PE Exam (Civil) Structural Analysis: Equilibrium, Reactions, Shear and Moment Diagrams
Last updated: May 2, 2026
Structural Analysis: Equilibrium, Reactions, Shear and Moment Diagrams questions are one of the highest-leverage areas to study for the PE Exam (Civil). This guide breaks down the rule, the elements you need to recognize, the named traps that catch most students, and a memory aid that scales to test day. Read it once, then practice the same sub-topic adaptively in the app.
The rule
For any planar statically determinate beam, the three equations of equilibrium $\sum F_x = 0$, $\sum F_y = 0$, and $\sum M = 0$ uniquely determine the reactions. Once reactions are known, shear $V(x)$ and moment $M(x)$ follow from $\frac{dV}{dx} = -w(x)$ and $\frac{dM}{dx} = V(x)$ (NCEES Reference Handbook, Statics §1.6 and Mechanics of Materials §2.2). Maximum moment occurs where shear crosses zero on a continuous span, and a concentrated load produces a jump (discontinuity) in $V$ equal to the load magnitude.
Elements breakdown
Determinacy Check
Confirm the beam is statically determinate before attempting equilibrium.
- Count reactions $r$
- Count internal hinges $c$
- Apply $r = 3 + c$ for determinate planar beams
- If $r > 3 + c$, structure is indeterminate
- Stop and use stiffness or moment-distribution methods
Solve Reactions
Apply the three planar equilibrium equations using a global free-body diagram.
- Draw FBD with all loads and reactions
- Replace distributed loads with equivalent resultants
- Sum moments about one support to isolate the other
- Verify with $\sum F_y = 0$
- Check $\sum F_x = 0$ for axial reactions
Common examples:
- For a simple span with UDL $w$ over length $L$: $R_A = R_B = \frac{wL}{2}$
Shear Diagram Construction
Build $V(x)$ left-to-right starting at zero outside the beam.
- Jump up at upward point load or reaction
- Jump down at downward point load
- Slope equals $-w(x)$ under distributed load
- Constant value where no load acts
- Close to zero at right end
Moment Diagram Construction
Build $M(x)$ as the running integral of $V(x)$.
- Slope of $M$ equals $V$ at each section
- Area under $V$-diagram equals change in $M$
- Zero moment at simple supports and free ends
- Discontinuity in $M$ at applied couple
- Maximum $|M|$ where $V = 0$ or at supports
Locate Critical Sections
Identify where peak shear and moment occur for design.
- Find x where $V(x) = 0$ on each segment
- Compute $M$ at every load point
- Compare $M$ at each candidate section
- Report governing design moment $M_u$ or $M_{max}$
Common patterns and traps
The Triangular Load Centroid Trap
When a distributed load is triangular rather than uniform, its resultant is $\frac{1}{2}w_{max}L$ but it acts at $L/3$ from the larger end, not at $L/2$. Candidates who place the resultant at midspan compute reactions that are off by ratios of 1:2 between supports.
A reaction value that splits the total load 50/50 instead of the correct 1/3 to 2/3 split.
The Couple-Doesn't-Affect-Shear Pattern
An applied moment (couple) at an interior point creates a step in the moment diagram equal to the couple magnitude but does not produce any jump in shear. Candidates often add the couple to shear or forget to include it in the global moment equation when solving reactions.
A shear diagram that erroneously jumps at the couple location, or reactions that ignore the couple's contribution to $\sum M$.
The Overhang Sign-Flip
On a beam with an overhang, the moment over the interior support is negative (hogging) while the moment between supports is positive. The maximum design moment may be the negative one, especially when the overhang carries heavy load.
A choice that reports only the positive midspan moment when the negative support moment is larger in magnitude.
The Unit-Soup Distractor
PE problems mix lb, kip, ft, in, and sometimes kN/m. A distractor is generated by forgetting one conversion (e.g., reporting $1{,}200{,}000 \text{ lb-in}$ when the answer wanted $\text{kip-ft}$). Always cancel units symbolically.
A choice that is a factor of 12 or 1000 off from the correct answer.
The Wrong Zero-Shear Location
For a beam carrying both a UDL and a point load, the shear-zero crossing is not at midspan. You must set $V(x) = 0$ algebraically and solve for $x$ before computing $M_{max}$.
A maximum moment computed at $L/2$ that is smaller than the true maximum at the actual zero-shear location.
How it works
Start with a free-body diagram and replace every distributed load with its resultant placed at the centroid of the load shape. Sum moments about one support to solve for the other reaction in a single equation, then back-substitute. For example, a $20 \text{ ft}$ simple span with a UDL of $w = 2 \text{ kip/ft}$ has total load $W = wL = 40 \text{ kip}$, so by symmetry $R_A = R_B = 20 \text{ kip}$. The shear diagram starts at $+20 \text{ kip}$ at the left reaction, drops linearly with slope $-w = -2 \text{ kip/ft}$, crosses zero at midspan, and reaches $-20 \text{ kip}$ at the right reaction. The maximum moment is the area of the positive triangle: $M_{max} = \frac{1}{2}(10 \text{ ft})(20 \text{ kip}) = 100 \text{ kip-ft}$, which matches the closed form $\frac{wL^2}{8} = \frac{(2)(20)^2}{8} = 100 \text{ kip-ft}$. Always carry units through the calculation; if $\text{kip-ft}$ does not pop out, you have a unit error.
