7.2 Statics and Stresses (Bending, Shear, Deflection)

Statics and Stresses in Structural Mechanics

Statics forms the bedrock of structural analysis. Before any member can be designed or any temporary support specified, the engineer must determine the internal forces and stresses acting within the system. The PE Construction exam heavily tests your ability to take a physical scenario, idealize it into a structural model, solve for the reactions using statics, and then calculate the resulting stresses and deflections. This section covers determinacy, shear and moment diagrams, and the core stress calculations for bending, shear, and axial loads.

Why This Topic Matters for the PE Construction Exam

In construction engineering, you frequently encounter beams, shores, walers, and struts. You must be able to quickly determine if a proposed timber beam is adequate to span a trench, or if a steel waler will yield under the pressure of a concrete pour. The exam requires fluency in calculating maximum moments to check bending stress, maximum shear to check shear stress, and maximum deflection to ensure the structure meets serviceability criteria. Mistakes in these fundamental statics and stress calculations cascade into catastrophic design errors.

Determinacy and Stability of Structures

Before analyzing a structure, you must establish if it is statically determinate, indeterminate, or unstable.

Equations of Equilibrium

For a planar structure, statics provides three equations of equilibrium:

1. $\sum F_x = 0$ (Sum of forces in the x-direction is zero)
2. $\sum F_y = 0$ (Sum of forces in the y-direction is zero)
3. $\sum M = 0$ (Sum of moments about any point is zero)

Evaluating Determinacy

A structure is statically determinate if all its support reactions and internal forces can be solved using only the equations of equilibrium. Let $r$ be the number of reaction components, $m$ be the number of members, and $j$ be the number of joints. For a simple frame or truss, a common check for determinacy is comparing $3m + r$ to $3j$.

• If $r = 3$, the structure is typically externally statically determinate.
• If $r > 3$, the structure is statically indeterminate. You need additional equations based on the compatibility of displacements to solve it.
• If $r < 3$, the structure is unstable.

Stability also requires that the reactions are not all parallel or concurrent (intersecting at a single point).

Shear and Moment Diagrams

Shear ($V$) and bending moment ($M$) diagrams are graphical representations of the internal forces along the length of a structural member. They are essential for identifying the locations and magnitudes of maximum shear and moment, which dictate the design.

The Mathematical Relationship

The relationships between load ($w$), shear ($V$), and moment ($M$) are fundamental:

• The change in shear between two points is equal to the area under the load diagram between those points: $\Delta V = \int w , dx$
• The change in moment between two points is equal to the area under the shear diagram between those points: $\Delta M = \int V , dx$

Worked Example: Simply Supported Beam with Uniform Load

Consider a simply supported beam of length $L$ with a uniform load $w$ (plf) across its entire length.

1. Reactions: By symmetry, the reactions at each support are $R = wL / 2$.
2. Shear Diagram: Starting at the left support, the shear jumps up to $wL/2$. It then decreases linearly at a rate of $w$ along the length, passing through zero at the midspan ($L/2$), and reaching $-wL/2$ at the right support.
3. Moment Diagram: The moment is zero at the pinned ends. The moment diagram is a parabola (integral of the linear shear). The maximum moment occurs where shear is zero (at midspan). The maximum moment is the area of the shear diagram triangle from the end to midspan: $M_{max} = \frac{1}{2} \times \frac{L}{2} \times \frac{wL}{2} = \frac{wL^2}{8}$.

Stresses in Structural Members

Once internal forces are known, you calculate stresses to compare against material allowables.

Bending Stress ($f_b$)

Bending moment causes tension on one side of the neutral axis and compression on the other. The flexure formula is:

$f_b = \frac{M y}{I}$

Where:

• $M$ is the bending moment.
• $y$ is the distance from the neutral axis to the fiber being evaluated (maximum stress occurs at the extreme fiber, $y = c$).
• $I$ is the moment of inertia of the cross-section.

Alternatively, $f_b = M / S$, where $S = I/c$ is the section modulus.

Shear Stress ($v$)

Transverse shear forces cause shear stresses within the cross-section. The general shear stress formula is:

$v = \frac{V Q}{I b}$

Where:

• $V$ is the shear force.
• $Q$ is the first moment of area of the portion of the cross-section above the point of interest, taken about the neutral axis.
• $I$ is the moment of inertia.
• $b$ is the width of the cross-section at the point of interest.

For a rectangular cross-section, the maximum shear stress occurs at the neutral axis and is $v_{max} = \frac{3V}{2A}$, where $A$ is the cross-sectional area.

Axial Stress ($f_a$)

For members subject to pure tension or compression (like truss members or columns), the axial stress is uniform across the cross-section:

$f_a = \frac{P}{A}$

Where $P$ is the axial force and $A$ is the cross-sectional area.

Combined Stress

Members often experience both axial forces and bending moments simultaneously (e.g., a column with an eccentric load). The stresses are combined using superposition:

$f = \frac{P}{A} \pm \frac{M y}{I}$

Deflection

Deflection is a serviceability requirement. Excessive deflection can cause structural damage (cracking partitions) or aesthetic concerns.

Deflection formulas depend on the load configuration and boundary conditions. For the simply supported beam with a uniform load $w$, the maximum deflection at midspan is:

$\Delta = \frac{5 w L^4}{384 E I}$

Where $E$ is the modulus of elasticity and $I$ is the moment of inertia. Codes dictate allowable deflection limits, often expressed as a fraction of the span, such as $L/360$ for live load deflection of floors.