You are likely familiar with tension forces. In our common experience, they are present in ropes, cables, and chains. They act to stretch a member along its length, illustrated in Figure 3.1 and Figure 3.2. In structures, we find tension forces in bracing (Figure 3.3), hangers, trusses (Figure 3.4), and beams and columns of lateral systems.
When we design tension members, it is important to follow the load as it transfers from connection to member and back to connection. This is known as load path, and can be visualized as a chain, each link a possible source of strength or weakness. Imagine that we pull on the chain shown in Figure 3.5. As we increase the tensile force, we expect the chain to break at some point. As the adage says, the chain is only as strong as the weakest link.
When designing a member, start with the load at one end and track all the elements and members it passes through. Then think of every way they may fail. Each of these failure modes may be considered as a link in
Figure 3.1 Foam tension member showing deformations
Figure 3.2 Cable under tension load showing uniform stress and deflection
Figure 3.3 Diagonal bracing carrying tension and compression, One Maritime Plaza, San Francisco, SOM
Figure 3.4 Truss carrying tension in bottom chord and diagonals, Harry S. Truman Bridge, Kansas City, Missouri
the chain. For a bolted connection, the failure modes are yield on gross, rupture on net, bolt bearing, tear-out, and tear-through, and finally failure of the bolt itself, illustrated in Figure 3.5. Find the weakest link and you have found the tension capacity of the system.
Many elements that make up a building undergo axial tension under different types of loading, shown in Figure 3.6. Some of these elements may be due to gravity loads, like a hanging column or truss chord.
Figure 3.5 Tension as the proverbial chain
Figure 3.6 Building elements that undergo tension
However, many tension elements occur due to lateral loading, which cause reversing tension and compression as the load changes direction. See Chapter 7 for a discussion on lateral load paths and the forces in diaphragms and frames.
3.1 Stability
The strength of a tension element is not a function of stability. We never worry about a tension element buckling. Thus, unlike in compression, the element slenderness is not a strength-related issue. However, AISC recommends the following slenderness ratio to prevent a tension member from becoming overly flexible (though this limit does not apply to rods).
| (3.1) |
where
- L = distance between bracing locations (in, mm)
- r = least radius of gyration (in, mm).
3.2 Capacity
Recalling our discussion about load paths, the following sections discuss the different tension failure modes. Each mode may be considered a link in our proverbial chain, and must be checked to determine the capacity of our system. We will focus on failure of the member itself in this chapter, specifically yield and rupture, illustrated in Figure 3.7. We also introduce connections, but cover these extensively in Chapter 8.
Figure 3.7 Yield on gross and rupture on net
3.2.1 Yield on Gross
Yielding on the gross cross-sectional area occurs when the force in the member is large enough to cause the entire area to reach the material yield point. This results in large member elongation with a small increase in force, shown in Figure 2.4.
To determine the nominal yield on gross capacity of an element, we use the following equation.
| (3.2) |
where
Fy = yield stress of material (k/in2, N/mm2)
Ag = gross area, in2 in2 (mm2)
φ = 0.9, strength reduction factor.
Multiplying Pn by φ, we get the design strength φPn. We compare this to the force in the member Tu, discussed further below.
3.2.2 Rupture on Net
For members with bolt holes, the gross area is reduced. While the gross area may have stresses less than yield, the stresses at the net area may exceed the ultimate (or rupture) strength. (Refer again to the stress-strain diagram shown in Figure 2.4.) We can allow the net area to yield, but we do not want it to rupture. The net area occurs over such a small length of the member that when it yields, the resulting elongation is relatively small.
To determine the nominal rupture on net capacity, we use the following equation.
| (3.3) |
where
Fu = ultimate stress of material, k/in2 (MN/m2)
Ae = effective area, in2 (mm2)
φ = 0.75.
It is of great importance that this equation uses the effective area rather than the net area, discussed below.
3.2.2.1 Net Area
Net area accounts for the loss of cross-sectional area due to holes in the member. For welded connections, the net area is typically equal to the gross area. (An exception to this would be plug or slot welds.) For bolted connections, the net area is equal to the gross area minus the area removed by the bolt holes.
