68402: Structural Design of Buildings II

68402: Structural Design of Buildings II
Design of Connections Monther Dwaikat Assistant Professor Department of Building Engineering An-Najah National University 68402/61420

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 68402/61420

Eccentric Connections
Types of Connections Simple Connections Eccentric Connections Bolted Connections Welded Connections Common Bolts High Strength Bolts Filet Weld Slip Critical Groove Weld Bearing Type 68402/61420

Eccentric Connections
Types of Connections Simple Connections Eccentric Connections Bolted Connections Welded Connections Elastic Analysis Ultimate Analysis Moment Resisting Elastic Analysis Ultimate Analysis Moment Resisting 68402/61420

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 68402/61420

Tension member Connection/ splice P P Beam end Simple shear connection 68402/61420

Hanger connection (Tension) Moment resisting connection 68402/61420

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. 68402/61420

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 A449 - High strength bolts with diameter > 1 ½”, anchor bolts, lifting hooks, tie-downs 68402/61420

Common Bolts ASTM A307 bolts
Common bolts are no longer common for current structural design but are still available 68402/61420

High Strength Bolts High strength bolts (HSB) are available as ASTM A 325 and ASTM A490 Slip Critical Bearing Type 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 68402/61420

High Strength Bolts Snug tight Pre-tensioned Slip critical
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 Bolts are tensioned using direct tension indicator, calibrated wrench or other methods (see AISC) Slip critical Bolts are pre-tensioned but surfaces shall be treated to develop specific friction. The main difference is in design, not installation. Load must be limited not to exceed friction capacity of the connection (Strength Vs. Serviceability!) Necessary when no slip is needed to prevent failure due to fatigue in bridges. 68402/61420

HSB – Bearing Type Connections
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 517 413 A490 330 A325 Type X Thread Type N Thread Type If the level of threads is not known, it is conservative to assume that the threads are type N. 68402/61420

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 68402/61420

Failure Modes of Bolted Connections
Bolt Shearing Tension Fracture Plate Bearing Block Shear 68402/61420

Actions on Bolt Lap Joint Butt Joint Shear, bearing, bending P
Bearing and single plane Shear Lap Joint P Bending Butt Joint Bearing and double plane Shear P/2 P 68402/61420

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 Single shear Double shear 68402/61420

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. 68402/61420

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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 68402/61420

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 68402/61420

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 68402/61420

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Spacing and Edge-distance requirements
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 68402/61420

Bolt Spacings & Edge Distances
- 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 68402/61420

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 68402/61420

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 68402/61420

Important Notes Lc – Clear distance 68402/61420

Design Provisions for Bolted Shear Connections
In a simple connection, all bolts share the load equally. 68402/61420

In a bolted shear connection, the bolts are subjected to shear and the connecting/connected plates are subjected to bearing stresses. 68402/61420

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. hjhkjkjolklktytvtvguu 68402/61420

AISC Design Provisions
Chapter J of the AISC Specifications focuses on connections. Section J3 focuses on bolts and threaded parts AISC Specification J3.3 indicates that the minimum distance (s) between the centers of bolt holes is 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. 68402/61420

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. 68402/61420

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 68402/61420

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 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 g p Edge distance p 68402/61420

Ex 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 120x15 mm 30 mm 60 mm 63 k 300 kN 30 mm 20 mm A325-N bolts 30 mm 60 mm 30 mm 68402/61420

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 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 = kN 68402/61420

Ex 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. 68402/61420

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 – ( )/2 = 19.2 Rn = 0.75 x (1.2 Lc t Fu) = 0.75 x (1.2 x19.2 x15x400)/1000 = kN But, Rn ≤ 0.75 (2.4 db t Fu) = 0.75 x (2.4 x 20x15x400)/1000 = 216 kN Therefore, Rn = kN at edge holes. At other holes, s = 60 mm, Lc = 60 – ( ) = 38.4 mm. Rn = 0.75 x (1.2 Lc t Fu) = 0.75x(1.2 x 38.4 x15 x400)/1000 = kN But, Rn ≤ 0.75 (2.4 db t Fu) = 216 kN. Therefore Rn = kN 68402/61420

Ex Design Strength Therefore, Rn = 216 kN at other holes Therefore, bearing strength at holes = 2 x x = kN Bearing strength at bolt holes in gusset plate (10 mm plate) At edges, Lc = 30 – hole diameter/2 = 30 – ( )/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. 68402/61420

