15.2 Structural Steel Compression Members and Beam Design

15.2 Structural Steel Compression Members and Beam Design

Compression members and flexural beams represent the primary load-resisting elements of building frames, industrial trusses, and bridges. Unlike tension members, steel elements subjected to axial compression or bending flexure are inherently prone to instability limit states—including overall flexural buckling, torsional-flexural buckling, local plate buckling, and lateral-torsional buckling (LTB). Understanding the mathematical transition between material yielding and geometric instability is fundamental to structural steel design under NSCP 2015 Sections 505 and 506 (based on AISC 360-10 Chapters E and F).

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1. Column Buckling Mechanics & Effective Length Factor ($K$)

An ideal, perfectly straight elastic column pinned at both ends buckles at the Euler critical load ($P_e$):

$P_e = \frac{\pi^2 E I}{(K L)^2} \implies F_e = \frac{P_e}{A_g} = \frac{\pi^2 E}{\left( \frac{K L}{r} \right)^2}$

Where:

• $E$ = modulus of elasticity of structural steel ($200,000\text{ MPa}$ or $29,000\text{ ksi}$).
• $I$ = moment of inertia about the axis of buckling ($I = A r^2$).
• $r$ = radius of gyration ($r = \sqrt{I/A}$).
• $L$ = actual unbraced length of the column.
• $K$ = effective length factor, representing the ratio of the distance between inflection points of zero moment to the actual unbraced length.
• $\frac{K L}{r}$ = slenderness ratio of the column.

Boundary Restraints and $K$-Values

In structural analysis, theoretical end conditions assume idealized frictionless pins or zero-rotation fixed supports. In actual construction, connection flexibility, foundation settlement, and gusset rotations prevent perfect fixity. The AISC Commentary table on effective length factors (Table C-A-7.1) gives both theoretical and recommended design $K$ values:

Maximum Slenderness Limit

AISC 360 (Section E2 user note), adopted in NSCP 2015, recommends that for members designed on the basis of compression, the slenderness ratio preferably not exceed:

$\frac{K L}{r} \le 200$

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2. Alignment Charts for Frames (Sway vs. Non-Sway)

For columns in continuous building frames, end rotational restraint depends on the relative flexural stiffnesses of the columns and framing girders meeting at joints $A$ and $B$. The rotational stiffness ratio ($G$) at a joint is defined as:

$G = \frac{\sum (E I / L)_{\text{columns}}}{\sum (E I / L)_{\text{girders}}}$

• If a column base is pinned to the foundation, theoretical $G = \infty$ (recommended design value $G = 10$).
• If a column base is rigidly fixed to the foundation, theoretical $G = 0$ (recommended design value $G = 1.0$).

Sidesway-Inhibited (Braced / Non-Sway) Frames

In braced frames (braced by diagonal trusses, shear walls, or moment cores), lateral drift is prevented. The effective length factor $K$ is always less than or equal to $1.0$ ($0.5 \le K \le 1.0$), governed by the transcendental equation:

$\frac{G_A G_B}{4} \left(\frac{\pi}{K}\right)^2 + \left(\frac{G_A + G_B}{2}\right) \left( 1 - \frac{\pi / K}{\tan(\pi / K)} \right) + 2 \left( \frac{\tan(\pi / 2K)}{\pi / K} \right) - 1 = 0$

Sidesway-Uninhibited (Unbraced / Sway) Frames

In unbraced moment frames, lateral stability depends solely on the flexural rigidity of the columns and beams. Sidesway buckling can occur, meaning the effective length factor $K$ is always greater than or equal to $1.0$ ($1.0 \le K < \infty$), governed by:

$\frac{G_A G_B (\pi / K)^2 - 36}{6 (G_A + G_B)} - \frac{\pi / K}{\tan(\pi / K)} = 0$

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3. AISC / NSCP 2015 Column Strength Curves

Real steel columns contain residual stresses (from differential cooling after hot-rolling or flame cutting) and initial geometric out-of-straightness (camber/sweep $\approx L/1000$). Consequently, columns with intermediate slenderness yield in their outer fibers before reaching the Euler buckling load, exhibiting inelastic buckling.

Slenderness Threshold

The boundary between inelastic buckling and elastic Euler buckling occurs at:

$\frac{K L}{r} = 4.71 \sqrt{\frac{E}{F_y}} \qquad \left( \text{equivalent to } \frac{F_y}{F_e} = 2.25 \right)$

• For standard A36 steel ($F_y = 248\text{ MPa}$, $E = 200,000\text{ MPa}$): $4.71 \sqrt{\frac{200,000}{248}} = 133.7$.
• For Grade 50 steel ($F_y = 345\text{ MPa}$, $E = 200,000\text{ MPa}$): $4.71 \sqrt{\frac{200,000}{345}} = 113.4$.

