Bolted Connections

 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
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
Simple Bolted Connections

P P
Tension member
Connection/ splice
P P

Beam end
Simple shear connection
Simple Bolted Connections
P P

Hanger connection Moment resisting

(Tension) connection
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.
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
Common Bolts
 ASTM A307 bolts
 Common bolts are no longer common for current structural
design but are still available

Pu   Rn   0.75

Rn  f v Abolt

f v  165 MPa
HSB – Bearing Type Connections
 The shear strength of bolts shall be determined as follows

Pu   Rn   0.75

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.
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
Failure Modes of Bolted
Connections

 Bolt Shearing

 Tension Fracture

 Plate Bearing

 Block Shear
 Actions on Bolt
Shear, bearing, bending
P P
P
P Bearing and single plane Shear
Lap Joint
P P
Bending

Butt Joint Bearing and double plane Shear

P/2 P/2
P P
P/2
P/2
Bolted Shear Connections
 Possible failure modes
 Failure of bolts: single or double shear
Single shear
PSingle Shear  f v Abolt P P

Double shear
P/2
PDouble Shear  2 f v Abolt 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
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) where fv = shear yield
stress = 0.6Fy
• 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.
Bolted Shear Connections
• The bearing strength is independent of the bolt material as it is failure in
the connected metal

PBearing  f p dbolt t
• The other possible common failure is shear end failure known as “shear
tear-out” at the connection end
Pu   Rn   0.75
Lc
Rn  1.2 Lc t Fu  2.4 d t Fu Lc

Shear limitation Bearing limitation

Bolted Shear Connections
Spacing and Edge-distance
requirements
 The AISC code gives guidance for edge distance and spacing to
avoid tear out shear

h
Lc  Le  AISC Table J3.4
2 Le
h is the hole diameter h  dbolt  1.6 mm
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

 Common we assume S  3 d bolt

 The AISC minimum spacing is S  2 23 d bolt

 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
 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
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
Important Notes

Lc – Clear distance
Design Provisions for Bolted Shear
Connections
 In a simple connection, all bolts share the load equally.

T/n T/n

T T/n T/n
T

T/n T/n
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.
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.
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 - deformation at the bolt holes is a
design consideration
 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 Edge
direction perpendicular to p
distance
member’s axis
 Pitch – ...parallel to member’s axis
 Edge Distance – Distance from
center of bolt to adjacent
edge of a member p
g
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.
A36 5 x ½ mm
120x15

301.25
mm
A36
60 2.50
mm

301.25
mm
65
63kkN
300 k

¾20
in.mm A325-N
bolts
bolts
30 1.25
mm 602.50
mm 1.25
30 mm
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
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.
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
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.
• 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
• 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.
Ex. 6.1 - Design Strength
• At other holes, s = 60 mm, Lc = 60 – (20 +1.6) = 38.4 mm.
• Rn = 0.75 x (1.2 Lc t Fu) = 0.75 x (1.2 x 38.4 x 10x 400)/1000 = 138.2
kN
• But, Rn ≤ 0.75 (2.4 db t Fu) = 144 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
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
Eccentrically-Loaded Bolted Connections
P P
P P
Pe

CG CG

Pe
e
e
Eccentricity in the plane of Eccentricity normal to the plane
the faying surface of the faying surface
Direct Shear + Additional Shear due to Direct Shear + Tension and Compression
moment Pe (above and below neutral axis)
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
Forces on Eccentrically-Loaded Bolts
with Eccentricity on the Faying Surface
ry M d y
pmx  r1 sin   1 1  CG 1 1
H1
d1 d1 d2 
y 1 r1
d1 M CG y1 Pe y1
pmx  
 
 V1 2
CG x1 d d2
M CG x1 Pe x1
pmy  

Total Forces in Bolt i:

 d2  d2

Pey i
-Horizontal Component =
 d2
P Pex i

-Vertical Component = n
 d2
Ex. 6.3 – Eccentric Connections –
Elastic Method
Determine the force in the most stressed bolt of the group
using elastic method
e P=140 kN
125 mm
Eccentricity wrt CG:
e = 125 + 50 = 175 mm
100 Direct Shear in each bolt:
mm
100 P/n = 140/8 = 17.5 kN
mm
CG Note that the upper right-hand and
100
mm
the lower right-hand bolts are the
most stressed (farthest from CG and
consider direction of forces)
100
mm
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)(50 2  150 2 )  120000
My (22500)(150)
pmx    30.6 kN
 d 2
120000
Mx (24500)(50)
pmy    10.2 kN
 d 2
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
 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
Fnt  1.3Fnt  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 620
A490 780
 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
Tension stress per bolt:
1073 kN 1
ft = Tu/NbAb=1073000/(8x380)= 353 MPa
2
Nominal Tension Strength Ft (Table J3.5)
Ft = 0.75[(1.3x620 – (620/310)x176.6) ≤ 620]
Eight 22 mm = 496 MPa ≤ 620]
A325X bolts = 496 MPa > ft = 353 MPa (OK)
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 Fillet weld

Fillet weld
P P
Introductory Concepts

Welding Process – Fillet Weld

Welded Connections
 Classification of welds
• According to type of weld

Fillet weld Groove 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
Introductory Concepts
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
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.
Design of Welded Connections

hypotenuse
L
Throat = a x cos45o
a = 0.707 a

a
root Failure Plane

L – length of the weld

a – size of the weld
Design of Welded Connections
 Shear stress in fillet weld of length L subjected to load P
P
= fv = If the ultimate shear strength of the weld = fw
0.707 a L w
 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.
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
 Fillet Weld
 Stronger in tension and compression than in shear
Convex Concave Unequal leg
Surface Surface fillet weld
Leg Leg

Throat 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.
 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

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.
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.
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.
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
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
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

• Automatic Submerged Arc Welding (SAW)

teff  a AISC – Section J2.2

Weld Symbols
(American Welding Society AWS)
10 200 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
12 75@125 symbol

Fillet weld, 12 mm size and 75 mm long intermitten

welds 125 on center, on the far side
6 200

Field fillet welds, 6 mm in size and 200 mm long, both

10 50@150 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

 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 ) Electrode FEXX (MPa)

  0.75 & f w  0.6 FExx E70XX 482

E80XX 551
• Base Metal Strength
Pu _ BM   ( tbase Lweld 0.6FY )
AISC Table J2.5   0.9 Yield Limit State
 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

 Minimum weld size

6 mm if tbase  6 mm
t weld _ max  tweld _ min  tthinner part
tbase  2 mm if tbasemetal  6 mm
AISC Table J2.4
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

Pu _ BM   y (tbaseLweld 0.6 FY )
Base metal governs

Angle ()
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
surrounding base metal. t = 15 mm

a = 6 mm 100 mm x 10 mm
125 mm

12 mm

12 mm

125 mm
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 and 38 mm - 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!
Ex. 7.6 – Design Strength of Welded
Connection
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 - tension yield
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.