Ex Design Strength At other holes, s = 60 mm, Lc = 60 – ( ) = 38.4 mm. Rn = 0.75 x (1.2 Lc t Fu) = 0.75 x (1.2 x 38.4 x 10x 400)/1000 = kN But, Rn ≤ 0.75 (2.4 db t Fu) = 144 kN Therefore, Rn = kN at other holes Therefore, bearing strength at holes = 2 x x = kN Bearing strength of the connection is the smaller of the bearing strengths = kN 68402/61420

Ex. 6.1 - Design Strength Connection Strength Shear strength = 311.2
Bearing strength (plate) = kN Bearing strength (gusset) = kN Connection strength (fRn) > applied factored loads (gQ). 311.2 > Therefore ok. Only connections is designed here Need to design tension member and gusset plate 68402/61420

Eccentrically-Loaded Bolted Connections
P Pe e P Pe e CG CG Eccentricity in the plane of the faying surface Direct Shear + Additional Shear due to moment Pe Eccentricity normal to the plane of the faying surface Direct Shear + Tension and Compression (above and below neutral axis) 68402/61420

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 68402/61420

Forces on Eccentrically-Loaded Bolts with Eccentricity on the Faying Surface
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 CG e P CG Pe P P/3 d1 d2 d3 r1 r2 r3 68402/61420

MCG = Pe = r1d1 + r2d2 + r3d3 Since the force on each bolt is proportional to its distance from the CG: Substitute into eqn. for MCG: 68402/61420

CG d1 r1 x1 y1 H1 V1  Total Forces in Bolt i: Horizontal Component = Vertical Component = 68402/61420

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 CG 125 mm Eccentricity wrt CG: e = = 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 68402/61420

Ex. 6.3 – Eccentric Connections – Elastic Method
Additional Shear in the upper and lower right-hand bolts due to moment M = Pe = 140x175 = 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 68402/61420

Eccentricity Normal to Plane of Faying Surface (a) Neutral Axis at CG 2rut 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 68402/61420

Eccentricity Normal to Plane of Faying Surface (b) Neutral Axis Not at CG Bolts above NA resist tension Bearing stress below NA resist compression X tf beff Depth d=Depth/6 y CG (tension group) X 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) 2rut NA 68402/61420

Check location of NA by equating the moment of the bolt area above the NA with the moment of the compression block area below the NA: Ab x y = beff x d x d/2 Ab = sum of areas of bolts above the NA y = distance from X-X to the CG of bolts above NA d = depth of compression block (adjust until satisfy) Once the NA has been located, the tensile force per bolt: rut = (PuecAb)/Ix c = distance from NA to most remote bolt in group Ix = combined moment of inertia of bolt group and compression block about NA 68402/61420

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:  = 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 68402/61420

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)= MPa Fnv=(0.75)(413)=310 MPa> fv = MPa (OK) Tension stress per bolt: ft = Tu/NbAb= /(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) Eight 22 mm A325X bolts 1200 kN 537 kN 1073 kN 1 2 68402/61420

Simple Welded Connections
Structural welding is a process by which the parts that are to be connected are heated and fused, with supplementary molten metal at the joint. A relatively small depth of material will become molten, and upon cooling, the structural steel and weld metal will act as one continuous part where they are joined. P P P P 68402/61420

Introductory Concepts
Welding Process – Fillet Weld 68402/61420

The additional metal is deposited from a special electrode, which is part of the electric circuit that includes the connected part. In the shielded metal arc welding (SMAW) process, current arcs across a gap between the electrode and the base metal, heating the connected parts and depositing part of the electrode into the molten base metal. A special coating on the electrode vaporizes and forms a protective gaseous shield, preventing the molten weld metal from oxidizing before it solidifies. The electrode is moved across the joint, and a weld bead is deposited, its size depending on the rate of travel of the electrode. 68402/61420

As the weld cools, impurities rise to the surface, forming a coating called slag that must be removed before the member is painted or another pass is made with the electrode. Shielded metal arc welding is usually done manually and is the process universally used for field welds. For shop welding, an automatic or semi automatic process is usually used. Foremost among these is the submerged arc welding (SAW), In this process, the end of the electrode and the arc are submerged in a granular flux that melts and forms a gaseous shield. There is more penetration into the base metal than with shielded metal arc welding, and higher strength results. 68402/61420

Other commonly used processes for shop welding are gas shielded metal arc, flux cored arc, and electro-slag welding. Quality control of welded connections is particularly difficult, because defects below the surface, or even minor flaws at the surface, will escape visual detection. Welders must be properly certified, and for critical work, special inspection techniques such as radiography or ultrasonic testing must be used. 68402/61420

