CELE Structural Theory & Analysis — Influence Lines and Moving LoadsDetailed Explanation
This is the "office hours" version of Influence Lines and Moving Loads for the CELE 2026. No shortcuts, no hand-waving — just a full unpacking of why Professional Regulation Commission (PRC) — Board of Civil Engineering cares about each concept and how the Structural Theory & Analysis section items tend to play out on exam day. Read this once, then hit the practice questions with real understanding.
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 Influence Lines and Moving Loads is the 5th 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.
Influence Lines and Moving Loads - Detailed Explanation
Influence lines are one of the most important and frequently tested topics in the PRC Civil Engineer Licensure Examination under Structural Theory & Analysis. Unlike the familiar shear and moment diagrams—where the load position is fixed and we study how internal forces vary along the beam—an influence line fixes a specific response function (a reaction, a shear, or a bending moment at a given section) and asks: how does that response change as a single unit load rolls across the entire span? This is precisely the question a bridge designer, crane-girder engineer, or floor-beam analyst must answer every day, because real structures carry moving vehicles, equipment, and people. Mastering influence lines means you can confidently determine the worst-case loading position for any structural response, a skill tested repeatedly in both the board exam and in professional practice. This chapter develops the theory from first principles, presents the Müller-Breslau principle for rapid IL sketching, derives all key formulas for simply supported beams, and systematically works through point-load, distributed-load, and multi-load scenarios—all in SI units and at board-exam depth.
Concepts
Definition and Fundamental Concept of an Influence Line
An influence line (IL) for a specified response function F (which may be a support reaction, an internal shear force V at a section, or an internal bending moment M at a section) is a graph that shows the value of F as a unit concentrated load (P = 1 kN or simply 1 unit) moves from one end of the structure to the other. The horizontal axis represents the position x of the unit load along the span; the vertical axis represents the corresponding value of F. Critical distinction — IL vs. shear/moment diagram: • Shear or moment DIAGRAM: load position is FIXED; the section varies along the beam. • INFLUENCE LINE: section (or support) is FIXED; the unit load position varies. These are completely different things. Confusing them is the single most common board-exam pitfall. To construct an IL by statics (the classical method): 1. Place a unit load at position x from the left support. 2. Write the equilibrium equation for the desired response F as a function of x. 3. Plot F(x) versus x — the resulting graph is the influence line. For a simply supported beam of span L with supports at A (x=0) and B (x=L): • Unit load at position x from A. • Sum of moments about B: R_A = (L - x)/L — a linear function declining from 1 at A to 0 at B. • Sum of moments about A: R_B = x/L — a linear function rising from 0 at A to 1 at B. These straight-line ILs for reactions are the simplest and form the building blocks for shear and moment ILs.
Examples
This is the most fundamental IL construction. Notice R_A is always positive (downward reaction, upward support) regardless of where the load sits — which makes physical sense for a simply supported beam. The IL is linear because statics gives a linear equation in x.
Scenario
A simply supported beam has span L = 10 m with pin at A (x=0) and roller at B (x=10 m). Construct the influence line for reaction R_A by statics.
Solution
Let the unit load P=1 kN be placed at distance x from A (0 ≤ x ≤ 10 m). Sum of moments about B (clockwise positive): R_A × 10 - 1 × (10 - x) = 0 R_A = (10 - x)/10 = 1 - x/10 Check: • At x = 0 (load at A): R_A = 1.0 ✓ (entire load goes to A) • At x = 5 (load at midspan): R_A = 0.5 ✓ (equal sharing) • At x = 10 (load at B): R_A = 0.0 ✓ (entire load goes to B) The IL for R_A is a straight line from ordinate +1 at A to ordinate 0 at B.
Applications
- Determining which supports carry the most load when a vehicle crosses a bridge
- Designing crane-runway girder supports for maximum reactions
- Finding the critical load position for footing design under moving equipment
- Verifying that influence-line ordinates match intuitive load-sharing behavior
Misconceptions
- Thinking an IL for moment has units of kN·m — it actually has units of meters (m), because it's multiplied by a force in kN to give kN·m.
- Believing the IL ordinate IS the actual response — it's the response per unit load, a multiplier.
- Confusing the IL for R_A (decreasing left to right) with the IL for R_B (increasing left to right).
- Thinking ILs can only be constructed for reactions — they apply to any response including internal shear and moment.
Related Concepts
- Shear and Moment Diagrams
- Statics and Equilibrium
- Müller-Breslau Principle
- Moving Loads and Load Patterns
Common Exam Questions
Example
A 12 m simple beam: what is the IL ordinate for R_A when the unit load is 4 m from A? Answer: (12-4)/12 = 0.667.
Approach
Use the statics-derived formula: R_A = (L-x)/L or R_B = x/L for reactions. Plug in the given load position x.
Question Type
Identify the IL ordinate at a given point
Example
An exam question shows a triangular diagram and asks whether it is an IL or a moment diagram — identify the context clue (fixed section vs. fixed load).
Approach
Ask: Is the section fixed and load moving? → IL. Is the load fixed and section varying? → Shear/Moment diagram.
Question Type
Distinguish IL from moment diagram
Key Points To Remember
- An influence line has a FIXED section/response and a MOVING unit load — opposite of a shear/moment diagram.
- The vertical ordinate of an IL at any point x equals the value of the response when P=1 is placed at x.
- For simply supported beams, all ILs are composed of straight-line segments.
- IL for reaction R_A: linear from 1 at A to 0 at B (slope = -1/L).
- IL for reaction R_B: linear from 0 at A to 1 at B (slope = +1/L).
- Units of IL ordinate: for reactions and shears (dimensionless or kN/kN); for moments (meters or m/kN).
- The IL ordinate is NOT a force — it is a MULTIPLIER (influence coefficient).
