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CIE 428 Module F Instructor: Andrew Whittaker

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MODULE F: SIMPLE CONNECTIONS

This module of CIE 428 covers the following subjects

Connector characterization

Failure modes of bolted shear connections

Detailing of bolted connections

Bolts: common and high-strength

Shear strength of high-strength bolts

Slip-critical connections

High-strength bolts in tension

Combined shear and tension in fasteners

Welded connections

Fillet welds

READING: Chapter 7 of Segui

AISC LRFD Manual of Steel Construction, 3rdEd.




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INTRODUCTION

For the purpose of this class, connections are characterized as

either

Simple

Eccentric

The figure below from Segui illustrates different types of bolted

and welded connections. Those connections in parts a and b of the

figure are termed simple

Line of action of resultant force passes through the center

of gravity of the connection

All parts of the connection share equally in resisting the

load

The other connections are termed eccentric and are discussed in

the following module. Eccentric connections are shown in parts cand d of the figure below.

Chapter J of the LRFD Specification addresses the design of

connections. The following connectors are covered

Bolts

Rivets

Welds

Only bolted and welded connections are considered in CIE 428.




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FAILURE MODES OF BOLTED SHEAR CONNECTIONS

Two types of bolted connector failure are considered in this section

Failure of the connector

Failure of the connected parts




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Connector failure

Consider first the figure shown below from Segui that presents

joints in single (one shear plane) and double (two shear planes)

shear.

In part a of the figure (single shear connection), the load (shear) on

the fastener is P and the connector failure is in the fastener.

Loading is not concentric on the fastener, the eccentricity

is small and it will be ignored.

The load resisted by the fastener can be written as

v P f A=

where v f is the average shearing stress and A is the cross-sectional

area.

In part b of the figure, the connector is in double shear and the load

is equal to

2 v P f A=




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Why?

Failure of the connected parts

The failure of the connected parts can be separated into twocategories.

Failure resulting from excessive tension, shear, or bending

in the parts being connected

For a tension member must consider tension on the net

area, tension on the gross area, and block shear

For beam-beam or beam-column connections, must

consider block shear

Gusset plates and framing angles must be checked for

P , M , and V

Failure of the connected part because of bearing exerted

by the fastener

If the hole is slightly larger than the fastener and the

fastener is assumed to be placed loosely in the hole

(rarely the case), contact between the fastener and the

connected part will exist over approximately 50% of the

circumference of the fastener. See the figure below for

information.




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The average bearing stress is

p

P f

dt =

where terms are defined in the figure. If the maximum

bearing pressure is known, the maximum load is easily

calculated by manipulating the above equation

The bearing problem is complicated a little by the edge

distance and bolt spacing; subjects that are discussed

below. See the figure below from Segui for an

illustration of these complications.




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DETAILING BOLTED CONNECTIONS

Bearing strength, bolt spacing and bolt edge distance are

considered in this section.

Bearing strength limits

One possible failure mode resulting from excessive bearing is

shown in the figure below from Segui. In this figure, the actual

failure surface is replaced by a failure surface in part b, which

simplifies the calculation.

Noting the total failure load is equal to the sum of the failure loads

on the two surfaces shown in part b of the figure:

2(0.6 ) 1.2n u c u c R F L t F L t = =

where




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0.6 u F = Shear fracture stress of the connected part

c L =Distance from the edge of the hole to the edge of

the connected part

t = Thickness of the connected part

The tear-out described above can also occur between bolt holes in

the direction of the applied load as shown in the figure below,

which provides an alternate definition for c L .

To prevent excessive elongation of the hole, an upper limit is

placed on the bearing load. This limit is

( )n u R CF dt =

where

C = Constant

d = Bolt diameter


The values assigned to C are 2.4, 3.0, and 2.0, depending on the

hole type and the acceptability of hole ovaling at service load. See

Equations J3-2a, -2b, and –2c of the LRFD Specification for




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details. Considering only standard (not slotted) holes, if excessive

deformation is a concern, C = 2.4. In this case,

If 2 , 1.2c n c u L d R L tF ≤ =

If 2 , 2.4c n u L d R dtF > =

The design bearing strength is equal to n Rφ , where 0.75φ = .

