Engineering story

A century of steel — from concept to skyline

Engineer reviewing blueprints against a steel-frame construction site at sunrise

The simplest structural action, and the least forgiving.

Tension members either hold — or they let go, all at once.

Tension diagonals, hangers, and bracing carry axial pull directly along their axis. Chapter D of AISC 360-22 governs yielding on the gross section, rupture on the effective net section, block shear at the connection, and slenderness for handling — the four checks that keep a tension member safely in service.

Iconic steel structures built on engineering excellence

Eads Bridge — wrought-iron tension chords
St. Louis1874
Silver Bridge collapse (eyebar fatigue)
Point Pleasant, WV1967
Hyatt Regency walkway — hanger-rod failure
Kansas City1981
AISC 360-22 Chapter D (current)
2022

Load pathApplied pullGross section (yielding)Net section (rupture)Connection (block shear)Reaction

02

Learning objectives

What you will be able to do after finishing Chapter 1 — and why each objective matters in practice

Objective 01

Check yielding on the gross section

Apply φt Pn = 0.90 · Fy · Ag as the ductile limit that governs away from the connection.

Why it matters: Yielding is the ductile limit — it warns before it breaks. Missing this check means you never verify that the member has the warning behaviour AISC expects.
Where it is used: Every tension diagonal, hanger, bracing chord, and truss element you will ever design.
Connects to: AISC 360-22 §D2(a); the same φ = 0.90 recurs for flexure and shear yielding.

Objective 02

Check rupture on the effective net area

Compute An = Ag − Σ(dh·t), apply the shear-lag factor U from Table D3.1 to get Ae, then φt Pn = 0.75 · Fu · Ae.

Why it matters: Rupture is brittle and sudden. A member that passes yielding but fails rupture will still tear at the bolt holes with no warning.
Where it is used: Any bolted or welded end connection where holes or partial connection reduce the effective area.
Connects to: AISC §D2(b) + §D3; shear lag reappears in Chapter J connection design.

Objective 03

Compute block-shear capacity

Rn = 0.60 Fu Anv + Ubs Fu Ant ≤ 0.60 Fy Agv + Ubs Fu Ant, with φ = 0.75.

Why it matters: Short end connections often let block shear govern — and the block-shear plane is missed by every non-D2 check. Skipping it is the classic first-year mistake.
Where it is used: Bolted angle bracing, coped beam webs, gusset plates on WT hangers.
Connects to: AISC §J4.3; the same tear-out kinematics appear in bolt bearing (§J3.10).

Objective 04

Navigate AISC Chapter D + Manual Part 5

Locate Ag, An, U, and rmin fluently in the AISC Manual dimensions and design tables.

Why it matters: Every number in a tension calc comes from the Manual — being slow there slows every design and every FE exam problem.
Where it is used: Every steel tension design you will ever produce.
Connects to: AISC Manual Part 5 (Design of Tension Members); Table D3.1 for U.

Objective 05

Screen slenderness L/r ≤ 300

The AISC recommended (not required) slenderness limit prevents sag, vibration, and handling damage.

Why it matters: A theoretically strong long slender rod can flap in the wind and fail in handling. The 300 screen catches this before fabrication.
Where it is used: Roof-bracing rods, long tie-rods, sag rods on purlins.
Connects to: AISC §D1 (User Note); compare with compression limit KL/r ≤ 200 (§E2).

Objective 06

Answer FE-style tension questions

Recognise which limit state controls, and pick the correct φ, area, and stress at a glance.

Why it matters: Tension members are a high-frequency FE Structural topic — direct, testable, and unforgiving.
Where it is used: FE Civil / FE Structural, PE Structural, plan-check reviews.
Connects to: The FE Prep panel at the end of this chapter drills these directly.

03

Engineering motivation

What each part of a steel-frame building actually does — and why it exists

Before you design any single member, you have to see the whole system. A steel-frame building is not a collection of independent shapes bolted together — it is a deliberate load path, engineered so that every kilonewton of gravity, wind, or seismic demand has a continuous route from where it starts to the ground where the earth can resist it.

The photograph below shows a typical steel framing detail. Drag each labelled chip onto the structural element it names — the drop is only accepted when it lands inside the correct element's outlined region. A correct answer locks in with a green outline and a short explanation; a wrong answer flashes the region red, tells you what you actually hit, and returns the chip so you can try again. Press Reveal expected placements to see the reference solution (that attempt is then marked as assisted).

Structural steel framing — identify each element by dragging the labels — Fig. 1.3 · Drop a chip inside the outlined element it names. Green = correct and locked; red flash = wrong element, chip returns.

04

Failure mechanisms

Why we design the way we do — six ways steel structures have failed, and what each disaster taught the profession

Every provision in AISC 360 is a scar. Behind each equation, load factor, and detailing rule is a bridge, a walkway, or a tower whose failure cost lives and rewrote the profession. The six case studies below trace the mechanisms that motivate the code you are about to learn.

Read each one as an engineer, not a spectator: identify the load, the limit state, the missing check, and the specific clause that exists today because that check was missed. When you meet those clauses again in Chapters 5–17, they will read as answers, not rules.

Failure mechanism

Block shear tear-out

A rectangular block of the connection plate or angle leg tears out through a combination of shear on the bolt lines and tension across the end row.

Root cause

Short connections concentrate demand on a few bolts. Rn = 0.60 Fu Anv + Ubs Fu Ant becomes small when the bolt group is compact.

Lesson learned

Always evaluate block shear per §J4.3 — it commonly governs for angles bolted through one leg with few bolts.

§AISC 360-22 §J4.3

Failure mechanism

Threaded-rod rupture

A hanger rod snaps at the threads where the net (root) area is smallest and the stress is amplified by the stress concentration in the thread.

Root cause

Designers sometimes use the nominal (gross) area to check the rod, forgetting that Ae is what carries the tensile force at the threads.

Lesson learned

Use the tensile stress area (root area) for threaded rods; check both yielding of the shank and rupture at the threads.

§AISC 360-22 §D2 · Manual Table 7-18 (threaded rods)

Failure mechanism

Hyatt Regency walkway (1981)

A field change doubled the demand on a single hanger-rod nut. The nut punched through the box beam under service load, dropping two walkways and killing 114 people.

Root cause

The as-built detail transferred twice the design load through a single bolted joint — a tension-connection error, not a member error.

Historical case

Kansas City, 17 July 1981.

Lesson learned

Every tension member must be traced to its connection. A rod is only as strong as the plate, nut, and bearing surface at each end.

§AISC 360-22 §J4 (connecting elements) · Engineering ethics case study

Failure mechanism

Fatigue cracking at bolt holes

Cyclic load ranges propagate a hairline crack from a drilled bolt hole edge through the net section until the remaining ligament fractures suddenly.

Root cause

Sharp-edged holes are stress raisers. Under cyclic tension the crack grows one Paris-law increment per cycle.

Lesson learned

For cyclically loaded tension members, apply AISC Appendix 3 fatigue provisions and prefer reamed / clean-drilled holes.

