AISC 360 Design Examples — Beam, Column & Connection Worked Solutions

AISC 360-22 is the governing standard for structural steel buildings in the United States. Whether you are preparing for the PE/SE exam, onboarding at a new firm, or verifying software outputs, having a set of annotated AISC 360 design examples that you can hand-check is essential. This guide provides six complete worked examples covering the most common design scenarios, with every intermediate value shown and every code clause referenced.

In this guide: Beam flexure (W-shape with LTB), column buckling (W-shape axial compression), combined axial + flexure interaction, bolted connection (bearing-type), fillet weld capacity, and column base plate design. All examples use LRFD methodology and ASTM A992 steel (Fy = 50 ksi, Fu = 65 ksi) unless noted.

PRELIMINARY — NOT FOR CONSTRUCTION. All examples are educational references only. Every structural design must be independently verified by a licensed Professional Engineer.

Example 1: Simply Supported Beam — Flexure and Shear

Design parameters

W21x50 beam, simply supported, 35 ft span. Uniformly distributed dead load = 0.75 kip/ft (including self-weight), live load = 1.5 kip/ft. Beam braced at third points (Lb = 11.67 ft). ASTM A992 steel.

Section properties (W21x50)

d = 20.8 in, bf = 6.53 in, tf = 0.535 in, tw = 0.380 in. Ix = 984 in^4, Sx = 94.5 in^3, Zx = 110 in^3. ry = 1.30 in. h/tw = 45.1. J = 1.14 in^4, Cw = 2510 in^6.

Factored loads

wu = 1.2 x 0.75 + 1.6 x 1.5 = 0.90 + 2.40 = 3.30 kip/ft. Mu = wu x L^2 / 8 = 3.30 x 35^2 / 8 = 505.3 kip-ft = 6064 kip-in. Vu = wu x L / 2 = 3.30 x 35 / 2 = 57.8 kips.

Step 1: Section classification (Table B4.1b)

Flange: lambda_f = bf/(2 x tf) = 6.53/(2 x 0.535) = 6.10. Compact limit = 0.38 x sqrt(E/Fy) = 0.38 x sqrt(29000/50) = 9.15. 6.10 < 9.15 -- flange is compact.

Web: lambda_w = h/tw = 45.1. Compact limit = 3.76 x sqrt(E/Fy) = 3.76 x 24.08 = 90.6. 45.1 < 90.6 -- web is compact.

Section is compact; plastic moment Mp applies.

Step 2: Lateral-torsional buckling (Chapter F2)

Lp = 1.76 x ry x sqrt(E/Fy) = 1.76 x 1.30 x sqrt(29000/50) = 5.50 ft.

Lr = 1.95 x ry x (E / (0.7 x Fy)) x sqrt((J x Cw) / (Sx x ry) x sqrt(1 + sqrt(1 + 6.76 x ((0.7 x Fy)/(E) x Sx/(J x Cw))^2))) Lr = 1.95 x 1.30 x (29000 / 35) x sqrt((1.14 x 2510) / (94.5 x 1.30) x sqrt(1 + sqrt(1 + 6.76 x (35/29000 x 94.5/(1.14 x 2510))^2))) = 15.8 ft.

Lp = 5.50 ft < Lb = 11.67 ft <= Lr = 15.8 ft -- inelastic LTB applies.

Mp = Zx x Fy = 110 x 50 = 5500 kip-in. Mr = 0.7 x Fy x Sx = 0.7 x 50 x 94.5 = 3308 kip-in.

Cb = 1.01 (uniform load, simply supported, braced at third points, from AISC Table 3-1).

Mn = Cb x [Mp - (Mp - Mr) x (Lb - Lp) / (Lr - Lp)] <= Mp Mn = 1.01 x [5500 - (5500 - 3308) x (11.67 - 5.50) / (15.8 - 5.50)] Mn = 1.01 x [5500 - 2192 x 6.17 / 10.3] Mn = 1.01 x [5500 - 1313] = 1.01 x 4187 = 4229 kip-in.

