CELE Structural Theory & Analysis — Analysis of Determinate StructuresDetailed Explanation
Detailed explanation of Analysis of Determinate Structures for the CELE 2026. Full depth, full reasoning — exactly what you need when Professional Regulation Commission (PRC) — Board of Civil Engineering tests this chapter with applied or scenario-based questions in the CELE Structural Theory & Analysis subtest.
Exam context
Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Structural Theory & Analysis section sits under a "Core" weighting, and Analysis of Determinate Structures is the 1st chapter in the 6-chapter CELE Structural Theory & Analysis rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Structural Theory & Analysis.
Analysis of Determinate Structures - Detailed Explanation
Structural analysis is the backbone of civil engineering design. Before any beam, frame, or arch can be sized or detailed, the engineer must find the support reactions and internal forces (axial force N, shear force V, and bending moment M) that the applied loads produce. The fundamental question asked at the very start of every analysis is: can these unknowns be found using the three equations of static equilibrium alone? If yes, the structure is statically determinate; if not, it is indeterminate and compatibility conditions must supplement equilibrium. This chapter covers the four pillars of determinate-structure analysis that appear repeatedly in the PRC Civil Engineer Licensure Examination: (1) classifying structures by their degree of indeterminacy, (2) computing support reactions using equilibrium, (3) drawing internal-force diagrams (AFD, SFD, BMD), and (4) analysing three-hinged arches. Mastery of these fundamentals is essential not only for the board exam but also for professional practice under RA 544 (Civil Engineering Law of the Philippines), which mandates that registered civil engineers possess competence in structural analysis and design.
Concepts
Determinacy and Stability of Structures
A structure is statically determinate when the number of independent equilibrium equations exactly equals the number of unknown reaction components. For a planar (2-D) structure, three global equilibrium equations are available: ΣFx = 0, ΣFy = 0, and ΣM = 0. If the structure contains internal hinges or other releases, each release provides one additional equation of condition (ec), because the bending moment (or other released force) at that point is known to be zero. The Degree of Indeterminacy (DI) quantifies how many extra unknowns exist beyond what equilibrium can solve: For BEAMS (shortcut formula): DI = r − (3 + ec) where r = total reaction components, ec = number of equations of condition. For FRAMES (general formula): DI = (3m + r) − (3n + ec) where m = number of members, r = total reaction components, n = number of joints (nodes including support joints), ec = equations of condition. Interpretation: DI = 0 → Statically Determinate (solvable by equilibrium alone) DI > 0 → Statically Indeterminate (DI = degree of indeterminacy) DI < 0 → Geometrically Unstable (mechanism, will collapse) Each internal hinge connecting exactly TWO members contributes ec = 1 (because M = 0 at that hinge, giving one extra equation). If k members meet at a single internal hinge, it contributes ec = k − 1 equations. STABILITY is separate from determinacy. A structure can have DI = 0 yet still be unstable if the reaction lines of action are: (a) Concurrent — all pass through one point (cannot resist rotation about that point), or (b) Parallel — all point in the same direction (cannot resist a force perpendicular to them). Always verify the geometric arrangement of supports, not just the equation count.
Examples
DI = −1 indicates the structure is geometrically unstable. A simply supported beam with only 3 reaction components cannot carry a general loading if one of those 'equations' is consumed by an internal hinge — there is one fewer independent equilibrium statement than unknowns. In practice, a Gerber beam needs at least r = 4 (e.g., pin + pin + roller + roller) with ec = 1 to be determinate: DI = 4 − (3+1) = 0.
Scenario
Classify a simply supported beam with a fixed-pin at A (r=2) and a roller at B (r=1), and one internal hinge between the two spans (ec=1). Total reactions r=3.
Solution
DI = r − (3 + ec) = 3 − (3 + 1) = 3 − 4 = −1.
A portal frame with pinned bases is singly indeterminate. The horizontal thrust H cannot be found from equilibrium alone — compatibility of deformation is needed. If the bases were fixed (r=3 each, total r=6): DI = (9+6) − 12 = 3, i.e., triply indeterminate. These are classic board-exam structures.
Scenario
A portal frame: two columns pinned at the base (r=2 each), one beam member connecting them. m=3 members (left column, beam, right column), n=4 joints (2 bases + 2 knee joints), r=4 (2 pins × 2 components each), ec=0.
Solution
DI = (3m + r) − (3n + ec) = (9 + 4) − (12 + 0) = 13 − 12 = 1.
The three-hinged arch is statically determinate. The crown hinge provides the fourth equation (ΣMC = 0 on one side), which allows the horizontal thrust H to be computed from equilibrium alone. This is the key advantage of the three-hinged arch over the two-hinged arch (DI=1) or the fixed arch (DI=3).
Scenario
Classify a three-hinged arch: supports A (pin, r=2) and B (pin, r=2) plus one crown hinge (ec=1). Treat as a beam-formula structure: r=4, ec=1.
Solution
DI = r − (3 + ec) = 4 − (3 + 1) = 0.
Applications
- Classifying bridges: simply supported spans (determinate) vs. continuous beams (indeterminate) — the choice affects analysis method and structural behaviour under support settlement.
- Gerber (compound) beams used in older Philippine bridge designs provide determinacy even over multiple spans.
- Three-hinged arches are used in roof structures (e.g., large gymnasium or arena roofs) because they are determinate and insensitive to support settlement.
- Portal frames with fixed bases (DI=3) are the standard for industrial buildings and require matrix stiffness or moment distribution methods — recognising DI=3 tells the engineer which method to use.