§AISC 360-22 Appendix 3 (Fatigue)

How failure propagates

The five-stage failure progression

1Applied load

2Elongation

3Yielding

4Rupture

5Fracture

Design codes intervene at the transition from yield to instability. Everything before yield is elastic and reversible; everything after instability is a race to collapse. LRFD keeps the demand well below the first transition.

05

Lecture notes

The full textbook chapter — figures, equations, and engineering narrative

Reflection · Think before you read

A single-angle tension member is bolted through only one leg. Why is the full gross area not effective, and how would you redesign the connection to recover the lost capacity?

Chapter 4 — Tension Members (AISC 360-22 Chapter D)

Chapter focus. Tension members — truss diagonals, hanger rods, cross-bracing — carry pure axial pull. They can fail two ways: gradual yielding over the full cross-section (ductile, φ = 0.90) or sudden rupture through the bolt holes (brittle, φ = 0.75). This chapter shows how to compute Ae with the shear-lag factor U, size a section against both limit states, and check block shear on the connection plates.

1. Behavior of Tension Members

A tension member transmits an axial pulling force along its longitudinal axis. Typical members: truss chords/webs, hangers, diagonal bracing in braced frames, sag rods. Because tension members do not buckle, their strength is governed purely by cross-section capacity, connection details, and stress concentrations near holes.

Load flows in tension members along a nearly uniform stress field except near bolt/weld connections, where load must funnel through fewer elements — producing the classic three limit states below.

Representative Structures

Where tension members appear in real engineering practice. In each photograph, the highlighted element carries essentially pure axial tension — its slender geometry is efficient because tension members do not buckle.

Golden Gate Bridge suspender ropes — Golden Gate Bridge — Suspender Ropes
Each vertical wire rope hangs the deck from the main cable. The deck weight flows straight up through the hanger as pure axial tension, then into the main catenary, then into the anchorages. Wire rope is the ideal geometry: enormous cross-sectional efficiency with no buckling concern.

Pont de Normandie cable-stayed bridge — Pont de Normandie — Cable-Stayed Bridge
Inclined stay cables run directly from each pylon to the deck. Every stay carries pure axial tension proportional to the deck load it supports. The pylon collects the horizontal components of the stays as compression — a clean separation of tension (stays) and compression (mast).

Sydney Harbour Bridge steel arch and diagonals — Sydney Harbour Bridge — Truss Diagonals
In the steel through-arch truss, alternating diagonals resist shear as tension while their neighbors act in compression. The lighter, slenderer members are typically the tension diagonals — sized by A_g·F_y (yielding) and the bolt-hole net section (rupture).

Electricity transmission tower with tension diagonals — Lattice Transmission Tower — Tension Diagonals
Single-angle steel diagonals stitch the four tower legs together against wind and conductor pull. Under a given load direction roughly half the diagonals go into tension. These are the canonical bolted single-angle tension members — governed by shear-lag (U < 1) and block shear at the gusset plates.

Hong Kong Stadium roof tension rods — Stadium Roof — Tension Rod Suspension
Long-span stadium roofs suspend the outer edge from raking masts through slender high-strength tension rods. Because rods do not buckle, their diameter is set by yielding (F_y·A_g) and by the threaded net area at the clevis end — a textbook tension design.

Guyed communication tower guy wires — Guyed Communication Tower — Guy Cables
Pretensioned steel guy cables triangulate the mast against wind, letting a very slender central column reach 300+ m. Each guy is a pure tension element — sized by yielding of the wire rope and by the swaged/threaded end fittings.

2. LRFD Design Inequality

φ_t · P_n ≥ P_u

P_u is the largest factored axial tension from ASCE 7 combinations. P_n is the smallest of three nominal capacities — gross yielding, net-section rupture, and block shear.

3. The Three Governing Limit States — Overview

Before we compute anything, tour the three failure modes side-by-side. Each has its own physics, its own AISC clause, and its own φ. The remainder of the chapter builds the tools you need (net area, shear-lag U, block-shear planes) and then applies each limit state in turn.

Three tension-member failure modes: (a) gross-section yielding, (b) net-section rupture, (c) block shear rupture

(a) Gross-section yielding — the whole cross-section stretches. Ductile. φ_t = 0.90 (§D2-a).
(b) Tensile rupture on the effective net section — fracture tears across the bolt holes at the end connection. Brittle. φ_t = 0.75 (§D2-b).
(c) Block shear — a chunk of steel around the bolt group tears out along combined shear + tension planes. φ = 0.75 (§J4.3).

Sections 4–8 build these three checks in the order you must compute them: yielding first (needs only A_g), then rupture (needs A_n and U — Sections 5 and 6 build them), then block shear (Section 8 identifies the failure planes).

4. Limit State 1 — Gross-Section Yielding (AISC D2-a)

Nominal strength

(1)

P_{n} = F_{y} \cdot A_{g}

ϕ_{t} = 0.90

This is the simplest check. Away from the end connections the full gross area A_g resists load and the member stretches ductilely as stress reaches F_y. Ductility is why φ = 0.90 (larger than the 0.75 used for brittle limit states).

Where it governs: long, unconnected regions of the member — away from holes and welds. It rarely controls when connections are heavy on holes; rupture usually does. But you must always check it.

Mini-Example 4A — Design Yield Strength of an A36 Channel Truss Chord

Problem. A structural-steel tension member in a roof truss consists of an A36 channel with gross area A_g = 8.0 in². Determine the LRFD design tensile yielding strength φP_n.

Given: A36 steel (F_y = 36 ksi); A_g = 8.0 in²; φ = 0.90 (tensile yielding, §D2-a).

Step 1 — Identify the design equation.

P_{n} = F_{y} \cdot A_{g}

X X X ϕ P_{n} = ϕ \cdot F_{y} \cdot A_{g}

where P_n is the nominal tensile strength, φ = 0.90 is the LRFD resistance factor for yielding, F_y is the minimum yield stress, and A_g is the gross cross-sectional area.

Step 2 — Substitute given values.

X X X ϕ P_{n} = 0.90 \cdot (36 ksi) \cdot (8.0 in^{2})

Step 3 — Compute the final strength.

X X X ϕ P_{n} = 0.90 \cdot 288.0 kips

X X X = 216.0 kips

✅ Design tensile yielding strength: φP_n = 216 kips.

Mini-Example 4B — Minimum Required Gross Area (A992 Diagonal, P_u = 180 k)

Problem. Method-of-joints analysis gives a factored tension of P_u = 180 k in a diagonal truss member fabricated from A992 steel. Based on the LRFD limit state of tensile yielding on the gross section, find the minimum required gross area A_g. Options: (A) 3.24 in², (B) 3.60 in², (C) 4.00 in², (D) 4.44 in².

Given: A992 (F_y = 50 ksi); φ = 0.90; design requirement φP_n ≥ P_u.

Step 1 — Design inequality.

X X X ϕ P_{n} \geq P_{u}

X X X ϕ \cdot F_{y} \cdot A_{g} \geq P_{u}

A_{g} \geq \frac{P _{u}}{ϕ \cdot F _{y}}

Step 2 — Substitute and reduce.