Critical Buckling Stress ($F_{cr}$)

1. Inelastic Buckling Regime: When $\frac{K L}{r} \le 4.71 \sqrt{\frac{E}{F_y}}$ (or $F_e \ge 0.44 F_y$): $F_{cr} = \left[ 0.658^{\frac{F_y}{F_e}} \right] F_y$

2. Elastic Buckling Regime: When $\frac{K L}{r} > 4.71 \sqrt{\frac{E}{F_y}}$ (or $F_e < 0.44 F_y$): $F_{cr} = 0.877 F_e$ (The factor $0.877$ accounts for initial column out-of-straightness).

Nominal Compressive Strength

For columns with compact and non-compact cross-sections (no local buckling):

$P_n = F_{cr} A_g$

$\phi_c P_n = 0.90 F_{cr} A_g \quad (\text{LRFD}), \qquad \frac{P_n}{\Omega_c} = \frac{F_{cr} A_g}{1.67} \quad (\text{ASD})$

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4. Local Buckling & Width-to-Thickness Limits

Before an entire column or beam reaches its overall buckling capacity, individual plate elements (flanges, webs, legs) may wrinkle or buckle locally under compressive stress. Cross-sections are classified per NSCP Section 502.4 based on their width-to-thickness ratio ($\lambda = b/t$ or $h/t_w$):

1. Compact Sections: Elements can develop full plastic moment capacity ($M_p$) and undergo extensive plastic rotation without local buckling ($\lambda \le \lambda_p$).
2. Non-Compact Sections: Elements can develop the yield stress in compression before local buckling occurs, but cannot achieve full plastic rotation capacity ($\lambda_p < \lambda \le \lambda_r$).
3. Slender Sections: Elements buckle locally in the elastic range before the material yield stress is reached ($\lambda > \lambda_r$). Slender columns require a reduction factor ($Q = Q_s Q_a$) that scales down $F_{cr}$.

For hot-rolled I-shaped columns:

• Unstiffened element (flange in compression): $\lambda = \frac{b_f}{2 t_f}$, $\lambda_r = 0.56 \sqrt{\frac{E}{F_y}}$.
• Stiffened element (web in axial compression): $\lambda = \frac{h}{t_w}$, $\lambda_r = 1.49 \sqrt{\frac{E}{F_y}}$.

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5. Steel Flexural Members (Beams): Plastic Moment & Shape Factor

When an I-shaped beam is subjected to bending moment about its major axis, the stress distribution progresses through three distinct stages:

1. Elastic Bending Stage: Stresses vary linearly from zero at the neutral axis to extreme fiber stress $f_b = M / S_x$. The yield moment ($M_y$) corresponds to first yield at the extreme fibers: $M_y = F_y S_x$ where $S_x = \frac{I_x}{c}$ is the elastic section modulus.

2. Inelastic Bending Stage: Outer fibers yield plastically while inner fibers remain elastic. The plastic zones propagate inward toward the neutral axis.

3. Fully Plastic Stage: The entire cross-section yields simultaneously in tension and compression ($f = F_y$). The neutral axis divides the section into equal areas of tension and compression ($A_T = A_C = A/2$). The plastic moment capacity ($M_p$) is: $M_p = F_y Z_x = F_y \sum |A_i y_i|$ where $Z_x$ is the plastic section modulus.

The Shape Factor

The ratio of the fully plastic moment to the elastic yield moment is defined as the Shape Factor ($SF$):

$SF = \frac{M_p}{M_y} = \frac{F_y Z_x}{F_y S_x} = \frac{Z_x}{S_x}$

• Standard Wide-Flange (W) beams: $SF \approx 1.10 \text{ to } 1.15$ (typically taken as $1.12$).
• Solid rectangular sections: $SF = 1.50$.
• Solid circular sections: $SF = 1.70$.
• Diamond sections (bending about diagonal): $SF = 2.00$.

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6. Lateral-Torsional Buckling (LTB) of Beams

When an I-beam bends about its major strong axis, the compression flange behaves like an unbraced column. If the compression flange lacks continuous lateral support, it tends to buckle laterally out of the plane of bending, while the tension flange remains stable. The cross-section twists simultaneously, a phenomenon termed Lateral-Torsional Buckling (LTB).

