Figure 8.1 Braced frame connection showing bolts, welds, and connected elements

Figure 8.1 Braced frame connection showing bolts, welds, and connected elements

Bolts, welds, and plates, like those shown in Figure 8.1, hold it all together. They transfer load between beams, columns, braces, chords, webs, and foundations to complete a structure. They allow individually fabricated pieces of steel to soar to the sky or span great lengths. In this chapter, we will learn the fundamentals of bolts, welds, and connected elements; looking at their properties, strength, detailing, and installation.

Regardless of the connector type, we need to ensure a continuous load path between the members being connected. Recalling Figure 3.5, we can look at a connection like a chain. Each time load moves from one element to another, there must be sufficient capacity. In the connection in Figure 8.1, the force moves from the beam through the weld into the gusset, then into the bolt, into the brace connection with the help of another weld, and finally the brace body. The analysis can be tedious, but necessary for a functional connection.

8.1 Connector Types

8.1.1 Bolts

In the first steel structures, connections were made of rivets, like those in Figure 8.2. Workers installed them red hot, and pounded them into shape, filling the hole. As the rivets cooled and contracted, they clamped the surfaces tightly together. Load transfer occurred by bearing and friction. As rivets were time consuming and dangerous to install, in the 1950s high-strength bolts found their way into use, and have completely replaced rivets.

Figure 8.2 Riveted connection in an industrial structure

Figure 8.2 Riveted connection in an industrial structure

There are two major classes of high-strength bolted connections—bearing and slip critical. Bearing bolts transfer force by contact between the plate and bolt, illustrated in Figure 8.3a. In slip critical joints, load is transferred through friction between the plates, with the bolt applying the clamping (normal) force (Figure 8.3b). Bearing bolts are common in smaller structures. Slip critical bolts are required in taller structures (over 125 ft, 38 m), bolted seismic systems, and those subject to fatigue and vibration.

Today, bolts are divided into three main divisions: Groups A and B, and A307. Groups A and B are high-strength bolts for structural connections. A307 bolts are what we commonly get at the hardware store, and are found in light connections like girts and handrails. Group A bolts include A325 and F1852. They have similar strength to SME J4291 Grade 5 bolts. Group B bolts include A490 and F2280, and have similar strength as Grade 8 bolts. Availability of different bolt materials and their use are shown in Table 2.6. Bolts are identified by markings on their heads, like those in Figure 8.4. Their use is governed by the Specification for Structural Joints using High-Strength Bolts.2

Figure 8.3 Force transfer in (a) bearing-type bolts, and (b) slip critical bolts

Figure 8.3 Force transfer in (a) bearing-type bolts, and (b) slip critical bolts

8.1.2 Welds

The earliest record of welding dates to 5500 BC, where the Egyptians made copper pipe from sheets by overlapping the edges and hammering them together. This is known as forge welding, and was utilized to create the intricate patterns in Damascus steel. Brazing became widespread by 3000 BC, and utilizes a filler metal with a lower melting point than the material it is joining. It is still used widely today to join pipes and manifolds, where high structural strength is not required.

Arc welding development began independently in the early 1800s by Sir Humphry Davy and Vasily Petrov. Loosely paralleling advances in electrical generation, arc welding development really took off in the last two decades of the 19th century and early 20th century. By World War I (1914), welding was used extensively in ship repair. In 1924, the first all-welded steel building was built in the US. Today welding is widespread in commercial, industrial, and bridge structures throughout the world.

Figure 8.4 Bolt heads showing identifying marks

Figure 8.4 Bolt heads showing identifying marks

Along with the development of welding came the need for standards to ensure the constructed product was properly built. The American Welding Society produced the first structural welding code in 1928. Since that time, they have developed the codes in the USA governing welding from reinforcing steel to aerospace components. The key welding codes for building and bridge structures are:

  • D1.1 Structural Welding Code—Steel3
  • D1.4 Structural welding (reinforcing steel)4
  • D1.5 Bridge Welding Code5
  • D1.8 Structural Welding Code—Seismic.6

Many arc welding processes exist today, yet they share common elements illustrated in Figure 8.5.

These are:

  • Base Metals—to be joined
  • Filler Metals—to fuse with the base metals into a solid weldment
  • Electric Circuit—to supply the electrical energy needed create an electric arc to melt the metals
  • Electrodes—to focus the electrical energy into the base metals
  • Shielding Media—flux and/or shielding gases to protect the molten metal from atmospheric contamination
  • Welding Power Sources—to create or rectify current and precisely control the voltage and the amperage flowing through the welding circuit.