Spacing and edge-distance requirements

Section J3.3 of the LRFD Specification sets a minimum center-to-

center spacing ( s) of bolts equal to 2.67d (and preferably not less

than 3d), where d is the fastener diameter. Such distances are

needed to maintain clearances between bolt nuts and maintain

clearance for wrench sockets. See the figure from Segui below for

details.

Section J3.4 of the LRFD Specification gives minimum edgedistances ( e L ) as a function of bolt-size and type of edge (sheared,

rolled, or gas cut). The table referenced in that section (Table J3.4)

is presented below.




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The example below from Segui illustrates the detailing

requirements introduced above.




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In the above example, the calculated bolt spacings and edge

distances are the same in the gusset and tension member. Only the

thickness will control in this case, and because the thickness of the

tension member is less than that of the gusset, the gusset willcontrol.

BOLTS: COMMON AND HIGH STRENGTH

Common bolts

Common bolts differ from high-strength bolts by

Material

Clamping force not accounted for with common bolts

Common bolts are designated as ASTM A307, with an ultimate

shearing stress of 24 ksi (see Table J3.2 of the AISC LRFD

Specification).

The design shear strength of A307 bolts is n Rφ , where φ is 0.75,

and the nominal shear strength is

24n v b b R F A A= =




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where

v F = Ultimate shearing stress

b = Cross-sectional area of the unthreaded part of the bolt (nominal bolt area)





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The example below from Segui illustrates the calculation of the

design strength of a tension connection with A307 bolts.




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High-strength bolts

High-strength bolts for structural joints come in two grades,

namely

ASTM A325

ASTM A490

Two conditions of bolt installation are used with high-strength

bolts

Snug-tight (producing a bearing connection)

Few impacts of an impact wrench

Full effort of a worker with an ordinary spud wrench




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Tensioned (producing a slip-critical connection)

Turn-of-nut method: specified number of rotations of

the nut from snug tight (nut rotations correlated to boltelongation)

Calibrated wrench tightening

Alternate design bolts: specially design bolts whose

tops twist off when the proper tension has been

achieved

Direct tension indicators: compress washer (under bolt

head or nut) with protrusions to a gap that is correlated

to bolt tension




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When high-strength bolts are to be tensioned, minimum limits are

set on the bolt tension. See Table J3.1 in the LRFD Specification.

Tension equal to 70% of the minimum tensile strength of

the bolt

Purpose of tensioning is to achieve the clamping force

shown in the figure below from Segui




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In reference to the figure above, as the bolt is tensioned and

elongates, the connected parts undergo compression. The total

compressive force on the parts is equal to the bolt tension.

If an external force P is applied, a friction force will develop between the connected parts, with a maximum value of

F N µ =

where µ is the coefficient of static friction between the connected

parts (a function of surface condition of the steel parts) and N is the

normal force shown in the figure above.

Each bolt in the connection is then capable of resisting a

force P = F before slippage. Prior to slippage, there is no

bearing or shear.

SHEAR STRENGTH OF HIGH-STRENGTH BOLTS

The design shear strength of high strength bolts, like A307 bolts, is

n Rφ , where φ is 0.75. Unlike A307 bolts, the shear strength of A325 and A490 bolts depends on whether the threads are in the

shear plane (threads not excluded from the shear plane) or not

(threads excluded from the shear plane).

A325-N: threads included in the shear plane

A325-X: threads excluded from the shear plane

Rather than using a reduced cross-sectional area when the threads

are not excluded from the shear plane, the ultimate shear stress is

multiplied by a factor of 0.75 in the LRFD Specification. Results

are presented in the last column of Table J3.2 of the Specification,

which was reproduced above. In summary,




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Fastener Threads n v b R F A=

A325 Included in shear plane 48 b

A325 Excluded from shear plane 60 b

A490 Included in shear plane 60 b

A490 Excluded from shear plane 75 b

The calculation of the design strength of a connection is illustrated

below. The bolts are 7/8 inch in diameter; threads are not in the

shear plane, and the steel is Grade 50 steel.




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SLIP-CRITICAL CONNECTIONS

As noted earlier, a connection with high-strength bolts is classified

as either a bearing or slip-critical connection.

In bearing connections, the bolt is brought to a snug-tightcondition so that the surfaces of the connected parts are in firm

contact.

Slippage is acceptable

Shear and bearing on the connector

In slip-critical connections, no slippage is permitted and the

friction force described earlier must not be exceeded.

Slippage is not acceptable

Slip-critical strength based on factored loads herein.