Misconceptions
- MISCONCEPTION: DI = 0 always means the structure is stable. TRUTH: DI = 0 is necessary but not sufficient for stability; the geometric arrangement of reactions must also be checked.
- MISCONCEPTION: Every internal hinge always gives ec = 1. TRUTH: A hinge where k members meet gives ec = k − 1. Three members meeting at one hinge give ec = 2.
- MISCONCEPTION: The frame formula and beam formula always give the same answer. TRUTH: For simple beams, the shortcut formula r − (3 + ec) is equivalent, but for complex frames with multiple closed loops, only the general formula (3m + r) − (3n + ec) is reliable.
- MISCONCEPTION: A structure with more reactions than equations is always stronger. TRUTH: Indeterminate structures redistribute loads but are sensitive to support settlement; determinate structures are not.
Related Concepts
- Support conditions and reaction components
- Equations of condition from internal hinges
- Compound (Gerber) beams
- Three-hinged arches
- Static equilibrium (ΣFx, ΣFy, ΣM)
Common Exam Questions
Example
A beam has a fixed support at one end (r=3), a pin support at an interior point (r=2), and a roller at the free end (r=1), with one internal hinge. DI = (3+2+1) − (3+1) = 6 − 4 = 2 → indeterminate to the 2nd degree.
Approach
Apply DI formula directly. Identify r, m, n, ec from the structure description. If a figure is given, count members between nodes, support conditions, and internal hinges.
Question Type
Multiple Choice — Classification
Example
Three vertical roller supports on a horizontal beam: r = 3, ec = 0, DI = 3 − 3 = 0 (seems determinate), but there is NO horizontal reaction component — unstable under any horizontal force.
Approach
After computing DI = 0, check if all reaction forces are parallel or concurrent. If a beam rests on three rollers all pointing vertically, it cannot resist horizontal loads — unstable even if DI = 0.
Question Type
Problem — Identify Stability
Key Points To Remember
- Beam formula: DI = r − (3 + ec); Frame formula: DI = (3m + r) − (3n + ec).
- Each internal hinge between 2 members gives ec = 1; k members at one hinge give ec = k − 1.
- DI = 0 → determinate; DI > 0 → indeterminate; DI < 0 → unstable (mechanism).
- A structure with DI = 0 can still be unstable if reactions are concurrent or parallel.
- A fixed support contributes r = 3 (Fx, Fy, M); a pin contributes r = 2 (Fx, Fy); a roller contributes r = 1.
- Internal hinges in Gerber (compound) beams are a classic board-exam topic.
- For frames, count ALL joints including support joints as part of n.
Support Reactions of Determinate Structures
Once a structure is confirmed to be statically determinate (DI = 0), its support reactions are found by applying the three global equilibrium equations plus any equations of condition: ΣFx = 0 (sum of all horizontal forces = 0) ΣFy = 0 (sum of all vertical forces = 0) ΣM_point = 0 (sum of moments about any convenient point = 0) STRATEGY FOR BEAMS: 1. Draw the complete Free Body Diagram (FBD) showing all applied loads and unknown reactions. 2. Choose a moment-sum point that eliminates the most unknowns (e.g., the point where two unknown reactions intersect). 3. Solve ΣM = 0 first to find one reaction, then use ΣFy and ΣFx for the rest. 4. Always verify with a fourth equation (e.g., moment about a different point) to check arithmetic. COMPOUND (GERBER) BEAMS — SUSPENDED-SPAN-FIRST METHOD: A Gerber beam is a multi-span beam that is made statically determinate by inserting internal hinges. The key rule: analyse the SUSPENDED (hanging) span first because it is simply supported at the internal hinge — it has no unknowns from the main span. Its reactions then become applied loads (in reverse) on the supporting span. STANDARD LOAD TYPES: - Concentrated load P: single force at a point. - Uniformly distributed load (UDL) w (kN/m): replaced by resultant W = wL at the midpoint of the loaded length. - Uniformly varying load (UVL / triangular): resultant W = wL/2 at the one-third point from the larger end. - Couple/moment M₀: creates equal-and-opposite reactions V = M₀/L (upward at one support, downward at the other) for a simply supported beam.
Examples
By symmetry, both reactions are equal at 60 kN each. The check equation confirms the result. This is the most basic board-exam reaction problem; the examiner often adds an inclined load or an internal hinge to increase difficulty.
Scenario
A simply supported beam AB, span L = 8 m. A UDL of w = 15 kN/m acts over the entire span. Find reactions at A (pin) and B (roller).
Solution
Total load W = 15 × 8 = 120 kN, acting at midspan (4 m from A). ΣM_A = 0: V_B(8) − 120(4) = 0 → V_B = 60 kN ↑ ΣF_y = 0: V_A + 60 − 120 = 0 → V_A = 60 kN ↑ ΣF_x = 0: H_A = 0 (no horizontal load) Check: ΣM_B = V_A(8) − 120(4) = 480 − 480 = 0 ✓
The suspended-span-first method is the correct procedure for Gerber beams. The reaction at the internal hinge B from the BC span (25 kN upward on BC) becomes a 25 kN downward load applied at B on the AD span. Notice that support C is NOT a fixed support — it is the free end of the overhang (or the end of the suspended span). Board exams frequently set up Gerber beams to test whether examinees know to analyse the free-hanging portion first.
