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68402: Structural Design of Buildings II

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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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- 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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Simple Connections

Eccentric Connections

Bolted Connections

Welded Connections

Common Bolts

High Strength Bolts

Filet Weld

Slip Critical

Groove Weld

Bearing Type

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Simple Connections

Eccentric Connections

Bolted Connections

Welded Connections

Elastic Analysis

Ultimate Analysis

Moment Resisting

Elastic Analysis

Ultimate Analysis

Moment Resisting

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- 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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P

P

Tension member

Connection/ splice

P

P

Beam end

Simple shear connection

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P

P

P

Hanger connection (Tension)

Moment resisting connection

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- 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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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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- ASTM A307 bolts
- Common bolts are no longer common for current structural design but are still available

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Slip Critical

Bearing Type

- High strength bolts (HSB) are available as ASTM A 325 and ASTM A490

Courtesy of Kao Wang Screw Co., Ltd.

- Advantages of HSB over A307 bolts
- Fewer bolts will be used compared to 307 è cheaper connection!
- Smaller workman force required compared to 307
- Higher fatigue strength
- Ease of bolt removal è changing connection

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- Snug tight
- All plies of the connection are in firm contact to each other: No pretension is used.
- Easer to install and to inspect

- Pre-tensioned
- Bolts are first brought to snug tight status
- Bolts are then tensioned to 70% of their tensile stresses

Courtesy of www.halfpricesurplus.com

- Bolts are tensioned using direct tension indicator, calibrated wrench or other methods (see AISC)

- 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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Type

Type N Thread

Type X Thread

A325

330

413

A490

413

517

- The shear strength of bolts shall be determined as follows

AISC Table J3.2

The table bellow shows the values of fv (MPa) for different types of bolts

- If the level of threads is not known, it is conservative to assume that the threads are type N.

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- 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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- Bolt Shearing
- Tension Fracture
- Plate Bearing
- Block Shear

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P

P

P

P

P

P

P/2

P/2

P

P

P/2

P/2

Shear, bearing, bending

Bearing and single plane Shear

Lap Joint

Bending

Bearing and double plane Shear

Butt Joint

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Single shear

Double shear

- Possible failure modes
- Failure of bolts: single or double shear
- 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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- 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.

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- 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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- 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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- The bearing strength is independent of the bolt material as it is failure in the connected metal
- The other possible common failure is shear end failure known as “shear tear-out” at the connection end

Shear limitation

Bearing limitation

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- The AISC code gives guidance for edge distance and spacing to avoid tear out shear

AISC Table J3.4

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
- The AISC minimum spacing is

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- 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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- 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

- 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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- 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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Lc – Clear distance

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- In a simple connection, all bolts share the load equally.

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- In a bolted shear connection, the bolts are subjected to shear and the connecting/connected plates are subjected to bearing stresses.

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- 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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- 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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- 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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- 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.

- 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

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p

p

Edge

distance

p

g

- 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 direction perpendicular to
member’s axis

- Pitch – ...parallel to member’s axis
- Edge Distance – Distance from
center of bolt to adjacent

edge of a member

p

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- 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

120x15 mm

30 mm

60 mm

63 k

300 kN

30 mm

20 mm A325-N bolts

30 mm

60 mm

30 mm

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- Step I. Shear strength of bolts
- The design shear strength of one bolt in shear = Fn Ab = 0.75 x 330 x p 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

- The design shear strength of one bolt in shear = Fn Ab = 0.75 x 330 x p x 202/4000 = 77.8 kN

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- 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.

- See Table J3.4M, minimum edge distance = 26 mm for rolled edges of plates

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- 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

- Bearing strength at bolt holes in connected part (120x15 mm plate)

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- Therefore, Rn = 216 kN at other holes
- Therefore, bearing strength at holes = 2 x 103.7 + 2 x 207.4 = 622.2 kN

- 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.

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- 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

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Connection strength (fRn) > applied factored loads (gQ).

311.2 > 300Therefore ok.