Since 4229 < Mp = 5500, Mn = 4229 kip-in = 352 kip-ft.

phi_b x Mn = 0.90 x 4229 = 3806 kip-in = 317 kip-ft.

Flexure check: Mu = 505.3 kip-ft > phi_b x Mn = 317 kip-ft. Ratio = 505.3 / 317 = 1.59 -- FAILS.

The W21x50 is undersized for this span with third-point bracing. Try W24x76 with closer bracing or a shorter span.

Step 3: Shear strength (Chapter G2)

h/tw = 45.1. kv = 5.34 (unstiffened web, a/h >> 3).

1.10 x sqrt(kv x E / Fy) = 1.10 x sqrt(5.34 x 29000 / 50) = 1.10 x 55.7 = 61.3.

Since h/tw = 45.1 < 61.3, Cv = 1.0 (shear yielding governs).

Vn = 0.6 x Fy x Aw x Cv = 0.6 x 50 x (20.8 x 0.380) x 1.0 = 237 kips. phi_v x Vn = 0.90 x 237 = 213 kips.

Shear check: Vu = 57.8 kips. Ratio = 57.8 / 213 = 0.27 -- PASS. Shear rarely governs for W-shapes under uniform load.

Step 4: Deflection (serviceability)

Live load deflection (unfactored): w_LL = 1.5 kip/ft.

Delta_LL = 5 x w_LL x L^4 / (384 x E x Ix) = 5 x (1.5/12) x (35 x 12)^4 / (384 x 29000 x 984) = 1.72 in.

Live load limit = L/360 = 35 x 12 / 360 = 1.17 in. Ratio = 1.72 / 1.17 = 1.47 -- FAILS.

Total load deflection: Delta_TL = 5 x (2.25/12) x 420^4 / (384 x 29000 x 984) = 2.58 in. Total load limit = L/240 = 420/240 = 1.75 in. Ratio = 2.58 / 1.75 = 1.47 -- FAILS.

Conclusion: W21x50 fails flexure and deflection for a 35 ft span. The design is governed by LTB (unbraced length too long for this section) and live load deflection. A deeper section (W24x68 or W27x84) with closer bracing spacing is required.

Example 2: Steel Column Design — Axial Compression

Design parameters

W12x65 column, 15 ft unbraced height. Fixed base, pinned top -- K = 0.80 per AISC Table C-A-7.1 (theoretical K = 2.0 for flagpole, but practical K = 0.80 for this end condition). Axial loads: dead = 180 kips, live = 140 kips. ASTM A992.

Section properties (W12x65)

A = 19.1 in^2, d = 12.1 in, bf = 12.0 in, tf = 0.605 in, tw = 0.390 in. rx = 5.28 in, ry = 3.02 in. bf/(2 x tf) = 9.92, h/tw = 25.3.

Factored axial load

Pu = 1.2 x 180 + 1.6 x 140 = 216 + 224 = 440 kips.

Step 1: Section classification

Flange: lambda_f = 9.92. Compact limit = 0.38 x sqrt(E/Fy) = 9.15. 9.92 > 9.15. Non-compact limit = 1.0 x sqrt(E/Fy) = 24.1. 9.92 < 24.1 -- flange is non-compact.

Web: lambda_w = 25.3. Compact limit for axial = 1.49 x sqrt(E/Fy) = 35.9. 25.3 < 35.9 -- web is compact.

Section with non-compact flange requires Qs reduction. For W12x65 flange: Qs = 1.0 (bf/(2tf) within non-compact range, Qs >= 0.942 calculated per Eq. E7-10). Use Q = Qs x Qa = 0.95.

Step 2: Flexural buckling (Section E3)

Weak axis governs: KLy = 0.80 x 15 x 12 = 144 in. KLy/ry = 144 / 3.02 = 47.7.

Fe = pi^2 x E / (KL/r)^2 = pi^2 x 29000 / 47.7^2 = 286,200 / 2275 = 125.8 ksi.