Figure 8.5 Required elements for arc welding

Figure 8.5 Required elements for arc welding

Common welding methods today include:

  • Shielded Metal Arc Welding (SMAW)—commonly called “stick” welding, utilizes a flux-coated metal wire that acts both as an electrode and as the filler metal (Figure 8.5). The flux melts with the wire in the heat of the welding arc. The shielding gas and slag that are created protect the weld. Additional alloying elements are added to the flux to provide specific operating and metallurgical characteristics. It is the most versatile and accessible arc welding process.
  • Submerged Arc Welding (SAW)—employs a continuous wire electrode that melts in the arc beneath a bed of granular flux. SAW is a high-deposition, automatic welding process for welding thick sections, Figure 8.5.
    Figure 8.6 Submerged arc weld

    Figure 8.6 Submerged arc weld

  • Gas–Metal Arc Welding (GMAW)—consumes a continuous wire electrode protected by shielding gas only, shown in Figure 8.7. Some wires are tubular and contain additional alloying elements to improve operating and metallurgical characteristics. This process is very versatile and commonly used for welded fabrication.
    Figure 8.7 Gas–metal arc weld

    Figure 8.7 Gas–metal arc weld

  • Flux-Cored Arc Welding (FCAW)—consists of a tubular, continuous wire electrode. Flux is formed into the tube creating shielding gas and slag in the welding arc, illustrated in Figure 8.8. It is like a “stick” electrode turned inside out. Some variations incorporate a secondary shielding gas for precision arc control and increased toughness.
  • Electroslag Arc Welding (ESG)—uses highly specialized welding equipment and copper cooling “shoes”, schematically shown in Figure 8.9. The process deposits large amounts of weld metal, joining base metals as thick as 4 in (100 mm) in a single pass. The cooling shoes pull the massive amount of heat out of the connection, mold the weld bead, and maintain the dimensional integrity of the connection.

Weldability of materials is fundamentally important to high-quality welds. Common structural steels, such as A36, A572, and A992 are all weldable. However, historic steels, high-strength low alloy, and reinforcing steels require more consideration. An effective way to estimate weldability of steel is to determine the carbon equivalentCE of the alloy. There are several equations used to determine the CE of a steel, depending upon the product form. Steel mills calculate the CE of common structural shapes and plate using Equation 8.1.

C E = C + ( M n + S i 6 ) + ( C r + M o + V 5 ) + ( N i + C u 15 ) (8.1)

Figure 8.8 Flux cored arc weld

Figure 8.8 Flux cored arc weld

Figure 8.9 Electroslag arc weld

Figure 8.9 Electroslag arc weld

The CE and actual chemical content are printed on the material test reports (MTRs) from the mill. Carbon equivalence of reinforcing steels is calculated using modifications of Equation 8.1.

If the CE is less than 0.4, the steel has very good weldability. As it increases, low-hydrogen filler metals will be required to prevent cracking. If the CE is even higher, preheat is required. Preheat not only depends upon the steel chemistry, but upon section thickness and initial base metal temperature. Preheating the steel slows down the rate of weld cooling to develop optimum steel grain structures and tough weldments.

Joint design is also fundamental to sound, economical welds. Construction codes will determine what joint designs are appropriate for a given load path. The best weld joint develops the required capacity with the least amount of weld metal and labor. Overwelding wastes weld metal, increases labor hours, causes distortion, and leaves higher residual stresses. Fillet welds require the least preparation but suffer a design penalty because of the inherent stress-riser at the unwelded portion of the joint. Complete joint penetration welds (CJP), like Figure 8.10, develop the full strength of the section but require the most preparation and weld time. Partial joint penetration welds (PJP) reduce preparation and welding time, but suffer the same design penalty for the unwelded portion of the joint. Common joint types and preparation are shown in Figure 8.11.

Welds are specified on drawings using welding symbols. The engineer of record specifies the weld size, length, and general joint preparation. It is best to not over specify the joint preparation. The steel detailer will then add more information to weld symbol and piece cuts, once the fabrication shop is chosen. Figure 8.12 shows common welding symbols and their meaning. Consult AISC7 for an expanded figure. Figure 8.13 shows example applications of these symbols.

