AISC 360-16 Chapter D — Design of Members for Tension

Complete guide to tension member design per AISC 360-16 — yielding, rupture, shear lag, block shear, and two fully solved examples in LRFD & ASD

Tension members are the simplest structural elements to understand conceptually — a bar pulled from both ends — yet their design requires careful attention to connection details, hole patterns, and shear lag effects. Chapter D of AISC 360-16 establishes two primary limit states: tensile yielding of the gross section and tensile rupture of the net section. The design strength (LRFD) or allowable strength (ASD) is the lesser of the two.

This article covers every aspect of Chapter D with the depth needed for practical design. We derive each formula, explain the physical meaning behind the safety factors, walk through the shear lag table case by case, and solve two complete numerical examples — a single angle and a W-shape — in both LRFD and ASD. We also cover the block shear rupture check from Chapter J, which frequently governs the design of tension connections.

D1 — Slenderness Limitations

Section D1 recommends that the slenderness ratio of tension members not exceed 300:

where $L$ is the length of the member between work points and $r$ is the governing (minimum) radius of gyration. This is a serviceability recommendation, not a mandatory limit — the Commentary explicitly states that members with $L/r > 300$ are not structurally inadequate but may experience objectionable sag under self-weight, vibration under dynamic loads, or difficulty in handling during erection. For rod bracing and other members designed for a small tensile force, exceeding 300 is common and acceptable provided the engineer accounts for these practical effects.

Unlike compression members, where slenderness directly reduces capacity via buckling, a slender tension member carries the same axial load regardless of $L/r$. The 300 limit is purely practical.

D2(a) — Tensile Yielding of the Gross Section

The first limit state is yielding across the gross cross-sectional area $A_g$. When the average stress on the gross section reaches the yield stress $F_y$, the entire member begins to elongate plastically. Although this does not cause immediate fracture, excessive elongation renders the structure unserviceable — a tension brace that has yielded by 1% will have permanently lengthened by $L/100$, causing visible sag and redistribution of forces to other members.

The nominal strength for this limit state is:

The design strength and allowable strength are:

The resistance factor $\phi_t = 0.90$ and safety factor $\Omega_t = 1.67$ reflect that yielding is a ductile limit state with significant reserve — the member does not fail catastrophically, it simply elongates. These factors correspond to a target reliability index $\beta \approx 3.0$.

This check uses the gross area $A_g$ because yielding occurs over the full length of the member, not just at the bolt holes. At the net section, strain hardening allows stresses beyond $F_y$ before the gross section yields uniformly.

D2(b) — Tensile Rupture of the Net Section

The second limit state is fracture through the net section — the cross-section reduced by bolt holes or other openings. Unlike yielding, rupture is a brittle, sudden failure with no warning. The member literally tears apart at the weakest plane.

The design strength and allowable strength are:

Notice the significantly lower resistance factor $\phi_t = 0.75$ (compared to 0.90 for yielding) and the higher safety factor $\Omega_t = 2.00$ (compared to 1.67). This reflects the higher consequence of a rupture failure: there is no ductile warning, no energy absorption, no redistribution. The target reliability index for rupture is $\beta \approx 4.0$, compared to $\beta \approx 3.0$ for yielding.

This check uses the ultimate tensile strength $F_u$ (not $F_y$) because rupture occurs at the peak stress the material can sustain. It uses the effective net area $A_e$ (not gross area) because fracture initiates at the reduced section through the bolt holes.

Net Area A_n

The net area is the gross area minus the material removed for bolt holes. Per AISC 360-16 Section B4.3b, the width of a standard bolt hole is taken as the nominal bolt hole diameter, which is the bolt diameter $d_b$ plus 1/16 in. for clearance, plus an additional 1/16 in. to account for damage to the surrounding material during punching or drilling:

For a single line of bolts with $n$ holes in the critical section:

where $t$ is the thickness of the element containing the holes.

Staggered Bolt Holes — The Cochrane Formula

When bolts are arranged in a staggered (zigzag) pattern, the critical failure path may run diagonally through multiple holes rather than straight across. In 1922, V.H. Cochrane proposed an elegant empirical correction: for each diagonal segment in the failure path, add back a quantity $s^2/(4g)$ to the net width, where:

• $s$ = longitudinal spacing (pitch) between consecutive bolts along the member axis
• $g$ = transverse spacing (gage) between bolt rows perpendicular to the member axis

The net width for any candidate failure path is:

Then $A_n = w_n \cdot t$. The engineer must check all possible paths (straight and diagonal) and use the one that gives the smallest net area.

The physical intuition behind $s^2/(4g)$ is that a diagonal path is longer than a straight path, requiring more energy to propagate the fracture. As the pitch $s$ increases relative to the gage $g$, the diagonal path becomes increasingly longer and less critical. When $s$ is large enough, the straight path through fewer holes governs instead.