The LRFD Specification writes rules for basing the

strength on service loads but such an approach is not

discussed in CIE 428. Refer to the Specification for

details.

Proper installation and tensioning is key

Connector not subjected to shear and bearing (in theory)




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Must have sufficient shear and bearing strength in the

event of overload that causes slip. See Section J3.8 of

the LRFD Specification for details.

AISC writes that slip-critical bolted (or welded) connections beused in specific circumstances. Part of Section J1.11 of the LRFD

Specification that lists these circumstances is reproduced below.

As introduced earlier, the resistance to slip is a function of the

product of the

Normal force (equal to the bolt tension)

Coefficient of static friction between the contact surfaces

The LRFD Specification writes that the design slip resistance per

bolt, str r φ , shall equal or exceed the required force due to factored

loads, where

1.13 str b sr T N µ =




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where 1.13 is a factor that accounts for the expected 13% increase

above the minimum specified preload provided by calibrated

torque wrench tightening procedures, and

µ = Mean slip coefficient of friction for Class A, B,or C surfaces

bT =Minimum fastener tension (see Table J3.1 of the

Specification)

s N = Number of slip planes (described earlier)

Guidance is provided for the mean slip coefficient (staticcoefficient of friction) associated with the three classes of surface

noted above.

For unpainted clean mill scale steel surfaces (Class A),

0.33 µ =

Assumed surface for CIE 428 u.n.o.

For unpainted blast-cleaned steel surfaces (Class B),

0.50 µ =

For hot-dip galvanized and roughened steel surfaces

(Class C), 0.35 µ =

The resistance factor varies as a function of the type of hole and its

orientation. The LRFD Specification writes that

For standard holes, 1.0φ =

Assumed hole geometry for CIE 428 u.n.o.




8/21/2002 12:12 PM 24

For oversized and short-slotted holes, 0.85φ =

For long-slotted holes transverse to the direction of load,

0.70φ =

For long-slotted holes parallel to the direction of load,

0.60φ =

The two examples from Segui below illustrate aspects of the

checking or design of high-strength, slip-critical bolted

connections.




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The following example presents a design procedure for slip-critical

connections. The C6x13 shown below is to resist a factored load of 108 kips. Bolts to be used are 7/8-inch diameter, A325; threads are

in the shear plane and slip of the connection is permitted.

Determine the number and layout of bolts so that h is a minimum.

A36 steel is used.




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Based on the above example, what are the steps in the design of a

slip-critical connection?

Below is an example from Segui of the design of a tension member

and its connection. This example is an extension of the last

example where the size of the tension member was given. In the

example below, the tension member must also be sized, based in part on the connection details.

Is this an appropriate consideration for design?




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HIGH-STRENGTH BOLTS IN TENSION

When a tensile force Q is applied to a non-tensioned (snug-tight)

bolt, the resulting tensile force in the bolt is equal to the applied

tensile force, Q.

When a tensile force Q is applied to a tensioned bolt with pre-

tension 0T , the resulting tensile force in the bolt is approximately

equal to 0T , where the applied force relieves the compression

(clamping forces) on the connected parts.

An example of a high-strength bolt in tension is shown below

(from Segui). Consider now the circled bolt in tension.




8/21/2002 12:12 PM 37

Consider the bolt and the connected part shown in the figure below

from Segui. Part a of the figure shows free-body diagrams of the

parts after tensioning. Note that the bolt and the connected part are

in equilibrium.

Shown in part a of the figure is the bolt tension 0T and the normal

clamping force 0 N (equal in magnitude to 0T ).




8/21/2002 12:12 PM 38

When an external tensile force F is applied (to one bolt), the forces

are those shown in part b of the figure above. Summing the forces

in part c of the figure gives the total tensile force in the bolt as

T F N = +

The application of the force F will increase the bolt tension and

cause it to elongate by an amount bδ . Compression in the flange of

the structural tee (assumed connection detail) will be reduced,

resulting in a distortion fl δ in the same sense as bδ . Assuming that

the clamping force N is applied over a uniform area of flange fl ,

and that the connected parts (the flanges) do not separate

( ) fl fl fl fl b b b b

bb fl b fl

E A A E A F T N E

L L L L

δ δ δ ∆ = ∆ − ∆ = − = −

where b L is the length of the bolt (in tension) and fl L is the

thickness of the flange. Note that these lengths are of the same

order and that fl is much greater than b :

N T ∆ ∆ and F N ∆ ≈ ∆

The ratio of T ∆ to N ∆ is approximately 0.05 to 0.10.