Figure 8.10 Completed CJP weld, with runoff tabs, shop access Courtesy S&S Steel fabrication

Figure 8.10 Completed CJP weld, with runoff tabs, shop access Courtesy S&S Steel fabrication

Quality is of great consideration for welded joints, and the source of much effort during fabrication and erection. Common weld discontinuities are incomplete fusion, lack of penetration, porosity, slag inclusions, undercutting, and cracking, illustrated in Figure 8.14. Weld inspection methods include visual, ultrasonic, magnetic particle (Figure 8.15), and radiographic examination. Weld inspection is extensively discussed in Chapter 10 of Special Structural Topics in this series.

Figure 8.11 Common weld joint and preparation types

Figure 8.11 Common weld joint and preparation types

You may be asked whether to specify bolting or welding on a project. The answer is yes. Yet there are times that one is preferred over the other. Strength may have little to do with the final decisions. Table 8.1 lists some advantages and disadvantages of each joining method.

8.1.3 Connected Elements

Connected elements are the connected portions of members and plate and angle used to make the connection. Figure 8.16 illustrates these in a column splice. In this example, we pay attention to the design of the plate, and flange and web with bolt holes.

Figure 8.12 Weld symbol parts and definitions

Figure 8.12 Weld symbol parts and definitions

Figure 8.13 Weld symbol examples

Figure 8.13 Weld symbol examples

Figure 8.14 Weld discontinuities

Figure 8.14 Weld discontinuities

Figure 8.15 Magnetic particle testing

Figure 8.15 Magnetic particle testing

Table 8.1 Advantages and disadvantages of bolting and welding

Bolting
Advantages Disadvantages
Require less inspection Reduces cross section
Field installation requires less skill Larger joints
Parts can be easily disassembled Require access to both sides of joint
Field installation requires less skill Can become loose in vibrating conditions
Welding
Advantages Disadvantages
Potentially smaller joints Require greater inspection
Join parts with unusual geometry Induces residual stresses into joint
Alterations are easier Field welding requires greater care
Figure 8.16 Column splice showing connecting elements

Figure 8.16 Column splice showing connecting elements

8.2 Capacity

8.2.1 Bolts

Bolts fail in tension, shear, bearing, or tearout in the base material, illustrated in Figure 8.17. These modes can exist for either tension or shear loads, depending on how we configure the joint. Remembering the chain analogy in Figure 3.5, we work our way through each member of a joint, designing each link in the load path.

We find bolt strength in tension and shear from Equation 8.2, denoting nominal capacity with Rn:

R n = F n A b (8.2)

where

Fn = nominal tensile stress Fnt , or shear stress Fnv , lb/in2 (MN/m2)

Ab = unthreaded area of the bolt, in2 (m2)

φ = 0.75 for bolt strength.

Figure 8.17 Bolt failure modes: (a) tension, (b) shear, (c) bearing and tearout

Figure 8.17 Bolt failure modes: (a) tension, (b) shear, (c) bearing and tearout

Table 8.2 Nominal bolt stresses

Fastener Type Nominal Strength, lb/in2)
Tension Shear1
Included Excluded
A307 45 27 27
(310) (188) (188)
Group A (A325) 90 54 68
(620) (372) (469)
Group B (A490) 113 68 84
(780) (469) (579)
Other2 0.75Fu 0.45Fu 0.563Fu

Source: AISC Steel Construction Manual, 14th Edition

Notes

1 Included, threads are in shear plane. Excluded, threads are not in shear plane.

2 Threaded parts meeting Section A3.4 of the Manual.

It may seem odd that we use the unthreaded bolt area. And it would be if the code didn’t account for it in the nominal stress values listed in Table 8.2. These values are reduced for bolts in tension and shear through the threads.

When bolts are loaded in shear, we need to also look at bearing and tearout, where the bolt body bears on or tears through the connection material. The nominal strength is given by Equations 8.3 and 8.4. Take the lower.

R n = 2.4 d t F u ( bearing ) (8.3)

R n = 1.2 l c t F u ( tearout ) (8.4)

where

lc = clear distance between hole edges or hole edge and plate edge, in the direction of force, illustrated in Figure 8.18, in (mm):

t = material thickness, in (mm)

Fu = ultimate strength, lb/in2 (MN/m2)

D = bolt diameter, in (mm)

φ = 0.75 for bolt bearing and tearout.

The equation above assumes that hole deformation at service loads is a concern, which it often is. When it is not, we increase the equation by 25%.

Remember to apply the strength reduction factor φ, to get φRn.

To speed up the design process, Tables 8.3, 8.4, and 8.5 present, respectively: bearing bolt shear strength, bolt tension strength, and bolt bearing strength.

