 mm
Try a = 7 mm
Step VI: Check the shear capacity of the base metal (the
angle controls):
•
•
•
Applied direct shear = fv = 362 N/mm
The shear yield strength of the angle leg is:
ΦRn = 0.9×0.6Fyt = (0.9)0.6(248)(13) = 1741 N/mm
The base metal shear strength is therefore:
1741 N/mm > 362 N/mm (OK).
68402/61420
Slide # 98
Ultimate Strength Analysis of
Eccentric Welded Connections

When comparing elastic analysis to experimental on eccentric welded
connections, it becomes obvious that elastic analysis is over conservative.

Load
90o
rs
30o
e
IC
0o
Elastic analysis
Deformation
68402/61420
Slide # 99
F
Ultimate Strength Analysis of
Eccentric Welded Connections



Similar to bolts, weld can be divided into segments
which rotate about an instantaneous centre (IC)
M
 R r dy   R r dx
LV
F

LV
Thus

IC

LH
R Cos dy 
e
rs
Instead of summing the forces we can integrate over
the length of the weld to get the basic equations of
equilibrium:
F
R Cos dx
LH
R  Rm (1  e    ) 
68402/61420
r
r


IC
Slide # 100

Ultimate Strength Analysis of
Eccentric Welded Connections

However, in weld: The force in each segment “R” is
also function of the angle  between the force
direction and the weld.
R  0.6FEXX (1  0.5 Sin1.5 )[ p(1.9  0.9 p)]0.3

e
rs
F
IC
s
p
m
Deformation of the segment
Deformation of the segment at max stress
m  0.209(  2)0.32 w
- Similar to bolts, the far weld element might have a higher proportion of force.
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Slide # 101
Ultimate Strength Analysis of
Eccentric Welded Connections
However, the critical weld is that of the smallest m/rs

m  0.209(  2)0.32 w
rs
IC
Determine the segment that has
( m / rs ) min
The ultimate deformation u happens for the segment with smallest m/rs
u  1.087 (  6)0.65 w  0.17w
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Slide # 102
e
F
Ultimate Strength Analysis of
Eccentric Welded Connections
u 
1.087(  6) 0.65 w  0.17 w
 m  0.209 (  2)0.32 w
In all equations “” is in radian ranges from zero to /2
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Slide # 103
Ultimate Strength Analysis of
Eccentric Welded Connections

Thus to estimate the force in the critical segment we do the following steps:


1- Divide the weld into segments and assume an IC
3- Compute the ratio m/r and determine rcrit
rcrit 

r
at which
e
rs
2- Calculate the deformation of each element
m  0.209(  2)0.32 w


IC
m
is min
r
4- For this critical segment compute the ultimate deformation u
u  1.087(  6)0.65 w  0.17w

5- Compute the deformation of each other segment
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s 
r
u
rcrit
Slide # 104
F
Ultimate Strength Analysis of
Eccentric Welded Connections

Steps continued:


6- Compute the stress in each segment
R  0.6FEXX (1  0.5Sin  )[ p(1.9  0.9 p)]
1.5
0.3
s
p
m

rs
e
IC
7- Check equilibrium equations
F
F
M
x
 zero
y
IC
 zero
 zero
≡
n
R
xi
i 1
n
≡
R
≡
R
i 1
n
i 1
yi
ni
 Fx
Eqn (1)
 Fy
Eqn (2)
.rni  F ( r0  e)
Eqn (3)
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Slide # 105
F
Extra Slides
68402/61420
Slide # 106
Slip-critical Bolted Connections


High strength (A325 and A490) bolts can be installed with
such a degree of tightness that they are subject to large
tensile forces.
These large tensile forces in the bolt clamp the connected
plates together. The shear force applied to such a
tightened connection will be resisted by friction as shown in
the Figure below.
P
P
Tightened
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Slide # 107
Slip-critical Bolted Connections
N =Tb
N =Tb
P
N =Tb
F=N
Tb
F=N
N =Tb
Tb
P
N = Tb
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Slide # 108
N = Tb
Slip-critical Bolted Connections

Thus, slip-critical bolted connections can be designed to
resist the applied shear forces using friction. If the
applied shear force is less than the friction that develops
between the two surfaces, then no slip will occur between
them.

However, slip will occur when the friction force is less
than the applied shear force. After slip occurs, the
connection will behave similar to the bearing-type bolted
connections designed earlier.