Scenario
A Gerber beam: Span AC = 10 m with a pin at A and a roller at D located 6 m from A. An internal hinge B is at 4 m from A. The suspended span BC has length = 6 m (from B at x=4 to C at x=10). A concentrated load of P = 50 kN acts at midspan of BC (i.e., 3 m from B, which is 7 m from A). Find all reactions.
Solution
STEP 1 — Analyse the suspended span BC (simply supported at B and C, length = 6 m, load P = 50 kN at 3 m from B): ΣM_B = 0: R_C(6) − 50(3) = 0 → R_C = 25 kN ↑ ΣF_y = 0: R_B_from_BC = 50 − 25 = 25 kN ↑ (this is the upward reaction at B from the BC span; it acts as a 25 kN downward load on the AD span at point B). STEP 2 — Analyse the main span AD (pin at A, roller at D at 6 m from A; load = 25 kN downward at B, 4 m from A): ΣM_A = 0: V_D(6) − 25(4) = 0 → V_D = 16.67 kN ↑ ΣF_y = 0: V_A = 25 − 16.67 = 8.33 kN ↑ ΣF_x = 0: H_A = 0
Applications
- Reaction calculation is the mandatory first step before any design check under NSCP 2015 Section 204 (load combinations) and structural analysis.
- Bridge girder reactions determine the bearing loads used to design bridge bearings and abutments — critical in Philippine bridge design practice.
- In multi-storey frame analysis, column axial forces are essentially the accumulated vertical reactions from beams above — correct reaction calculation is essential.
- Compound (Gerber) beams were used historically in Philippine timber bridges to ensure determinacy and avoid overstress from uneven pier settlement.
Misconceptions
- MISCONCEPTION: For a Gerber beam, you can take the entire beam as one FBD and sum moments. TRUTH: This is correct for finding the total reactions, but you still need the hinge condition (M=0 at hinge) as an extra equation — and the suspended-span-first method is the most systematic approach.
- MISCONCEPTION: Roller supports always provide only a vertical reaction. TRUTH: A roller is constrained to move along a specific surface. If the surface is inclined, the roller reaction is perpendicular to that surface (i.e., inclined), not necessarily vertical.
- MISCONCEPTION: The location of the UDL resultant is always at the centroid of the beam. TRUTH: The resultant of a UDL acts at the centroid of the load diagram, which is the midpoint of the LOADED LENGTH, not the whole beam.
Related Concepts
- Static equilibrium equations
- Free body diagrams
- Internal hinges and equations of condition
- Three-hinged arches (special case of reaction calculation)
- Load types: concentrated, UDL, UVL, applied moment
Common Exam Questions
Example
A 12-m simply supported beam carries a 30-kN/m UDL over the left 4 m and a 60-kN concentrated load at 9 m from A. Find V_A and V_B.
Approach
1. Resolve all inclined loads. 2. Replace distributed loads with resultants. 3. Take moment about a support to find the far reaction. 4. Use ΣFy for the near reaction. 5. Verify.
Question Type
Problem — Find reactions with mixed loads
Example
A three-span Gerber beam with two internal hinges — identify which portion is suspended, solve it, then proceed leftward or rightward on the main span.
Approach
Identify internal hinges to locate the suspended span. Solve the suspended span first (simple beam). Transfer the hinge reaction as a load on the main span. Solve the main span.
Question Type
Problem — Gerber beam reactions
Key Points To Remember
- Take moments about a pin/hinge joint to eliminate two unknowns at once.
- UDL resultant acts at the midpoint; UVL (triangular) resultant acts at L/3 from the larger end.
- For Gerber beams: always analyse the suspended span first, then treat its reactions as loads on the supporting span.
- At an internal hinge, the bending moment M = 0 on each side — use ΣM = 0 about the hinge for one side of the structure as the extra equation.
- Always verify reactions with a check equation (ΣM about a different point should equal zero).
- Sign convention: upward forces and counterclockwise moments are typically taken as positive.
- Inclined loads must be resolved into horizontal (Fx) and vertical (Fy) components before applying equilibrium.
Internal Forces: Shear and Moment Diagrams
After computing support reactions, the next step is to find the internal forces at every cross-section. The three internal force quantities in a planar structure are: N = Axial force (tension positive, compression negative) V = Shear force M = Bending moment METHOD OF SECTIONS: 1. Make an imaginary cut at the section of interest. 2. Choose either the LEFT or RIGHT free body — both must give the same result. 3. Apply equilibrium to the chosen free body to find N, V, and M at the cut. SIGN CONVENTION (standard structural engineering): Positive shear V: left face forces upward, right face forces downward (or the section tends to rotate clockwise). Positive moment M: sagging (beam bends concave upward, tension on bottom fiber). Positive axial N: tension (member being stretched). RELATIONSHIPS BETWEEN LOAD, SHEAR, AND MOMENT: These differential relationships allow rapid construction of diagrams without computing V and M at every point: dV/dx = −w(x) (slope of shear diagram = negative of distributed load intensity) dM/dx = V(x) (slope of moment diagram = shear) ΔV = −(area of load diagram between two sections) ΔM = area of shear diagram between two sections Consequences: - Under a UDL, shear is linear and moment is parabolic. - Under no load, shear is constant and moment is linear. - At a concentrated load P, shear jumps by P. - At an applied couple M₀, moment jumps by M₀. - Maximum moment occurs where V = 0 (or changes sign). FRAME INTERNAL FORCES: For frames, each member is treated as a beam aligned along its own axis. Draw AFD, SFD, and BMD for each member separately using the same method-of-sections approach. Maintain consistent sign convention across the entire frame.
