AISC 360-22 Tension Member Design — W10x49 Full Worked Example

Complete step-by-step tension member design following AISC 360-22 LRFD provisions. This worked example covers a W10x49 diagonal brace member with bolted gusset plate connections: tensile yielding (Section D2), tensile rupture (Section D2/D3), shear lag effects (Table D3.1), and block shear rupture (Section J4.3). Every intermediate substitution and code clause is shown with actual numbers.

Problem Statement

PRELIMINARY — NOT FOR CONSTRUCTION. All results presented here are for educational and reference use only. Values must be independently verified by a licensed Professional Engineer (PE) or Structural Engineer (SE) before use in any design or construction.

A W10x49 diagonal brace in a concentrically braced frame carries a factored tension force of Tu = 290 kips. The brace is connected to a 3/8 in. gusset plate at each end using eight 7/8 in. diameter A325 bolts in a four-by-two staggered pattern. The brace flanges are connected by two lines of four bolts passing through the gusset plate. The connection length from the first to last bolt is 9 in. Steel grade is ASTM A992 (Fy = 50 ksi, Fu = 65 ksi).

Design parameters:

• Member: W10x49, ASTM A992
• Factored axial tension force: Tu = 290 kips
• Bolts: 8 × 7/8 in. diameter A325-N, standard holes
• Connection length (first to last bolt): 9 in.
• Gusset plate thickness: 3/8 in.

Section Properties (W10x49)

Section properties obtained from AISC Steel Construction Manual Table 1-1 for the W10x49 wide-flange shape. The W10x49 is a medium-weight section commonly used for brace members in lateral-force-resisting systems. The wide flange provides good buckling resistance in compression, but here we are checking only the tension limit states.

Step 1: Tensile Yielding (AISC 360-22 Section D2)

AISC Equation D2-1: The design tensile yielding strength is:

phi_t * Pn = phi_t * Fy * Ag

where phi_t = 0.90 for yielding (AISC D2). Substituting values:

phi_t * Pn = 0.90 * 50 ksi * 14.4 in^2
           = 0.90 * 720
           = 648 kips

Check: Tu = 290 kips < phi_t*Pn = 648 kips. Tensile yielding OK. Utilization ratio = 290/648 = 0.448.

Tensile yielding is rarely the governing limit state for bolted tension members because the gross area Ag is the full cross-section without any reduction. However, it serves as an important upper bound and must always be checked.

Step 2: Tensile Rupture in the Net Section (AISC 360-22 Section D2)

AISC Equation D2-2: The design tensile rupture strength is based on the effective net area Ae with consideration of shear lag:

phi_t * Pn = phi_t * Fu * Ae

where phi_t = 0.75 for rupture (AISC D2) and Ae = An * U (effective net area accounting for shear lag).

Step 2a: Net Area An Calculation

For the bolted connection, bolt holes remove material from the gross section. The hole diameter for a 7/8 in. bolt in a standard hole per AISC Table J3.3 is:

dh = d_bolt + 1/16 in. = 7/8 + 1/16 = 0.9375 in.

The W10x49 flanges each have two bolts in the critical cross-section (the first transverse line). The total area removed by holes from both flanges is:

A_holes = 4 * dh * tf = 4 * 0.9375 * 0.560 = 2.100 in^2

Net area:

An = Ag - A_holes = 14.4 - 2.100 = 12.30 in^2

Step 2b: Shear Lag Factor U (AISC Table D3.1)

For a W-shape connected through both flanges with at least three fasteners per line in the direction of loading, AISC Table D3.1 Case 2 applies. The connection eccentricity x̅ is the distance from the face of the connection element to the centroid:

x_bar = bf / 2 = 10.0 / 2 = 5.0 in.

Connection length L = 9 in. (center of first bolt to center of last bolt).

Shear lag factor per AISC D3:

U = 1 - x_bar / L = 1 - 5.0 / 9.0 = 1 - 0.556 = 0.444

However, AISC Table D3.1 footnotes specify that when the section is connected through both flanges and there are four or more fasteners per line in the direction of loading, the calculated U need not be taken less than 0.9. But with two bolts per line (four total per flange line), let us check: The connection has four bolts in a 2×2 pattern per flange. In the direction of load, there are only two bolt lines per flange. AISC Table D3.1 specifies for W, M, S, or HP shapes connected at both flanges: use Case 2 or Case 5. Case 5 for both flanges connected using at least three fasteners per line in the direction of loading gives U = 1.0.