LTB Regimes Based on Unbraced Length ($L_b$)

NSCP Section 506 divides flexural behavior into three distinct zones depending on the lateral unbraced length ($L_b$) of the compression flange:

1. Zone 1: Plastic Yielding ($L_b \le L_p$): Full plastic moment is achieved; no LTB can occur: $M_n = M_p = F_y Z_x$ Where the limiting unbraced length for full plastic strength is: $L_p = 1.76 r_y \sqrt{\frac{E}{F_y}}$

2. Zone 2: Inelastic LTB ($L_p < L_b \le L_r$): Buckling initiates after part of the compression flange has yielded: $M_n = C_b \left[ M_p - (M_p - 0.70 F_y S_x) \left( \frac{L_b - L_p}{L_r - L_p} \right) \right] \le M_p$

3. Zone 3: Elastic LTB ($L_b > L_r$): Buckling occurs in the purely linear elastic range: $M_n = F_{cr} S_x = C_b \frac{\pi^2 E}{(L_b / r_{ts})^2} \sqrt{1 + 0.078 \frac{J c}{S_x h_0} \left( \frac{L_b}{r_{ts}} \right)^2} S_x \le M_p$

Moment Gradient Factor ($C_b$)

The theoretical derivation of LTB assumes a uniform bending moment throughout the unbraced length (the worst-case condition, $C_b = 1.0$). When the moment varies along the unbraced length, non-uniform compression relieves buckling propensity, accounted for by the moment gradient factor ($C_b$):

$C_b = \frac{12.5 M_{\max}}{2.5 M_{\max} + 3 M_A + 4 M_B + 3 M_C} R_m$

Where:

• $M_{\max}$ = absolute value of maximum moment in the unbraced segment.
• $M_A$ = absolute moment at quarter-point ($L_b / 4$).
• $M_B$ = absolute moment at mid-point ($L_b / 2$).
• $M_C$ = absolute moment at three-quarter point ($3 L_b / 4$).
• $R_m = 1.0$ for doubly symmetric members.
• Note: In all cases, $C_b M_n$ cannot exceed the plastic moment $M_p$.

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7. Beam Shear Strength Design

Under NSCP 2015 Section 507, the nominal shear strength of unstiffened or stiffened webs of I-shaped members is governed by shear yielding or shear buckling:

$V_n = 0.60 F_y A_w C_v$

$\phi_v V_n = 1.00 V_n \quad (\text{LRFD for most hot-rolled standard shapes}), \qquad \frac{V_n}{\Omega_v} = \frac{V_n}{1.50} \quad (\text{ASD})$

Where:

• $A_w = d \cdot t_w$ is the overall web area (overall beam depth $\times$ web thickness).
• $C_v$ = web shear coefficient. For virtually all standard hot-rolled W-beams satisfying $\frac{h}{t_w} \le 2.24 \sqrt{\frac{E}{F_y}}$, full shear yielding develops without shear buckling, so $C_v = 1.0$ and $\phi_v = 1.00$.

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8. Members Under Combined Axial Load and Bending (Beam-Columns)

The PSAD TOS item "solve for the stresses in steel beams and columns due to eccentric loads" covers members carrying axial force and moment together. Examples are columns with eccentric beam reactions, rafters, and truss chords with transverse loads.

Elastic stress check (allowable stress approach). For an axial force $P$ at eccentricity $e$, the extreme fiber stresses are:

$f = \frac{P}{A} \pm \frac{P e c}{I} = \frac{P}{A} \pm \frac{M}{S}$

AISC 360 interaction (H1-1). NSCP 2015 Section 508 is based on AISC 360-10 Chapter H. With required strengths $P_r$, $M_{rx}$, $M_{ry}$ and available strengths $P_c = \phi_c P_n$, $M_{cx} = \phi_b M_{nx}$, $M_{cy} = \phi_b M_{ny}$:

$\text{If } \frac{P_r}{P_c} \ge 0.2: \quad \frac{P_r}{P_c} + \frac{8}{9}\left( \frac{M_{rx}}{M_{cx}} + \frac{M_{ry}}{M_{cy}} \right) \le 1.0$

$\text{If } \frac{P_r}{P_c} < 0.2: \quad \frac{P_r}{2 P_c} + \left( \frac{M_{rx}}{M_{cx}} + \frac{M_{ry}}{M_{cy}} \right) \le 1.0$

The required moments must include second-order ($P$-$\delta$ and $P$-$\Delta$) effects, from a second-order analysis or from amplification factors.

Example. A column has $\phi_c P_n = 1{,}237\text{ kN}$ and $\phi_b M_{nx} = 300\text{ kN}\cdot\text{m}$, and carries $P_u = 600\text{ kN}$ with $M_{ux} = 120\text{ kN}\cdot\text{m}$. Since $600/1{,}237 = 0.485 \ge 0.2$, use H1-1a: $0.485 + (8/9)(120/300) = 0.485 + 0.356 = 0.841 \le 1.0$. The column is adequate.