A_{g} \geq \frac{180 kips}{0.90 \cdot 50 ksi}

X X X = \frac{180}{45}

X X X = 4.00 in^{2}

✅ Answer: C — A_g,min = 4.00 in² for tensile yielding on the gross section.

Interpretation. Any A992 shape with A_g ≥ 4.00 in² satisfies yielding; rupture and block-shear (Sections 5 – 6) still have to be checked at the end connection.

5. Limit State 2 — Tensile Rupture on the Effective Net Section (§D2-b)

Nominal strength

(2)

P_{n} = F_{u} \cdot A_{e}

A_{e} = U \cdot A_{n}

ϕ_{t} = 0.75

Fracture tears across the bolt holes at the end connection. Uses the ultimate stress F_u (not F_y) because the failure is brittle; the lower φ = 0.75 compensates for the loss of ductility.

Two ingredients feed the effective area A_e: the net area A_n (gross area minus bolt-hole material, §5.1) and the shear-lag factor U (fraction of the net area that actually engages, §5.2). The two are multiplied in §5.3 to give A_e, then φP_n is compared to P_u.

5.1 Net Area A_n — Hole Deduction and Staggered Holes (§D3)

Before we can check rupture (this limit state) or block shear (§6), we need the net area — the gross area minus the material removed by bolt holes on the critical failure path.

Hole size rule (D3.2): the actual punched hole is d_b + 1/16″, but AISC adds another 1/16″ for damage around the hole edge, so deduct d_h = d_b + 1/8″ when computing A_n.
Straight failure path: A_n = A_g − Σ(d_h · t) over the holes the path crosses.
Staggered holes: when the failure path must zig-zag diagonally between offset holes, add s²/(4g) for every diagonal segment (see below).
Governing net area: when multiple failure paths are possible, compute A_n for each and take the smallest.

5.1.1 Staggered Holes — the s²/4g Rule (D3.2)

When bolt holes are offset row-to-row (staggered), the failure path can zig-zag from one hole to the next diagonal one. That diagonal segment is partly in the shear direction, so a straight subtraction over-penalizes the net area. AISC D3.2 credits the diagonal path by adding s²/4g for every diagonal segment:

Two candidate failure paths — check both; smaller A_n governs.

Procedure:

Sketch every possible tear path from one edge of the plate to the other, going through as many holes as physically possible.
Compute the length of each path along the width (perpendicular to load) by adding straight and staggered segments.
For each hole crossed, subtract (d_b + 1/8″).
Add s²/4g for every diagonal segment between two holes offset by pitch s and gage g.
Multiply the resulting net width by t. The smallest A_n from all candidate paths governs.

5.1.2 Worked Mini-Example — Net Area of a Staggered-Bolt Plate

Given. 8″ wide plate, t = ½″, three ¾″ bolt holes on two gage lines with gage g = 3″ and pitch s = 2″. Two of the holes (A on Row 1 and B on Row 2) are vertically aligned across the two rows at the same x, and the third hole (C) is offset by the pitch s along Row 1. Find the governing A_n.

Holes A and B are vertically aligned across the two gage lines (same x); hole C is offset from A by the pitch s along Row 1. Path 1 is a straight cut at x=220 crossing the aligned pair A–B (2 holes, no s²/4g). Path 2 zig-zags through all 3 holes: A→B is straight (aligned, no diagonal credit) and B→C is the single diagonal segment (one s²/4g credit).

Step 1 — Hole deduction.

Formula

(3)

d_{h} = d_{b} + 1/8″

d_h = 0.75 + 0.125 = 0.875 in

Step 2 — Path 1 (straight, 2 holes).

Net width

(4)

w_{n, 1} = w - n_{1} \cdot d_{h}

w_n,1 = 8 − 2 · 0.875 = 6.25 in

Net area

(5)

A_{n, 1} = w_{n, 1} \cdot t

A_n,1 = 6.25 · 0.5 = 3.125 in²

Step 3 — Path 2 (zig-zag, 3 holes, one diagonal segment).

Diagonal credit (D3.2)

(6)

Δ = s^{2} / (4 g)

Δ = 2² / (4 · 3) = 4 / 12 = 0.333 in

Net width

(7)

w_{n, 2} = w - n_{2} \cdot d_{h} + Σ (s^{2} /4 g)

w_n,2 = 8 − 3(0.875) + 1(0.333) = 8 − 2.625 + 0.333 = 5.708 in

Net area

(8)

A_{n, 2} = w_{n, 2} \cdot t

A_n,2 = 5.708 · 0.5 = 2.854 in²

Step 4 — Governing net area (smaller controls).

(9)

A_{n} = min (A_{n, 1}, A_{n, 2})

A_n = min(3.125, 2.854) = 2.85 in² — Path 2 governs.

5.2 Shear-Lag Factor U (Table D3.1)

Why U exists: when only part of a cross-section is bolted (e.g. only the flanges of a W-shape, or one leg of an angle), the load must "shear-lag" its way into the unconnected part. Stress is non-uniform across the section near the connection, so the effective area is less than the full net area. AISC Table D3.1 gives U for the eight most common configurations.

Understanding Shear Lag and Its Impact

What is shear lag? Shear lag is a phenomenon that occurs when some, but not all, parts of a structural-steel cross-section are directly connected to a gusset plate or adjoining member. For instance, if only one leg of an angle — or only the web of a channel — is bolted down, the tensile force cannot distribute instantly across the entire profile. The stress must "flow" from the connected parts into the unconnected parts through longitudinal shear. Because the steel transitions this load over a finite length (the connection length), the stress "lags behind" in the unconnected components.

How does it impact net-section rupture? Near the connection, the stress distribution across the net area is non-uniform, producing highly localized stress concentrations around the fasteners. This non-uniformity reduces the cross-section's effectiveness at resisting fracture. To account for the inefficiency, the AISC Specification introduces the reduction factor U — the shear-lag factor — which modifies the physical net area A_n to give a smaller effective net area A_e:

A_e = A_n · U

The standard formulation for U is based on the connection geometry:

U = 1 − x̄ / L

x̄ — the distance from the connection plane to the centroid of the member segment resisting the force (the connection eccentricity).
L — the length of the connection along the line of load (distance from the first bolt to the last bolt, or the average weld length).

A shorter connection length (L) or a larger eccentricity (x̄) yields a smaller U, which reduces the calculated net-section rupture capacity. Long, well-distributed connections drive U toward 1.0; short, eccentric ones penalize it.

