Tension Members

Yielding, rupture, shear lag, and block shear per AISC Chapter D.

AISC Reference Box

AISC 360-22 §D2 — Tensile Strength
AISC 360-22 §D3 — Effective Net Area (shear lag)
AISC 360-22 §J4.3 — Block Shear Strength
AISC Manual Part 5 — Design of Tension Members

Lecture Notes

Tension members carry axial loads that try to pull them apart. The classic examples are truss diagonals, hanger rods, and bracing. Because failure is gradual when controlled by yielding, but sudden when controlled by rupture, AISC 360-22 Chapter D requires checking BOTH limit states.

Yielding on the gross section uses φt = 0.90 and the full Ag. Rupture on the effective net section uses φt = 0.75 and Ae = U·An, where U is the shear-lag factor from Table D3.1. Block shear (J4.3) governs when only a few bolts connect the member.

Slenderness L/r is recommended (not required) to be ≤ 300 for tension members to control vibration and handling damage.

Project case study — Cardinal Square — 4-story braced-frame office

Every chapter's worked example is one step in the design of the same building: Plan: 4 bays N–S × 3 bays E–W, each 30 ft × 30 ft. Stories: 4 @ 13 ft (52 ft roof). Composite floor: 4.5 in NW concrete on 3 VLI20 deck. Roof: 1.5 in B-deck + insulation + membrane. Materials: Wide-flange members A992 (Fy = 50 ksi, Fu = 65 ksi). Plates A572 Gr. 50. HSS bracing A500 Gr. C. Bolts A325-N 7/8 in dia. Welds E70XX. Concrete f'c = 4 ksi. Anchor rods F1554 Gr. 36.

Chapter 4 — Tension members (CBF diagonals)

Typical first-story CBF diagonal — HSS or 2L back-to-back

Demand carried forward

From a story shear V = 80 kips on a 30 × 13 ft braced bay: Tu ≈ V · L / (2·h) ≈ 80 · 32.8 / 26 ≈ 100 kips per brace (tension half of an X-brace).

This chapter contributes

Sizes the brace by checking gross-section yielding (φ = 0.90) and effective-net-section rupture (φ = 0.75) with shear-lag factor U from Table D3.1. Verifies block shear at the gusset.

Feeds into next chapter

Brace force Tu and gusset geometry feed Chapter 12 (bolts) and Chapter 16 (column base plates that resist the brace uplift).

AISC Chapter D limit states: gross-section yielding, net-section rupture (shear lag), and block-shear tear-out.

Formula Sheet

Name	Equation	AISC Ref
Yielding (gross)	φt Pn = 0.90 · Fy · Ag	AISC §D2(a)
Rupture (net)	φt Pn = 0.75 · Fu · Ae	AISC §D2(b)
Effective net area	Ae = U · An	AISC §D3
Net area	An = Ag − Σ(dh · t)	AISC §B4.3
Block shear	Rn = 0.60 Fu Anv + Ubs Fu Ant ≤ 0.60 Fy Agv + Ubs Fu Ant (φ = 0.75)	AISC §J4.3

Worked Example

Design a tension diagonal in a roof truss

Given

Service dead load Pd = 30 kips, service live load Pl = 70 kips. Member is an A36 angle (Fy = 36 ksi, Fu = 58 ksi). Connected by a single line of 7/8" diameter bolts (one bolt hole at each end, U = 0.80).

Load combination

Controlling LRFD combo: 1.2D + 1.6L (ASCE 7).

Required strength

Pu = 1.2(30) + 1.6(70) = 36 + 112 = 148 kips.

Limit states

Yielding on the gross section (AISC D2(a))
Rupture on the effective net section (AISC D2(b))
Block shear at the connection (AISC J4.3)