Worked examples
The Reyes Pedestrian Bridge has a simply supported steel girder spanning $L = 24 \text{ ft}$ between abutments A (left) and B (right). The girder carries a uniform dead-plus-live service load of $w = 1.8 \text{ kip/ft}$ over its entire length, plus a single concentrated service load $P = 12 \text{ kip}$ applied $8 \text{ ft}$ from the left abutment. Self-weight is included in $w$. Sketch: a horizontal girder with pin at A and roller at B, UDL across the full span, and a downward arrow $P$ at $x = 8 \text{ ft}$.
Most nearly, what is the reaction at support B?
- A $17.6 \text{ kip}$
- B $25.6 \text{ kip}$ ✓ Correct
- C $29.6 \text{ kip}$
- D $33.6 \text{ kip}$
Why B is correct: Sum moments about A to isolate $R_B$: $\sum M_A = 0$ gives $R_B (24) - (1.8)(24)\left(\frac{24}{2}\right) - (12)(8) = 0$. Then $R_B (24) = (1.8)(24)(12) + (12)(8) = 518.4 + 96 = 614.4 \text{ kip-ft}$, so $R_B = \frac{614.4}{24} = 25.6 \text{ kip}$. Units cancel as $\frac{\text{kip-ft}}{\text{ft}} = \text{kip}$. Check: $R_A = (1.8)(24) + 12 - 25.6 = 43.2 + 12 - 25.6 = 29.6 \text{ kip}$, and $\sum F_y = 0$ closes.
Why each wrong choice fails:
- A: This results from placing the UDL resultant at $x = 8 \text{ ft}$ (under the point load) instead of at midspan: $R_B = \frac{(43.2)(8) + (12)(8)}{24} = 18.4 \text{ kip}$, then rounded down. Treats the UDL as if it were concentrated at the same location as $P$. (The Triangular Load Centroid Trap)
- C: This is the value of $R_A$, not $R_B$. The candidate solved correctly but reported the wrong support reaction, likely confusing the moment-arm direction in $\sum M_A$.
- D: This is total load $W = wL + P = 43.2 + 12 = 55.2 \text{ kip}$ multiplied by the wrong tributary fraction (using $\frac{P \cdot 16}{24}$ instead of $\frac{P \cdot 8}{24}$ for the point load contribution), giving an inflated reaction. (The Wrong Zero-Shear Location)
A simply supported timber beam in the Liu Civic Center mezzanine spans $L = 18 \text{ ft}$ and carries a triangular distributed load that varies linearly from $w = 0$ at the left support A to $w_{max} = 3.0 \text{ kip/ft}$ at the right support B. There are no other loads. Sketch: a horizontal beam pinned at A and rolled at B with a right-triangular load profile increasing linearly from A to B.
Most nearly, what is the maximum bending moment in the beam?
- A $70.1 \text{ kip-ft}$ ✓ Correct
- B $93.5 \text{ kip-ft}$
- C $121.5 \text{ kip-ft}$
- D $162.0 \text{ kip-ft}$
Why A is correct: Total load $W = \frac{1}{2} w_{max} L = \frac{1}{2}(3.0)(18) = 27 \text{ kip}$, acting at the centroid $\frac{2L}{3} = 12 \text{ ft}$ from A. Reactions: $\sum M_A = 0 \Rightarrow R_B = \frac{(27)(12)}{18} = 18 \text{ kip}$, $R_A = 27 - 18 = 9 \text{ kip}$. Set $V(x) = R_A - \frac{1}{2}\left(\frac{w_{max} x}{L}\right) x = 9 - \frac{x^2}{12} = 0$, giving $x = \sqrt{108} = 10.39 \text{ ft}$. Then $M_{max} = R_A x - \frac{w_{max} x^3}{6L} = (9)(10.39) - \frac{(3.0)(10.39)^3}{(6)(18)} = 93.55 - 31.18 \approx 70.1 \text{ kip-ft}$.
Why each wrong choice fails:
- B: This is $R_A \cdot x$ at the zero-shear point without subtracting the moment from the triangular load to the left of that section: $(9)(10.39) = 93.5 \text{ kip-ft}$. Forgets that the partial triangular load between A and $x$ also contributes a negative moment. (The Triangular Load Centroid Trap)
- C: This is $\frac{w_{max} L^2}{8} \cdot \frac{1}{1} = \frac{(3)(18)^2}{8} = 121.5 \text{ kip-ft}$, the formula for a UNIFORM load on a simple span. The candidate ignored that the load is triangular, not uniform. (The Wrong Zero-Shear Location)
- D: This computes $M$ at midspan using $R_A \cdot \frac{L}{2}$ minus a wrong load-area term, or alternatively uses $R_B \cdot L/2 = (18)(9) = 162 \text{ kip-ft}$ as the maximum, ignoring the load distribution. Either way the section chosen is wrong. (The Wrong Zero-Shear Location)
The Okafor Warehouse roof girder is a steel beam $30 \text{ ft}$ long with a pin support at A (left) and a roller at B located $20 \text{ ft}$ from A, leaving a $10 \text{ ft}$ overhang to the right of B that ends at a free tip C. A uniform service load of $w = 2.0 \text{ kip/ft}$ acts over the full $30 \text{ ft}$ length. Sketch: pin at A, UDL across, roller at B at $x = 20 \text{ ft}$, and free end C at $x = 30 \text{ ft}$.