Per AISC Table D3.1 Case 5: Both flanges connected with four or more fasteners per line in the direction of loading, U = 1.0. With the connection having at least four fasteners in the bolt line direction (assuming two rows along the brace longitudinal axis with four bolts each): U = 1.0 for four or more bolts per line.

Using U = 1.0:

Ae = An * U = 12.30 * 1.0 = 12.30 in^2

Step 2c: Design Tensile Rupture Strength

phi_t * Pn = 0.75 * Fu * Ae
           = 0.75 * 65 * 12.30
           = 0.75 * 799.5
           = 599.6 kips

Check: Tu = 290 kips < 599.6 kips. Tensile rupture OK. Utilization = 290/599.6 = 0.484.

Step 3: Block Shear Rupture (AISC 360-22 Section J4.3)

Block shear must be checked for tear-out at the connection end. Consider the potential failure block at one flange connection. The block is defined by:

• Shear planes: parallel to the applied force, along two vertical bolt lines
• Tension plane: perpendicular to the applied force, across the last line of bolts

Geometry of potential block shear failure at one flange:

• Shear area: Two vertical planes through the gusset plate along the bolt lines, each extending from the end of the beam through the centers of four bolts. Shear length per plane L_sh = 9 in. (first to last bolt center) + 1.5 in. (end distance from first bolt to beam end) = 10.5 in. per shear plane.
• Tension area: Across the flange width at the last bolt line. Tension length = bf = 10.0 in.

Block shear strength (AISC Equation J4-5):

Rn = 0.60 * Fu * Anv + Ubs * Fu * Ant  <=  0.60 * Fy * Agv + Ubs * Fu * Ant

where:

• Agv = gross shear area (two planes)
• Anv = net shear area after subtracting bolt holes
• Ant = net tension area
• Agt = gross tension area
• Ubs = 1.0 (uniform tension stress, as the tension plane is perpendicular to the applied load)

Shear plane areas (both flanges):

Gross shear area:

Agv_total = 2 flanges * 2 shear planes * L_sh * tf
          = 2 * 2 * 10.5 * 0.560 = 23.52 in^2

Net shear area: Subtract area of bolt holes in shear planes. Each shear plane passes through four bolt holes:

Holes per plane = 4
Total hole reduction = 2 flanges * 2 planes * 4 holes * dh * tf
                     = 2 * 2 * 4 * 0.9375 * 0.560 = 8.400 in^2

Anv_total = 23.52 - 8.400 = 15.12 in^2

Tension plane area:

Gross tension area:

Agt_total = 2 flanges * bf * tf = 2 * 10.0 * 0.560 = 11.20 in^2

Net tension area: Subtract area of the two bolt holes crossing the tension plane:

Ant_total = 11.20 - (2 flanges * 2 holes * dh * tf)
          = 11.20 - (4 * 0.9375 * 0.560)
          = 11.20 - 2.100 = 9.10 in^2

Block shear nominal strength:

Term 1: 0.60 _ Fu _ Anv + Ubs _ Fu _ Ant

Term1 = 0.60 * 65 * 15.12 + 1.0 * 65 * 9.10
       = 589.7 + 591.5
       = 1,181.2 kips

Term 2 (upper bound): 0.60 _ Fy _ Agv + Ubs _ Fu _ Ant

Term2 = 0.60 * 50 * 23.52 + 1.0 * 65 * 9.10
       = 705.6 + 591.5
       = 1,297.1 kips

Block shear nominal strength:

Rn = min(1,181.2, 1,297.1) = 1,181.2 kips

Design block shear strength (phi = 0.75 per AISC J4.3):

phi * Rn = 0.75 * 1,181.2 = 885.9 kips

Check: Tu = 290 kips < 885.9 kips. Block shear OK. Utilization = 290/885.9 = 0.327.

Block shear does not govern for this configuration because the W10x49 flanges are relatively thick (0.560 in.), providing ample shear area at the two failure planes.

Step 4: Summary of Design Checks

All limit states pass. Tensile rupture governs with a utilization of 0.484. The W10x49 provides adequate tension capacity for the 290 kip demand with substantial reserve. A lighter section (W10x39 or W8x40) could be considered if additional checks confirm adequacy.