Influence Lines for Shear and Moment at a Section
For a simply supported beam of span L, with section C located at distance a from A and b from B (so a + b = L), the influence lines for internal shear V_C and bending moment M_C are derived by cutting the beam at C and applying statics. --- IL FOR SHEAR V_C --- Sign convention: positive shear on left face acts downward (or equivalently, R_A upward on the left free body). Case 1 — unit load to the LEFT of C (0 ≤ x ≤ a): V_C = R_A - 1 = (L-x)/L - 1 = -x/L At x=0: V_C = 0; at x=a (just left of C): V_C = -a/L Case 2 — unit load to the RIGHT of C (a ≤ x ≤ L): V_C = R_A = (L-x)/L At x=a (just right of C): V_C = (L-a)/L = b/L; at x=L: V_C = 0 Result: The IL for V_C consists of two parallel sloping lines with a UNIT JUMP at C: • Left of C: negative segment from 0 at A to -a/L just left of C • Right of C: positive segment from +b/L just right of C to 0 at B The jump at C equals (-a/L) to (+b/L), a total change of 1 (the unit load). --- IL FOR MOMENT M_C --- Case 1 — unit load to the LEFT of C (0 ≤ x ≤ a): Taking moments about the right free body: M_C = R_B × b = (x/L) × b = bx/L At x=0: M_C=0; at x=a: M_C = ab/L Case 2 — unit load to the RIGHT of C (a ≤ x ≤ L): Taking moments about the left free body: M_C = R_A × a = [(L-x)/L] × a = a(L-x)/L At x=a: M_C = a(L-a)/L = ab/L; at x=L: M_C = 0 Result: The IL for M_C is a TRIANGLE peaking at C with: • Peak ordinate = ab/L (at the section C) • Base = full span L, descending linearly to 0 at both A and B This triangular shape is the most important IL for board-exam moment calculations.
Examples
This problem illustrates the three most common IL applications for a section. Note that placing the load AT the section gives maximum moment, while placing it to one side of the section gives maximum shear of that sign. The sign convention for shear must be consistent throughout.
Scenario
A simple beam of span 12 m has section C at 4 m from A (a=4 m, b=8 m). A 75 kN load moves across. Find: (a) max positive shear at C, (b) max negative shear at C, (c) max moment at C.
Solution
(a) Max positive V_C: Place the 75 kN load just to the RIGHT of C. IL ordinate = +b/L = 8/12 = +0.667 V_C(max positive) = 75 × 0.667 = +50.0 kN (b) Max negative V_C: Place the 75 kN load just to the LEFT of C. IL ordinate = -a/L = -4/12 = -0.333 V_C(max negative) = 75 × (-0.333) = -25.0 kN (c) Max M_C: Place the 75 kN load AT section C. IL peak ordinate = ab/L = (4)(8)/12 = 32/12 = 2.667 m M_C(max) = 75 × 2.667 = 200 kN·m Verification by statics (load at C): R_A = 75(8)/12 = 50 kN M_C = R_A × a = 50 × 4 = 200 kN·m ✓
For a UDL longer than the span, the load covers the entire span, and the response equals w times the total IL area. The triangular area formula gives the result directly without need for reaction calculations.
Scenario
For the same 12 m beam, section C at 4 m from A: what is the maximum moment at C if a uniformly distributed live load w = 20 kN/m (longer than the span) crosses the beam?
Solution
The IL for M_C is a triangle with base 12 m and peak ordinate 2.667 m. To maximize M_C, load the ENTIRE span (all ordinates are positive). Area of triangle = (1/2)(base)(height) = (1/2)(12)(2.667) = 16.0 m² M_C(max) = w × A_IL = 20 × 16.0 = 320 kN·m
Applications
- Bridge design: finding maximum shear and moment at critical sections under truck loading
- Floor beam design: worst-case moment under moving equipment loads
- Crane girder design: maximum shear at the wheel load position
- Checking whether a section is in positive or negative shear under a given load pattern
Misconceptions
- Placing the load at the section to maximize SHEAR — maximum shear occurs with the load ADJACENT to the section, not at it.
- Using the moment formula M_C = Pab/L for any load position — it is only valid when the load is AT position C (a from A, b from B).
- Forgetting the sign of the shear IL left of C is NEGATIVE — loading the left segment maximizes negative shear.
- Assuming the moment IL can have negative ordinates for a simple beam — it is always non-negative (all positive triangle).
- Using the full span area for a UDL shorter than the span — only the loaded length contributes.
Related Concepts
- Shear Force and Bending Moment Diagrams
- Free Body Diagram Method
- UDL Response Using IL Area
- Müller-Breslau Principle
Common Exam Questions
Example
8 m beam, C at 3 m from A, unit load at 6 m from A: V_C = (L-x)/L = (8-6)/8 = 0.25 (load is right of C).
Approach
Identify whether the load is left or right of the section. Apply the appropriate formula: V_C left = -x/L, V_C right = (L-x)/L, M_C = bx/L (load left) or a(L-x)/L (load right).
Question Type
Compute IL ordinate at a specific section for a given load position
Example
60 kN load, span 10 m, C at 4 m from A: M_C(max) = 60(4)(6)/10 = 144 kN·m.
Approach
Place load AT the section. Use M_C(max) = P × ab/L.
Question Type
Maximum moment at a section under a single moving point load
Example
If the UDL covers only part of the span, identify which segment maximizes the IL area for the desired sign.
Approach
Load only the positive or negative region of the shear IL. Compute the trapezoidal or triangular area under the loaded portion.
Question Type
Maximum shear from a UDL shorter than the span
Key Points To Remember
- IL for V_C: two parallel sloping lines, negative (left of C) and positive (right of C), with a unit jump at C.
- IL ordinate for V_C just LEFT of C = -a/L (negative shear).
- IL ordinate for V_C just RIGHT of C = +b/L (positive shear).
- IL for M_C: triangle peaking at C with ordinate ab/L.