Slip critical bolt strength is out of the scope of this book. However, their shear capacities are approximately 45% less than bearing-type bolts.

Figure 8.18 Bolt hole geometry for bearing calculations

Figure 8.18 Bolt hole geometry for bearing calculations

Table 8.3 Bearing bolt shear strength

Table 8.3 Bearing bolt shear strength

Table 8.4 Bolt tension strength

Table 8.4 Bolt tension strength

Table 8.5 Bolt tension strength

tab8_5.jpg.

Table 8.5m Bolt tension strength

tab8_5m.jpg.

Prying is an important consideration for bolts in tension. Where the connected material is thin enough, it will deform and put a bending moment on the bolt body, shown in Figure 8.19. The easiest way to handle prying is to make the plate thick enough so it does not bend the bolt. We do this with Equation 8.5:

t min = 4 T u b φ p F u (8.5)

where

Tu = tension force in the bolt, k (kN)

b’ = material cantilever, in (mm), defined in Figure 8.19.

φ = prying strength reduction factor

p = tributary length to the fastener, in (mm)

Fu = ultimate strength, lb/in2 (MN/m2).

Where it is not possible to thicken the material, AISC provides additional equations to determine the additional tension force from prying.

Figure 8.19 Prying in a bolted hanger connection

Figure 8.19 Prying in a bolted hanger connection

8.2.2 Welds

Weld strength is a function of the filler metal, base metal, and joint type. For complete joint penetration welds, the strength is determined by the base metal—discussed in the next section—providing the weld uses matching filler metal. For partial joint penetration, fillet, and plug welds, the base equation for capacity is shown by Equation 8.6:

R n = F n w A w e (8.6)

where

Fnw = nominal weld metal strength, lb/in2 (MN/m2)

  = 0.6 FEXX for tension and shear joints

  = 0.9 FEXX for compression joints

FEXX = filler metal classification strength, lb/in2 (MN/m2)

Awe = effective weld area, in2 (m2)

  = alw , shown in Figure 8.20.

φ = 0.75 for welds in shear, and 0.80 for partial penetration welds in tension.

The filler classification strength FEXX is typically 60, 70, or 80 k/in2 (414, 483, 552 MN/m2), with the middle value being very common.

To help with fillet weld sizing, Table 8.6 provides weld strength per inch (mm) for three filler metal strengths. We can multiply these values by weld length to get capacity. Or, we can divide a concentric joint load by the weld unit strength, to get the required length. You can also use it to check your homework.

Figure 8.20 Weld throat determination

Figure 8.20 Weld throat determination

Table 8.6 Fillet weld capacity

fig8_6_t.jpg.
Imperial Measures
Leg Throat Weld Strength per inch, φrn (k/in)
a t2)
(in) (in) 60 70 80
1/8 0.1 2.39 2.78 3.18
3/16 0.1 3.58 4.18 4.77
1/4 0.2 4.77 5.57 6.36
5/16 0.2 5.97 6.96 7.95
3/8 0.3 7.16 8.35 9.54
7/16 0.3 8.35 9.74 11.14
1/2 0.4 9.54 11.1 12.7
9/16 0.4 10.7 12.5 14.3
5/8 0.4 11.9 13.9 15.9
11/16 0.5 13.1 15.3 17.5
3/4 0.5 14.3 16.7 19.1 

Note

1) See Table 8.12 for limits on fillet weld size

Table 8.6m Fillet weld capacity

fig8_6_t.jpg.
Metric Measures
Leg Throat Weld Strength per mm, φrn (kN/mm)
a te Filler Strength, F2)
(mm) (mm) 414 483 552
3 2.1 0.395 0.461 0.527
4 2.8 0.527 0.615 0.702
5 3.5 0.659 0.768 0.878
6 4.2 0.790 0.922 1.05
7 4.9 0.922 1.08 1.23
8 5.7 1.05 1.23 1.40
10 7.1 1.32 1.54 1.76
12 8.5 1.58 1.84 2.11
14 9.9 1.84 2.15 2.46
16 11.3 2.11 2.46 2.81
18 12.7 2.37 2.77 3.16 

Note

1) See Table 8.12 for limits on fillet weld size

8.2.3 Connected Elements

Connected elements have failure modes that parallel members; tension, shear, and compression. However, we add a fourth, block shear. We will discuss each of these modes in the paragraphs that follow:

For tension, we follow the equations in Sections 3.2.1 and 3.2.2. To be technically correct, we substitute Pn with Rn, to remind us we are designing connections.