Most of the applied load relieves the compression on the

connected parts.

To estimate the magnitude of the applied load required to

overcome the clamping effect (for the parts to separate), consider

the figure below from Segui.




8/21/2002 12:12 PM 39

For the connected parts to separate, the change in the compressive

force must equal the initial pre-compression, namely, 0 N N ∆ = .

When the parts have separated,

0 0 0 0 00.1 0.1 1.1 F T T T N T N T = + ∆ = + ∆ = + =

That is, at the point of separation, the bolt tension is approximately

10% larger than the pretension.

What happens once the parts have separated?

Any increase in external load F will be resisted entirely by an

increase in bolt tension.

To avoid separation (and use the design strengths given in Table

J3.2 of the LRFD Specification), high-strength bolts subject to

direct tension must be pretensioned to the values given in Table

J3.1, regardless of whether the connection is slip-critical or not.

Calculate the tension force in the bolt assuming that there is

no initial tension.

Prying Action

In connections in which the fasteners are subjected to tension

forces, the flexibility of the connected parts can lead to

deformations that increase the tensile force applied to the fastener.




8/21/2002 12:12 PM 40

Force is known as prying force

Shown in the figure below from Segui

Before the external load is applied, the normal compressive force

0 N is centered on the bolt. As the external load is applied and if

the flange is sufficiently flexible to deform as shown above, thecompressive forces will migrate towards the edges of the flanges.

This redistribution will change the relationship between all of the

forces and the bolt tension will increase.

If the flange is rigid, no redistribution will occur.

How stiff is rigid in this case?

The maximum prying force will be reached when only the corners

of the flange remain in contact with the connected part. See the

figure above for details.




8/21/2002 12:12 PM 41

For connections of the type shown above, Section J3.6 of the

LRFD Specification writes that the applied load shall be the sum of

the factored loads and any tension resulting from prying action

produced by deformation of the connected parts.

A procedure for determining prying forces is given in Part 9 of the

LRFD Specification. The presentation of Segui is similar and gives

identical results. Consider the part of a structural tee shape shown

below. The presentation is for one fastener only. The force T is the

external factored tension force applied to one bolt, Q is the prying

force applied to one bolt, andc

B is the total bolt force. The prying

force has shifted to the tip of the flange of the tee and is at itsmaximum value.




8/21/2002 12:12 PM 42

From consideration of force and moment equilibrium, the

following equations are derived:

; ;a a b b c

Tb M Qa M Qa B T Q− −

− = = = +

These equations can be combined into a single equation for the

bolt force, including the effects of prying. First, a variable is

defined as the ratio of the moment per unit length along the bolt

line to the moment per unit length at the face of the stem

b b

a a

M

M

α

δ

−

−

=

where

δ = Net area at bolt line divided by gross area at

stem = 1 /d p′−

d ′ = Diameter of the bolt hole

p = Length of flange tributary to one bolt

a aM − = Design strength at a-a = 2( / 4)b f y

pt F φ

The total bolt forcec

B can thus be calculated as

[1 ](1 )c

b B T

a

δα

δα = +

+

At this level of loading, the resultant of the tensile force in the bolt

does not coincide with the axis of the bolt. For better agreement

with test results, the force resultant is shifted toward the stem of

the tee by d /2, where d is the bolt diameter, that is




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0.5 ; 0.5b b d a a d ′ ′= − = +

and

[1 ](1 )c

b B T

a

δα

δα

′= +

′+

If the bolt forcec

B is set equal to the design tensile strength,

denoted B,

[( / ) 1]( / )

{1 [ / ) 1]( / )}

B T a b

B T a b

α

δ

′ ′−=

′ ′− −

Two limit states are possible

Tensile failure of the bolt

Bending failure of the tee

Plastic moments form at lines a-a and b-b producing amechanism

If the absolute value of α is less than 1.0, the moment at the bolt

line per unit length is less than that at the face of the stem,

indicating that the mechanism has not formed

Tensile failure of the bolt controls, andc

B B=

If the absolute value of α is greater than or equal to 1.0, plastic

hinges will have formed at both a-a and b-b

Controlling limit state is bending failure of the tee




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Since the moments at a-a and b-b are limited to the plastic

moment, α should be set equal to 1.0.