Examples
The maximum moment of 120 kN·m occurs at the load point (where V crosses zero). The BMD is a triangle peaking at 120 kN·m. On a board exam, you may be asked for: (1) the maximum moment, (2) the location of maximum moment, or (3) the moment at a specified section — all read directly from this diagram.
Scenario
A simply supported beam AB, span = 6 m. Concentrated load P = 90 kN at 2 m from A. Construct the SFD and BMD.
Solution
Reactions: ΣM_A = 0 → V_B(6) = 90(2) → V_B = 30 kN ↑; V_A = 90 − 30 = 60 kN ↑. SHEAR DIAGRAM (SFD): Just right of A: V = +60 kN Just left of load (2 m from A): V = +60 kN (no load in between) Just right of load: V = 60 − 90 = −30 kN At B: V = −30 + 30 = 0 ✓ BENDING MOMENT DIAGRAM (BMD): At A: M = 0 (pin support) At load point (2 m): M = V_A × 2 = 60 × 2 = +120 kN·m (maximum, since V changes sign here) At B: M = 0 (roller support) Between A and load: M increases linearly from 0 to 120 kN·m Between load and B: M decreases linearly from 120 to 0 kN·m
For cantilevers, maximum shear and maximum moment both occur at the fixed support. The parabolic BMD (because w is uniform and moment is proportional to x²) is characteristic of UDL loading. Sign convention here uses negative for hogging (tension on top). Board exams may ask for the moment at an intermediate section — use M = −10x² with the appropriate x value.
Scenario
A cantilever beam AB, length = 4 m, fixed at A, free end B. UDL w = 20 kN/m over entire length. Find the fixed-end reactions and draw SFD and BMD.
Solution
Reactions at A: V_A = 20 × 4 = 80 kN ↑; M_A = 20 × 4 × 2 = 160 kN·m (clockwise, resisting the load). SHEAR DIAGRAM (measuring x from free end B): At B (x=0): V = 0 (free end) At x: V = 20x (shear increases linearly from B toward A) At A (x=4 m): V = 80 kN ✓ (matches reaction) BENDING MOMENT DIAGRAM: At B: M = 0 (free end) At x: M = −20x(x/2) = −10x² (hogging, parabolic) At A (x=4): M = −10(16) = −160 kN·m (maximum hogging at fixed end) ✓
Applications
- SFD and BMD are the basis for all beam design checks: maximum moment determines the required section modulus (flexure); maximum shear determines the required shear capacity.
- Under NSCP 2015 Section 406 (ACI 318-14 equivalent), the design moment Mu and design shear Vu come directly from factored load SFD and BMD.
- Point of contraflexure (where M changes sign, i.e., M = 0) is used in detailing: longitudinal reinforcement can be terminated at a code-specified distance past the contraflexure point.
- Deflection calculations (using moment-area or conjugate-beam methods) use the M/EI diagram, which is derived directly from the BMD.
Misconceptions
- MISCONCEPTION: The maximum bending moment always occurs at the midspan. TRUTH: Midspan moment is maximum only for symmetrically loaded simply supported beams. For unsymmetric loading, the maximum is where V = 0, which may not be at midspan.
- MISCONCEPTION: At a pin support, all internal forces are zero. TRUTH: Only the bending moment M = 0 at a pin support. Shear V and axial force N can be non-zero (they equal the reaction components).
- MISCONCEPTION: A concentrated load causes a jump in the bending moment diagram. TRUTH: A concentrated LOAD causes a kink (slope change) in the BMD, not a jump. A concentrated MOMENT/COUPLE causes a jump in the BMD.
- MISCONCEPTION: Positive bending moment means the beam is safe. TRUTH: Both positive (sagging) and negative (hogging) moments cause stress; the beam must be designed for the maximum absolute value, and reinforcement must be placed on the tension side (bottom for sagging, top for hogging).
Related Concepts
- Support reactions (prerequisite)
- Differential relationships: dV/dx = -w, dM/dx = V
- Beam deflection and elastic curve
- Flexure formula: f = Mc/I (NSCP 2015)
- Shear stress formula: fv = VQ/(Ib)
Common Exam Questions
Example
A 10-m simply supported beam with a 50-kN/m UDL over the left 6 m. Find the maximum positive bending moment and its location.
Approach
Find reactions, draw SFD, locate where V = 0 (or changes sign), then compute M at that location using the area under the SFD up to that point.
Question Type
Problem — Maximum bending moment
Example
Find the bending moment at 5 m from the left support of a simply supported beam with mixed UDL and concentrated loads.
Approach
Cut at the section, isolate one side, apply equilibrium. ΣM about the cut = M at the cut.
Question Type
Problem — Moment at a specific section
Example
Which statement about the BMD of a cantilever with a tip load is correct? (a) Linear (b) Parabolic (c) Cubic — Answer: (a) Linear, because the only load is a concentrated force at the tip.
Approach
Use the load-shear-moment relationships: no load → constant shear, linear moment; UDL → linear shear, parabolic moment; concentrated load → constant shear between loads, kink in BMD.
Question Type
Conceptual — Shape of diagrams
Key Points To Remember
- dV/dx = −w and dM/dx = V are the governing differential relationships — memorise these.
- Maximum (or minimum) bending moment occurs where the shear force is zero.
- A concentrated load causes a sudden jump in the shear diagram equal to the load magnitude.
- An applied moment/couple causes a sudden jump in the moment diagram equal to the moment magnitude.
- Under UDL: shear is linear (1st degree) and moment is parabolic (2nd degree).