For bending, follow the provisions of Chapter 4. However, flexure is an inefficient method to transfer connection forces and should be avoided when possible.

For shear, we parallel the direct shear method in Section 5.2.1, with the following slight modifications for strength (Equations 8.7 and 8.8):

R n = 0.6 F y A g v for shear yielding (8.7)

R n = 0.6 F u A n v for shear rupture (8.8)

where

Fy = tension yield strength, lb/in2 (MN/m2)

Fu = tension ultimate strength, lb/in2 (MN/m2)

Agv = gross shear area, in2 (m2)

Anv = net shear area, in2 (m2)

φ = 1.0 for shear yield, and 0.75 for shear rupture.

For compression, if Lc/r ≤ 25, we don’t need to worry about buckling and can use Equation 8.9:

R n = F y A g (8.9)

where

Fy = yield strength, lb/in2 (MN/m2)

Ag = gross compression area, in2 (m2)

φ = 0.90 for compression.

When Lc/r > 25, we use the buckling equations in Chapter 6.

Block shear occurs where a portion of a connected element tears out in a combination of shear and tension, illustrated in Figure 8.21. We determine block shear capacity as the lesser of the following two equations (8.10). The first checks shear and tension rupture, the second checks shear yield and tension rupture.

Figure 8.21 Block shear in angle brace

Figure 8.21 Block shear in angle brace

R n = 0.6 F u A n v + U b s F u A n t R n = 0.6 F y A g v + U b s F u A n t (8.10)

where

Fu = tension ultimate strength, lb/in2 (MN/m2)

Fy = tension yield strength, lb/in2 (MN/m2)

Anv = net shear area, in2 (m2)

Agv = gross shear area, in2 (m2)

Ant = net tension area, in2 (m2)

Ubs = shear lag factor,

  = 1.0 for uniform tension stress, 0.5 for non-uniform stress

φ = 0.75 for block shear.

8.3 Demand vs Capacity

Once we have the demand (from structural analysis) and capacity, we compare the two, just like we did for member failure modes. When the capacity φRn, is greater than demand Tu or Pu, we know our connection has adequate strength. If not, we increase the number of bolts, length of weld, or thickness of plate and angle. Alternatively, we can reconfigure the joint, add more connectors, or change the structural geometry.

8.4 Detailing

Connections have numerous detailing requirements to ensure they perform the way we intend. This section provides the basic detailing requirements for bolts and welds. To get started, the following figures are examples of common structural connections. See if you can figure out how the requirements in this section apply to the photos in Figures 8.228.25.

8.4.1 Bolts

Tables 8.78.11 provide dimensions and detailing requirements for bolted joints.

Figure 8.22 Heavy base plate and shear lug in shop

Figure 8.22 Heavy base plate and shear lug in shop

Shop access provided by S&S Steel

Figure 8.23 Double angle shear connection

Figure 8.23 Double angle shear connection

Figure 8.24 Welded pipe brace

Figure 8.24 Welded pipe brace

Courtesy Mark Steel

Figure 8.25 Bolted truss splice and web connection

Figure 8.25 Bolted truss splice and web connection

Table 8.7 Bolt dimensions

Table 8.7 Bolt dimensions

Table 8.7m Bolt dimensions

Table 8.7m Bolt dimensions

Table 8.8 Bolt areas

Table 8.8 Bolt areas

Table 8.8m Bolt areas

Table 8.8m Bolt areas

Table 8.9 Tightening clearances

Table 8.9 Tightening clearances

Table 8.10 Hole and slot standard sizes

Imperial Measures (in)
  Hole Dimensions (in)
Bolt Diameter Standard Diameter Oversize Diameter Short Slot Width × Length Long Slot Width × Length
1/2 9/16 5/8 9/16 × 11/16 9/16 × 1 1/4
5/8 11/16 13/16 11/16 × 7/8 11/16 × 1 9/16
3/4 13/16 15/16 13/16 × 1 13/16 × 1 7/8
7/8 15/16 1 1/16 15/16 × 1 1/8 15/16 × 2 3/16
1 1 1/16 1 1/4 1 1/16 × 1 5/16 1 1/16 × 2 1/2
≥1 1/8 d+1/16 d+5/16 (d+1/16) × (d+3/8) (d+1/16) × (d+2.5d)