There are four common types of bolt holes. Other sizes may be used, but must be dimensioned by the designer. The four types are; standard and oversized holes, and short and long slots. Hole and slot sizes are provided in Chapter 8.
Standard holes are used everywhere except where the designer specifies otherwise. The standard hole is 1/16 in (2–3 mm) larger than the bolt diameter. Oversized holes can be used with slip-critical bolts, but not for bearing connections. Oversized holes vary from 1/8 to 5/16 in (4–8 mm) larger than the bolt. Short and long-slotted holes are the same as a standard hole, but elongated in one direction. They can be used for bearing and slip critical connections, but the slot must be perpendicular to the force in bearing connections.
Figure 3.8 Staggered hole failure mode
When calculating net area, we take the size of the bolt hole plus 1/16 in (2 mm) to determine the area that needs to be subtracted from the gross area. The additional 1/16 in accounts for drilling imperfections.
When bolt holes are staggered, the critical failure plane may not be a straight line, as illustrated in Figure 3.8. This is further discussed in section B2 of AISC-360.1
3.2.2.2 Effective Area
Effective area accounts for how a tensile force is distributed from the full cross-section of the member to the area where the connection actually occurs. Also known as shear lag, it accounts for force flowing from one part of a member to another, shown in Figure 3.9.
Figure 3.9 Force transfer utilizing shear lag
Effective area is related to net area through the shear lag factor U, as follows:
| (3.4) |
where
An = net area, in2 (mm2)
U = shear lag factor (unitless).
The shear lag factor is 1.0, when each portion of a cross-section is connected, like that shown in Figure 3.10. It is less than one when only some portions of a member are connected, like Figure 3.11. Calculation of the shear lag factor can be effort intensive. Table 3.1 provides them for a number of conditions.
Rupture controls for the conditions, for the given strengths
- A992 when Ae <0.923 Ag
- A36 when Ae <0.745 Ag
- A500 Grade B when Ae <0.952 Ag.
Figure 3.10 Tension transmitted directly to all cross-sectional elements
Figure 3.11 Tension transmitted to only some of cross-sectional elements with bolts
3.2.3 Connection Failure Modes
Members must be connected to each other to transfer tension forces, otherwise our proverbial chain would end, as would our load path. Connection failure modes that must be considered when looking at tension load paths include welds, bolts, plates, angles, and connected portions of connected members. These failure modes are extensively covered in Chapter 8, and illustrated in the example in this chapter.
3.2.3.1 Initial Tension Sizing
An effective rule of thumb for initial member sizing is to take the demand and work backwards to area. This will give us an idea of the required section size. Using the following equation, and previously defined terms above, we find the required area as:
Table 3.1 Shear lag factors for various connection types
| Case | Description | Shear Lag Factor | Sketch |
| 1 | Force is transferred to each portion of a member cross section | U = 1.0 |  |
| 2 | Force is transferred to only some of the cross section | ||
| 3 | Force transmitted by weld perpendicular to force, and area based on connected part (bt) | U = 1.0 | |
| 4 | Plate where force is transmitted by welds parallel to force | l ≥ 2w, U = 1.0`2w > l ≥ 1.5w, U = 0.87`1.5w > l ≥ 1.5w, U = 0.87 | |
| 5 | Round HSS with middle gusset | l ≥ 1.3D, U = 1.0`D ≤ l < 1.3D, | |
| 6 | Rectangular HSS with middle gusset | l ≥ H, | |
| 7a | W, M, S, HP, or WT with 3 or more rows of fasteners in flanges | U = 0.85 | |
| 7b | W, M, S, HP, or WT with 4 or more rows of fasteners in web | U = 0.7 | |
| 8a | Angles with 4 or more fasteners parallel to load | U = 0.8 | |
| 8b | Angles with 3 fasteners parallel to load | U = 0.6 |
Source: AISC 360–16
| (3.5) |
Another method is to use the tension yield strengths provided in Table 3.2. These can be compared directly with tension force to give an idea of required area. Regardless of approach, we will need to check rupture on net once we select our member and connections.
