contact | structural element calculators | paperback and pdf | « previous section | next section » | contents

Contents | 1. Introduction to structural design | 2. Loads | 3. Wood |

Introduction to steel | Material properties | Sectional properties | Design approaches | Construction systems | Tension elements | Columns | Beams |

Steel structural elements are typically connected to each other using high-strength bolts or welds. Especially in so-called field connections — those that take place at the construction site — bolts are preferred, as they are easier, and generally less expensive, to execute in such contexts (outdoors, with unpredictable weather conditions, and without convenient access to welding equipment). Often, when welding is found to be either necessary or expedient, it occurs at the fabricating shop, although field welding is sometimes unavoidable.

Steel connections are designated according to the types of forces and/or bending moments
that are intended to be resisted, and that are symbolized by the hinges, rollers, or fixed constraints
that populate load diagrams in statics texts (see Figure 1.14). In practice (see Figure 4.31), hinges
and rollers become *simple* connections (previously designated as Type 2); fixed joints become *fully restrained*, or FR, connections (previously designated as Type 1); and the intermediate conditions between simple and fully restrained, become *partially restrained*, or PR, connections (previously designated as Type 3).

Figure 4.31: Typical bolted connections for steel members: (*a*) simple column-beam connection (formerly Type 2); (*b*) simple beam-girder connection (formerly Type 2); and (c) fully restrained, rigid frame connection between column and girder (formerly Type 1)

High-strength bolts typically used to connect steel elements are stronger than the bolts most often used to connect wood elements: the two most commonly specified bolts used in steel structures in the U.S. are designated *Group A* (including A325 bolts with an ultimate strength, *F _{u}* = 120 ksi) and

Figure 4.32: Bolts stressed in (*a*) shear; (*b*) tension; and (c) shear and tension

For shear connections, most bolts are designed so that they "bear" against the edge of the
bolt holes into which they are inserted. These are bearing-type, or "snug-tightened" joints, and a
small amount of slip of the bolt within the slightly-larger bolt hole is permitted. In the less common
cases where no slip is desired — for example, in structures subjected to repeated stress reversals — so-called *slip-critical* connections are designed on the basis of the clamping force that the bolts place on the steel elements being joined, so that friction between the surfaces clamped together resists the tendency of the bolts to slip within the bolt holes. In either case (bearing or slip-critical bolt design), two separate strength criteria must be satisfied: (1) the shear strength of the bolt itself; and (2) the compressive capacity of the elements being joined, as the bolts "bear" on the inside surface of the bolt holes.

* Shear capacity.* The nominal bolt shear stress can be taken as 68 ksi for

* Bearing capacity.* The nominal bearing capacity of a bolt,

Dividing the nominal bearing capacity by the safety factor for bearing, Ω = 2.0, we get the available
strength for a bolt in bearing, *R _{n}*/Ω = 1.5

* Minimum and maximum spacing.* Bolts that are used to connect steel elements are also subjected to minimum and maximum spacing rules. The basic suggested minimum centerline spacing between
bolts is 3 times the nominal bolt diameter,

* Tension, shear and block shear.* Where bolt holes reduce the cross-sectional area of a tension element, the design of the tension element itself must account for this reduced net, or effective net, area, as described previously for steel tension elements. For coped beams bolted to the webs of girders, block shear must be checked, as described previously for steel beams.

** Problem definition.** Examine the W8 × 24 wide-flange shape used as a tension element in a steel truss (the section's capacity was determined to be 148 kips in Example 4.1 when using two lines of 3/4 in. diameter bolts). Find the required number of bolts so that their available strength is no less than the beam's tension capacity (Figure 4.33). Assume A36 steel for the W8 × 24 section, and A490 high-strength bearing-type bolts (threads included in the shear plane).

Figure 4.33: Connection of W8 × 24 tension element using two lines per flange of high-strength bolts for
Example 4.8

** Solution overview..** Find the required number of bolts based on bolt shear; check for bolt bearing.

*Problem solution*

**1.** Required number of bolts (design based on shear): From Appendix Table A-4.18 part *A*, for A490 bolts and 3/4 in. bolt diameter, the shear capacity per bolt is 18.6 kips, assuming threads excluded from the single shear plane. Based on Note 1 (for threads included in shear plane), this value is reduced by 80%, so the capacity per bolt becomes 0.80 × 18.6 = 14.88 kips per bolt. The required number of bolts is equal to the total capacity divided by the capacity per bolt, or 148/14.88 = 9.95 bolts. Clearly, this number must be rounded up to an integer that is divisible by 4, so that the four lines of bolts distributed on the two flanges all have the same number:
therefore, we provisionally select 12 bolts, as shown in Figure 4.33.