Case	Member / Connection	U
1	Tension load transmitted directly to all cross-sectional elements by fasteners or welds.	1.0
2	Tension transmitted to some but not all elements. General case — use when no simpler case applies.	1 − x̄/L
3	Members with transverse welds only.	1.0 (A_n = connected elements only)
4	Plates / tongue plates with longitudinal welds along both edges, length L, width w.	L≥2w: 1.0; 1.5w≤L<2w: 0.87; w≤L<1.5w: 0.75
5	Round HSS with single concentric gusset, L ≥ 1.3D.	1.0 (else 1 − x̄/L)
6	Rectangular HSS with single concentric gusset.	1 − x̄/L
7	W, M, S, HP, or T — flange connected, ≥ 3 bolts/line, b_f/d ≥ 2/3.	0.90
7	Same, but b_f/d < 2/3.	0.85
7	Same shapes, web connected, ≥ 4 bolts in load direction.	0.70
8	Single & double angles, ≥ 4 bolts in line.	0.80
8	Single & double angles, exactly 3 bolts in line.	0.60

How to pick U — decision flowchart:

Is the entire cross-section connected? → Case 1, U=1.0. Done.
Plate with longitudinal edge welds? → Case 4, use L/w ratio.
W, M, S, HP or T? → Case 7 (flange vs web, count bolts, check b_f/d).
Single or double angle? → Case 8 (count bolts in line).
HSS with gusset plate? → Case 5 (round) or 6 (rectangular).
Anything else → Case 2 general formula U = 1 − x̄/L. x̄ = distance from connection plane to centroid of connected element (tabulated in Manual); L = distance between first & last bolt (or average weld length).

5.3 Putting It Together — Compute A_e and φP_n

How to compute rupture capacity, in order:

Find the governing A_n across all candidate failure paths (§5.1).
Pick U from Table D3.1 using the decision flowchart (§5.2).
A_e = U · A_n.
φ_tP_n = 0.75 · F_u · A_e — must be ≥ P_u.

Where it governs: almost always the critical check for bolted end connections, especially angles/tees/channels where U < 1.0 amplifies the hole penalty.

6. Limit State 3 — Block Shear (§J4.3)

Nominal strength (J4-5)

(10)

R_{n} = min [0.60 F_{u} A_{n v}, 0.60 F_{y} A_{g v}] + U_{b s} \cdot F_{u} \cdot A_{n t}

ϕ = 0.75

A block of material at the connection tears out via shear on one or two planes plus tension on a perpendicular plane. U_bs = 1.0 for uniform tension across the tension plane (single bolt row perpendicular to load), or 0.5 for non-uniform tension (two rows, unequal edge distances).

6.1 How to Find A_gv, A_nv, A_nt — the Four-Step Procedure

Block shear is the single biggest source of confusion in tension-member design because you must identify failure planes, not just cross-sections. The block that tears out has a physical outline you can trace with a pencil directly on the connection sketch. Follow this four-step procedure every time:

The tear-out block (yellow) with two red shear planes and one blue tension plane.

Step 1 — Draw the tear-out block. Start at the loaded edge of the plate. Follow the bolt line toward the load through the outermost bolts on top and bottom rows, then jump across between the innermost bolts to close the block. The interior of that polygon is the material that would fall out if the connection failed.

Step 2 — Identify the planes.

Shear planes (red, dashed) = the sides of the block parallel to the applied load. There are 1 or 2 shear planes depending on the connection geometry.
Tension plane (blue, solid) = the side of the block perpendicular to the applied load, at the far end (the "back wall" of the block).

Step 3 — Compute the three areas. Multiply each plane's linear length by the plate thickness t.

Area	What it is	Formula (per plane; sum both planes if 2)
A_gv gross shear	Full shear-plane length × t. No hole deductions.	A_gv = [L_ev + (n−1)·s] · t
A_nv net shear	Shear-plane length minus the shear-plane portion of holes crossed (typically n − 0.5 holes; the last hole is half in the shear plane and half in the tension plane).	A_nv = A_gv − (n − 0.5)·(d_b+1/8″)·t
A_nt net tension	Tension-plane length minus the tension-plane portion of hole(s) crossed (usually 1 hole for two shear planes, 0.5 hole for one shear plane).	A_nt = [g − n_t·(d_b+1/8″)]·t

n = bolts along one shear plane; s = bolt spacing; L_ev = edge distance from last bolt to loaded edge; g = gage (perpendicular distance between rows); d_b = bolt diameter; d_b+1/8″ = deducted hole width per D3.2.

Step 4 — Plug into J4-5 and take the smaller shear term:

(11)

R_{n} = min [0.60 F_{u} A_{n v}, 0.60 F_{y} A_{g v}] + U_{b s} F_{u} A_{n t}

The bracket compares shear rupture (F_u·A_nv) with shear yielding (F_y·A_gv) — smaller governs. The tension term is always full rupture (F_u·A_nt). Then apply φ = 0.75.

6.2 Worked Mini-Example — Block Shear on a 3/8″ Gusset Plate

Given. 3/8″ gusset plate, single row of three ¾″ A325 bolts, F_y = 36 ksi, F_u = 58 ksi, pitch s = 3″, edge L_ev = 1.5″, tension-side edge L_eh = 1.5″, one shear plane (gusset edge on one side of the row).

Failure block: red dashed = shear plane (A_gv, A_nv); blue = tension plane (A_nt).

Step 1 — Hole width and shear-plane length.

(12)

d_{h} = d_{b} + 1/8″

d_h = 3/4 + 1/8 = 0.875 in

(13)

L_{g v} = L_{e v} + (n - 1) \cdot s

L_gv = 1.5 + 2·3 = 7.5 in

Step 2 — Gross shear area A_gv.

(14)

A_{g v} = L_{g v} \cdot t

A_gv = 7.5 · 0.375 = 2.81 in²

Step 3 — Net shear area A_nv. Shear plane crosses (n − 0.5) = 2.5 holes.

(15)

A_{n v} = [L_{g v} - (n - 0.5) \cdot d_{h}] \cdot t

A_nv = (7.5 − 2.5·0.875) · 0.375 = 5.3125 · 0.375 = 1.99 in²

Step 4 — Net tension area A_nt. Tension plane crosses 0.5 hole (one shear plane).

(16)

A_{n t} = [L_{e h} - 0.5 \cdot d_{h}] \cdot t

A_nt = (1.5 − 0.5·0.875) · 0.375 = 1.0625 · 0.375 = 0.398 in²

Step 5 — Shear-term governor (rupture vs. yield).

(17)

0.60 F_{u} A_{n v} v s .0.60 F_{y} A_{g v}

t ak e t h es ma l l er

0.60 · 58 · 1.99 = 69.3 kip (rupture)
0.60 · 36 · 2.81 = 60.7 kip (yield) ← governs

Step 6 — Tension term.

(18)

U_{b s} F_{u} A_{n t}

1.0 · 58 · 0.398 = 23.1 kip

Step 7 — Nominal and design block-shear.

(19)

R_{n} = min (0.60 F_{u} A_{n v}, 0.60 F_{y} A_{g v}) + U_{b s} F_{u} A_{n t}

R_n = 60.7 + 23.1 = 83.8 kip

(20)

ϕ R_{n} = 0.75 \cdot R_{n}

φR_n = 0.75 · 83.8 = 62.9 kip

7. Slenderness Preference (§D1)

(21)

L / r \leq 300 (p r e f er r e d, n o t man d a t or y f or t e n s i o n)

Limits vibration and sag during shipping/erection. Applies to rods and slender rolled members; does not apply to rods in tension-only bracing where sag rods are provided.

