Steel Beam Design Example — AISC 360-22 Step-by-Step

Steel beam design is the most fundamental task in structural engineering. Every floor system, every roof, every transfer girder starts with a beam design check. The AISC 360-22 Specification provides a methodical five-step procedure: classify the section, check flexure (including lateral-torsional buckling), check shear, and verify deflection.

In this guide: We design a W12x65 beam spanning 20 feet under a uniform dead load of 0.5 kip/ft and live load of 1.5 kip/ft, using AISC 360-22 LRFD. Every equation is shown with actual numbers.

PRELIMINARY — NOT FOR CONSTRUCTION. All results discussed are for educational and reference use only. Must be independently verified by a licensed Professional Engineer or Structural Engineer before use in any project.

Design Problem Statement

Required moment (LRFD): $$w_u = 1.2D + 1.6L = 1.2(0.5) + 1.6(1.5) = 0.6 + 2.4 = 3.0\text{ kip/ft}$$

$$M_u = \frac{w_u L^2}{8} = \frac{3.0 \times 20^2}{8} = \frac{1200}{8} = 150\text{ kip-ft}$$

Required shear (LRFD): $$V_u = \frac{w_u L}{2} = \frac{3.0 \times 20}{2} = 30\text{ kips}$$

Step 1: Section Properties — W12x65

From AISC Manual Table 1-1:

For quick lookup of section properties without the AISC Manual, see the Section Properties Database — browse 500+ W, HSS, C, L, and WT sections with dimensions, Ix, Sx, Zx, ry, J, Cw, and classification data.

Step 2: Section Classification — AISC 360 Table B4.1b

For flexure, classify both the flange and web.

Flange (unstiffened element, Case 1): $$\lambda_f = \frac{b_f}{2t_f} = \frac{12.0}{2 \times 0.605} = 9.92$$

$$\lambda_{pf} = 0.38\sqrt{\frac{E}{F_y}} = 0.38\sqrt{\frac{29,000}{50}} = 9.15$$

Since $\lambda_f = 9.92 > \lambda_{pf} = 9.15$, the flange is noncompact.

Check $\lambda_{rf}$: $$\lambda_{rf} = 1.0\sqrt{\frac{E}{F_y}} = 1.0 \times 24.1 = 24.1$$

Since $\lambda_f = 9.92 < \lambda_{rf} = 24.1$, the flange is noncompact (not slender). OK.

Web (stiffened element, Case 10): $$\lambda_w = \frac{h}{t_w} = \frac{d - 2t_f}{t_w} = \frac{12.1 - 2(0.605)}{0.390} = \frac{10.89}{0.390} = 27.9$$

$$\lambda_{pw} = 3.76\sqrt{\frac{E}{F_y}} = 3.76 \times 24.1 = 90.6$$

Since $\lambda_w = 27.9 < \lambda_{pw} = 90.6$, the web is compact.

Overall classification: The section is noncompact due to the flange. Flexural capacity is limited to $M_n$ between $M_p$ and $M_y$.

Step 3: Flexural Strength — AISC 360 Chapter F2

3a. Yielding (Plastic Moment)

$$M_p = F_y \times Z_x = 50 \times 96.8 = 4840\text{ kip-in} = 403.3\text{ kip-ft}$$

This is the theoretical upper bound assuming full plastification with no instability.

3b. Lateral-Torsional Buckling (LTB)

The unbraced length $L_b = 20\text{ ft} = 240\text{ in}$.

Limiting lengths per AISC 360 F2-5 and F2-6:

$$L_p = 1.76 \times r_y \times \sqrt{\frac{E}{F_y}} = 1.76 \times 3.02 \times \sqrt{\frac{29,000}{50}} = 1.76 \times 3.02 \times 24.08 = 128.1\text{ in} = 10.7\text{ ft}$$

$$L_r = \text{(calculate from F2-6)}$$

For LTB calculation, use AISC 360 Equation F2-2 with $C_b = 1.0$ (conservative for uniform moment):