The three equilibrium equations presented at the top of page 42 can

be combined into a single equation to establish the required flangethickness,

f t , as follows:

2

( ) ( )4

a a b b

a a

f y

b p b

Tb M M

M

pt F

M

δα

δα φ δα φ

− −

−

′ − =

=

= =

and for 0.90b

φ = ,

4.444

(1 ) f y

Tbt

pF δα

′=

+

The design of connections subjected to prying forces involvesiteration.

Always make an allowance for prying force in design

Select a trial thickness and bolt geometry

Calculate f t and

c B using the above equations

If actual f t is different from the required value, the

actual values of α andc

B will be different from those

previously calculated




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If the actual bolt force, which includes the prying force, is desired,

α must be recomputed. Using b′ instead of b, and setting

a a b pM φ

−= , then

2

1 4.444( 1)

f y

Tb

pt F α

δ

′= −

The total bolt force can now be found using the boxed equation on

page 43.

What if the flange thickness is inadequate?

Increase the flange thickness by using another tee shape

Use more bolts to reduce T

The example below from Segui illustrates most of the concepts

introduced above.




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COMBINED SHEAR AND TENSION IN FASTENERS

Most connections involving simultaneous shear and tension are

eccentric connections, which will be discussed in the next module.

One exception is the connection shown below, where, because theline of action passes through the center of gravity of the

connection, each fastener can be assumed to resist an equal share

of each component (V and T ) of the load.

An elliptical interaction curve is used for the case of combined

loading in bearing-type connections, as shown in the figure below.

The equation for this curve is given by the following equation

2 2[ ] [ ] 1.0( ) ( )

u u

n t n v

T V

R Rφ φ + =

where




8/21/2002 12:12 PM 49

uT = Factored tensile load on the bolt

( )n t Rφ = Design strength of the bolt in tension

uV = Factored tensile load on the bolt

( )n v Rφ = Design strength of the bolt in shear

For performance checking, the above equation is written as

2 2[ ] [ ] 1.0

( ) ( )

u u

n t n v

T V

R Rφ φ

+ ≤

This equation can be re-packaged to permit calculation of the

required bolt area in a connection as follows

2 2( ) ( )u ub

t v

T V A

F F φ φ ≥ +∑

The 3rd Ed of the LRFD presents the interaction equations in two

forms. The interaction equations are listed in Table A-J3.1 of theSpecification. The body of the specification uses a different format

and parses the elliptical interaction curve into three straight-line

segments as shown in the figure below from the Specification.




8/21/2002 12:12 PM 50

The Specification writes that in combined shear and tension in a

bearing type connection, the design strength of the bolt is given by

t b F Aφ

where

t F =

Nominal tension stress computed using the

equations in Table J3.5 of the LRFD

Specification (see below) as a function of the

shear stress produced by the factored loads. Note

that the design shear strength, v F φ , must equal

or exceed the shear stress v f .

b = Nominal unthreaded body area of the bolt

φ = 0.75

For slip-critical connections in which the bolts are subjected to

shear and tension, the effect of the applied tensile force is to reduce

the clamping force, thereby reducing the allowable friction force.




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The LRFD Specification reduces the slip-critical shear strength in

this case by multiplying the slip-critical shear strength ( str r φ ) by

[1 ]1.13u

m b

T

T N −

where

uT = Factored tensile load on the bolt

mT = Prescribed initial bolt tension

b N = Number of bolts in the connection

The uses of some of the above design equations are illustrated

below using an example from Segui.




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WELDED CONNECTIONS

Structural welding is a process whereby the parts to be connected

are heated and fused with a molten filler metal. The figure below

from Segui illustrates two fillet-welded connections.