8. Design Procedure (put it all together)

Compute P_u from ASCE 7 combinations.
Required A_g ≥ P_u/(0.90 F_y); required A_e ≥ P_u/(0.75 F_u).
Estimate A_n ≈ 0.85 A_g, U from Table D3.1 → required A_g for rupture.
Select trial section from AISC Manual, satisfy L/r ≤ 300.
Verify all three limit states (yielding, rupture, block shear).

⚠ Common mistakes

Using φ = 0.90 for rupture — it is 0.75.
Deducting only d_b + 1/16″ instead of d_b + 1/8″ for the hole width.
Forgetting to apply U to A_n.
Skipping block shear when only 2–3 bolts are present.

Worked Example 4.1 — Single-Plate Tension Member (LRFD, All Three Limit States)

Given — connection parameters.

Member: A36 steel plate, 1/2 in. thick × 6 in. wide (t = 0.5 in., w = 6.0 in.).
Material: Yield stress F_y = 36 ksi; ultimate stress F_u = 58 ksi.
Fasteners: Single row of 3 bolts along the load line; bolt diameter d_b = 3/4 in.
Connection geometry:
- Bolt-hole design dimension: d_h = d_b + 1/8 in. = 7/8 in. (AISC standard allowance for damage/clearance).
- End distance (edge to first bolt center) = 2.0 in.
- Bolt spacing (pitch between centers) = 3.0 in.
- Because it is a flat plate with all elements connected, there is no eccentricity, so U = 1.0.

A36 steel plate 1/2 in. thick × 6 in. wide, single row of three 3/4 in. bolts at 2.0-3.0-3.0 in. spacing, bolted to gusset plate

Step A — Gross-section Yielding (§D2-a)

First, determine the capacity based on the unreduced cross-sectional profile stretching under uniform tension.

A.1 Calculate Gross Area (A_g)

Formula

(1)

A_{g} = w \cdot t

A_g = 6.0 in. · 0.5 in. = 3.0 in²

A.2 Calculate Nominal Yield Strength (P_n)

Formula

(2)

P_{n} = F_{y} \cdot A_{g}

P_n = 36 ksi · 3.0 in² = 108.0 kips

A.3 Apply LRFD Resistance Factor (φ_y = 0.90)

Design

(3)

ϕ_{t} P_{n} = 0.90 \cdot P_{n}

φ_t P_n = 0.90 · 108.0 = 97.2 kips — gross-yielding design capacity.

Step B — Net-section Rupture (§D2-b)

Next, evaluate fracture at the bolt line where cross-sectional material is subtracted.

B.1 Calculate Net Area (A_n)

Since there is a single line of bolts, one bolt-hole path is subtracted:

Formula

(4)

A_{n} = A_{g} - (1 \cdot d_{h} \cdot t)

A_n = 3.0 in² − (1 · 0.875 in. · 0.5 in.) = 3.0 − 0.4375 = 2.5625 in²

B.2 Determine Effective Net Area (A_e)

Because a flat plate connected to flat fasteners experiences no out-of-plane load transitions, U = 1.0.

Formula

(5)

A_{e} = A_{n} \cdot U

A_e = 2.5625 in² · 1.0 = 2.5625 in²

B.3 Calculate Nominal Fracture Strength (P_n)

Formula

(6)

P_{n} = F_{u} \cdot A_{e}

P_n = 58 ksi · 2.5625 in² = 148.63 kips

B.4 Apply LRFD Resistance Factor (φ_t = 0.75)

Design

(7)

ϕ_{t} P_{n} = 0.75 \cdot P_{n}

φ_t P_n = 0.75 · 148.63 = 111.5 kips — net-section rupture design capacity.

Step C — Block-Shear Rupture (§J4.3)

For a single row of bolts in a plate, block shear involves tearing out an L-shaped block from the end of the plate. The failure path is defined by:

A longitudinal shear line spanning from the innermost bolt to the edge of the plate.
A transverse tension line extending from the innermost bolt center out to the adjacent long edge.

C.1 Define Paths and Geometric Dimensions

Total shear line length = end distance + 2 × spacing = 2.0 in. + (2 × 3.0 in.) = 8.0 in.
Number of bolt holes on the shear path = 2.5 holes (the line passes through 2 full holes and terminates halfway into the 3rd hole).
Total tension line length = distance from bolt center-line to plate edge. Assuming the bolt line is centered at 3.0 in. from the edge: 3.0 in.
Number of bolt holes on the tension path = 0.5 holes (the line starts exactly at the center of the 3rd bolt hole).

C.2 Calculate the Sectional Areas

Gross shear area

(8)

A_{g v} = 8.0 in . \cdot 0.5 in .

A_gv = 4.0 in²

Net shear area

(9)

A_{n v} = [8.0 in . - (2.5 \cdot 0.875 in .)] \cdot 0.5 in .

A_nv = [8.0 − 2.1875] × 0.5 = 5.8125 × 0.5 = 2.906 in²

Net tension area

(10)

A_{n t} = [3.0 in . - (0.5 \cdot 0.875 in .)] \cdot 0.5 in .

A_nt = [3.0 − 0.4375] × 0.5 = 2.5625 × 0.5 = 1.281 in²

C.3 Compute Nominal Block-Shear Strength

Per AISC §J4.3, the nominal block-shear strength R_n is governed by the combination of tension fracture and shear yielding/rupture:

Eq. J4-5

(11)

R_{n} = 0.6 F_{u} A_{n v} + U_{b s} F_{u} A_{n t} \leq 0.6 F_{y} A_{g v} + U_{b s} F_{u} A_{n t}

Using a uniform stress distribution factor U_bs = 1.0:

Left side (shear rupture + tension fracture):
0.6 × 58 ksi × 2.906 in² + 1.0 × 58 ksi × 1.281 in² = 101.13 + 74.30 = 175.43 kips

Right-side upper bound (shear yielding + tension fracture):
0.6 × 36 ksi × 4.0 in² + 1.0 × 58 ksi × 1.281 in² = 86.40 + 74.30 = 160.70 kips

The lower value controls the nominal threshold: R_n = 160.70 kips.

C.4 Apply LRFD Resistance Factor (φ = 0.75)

Design

(12)

ϕ R_{n} = 0.75 \cdot R_{n}

φ R_n = 0.75 × 160.70 = 120.5 kips — block-shear design capacity.

Governing Limit State — Summary.

Gross-section yielding: φ_t P_n = 97.2 kips
Net-section rupture: φ_t P_n = 111.5 kips
Block-shear rupture: φ R_n = 120.5 kips

Design tensile capacity φP_n = 97.2 kips — gross-section yielding governs.

Worked Example 4.2 — Tension Diagonal in a Braced Frame

Given: Diagonal brace of the Cardinal Square 4-story building. Service loads P_D = 30 k, P_L = 70 k. Steel: A36 (F_y = 36 ksi, F_u = 58 ksi). Connection: two rows of 7/8″ A325 bolts on the long leg of an angle; pitch s = 2.5″, gage g = 2.5″, stagger 1.25″. Trial: L6×4×1/2 (A_g = 4.75 in², x̄ = 1.03″), connection length L = 5″.