Metric Measures (mm)
  Hole Dimensions (mm)
Bolt Diameter Standard Diameter Oversize Diameter Short Slot Width × Length Long Slot Width × Length
M16 18 20 18 × 22 18 × 40
M20 22 24 22 × 26 22 × 50
M22 24 28 24 × 30 24 × 55
M24 27 30 27 × 32 27 × 60
M27 30 35 30 × 37 30 × 67
M30 33 38 33 × 40 33 × 75
≥M36 d+3 d+8 (d+3) × (d+10) (d+3) × (d+2.5d)

Source: AISC Steel Construction Manual, 14th Edition

Table 8.11 Spacing and edge distance requirements

fig8_11_t.jpg.
Imperial Measures (in)
  Spacing Edge Distance
Bolt Diameter Preferred1 Minimum Standard Hole Oversize Hole
1/2 2 1.33 3/4 13/16
5/8 3 1.67 7/8 15/16
3/4 3 2.00 1 1 1/16
7/8 3 2.33 1 1/8 1 3/16
1 3 2.67 1 1/4 1 3/8
1 1/8 3 3/8 3.00 1 1/2 1 11/16
1 1/4 3 3/4 3.33 1 5/8 1 13/16
≥ 1 1/4 3d (2 2/3)d 1.25d 1.25d+3/16
Metric Measures (mm)
Spacing Edge Distance
Bolt Diameter Preferred1 Minimum Standard Hole Oversize Hole
16 50 43 22 24
20 75 54 26 28
22 75 59 28 30
24 75 64 30 33
27 80 72 34 37
30 90 80 38 41
34 100 91 46 49
≥ 36 3d (2 2/3)d 1.25d 1.25d+3

Note

1) Based on AISC recommendations and common shop practice

Source: AISC Steel Construction Manual, 14th Edition

8.4.2 Welds

Welds have limits on their minimum and maximum sizes. Table 8.12 provides these for fillet welds. Partial penetration welds have a similar table in the Steel Manual.

8.5 Design Steps

Now that we have the basics, the following steps will guide you when designing a connection.

Step 1:Determine the force in the connector group and/or individual connector.

Step 2:Choose the fastener type. This decision is a function of load magnitude, constructability, contractor preference, and aesthetics.

Step 3:Configure the joint geometry.

Step 4:Find the connector strength by equation or from the applicable table in this chapter.

Step 5:Compare the strength to the connection load.

Step 6:Sketch the final connection geometry.

8.6 Design Example

In this example, we will assume the architect has decided that the 62 ft (18.9 m) beam in the example in Chapter 5 should be a truss to allow ductwork to pass between the webs.

Step 1: Determine Forces

From a truss analysis, we determined the axial tension force in the connection is

T u = 48 k

T u = 214 kN

See Chapter 10 of Introduction to Structures, in this series, for additional information on truss analysis.

Step 2: Choose Fastener Type

We will use bearing bolts for this connection.

Table 8.12 Fillet weld size limitations

fig8_12_t.jpg.
Minimum Fillet Weld Sizes
Imperial Measures Metric Measures
Material Thickness (in) Minimum Weld Leg, a (in) Material Thickness (mm) Minimum Weld Leg, a (mm)
1/4 1/8 6.0 3.0
Over 1/4 to 1/2 3/16 Over 6 to 13 5.0
Over 1/2 to 3/4 1/4 Over 13 to 19 6.0
Over 3/4 5/16 Over 19 8.0

Maximum Fillet Weld Sizes
Imperial Measures
Material Edge Thickness Maximum Weld Leg, a
less than 1/4 in Thickness of Material
1/4 in or greater Thickness of Material- 1/16 in

Metric Measures
Material Edge Thickness Maximum Weld Leg, a
less than 6 mm Thickness of Material
6 mm or greater Thickness of Material- 2 mm

Source: AISC Steel Construction Manual, 14th Edition

Step 3: Configure the Joint Geometry

We sketch out the joint geometry, considering member type, bolt spacing, and edge distance. Figure 8.26a shows our initial joint and pertinent information.

Figure 8.26 Example truss joint configuration

Figure 8.26 Example truss joint configuration

Step 4: Find Connector Strength

We have several failure modes to check. Starting at the member and moving to the gusset plate, we will look at:

  • Net Section Rupture in the angle
  • Bolt Bearing
  • Bolt Shear
  • Block Shear.

We don’t need to check gross yield on the member, as we would have checked this when we sized it.

Step 4a Net Section Rupture

From our sketch, we see we only have one bolt hole in the cross-section.

The base equation is:

R n R = A e F u

To find effective area Ae we will need the following information on the member and bolt sizes.