AISC reference

AISC 360-22 Chapter D — Design of Members for Tension

Solution steps

1. Required strength
Pu = 148 kips (from 1.2D + 1.6L).
2. Try a section
Try a single L4x4x3/8 angle: Ag ≈ 2.86 in² Instructor should verify Ag from official AISC Manual Part 1.
3. Gross-section yielding
φt Pn = 0.90 · Fy · Ag = 0.90 · 36 · 2.86 = 92.7 kips NG — need a larger section.
4. Re-try L5x5x3/8
Ag ≈ 3.61 in². φt Pn = 0.90 · 36 · 3.61 = 117.0 kips Still NG Try L5x5x1/2: Ag ≈ 4.75 in². φt Pn = 0.90 · 36 · 4.75 = 153.9 kips ≥ 148 ✓.
5. Effective net area
An = Ag − (dh · t) = 4.75 − (15/16)(1/2) = 4.28 in² Ae = U · An = 0.80 · 4.28 = 3.42 in².
6. Tensile rupture
φt Pn = 0.75 · Fu · Ae = 0.75 · 58 · 3.42 = 148.8 kips ≥ 148 ✓ (just barely — check block shear too).
7. Block shear (J4.3)
Compute Rn = 0.60 Fu Anv + Ubs Fu Ant ≤ 0.60 Fy Agv + Ubs Fu Ant apply φ = 0.75 Verify against connection geometry.

Final design decision

Use L5x5x1/2, A36, with U = 0.80. All three LRFD limit states satisfied.

Common mistakes in this example

Forgetting to add 1/16" to the bolt diameter when computing hole size (use dh = db + 1/8" for standard holes).
Using φ = 0.90 for the rupture limit state instead of 0.75.
Skipping the shear-lag factor U and using An directly.
Ignoring block shear when the connection has few bolts.

FE-Style Worked Examples (7)

Each example mirrors the NCEES FE Civil Reference Handbook style: brief givens, a labeled figure, AISC section reference, step-by-step numeric solution, and a single boxed answer.

Given

L4×4×3/8, A36, Ag = 2.86 in².

AISC Reference

AISC §D2(a)

Step-by-step solution

φPn
0.90 × 36 × 2.86 = 92.7 k

Answer φPn (yielding) = 92.7 k.

Course Materials — Lecture & Worked Examples

The following lecture highlights and worked examples are drawn directly from the instructor's uploaded course materials (Lecture 2 — Tension Members, CE 436 Week 2 notes, Tension Members example sets, and the Block-Shear presentation). All figures are page extracts from those documents.

Lecture highlights (from instructor notes)

Tension members carry axial pull only — trusses (top/bottom chords, diagonals), bracing (X-brace, K-brace), sag rods, hangers, cables, and lap-splice plates.
Stress in an axially-loaded member is f = P/A. The distribution is uniform except (a) near the point of load application and (b) at any cross section with holes or discontinuities.
Two strength limit states (AISC 360 §D2): gross-section YIELDING — φt = 0.90, Pn = Fy·Ag; and effective-net-section RUPTURE — φt = 0.75, Pn = Fu·Ae, with Ae = U·An.
Hole-size rule (AISC B4.3b): dh = db + 1/16″ for the punched hole, and add another 1/16″ for the roughened edge → use (db + 1/8″) when computing An.
Shear-lag factor U from AISC Table D3.1 accounts for only part of the cross-section being connected. U = 1 − x̄/L ≤ 0.90 (and never less than the tabulated case value — use the larger).
Staggered bolt holes add the s²/(4g) correction term: An = Ag − Σ(db + 1/8)·t + Σ s²/(4g)·t. The critical path is the one giving the smallest net width.
Block shear (§J4.3) tears a block of material from the connection: Rn = 0.60 Fu Anv + Ubs Fu Ant ≤ 0.60 Fy Agv + Ubs Fu Ant, with φ = 0.75. Ubs = 1.0 for uniform tension, 0.5 for non-uniform.
Serviceability: AISC §D1 recommends KL/r ≤ 300 for tension members to avoid sag, flutter, vibration, and handling damage.

Given

PL ½″ × 12″, A36 steel. Single 7/8″-diameter bolt hole on the cross section.

AISC Reference

AISC 360-22 §B4.3b (hole size), §D3

Step-by-step solution

1. Gross area
Ag = 12 × ½ = 6.0 in²
2. Hole allowance
Use (db + 1/8) = 7/8 + 1/8 = 1.0 in
3. Net area
An = Ag − t·Σ(db + 1/8) = 6.0 − (½)(1)(1.0) = 5.50 in²

Answer An = 5.50 in²

Textbook — Aghayere & Vigil (2009)(6 worked examples with figures + numerical answers)

Worked examples scanned directly from the CEGR 436 course textbook. Each card shows the original page (figure + full step-by-step solution) and adds an FE-style numerical multiple-choice prompt with answer key.