AISC 360 Section B4.3b: for angles, the gage for holes in opposite legs is measured along the back of the angle, minus the leg thickness. This "unfolding" technique allows treating the angle as a flat plate.

Effective Net Area A_e and the Shear Lag Factor U

When tension is transmitted through only some elements of the cross-section (for example, an angle connected by one leg, or a W-shape connected by its flanges only), the stress distribution across the section is non-uniform. The connected elements carry higher stress while the unconnected elements carry lower stress — they "lag" behind because the load must transfer via shear through the cross-section. This phenomenon is called shear lag.

The effective net area accounts for shear lag:

where $U$ is the shear lag factor ($0 < U \leq 1.0$). When all elements are connected, there is no shear lag and $U = 1.0$. When only some elements are connected, $U < 1.0$ reduces the effective area.

Alternative Formula: U = 1 - x̄/L

The general analytical expression for the shear lag factor is:

where $\bar{x}$ is the connection eccentricity — the perpendicular distance from the centroid of the connected element(s) to the plane of connection — and $L$ is the length of the connection, measured as the distance between the first and last bolts in the line, or the length of the weld. This formula was derived by Munse and Chesson (1963) from regression analysis of over 1,000 test specimens.

Table D3.1 — Shear Lag Factor for Bolted and Welded Connections

AISC 360 Table D3.1 provides tabulated values of $U$ for common connection geometries. The engineer may use the larger of the table value or the formula value:

The specification permits taking the larger of the formula value ($1 - \bar{x}/L$) and the applicable table value. In many cases the formula gives a higher (less conservative) result, especially for longer connections. Always compute both.

For welded connections, additional cases in Table D3.1 cover transverse welds (U = 1.0 when all elements are connected with transverse welds), longitudinal welds only, and plate connections. For members connected by longitudinal welds only, the specification provides U values that depend on the weld length relative to the member width.

D5 — Pin-Connected Members

Pin-connected members transfer tension through a single pin bearing on the member in a hole. The design must check tensile rupture on the net section through the pin hole, shear rupture on the effective area beyond the pin hole, bearing on the projected area of the pin, and yielding on the gross section. The nominal strength for rupture on the net section is:

where $b_{\text{eff}} = 2t + 16 \text{ mm (0.63 in.)}$, but not more than the actual distance from the edge of the hole to the edge of the member. The resistance factor is $\phi_t = 0.75$ (LRFD) or$\Omega_t = 2.00$ (ASD). Dimensional requirements include a minimum net area beyond the pin hole of $2t \cdot b_{\text{eff}}$ on each side, and the pin hole diameter must not exceed the pin diameter by more than 1/32 in. (1 mm).

D6 — Eyebars

Eyebars are a specialized pin-connected tension member with a circular enlarged head forged or flame-cut from the same plate as the body. They were common in historic truss bridges and are still used in some long-span roof trusses. The nominal strength is limited to yielding:

where $A_b$ is the cross-sectional area of the body. The resistance factor is $\phi_t = 0.90$ (LRFD) or $\Omega_t = 1.67$ (ASD). Prescriptive geometric requirements in Section D6 ensure that the head is strong enough to prevent rupture: the width of the head through the pin hole must be between $\tfrac{2}{3} w$ and $\tfrac{3}{4} w$ where $w$ is the body width, and the head transition radius must be at least the head width.

Solved Example 1 — Single Angle L4×4×1/4, A36 Steel

Given: A single angle L4×4×1/4 is used as a tension brace connected by one leg with 3 bolts in a single line, 3/4 in. diameter (A325-N), spaced at 3 in. on center. The member is 10 ft long. Steel is ASTM A36 ($F_y = 36$ ksi, $F_u = 58$ ksi).

Section properties: $A_g = 1.94$ in.², $r_{\min} = 0.776$ in., $\bar{x} = 1.18$ in. (distance from the centroid to the back of the connected leg), $t = 0.25$ in.

Solved Example 2 — W10×33, A992, Flange Connection

Given: A W10×33 tension member (ASTM A992, $F_y = 50$ ksi, $F_u = 65$ ksi) is connected through both flanges using 4 bolts per flange (8 bolts total), 7/8 in. diameter (A325-N), arranged in 2 rows per flange. The web is not connected. Bolt spacing is 3 in. in the direction of load; gage in each flange is 5-1/2 in.

Section properties: $A_g = 9.71$ in.², $b_f = 7.96$ in., $d = 9.73$ in., $t_f = 0.435$ in., $t_w = 0.290$ in.

Block Shear Rupture (AISC 360 Section J4.3)

Block shear is a combined shear and tension failure mode where a "block" of material tears out of the member along the bolt pattern. It is not part of Chapter D, but it frequently governs the design of tension connections and must always be checked. The failure involves shear along one or more planes parallel to the load and tension on a perpendicular plane.