3.3 Demand vs Capacity
The LRFD equation for demand versus capacity uses factored load combinations, which produce demand (load) Tu, and compares this with the nominal resistance multiplied by the strength reduction factor φ, yielding capacity φPn. When the capacity is greater than the demand, our design works, as shown in Equation 3.6.
| (3.6) |
The strength reduction factor is ϕ 0.9 for yielding and 0.75 for limit states involving rupture, like rupture on net.
Looking at demand more in-depth, we see demand Tn is defined as in Equation 3.7:
| (3.7) |
where
- γI = Load Factor
- Qi = Nominal Load, k (kN).
Load factors, which are a part of load combinations, are discussed further in Introduction to Structures in this series, and ASCE7.2
3.4 Deflection
There are two main classes of limit states, strength and serviceability. We use ASD load combinations when checking serviceability. Deflection is the main serviceability limit state we are concerned with when looking at tension element design.
The code does not have any specific tension deflection requirements, but they can be determined by considering the types of systems they are in and any adverse effects that would result from excessive deformation. For example, if a brace in a vertical frame were to elongate excessively,
Table 3.2 Yield on gross capacities based on area
| Gross Area (in2) | Tension
Strength ϕPn (k) Yield Strength fy (k/in2) |
||
| 36 | 46 | 50 | |
| 0.500 | 16.2 | 20.7 | 22.5 |
| 0.800 | 24.3 | 31.1 | 33.8 |
| 1.00 | 32.4 | 41.4 | 45.0 |
| 1.50 | 48.6 | 62.1 | 67.5 |
| 2.00 | 64.8 | 82.8 | 90.0 |
| 3.00 | 97.2 | 124 | 135 |
| 4.00 | 130 | 166 | 180 |
| 5.00 | 162 | 207 | 225 |
| 7.50 | 243 | 311 | 338 |
| 10 | 324 | 414 | 450 |
| 12.5 | 405 | 518 | 563 |
| 15.0 | 486 | 621 | 675 |
| 20.0 | 648 | 828 | 900 |
| 30.0 | 972 | 1,242 | 1,350 |
| 40.0 | 1,296 | 1,656 | 1,800 |
| 50.0 | 1,620 | 2,070 | 2,250 |
| 70.0 | 2,268 | 2,898 | 3,150 |
| 100 | 3,240 | 4,140 | 4,500 |
| 125 | 4,050 | 5,175 | 5,625 |
| 150 | 4,860 | 6,210 | 6,750 |
| 175 | 5,670 | 7,245 | 7,875 |
| 200 | 6,480 | 8,280 | 9,000 |
| 250 | 8,100 | 10,350 | 11,250 |
| 300 | 9,720 | 12,420 | 13,500 |
| 400 | 12,960 | 16,560 | 18,000 |
| 500 | 16,200 | 20,700 | 22,500 |
Table 3.2m Yield on gross capacities based on area
Note 1) Check net section rupture for members with holes
the resulting story drift may be unacceptable. Similarly, the horizontal deflection limits discussed in Chapter 4 for beams, can be applied to truss deflections, which are the result of tension and compression deflections of the chords and webs.
From strengths and materials, we know axial deflection is given as in Equation 3.8.
| (3.8) |
where
P = force, k (kN)
L = length of member, in (mm)
Ag = gross area of the element, in2 (mm2)
E = modulus of elasticity, k/in2 (MN/mm2).
Note this equation only holds true in the elastic range. As discussed in Chapter 2, once a stress exceeds the yield strength the elongation is no longer linearly related to deflection.
3.5 Detailing
When detailing tension elements, keep in mind the effective area of the connection, which is a function of net area and shear lag. We balance bolt size with member size, so we have about a 70% reduction in area. For shear lag, it is best to connect each portion of the member—both legs of an angle or both flanges and web of wide flanges. When this is not feasible or economical, make the length of the connection sufficiently long to avoid large reductions in effective area, and thus connection capacity. Connection detailing is further discussed in Chapter 8.