The two most common types of welds are the fillet weld and the groove weld. Fillet weld examples: lap joint – fillet welds placed in the corner formed by two plates Tee joint – fillet welds placed at the intersection of two plates. Groove welds – deposited in a gap or groove between two parts to be connected e.g., butt, tee, and corner joints with beveled (prepared) edges Partial penetration groove welds can be made from one or both sides with or without edge preparation. 68402/61420

Welded Connections 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 68402/61420

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Weld Limit States The only limit state of the weld metal in a connection is that of fracture Yielding is not a factor since any deformation that might take place will occur over such a short distance that it will not influence the performance of the structure 68402/61420

Design of Welded Connections
Fillet welds are most common and used in all structures. Weld sizes are specified in 1 mm increments A fillet weld can be loaded in any direction in shear, compression, or tension. However, it always fails in shear. The shear failure of the fillet weld occurs along a plane through the throat of the weld, as shown in the Figure below. 68402/61420

hypotenuse root L – length of the weld a – size of the weld 68402/61420

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. 68402/61420

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  MPa, use E70XX electrode. If yield stress (y) of the base metal is  MPa, use E80XX electrode. E70XX is the most popular electrode used for fillet welds made by the SMAW method. E – electrode 70 – tensile strength of electrode (ksi) = 482 MPa XX – type of coating 68402/61420

Fillet Weld 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. Leg Throat Concave Surface Leg Throat Convex Surface Unequal leg fillet weld 68402/61420

Stress in fillet weld = factored load/eff. throat area
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 Design Strength: For equal leg fillet weld: 68402/61420

Table J2.5 in the AISC Specifications gives the weld design strength fw = 0.60 FEXX For E70XX, fw = 0.75 x 0.60 x 482 = 217 MPa Additionally, the shear strength of the base metal must also be considered: Rn = 0.9 x 0.6 Fy x area of base metal subjected to shear where, Fy is the yield strength of the base metal. 68402/61420

For example Strength of weld in shear = 0.75 x x a x Lw x fw In weld design problems it is advantageous to work with strength per unit length of the weld or base metal. 68402/61420

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 Intermittent fillet welds: Lw-min = 4 a and 38 mm. 68402/61420

Limitations on Weld Size – AISC Specifications J2.2b Page 16.1-95
The minimum length of fillet weld may not be less than 4 x the weld leg size. If it is, the effective weld size must be reduced to ¼ of the weld length The maximum size of a fillet weld along edges of material less than 6 mm thick equals the material thickness. For material thicker than 6 mm, the maximum size may not exceed the material thickness less 2 mm. (to prevent melting of base material) The minimum weld size of fillet welds and minimum effective throat thickness for partial-penetration groove welds are given in LRFD Tables J2.4 and J2.3 based on the thickness of the base materials (to ensure fusion and minimize distortion) Minimum end return of fillet weld  2 x weld size 68402/61420

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 – Lw/a) If Lw > 300 a, the effective weld length (Lw-eff) = 0.6 Lw Weld Terminations - read AISC J2.2b Lap joint – fillet welds terminate at a distance > a from edge. Weld returns around corners must be > 2 a 68402/61420

Guidelines for Fillet Weld design
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 68402/61420

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 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 6 200 10 68402/61420

Fillet weld design can be governed by the smaller value of Weld material strength Base Metal Strength 551 E80XX 482 E70XX FEXX (MPa) Electrode & AISC Table J2.5 Yield Limit State 68402/61420

The weld strength will increase if the force is not parallel to the weld & Maximum weld size Minimum weld size AISC Table J2.4 68402/61420

Capacity of Fillet Weld
The weld strength is a function of the angle q Weld governs Base metal governs Strength w = weld size Angle (q) 68402/61420

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 68402/61420

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 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! 68402/61420

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Length/weld size = 125/6 = 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 x a x 0.60 x FEXX x Lw = 0.75 x 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 hjhkjkjolklktytvtvguu 68402/61420

Rn = 0.75 x Ae x tension fracture Ae = U A Ae = Ag = 100 x 10 = 1000 mm Therefore, Rn = 336 kN The design strength of the member-connection system = 230 kN. Weld strength governs. The end returns at the corners were not included in the calculations. 68402/61420