Limit = 4.71 x sqrt(E / (Q x Fy)) = 4.71 x sqrt(29000 / (0.95 x 50)) = 4.71 x sqrt(610.5) = 4.71 x 24.71 = 116.4.

Since KL/r = 47.7 < 116.4, inelastic buckling applies (Eq. E7-2):

Fcr = Q x (0.658^(Q x Fy / Fe)) x Fy = 0.95 x (0.658^(0.95 x 50 / 125.8)) x 50 = 0.95 x (0.658^0.378) x 50 = 0.95 x 0.857 x 50 = 40.7 ksi.

Pn = Fcr x A = 40.7 x 19.1 = 777 kips. phi_c x Pn = 0.90 x 777 = 699 kips.

Axial check: Pu = 440 kips. Ratio = 440 / 699 = 0.63 -- PASS. The column has 37% reserve capacity in pure compression.

Example 3: Combined Axial + Flexure Interaction

Design parameters

Same W12x65 column from Example 2, now with a first-order end moment Mnt = 85 kip-ft from frame action (no sidesway, B2 = 1.0 for braced frame). Pu = 440 kips from Example 2.

Step 1: Second-order moment amplification

Cm = 0.60 (braced frame, transverse load between supports, ends restrained).

Pe1 = pi^2 x E x Ix / (K1 x L)^2 = pi^2 x 29000 x 533 / (1.0 x 180)^2 = 47,200 kips (strong-axis buckling load).

B1 = Cm / (1 - Pu/Pe1) = 0.60 / (1 - 440/47200) = 0.60 / (1 - 0.0093) = 0.60 / 0.9907 = 0.606.

Wait -- with Pu/Pe1 only 0.93%, B1 is close to Cm. Let me recalculate with the actual formula.

Actually, for a column with Pu/Pe1 = 0.0093, B1 = 0.60 / (1 - 0.0093) = 0.61 (minimal amplification since the axial load is far below the Euler load).

Mux = B1 x Mnt = 0.61 x 85 = 51.9 kip-ft = 623 kip-in.

Wait, this seems off. Let me re-check. For the strong axis:

Pe1x = pi^2 x E x Ix / (K1 x L)^2. Ix for W12x65 = 533 in^4. Pe1x = pi^2 x 29000 x 533 / (1.0 x 180)^2 = 9.87 x 29000 x 533 / 32400 = 152,600,000 / 32400 = 4710 kips.

B1 = 0.60 / (1 - 440/4710) = 0.60 / (1 - 0.0934) = 0.60 / 0.9066 = 0.662.

Mrx = B1 x Mnt = 0.662 x 85 = 56.3 kip-ft = 676 kip-in.

Step 2: Flexural strength

For strong-axis bending of W12x65: phi_b x Mnx = phi_b x Mp = 0.90 x Zx x Fy = 0.90 x 96.8 x 50 = 4356 kip-in = 363 kip-ft (compact, Lb = 0 for strong axis in column application).

Step 3: Interaction check (Section H1.1)

Pr/Pc = Pu / (phi_c x Pn) = 440 / 699 = 0.630.

Since Pr/Pc = 0.630 > 0.2, use Eq. H1-1a:

Pr/Pc + 8/9 x (Mrx/Mcx + Mry/Mcy) <= 1.0

0.630 + 8/9 x (56.3/363 + 0) = 0.630 + 0.889 x 0.155 = 0.630 + 0.138 = 0.768.

Interaction check: 0.768 <= 1.0 -- PASS. The combined loading is well within capacity.

Key insight: For this lightly loaded column, the moment amplification is small (<10%) because Pu/Pe = 0.093. For columns loaded near their buckling capacity, B1 can exceed 2.0, dramatically increasing the required flexural strength.

Example 4: Bolted Connection — Bearing-Type Shear

Design parameters

Two A36 plates (3/8 in thick each) joined by 3 x 3/4 in diameter ASTM A325-N bolts in a single line. Bolt spacing = 3 in, edge distance = 1.5 in (sheared edge). Design shear per bolt = 12.5 kips. Threads NOT excluded from shear plane (N condition).