- For maximum positive V_C: place load just to the right of C; for maximum negative V_C: place just to the left.
- For maximum M_C under a single point load: place the load AT the section C.
- The moment IL ordinate has units of length (meters), not force.
- At midspan (a = b = L/2): peak moment IL ordinate = L/4.
Müller-Breslau Principle
The Müller-Breslau principle (also called the virtual work or deflected-shape method) provides a powerful geometric shortcut for sketching influence lines without performing full statics calculations: 'The influence line for any force (or moment) response F at a point in a structure has the SAME SHAPE as the deflected profile of the structure when the restraint corresponding to F is removed and a unit displacement (or unit rotation) is introduced at that point in the direction of F.' --- HOW TO APPLY IT --- Step 1: Identify the response F (reaction, shear, or moment at a section). Step 2: Remove the restraint that F represents: • For a reaction at a support: remove that support (replace with a roller/pin then remove). • For shear at a section: cut the beam at that section and add a shear-release mechanism (a vertical slider — allows relative vertical movement but no rotation change). • For moment at a section: cut the beam at that section and insert a moment-release mechanism (a pin — allows relative rotation but no relative vertical movement). Step 3: Apply a unit displacement (for reactions and shears) or unit rotation (for moments) in the positive direction of F. Step 4: Sketch the deflected shape — this IS the influence line. --- RESULTS FOR SIMPLE BEAMS --- • IL for R_A: remove support at A, push up by 1 unit → straight line from 1 at A to 0 at B. • IL for V_C: insert shear release at C, impose unit shear displacement → two parallel sloping lines with unit jump at C. • IL for M_C: insert pin at C, impose unit relative rotation → triangle peaking at C. The Müller-Breslau principle is especially powerful for INDETERMINATE structures, where statics alone cannot directly give IL shapes. For determinate beams, it confirms what statics gives; for indeterminate beams, it tells us the SHAPE (though not the ordinates, which still require additional analysis).
Examples
Müller-Breslau turns an algebra problem into a geometry problem. For a simply supported beam (all straight-line segments because supports are fixed), the deflected shape under the pin constraint is simply a triangle. The peak ordinate is derived from geometry alone — no equilibrium equations needed.
Scenario
Use Müller-Breslau to sketch the IL for moment M_C at the midpoint of a 10 m simply supported beam.
Solution
Step 1: Section C is at midspan (a = b = 5 m). Response is M_C. Step 2: Insert a pin (moment release) at C. The beam now consists of two rigid segments AC and CB connected by the pin at C. Step 3: Apply a unit relative rotation at the pin — the two segments rotate about A and B respectively, with the pin acting as the hinge. Step 4: The left segment rotates by angle θ_1 upward at C; the right segment rotates by angle θ_2 downward at C. For unit relative rotation: θ_1 + θ_2 = 1 rad. By geometry, the deflection at C must be equal from both sides: h = a·θ_1 = b·θ_2 = 5θ_1 = 5θ_2, so θ_1 = θ_2 = 0.5 rad and h = 5(0.5) = 2.5 m. The deflected shape is a triangle with peak 2.5 m at midspan. This confirms: peak IL ordinate = ab/L = 5×5/10 = 2.5 m ✓
Applications
- Rapid sketching of IL shapes for exam problems without full statics derivation
- Qualitative analysis of ILs for indeterminate structures (continuous beams)
- Identifying which region to load for maximum positive or negative response
- Understanding why moment ILs are triangular and shear ILs have jumps
Misconceptions
- Thinking Müller-Breslau applies only to determinate structures — it is valid for ALL structures.
- Forgetting that for indeterminate beams the deflected shape is CURVED, not straight-line segments.
- Confusing the type of release: shear release (vertical slider) vs. moment release (pin).
- Applying unit load instead of unit deformation — Müller-Breslau imposes a unit deformation, not a unit force.
Related Concepts
- Virtual Work Principle
- Betti's Reciprocal Theorem
- Influence Lines for Indeterminate Structures
- Deflected Shape Analysis
Common Exam Questions
Example
The IL for moment at an interior support of a two-span continuous beam: remove the rotational restraint → the shape curves — not a simple triangle. This eliminates several wrong choices.
Approach
Apply Müller-Breslau mentally: what restraint is removed? What is the resulting deflected shape? Match to the given options.
Question Type
Identify the correct IL shape from a set of choices
Example
Beam 8 m, section at 3 m from A: peak M_C IL = (3)(5)/8 = 1.875 m. Confirmed by geometry.
Approach
Insert pin, apply unit rotation. Use θ_1 = b/(aL) × (aL... use similar triangles: h = a·θ_1 and the unit rotation constraint θ_1 + θ_2 = 1, giving h = ab/L.
Question Type
Find the peak ordinate of a moment IL using Müller-Breslau geometry
Key Points To Remember
- Müller-Breslau: IL shape = deflected shape after removing the restraint and imposing unit deformation.
- For reactions: remove support, apply upward unit displacement — IL is the deflected shape.
- For shear at C: insert vertical slider at C, impose unit vertical shear displacement — IL has two parallel segments with unit jump.
- For moment at C: insert pin at C, impose unit rotation — IL is triangular with peak at C.
- For determinate beams, the shapes are straight lines (rigid body kinematics — no elastic curvature needed).
- For indeterminate structures, the IL shapes are CURVED — Müller-Breslau identifies the shape but not the ordinates directly.
- This principle is based on the virtual work theorem (Betti's reciprocal theorem).