Elastic Analysis of Eccentric Welded Connections
It is assumed here that the rotation of the weld at failure occur around the elastic centre (EC) of the weld. The only difference from bolts is we are dealing with unit length of weld instead of a bolt 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 68402/61420

Elastic Analysis of Eccentric Welded Connections – Shear & Torsion
stresses due to torsional moment “M” is - Calculation shall be done for teff - Or for teff = 1 mm 68402/61420

Elastic Analysis of Eccentric Welded Connections – Shear & Torsion
Forces due to direct applied force is Total stress in the weld is 68402/61420

Ex. 7.7 – Design Strength of Welded Connection – Shear and Torsion
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 68402/61420

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 = – 57.1 =  M = Pe = 252(392.9)=99011 kN-mm. 68402/61420

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)( )2 }+ 300(57.1)2=3.05×106 mm4 J = Ix + Iy = ( )×106 = 14.3×106 mm4 Step VI: Compute stresses at critical location: 68402/61420

Step VII: Check the shear strength of the base metal The shear yield strength of the angle leg is: ΦRn = (0.9)0.6Fyt = 0.9(0.6)(248)(15) = 2009 N/mm The base metal shear strength is therefore: 2009 N/mm > 1703 N/mm (OK). Step VIII: Calculate the weld size, assuming Fw = 0.6FEXX  Use 12 mm Answer: Use a 12-mm fillet weld, E70 electrode. 68402/61420

Elastic Analysis of Eccentric Welded Connections – Shear & Tension
68402/61420

Elastic Analysis of Eccentric Welded Connections – Shear & Tension
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 68402/61420

Ex. 7.8 – Design Strength of Welded Connection – Shear & Tension
An L6x4x1/2 is used in a seated beam connection, as shown in the figure. It must support a service load reaction of 25 kN dead load and a 50 kN live load. The angles are A36 and the columns in A992. E70XX electrodes are to be used. What size fillet weld are required for the connection to the column flange? 152 mm 20 mm 20 mm 82 mm 68402/61420

Step I: calculate the eccentricity of the reaction with respect to the weld is: e = /2 = 61 mm Step II: Calculate the moment of inertia for the weld configuration: I = 2(1)(152)3 / 12 = mm4 c = 152/2 = 76 mm Step III: Calculate the factored-load reaction is: Pu = 1.2D + 1.6L = 1.2(25)+1.6(50) = 110 kN Mu = Pue = 110(61) = 6710 kN-mm 68402/61420

Step III: Calculate the factored-load reaction is: Step IV: The required weld size a a = 943/(0.9x0.707x0.6x482) = 6.2 mm 68402/61420

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 = = 11 mm Try a = 7 mm Step VI: Check the shear capacity of the base metal (the angle controls): Applied direct shear = fv = 362 N/mm The shear yield strength of the angle leg is: ΦRn = 0.9×0.6Fyt = (0.9)0.6(248)(13) = 1741 N/mm The base metal shear strength is therefore: 1741 N/mm > 362 N/mm (OK). 68402/61420

Ultimate Strength Analysis of Eccentric Welded Connections
When comparing elastic analysis to experimental on eccentric welded connections, it becomes obvious that elastic analysis is over conservative. Deformation Load 30o 90o 0o Elastic analysis 68402/61420

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 68402/61420

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. 68402/61420

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 68402/61420

In all equations “q” is in radian ranges from zero to p/2 68402/61420

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 68402/61420

Steps continued: 6- Compute the stress in each segment 7- Check equilibrium equations ≡ Eqn (1) Eqn (2) Eqn (3) 68402/61420

Extra Slides 68402/61420

Slip-critical Bolted Connections
High strength (A325 and A490) bolts can be installed with such a degree of tightness that they are subject to large tensile forces. These large tensile forces in the bolt clamp the connected plates together. The shear force applied to such a tightened connection will be resisted by friction as shown in the Figure below. 68402/61420

68402/61420

Thus, slip-critical bolted connections can be designed to resist the applied shear forces using friction. If the applied shear force is less than the friction that develops between the two surfaces, then no slip will occur between them. However, slip will occur when the friction force is less than the applied shear force. After slip occurs, the connection will behave similar to the bearing-type bolted connections designed earlier. Table J3.1 summarizes the minimum bolt tension that must be applied to develop a slip-critical connection. 68402/61420

Slip-Critical Connections
Loads to be transferred  Frictional Resistance (tension force in bolt x coefficient of friction )  No slippage between members  No bearing and shear stresses in bolt LRFD J3.10 requires bearing strength to be checked for both Bearing-Type connections and Slip-Critical connections (even though there is supposed to be little or no bearing stresses on the bolts in Slip-Critical connections) 68402/61420