Angle geometry & bolt pattern

Step 1 — Required strength (LRFD)

Formula

(13)

P_{u} = 1.2 D + 1.6 L

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

Step 2 — Gross-section yielding (D2-a)

Formula

(14)

ϕ_{t} P_{n} = 0.90 \cdot F_{y} \cdot A_{g}

φ_t P_n = 0.90 · 36 · 4.75 = 153.9 k ≥ 148 k ✓

Step 3 — Effective net area (D3)

Hole deduction

(15)

d_{h} = d_{b} + 1/8″ = 7/8 + 1/8 = 1.0″

Staggered path An

(16)

A_{n} = A_{g} - Σ (d_{h} \cdot t) + Σ (s^{2} /4 g) \cdot t

Path 1 (one hole): A_n1 = 4.75 − (1.0)(0.5) = 4.25 in²
Path 2 (two holes with stagger): A_n2 = 4.75 − 2(1.0)(0.5) + (2.5²/(4·2.5))(0.5) = 4.75 − 1.00 + 0.3125 = 4.06 in²
Governing A_n = 4.06 in²

Shear lag (Table D3.1 Case 2)

(17)

U = 1 - \overset{x}{ˉ} / L = 1 - 1.03/5 = 0.794

A_e = U · A_n = 0.794 · 4.06 = 3.22 in²

Step 4 — Tensile rupture (D2-b)

(18)

ϕ_{t} P_{n} = 0.75 \cdot F_{u} \cdot A_{e}

φ_t P_n = 0.75 · 58 · 3.22 = 140.1 k < 148 k ✗ — rupture governs and section is inadequate. Try L6×4×5/8: A_g = 5.86 in², t = 0.625.
A_n = 5.86 − 2(1.0)(0.625) + (2.5²/10)(0.625) = 5.86 − 1.25 + 0.391 = 5.00 in²; A_e = 0.794 · 5.00 = 3.97 in².
φ_tP_n = 0.75 · 58 · 3.97 = 172.7 k ≥ 148 k ✓

Step 5 — Block shear (J4.3)

Given L_ev = 1.5″, L_eh (end distance to first bolt, in load direction) = 1.5″, t = 0.625″, d_h = 0.875 + 0.125 = 1.0″, n = 3 bolts per row, s = 2.5″, g = 2.5″, F_y = 36, F_u = 58 ksi, U_bs = 1.0.

Failure block for the long-leg connection: two red-dashed shear planes (top & bottom of the block) plus one blue tension plane across the gage g.

5a — Gross shear area (both planes)

Formula

(19)

A_{g v} = n_{p} \cdot [L_{e v} + (n - 1) \cdot s] \cdot t

A_gv = 2 · (1.5 + 2·2.5) · 0.625 = 2 · 6.5 · 0.625 = 8.125 in²

5b — Net shear area

Formula

(20)

A_{n v} = A_{g v} - n_{p} \cdot (n - 0.5) \cdot d_{h} \cdot t

A_nv = 8.125 − 2 · 2.5 · 1.0 · 0.625 = 8.125 − 3.125 = 5.00 in²

5c — Net tension area

Formula

(21)

A_{n t} = [g - 1.0 \cdot d_{h}] \cdot t (o n e f u l l h o l eo n t h e t e n s i o n pl an e, tw os h e a r pl an es)

A_nt = (2.5 − 1.0·1.0) · 0.625 = 1.5 · 0.625 = 0.938 in²

5d — Compare shear rupture vs. shear yield

Rupture

(22)

0.60 F_{u} A_{n v}

0.60 · 58 · 5.00 = 174.0 k

Yield

(23)

0.60 F_{y} A_{g v}

0.60 · 36 · 8.125 = 175.5 k → rupture (174.0) is smaller and governs the shear term.

5e — Nominal and design block-shear strength

Eq. J4-5

(24)

R_{n} = min (0.60 F_{u} A_{n v}, 0.60 F_{y} A_{g v}) + U_{b s} F_{u} A_{n t}

R_n = 174.0 + 1.0 · 58 · 0.938 = 174.0 + 54.4 = 228.4 k

Design

(25)

ϕ R_{n} = 0.75 \cdot R_{n}

φR_n = 0.75 · 228.4 = 171.3 k ≥ 148 k ✓

Step 6 — Slenderness check

Brace length L = 14 ft = 168 in; r_min ≈ 0.86 in → L/r = 195 < 300 ✓

DESIGN: L6×4×5/8 (A36), long leg connected with two rows of three 7/8″ A325 bolts, pitch s = 2.5″, gage g = 2.5″, edge L_ev = L_eh = 1.5″. Governing limit state: tensile rupture with φ_tP_n = 172.7 k; block-shear φR_n = 171.3 k; both ≥ P_u = 148 k ✓.

Engineering tips

Two strength limit states. The smaller φt (0.75) plus the smaller area (Ae) doesn't automatically mean rupture governs — always compute both.
Shear lag and the U factor. Longer connections (more bolts, longer welds) let U approach 1.0 — the whole section becomes effective.
Block shear at the connection. Compute both terms and take Ubs = 1.0 for uniform tension across the tension plane.
Slenderness screen. For long tension rods, add turnbuckles and pre-tension to control sag.

Common mistakes

Two strength limit states. Using Ag in the rupture equation, or applying φ = 0.90 to Fu·Ae.
Shear lag and the U factor. Assuming U = 1.0 for a single-angle member bolted through one leg — the correct U is well below 1.0.
Block shear at the connection. Ignoring block shear when the member passes §D2.
Slenderness screen. Confusing L/r ≤ 300 (tension) with KL/r ≤ 200 (compression).

Real-world use

Two strength limit states. Every hanger, brace, and truss chord in every steel building.
Shear lag and the U factor. Single-angle bracing, WT hangers, C-shapes bolted on the flange.
Block shear at the connection. Bolted angle bracing, gusset plates, coped beam ends acting as tension links.
Slenderness screen. Roof-bracing sag rods, long tie-rods, X-bracing in low-rise frames.

06

Professional practice, safety & ethics

Tension member practice

Professional practice

Show all hole patterns, edge distances, and gage lines on the drawing — shear lag and net area depend on the detailer's layout, not yours.
State the shear-lag factor U and the assumed connection length in the calculation; the fabricator may change the bolt count.
Coordinate turnbuckles / rod bracing pretension with the erector; slack rods are not tension members.

Safety in design & construction

Tension members fail suddenly by rupture — there is no plastic warning at the net section. Treat rupture as a brittle, non-redundant limit state.
Block shear governs surprisingly often in angles and gusset plates; never skip it.
Field-drilled holes reduce An — require EOR approval before any hole is added.

Engineering ethics

The 2007 I-35W collapse traced to undersized gusset plates: reviewing your own connection assumptions is an ethical duty, not optional QA.
Do not accept 'it worked last time' details for a new load level.
Document any reliance on a fabricator's standard connection in writing.

Ironworkers bolting a steel beam connection while tied off at height — Erection safety: OSHA Subpart R fall protection and stable temporary bracing.