**1.** Check required number of bolts (based on bearing capacity): From Appendix Table A-4.19, the
bearing capacity per bolt, per inch of A36 material thickness, is 65.3 kips. As can be seen from
Appendix Table A-4.3, the flange thickness of a W8 × 24 section is 0.40 in. Therefore, the capacity of a
single bolt, based on bearing on the flange thickness, is 0.40 × 65.3 = 26.12 kips. The total capacity of the 12-bolt connection, again based on bearing, is 12 × 26.12 = 313 kips. Since this
capacity is no smaller than the capacity determined in step 1 for shear, the provisional selection
of 12 bolts is satisfactory. For a bearing capacity less than that determined for shear, the number
of bolts would need to be increased accordingly, and the bolt design would be governed by bearing instead of shear.

Two pieces of steel may be welded together, not by directly melting one piece into the other, but
rather by depositing melted steel contained in a separate electrode along the surfaces of the two
steel pieces to be joined. Naturally, some melting of the joined pieces occurs as the "weld" steel is
deposited; however, the weld and adjacent surfaces rapidly cool and harden as the electrode moves
along the weld line, effectively connecting the pieces together. While there are numerous types of
weld geometries — including groove welds, plug welds, and slot welds — the most common is the triangular fillet weld (pronounced * fill*-it). In what follows, we will discuss the strength of fillet welds subjected to loads parallel, perpendicular, or angled to the weld line.

As can be seen in Figure 4.34, a fillet weld is assumed to fail along the surface defined by its
throat, labeled *t* in Figure 4.34a, whether the weld itself is stressed in tension, compression, or shear.

Figure 4.34: Three views of a typical fillet weld illustrating (*a*) the root, size (leg length), *w*, and throat dimension, *t*, as well as two modes of failure on the throat surface, based on either (*b*) tension or (*c*) shear

With symmetrical welds angled at 45° to the surfaces being joined, it can be seen that the
throat dimension, *t*, equals 0.707*w* (where *w* is the hypotenuse of a 45° right triangle with both legs equal to *t*). For a weld of length, *Lt*, the surface area resisting either tension, compression, or shear is therefore *A _{w}* =

The general equation for all fillet welds, loaded longitudinally as shown in Figure 4.34*c*, transversely as shown in Figure 4.34*b*, or at any angle in between, is:

4.15

where

*R _{n}*/Ω = the available strength of a one-inch-long weld (kips).

θ = the angle (from 0° to 90°) between the weld line and the direction of load.

*w* = the weld size, or leg length (in.).

It can be seen that for longitudinal welds, with θ = 0°, the parenthetical term drops out, and Equation 4.15 is as derived earlier. For θ = 90° (a transverse weld), the capacity increases by a factor of (1.0 + 0.50 sin^{1.5} 90°) = 1.5. The available strengths for longitudinal and transverse welds are therefore as follows: for a one-inch-long longitudinal weld, we get

4.16

while the available strength for a one-inch-long transverse weld is

4.17

In these equations, *R _{wl}* /Ω and

The minimum length of a fillet weld is required to be at least four times its leg size. Otherwise the effective size of the weld, used in calculations, must be taken as no more than one-fourth of the weld length. For example, the minimum weld length for a 1/2 in. leg size is 4 × 1/2 = 2 in. If a 1/2 in. weld size is used with a shorter weld length — say 1 in. — the effective weld size used in calculating the available strength of the weld would be no more than the actual length (1 in.) divided by 4, or 1/4 in, even though the actual weld size is 1/2 in.

* Longitudinal welds.* For symmetrical and parallel longitudinal welds, the weld length,

Figure 4.35: Parallel, longitudinal welds

Where such welds transmit force to the "end" of an element subject to tension or compression (that is, through an "end-loaded" weld), an effective length *L _{e}* = β

0.6 ≤ β = 1.2 – 0.002(*L/w*) ≤ 1.0

4.18

In other words, where the ratio of weld length to weld size, *L/w* ≤ 100, β = 1.0, and the effective length equals the actual weld length. Otherwise, β = 1.2 – 0.002(*L/w*), with a lower limit of β = 0.60 where the ratio of weld length to weld size, *L/w* ≥ 300.