- At a fixed support, both shear and moment are generally non-zero; at a pin/roller, moment = 0.
- At a free end with no load, both V = 0 and M = 0.
- Use the area method (ΔM = ∫V dx) to quickly compute moment values without integration.
Three-Hinged Arches
An arch carries loads primarily through axial compression, making it structurally efficient for long spans. A three-hinged arch has hinges at both supports (A and B) and at one interior point (typically the crown C), giving it four unknown reaction components (Ax, Ay, Bx, By) and four available equations (three global equilibrium + one condition at C). It is therefore statically DETERMINATE. ARCH GEOMETRY: Let the supports A and B be at the same elevation (horizontal chord). The span is L and the rise (height of crown above the chord) is h. The crown hinge C is typically at midspan (x = L/2). Arch profile equation (parabolic arch): y = 4h·x(L−x)/L² This gives the elevation y at any horizontal distance x from A. REACTION CALCULATION — FOUR-EQUATION APPROACH: Step 1: ΣFx = 0: Ax = Bx = H (horizontal thrust, equal and opposite at both supports for a symmetric arch under vertical loads). Step 2: ΣMA = 0: Find V_B (vertical reaction at B). Step 3: ΣFy = 0: Find V_A. Step 4: CROWN CONDITION — ΣMC = 0 for either the left segment (A to C) or right segment (C to B): Find H. CROWN CONDITION (left segment): Take moments about C for the portion from A to C: V_A(L/2) − H(h) − Σ(applied loads between A and C × their moment arms about C) = 0 Solve for H. HORIZONTAL THRUST SIGNIFICANCE: For a simply supported beam of the same span carrying the same loads, the bending moment at midspan would be M_beam. In the arch, the bending moment at any section is: M_arch = M_beam_equiv − H·y where y is the rise of the arch at that section. The term H·y reduces the arch moment compared to the equivalent beam. This is the fundamental advantage of arch action. INTERNAL FORCES IN THE ARCH RIB: At any section at angle θ from the horizontal: Axial thrust: N = H·cos θ + V·sin θ Shear: V_s = V·cos θ − H·sin θ where V is the net vertical force on the free body to one side, and H is the horizontal thrust.
Examples
The horizontal thrust H = 30 kN is what makes the arch efficient. The equivalent simply supported beam under the same loading would have a midspan moment of V_A(10) − 60(5) = 450 − 300 = 150 kN·m. In the arch, M_arch at crown = 0 (by definition, it is a hinge). At x=5 (load point): M_arch = V_A(5) − H(y at x=5). Arch rise at x=5: y = 4(5)(5)(15)/400 = 3.75 m. M_arch = 45(5) − 30(3.75) = 225 − 112.5 = 112.5 kN·m vs M_beam = 45(5) = 225 kN·m — moment is cut by half! This is arch advantage.
Scenario
A three-hinged arch: span L = 20 m, rise h = 5 m (crown at midspan). Supports A and B at the same level. A vertical concentrated load of 60 kN acts at 5 m from A (x = 5 m). Find reactions Ay, By, and horizontal thrust H.
Solution
STEP 1 — Global vertical reactions: ΣM_A = 0: V_B(20) = 60(5) → V_B = 15 kN ↑ ΣF_y = 0: V_A = 60 − 15 = 45 kN ↑ STEP 2 — Crown condition (take left segment A to C, where C is at x=10, y=5): Loads between A and C: the 60-kN load at x=5 m is in this segment. ΣM_C (left segment) = 0: V_A(10) − H(5) − 60(10−5) = 0 45(10) − 5H − 60(5) = 0 450 − 5H − 300 = 0 5H = 150 → H = 30 kN STEP 3 — Check: ΣF_x = 0: Ax = Bx = H = 30 kN ✓ Final reactions: A_y = 45 kN ↑, B_y = 15 kN ↑, H = 30 kN (inward thrust at both supports).
The zero shear result makes sense for a segment of the arch between the load and a support that carries only the direct compressive thrust along the arch axis. Arches are designed to carry primarily axial compression, and the shear being nearly zero confirms efficient arch action. In practice, eccentricities and non-ideal load patterns introduce some bending and shear.
Scenario
For the same arch above, find the axial force N and shear Vs in the arch rib at x = 5 m (load point), just to the RIGHT of the load. Arch slope: tan θ = dy/dx. For parabolic arch y = 4hx(L−x)/L², dy/dx = 4h(L−2x)/L².
Solution
At x = 5 m: dy/dx = 4(5)(20−10)/400 = 4(5)(10)/400 = 0.5; θ = arctan(0.5) = 26.57° cos θ = 0.894, sin θ = 0.447 For the section just RIGHT of load at x=5, using the RIGHT free body (from x=5 to B): Net vertical force on right FBD = V_B = 15 kN ↑ (upward, toward section) Net horizontal force = H = 30 kN (inward, toward section) Axial force (compressive positive for arches): N = H·cos θ + V_B·sin θ = 30(0.894) + 15(0.447) = 26.82 + 6.71 = 33.53 kN (compression) Shear: Vs = V_B·cos θ − H·sin θ = 15(0.894) − 30(0.447) = 13.41 − 13.41 = 0 kN
Applications
- Three-hinged arches are used in gymnasium and arena roofs, aircraft hangars, and auditorium structures in the Philippines (e.g., large covered courts) because they are determinate and tolerant of foundation settlement.
- Stone and concrete arch bridges are historical examples of arch action — the horizontal thrust is transferred to massive abutments.