Elastic LTB stress (Eq. F2-4):

$$F_{cr} = \frac{C_b \pi^2 E}{\left(\frac{L_b}{r_{ts}}\right)^2}\sqrt{1 + 0.078\frac{Jc}{S_x h_o}\left(\frac{L_b}{r_{ts}}\right)^2}$$

$$\frac{L_b}{r_{ts}} = \frac{240}{3.37} = 71.2$$

For a doubly symmetric I-shape, $c = 1.0$:

$$F_{cr} = \frac{1.0 \times \pi^2 \times 29,000}{(71.2)^2}\sqrt{1 + 0.078\frac{2.01 \times 1.0}{87.9 \times 11.5}(71.2)^2}$$

$$F_{cr} = \frac{286,180}{5070}\sqrt{1 + 0.078 \times \frac{2.01}{87.9 \times 11.5} \times 5070}$$

$$F_{cr} = 56.44 \times \sqrt{1 + 0.078 \times 0.00199 \times 5070}$$

$$F_{cr} = 56.44 \times \sqrt{1 + 0.787} = 56.44 \times \sqrt{1.787} = 56.44 \times 1.337 = 75.45\text{ ksi}$$

Since $L_b = 20\text{ ft} > L_r$ (calculation confirms), elastic LTB governs.

Nominal flexural strength:

$$M_n = F_{cr} \times S_x = 75.45 \times 87.9 = 6630\text{ kip-in} = 552.5\text{ kip-ft}$$

But $M_n \leq M_p = 403.3\text{ kip-ft}$. Since $M_n$ from the elastic LTB equation exceeds $M_p$, we cap at $M_p$.

Wait — this requires a check on $L_r$:

$$L_r = 1.95 \times r_{ts} \times \frac{E}{0.7F_y}\sqrt{\frac{Jc}{S_x h_o} + \sqrt{\left(\frac{Jc}{S_x h_o}\right)^2 + 6.76\left(\frac{0.7F_y}{E}\right)^2}}$$

For simplicity, from AISC Manual Table 3-10 for W12x65:

• $L_p = 10.7$ ft
• $L_r = 38.4$ ft

Since $L_p < L_b = 20\text{ ft} < L_r = 38.4\text{ ft}$, we are in the inelastic LTB range. Use Eq. F2-2:

$$M_n = C_b\left[M_p - (M_p - 0.7F_y S_x)\left(\frac{L_b - L_p}{L_r - L_p}\right)\right] \leq M_p$$

$$0.7F_y S_x = 0.7 \times 50 \times 87.9 = 3076\text{ kip-in} = 256.3\text{ kip-ft}$$

$$M_n = 1.0\left[403.3 - (403.3 - 256.3)\left(\frac{20 - 10.7}{38.4 - 10.7}\right)\right]$$

$$M_n = 403.3 - 147.0 \times \frac{9.3}{27.7} = 403.3 - 147.0 \times 0.336 = 403.3 - 49.4 = 353.9\text{ kip-ft}$$

LRFD design flexural strength: $$\phi_b M_n = 0.90 \times 353.9 = \mathbf{318.5\text{ kip-ft}}$$

Check: $\phi_b M_n = 318.5 > M_u = 150$ kip-ft. Flexure OK. Demand/capacity ratio = 150/318.5 = 0.471.

Step 4: Shear Strength — AISC 360 Chapter G2

Web shear coefficient: $$C_v = 1.0 \text{ when } \frac{h}{t_w} \leq 1.10\sqrt{\frac{k_v E}{F_y}}$$

For unstiffened webs, $k_v = 5.34$:

$$1.10\sqrt{\frac{5.34 \times 29,000}{50}} = 1.10 \times 55.7 = 61.3$$

Our $h/t_w = 27.9 < 61.3$, so $C_v = 1.0$. Shear yielding governs.

Nominal shear strength: $$V_n = 0.6F_y A_w C_v = 0.6 \times 50 \times (12.1 \times 0.390) \times 1.0$$ $$V_n = 0.6 \times 50 \times 4.719 = 141.6\text{ kips}$$

LRFD design shear strength: $$\phi_v V_n = 0.90 \times 141.6 = \mathbf{127.4\text{ kips}}$$

Check: $\phi_v V_n = 127.4 > V_u = 30.0$ kips. Shear OK. Demand/capacity ratio = 30/127.4 = 0.236.

Since $V_u < 0.5\phi_v V_n = 63.7$ kips, shear-flexure interaction does not need to be checked per AISC 360 G2.

