7.3 Beams, Columns, and Slabs

Beams, Columns, and Slabs

While statics tells us the forces within a structure, member design involves sizing the components (beams, columns, slabs) to safely resist those forces. The PE Construction exam tests your understanding of the behavioral differences between materials like concrete and steel, and how different structural members fail. Beams fail in flexure or shear, columns fail by buckling or crushing, and slabs must distribute loads appropriately based on their support conditions. This section details the fundamental mechanics governing these structural elements.

Why This Topic Matters for the PE Construction Exam

Construction engineers are constantly evaluating the adequacy of existing structures to handle construction loads, or designing temporary shoring and scaffolding systems. You must know how to check if a steel W-section acting as a beam has adequate flexural capacity, or if a timber post used for shoring will buckle under the load of a wet concrete pour above. Understanding the mechanics of one-way versus two-way slabs is crucial when analyzing existing structures for demolition or modification, or when designing formwork.

Flexural Design Basics: Beams

Beams are members primarily subjected to bending moments and shear forces.

Concrete Flexural Design

Reinforced concrete is a composite material. Concrete is strong in compression but weak in tension. Steel rebar is placed in the tension zone to resist the tensile forces caused by bending.

To simplify the calculation of the compressive force in the concrete, engineers use the Whitney Stress Block. This model assumes a uniform compressive stress of $0.85 f'_c$ acting over a depth $a$, which is a fraction of the distance to the neutral axis ($c$).

The nominal moment capacity ($M_n$) of a singly reinforced rectangular concrete beam is based on ensuring the steel yields before the concrete crushes (a ductile failure mode):

$M_n = A_s f_y \left( d - \frac{a}{2} \right)$

Where:

• $A_s$ is the area of tension steel.
• $f_y$ is the yield strength of the steel.
• $d$ is the effective depth from the extreme compression fiber to the centroid of the tension steel.
• $a$ is the depth of the equivalent rectangular stress block.

Steel Flexural Design

Steel beams are isotropic and strong in both tension and compression. Design focuses on the yield moment and the plastic moment. When the extreme fibers reach the yield stress ($f_y$), the yield moment is reached: $M_y = F_y S$. If the beam is adequately braced against lateral-torsional buckling, the entire cross-section can yield, reaching the plastic moment capacity: $M_p = F_y Z$, where $Z$ is the plastic section modulus.

Shear Design

Shear failure in beams is often sudden and brittle, making it a critical design consideration.

Concrete Shear

In concrete, high shear stresses combine with bending stresses to create diagonal tension cracks. When the shear force ($V_u$) exceeds the concrete's shear capacity ($\phi V_c$), steel reinforcement (stirrups) must be added. The shear capacity provided by stirrups is $V_s$. The total nominal shear capacity is $V_n = V_c + V_s$. Stirrups act like vertical tension members in a truss analog of the beam, holding the cracks closed.

Steel Shear

In steel I-beams, the web carries the vast majority of the shear force. Shear failure occurs through web yielding or web crippling. The shear capacity is related to the web area ($d \times t_w$) and the steel yield strength.

Column Buckling

Columns are compression members. While short columns fail by material crushing, slender columns fail by buckling—a sudden lateral deflection that occurs at a critical load, often well below the material's yield strength.

Euler's Buckling Formula

The critical elastic buckling load ($P_{cr}$) for an ideal column is given by Euler's formula:

$P_{cr} = \frac{\pi^2 E I}{(K L)^2}$

Where:

• $E$ is the modulus of elasticity.
• $I$ is the minimum moment of inertia of the cross-section.
• $L$ is the unbraced length of the column.
• $K$ is the effective length factor.

Effective Length Factor (K)

The effective length ($KL$) represents the distance between inflection points (points of zero moment) on the buckled shape. It depends entirely on the end support conditions:

• Pinned-Pinned: $K = 1.0$
• Fixed-Free (Flagpole): $K = 2.1$
• Fixed-Pinned: $K = 0.7$
• Fixed-Fixed: $K = 0.5$ (theoretical), $0.65$ (recommended for design)

Worked Calculation: Column Buckling Load

Calculate the critical buckling load for a pinned-pinned ($K=1.0$) timber column. The column is a 4x4 (actual dimensions 3.5" x 3.5"), 10 feet long, with a modulus of elasticity $E = 1,600,000$ psi.

1. Calculate $I$: For a square cross section, $I = \frac{b h^3}{12} = \frac{3.5 \times 3.5^3}{12} = 12.5$ in$^4$.
2. Calculate Length in inches: $L = 10 \text{ ft} \times 12 \text{ in/ft} = 120$ inches.
3. Apply Euler's Formula: $P_{cr} = \frac{\pi^2 \times 1,600,000 \times 12.5}{(1.0 \times 120)^2}$ $P_{cr} = \frac{197,392,088}{14,400} = 13,707$ pounds.

Slab Design

Slabs are planar elements that carry loads primarily by flexure to supporting beams, walls, or columns.

One-Way Slabs

A one-way slab is supported on two opposite sides, or is supported on all four sides but has a length-to-width ratio greater than 2. It bends primarily in one direction, acting effectively as a wide, shallow beam spanning between the supports. Main reinforcement is placed parallel to the short direction, while shrinkage and temperature steel is placed in the long direction.

Two-Way Slabs

A two-way slab is supported on all four sides with a length-to-width ratio less than or equal to 2. It bends significantly in both directions, distributing loads to all four supporting edges. It requires main structural reinforcement in both orthogonal directions. Two-way slabs are more complex to analyze and can include flat plates (supported directly by columns) and flat slabs (with drop panels or column capitals).