Using Influence Lines with Point Loads and Distributed Loads
Once an influence line has been constructed, it is used to find the actual response under real loads — which may be single point loads, multiple point loads, or uniformly distributed loads. --- SINGLE POINT LOAD P --- The response F = P × η_x where η_x is the IL ordinate directly under the load at position x. To MAXIMIZE the response: • Positive maximum: place P at the PEAK POSITIVE ordinate of the IL. • Negative maximum (minimum): place P at the PEAK NEGATIVE ordinate. --- MULTIPLE POINT LOADS --- For loads P_1, P_2, ..., P_n at positions x_1, x_2, ..., x_n: F = P_1·η_1 + P_2·η_2 + ... + P_n·η_n = Σ P_i·η_i Superposition applies because the IL is linear. To find the worst combination, trial positions are used — typically by series of load positions where each load sits at the IL peak alternately. --- UNIFORMLY DISTRIBUTED LOAD (UDL) --- For a UDL of intensity w (kN/m) over a segment from x_1 to x_2: F = w × (area of IL between x_1 and x_2) = w × ∫[x1 to x2] η(x) dx Physical interpretation: each infinitesimal load element w·dx acts like a point load w·dx at position x, contributing w·dx·η(x) to the response. Integrating gives w times the IL area under the loaded region. To MAXIMIZE the response: • Positive maximum: load ALL regions where IL ordinate is positive. • Negative maximum (minimum): load ALL regions where IL ordinate is negative. • If load length is fixed (shorter than span): position the load where the IL area under it is maximum. --- KEY FORMULAS FOR SIMPLE BEAMS --- Max R_A under UDL (full span): = w × (1/2)(L)(1) = wL/2 [triangular IL area] Max M_C under UDL (full span): = w × (1/2)(L)(ab/L) = wab/2 Max M at midspan under UDL: = w × (1/2)(L)(L/4) = wL²/8 ✓ (standard formula) Max V at C under UDL (full span, positive only): = w × (1/2)(b)(b/L) = wb²/(2L)
Examples
When multiple loads are present, the maximum response is generally obtained by placing the HEAVIEST load at the IL peak. Always verify by checking all reasonable trial positions. Here, placing the 60 kN load at midspan (the peak) gives a significantly higher response.
Scenario
A simple beam, span L = 10 m, has two moving loads: P_1 = 40 kN and P_2 = 60 kN spaced 3 m apart (P_1 leading). Find the maximum moment at midspan (C at 5 m from A).
Solution
IL for M_C: triangle, peak = ab/L = (5)(5)/10 = 2.5 m at midspan. The IL is linear, so at any position x from A: Left of midspan (x ≤ 5): η(x) = x/2 [rising line from 0 to 2.5] Right of midspan (x > 5): η(x) = (10-x)/2 [falling line from 2.5 to 0] Trial 1 — P_1 at midspan (x_1 = 5 m), P_2 at x_2 = 8 m: η_1 = 2.5 m; η_2 = (10-8)/2 = 1.0 m M_C = 40(2.5) + 60(1.0) = 100 + 60 = 160 kN·m Trial 2 — P_2 at midspan (x_2 = 5 m), P_1 at x_1 = 2 m: η_2 = 2.5 m; η_1 = 2/2 = 1.0 m M_C = 40(1.0) + 60(2.5) = 40 + 150 = 190 kN·m ← GOVERNS Maximum M_C = 190 kN·m (heavier load P_2 placed at the IL peak).
When a UDL is shorter than the span, systematically check loading positions that include the IL peak. The position that maximizes the area under the 5 m window is found by centering the window at (or near) the peak and comparing areas.
Scenario
A simple beam of span 8 m carries a moving UDL w = 25 kN/m that is 5 m long (shorter than the span). Find the maximum moment at C located 3 m from A (a=3, b=5).
Solution
IL for M_C: triangle, peak = ab/L = (3)(5)/8 = 1.875 m at x=3 m. The IL rises linearly from 0 at A (x=0) to peak 1.875 m at C (x=3), then falls to 0 at B (x=8). The 5 m UDL should be positioned to cover the maximum IL area. Since the peak is at x=3 and the right side has slope (−1.875/5) while left side has slope (+1.875/3), the right side descends more slowly — the area under the right segment (length 5) is larger than under the left (length 3). Best position: load from x=3 m to x=8 m (C to B, covering the entire right side). IL ordinates: from 1.875 m at x=3 to 0 at x=8. Area (triangle) = (1/2)(5)(1.875) = 4.6875 m² M_C = 25 × 4.6875 = 117.2 kN·m Alternative: load from x=0 to x=5 m (includes peak and left portion): Area from 0 to 3: (1/2)(3)(1.875) = 2.8125 m² Area from 3 to 5: ordinate at x=5: η(5) = (8-5)(3)/(8×... use right side formula: η = (8-x)(3/8) → η(5) = 3(3/8) = 1.125 m Area from 3 to 5 (trapezoid): (1/2)(2)(1.875+1.125) = 3.0 m² Total area = 2.8125 + 3.0 = 5.8125 m² M_C = 25 × 5.8125 = 145.3 kN·m ← LARGER, GOVERNS Optimal position is the load from x=0 to x=5 m (covering the peak and spanning from A toward B), M_C(max) = 145.3 kN·m.
Applications
- AASHTO/DPWH truck loading: computing reactions and moments from moving axle loads
- Live load pattern selection for maximum span moments in continuous beams
- Crane rail design: maximum rail bending under moving wheel loads
- Pedestrian bridge design: worst-case crowd loading patterns
Misconceptions
- Multiplying w by the IL ordinate (not the area) — UDL uses AREA, not ordinate.
- Loading negative IL regions when maximizing positive response — always load only the sign-matching region.
- Placing multiple loads without checking all trial positions — always trial the heavy load at the peak.
- Using the midspan moment formula wL²/8 for a section NOT at midspan — it only applies exactly at midspan.
Related Concepts
- Superposition Principle
- Area Under Influence Line
- Load Combinations (NSCP 2015 Section 203)
- Pattern Loading for Continuous Beams
Common Exam Questions
Example
P=80 kN, L=12 m, section at 4 m from A: M_C(max) = 80(4)(8)/12 = 213.3 kN·m.
Approach
Use M_C(max) = P × ab/L with load placed at C.