Table J3.1 summarizes the minimum bolt tension that
must be applied to develop a slip-critical connection.
68402/61420
Slide # 109
Slip-Critical Connections
Loads to be transferred  Frictional
Resistance (tension force in bolt x
coefficient of friction )  No slippage
between members
 No bearing and shear stresses in bolt
LRFD J3.10 requires bearing strength to be checked for both
Bearing-Type connections and Slip-Critical connections
(even though there is supposed to be little or no bearing
stresses on the bolts in Slip-Critical connections)
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Slide # 110
Slip-critical Bolted Connections

The shear resistance of fully tensioned bolts to slip at
factored loads & service loads is given by AISC
Specification J3.8
Shear resistance at factored load = Rn = (1.13 hscTb Ns)
 - 0.85 for factored loads & 1.00 for service loads
 - friction coefficient
Tb - minimum bolt tension given in Table J3.1
hsc – hole factor determined as:
For standrad size holes
For oversized and short-slotted holes
For long-slotted holes
hsc = 1.0
hsc = 0.85
hsc = 0.7
Ns - number of slip planes
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Slide # 111
Slip-Critical Connections
Slip Coefficients (LRFD J3.8)
Surface

Class A (unpainted clean mill scale or
surfaces with class A coating on blastcleaned steel)
0.35
0.50
Class B (unpainted blast-cleaned surfaces
or surfaces with Class B coating on blastcleaned steel
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Slide # 112
Slip-critical Bolted Connections
•
When the applied shear force exceeds the Rn value stated above,
slip will occur in the connection.

The final strength of the connection will depend on the
shear strength of the bolts and on the bearing strength of
the bolts. This is the same strength as that of a bearing
type connection.

Slip critical connections shall still be checked as bearing
type in case slip occurs as a result of overload.
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Slide # 113
Ex. 6.2 - Slip-critical Connections

Design a slip-critical splice for a tension member
subjected to 600 kN of tension loading. The tension
member is a W8 x 28 section made from A36 material.
The unfactored dead load is equal to 100 kN and the
unfactored live load is equal to 300 kN. Use A325 bolts.
The splice should be slip-critical at service loads.
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Slide # 114
Ex. 6.2 - Slip-critical Connections

Step I. Service and factored loads
•
•
•

Service Load = D + L = 400 kN.
Factored design load = 1.2 D + 1.6 L = 600 kN
Tension member is W8 x 28 section made from A36 steel. The
tension splice must be slip critical (i.e., it must not slip) at service
loads.
Step II. Slip-critical splice connection (service load)
•
Rn of one fully-tensioned slip-critical bolt = (1.13 hscTb Ns)
(See Spec. J3.8)
68402/61420
Slide # 115
Ex. 6.2 - Slip-critical Connections
• Assume db = 20 mm.
• Rn of one bolt = 1.0 x 1.13 x 0.35 x 1.0x142x1 = 56.2 kN
• Note, Tb = 142 kN from Table J3.1M
• Rn of n bolts = 56.2 x n > 400 kN (splice must be slip-critical at
service)
• Therefore, n > 7.12
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Slide # 116
Ex. 6.2 - Slip-critical Connections

Step III. Layout of splice connection
•
Flange-plate splice connection
Splice plate
W8 x 28
W8 x 28
Splice plate
C.L.
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Slide # 117
Ex. 6.2 - Slip-critical Connections
•
•
•
To be symmetric about the centerline, need the number of bolts to
be a multiple of 8.
Therefore, choose 16 fully tensioned 20 mm A325 bolts with layout
as shown above.
Minimum edge distance (Le) = 34 mm from Table J3.4M
•
•
Minimum spacing = s = (2+2/3) db = 2.67 x 20 = 53.4 mm.
(Spec. J3.3)
•
•
•
Design edge distance Le = 40 mm.
Preferred spacing = s = 3.0 db = 3.0 x 20 = 60 mm (Spec. J3.3)
Design s = 60 mm.
Assume 10 mm thick splice plate
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Slide # 118
Ex. 6.2 - Slip-critical Connections

Step IV. Connection strength at factored loads
•
•
•
•
•
•
•
The splice connection should be designed as a normal
shear/bearing connection beyond this point for the factored load of
600 kN.
Shear strength of a bolt = 77.8 kN (see Example 7.1)
The shear strength of bolts = 77.8 kN/bolt x 8 = 622.4 kN
Bearing strength of 20 mm bolts at edge holes (Le = 30 mm) = 69.1
kN (see Example 7.1)
Bearing strength of 20 mm bolts at non-edge holes (s = 60 mm) =
138.2 kN (see Example 7.1)
Bearing strength of bolt holes in flanges of wide flange section
= 4 x 69.1 + 4 x 138.2 = 829.2 kN > 600 kN
68402/61420
OK
Slide # 119
Ex. 6.2 - Slip-critical Connections