8/21/2002 12:12 PM 54

Upon cooling, the structural steel (parent metal) and weld or filler

metal will act as one continuous part. The filler metal is deposited

from a special electrode. A number of welding processes are used,

depending on the application

Field welds

Shop welds

Shown below is a figure from Segui that illustrates shielded metal

arc welding (SMAW):

Current arcs across the gap between the electrode and the base metal

Connected parts are heated and part of the filler metal is

deposited into the molten base metal

Coating on the electrode vaporizes and forms a protective

gaseous shield, preventing the molten metal from oxidizing

before it solidifies




8/21/2002 12:12 PM 55

The electrode is moved across the joint and a weld bead is

deposited

Size of the weld bead depends on the rate of travel

As the weld cools, impurities rise to the surface and form a coating

called slag

Slag must be removed before the next pass or the weld is

painted

Shielded metal arc welding is normally done manually and is

widely used for field welding

Self-shielded flux core

Gas shielded flux core

These and other processes are used for shop welding. Shop

processes are automated or semi-automated in many cases. Other

processes include

Submerged arc welding: end of the electrode and the arc are

submerged in a granular flux that melts and forms a gaseous

shield.

Electroslag

As noted earlier, the two common types of welds are

Fillet welds

Welds placed in a corner formed by two parts in contact

See figure on page 54




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Groove welds

Welds deposited in a gap between two parts; see figure below

Single bevel groove welds shown in the figure

Backing bar

One of both of the parts must be prepared

Plug welds are shown in the figure below from Segui

Circular or slotted hole that is filled with weld metal

Used sometimes when more weld length is needed than is

available




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Panel zone doubler plates of seismic moment frames

Of the two key types of welds, fillet welds are the most common

and are considered in the following section. Fillet welds are

cheaper than groove welds.

Why?

The design of complete joint penetration (CJP) groove welds is

generally straightforward. The filler metal is equally or

overmatched to the parent metal (equal strength or greater)

Connected parts considered continuous through the weld

FILLET WELDS

The design and analysis of fillet welds is based on the assumption

that the geometry of the weld is a 45-degree right triangle as shown

below in the figure from Segui.

Shown in the above figure are the leg length (w) and the throat

thickness (t ).




8/21/2002 12:12 PM 58

Standard weld sizes are expressed in sixteenths of an inch.

Failure of fillet welds is assumed to occur in shear on the throat.

That failure plane is also shown in the figure above.

The strength of a fillet weld depends on the strength of the filler or

electrode metal used. The strength of an electrode is given in terms

of its tensile strength in ksi. Strengths of 60, 70, 80, 90, 100, 110,

and 120 ksi are available.

The standard notation for an electrode is E**XX where ** indicate

the tensile strength in ksi and XX denotes the type of coating used.

Usually XX is the focus of design

E70XX is an electrode with a tensile strength of 70 ksi

Electrodes should be chosen to match the base metal.

Use E70XX electrodes for use with steels that have a

yield stress less than 60 ksi

Use E80XX electrodes that have a yield stress of 60 ksi

or 65 ksi

The critical shearing stress on a weld of length L is given by

0.707v

P f

wL=

where all terms are defined above. If the ultimate shearing stress in

the weld is termedW

F , the nominal design strength of the weld

can be written as




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0.707 ( ) 0.707 (0.75[0.6 ]) 0.32n w EXX EXX

R wL F wL F wLF φ φ = = =

For E70XX and E80XX electrodes, the design stresses areW

F φ , or

31.5 ksi and 36 ksi, respectively.

In addition, the factored load shear on the base metal shall not

produce a stress in excess of BM

F φ , where BM

F is the nominal

shear strength of the connected material. The factored load on the

connection is thus subjected to the limit of

0.90(0.6 ) 0.54n BM g y g y g

R F A F A F Aφ φ = = =

where g

is the area subjected to shear. When the load is in the

same direction as the axis of the weld, such as that shown in the

figure below from Segui, the base metal must also be examined

using the above equation.

Putting aside the eccentricity of the connection, the design strength

of the welds (2 in total) is 2(0.707 )W

w F φ and the shear capacity of

the bracket is BM t F φ .

The following table from the LRFD Specification presents the

design strength of welds. The example on the page following the

table, from Segui, illustrates the design approaches presented

above.




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8/21/2002 12:12 PM 61






8/21/2002 12:12 PM 63

When a weld extends to the corner of a member, it must be

continued around the corner (an end return) as shown in the figure

below from Segui.

Prevent stress concentrations at the corner of the weld

Minimum length of return is 2w

Welds are shown on structural drawings by standard symbols.Some standard symbols are shown below in a figure from Segui.

Symbols from the LRFD Specification are shown on the next page.

Near side, far side, weld size, weld length




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8/21/2002 12:12 PM 65

Below is an example from Segui that illustrates some of the

concepts introduced above.




- END OF MODULE -

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