Step 1: Bolt shear strength (Section J3.6)

A325-N bolt, Fnv = 54 ksi (Table J3.2). Nominal area Ab = pi x (0.75)^2 / 4 = 0.442 in^2.

Rn = Fnv x Ab = 54 x 0.442 = 23.9 kips per bolt per shear plane (single shear). phi x Rn = 0.75 x 23.9 = 17.9 kips.

Shear check per bolt: 12.5 / 17.9 = 0.70 -- PASS.

Step 2: Bolt bearing at bolt holes (Section J3.10)

For the 3/8 in plate (Fu = 58 ksi), bolt hole diameter dh = 13/16 in (standard hole for 3/4 in bolt).

Edge bolts (outer two): Lc = Le - dh/2 = 1.5 - 0.406 = 1.094 in. Rn_edge = 1.2 x Lc x t x Fu <= 2.4 x d x t x Fu Rn_edge = 1.2 x 1.094 x 0.375 x 58 = 28.6 kips

Capacity limit: 2.4 x 0.75 x 0.375 x 58 = 39.2 kips. 28.6 < 39.2 -- edge distance governs. phi x Rn_edge = 0.75 x 28.6 = 21.5 kips.

Interior bolt (middle): Lc = s - dh = 3.0 - 0.813 = 2.187 in. Rn_int = 1.2 x 2.187 x 0.375 x 58 = 57.1 kips > 39.2 kips. Bearing limit controls: Rn = 39.2 kips. phi x Rn_int = 0.75 x 39.2 = 29.4 kips.

Bearing check: 12.5 / 21.5 = 0.58 for edge bolts (governs) -- PASS.

Step 3: Block shear rupture (Section J4.3)

For the connected plate, check the block of material around the bolt group:

Agv = (1.5 + 3.0 + 3.0) x 0.375 = 7.5 x 0.375 = 2.81 in^2. Anv = Agv - 2.5 x dh x t = 2.81 - 2.5 x 0.813 x 0.375 = 2.81 - 0.762 = 2.05 in^2. Ant = (1.5 - 0.5 x dh) x t = (1.5 - 0.406) x 0.375 = 0.41 in^2.

Ubs = 1.0 (uniform tension stress).

Rn = 0.6 x Fu x Anv + Ubs x Fu x Ant = 0.6 x 58 x 2.05 + 1.0 x 58 x 0.41 = 71.3 + 23.8 = 95.1 kips. Upper limit = 0.6 x Fy x Agv + Ubs x Fu x Ant = 0.6 x 36 x 2.81 + 23.8 = 60.7 + 23.8 = 84.5 kips.

Rn = min(95.1, 84.5) = 84.5 kips. phi x Rn = 0.75 x 84.5 = 63.4 kips.

Block shear check: Total shear on block = 3 x 12.5 = 37.5 kips. Ratio = 37.5 / 63.4 = 0.59 -- PASS.

Summary: All three bolt checks pass. Edge distance bearing governs at 58% utilization.

Example 5: Fillet Weld Design

Design parameters

5/16 in fillet weld, E70XX electrode (FEXX = 70 ksi). Weld length = 6 in on each side of a lap joint (two weld lines). Applied shear = 45 kips parallel to weld axis. A36 base metal (plate thickness = 1/2 in).

Step 1: Weld metal capacity (Section J2.4)

Effective throat: tt = leg x 0.707 = 0.3125 x 0.707 = 0.221 in.

Rn per inch = 0.60 x FEXX x tt = 0.60 x 70 x 0.221 = 9.28 kips per inch. phi x Rn per inch = 0.75 x 9.28 = 6.96 kips/in.

Total capacity for two 6-in welds: 6.96 x 6 x 2 = 83.5 kips.

Weld metal check: 45 / 83.5 = 0.54 -- PASS.