Question Type
Compute maximum moment at a section under a single moving load
Example
w=30 kN/m, L=10 m: R_A(max) = 30 × 5 = 150 kN (= wL/2 ✓).
Approach
IL for R_A is a triangle of area = (1/2)(L)(1) = L/2. Response = w × L/2.
Question Type
Maximum reaction under a moving UDL (longer than span)
Example
w=20 kN/m, L=8 m: M_mid(max) = 20(64/8) = 160 kN·m (= wL²/8 ✓).
Approach
IL area for midspan moment = (1/2)(L)(L/4) = L²/8. Response = w × L²/8.
Question Type
Maximum midspan moment under full UDL
Key Points To Remember
- Point load response: F = P × (IL ordinate at load position).
- UDL response: F = w × (IL area under loaded region).
- Multiple loads: F = ΣP_i × η_i (superposition).
- For maximum POSITIVE response: load where IL is POSITIVE.
- For maximum NEGATIVE response: load where IL is NEGATIVE.
- The standard formula wL²/8 for midspan moment can be derived directly from the IL area.
- For a UDL shorter than the span, position it to maximize the IL area under the load.
- IL area for a triangle of base L and height h: A = hL/2.
Absolute Maximum Moment Under Moving Loads
The previous sections found the maximum moment AT A SPECIFIC SECTION. The absolute maximum moment question asks: over the ENTIRE beam, what is the largest bending moment that can occur, and where does it occur? --- SINGLE MOVING LOAD --- For a single load P on a simple beam of span L, the moment at any section x (where load also sits at x) is: M(x) = P × x(L-x)/L Maximizing dM/dx = 0 gives x = L/2 (midspan). Absolute maximum moment = PL/4 (at midspan). This is a well-known result and a frequent board-exam formula. --- SERIES OF MOVING LOADS --- When a series of concentrated loads (e.g., truck axle loads) moves across the beam, the absolute maximum moment occurs under ONE OF THE LOADS — specifically, the load closest to the resultant of all loads on the span. CRITERION (The Load-Resultant Straddling Rule): The absolute maximum moment occurs when the beam centerline bisects the distance between: (1) the load under which the maximum occurs (call it the 'critical load'), AND (2) the resultant R of all the loads currently on the span. PROCEDURE: 1. Compute the resultant R = ΣP_i and its position relative to the loads. 2. Identify the candidate critical load (usually the heaviest load nearest the resultant). 3. Position the load group so that the midpoint of the beam (L/2 from A) falls midway between the critical load and the resultant. → Let d = distance from critical load to resultant. → Critical load position from A: x_cr = L/2 - d/2 4. Compute reactions R_A or R_B for this load position. 5. Compute moment under the critical load (by summing moments from A to that point, excluding the load itself if it straddles a free-body cut — but it is directly on the cross-section, so use R_A × x_cr minus any loads between A and the critical load). 6. Check adjacent loads as candidates — the governing position gives the absolute maximum. IMPORTANT: As the load group moves across the span, some loads may exit the beam. Always check that all loads in your analysis are actually ON the span for the position you compute.
Examples
This problem demonstrates the key principle: the absolute max occurs at 4.5 m, not at 5 m (midspan). The difference here is small (162 vs 160 kN·m), but for unequal loads the difference can be significant. Always apply the straddling rule rather than assuming midspan governs.
Scenario
Two equal loads P = 40 kN spaced 2 m apart move across a simply supported beam of span L = 10 m. Find the absolute maximum bending moment.
Solution
Step 1 — Resultant: R = 40 + 40 = 80 kN Resultant location: midway between the two loads (equal loads), so 1 m from each load. Step 2 — Candidate critical loads: either load (by symmetry, try the LEFT load as critical). d = distance from left load to resultant = 1 m (resultant is to the right of left load). Step 3 — Position the load group: Beam midpoint at L/2 = 5 m from A. Midpoint between left load and resultant = (x_left + x_resultant)/2 = 5 m (beam center). x_left + (x_left + 1) = 10 → 2x_left = 9 → x_left = 4.5 m from A x_resultant = 5.5 m from A ✓ (midspan bisects the 1 m gap between 4.5 m and 5.5 m) Step 4 — Reaction at A: R_A = [80(10 - 5.5)]/10 = 80(4.5)/10 = 36 kN (or: R_A = [40(10-4.5) + 40(10-6.5)]/10 = [40(5.5)+40(3.5)]/10 = [220+140]/10 = 36 kN ✓) Step 5 — Moment under the left load (at x = 4.5 m, no other loads between A and this point): M_left = R_A × 4.5 = 36 × 4.5 = 162 kN·m Step 6 — Check: right load as critical (by symmetry of the problem, gives same result → 162 kN·m). Absolute maximum M = 162 kN·m at 4.5 m from A. Comparison: midspan moment (both loads on beam, load group centered): x_left = 4 m, x_right = 6 m; R_A = (40×6 + 40×4)/10 = 40 kN; M_mid = 40×5 - 40×1 = 160 kN·m. 162 > 160 ✓ — the absolute max is indeed slightly off-center.
For three or more loads, always check the heaviest load and its neighbors as candidates. The heaviest load (P_2 = 50 kN) is the critical one here, giving 380 kN·m. Checking P_3 gives a lower value (346.6 kN·m), confirming P_2 governs.
Scenario
Three axle loads: P_1 = 30 kN, P_2 = 50 kN, P_3 = 40 kN spaced 2 m apart (P_1–P_2: 2 m, P_2–P_3: 2 m) cross a 15 m simple beam. Find the absolute maximum bending moment.