3.6 Design Steps
To design a tension member, we must determine the following information:
- Step 1Determine the structural layout
- Step 2Determine the loads
- Maximum service load
- Maximum factored load
- Step 3Determine material parameters
- Step 4Determine initial size
- Step 5Check All Applicable
Strength Limit States
- Yield on Gross
- Rupture on Net
- Bolt Bearing
- Bolt Tear-out
- Block Shear
- Step 6Check Member Serviceability Limit States
- Check slenderness requirements
- Check deflection requirements
- Step 7Summarize the Final Results
3.7 Example: Tension Member Design
Step 1 Determine the Structural Layout
Our structural configuration is shown in Figure 3.12. We will design the steel beam for tension, which collects lateral seismic forces and drags it into an adjacent steel braced frame. Known as a drag strut, it is 28 ft (8.5 m) long, and connected with plates at the top and bottom flanges, with 3/4 in (20 mm) diameter bolts in standard holes, illustrated in Figure 3.12.
We will limit this example to the tension design of the drag strut. Perhaps once you have finished the book, you can check the bending and compression capacity of the beam, and failure modes of the bolts, drag plates, welds. Thus we will only be looking at one section of the chain, and its associated links. In reality, the chain extends from the surface of the element where the load is first applied all the way to the footings where it is resolved by the soil beneath the building. It simplifies our design to break the load path into smaller discrete sections that can be evaluated.
Step 2 Determine the Loads
Axial loads in our drag strut are only created by seismic and wind. From our structural analysis, we determine that the seismic and wind axial loads are shown in Equation 3.9.
Determine the maximum factored load.
We will check the seismic and wind combinations, since this is where our load is coming from. For seismic:
Figure 3.12 Design example
For wind we have
It looks like wind controls.
Step 3 Determine Material Parameters
The typical material for a wide flange beam is A992. It has a yield strength Fy = 50 k/in2 (345 N/mm2) and an ultimate strength Fu = 65 k/in2 (450 N/mm2).
Step 4 Determine Initial Size
Because this member will also take bending forces, we estimate its depth as half the beam span in feet, giving us 14 in (356 mm).
Additionally, the tension attachments are plates bolting to the top flange of the beam. The beam flange will need to be wide enough for proper bolt clearances. The required clearance is 1.25 in (31.8 mm), from Table 8.11. The nut diameter is 1.5 in (38.1 mm) and we can approximate the rounding at the corner between the flange and the web as 0.75 in (19.1 mm). Thus, the minimum flange width is:
Based on these criteria, we choose a W16 × 31 (W410 × 46.1), which has a beam width of 5.53 in (140 mm).
Step 5 Check Strength Limit States
Step 5a Yield on Gross
From the steel beam properties table for a W16 × 31, the area is.
The tensile yield strength follows as
This is quite a bit larger than our demand. Let’s check rupture.
Step 5B Rupture on Net
We begin by finding the net area, given as:
Because only the flanges are connected, we need to determine the shear lag factor U. From Table 3.1, line 7a, we get U = 0.85 for three bolts in a row. The effective area then becomes
Calculating the axial tension rupture capacity we get
Step 5C Compare Strength Limit States
Because both the yield and rupture strengths ϕPn are greater than the demand Tu, we know our member is OK.
Step 6 Check Deflection
We will skip the deflection check of this member for now. When we check drift in the structural analysis, it will capture the effect of the drag struts axial extension on the frame system.
Step 7 Summarize the Final Results
The final tension design is a W16 × 31 (W410 × 46.1), A992 steel beam 28 ft (8.5 m) long connected with (12) 3/4 in (20 mm) diameter bolts in standard holes connected to the beam flanges, sketched in Figure 3.12. Remember, we will need to also size this member for combined tension and bending, and compression and bending, in addition to checking the connection.
3.8 Where We Go From Here
From here we will get further into connection failure modes, and their influence on member capacity. Key to this is block shear, where a portion of a connected member fails in a combination of tension and shear, discussed in Chapter 8.
For additional design aids, the AISC Steel Construction Manual.3 includes numerous tables for quick tension design of common shapes.
Notes
1. AISC, Steel Construction Manual, 14th Edition (Chicago: American Institute of Steel Construction, 2011).
2. ASCE, Minimum Design Loads for Buildings and Other Structures, ASCE/SEI 7–16 (Reston, VA: American Society of Civil Engineers, 2016).
3. AISC, Steel Construction Manual, 14th Edition (Chicago: American Institute of Steel Construction, 2011).