- In parabolic arches under UDL (e.g., the weight of a uniformly distributed deck), the bending moment is theoretically zero throughout — purely axial compression. This is why parabolic arch form is the 'ideal' arch shape for uniform loads.
- Cable structures are the tension analogue of arches — the catenary cable under UDL takes a parabolic shape, mirroring the parabolic arch.
Misconceptions
- MISCONCEPTION: The horizontal thrust H acts outward (pushing out the supports). TRUTH: The horizontal thrust is an inward thrust — both supports push toward the arch. Abutments must be designed to resist this inward force.
- MISCONCEPTION: For a three-hinged arch under UDL, the reactions at A and B include a vertical component only. TRUTH: Both supports provide both vertical and horizontal reactions; for level supports under vertical UDL, H_A = H_B = H (equal and inward).
- MISCONCEPTION: The crown condition is ΣM = 0 for the whole arch about point C. TRUTH: The crown condition is ΣM = 0 for ONE SIDE (either left or right of C). If you sum moments for the whole arch about C, you just get the global moment equation — not the extra condition needed.
- MISCONCEPTION: A parabolic arch always has zero bending moment. TRUTH: Only under a uniformly distributed load does a parabolic arch have zero moment (pure compression). Any other load pattern will produce bending in a parabolic arch.
Related Concepts
- Support reactions via equilibrium
- Equations of condition (internal hinges)
- Arch rise and geometry (parabolic, circular, catenary profiles)
- Equivalent beam concept
- Internal forces: N, V, M in curved members
Common Exam Questions
Example
A three-hinged arch, L=24 m, h=6 m, 80 kN at quarter point (x=6 m from A). Find H.
Approach
1. Find V_A and V_B from global equilibrium. 2. Apply ΣMC = 0 for the segment to the left (or right) of the crown hinge. 3. Solve for H. NOTE: include all loads between A and C in the left-segment FBD.
Question Type
Problem — Find horizontal thrust H
Example
For the arch above, find the bending moment in the arch rib at x = 8 m from A.
Approach
Compute M_arch = M_beam_equivalent − H·y_arch at the section of interest. First find y_arch from the arch geometry equation.
Question Type
Problem — Arch moment at an intermediate section
Example
Compare the maximum bending moment in a 20-m simply supported beam vs. a 20-m three-hinged parabolic arch, both carrying a 30 kN/m UDL.
Approach
Explain that H·y reduces the arch moment below the equivalent beam moment. Under UDL on a parabolic arch, M_arch = 0 everywhere (pure compression).
Question Type
Conceptual — Why are arches efficient?
Key Points To Remember
- Three-hinged arch has 4 unknowns (Ax, Ay, Bx, By) and 4 equations (3 equilibrium + 1 crown condition).
- The crown condition is: ΣMC = 0 for one side of the arch (left or right of the crown hinge).
- For vertical loads and level supports: Ax = Bx = H (horizontal thrust is the same at both supports).
- Arch moment = Equivalent beam moment − H × rise at that section (arch is more efficient than a beam).
- The horizontal thrust H is what makes arches more efficient — it reduces bending dramatically.
- For a parabolic arch under UDL: M = 0 throughout the arch (pure compression, no bending).
- Always identify the arch profile equation before computing internal forces at intermediate sections.
- The internal axial force N in the arch rib is compressive — this is why arches are built of masonry or concrete (weak in tension).
Practice Problems
Case (d) is a classic trap: a simply supported beam has exactly 3 reactions, which satisfies 3 equilibrium equations. Adding an internal hinge gives one MORE equation requirement (ec=1), so now you need 4 equations for 3 unknowns — meaning the structure is deficient (unstable, not just underdetermined). Case (b) shows how Gerber beams achieve determinacy: 5 reactions with 2 hinges gives exactly DI=0.
Problem
PROBLEM 1 — DETERMINACY CLASSIFICATION Classify each of the following structures. State whether each is statically determinate, indeterminate (and to what degree), or geometrically unstable. (a) A propped cantilever: fixed at A (r=3) and roller at B (r=1). No internal hinges. (b) A continuous beam with 5 reaction components and 2 internal hinges. (c) A portal frame: m=3 members, fixed at both bases (r=3 each, total r=6), 4 joints (n=4), no releases. (d) A simply supported beam (r=3) with one internal hinge (ec=1).
Solution
(a) Propped cantilever: Using beam shortcut: DI = r − (3 + ec) = (3+1) − (3+0) = 4 − 3 = 1. INDETERMINATE to the 1st degree. (b) Continuous beam: DI = r − (3 + ec) = 5 − (3+2) = 5 − 5 = 0. STATICALLY DETERMINATE. (c) Portal frame with fixed bases: DI = (3m+r) − (3n+ec) = (9+6) − (12+0) = 15 − 12 = 3. INDETERMINATE to the 3rd degree. (d) Simply supported with one hinge: DI = 3 − (3+1) = −1. GEOMETRICALLY UNSTABLE.
The negative sign for R_C means the hinge reaction on CDE is downward — physically, the cantilever portion of the main span HELPS hold up the CDE portion, pulling it down at C. This reversal is a common source of error. Always let the math determine the sign and interpret physically afterward. The 10.5 kN downward reaction at A means the beam lifts off a roller at A — if A is a roller (not a pin), this indicates uplift. Since A is a pin here, it can take both upward and downward forces.