Engineers reviewing sealed structural drawings across a conference table — Design review: documenting assumptions before the drawings are sealed.

ABET / licensure link. These points map to ABET Student Outcomes 2 and 4 — engineering design within realistic constraints, and recognition of ethical and professional responsibilities. Expect NCEES FE and PE exam questions on the NSPE Code of Ethics, OSHA construction requirements, and the engineer's standard of care.

07

Cost analysis

Tension member and connection cost

Approach

The member is cheap; the connection is not. Count bolts, holes, and gusset plate area — each hole is a shop operation.
Reducing bolts to save cost can push the governing limit state to rupture or block shear — re-check before accepting.
Rod/cable bracing is cheap in material but expensive in end fittings and field adjustment.

Worked cost example — WT hanger with a bolted end connection

Basis: Tu = 120 kip, 3/4-in A325 bolts

Line item	Qty	Rate	Cost
WT member, 22 lb/ft × 14 ft	0.154 ton	$1,150	$177
Gusset plate 1/2″ × 10″ × 14″	1 ea	$85	$85
Punched holes	12 holes	$7	$78
Field bolting labor	6 bolts	$18	$108
Estimated total			$448

Takeaway. Connections dominate: doubling bolt count to ‘be safe’ costs more than upsizing the member one section.

Unit rates are representative US averages for teaching purposes. On a real project, price with current local rates (RSMeans, fabricator quotes, or contractor pricing) and state the estimate date.

08

Animated concepts

Key mechanics visualised — watch the strain profile, stress block, or buckled shape evolve

Shear lag around bolt holes

Tensile stress detours around holes, so not all of An is fully effective — hence Ae = U·An in D3.

09

Engineering figures

Full-page reference diagrams — the visual vocabulary you will use for the rest of the course

§4.6.1

End connections

Fig. 4.1Bolted gusset connection
High-strength bolts arranged in staggered rows on a gusset plate — the geometry that drives An, Ae, and block-shear planes.

§4.6.2

Shear-lag geometry

Fig. 4.2Single-leg connection
Angle connected by one leg only. The offset between bolt line and section centroid is the length x̄ used to compute U = 1 − x̄/L in Table D3.1 Case 2.

§4.6.3

Net section

Fig. 4.3Bolt-hole pattern at the end of a W-shape
Rows of drilled holes reduce the section from Ag to An = Ag − Σ(dh·t). Hole diameter dh = bolt diameter + 1/16″ + 1/16″ per AISC §B4.3.

§4.6.4

Hanger application

Fig. 4.4Threaded rod hanger for MEP equipment
Classic tension application: a threaded rod hangs a mechanical unit from a slab. Check rupture on the tensile stress area of the threaded portion.

10

Worked examples

Full textbook solutions — problem, theory, step-by-step, verification, interpretation

Example 4.1

Critical net area — ½″ plate with staggered 3/4″ bolts

Determine the critical net area of the ½-in-thick plate (11 in wide) shown in Fig. 3.5 using AISC Specification §D3.2. Holes are punched for 3/4-in bolts (dh = 3/4 + 1/8 = 7/8 in).

Problem statement

Determine the critical net area of the ½-in-thick × 11-in-wide plate shown in Fig. 3.5 using the AISC Specification (§D3.2). The holes are punched for 3/4-in-diameter bolts. Investigate paths $A B C D$ , $A B C E F$ , and $A B E F$ .

Bolt pattern: two vertical gages 3 in apart between top pair and 3 in between the third and fourth rows, outer edge distances 2½ in top & bottom; staggered pitch $s = 3 in$ and gage $g = 3 in$ .

Fig. 3.5 — ½″ × 11″ plate with staggered 3/4″ bolts — Fig. 3.5 (Salmon & Johnson) — staggered bolt pattern in a ½-in plate.

DIMStaggered pattern — s = 3″ along load, g = 3″ perpendicular. Route ABCEF crosses three holes with two diagonal segments.

Given

Plate ½ in × 11 in (A_g = 5.50 in²)
Bolts: 3/4 in punched → d_h = 3/4 + 1/8 = 7/8 in (§B4.3b)
Pitch s = 3 in, gage g = 3 in
Three candidate paths: ABCD, ABCEF, ABEF

Find

Critical (minimum) net width and net area A_n

Assumptions

Punched holes → deduct d_b + 1/8 in for damage (AISC §B4.3b).
s²/(4g) credit applied once for every diagonal segment on the failure path.
Tension is uniform across the plate width.

Code references

AISC 360-22 §B4.3b — Effective hole diameter
AISC 360-22 §D3 — Effective net area

Step-by-step solution

1
Effective hole diameter (AISC §B4.3b)
$d_{h} = d_{b} + \frac{1}{8}$
$X = \frac{3}{4} + \frac{1}{8} = \frac{7}{8} in$
2
Route ABCD — two holes, no stagger
Formula§B4.3b
w_n = w_g − Σ d_h
$w_{A B C D} = 11 - (2) (\frac{7}{8})$
$X = 11 - 1.75 = 9.25 in$
3
Route ABCEF — three holes, one diagonal segment (C→E)
Formula§B4.3b
w_n = w_g − Σ d_h + Σ s²/(4g)
$w_{A B C E F} = 11 - (3) (\frac{7}{8}) + \frac{( 3 ) ^{2}}{( 4 ) ( 3 )}$
$X = 11 - 2.625 + 0.750$
$X = 9.125 in (controls)$ smallest w_n
4
Route ABEF — two holes, one diagonal segment (B→E)
$w_{A B E F} = 11 - (2) (\frac{7}{8}) + \frac{( 3 ) ^{2}}{( 4 ) ( 6 )}$
$X = 11 - 1.75 + 0.375 = 9.625 in$
5
Critical net area
$A_{n} = w_{n, m i n} \cdot t$
$X = (9.125) (\frac{1}{2})$
$X = 4.56 in^{2}$

Verification

Only two holes lie on $A B C D$ and $A B E F$ , while $A B C E F$ loses three; even with two $s^{2} /4 g$ credits, $A B C E F$ still governs because the third hole dominates the two small stagger credits. As Salmon & Johnson note, checking $A B E F$ is a waste of time once $A B C D$ is known (both routes cut two holes but $A B C D$ is straight and therefore shorter).

Final answer

A_n = 4.56 in² — route ABCEF controls.

Design interpretation

Each stagger buys back $s^{2} / (4 g)$ of net width, but never enough to erase the loss of an entire extra hole unless $s$ is very large or $g$ is very small. Always check every zig-zag path that crosses the section — the controlling route is not always visually obvious.

Common mistakes

Deducting d_b instead of d_h = d_b + 1/8 in (punched holes) — under-predicts hole loss.
Adding one s²/(4g) credit for the whole group instead of one per diagonal segment.
Skipping paths that seem 'obviously worse' — always tabulate every candidate.