We have selected a single L4 × 4 × 3/8 (L102 × 102 × 9.5), and 3/4 in (M20) bolt. This yields

t = 0.375 in d = 3 / 4 in

t = 9.5 mm d = 20 mm

d hole = d + 1 / 8 = 0.75 in + 1.8 in = 0.875 in

d hole = d + 2 = 20 mm + 2 mm = 22 mm

From Table A1.3 we find the gross area of the angle as

A g = 2.86 in 2

A g = 1 , 850 mm 2

The effective area is given by

A e = U A net = U ( A g A hole )

A hole = t d hole = 0.375 in ( 0.875 in ) = 0.328 in 2

= 9.5 mm ( 22 mm ) = 209 mm 2

With three bolts, and using Table 3.1, U = 0.6

A e = 0.6 ( 2.86 in 2 0.328 in 2 ) = 1.52 in 2
A e = 0.6 ( 1 , 850 mm 2 209 mm 2 ) = 985 mm 2

Knowing

F u = 58 k / in 2

F u = 400 MN / m 2

we solve for the nominal rupture capacity, denoted by a subscript R.

R n R = A e F u = 1.52 in 2 ( 58 k / in 2 ) = 88.2 k

= 985 mm 2 ( 400 MN / m 2 ) ( 1 m 1000 mm ) 2 ( 1000 kN 1 MN ) = 394 kN

Multiplying by φ = 0.75, we get

φ R n R = 0.75 R n R = 0.75 ( 88.2 k ) = 66.2 k

= 0.75 ( 394 kN ) = 296 kN

Since this is greater than the demand Tu, we are OK.

Step 4b Bolt Bearing

The load now moves from the brace into the interface with the bolt, known as bolt bearing. We take the lower of the following equations:

R n B l = 1.2 l c t F u R n B d = 2.4 d t F u

From our sketch, we see

l c = 1.25 in 13 16 / 2 in = 0.844 in

l c = 31.8 mm 22 2 mm = 20.8 mm

Knowing the other inputs, we get

R n B l = 1.2 ( 0.844 in ) 0.375 in ( 58 k / in 2 ) = 22.0 k

R n B l = 1.2 ( 20.8 mm ) 9.5 mm ( 0.4 kN / mm 2 ) = 94.9 kN

Multiplying by φ = 0.75, we get

φ R n B l = 0.75 ( 22.0 k ) = 16.5 k

R n B l = 0.75 ( 94.9 kN ) = 71.2 kN

Checking the bearing limit, based on diameter, we get

φ R n B d = 0.75 ( 2.4 d t F u ) = 0.75 [ 2.4 ( 0.75 in ) 0.375 in ( 58 k / in 2 ) ] = 29.4 k

= 0.75 [ 2.4 ( 20 mm ) 9.5 mm ( 0.4 kN / mm 2 ) ] = 137 kN

The first equation is controlling. Based on this, we can determine how many bolts we need to use:

n = T u φ R n B l = 48 k 16.5 k / bolt = 2.9 bolts

= 214 kN 71.2 k N / bolt = 3.01 bolts

We will use three bolts.

Step 4c Bolt Shear

The force now causes a shear force through the body of the bolt. Assuming the threads are in the shear plane, we get the bolt capacity from Table 8.3.

φ R n V = 17.9 k

φ R n V = 87.7 kN

Since this is greater than the limiting bearing strength φRnBl, we know the number of bolts we chose is sufficient.

Cool!

Step 4d Block Shear

The last thing we need to check is block shear, where a piece of the angle could break out, shown in Figure 8.26. We take the minimum of the following two equations.

R n U = 0.6 F u A n v + U b s F u A n t R n Y = 0.6 F y A g v + U b s F u A n t

F y = 36 k / in 2

F y = 248 MN / m 2

Using Figure 8.26, we find the following lengths and areas.

l v = 1.25 in + 2 ( 3 in ) = 7.25 in l t = 1.5 in

l v = 31.8 mm + 2 ( 76.2 mm ) = 184 mm l t = 38.1 mm

A g v = l v t = 7.25 in ( 0 .375 in ) = 2.72 in 2

= 184 mm ( 9 .5 mm ) = 1 , 784 mm 2

Knowing we have 2 1/2 holes reducing our area, our net shear area is

A n v = A g v 2.5 d hole t = 2.72 in 2 2.5 ( 0.875 in ) 0.375 in = 1 .9 in 2

= 1 , 748 mm 2 2.5 ( 22 mm ) 9.5 mm =1,225 mm 2

Next, we get the net tension area:

A n t = t ( l t d hole 2 ) = 0.375 in ( 1 .5 in 0.875 in 2 ) = 0.398 in 2

= 9.5 mm ( 38 .1 mm 22 mm 2 ) = 258 mm 2

We now calculate the two block shear equations, taking Ubs = 1.0, since we will have uniform stress in the tension portion

R n U = 0.6 F u A n v + U b s F u A n t

R n U = 0.6 ( 58 k / in 2 ) 1.9 in 2 + 1.0 ( 58 k / in 2 ) 0.39 in 2 = 89.2 k

R n U = 0.6 ( 0.4 kN / mm 2 ) 1 , 225 mm 2 + 1.0 ( 0.4 kN / mm 2 ) 258 mm 2 = 397 kN

R n Y = 0.6 F y A g v + U b s F u A n t

R n Y = 0.6 ( 36 k / in 2 ) 2.72 in 2 + 1.0 ( 58 k / in 2 ) 0.398 in 2 = 81.8 k

R n Y = 0.6 ( 0.248 kN / mm 2 ) 1 , 748 mm 2 + 1.0 ( 0.4 kN / mm 2 ) 258 mm 2 = 363 kN

The second equation controls. Multiplying by φ = 0.75, we get

φ R n Y = 0.75 R n Y = 0.75 ( 81.8 k ) = 61.4 k

= 0.75 ( 363 kN ) = 272 kN

This is higher than Tu, which tells us we have adequate block shear strength.

Step 5: Compare Strength to Load

We see that each load transfer mechanism (failure mode) has greater capacity than demand. Our joint has adequate strength.

Step 6: Sketch Final Connection

Figure 8.26a shows our final connection geometry.

For interest sake, let’s look at how much fillet it would take to carry the required force. From Table 8.12, we see we need to use at least a 3/16 in (5 mm) weld on a 3/8 in (9.5 mm) thick angle. And from Table 8.6, we see the capacity per inch for a 70 k/in2 (483 MN/m2) weld metal is

φ r n W = 4.18 k / in

φ r n W = 0.768 kN / mm

Dividing the total force by the weld strength per inch, we get the required weld length:

l w = T u φ r n W = 48 k 4.18 k / in = 11.5 in

= 214 kN 0.768 kN / mm = 279 mm

We’ll round up to 12 in and 300 mm, and place half of this on each side of the angle, shown in Figure 8.26b.

And there you have it.

As a note, we have not yet designed the gusset plate. We need to check tension yield and rupture, along with bolt bearing. Multiplying lG in Figure 8.26a by the gusset thickness, we get the gross area that is effective in carrying the tension force, and work from there.

8.7 Where We Go From Here

In this chapter, we have looked at the basic structural steel connections; bolts and welds, and their connecting elements. There are myriad iterations on how we use this information that allow us to design simple and complex connections. Looking at Parts 7 through 15 of the Manual, we see a great expansion on these ideas and their application to specific connection types.

Welding, like many topics in structural engineering, can consume a lifetime of study and practice. When forces acting on a weld become complex—say with tension, shear, moment, and torsion—it is helpful to treat the weld as a line. This method is documented in Design of Welded Structures,8 by Omar Blodgett. Based on classical mechanics, this method computes weld section properties (A, S, J) as though the weld has no thickness. One then divides force, moment, and torsion by the appropriate section property, combines stresses, and backs out weld thickness. It is somewhat conservative compared to the inelastic methods now permitted in the code, but it is the only reasonable method for complex loadings on welds.

Notes

1. SAE. Mechanical and Material Requirements for Externally Threaded Fasteners, J429 (Warrendale: SAE International, 2014).

2. AISC. Specification for Structural Joints using High-Strength Bolts (Chicago: American Institute of Steel Construction, 2009).

3. AWS. Structural Welding Code—Steel, D1.1 (Miami: American Welding Society, 2015); X AWS. Structural Welding Code—Sheet Metal, D1.3 (Miami: American Welding Society, 2008).

4. AWS. Structural Welding Code—Reinforcing Steel, D1.4 (Miami: American Welding Society, 2011).

5. AWS. Bridge Welding Code, D1.5 (Miami: American Welding Society, 2015).

6. AWS. Structural Welding Seismic, D1.8 (Miami: American Welding Society, 2016)

7. AISC, Specification for Structural Steel Buildings, AISC 360 (Chicago: American Institute of Steel Construction, 2016).

8. Blodgett, O.W. Design of Welded Structures, 12th Printing (Ohio: Lincoln Arc Welding Foundation, 1982).