- Only connections is designed here
- Need to design tension member and gusset plate

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P

P

P

P

Pe

Pe

e

e

Eccentricity in the plane of the faying surface

Direct Shear + Additional Shear due to moment Pe

CG

CG

Eccentricity normal to the plane of the faying surface

Direct Shear + Tension and Compression (above and below neutral axis)

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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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P

P

e

Pe

r3

r1

d1

d3

P/3

P/3

P/3

CG

CG

d2

r2

- Elastic Method
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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MCG = Pe = r1d1 + r2d2 + r3d3

Since the force on each bolt is proportional to its distance from the CG:

Substitute into eqn. for MCG:

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H1

r1

d1

y1

V1

x1

CG

- Total Forces in Bolt i:
- Horizontal Component =
- Vertical Component =

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CG

Determine the force in the most stressed bolt of the group using elastic method

P=140 kN

e

125 mm

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)

100 mm

100 mm

100 mm

100 mm

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Additional Shear in the upper and lower right-hand bolts due to moment M = Pe = 140x175 = 24500 kN.mm:

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

17.5 kN

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2rut

- Eccentricity Normal to Plane of Faying Surface
(a) Neutral Axis at CG

Shear force per bolt due to concentric force Pu

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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tf

CG (tension group)

d=Depth/6

Depth

y

X

X

beff

- Eccentricity Normal to Plane of Faying Surface
(b) Neutral Axis Not at CG

Shear per bolt due to concentric 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, S-shapes, welded plates and angles)

Bolts above NA resist tension

Bearing stress below NA resist compression

2rut

NA

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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 = beffx 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

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- 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:
- = 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

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537 kN

1200 kN

1073 kN

Eight 22 mm

A325X bolts

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)

Tension stress per bolt:

ft = Tu/NbAb=1073000/(8x380)= 353 MPa

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)

1

2

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- 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

P

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Welding Process – Fillet Weld

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- 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.

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- 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.

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- 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.

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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.

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- Classification of welds
- According to type of weld
- According to weld position
- According to type of joint
- Butt, lap, tee, edge or corner

- According to the weld process
- SMAW, SAW

Groove weld

Fillet weld

Flat, Horizontal, vertical or overhead weld

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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

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- 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.

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hypotenuse

root

L – length of the weld

a – size of the weld

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- Shear stress in fillet weld of length L subjected to load P
= fv = If the ultimate shear strength of the weld = fw

- Rn =
- Rn = 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.

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- The two digits "XX" denote the type of coating.

- 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.

- E – electrode70 – tensile strength of electrode (ksi) = 482 MPa
- XX – type of coating

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Concave Surface

Convex Surface

Leg

Leg

Leg

Leg

Throat

Throat

- Stronger in tension and compression than in shear
- 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.

Unequal leg fillet weld

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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

Design Strength:

For equal leg fillet weld:

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- 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.

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- For example

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.

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- 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 aotherwise, aeff = Lw / 4 a = weld size
- Read J2.2 b page 16.1-95
- Intermittent fillet welds:Lw-min = 4 a and 38 mm.

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- 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

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- 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

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- Two types of fillet welds can be used
- Shielded Metal Arc Welding (SMAW)
- Automatic Submerged Arc Welding (SAW)

Shear failure plane

AISC – Section J2.2

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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 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

12

75@125

6

200

10

50@150

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Electrode

FEXX (MPa)

E70XX

482

E80XX

551

- Fillet weld design can be governed by the smaller value of
- Weld material strength
- Base Metal Strength

&

Yield Limit State

AISC Table J2.5

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- The weld strength will increase if the force is not parallel to the weld

&

- Maximum weld size

- Minimum weld size

AISC Table J2.4

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Weld governs

Base metal governs

- The weld strength is a function of the angle q

Strength

w = weld size

Angle (q)

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- 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

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- 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 mmand38 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!

- tmin = 10 mm (member)

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- 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!

- 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

- Rn = 0.9 x 344 x 100 x 10/1000 = 310 kN- tension yield

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- 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.

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- 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

- The shear stress in weld due to torsion moment M is

- 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

AISC Manual Part 8

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- stresses due to torsional moment “M” is

- Calculation shall be done for teff

- Or for teff = 1 mm

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- Forces due to direct applied force is

- Total stress in the weld is

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250 mm

- 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.

D = 50 kN

L = 120 kN

300 mm

15 mm PL

200 mm

Calculations are done for teff = 25 mm

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- 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:
- Step III: Compute the location of the centroid:
- Step IV: Compute the torsional moment:
e = 250+ 200 – 57.1 = 392.9 M = Pe = 252(392.9)=99011 kN-mm.