AISC 360-22 Tension Member Design Overview

AISC 360-22 Chapter D governs the design of members subject to axial tension. The chapter is organized into three sections:

• Section D1 — General Provisions: Defines the scope and references connection design requirements.
• Section D2 — Tensile Strength: Provides the two primary limit state equations for yielding (Eq. D2-1) and rupture (Eq. D2-2). The tensile strength is the minimum of yielding on the gross section and rupture on the effective net section.
• Section D3 — Effective Net Area: Addresses shear lag through the reduction factor U, which accounts for portions of the cross-section that are not directly connected. Table D3.1 provides U values for various connection configurations including plates, angles, channels, W-shapes, and HSS members.

In addition, connection limit states in Chapter J must be satisfied:

• Section J4.3 — Block Shear Strength: Checks against tear-out failure in connected elements where a block of material separates along shear and tension planes.
• Section J3 — Bolts and Threaded Parts: Governs bolt shear, bearing, and tension capacity.

Shear Lag Deep Dive

Shear lag is the reason the effective net area Ae is often less than the net area An. When only part of a cross-section is connected (such as one leg of an angle, or the web only of a channel), the tensile stress must transfer from the connected element to the unconnected elements through shear. This results in a non-uniform stress distribution and reduces the effective area available for tension resistance.

For W-shapes connected through both flanges, shear lag effects are minimal because both flanges are directly loaded. The AISC Table D3.1 reflects this: Case 5 for W-shapes with both flanges connected using at least three fasteners per line assigns U = 1.0. For connections with fewer fasteners or only the web connected, U can drop to as low as 0.6, significantly reducing the effective net area.

In our example with a W10x49 brace, the full flange connection eliminates shear lag concerns and allows effective use of the net section area.

Key Takeaways

• Always check all three limit states: Tensile yielding, tensile rupture, and block shear. Each can govern under different connection configurations.
• Net area matters for rupture: Bolt holes reduce the available cross-section. The net area An subtracts the projected hole area from the gross area.
• Shear lag reduces effective area: When only part of the section is connected (Case 1, 2, 3, 4, 7, 8 in Table D3.1), the effective net area Ae is less than the net area An.
• Block shear can be critical: For short connections, thin flange elements, or angles connected by one leg, block shear frequently governs over tensile rupture.
• Connection length affects U: The shear lag factor U depends on the connection length L. Longer connections improve stress distribution and increase U.

Frequently Asked Questions

What are the primary limit states for tension members under AISC 360-22?

The three primary limit states per AISC 360-22 Chapter D are: tensile yielding in the gross section (Section D2), tensile rupture in the net section at bolt holes (Section D2), and block shear rupture where a block of material tears out at the connection end (Section J4.3). Shear lag effects from partial connection of cross-section elements must also be considered per Section D3 and Table D3.1.

How is the net area calculated in a bolted tension connection?

The net area An is the gross area Ag minus the area removed by bolt holes. For each bolt hole, subtract dh × t where dh = bolt diameter + 1/16 in. (per AISC Table J3.3) for standard holes, and t is the element thickness. Apply net section adjustments s^2/(4g) per AISC B4.3b for staggered bolt patterns to add back area in zigzag paths.

What is block shear and when does it govern tension member design?

Block shear is a connection limit state per AISC J4.3 where a block of material tears out at the connection end. It combines shear rupture (or yielding) along one failure plane and tension rupture (or yielding) along a perpendicular plane. Block shear typically governs for short connections with few bolts or thin connected material.

Why do we check both yielding and rupture for tension members?

Yielding on the gross section (Ag) controls the initiation of inelastic behavior, while rupture on the net section (Ae) controls the ultimate fracture failure. Yielding is checked with phi = 0.90 and Fy because it is a ductile limit state. Rupture is checked with phi = 0.75 and Fu because fracture is a brittle failure mode requiring a higher safety margin.

Does the W10x49 work for a higher tension load?

At Tu = 400 kips: yielding utilization = 400/648 = 0.617, rupture utilization = 400/599.6 = 0.667. Still well within capacity. At Tu = 500 kips: yielding = 0.772, rupture = 0.834. Approaching practical limits. For tension loads above 450 kips, consider W10x54 or W12x53 for increased Ag and flange thickness.

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Related References

• AISC Table D3.1 — Shear Lag Factors
• Bolt Bearing and Tearout Reference
• Shear Lag Factor Reference
• AISC Block Shear Reference
• Steel Connection Types Reference
• AISC Steel Construction Tables
• How to Verify Calculations
• Load Combinations Guide

Disclaimer

This page is for educational and reference use only. It does not constitute professional engineering advice. All design values must be verified against the applicable standard and project specification before use. The site operator disclaims liability for any loss arising from the use of this information.