11

Guided practice

Compute the governing variables — hints unlock as you need them

A 1″ diameter A36 threaded hanger rod (Fy = 36 ksi, Fu = 58 ksi, Ae ≈ 0.606 in²) supports a service dead load PD = 8 kip and live load PL = 12 kip. Compute the LRFD demand Pu and the design strength φt Pn (rupture on threaded section, φ = 0.75). Is the rod adequate?

Your turn

P_u =kipphi_{t*P_n} =kip

Hints

1.LRFD combination 2 governs for gravity: Pu = 1.2 PD + 1.6 PL.

12

Independent practice

Solve the chapter's design task — compute each governing variable

Design task

An A36 plate PL½″ × 6″ (Fy = 36 ksi, Fu = 58 ksi) is connected by two transverse rows of two ¾″ bolts (hole diameter dh = ⅞″). Shear-lag factor U = 1.0. Compute An and the design tensile strength φt Pn; identify whether yielding or rupture governs.

Given

Plate: b = 6 in, t = 0.5 in → Ag = 3.00 in²
2 bolt holes per critical section, dh = 7/8 in
U = 1.0, φy = 0.90, φr = 0.75

Approach

An = Ag − n·dh·t on the critical (transverse) section.
Ae = U·An.
φt Pn_yield = 0.90·Fy·Ag; φt Pn_rupture = 0.75·Fu·Ae. Governing = min.

Submit your answer

A_n =in^2phi_{t*P_n} =kip

13

Mini design challenge

Select the option that satisfies every code and serviceability requirement in the brief

Brief

Select the lightest single-angle A36 tension diagonal for a brace: Pu = 120 kips, connected by 4 bolts of ¾″ in one leg. Verify yielding, rupture (with shear-lag U from Table D3.1), and slenderness L/r ≤ 300 for L = 14 ft.

Requirements

Factored demand Pu = 120 kips
A36 steel (Fy = 36 ksi, Fu = 58 ksi)
Bolted through one leg — expect U ≈ 0.85 (Case 8, four bolts in line)
Slenderness L/r ≤ 300 over L = 14 ft = 168 in → rmin ≥ 0.56 in
Lightest AISC-listed angle satisfying all three checks wins

Section

Wt (lb/ft)

Δ (in)

Ru/Rn

Cost

Pick

14

Chapter summary

A mind map of how every concept connects

Graded Chapter Quiz(15 FE-style questions · AISC Manual required)

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16

FE exam preparation

NCEES-style practice with timer, equation sheet, and mastery tracking

Exam mode

30:00 Calculator

Question 1 / 10

An ASTM A36 steel hanger bar ( $F_{y} = 36 ksi$ , $F_{u} = 58 ksi$ ) is used as a tension link. The plate has a thickness of $0.375 in$ and an effective net area $A_{e} = 2.50 in^{2}$ . Determine the LRFD design tensile rupture strength $ϕ P_{n}$ .

◆ EasyAISC 360-22 §D2(b)

Fig. A36 hanger bar — tensile rupture check.

A. 90.0 kips
B. 108.8 kips
C. 130.5 kips
D. 145.0 kips

Tension Members

Check yielding on the gross section

Check rupture on the effective net area

Compute block-shear capacity

Navigate AISC Chapter D + Manual Part 5

Screen slenderness L/r ≤ 300

Answer FE-style tension questions

Block shear tear-out

Threaded-rod rupture

Hyatt Regency walkway (1981)

Fatigue cracking at bolt holes

Chapter 4 — Tension Members (AISC 360-22 Chapter D)

1. Behavior of Tension Members

Representative Structures

2. LRFD Design Inequality

3. The Three Governing Limit States — Overview

4. Limit State 1 — Gross-Section Yielding (AISC D2-a)

Mini-Example 4A — Design Yield Strength of an A36 Channel Truss Chord

Mini-Example 4B — Minimum Required Gross Area (A992 Diagonal, Pu = 180 k)

5. Limit State 2 — Tensile Rupture on the Effective Net Section (§D2-b)

5.1 Net Area An — Hole Deduction and Staggered Holes (§D3)

5.1.1 Staggered Holes — the s²/4g Rule (D3.2)

5.1.2 Worked Mini-Example — Net Area of a Staggered-Bolt Plate

5.2 Shear-Lag Factor U (Table D3.1)

Understanding Shear Lag and Its Impact

5.3 Putting It Together — Compute Ae and φPn

6. Limit State 3 — Block Shear (§J4.3)

6.1 How to Find Agv, Anv, Ant — the Four-Step Procedure

6.2 Worked Mini-Example — Block Shear on a 3/8″ Gusset Plate

7. Slenderness Preference (§D1)

8. Design Procedure (put it all together)

⚠ Common mistakes

Worked Example 4.1 — Single-Plate Tension Member (LRFD, All Three Limit States)

Step A — Gross-section Yielding (§D2-a)

A.1 Calculate Gross Area (Ag)

A.2 Calculate Nominal Yield Strength (Pn)

A.3 Apply LRFD Resistance Factor (φy = 0.90)

Step B — Net-section Rupture (§D2-b)

B.1 Calculate Net Area (An)

B.2 Determine Effective Net Area (Ae)

B.3 Calculate Nominal Fracture Strength (Pn)

B.4 Apply LRFD Resistance Factor (φt = 0.75)

Step C — Block-Shear Rupture (§J4.3)

C.1 Define Paths and Geometric Dimensions

C.2 Calculate the Sectional Areas

C.3 Compute Nominal Block-Shear Strength

C.4 Apply LRFD Resistance Factor (φ = 0.75)

Worked Example 4.2 — Tension Diagonal in a Braced Frame

Step 1 — Required strength (LRFD)

Step 2 — Gross-section yielding (D2-a)

Step 3 — Effective net area (D3)

Step 4 — Tensile rupture (D2-b)

Step 5 — Block shear (J4.3)

5a — Gross shear area (both planes)

5b — Net shear area

5c — Net tension area

5d — Compare shear rupture vs. shear yield

5e — Nominal and design block-shear strength

Step 6 — Slenderness check

End connections

Shear-lag geometry

Net section

Hanger application

Critical net area — ½″ plate with staggered 3/4″ bolts

Effective hole diameter (AISC §B4.3b)

Route ABCD — two holes, no stagger

Route ABCEF — three holes, one diagonal segment (C→E)

Route ABEF — two holes, one diagonal segment (B→E)

Critical net area

Graded Chapter Quiz(15 FE-style questions · AISC Manual required)

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Mini-Example 4B — Minimum Required Gross Area (A992 Diagonal, P_u = 180 k)

5.1 Net Area A_n — Hole Deduction and Staggered Holes (§D3)

5.3 Putting It Together — Compute A_e and φP_n

6.1 How to Find A_gv, A_nv, A_nt — the Four-Step Procedure

A.1 Calculate Gross Area (A_g)

A.2 Calculate Nominal Yield Strength (P_n)

A.3 Apply LRFD Resistance Factor (φ_y = 0.90)

B.1 Calculate Net Area (A_n)

B.2 Determine Effective Net Area (A_e)

B.3 Calculate Nominal Fracture Strength (P_n)

B.4 Apply LRFD Resistance Factor (φ_t = 0.75)