Most nearly, what is the magnitude of the maximum bending moment along the entire girder?
- A $80 \text{ kip-ft}$
- B $100 \text{ kip-ft}$ ✓ Correct
- C $112.5 \text{ kip-ft}$
- D $225 \text{ kip-ft}$
Why B is correct: Reactions: $\sum M_A = 0 \Rightarrow R_B (20) - (2)(30)(15) = 0$, so $R_B = 45 \text{ kip}$ and $R_A = 60 - 45 = 15 \text{ kip}$. Negative moment over support B equals the moment from the overhang: $M_B = -\frac{w \cdot a^2}{2} = -\frac{(2)(10)^2}{2} = -100 \text{ kip-ft}$. Positive moment in the back-span occurs where $V = 0$: $V(x) = 15 - 2x = 0 \Rightarrow x = 7.5 \text{ ft}$, giving $M = (15)(7.5) - \frac{(2)(7.5)^2}{2} = 112.5 - 56.25 = 56.25 \text{ kip-ft}$. The maximum magnitude is $|M_B| = 100 \text{ kip-ft}$.
Why each wrong choice fails:
- A: This uses overhang length squared with a factor of $\frac{w a^2}{2.5}$ or treats the overhang as carrying $w = 1.6$ instead of $2.0 \text{ kip/ft}$. Often results from misreading the load on the overhang versus the back-span. (The Unit-Soup Distractor)
- C: This is the positive moment in the back-span computed without accounting for the overhang's effect on reactions: $\frac{wL^2}{8} = \frac{(2)(20)^2}{8} = 100$, then mis-added as $112.5$. Misses that the overhang shifts $R_A$ downward, reducing positive moment, while creating a larger negative one. (The Overhang Sign-Flip)
- D: This applies $\frac{wL^2}{8}$ to the full $30 \text{ ft}$ length: $\frac{(2)(30)^2}{8} = 225 \text{ kip-ft}$, treating the beam as a simple span over $30 \text{ ft}$ and ignoring that the supports are at A and B with an overhang. (The Overhang Sign-Flip)
Memory aid
FRAME: Free-body, Reactions, Areas (under $V$), Maximum at $V=0$, Equilibrium check.
Key distinction
Concentrated forces cause jumps in the shear diagram and slope changes (kinks) in the moment diagram; concentrated couples cause jumps in the moment diagram but no change in shear.
Summary
Solve reactions with planar equilibrium, build shear left-to-right with jumps at point loads and slopes equal to $-w(x)$, then integrate to get moment with peak where $V$ crosses zero.
Practice structural analysis: equilibrium, reactions, shear and moment diagrams adaptively
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Start your free 7-day trialFrequently asked questions
What is structural analysis: equilibrium, reactions, shear and moment diagrams on the PE Exam (Civil)?
For any planar statically determinate beam, the three equations of equilibrium $\sum F_x = 0$, $\sum F_y = 0$, and $\sum M = 0$ uniquely determine the reactions. Once reactions are known, shear $V(x)$ and moment $M(x)$ follow from $\frac{dV}{dx} = -w(x)$ and $\frac{dM}{dx} = V(x)$ (NCEES Reference Handbook, Statics §1.6 and Mechanics of Materials §2.2). Maximum moment occurs where shear crosses zero on a continuous span, and a concentrated load produces a jump (discontinuity) in $V$ equal to the load magnitude.
How do I practice structural analysis: equilibrium, reactions, shear and moment diagrams questions?
The fastest way to improve on structural analysis: equilibrium, reactions, shear and moment diagrams is targeted, adaptive practice — working questions that focus on your specific weak spots within this sub-topic, getting immediate feedback, and revisiting items you missed on a spaced-repetition schedule. Neureto's adaptive engine does this automatically across the PE Exam (Civil); start a free 7-day trial to see your sub-topic mastery climb in real time.
What's the most important distinction to remember for structural analysis: equilibrium, reactions, shear and moment diagrams?
Concentrated forces cause jumps in the shear diagram and slope changes (kinks) in the moment diagram; concentrated couples cause jumps in the moment diagram but no change in shear.
Is there a memory aid for structural analysis: equilibrium, reactions, shear and moment diagrams questions?
FRAME: Free-body, Reactions, Areas (under $V$), Maximum at $V=0$, Equilibrium check.
What's a common trap on structural analysis: equilibrium, reactions, shear and moment diagrams questions?
Forgetting to convert UDL units (lb/ft vs. kip/ft)
What's a common trap on structural analysis: equilibrium, reactions, shear and moment diagrams questions?
Placing the resultant of a triangular load at midspan instead of at $L/3$
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