The nominal block shear rupture strength is:

where:

• $A_{gv}$ = gross area subject to shear (along the bolt line)
• $A_{nv}$ = net area subject to shear (shear planes minus bolt holes)
• $A_{nt}$ = net area subject to tension (perpendicular tension plane minus bolt holes)
• $U_{bs}$ = 1.0 when the tension stress is uniform, 0.5 when it is non-uniform

The left side of the inequality represents shear rupture + tension rupture. The right side caps the shear component at shear yielding — because if the shear planes are long relative to the tension plane, the shear planes yield before they can reach the full rupture stress. The design strength is $\phi = 0.75$ (LRFD) or $\Omega = 2.00$ (ASD).

U_bs — Uniform vs. Non-Uniform Tension

$U_{bs} = 1.0$ applies when the tension stress on the perpendicular plane is essentially uniform — as in a simple single-row bolt pattern where the load is concentric. $U_{bs} = 0.5$ applies when the tension stress is highly non-uniform, typically in coped beam connections or gusset plates where eccentricity causes one end of the tension plane to be more highly stressed. For the single angle in Example 1, the tension is reasonably uniform across the connected leg, so $U_{bs} = 1.0$.

Block Shear Check for Example 1 (L4×4×1/4)

Decision Tree — What Controls?

Understanding which limit state governs helps engineers develop intuition for efficient tension member design. The following guidelines apply for LRFD:

When Yielding Governs

Yielding governs when the effective net area is large relative to the gross area. Setting the two limit states equal:

For A36 steel ($F_y/F_u = 36/58$): $A_e/A_g > 0.90 \times 36 / (0.75 \times 58) = 0.745$. For A992 steel ($F_y/F_u = 50/65$): $A_e/A_g > 0.90 \times 50 / (0.75 \times 65) = 0.923$. Higher-strength steels have a smaller gap between $F_y$ and $F_u$, making it harder for yielding to govern. Practically, yielding controls when:

• All elements are connected ($U = 1.0$)
• Few bolt holes relative to the gross area
• Lower-strength steel (A36 rather than A992)
• Welded connections (no bolt hole deductions)

When Rupture Governs

Rupture governs in most practical bolted connections, especially when:

• Only some elements are connected (angles, tees, channels connected by one element)
• Many bolt holes reduce the net area significantly
• Higher-strength steels with $F_y/F_u$ close to 1.0
• Short connections that increase shear lag ($U$ decreases)

When Block Shear Governs

Block shear tends to control when:

• Few bolts (2-3) with short edge distances
• Single-row connections in angles and tees
• Small gage distances that create a narrow block
• Coped beams and gusset plate connections

As a rule of thumb: for single angles with 2-3 bolts, always check block shear first — it will frequently be the controlling limit state, as demonstrated in Example 1.

Comparison: AISC 360 vs. Eurocode 3 vs. NBR 8800

The three most widely used steel design standards handle tension members with the same fundamental philosophy but differ in safety format, shear lag treatment, and hole deduction rules:

Key Differences

Eurocode 3 applies a 0.9 reduction factor to the net area in the rupture check, which AISC and NBR 8800 do not. This accounts for local stress concentrations around holes that may reduce ductility. However, the Eurocode uses $\gamma_{M2} = 1.25$ for rupture, which is less conservative than AISC's equivalent $1/\phi_t = 1/0.75 = 1.33$. The net effect is that Eurocode 3 and AISC 360 produce similar rupture capacities for most practical sections.

NBR 8800 is directly derived from AISC 360. The shear lag factor $C_t$ is identical to $U$, the Cochrane formula is the same, and the block shear equation is the same. The only differences are the safety format (partial safety factors instead of resistance/safety factors) and the hole deduction rule (metric: $d_b + 2$ mm vs. imperial: $d_b + 1/8$ in., which are approximately equivalent).

Tension Member Design Checklist

1. Verify slenderness: $L/r \leq 300$ (recommendation)
2. Compute gross yielding: $\phi_t P_n = 0.90 \, F_y \, A_g$
3. Determine hole diameter: $d_h = d_b + 1/8$ in. (standard holes)
4. Compute net area $A_n$ — check all possible failure paths including staggered
5. Determine shear lag factor $U$ from Table D3.1 and from $1 - \bar{x}/L$; use the larger
6. Compute effective net area: $A_e = U \cdot A_n$
7. Compute net section rupture: $\phi_t P_n = 0.75 \, F_u \, A_e$
8. Compute block shear rupture: $\phi R_n$ per J4.3
9. Design strength = minimum of steps 2, 7, and 8
10. Verify: $P_u \leq \phi_t P_n$ (LRFD) or $P_a \leq P_n / \Omega_t$ (ASD)