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- 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:

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- 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, assumingFw = 0.6FEXX
Use 12 mm

Answer: Use a 12-mm fillet weld, E70 electrode.

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- stresses due to torsion moment “M” is

- Calculation shall be done for teff

- Or for teff = 1 mm

- 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

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- 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?

152 mm

20 mm

20 mm

82 mm

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- 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

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- Step III: Calculate the factored-load reaction is:

- Step IV: The required weld size a
- a = 943/(0.9x0.707x0.6x482) = 6.2 mm

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- 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).

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Load

90o

30o

0o

Elastic analysis

Deformation

- When comparing elastic analysis to experimental on eccentric welded connections, it becomes obvious that elastic analysis is over conservative.

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- Similar to bolts, weld can be divided into segments which rotate about an instantaneous centre (IC)
- Instead of summing the forces we can integrate over the length of the weld to get the basic equations of equilibrium:

Thus

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- However, in weld: The force in each segment “R” is also function of the angle q between the force direction and the weld.

Deformation of the segment

Deformation of the segment at max stress

- Similar to bolts, the far weld element might have a higher proportion of force.

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However, the critical weld is that of the smallest Dm/rs

Determine the segment that has

The ultimate deformation Du happens for the segment with smallest Dm/rs

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In all equations “q” is in radian ranges from zero to p/2

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- Thus to estimate the force in the critical segment we do the following steps:
- 1- Divide the weld into segments and assume an IC
- 2- Calculate the deformation of each element
- 3- Compute the ratio Dm/r and determine rcrit
- 4- For this critical segment compute the ultimate deformation Du
- 5- Compute the deformation of each other segment

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≡

Eqn (1)

≡

Eqn (2)

≡

Eqn (3)

- Steps continued:
- 6- Compute the stress in each segment
- 7- Check equilibrium equations

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- 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.

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- 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.

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Loads to be transferredFrictional 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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- 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 holeshsc = 1.0

For oversized and short-slotted holeshsc = 0.85

For long-slotted holeshsc = 0.7

Ns - number of slip planes

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Slip Coefficients (LRFD J3.8)

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- When the applied shear force exceeds the Rn value stated above, slip will occur in the connection.

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- 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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- 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)

- Rn of one fully-tensioned slip-critical bolt = (1.13 hscTb Ns)

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- 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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- Step III. Layout of splice connection
- Flange-plate splice connection

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- 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
- Design edge distance Le = 40 mm.

- Minimum spacing = s = (2+2/3) db = 2.67 x 20 = 53.4 mm.(Spec. J3.3)
- 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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- 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 OK

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- 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.

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- Experimental study by Crawford and Kulak (1971) showed:

- The load-deformation relationship of any bolt is non-linear

AISC Manual Part 7

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- 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

≡

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Measured at the elastic centroid

- At failure

≡

Eqn (1)

≡

Eqn (2)

≡

Eqn (3)

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- 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´

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Pu

e

e’

R1

1

2

d1

d2

R2

CG

IC

d4

d3

3

4

R3

R4

- Ultimate Strength Method (Instantaneous Center of Rotation Method)
R = Rult(1 – e-0.394)0.55

- R = Nominal shear strength of 1 bolt at a deformation , k
- Rult= Ultimate shear strength of 1 bolt, kN
- = 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

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- Trial and error:
- Assume e’
- Compute i = dimax/dmax(maxis assumed for bolt at farthest distance from IC)
- Compute Ri=Rult(1- e-0.394i)0.55
- Check for: Pu=( Rd)/(e’+e)
- If not satisfied, repeat with another e’

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R1

1

2

R2

CG

IC

3

4

R3

R4

Determine the largest eccentric force Pufor which the design shear strength of the bolts in the connection is adequate using the IC method. Use bearing-type 20 mm A325X bolts

Pu

e = 100 mm

e’=60 mm

- - Design shear strength per bolt (Ex. 7-1)
- Ru = Rn= 77.8 kN
- After several trials, assume e’= 60 mm. Bolts 2 and 4 are furthest from the IC, therefore 2 = 4 = max = 8.6 mm
- Compute iand Ri in tabulated form:

75

mm

d1

d2

75

mm

d4

d3

75 mm

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Check:

Pu= (Rd)/(e’+e) = (26554/(60+100))

= 166 kN ~ Ry = 165.14 kN (OK)

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