Step 2: Base metal capacity (Section J4)

Shear yielding in the 1/2 in plate adjacent to weld: Rn = 0.60 x Fy x Ag = 0.60 x 36 x (0.5 x 6 x 2) = 0.60 x 36 x 6.0 = 129.6 kips. phi x Rn = 0.90 x 129.6 = 116.6 kips (LRFD phi = 0.90 for base metal yielding, not 0.75).

Actually, per AISC J2.4, the base metal check at welds uses phi = 0.75 (same as welds) per Section J2.4. Let me correct:

Rn_base = 0.60 x Fy x t_plate x Lw = 0.60 x 36 x 0.5 x 6 = 64.8 kips per weld line. Total for two sides = 129.6 kips. phi x Rn_base = 0.75 x 129.6 = 97.2 kips.

Base metal check: 45 / 97.2 = 0.46 -- PASS. Weld metal governs over base metal.

Step 3: Minimum weld size check (Table J2.4)

For 1/2 in base metal, minimum fillet weld size = 3/16 in. 5/16 in > 3/16 in -- OK.

Summary: 5/16 in fillet weld x 6 in long on each side is adequate at 54% utilization.

Example 6: Column Base Plate Design

Design parameters

W10x49 column (d = 10.0 in, bf = 10.0 in), axial compression Pu = 300 kips. A36 base plate (Fy = 36 ksi). Concrete f'c = 4 ksi. Pedestal dimensions = 22 in x 22 in (geometrically similar and concentric with base plate). No moment.

Step 1: Required bearing area (AISC 360 J8)

phi_c x Pp_max = phi_c x 0.85 x f'c x A1 x sqrt(A2/A1)

With sqrt(A2/A1) <= 2.0, and assuming full confinement:

A1_req = Pu / (phi_c x 0.85 x f'c x 2.0) = 300 / (0.65 x 0.85 x 4 x 2.0) = 300 / 4.42 = 67.9 in^2.

Step 2: Plate dimensions (AISC DG1)

Delta = (0.95 x d - 0.8 x bf) / 2 = (0.95 x 10.0 - 0.8 x 10.0) / 2 = (9.5 - 8.0) / 2 = 0.75 in.

N_min = sqrt(A1_req) + Delta = sqrt(67.9) + 0.75 = 8.24 + 0.75 = 8.99 in. B_min = A1_req / N_min = 67.9 / 8.99 = 7.55 in.

Try N = 14 in, B = 14 in (A1 = 196 in^2).

Verify sqrt(A2/A1): A2 = 22 x 22 = 484 in^2. sqrt(484/196) = sqrt(2.47) = 1.57 <= 2.0 -- OK.

phi_c x Pp = 0.65 x 0.85 x 4 x 196 x 1.57 = 679 kips > 300 kips. Bearing utilization = 300/679 = 0.44 -- PASS.

Step 3: Plate thickness (AISC DG1)

Cantilever dimensions: m = (N - 0.95 x d) / 2 = (14 - 0.95 x 10.0) / 2 = (14 - 9.5) / 2 = 2.25 in. n = (B - 0.8 x bf) / 2 = (14 - 0.8 x 10.0) / 2 = (14 - 8.0) / 2 = 3.00 in.

Critical cantilever = max(m, n, lambda_n') = max(2.25, 3.00) = 3.00 in.

Bearing pressure: fp = Pu / (N x B) = 300 / 196 = 1.531 ksi.

Plate bending moment per inch width: Mpl = fp x n^2 / 2 = 1.531 x 3.00^2 / 2 = 6.89 kip-in/in.

Required thickness: t_req = sqrt(4 x Mpl / (phi_b x Fy)) = sqrt(4 x 6.89 / (0.90 x 36)) = sqrt(27.56 / 32.4) = sqrt(0.851) = 0.92 in.

Use 1 in thick base plate. 14 x 14 x 1 in A36 plate.

Summary: 14 x 14 x 1 in A36 base plate is adequate for 300 kip axial compression on 4 ksi concrete.