Chapter summary

Chapter 4 (Tension Members) walks through (a) gross-section yielding, (b) net-section rupture with shear-lag U, and (c) block shear. Examples 4-1 through 4-9 cover plates, single and double angles, and W-shape chords with bolted and welded connections.

φt Pn = 0.90 Fy Ag (yielding) vs 0.75 Fu Ae (rupture) — always check both.
An = Ag − Σ(dh · t); dh = db + 1/8" for standard holes.
Staggered bolts: add s²/(4g) for each diagonal segment when computing An (AISC §B4.3b).
Shear lag U: pick from Table D3.1 (Case 2: 1 − x̄/L for W/Tee/angles connected by one element).
Block shear: Rn = 0.60 Fu Anv + Ubs Fu Ant ≤ 0.60 Fy Agv + Ubs Fu Ant ; φ = 0.75 (§J4.3).
L/r ≤ 300 recommended for tension members.

Setup

PL 1/2×8 (Ag = 4.0 in²), A36 steel (Fy = 36 ksi). Find φPn for gross-section yielding.

AISC Reference

AISC §D2(a)

Numerical practice

φPn (yielding)?

A. 108 kipsB. 118 kipsC. 130 kipsD. 144 kips

Textbook page 153 — Yielding capacity of a tension plate — Aghayere & Vigil (2009), p. 153 — full worked solution & sketch.

FE Practice Bank (6)

Interactive Calculator

Tension Member Design

AISC §D2, §D3

FyksiFuksiAg (gross area)in²An (net area)in²U (shear lag, Table D3.1)Pu (required)kips

Ae = U·An3.42 in²

φ Pn (yielding) = 0.90 Fy Ag213.8 kips

φ Pn (rupture) = 0.75 Fu Ae166.9 kips

Governing φ Pn166.9 kipsOK

Practice Problems

[E] State AISC §D2 yielding and rupture φ-factors (0.90 and 0.75).
[E] Compute dh for a 7/8 in. bolt in a standard hole (dh = db + 1/8).
[E] Define shear-lag factor U and where it comes from (AISC Table D3.1).
[E] State recommended slenderness for tension members (L/r ≤ 300).
[E] List the three limit states for a bolted tension angle.
[M] Design a W8 tension chord for Pu = 220 k, A992, two lines of 3/4 in. bolts.
[M] A 1 in. dia. A36 rod hangs Pu = 18 k. Check yield + rupture; recommend thread size.
[M] L6x4x1/2, A36, one line of 7/8 in. bolts. Compute φPn; identify governing limit state.
[M] Plate 1/2 x 8 in., A36, two staggered lines of 3/4 in. bolts; pitch s = 2.5 in., gauge g = 3 in. Compute An using s²/4g.
[M] WT5x16.5 connected by both flanges with three bolts each side. Compute U using Table D3.1 Case 2.
[H] Design a single L5x5 tension diagonal for Pu = 165 k, A36, 7/8 in. bolts in one leg.
[H] Block-shear check: 3/8 in. plate, A572 Gr 50, 4-bolt single line. Compute Rn per §J4.3.
[H] Double-angle 2L4x4x3/8 LLBB welded to gusset, Lw = 8 in. Compute U and An.
[H] Build full LRFD design table (yield, rupture, block shear) for L4x3x3/8 with two 3/4 in. bolts in long leg, A36.
[H] Threaded-rod connection: 1-1/4 in. A325 anchor rod with cut threads in shear plane. Compute φRn shear, tension, and combined per §J3.7.

Structured Clues

Check yielding AND rupture AND block shear — the smallest governs.
dh = db + 1/8 for standard holes; Ae = U·An (Table D3.1).
For threaded rods use Ae = 0.75·Ag and Fu (not Fy).

Code References

AISC 360-22 §D2, D3, J3.6, J4.3
AISC Manual Part 5, 7

Quiz (6 FE-bank + 4 concept)

Common Student Mistakes

Using gross area for the rupture check.
Using φ = 0.90 instead of 0.75 for rupture.
Forgetting shear lag U.
Skipping block shear at the connection.
Computing hole size as the bolt diameter (forgetting the +1/16" + 1/16" allowance).

"Professor Explains" Script

Today we're talking about tension members. Think of this topic as one step in the LRFD workflow: identify the demand, identify the limit states from the relevant AISC chapter, then check that φ·Rn is at least equal to Ru. We'll walk through the failure modes, the equations, and a worked example. Pay close attention to where the resistance factor changes — that's where students lose points on exams.