* Fillet weld terminations.* In certain cases, fillet welds should be terminated before reaching the edge of the steel elements they are connecting, to prevent damage (notching, gouging) of the
element's edge. Figure 4.36

Figure 4.36: Termination of fillet welds where (*a*) welds occur on opposite sides of a common plane; and (*b*) a lap joint extends beyond a tension element

* Shear strength of connecting elements.* Where welded connecting elements such as gussets, angles, or other plates are subjected to shear, the required thickness,

4.19

** Problem definition.** Find the capacities of the 6-in.-wide, 7/8-in.-thick plates shown in Figure 4.37, welded to (

Figure 4.37: Connector plate capacity for Example 4.9 using (*a*) transverse welds and (*b*) longitudinal welds

** Solution overview..** Find the capacity of the welds; confirm that the tensile capacity of the plates is no smaller than the weld capacity.

*Problem solution*

**1.** Based on Equations 4.16 and 4.17, we can express the capacity of the transverse and longitudinal welds as follows:

For the transverse weld, the unit capacity, *R _{wt}* /Ω = 1.5(14.85

For the longitudinal weld, the unit capacity, *R _{wl}* /Ω = 14.85

**2.** The tensile capacity of both plates is based on the smaller of the following: either the capacity
to resist tensile yielding on the gross area or to resist rupture on the net area. The capacity
based on yielding (see previous section in this chapter) is 0.6*F _{y}*

**3.** *Conclusion:* the capacity of Plate *a*, *P _{t}* , equals 100.2 kips; and the capacity of Plate

** Problem definition.** Find the capacity of the 1/2-in.-thick plate shown in Figure 4.38, welded to a wide-flange column shape. Assume that the plate is fabricated from A36 steel, and that the weld size is 3/16 in., on both sides of the plate. Use an E70XX electrode with

Figure 4.38: Connector plate capacity for Example 4.10: a gusset plate welded to a W-shape is shown (*a*) in elevation; (*b*) in section; and (*c*) in a schematic "cut-away" view showing the potential shear failure planes for the plate and fillet welds

** Solution overview..** Confirm that the shear capacity of the plate is greater than the capacity of the weld; compute the available strength of the weld.

*Problem solution*

**1.** From Equation 4.19, the required thickness of the plate (that is, the plate thickness consistent with the maximum available shear strength of a weld on both sides of a connector plate) is *t _{min}* = 1.71

**2.** Since the actual plate thickness of 1/2 in. is larger than the required thickness, *t _{min}* = 0.32 in., the weld will fail in shear before the plate does. For this reason, we can find the capacity (available strength) of the connector by determining the available strength of the weld, per inch of length, according to Equation 4.15:

**3.** The total weld length is 6 × 2 = 12 in., so the total available strength of the connector, *P* = 12(3.91) = 46.9 kips.

** Problem definition.** Find the required longitudinal weld length,

Figure 4.39: View of welded plate connectors for Example 4.11

** Solution overview..** Confirm that capacity of both plates is no less than 80 kips; find the required longitudinal weld length so that the total weld capacity is no less than 80 kips.

*Problem solution*

**1.** The tensile capacity of both plates is based on the smaller of the following: either the capacity to resist tensile yielding on the gross area or to resist rupture on the net area. The capacity based on yielding (see previous section in this chapter) is 0.6*F _{y}*

**2.** From Appendix Table A-4.21, for a 1/2-in.- thick plate, the minimum weld size is 3/16 in., and the maximum weld size is 1/2 – 1/16 = 7/16 in. For this example, we will choose a weld size of *w* = 3/16 in.

**3.** Based on Equations 4.16 and 4.17, we can express the capacity of the longitudinal and transverse welds as follows:

For the longitudinal weld, the unit capacity, *R _{wl}* /Ω = 14.85

For the transverse weld, the unit capacity, *R _{wt}* /Ω = 1.5(14.85

**4.** Where both longitudinal and transverse welds occur in the same connection, the available strength is taken as either (*a*) *R _{wl}* /Ω +

*R _{wl}* /Ω +

0.85*R _{wl}* /Ω + 1.5

**5.** Since the *greater* capacity of the two alternatives may be used, the smaller length, *L* = 3.16 in. is acceptable. Looked at another way, if the length for both alternatives were set at *L* = 3.16 in., case (a) would have a capacity smaller than 80 kips, while case (b) would have a capacity exactly equal to 80 kips; it can be seen that case (b) has the greater capacity and therefore would govern the design. Increasing the length to 4.18 in. found in case (a) is not required. We round up the required length for the longitudinal weld to 3-1/2 in.

© 2020 Jonathan Ochshorn; all rights reserved. This section first posted November 15, 2020; last updated November 15, 2020.