9. Comprehensive Worked Examples

Worked Example 1: Compressive Strength of a W-Shape Column

Problem: A W12x50 column ($A_g = 9,420\text{ mm}^2$, $r_x = 131.6\text{ mm}$, $r_y = 49.8\text{ mm}$) of A36 steel ($F_y = 248\text{ MPa}$, $E = 200,000\text{ MPa}$) has an unbraced length of $L = 5.0\text{ m}$. Both ends are pinned ($K_x = K_y = 1.0$). Check local buckling, determine the governing slenderness ratio, and compute the design compressive strength $\phi_c P_n$ (LRFD).

Solution:

• Step 1: Slenderness Ratio: Buckling governs about the weak $y$-axis because $r_y < r_x$: $\left( \frac{K L}{r} \right)_y = \frac{1.0 \times 5,000\text{ mm}}{49.8\text{ mm}} = 100.40$ (Check slenderness limit: $100.40 \le 200$. Satisfied).

• Step 2: Slenderness Threshold: $4.71 \sqrt{\frac{E}{F_y}} = 4.71 \sqrt{\frac{200,000}{248}} = 133.7$ Since $\frac{K L}{r} = 100.40 \le 133.7$, buckling is inelastic.

• Step 3: Euler Elastic Buckling Stress ($F_e$): $F_e = \frac{\pi^2 E}{(K L / r)^2} = \frac{\pi^2 \times 200,000}{(100.40)^2} = \frac{1,973,921}{10,080.16} = 195.82\text{ MPa}$

• Step 4: Critical Stress ($F_{cr}$): $\frac{F_y}{F_e} = \frac{248}{195.82} = 1.2665$ $F_{cr} = \left[ 0.658^{\frac{F_y}{F_e}} \right] F_y = \left[ 0.658^{1.2665} \right] \times 248$ $\ln(0.658) = -0.41855 \implies 1.2665 \times (-0.41855) = -0.53009 \implies e^{-0.53009} = 0.58855$ $F_{cr} = 0.58855 \times 248 = 145.96\text{ MPa}$

• Step 5: Nominal and Design Compressive Strength: $P_n = F_{cr} A_g = 145.96\text{ MPa} \times 9,420\text{ mm}^2 = 1,374,943\text{ N} = 1,374.9\text{ kN}$ $\phi_c P_n = 0.90 \times 1,374.9\text{ kN} = 1,237.4\text{ kN}$

Worked Example 2: Flexural Plastic Capacity & Shear of a Beam

Problem: A simply supported rolled wide-flange beam has a span of $6.0\text{ m}$. The beam properties are: $d = 413\text{ mm}$, $t_w = 9.65\text{ mm}$, $S_x = 1,180 \times 10^3\text{ mm}^3$, $Z_x = 1,340 \times 10^3\text{ mm}^3$, $F_y = 345\text{ MPa}$ (Grade 50). Full lateral bracing is provided to the compression flange ($L_b = 0$). Calculate: (a) the yield moment $M_y$, (b) the plastic moment $M_p$, (c) the shape factor, and (d) the design shear strength $\phi_v V_n$.

Solution:

• Step 1: Yield Moment: $M_y = F_y S_x = 345\text{ MPa} \times (1,180 \times 10^3\text{ mm}^3) = 407.1 \times 10^6\text{ N}\cdot\text{mm} = 407.1\text{ kN}\cdot\text{m}$

• Step 2: Plastic Moment: $M_p = F_y Z_x = 345\text{ MPa} \times (1,340 \times 10^3\text{ mm}^3) = 462.3 \times 10^6\text{ N}\cdot\text{mm} = 462.3\text{ kN}\cdot\text{m}$

• Step 3: Shape Factor: $SF = \frac{M_p}{M_y} = \frac{462.3}{407.1} = 1.136$

• Step 4: Design Shear Strength: Web area $A_w = d \cdot t_w = 413\text{ mm} \times 9.65\text{ mm} = 3,985.45\text{ mm}^2$. For standard hot-rolled wide flanges with $C_v = 1.0$ and $\phi_v = 1.00$: $V_n = 0.60 F_y A_w = 0.60 \times 345\text{ MPa} \times 3,985.45\text{ mm}^2 = 824,988\text{ N} = 825.0\text{ kN}$ $\phi_v V_n = 1.00 \times 825.0\text{ kN} = 825.0\text{ kN}$

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10. Licensure Exam Pitfalls & Review Notes