Solution
Step 1 — Resultant: R = 30 + 50 + 40 = 120 kN Take P_1 as origin; positions: P_1 at 0, P_2 at 2, P_3 at 4. x_R = (30×0 + 50×2 + 40×4)/120 = (0 + 100 + 160)/120 = 260/120 = 2.167 m from P_1 Step 2 — Candidate critical load: P_2 (heaviest, nearest to resultant at 2.167 m; P_2 is at 2.0 m, distance = 0.167 m). d = 2.167 - 2.0 = 0.167 m (resultant is 0.167 m to the right of P_2). Step 3 — Position: Beam midpoint at 7.5 m from A. x_P2 = 7.5 - 0.167/2 = 7.5 - 0.083 = 7.417 m from A x_P1 = 7.417 - 2 = 5.417 m from A x_P3 = 7.417 + 2 = 9.417 m from A All loads are within 0 to 15 m ✓ Step 4 — Reaction: x_R from A = 7.417 + 0.167 = 7.583 m R_A = R(L - x_R)/L = 120(15 - 7.583)/15 = 120(7.417)/15 = 59.33 kN Step 5 — Moment under P_2 (sum moments from A to P_2, subtracting P_1 contribution): M_P2 = R_A × x_P2 - P_1 × (x_P2 - x_P1) = 59.33 × 7.417 - 30 × 2 = 440.0 - 60.0 = 380.0 kN·m Step 6 — Check P_3 as critical (d from P_3 to resultant = 4 - 2.167 = 1.833 m from P_1 side; resultant is LEFT of P_3): d = 2.167 - 0 ... recalculate: from P_3: P_3 at 4 m from P_1; x_R = 2.167 m from P_1 → distance from P_3 to resultant = 4 - 2.167 = 1.833 m to LEFT. x_P3 = 7.5 - (-1.833/2) ... resultant is left of P_3, so x_P3 = 7.5 + 1.833/2 = 7.5 + 0.917 = 8.417 m from A. x_P1 = 8.417 - 4 = 4.417 m (OK, on span); x_P2 = 6.417 m. x_R from A = 8.417 - 1.833 = 6.583 m. R_A = 120(15 - 6.583)/15 = 120(8.417)/15 = 67.33 kN. M_P3 = R_A(8.417) - P_1(8.417-4.417) - P_2(8.417-6.417) = 67.33(8.417) - 30(4) - 50(2) = 566.6 - 120 - 100 = 346.6 kN·m < 380.0 kN·m. Absolute maximum M = 380.0 kN·m under P_2 at 7.417 m from A.
Applications
- Bridge live-load design: AASHTO HL-93 truck positioning for maximum span moment
- Railway bridge design: locomotive axle load positioning per PNR/DPWH standards
- Overhead crane girder design: tandem wheel load positioning
- Highway overpass design under DPWH Standard Truck Loading
Misconceptions
- Assuming absolute max moment always occurs at midspan — TRUE only for a SINGLE load; for multiple loads it is slightly off-center.
- Using the resultant position directly as the section for maximum moment — the max is under the CRITICAL LOAD, not the resultant.
- Forgetting to check that all loads are still on the span for the computed position.
- Not checking adjacent loads — sometimes the load adjacent to the heaviest governs.
- Confusing the straddling rule: it is the BEAM CENTERLINE that bisects the gap, not the average of load positions.
Related Concepts
- Resultant of Force Systems
- Influence Lines for Moment at a Section
- AASHTO HL-93 Truck Loading
- Live Load Positioning in NSCP 2015
Common Exam Questions
Example
P = 100 kN, L = 20 m: M_abs_max = 100(20)/4 = 500 kN·m at midspan.
Approach
Direct formula: M_abs_max = PL/4 at midspan.
Question Type
Single moving load: absolute maximum moment
Example
P_1=60 kN, P_2=80 kN, spacing=3 m, L=12 m — identify critical load (P_2, heavier), compute d, position, solve.
Approach
Find resultant position, apply straddling rule, compute R_A, find M under critical load.
Question Type
Two-load system: find absolute max moment
Example
Two equal loads 4 m apart, L=16 m: d=2 m, x_cr = 8 - 1 = 7 m from A.
Approach
Apply x_cr = L/2 - d/2 where d = distance from critical load to resultant.
Question Type
Position of absolute max moment from support A
Key Points To Remember
- Single moving load: absolute max M = PL/4 at midspan.
- Series of loads: max M occurs UNDER ONE OF THE LOADS (the critical load near the resultant).
- Positioning rule: beam centerline bisects the gap between the critical load and the resultant.
- d = distance from critical load to resultant; critical load at x = L/2 - d/2 from A.
- Always verify all loads are on the span for the computed position.
- Check both the primary candidate and adjacent loads for the absolute maximum.
- For equal spaced equal loads, the max occurs under the middle load (by symmetry).
- The absolute max M is always slightly greater than the midspan moment for a given load group.
Practice Problems
Parts (b) and (c) confirm that the SIGN of maximum shear depends on which side of the section the load is placed. Part (d) shows the UDL area method — only the positive IL region is loaded to maximize positive shear. Note that wL/2 = 108 kN is the max reaction, which is different from max shear at C.
Problem
PROBLEM 1 (IL for Shear — Board Exam Type) A simply supported beam has a span of 12 m. A section C is located 3 m from the left support A. (a) Draw the influence line for shear V_C. (b) Determine the maximum positive shear at C due to a 90 kN moving load. (c) Determine the maximum negative shear at C due to the same 90 kN moving load. (d) A moving UDL of w = 18 kN/m (longer than the span) crosses the beam. Find the maximum positive shear at C.