Step V. Design the splice plate
•
•
•
•
•

Tension yielding: 0.9 Ag Fy > 300 kN; Therefore, Ag > 1344 mm2
Tension fracture: 0.75 An Fu > 300 kN
Therefore, An =Ag - 2 x (20 +3.2) x 10 > 1000 mm2
Beam flange width = 166 mm
Assume plate width 160 mm x 10 mm which has Ag = 1660 mm2
Step VI. Check member strength
•
Student on his/her own.
68402/61420
Slide # 120
Ultimate Strength Analysis of
Eccentric Bolted Connections
F

Experimental study by Crawford and Kulak (1971) showed:
e
- The load-deformation relationship
of any bolt is non-linear
AISC Manual Part 7
68402/61420
Slide # 121
Ultimate Strength Analysis of
Eccentric Bolted Connections
F
e

The following conclusions were also shown:

Failure rotation does not happen around the elastic center but
around an instantaneous centre (IC)

The IC does not coincide with the EC

The deformation of each bolt is proportional to its distance from
the IC

Similar to the elastic analysis, the connection capacity is
governed by the force in the farthest bolt
ri
M
≡
EC
IC
F
e
68402/61420
e
Slide # 122
Ultimate Strength Analysis of
Eccentric Bolted Connections
M
Fe
ri
EC
Measured at the elastic centroid
b 
rb
 max
rmax
IC
F
Experimentally  8.6 mm

e
e
At failure
F
F
M
x
 zero
y
IC
 zero
 zero
≡
≡
≡
n
R
i 1
xi
n
R
i 1
yi
 Fx
Eqn (1)
 Fy
Eqn (2)
n
 Rni .rni  F ( e  e )
i 1
68402/61420
Eqn (3)
Slide # 123
Ultimate Strength Analysis of
Eccentric Bolted Connections
ri
EC
IC
F
e
e

Therefore, getting the maximum force in the farthest bolt requires
determining the unknown “e´”

Because of the non-linear relationship, e´ can be determined by trial
and error

A spreadsheet can be used to determine e´
68402/61420
Slide # 124
Forces on Eccentrically-Loaded Bolts
with Eccentricity on the Faying Surface
Ultimate Strength Method (Instantaneous Center of
Rotation Method)

e’
1
d1
R3
3
d3
R = Nominal shear strength of 1 bolt at a deformation , k
Rult= Ultimate shear strength of 1 bolt, kN
2

R2
CG
IC
d4
Pu
e
R1
d2
R = Rult(1 – e-0.394)0.55
4
R4
= Total deformation, including shear, bearing and
bending deformation in the bolt and bearing
deformation of the connected elements, in. (max = 8.6
mm for 20 mm ASTM A325 bolt)
1/d1 = 2/d2 = … = max/dmax
e = 2.718…base of the natural logarithm
68402/61420
Slide # 125
Ultimate Strength Method (Instantaneous
Center of Rotation Method)

Trial and error:
•
•
•
•
•
Assume e’
Compute i = dimax/dmax (max is assumed for bolt at farthest
distance from IC)
r
Compute Ri=Rult(1-
e-0.394i)0.55
Check for: Pu=( Rd)/(e’+e)
If not satisfied, repeat with another e’
68402/61420
b 
b
rmax
 max
Rb  Rult (1  e  b )
Slide # 126
Ex. 6.4 – Eccentric Connections –
Ultimate Method
Determine the largest eccentric force Pu for which the design
shear strength of the bolts in the connection is adequate using
the IC method. Use bearing-type 20 mm A325X bolts
e = 100 Pu
mm
e’=60 mm
- Design shear strength per bolt (Ex. 7-1)
Ru =  Rn= 77.8 kN
1
d1
R1
d2
2
therefore 2 = 4 = max = 8.6 mm
CG
IC
d4
3
d3
R3
75 -After several trials, assume e’= 60 mm.
R2 mm Bolts 2 and 4 are furthest from the IC,
4
75
mm - Compute i and Ri in tabulated form:
R4
75 mm
68402/61420
Slide # 127
Ex. 6.4 – Eccentric Connections –
Ultimate Method
Bolt
#
h
v
(mm) (mm)
d (mm)

(mm)
R (kN)
Ry (kN)
Rd
(kN.mm)
1
22.5
75
78.3
5.47
72.7
20.9
5692
2
97.5
75
123
8.6
77.8
61.67
7585
3
22.5
75
78.3
5.47
72.7
20.9
5692
4
97.5
75
123
8.6
77.8
61.67
7585
=
165.14
 = 26554
Check:
Pu= (Rd)/(e’+e) = (26554/(60+100))
= 166 kN ~ Ry = 165.14 kN (OK)
68402/61420
Slide # 128