Step 5: Deflection — AISC 360 Chapter L (Serviceability)

Live load deflection (L/360 limit):

$$w_L = 1.5\text{ kip/ft} = 0.125\text{ kip/in}$$

$$\Delta_L = \frac{5w_L L^4}{384EI_x} = \frac{5 \times 0.125 \times (240)^4}{384 \times 29,000 \times 533}$$

$$\Delta_L = \frac{5 \times 0.125 \times 3.318 \times 10^9}{384 \times 29,000 \times 533} = \frac{2.074 \times 10^9}{5.935 \times 10^9} = 0.349\text{ in}$$

Allowable: $L/360 = 240/360 = 0.667$ in.

Check: 0.349 in < 0.667 in. Deflection OK.

Total load deflection (L/240 limit):

$$w_{total} = D + L = 2.0\text{ kip/ft} = 0.167\text{ kip/in}$$

$$\Delta_{total} = 0.349 \times \frac{2.0}{1.5} = 0.466\text{ in}$$

Allowable: $L/240 = 240/240 = 1.0$ in.

Check: 0.466 in < 1.0 in. Total deflection OK.

Summary of Results — W12x65, 20ft, 3.0 kip/ft (factored)

The W12x65 is adequate for this loading condition. With a demand/capacity ratio of 0.471 in flexure, the beam has significant reserve capacity. A W12x50 or W12x53 could be considered for a more economical design, subject to a repeat of this analysis.

Practical Application — Using Our Beam Calculator

Manual beam design with LTB calculations involves multiple iterative steps with long-form equations. Our free Beam Capacity Calculator handles all AISC 360 limit states automatically:

• Beam Capacity Calculator — Enter your beam section, span, loading, and unbraced length. The calculator runs flexure, LTB, shear, and deflection checks per AISC 360, AS 4100, EN 1993, and CSA S16. Results shown side-by-side across all codes.

The calculator runs entirely in your browser with no signup required.

FAQ

What are the steps to design a steel beam per AISC 360?

Steel beam design per AISC 360-22 follows five sequential steps: (1) Determine required strength from factored loads (LRFD) or service loads (ASD). (2) Classify the section as compact, noncompact, or slender per Table B4.1b. (3) Check flexural strength per Chapter F, including lateral-torsional buckling for unbraced segments. (4) Check shear strength per Chapter G. (5) Check serviceability per Chapter L, primarily deflection. Each step must be satisfied before proceeding.

What is lateral-torsional buckling (LTB)?

Lateral-torsional buckling (LTB) is a stability limit state where a beam under flexure buckles out-of-plane — the compression flange displaces laterally while the section twists. LTB is governed by the unbraced length $L_b$. Compact beams with short unbraced lengths reach the plastic moment $M_p$. As $L_b$ increases, the capacity transitions to the inelastic LTB range, then the elastic LTB range where the critical moment $M_{cr}$ is calculated per AISC 360 Equation F2-4. Continuous lateral bracing of the compression flange eliminates LTB.

What is the difference between LRFD and ASD in beam design?

LRFD (Load and Resistance Factor Design) uses factored loads with resistance factors ($\phi$) applied to nominal strengths. ASD (Allowable Strength Design) uses service-level loads with safety factors ($\Omega$) applied to nominal strengths. For beam flexure per AISC 360, $\phi_b = 0.90$ and $\Omega_b = 1.67$, so LRFD provides approximately 1.5x the design strength of ASD for the same nominal capacity. Both methods are valid per AISC 360.

How do I check deflection per AISC 360?

Deflection is a serviceability check under Chapter L of AISC 360-22, using service-level (unfactored) loads. Common deflection limits are L/360 for live load on floor beams, L/240 for total load, and L/180 for roof beams. The beam stiffness EI is based on the moment of inertia $I_x$ about the axis of bending. For a simply supported beam with uniform load, $\Delta_{max} = 5wL^4/(384EI)$.

Related Calculators

• Beam Capacity Calculator — Complete beam design across four codes
• Beam Span Calculator — Preliminary sizing based on span and loading
• Section Properties Guide — How to read and use A, Ix, Sx, Zx
• Steel Beam Deflection Guide — Deflection checks and limits explained
• Steel Beam Span Tables — Pre-calculated span ranges for common sections