Solution
Given: L=12 m, a=3 m (AC), b=9 m (CB) (a) IL for V_C: • Left of C (0 ≤ x ≤ 3 m): η_V = -x/12, ranging from 0 at A to -3/12 = -0.25 at x=3⁻ • Right of C (3 m ≤ x ≤ 12 m): η_V = (12-x)/12, ranging from 9/12 = +0.75 at x=3⁺ to 0 at B • Jump at C: from -0.25 to +0.75 (total = 1.0 ✓) (b) Maximum positive V_C: Place 90 kN just to the RIGHT of C. η = +b/L = +9/12 = +0.75 V_C(max+) = 90 × 0.75 = +67.5 kN (c) Maximum negative V_C: Place 90 kN just to the LEFT of C. η = -a/L = -3/12 = -0.25 V_C(max−) = 90 × (−0.25) = −22.5 kN (d) Maximum positive V_C under UDL: Load ONLY the positive IL region (right of C, from x=3 to x=12). Positive IL region: triangle with base 9 m and peak 0.75 at x=3⁺. Area = (1/2)(9)(0.75) = 3.375 m² V_C(max+) = 18 × 3.375 = 60.75 kN
This problem tests two different questions: max moment AT A SPECIFIED SECTION (part a, using IL area = wab/2) versus ABSOLUTE MAX MOMENT IN THE ENTIRE BEAM (part b, using wL²/8 at midspan). These are distinct questions and frequently confused in board exams. For a single UDL covering the full span, the absolute max is always at midspan.
Problem
PROBLEM 2 (IL for Moment — Board Exam Type) A 16 m simply supported beam carries a moving live load of w = 22 kN/m. Find: (a) The maximum bending moment at a section 6 m from the left support. (b) The absolute maximum bending moment in the beam.
Solution
Given: L=16 m, a=6 m, b=10 m, w=22 kN/m (longer than span) (a) Maximum moment at C (a=6, b=10): IL for M_C: triangle, peak = ab/L = (6)(10)/16 = 3.75 m Load the ENTIRE span (positive IL region = full span for a simply supported beam). Area of IL = (1/2)(L)(peak) = (1/2)(16)(3.75) = 30.0 m² M_C(max) = w × A_IL = 22 × 30.0 = 660 kN·m Verification: M_C = wab/2 = 22(6)(10)/2 = 660 kN·m ✓ (b) Absolute maximum bending moment: For a UDL, the absolute maximum moment occurs at MIDSPAN: M_abs_max = wL²/8 = 22(16)²/8 = 22(256)/8 = 704 kN·m at x = 8 m Alternatively using IL: IL peak at midspan = L/4 = 4 m; Area = (1/2)(16)(4) = 32 m²; M = 22×32 = 704 kN·m ✓
This problem mirrors actual bridge design practice using HS-20 truck loads (scaled to metric). The procedure is systematic: find resultant, identify critical load (heaviest near resultant), apply straddling rule, compute reaction and moment. Checking both P_2 and P_3 confirms P_2 governs. In Philippine practice, DPWH and NSCP 2015 specify truck loads similar to AASHTO standards for bridge design.
Problem
PROBLEM 3 (Absolute Maximum Moment — Series of Moving Loads) A truck crosses a simply supported bridge of span L = 20 m. The axle loads are: • Front axle: P_1 = 35 kN • Middle axle: P_2 = 145 kN (spacing: 4.3 m from P_1) • Rear axle: P_3 = 145 kN (spacing: 4.3 m from P_2) This approximates the AASHTO HS20-44 truck. Find the absolute maximum bending moment in the bridge girder.
Solution
Step 1 — Resultant: R = 35 + 145 + 145 = 325 kN Take P_1 as origin (positions: P_1 at 0, P_2 at 4.3, P_3 at 8.6 m) x_R = (35×0 + 145×4.3 + 145×8.6)/325 = (0 + 623.5 + 1247)/325 = 1870.5/325 = 5.756 m from P_1 Step 2 — Candidate critical load: P_2 at 4.3 m from P_1; distance from P_2 to R = 5.756 - 4.3 = 1.456 m (R is to the right of P_2). P_3 at 8.6 m from P_1; distance from P_3 to R = 8.6 - 5.756 = 2.844 m (R is to the left of P_3). P_2 is heavier AND closer to the resultant → P_2 is the critical load. d = 1.456 m Step 3 — Position: Beam midpoint at L/2 = 10.0 m from A. x_P2 = 10.0 - 1.456/2 = 10.0 - 0.728 = 9.272 m from A x_P1 = 9.272 - 4.3 = 4.972 m from A ✓ (on span) x_P3 = 9.272 + 4.3 = 13.572 m from A ✓ (on span) x_R from A = 9.272 + 1.456 = 10.728 m Step 4 — Reaction at A: R_A = R(L - x_R)/L = 325(20 - 10.728)/20 = 325(9.272)/20 = 150.67 kN Step 5 — Moment under P_2: M_P2 = R_A(x_P2) - P_1(x_P2 - x_P1) = 150.67(9.272) - 35(9.272 - 4.972) = 1396.4 - 35(4.3) = 1396.4 - 150.5 = 1245.9 kN·m Step 6 — Check P_3 as critical: d from P_3 to R = 2.844 m; x_P3 = 10.0 - 2.844/2 = 10.0 - 1.422 = 8.578 m (with P_3 to the right of beam center since R is left of P_3) Wait — recheck: R is to the LEFT of P_3, so P_3 should be positioned to the RIGHT of beam center. x_P3 = 10.0 + 1.422 = 11.422 m from A x_P2 = 11.422 - 4.3 = 7.122 m; x_P1 = 7.122 - 4.3 = 2.822 m ✓ (all on span) x_R from A = 11.422 - 2.844 = 8.578 m R_A = 325(20 - 8.578)/20 = 325(11.422)/20 = 185.6 kN M_P3 = R_A(11.422) - P_1(11.422 - 2.822) - P_2(11.422 - 7.122) = 185.6(11.422) - 35(8.6) - 145(4.3) = 2119.9 - 301 - 623.5 = 1195.4 kN·m < 1245.9 kN·m Absolute maximum M = 1246 kN·m under P_2 at 9.272 m from A.
In practice, dead loads are always present while live loads are positioned for worst case. Superposition is applied: DL effect (computed from standard formulas or IL area) + LL effect (from IL). Note that the dead load shear at C was found by simple statics, then the live load component was added via IL. This combined approach is the standard board-exam method.