Code comparison: AISC 360 vs other standards

Check Type AISC 360-22 Phi Factor AS 4100 Phi Factor EN 1993 Gamma_M Notes
Beam flexure 0.90 0.90 1.00 (gamma_M0) Similar factors across codes
Column compression 0.90 0.90 1.00 EN uses multiple buckling curves
Bolt shear (bearing) 0.75 0.80 1.25 (gamma_M2) AS 4100 gives slightly higher bolt cap.
Fillet weld 0.75 0.80 (SP) 1.25 AS 4100 SP = 0.80, GP = 0.60
Concrete bearing 0.65 0.60 1.50 AISC gives highest bearing capacity
Block shear 0.75 0.75 1.25 Similar theoretical basis

The resistance factors are broadly comparable, but mixing factors between codes produces incorrect results. Always use the correct factors for the governing standard of your project.

Common mistakes in AISC 360 design

  1. Forgetting to check LTB: Many engineers size beams for Mp and forget that unbraced length reduces capacity significantly (Example 1 ratio went from 0.92 to 1.59 when LTB was considered).
  2. Using wrong Cb: Cb = 1.0 is conservative for all cases. Using the actual Cb (1.14 for uniform load with 5 braces) increases Mn by up to 14%.
  3. Confusing LRFD and ASD load combinations: Mixing ASD service loads with LRFD phi factors is the most common spreadsheet error.
  4. Ignoring non-compact flange reductions: W12x65 flange is non-compact; ignoring Qs overestimates column capacity by ~5%.
  5. Missing the Pr/Pc > 0.2 threshold: The H1 interaction equation changes form at Pr/Pc = 0.2. Using the wrong equation can be unconservative.
  6. Not checking minimum weld sizes: A 1/8 in weld on 3/4 in plate violates J2.4 minimum weld size (need 1/4 in minimum).
  7. Edge distance violations: AISC Table J3.4 specifies minimum edge distances. Sheared edges require larger distances than rolled edges.

Frequently Asked Questions

Do I need to check both LRFD and ASD? No. Choose one methodology per project. LRFD is standard for building design in the US. ASD persists in some industrial and bridge applications where service-load design is preferred.

How do I know if my section is compact? Check Table B4.1b. For W-shapes in A992: flange lambda_r = 1.0 x sqrt(E/Fy) = 24.1 for flexure. Web lambda_r = 5.70 x sqrt(E/Fy) = 137 for flexure. Most rolled W-shapes in A992 are compact for flexure.

What is the most commonly failed check in AISC 360 design? Deflection consistently governs over strength for long-span beams. For columns, weak-axis buckling (KL/ry) governs unless the column is braced in the weak direction.

Can these examples be used for the PE exam? Yes, the methodology matches what is tested. However, the PE exam favors AISC Manual tables for standard cases. Practice with Parts 3-6 of the Manual for speed.

What changed from AISC 360-16 to AISC 360-22? Key changes include updated bolt strength values in Table J3.2 (higher strengths for Group A and B bolts), revised composite beam provisions, and new HSS connection design provisions aligned with AISC Design Guide 24.

Key Takeaways

Run These Calculations

Beam Capacity Calculator -- W-shape moment, shear, and deflection checks per AISC 360 with LTB and Cb parameters.

Column Capacity Calculator -- Axial compression and buckling check per AISC 360 Chapter E with K-factor selection.

Bolted Connection Calculator -- Bearing-type and slip-critical bolt groups. Shear, bearing, tension, and block shear.

Welded Connection Calculator -- Fillet and groove weld capacity per AISC 360 J2.4 with directional strength enhancement.

Base Plate & Anchors Calculator -- Bearing, plate bending, anchor tension and shear per AISC 360 and ACI 318.

Section Properties Database -- Browse 500+ W, HSS, C, L, and WT sections with dimensions, Ix, Sx, Zx, ry, J, Cw, and classification limits per AISC 360.

Further Reading

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