Problem
PROBLEM 2 — REACTIONS OF A GERBER BEAM A Gerber beam ABCDE has the following configuration (measuring from left, A): A: pin support (r=2) B: roller support at x=4 m (r=1) C: internal hinge at x=7 m D: roller support at x=10 m (r=1) E: free end at x=14 m Loads: UDL of 12 kN/m from C to E (7 m to 14 m, length = 7 m). Find all reactions.
Solution
Verify determinacy: r = 2+1+1 = 4, ec = 1 (hinge at C). DI = 4 − (3+1) = 0. ✓ Determinate. STEP 1 — Identify suspended span: The portion CDE (from hinge C at x=7 to free end E at x=14) is the suspended span. It is supported by: (i) internal hinge at C (reaction from main span), (ii) roller at D (x=10 m). Analyse CDE (span from C to E = 7 m, with roller at D which is 3 m from C): UDL = 12 kN/m over 7 m; Total W = 84 kN at centroid 3.5 m from C. ΣM_C = 0: R_D(3) − 84(3.5) = 0 → R_D = 98 kN ↑ ΣF_y = 0: R_C = 84 − 98 = −14 kN → R_C = 14 kN ↓ (downward! The hinge pulls down on the CDE portion, meaning it pushes UP on the main span ABC). STEP 2 — Analyse main span ABCD (A: pin at x=0, B: roller at x=4; applied loads: 14 kN ↑ at C from hinge reaction): ΣM_A = 0: R_B(4) − 14(7) = 0 → R_B = 24.5 kN ↑ ΣF_y = 0: R_A_y = 14 − 24.5 = −10.5 kN → R_A_y = 10.5 kN ↓ ΣF_x = 0: R_A_x = 0 FINAL REACTIONS: A_y = 10.5 kN ↓, A_x = 0, R_B = 24.5 kN ↑, R_D = 98 kN ↑.
Key observations: (1) The applied clockwise moment M₀=60 kN·m causes a JUMP in the BMD at x=9 m — the moment goes from +55 kN·m just before to −5 kN·m just after. This sign change means there is a point of contraflexure near x=9. (2) The shear reverses sign somewhere between x=5 and x=7 — setting V=0: 1.67 − 10(x−5) = 0 → x−5 = 0.167 → x = 5.167 m from A. The maximum positive moment occurs at x=5.167 m. This problem combines three different load types — typical of PRC board exam 3-in-1 questions.
Problem
PROBLEM 3 — SFD AND BMD FOR A SIMPLY SUPPORTED BEAM A simply supported beam AB, span = 12 m. Loads: (1) Concentrated load P = 40 kN at x = 3 m from A; (2) UDL w = 10 kN/m from x = 5 m to x = 9 m (length = 4 m); (3) Applied clockwise moment M₀ = 60 kN·m at x = 9 m. Find V_A, V_B. Then find M at x = 3 m, x = 7 m, and x = 9 m⁻ and 9⁺.
Solution
REACTIONS: Resultant of UDL: W = 10 × 4 = 40 kN at centroid x = 5 + 2 = 7 m from A. ΣM_A = 0: V_B(12) = 40(3) + 40(7) + 60 = 120 + 280 + 60 = 460 → V_B = 38.33 kN ↑ ΣF_y = 0: V_A = 40 + 40 − 38.33 = 41.67 kN ↑ SHEAR (using left FBD, measuring x from A): x = 0⁺: V = +41.67 kN x = 3⁻: V = +41.67 kN x = 3⁺: V = 41.67 − 40 = +1.67 kN (jump at concentrated load) x = 5: V = +1.67 kN (UDL starts) x = 7: V = 1.67 − 10(2) = 1.67 − 20 = −18.33 kN x = 9: V = 1.67 − 10(4) = 1.67 − 40 = −38.33 kN At B (x=12): V = −38.33 + 38.33 = 0 ✓ MOMENTS: M(x=3) = V_A(3) = 41.67(3) = 125 kN·m M(x=7) = V_A(7) − 40(4) − 10(2)(1) = 291.67 − 160 − 20 = 111.67 kN·m [Alternatively: M(3) + area under SFD from 3 to 7 = 125 + 1.67(2) + 0.5(1.67+(-18.33))(2)... use exact trapezoidal area] M(x=9⁻) = V_A(9) − 40(6) − 40(2) = 375 − 240 − 80 = 55 kN·m M(x=9⁺) = M(9⁻) − M₀ = 55 − 60 = −5 kN·m (moment JUMPS at applied couple) Check M at B: M(12) = 0 ✓
The equivalent beam moment at x=6 m is 360 kN·m. The arch moment is only 180 kN·m — exactly half, because H×y = 40×4.5 = 180 kN·m relieves the bending. This illustrates why arches are structurally efficient: the horizontal thrust dramatically reduces bending, allowing the structure to carry loads primarily in compression. Note that M_arch = 0 at x=0 (A), x=12 (C, by definition of the hinge), and... check at B (x=24): M_beam = V_A(24)−80(18) = 1440−1440 = 0 ✓, so M_arch at B = 0−H(0) = 0 ✓.
Problem
PROBLEM 4 — THREE-HINGED ARCH A three-hinged parabolic arch has span L = 24 m and rise h = 6 m. Crown hinge C is at midspan. Supports A and B are at the same level. The arch carries an 80-kN vertical load at the quarter point (x = 6 m from A). Find: (a) vertical reactions V_A and V_B, (b) horizontal thrust H, (c) bending moment in the arch rib at x = 6 m.