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)  for factored loads & 1.00 for service loads  - friction coefficient Tb - minimum bolt tension given in Table J3.1 hsc – hole factor determined as: For standrad size holes hsc = 1.0 For oversized and short-slotted holes hsc = 0.85 For long-slotted holes hsc = 0.7 Ns - number of slip planes 68402/61420

Slip-Critical Connections
Slip Coefficients (LRFD J3.8) Surface  Class A (unpainted clean mill scale or surfaces with class A coating on blast-cleaned steel) Class B (unpainted blast-cleaned surfaces or surfaces with Class B coating on blast-cleaned steel 0.35 0.50 68402/61420

When the applied shear force exceeds the Rn value stated above, slip will occur in the connection. The final strength of the connection will depend on the shear strength of the bolts and on the bearing strength of the bolts. This is the same strength as that of a bearing type connection. Slip critical connections shall still be checked as bearing type in case slip occurs as a result of overload. 68402/61420

Ex. 6.2 - Slip-critical Connections
Design a slip-critical splice for a tension member subjected to 600 kN of tension loading. The tension member is a W8 x 28 section made from A36 material. The unfactored dead load is equal to 100 kN and the unfactored live load is equal to 300 kN. Use A325 bolts. The splice should be slip-critical at service loads. 68402/61420

Step I. Service and factored loads Service Load = D + L = 400 kN. Factored design load = 1.2 D L = 600 kN Tension member is W8 x 28 section made from A36 steel. The tension splice must be slip critical (i.e., it must not slip) at service loads. Step II. Slip-critical splice connection (service load) Rn of one fully-tensioned slip-critical bolt = (1.13 hscTb Ns) (See Spec. J3.8) 68402/61420

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 68402/61420

Step III. Layout of splice connection Flange-plate splice connection 68402/61420

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 68402/61420

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 = 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) = kN (see Example 7.1) Bearing strength of bolt holes in flanges of wide flange section = 4 x x = kN > 600 kN OK 68402/61420

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 ( ) x 10 > 1000 mm2 Beam flange width = 166 mm Assume plate width 160 mm x 10 mm which has Ag = 1660 mm2 Step VI. Check member strength Student on his/her own. 68402/61420

Ultimate Strength Analysis of Eccentric Bolted Connections
Experimental study by Crawford and Kulak (1971) showed: - The load-deformation relationship of any bolt is non-linear AISC Manual Part 7 68402/61420

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 ≡ 68402/61420

Measured at the elastic centroid At failure ≡ Eqn (1) ≡ Eqn (2) ≡ Eqn (3) 68402/61420

Therefore, getting the maximum force in the farthest bolt requires determining the unknown “e´” Because of the non-linear relationship, e´ can be determined by trial and error A spreadsheet can be used to determine e´ 68402/61420

Ultimate Strength Method (Instantaneous Center of Rotation Method) R = Rult(1 – e-0.394)0.55 CG IC 1 2 3 4 e e’ Pu d1 d2 d3 d4 R1 R2 R4 R3 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 68402/61420

Ultimate Strength Method (Instantaneous Center of Rotation Method)
Trial and error: Assume e’ Compute i = dimax/dmax (max is assumed for bolt at farthest distance from IC) Compute Ri=Rult(1- e-0.394i)0.55 Check for: Pu=( Rd)/(e’+e) If not satisfied, repeat with another e’ 68402/61420

Ex. 6.4 – Eccentric Connections – Ultimate Method
Determine the largest eccentric force Pu for which the design shear strength of the bolts in the connection is adequate using the IC method. Use bearing-type 20 mm A325X bolts 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 i and Ri in tabulated form: CG IC 1 2 3 4 R1 R2 R4 R3 75 mm d1 d2 75 mm d4 d3 75 mm 68402/61420

Ex. 6.4 – Eccentric Connections – Ultimate Method
Bolt # h (mm) v (mm) d (mm)  (mm) R (kN) Ry (kN) Rd (kN.mm) 1 22.5 75 78.3 5.47 72.7 20.9 5692 2 97.5 123 8.6 77.8 61.67 7585 3 4  =  = 26554 Check: Pu= (Rd)/(e’+e) = (26554/(60+100)) = 166 kN ~ Ry = kN (OK) 68402/61420

68402: Structural Design of Buildings II

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