Problem
PROBLEM 4 (Combined IL Application — Beam with Overhang Concept) A simply supported beam of span L = 10 m supports section C at the quarter point (2.5 m from A). A moving load of P = 50 kN and a fixed dead load of 10 kN/m act simultaneously. Using influence lines: (a) Find the maximum TOTAL moment at C due to the combination. (b) Find the maximum TOTAL shear at C due to the combination.
Solution
Given: L=10 m, a=2.5 m, b=7.5 m, P=50 kN, w_DL=10 kN/m (fixed, full span) Dead Load Effects (fixed, full span): M_C(DL) = w×a×b/2 = 10(2.5)(7.5)/2 = 93.75 kN·m V_C(DL): reaction R_A = wL/2 = 50 kN; V_C = R_A - w×a = 50 - 10(2.5) = +25 kN Live Load Effects (moving P = 50 kN): (a) Max M_C(LL): place P at C. M_C(LL) = P×ab/L = 50(2.5)(7.5)/10 = 93.75 kN·m Max TOTAL M_C = 93.75 + 93.75 = 187.5 kN·m (b) Max positive V_C(LL): place P just right of C. V_C(LL+) = P×b/L = 50(7.5)/10 = +37.5 kN Total V_C(max+) = 25 + 37.5 = 62.5 kN Max negative V_C(LL): place P just left of C. V_C(LL−) = −P×a/L = −50(2.5)/10 = −12.5 kN Total V_C(min) = 25 + (−12.5) = +12.5 kN (still positive, meaning shear reversal does NOT occur here) The critical combinations are: Max M_C = 187.5 kN·m; Max V_C = 62.5 kN (positive, load right of C).
Exam Preparation Tips
- MEMORIZE THE FOUR KEY IL FORMULAS for simple beams: (1) IL ordinate for R_A = (L-x)/L; (2) IL ordinate for V_C right of section = b/L; (3) IL ordinate for V_C left of section = -a/L; (4) IL peak ordinate for M_C = ab/L. These appear in nearly every board exam problem.
- ALWAYS DRAW THE IL FIRST before computing — even a rough sketch helps you identify peak positions, sign regions, and whether to use ordinate or area. Examiners often award partial credit for correct IL sketches.
- UNIT CHECK: Reaction and shear IL ordinates are dimensionless (or m/m). Moment IL ordinates have units of METERS (not kN·m). When you multiply by a force P (kN), you get a moment in kN·m. This is a common source of dimensional errors.
- DISTINGUISH THE TWO TYPES OF PROBLEMS: 'Maximum moment AT a section' uses the IL peak ordinate at that section. 'Absolute maximum moment in the beam' requires the load-resultant straddling rule for multiple loads, or simply PL/4 for a single load.
- FOR UDL PROBLEMS: If the UDL is LONGER than the span, cover the entire span for maximum positive moment (all IL ordinates for M are positive). For maximum positive SHEAR at a section, cover only the positive shear region (right of the section). If the UDL is SHORTER than the span, position it where the IL area is maximum.
- THE STRADDLING RULE for absolute maximum moment: The beam's midpoint bisects the distance between the critical load and the system resultant. Memorize: x_critical = L/2 - d/2, where d = distance from critical load to resultant (with sign based on which side).
- MÜLLER-BRESLAU AS A QUICK CHECK: Use it to verify your IL shape — if the sketched IL doesn't match the deflected shape after releasing the constraint, recheck your work. For determinate beams, all IL segments are straight lines.
- SUPERPOSITION: Dead load response (from standard formulas) + Live load response (from IL) = total response. The live load is positioned for worst case; the dead load is always over the full span.
- WATCH FOR SIGN CONVENTIONS in shear: The negative shear region of the shear IL is LEFT of the section. When maximizing NEGATIVE shear, place the load to the LEFT of the section. This is opposite to what many students intuitively assume.
- PRACTICE THE THREE-STEP PROCESS for every IL problem: (1) Identify the response and draw its IL; (2) Determine the worst load position (peak for point load, sign-matching area for UDL); (3) Compute response using F = P×η or F = w×Area. Consistent application of this process eliminates most errors.
In summary
Influence lines are the fundamental tool for structural analysis under moving loads — a topic that appears consistently in the PRC Civil Engineer Licensure Examination and is essential for professional practice in bridge, crane-girder, and floor-beam design. The core insight to carry into the exam room is this: an influence line is defined by a FIXED section and a MOVING unit load, the exact opposite of a shear or moment diagram. From this single distinction flows everything else. For simply supported beams, the three essential IL shapes are: (1) the reaction IL — a straight line from 1 at the support to 0 at the other end; (2) the shear IL at a section — two parallel sloping lines with a unit jump at the section, negative on the left and positive on the right; and (3) the moment IL at a section — a triangle peaking at the section with ordinate ab/L. The Müller-Breslau principle confirms these shapes geometrically by showing that each IL equals the deflected form of the beam after releasing the corresponding restraint. Applying ILs to real loads follows two rules: for a point load, multiply P by the IL ordinate at the load position (maximum when load is at the IL peak); for a UDL, multiply w by the IL area over the loaded region (load only the sign-matching region for maximum response). For the absolute maximum moment under a series of moving loads — the type of problem that appears in bridge design and on the board exam — the straddling rule is paramount: position the critical load (nearest to the resultant) and the resultant symmetrically about the beam's midpoint, then compute the moment under that load. Mastery of this chapter requires not just formula memorization but genuine understanding of what influence lines represent physically. When you can look at a moving truck on a bridge and immediately visualize which axle position puts the maximum moment in the midspan section, you have achieved the professional-level competence that the PRC examination demands. Practice systematically with board-style numerical problems, always draw the IL before computing, and consistently verify your answers using direct statics. Good luck on your examination.
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