Solution
(a) VERTICAL REACTIONS: ΣM_A = 0: V_B(24) = 80(6) → V_B = 20 kN ↑ ΣF_y = 0: V_A = 80 − 20 = 60 kN ↑ (b) HORIZONTAL THRUST (crown condition, left segment A to C, x=0 to x=12): The 80-kN load is in the left segment (x=6 m, which is between A at 0 and C at 12). Rise at crown (C): y_C = 6 m (given). ΣM_C (left segment) = 0: V_A(12) − H(6) − 80(12−6) = 0 60(12) − 6H − 80(6) = 0 720 − 6H − 480 = 0 6H = 240 → H = 40 kN (c) BENDING MOMENT AT x = 6 m: Arch rise at x=6 m: y = 4h·x(L−x)/L² = 4(6)(6)(18)/576 = 2592/576 = 4.5 m Equivalent beam moment at x=6 m: M_beam = V_A(6) = 60(6) = 360 kN·m Arch bending moment: M_arch = M_beam − H·y = 360 − 40(4.5) = 360 − 180 = 180 kN·m Answer: V_A = 60 kN, V_B = 20 kN, H = 40 kN, M_arch(x=6) = 180 kN·m.
Exam Preparation Tips
- MASTER THE DI FORMULA FIRST: The very first question in a structural analysis problem is always 'Is the structure determinate?' Use DI = r−(3+ec) for beams and DI = (3m+r)−(3n+ec) for frames. Practice classifying at least 20 different structures until it becomes automatic.
- ALWAYS DRAW A COMPLETE FREE BODY DIAGRAM: Every reaction and every applied load must be shown. Missing a reaction component or a load resultant is the single most common source of error in board-exam structural problems.
- MOMENT ABOUT A STRATEGIC POINT: Take moments about the point where the most unknowns intersect (e.g., a pin support eliminates both horizontal and vertical reactions at that joint). This often gives a single equation with a single unknown.
- MEMORISE THE dV/dx AND dM/dx RELATIONSHIPS: These allow you to check the shape and slope of SFD and BMD without full integration. Under UDL: shear is linear, moment is parabolic. Under no load: shear is constant, moment is linear.
- GERBER BEAMS — ALWAYS START FROM THE SUSPENDED SPAN: Identify the portion of the beam that is 'hanging' from the internal hinge. Analyse it first as a simple beam. Its hinge reaction becomes a load on the main span.
- THREE-HINGED ARCH PROTOCOL — FOUR STEPS: (1) Find V_A and V_B from global ΣM_A=0 and ΣFy=0. (2) Apply crown condition ΣMC=0 to the left OR right segment (not both) to find H. (3) Check ΣFx=0. (4) Use M_arch = M_beam − H·y for moments at intermediate sections.
- SIGN CONVENTION — BE CONSISTENT: Choose your sign convention at the start and maintain it throughout. Mixing conventions (positive shear on left face vs right face) is a frequent mistake that leads to wrong diagram shapes.
- VERIFY WITH CHECK EQUATIONS: After computing all reactions, verify with an independent equation (e.g., ΣM about a different support should equal zero). This catches arithmetic errors before they propagate into the SFD and BMD.
- KNOW THE CLASSIC STRUCTURE TYPES FOR THE BOARD EXAM: Simply supported beams, cantilevers, propped cantilevers, portal frames (pinned and fixed bases), three-hinged arches, and Gerber beams — these appear most frequently in the PRC exam.
- USE SI UNITS CONSISTENTLY: Forces in kN, distances in m, moments in kN·m. Never mix N and kN in the same calculation. The PRC exam is entirely in SI units.
- FOR FRAMES: Analyse member by member. At each joint, apply equilibrium to find how forces transfer from one member to the next. Draw the AFD, SFD, and BMD for EACH member, not the whole frame as one unit.
- ARCH GEOMETRY: For a parabolic arch, the rise at any point x is y = 4h·x(L−x)/L². Memorise this formula — you will need it to find the rise at the section where you are computing the arch moment.
In summary
Analysis of determinate structures is the essential foundation for every structural design task performed by a licensed civil engineer in the Philippines. The four competencies covered in this chapter — classifying structural determinacy, computing support reactions, constructing shear and moment diagrams, and analysing three-hinged arches — appear in virtually every PRC Civil Engineer Licensure Examination and underpin every subsequent structural analysis and design topic you will encounter in professional practice. The logical sequence never changes: (1) Check determinacy using DI = r−(3+ec) or DI = (3m+r)−(3n+ec). (2) If DI = 0 and the structure is geometrically stable, draw the complete Free Body Diagram and apply the three equilibrium equations plus any equations of condition. (3) Use the method of sections — cut, isolate, equilibrate — to find N, V, and M at any section. (4) For three-hinged arches, the fourth equation is always the crown condition ΣMC = 0 for one side, which yields the horizontal thrust H that defines arch behaviour. As you prepare for the PRC board exam, practise classifying and solving at least 50 determinate structure problems across all types: simply supported beams, cantilevers, compound Gerber beams, portal frames, and three-hinged arches. Focus particularly on mixed-load problems (UDL + concentrated load + applied moment together), as these are the most common format at the professional licensure level. Verify every reaction with a check equation, and verify every SFD and BMD by confirming that V = 0 and M = 0 at all free ends, and M = 0 at all internal hinges and pin/roller supports. Mastery of determinate structures is not merely an exam requirement — it is a professional obligation under RA 544, the Civil Engineering Law of the Philippines, which mandates that registered civil engineers possess the competence to analyse and design safe structures that protect public welfare. These methods are the tools that allow you to fulfil that mandate with confidence and precision.
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