CELE Structural Theory & Analysis — Influence Lines and Moving LoadsRevision Notes
Revision notes for CELE Structural Theory & Analysis — Influence Lines and Moving Loads. Short, focused, and designed for the week before exam day. Use these when you are already familiar with the chapter and need a quick refresh on the high-yield items Professional Regulation Commission (PRC) — Board of Civil Engineering tests.
Exam context
The Civil Engineer Licensure Examination is conducted by Professional Regulation Commission (PRC) — Board of Civil Engineering and is scheduled for May and November 2026. The Structural Theory & Analysis subtest is marked as "Core" in the official pattern, and Influence Lines and Moving Loads appears in position 5th of 6 in the CELE Structural Theory & Analysis review rotation. Passing mark: 70% weighted average, no sub-test below 50%. Recent CELE 2026 papers have drawn roughly a meaningful share of questions from this subject.
Influence Lines and Moving Loads - Revision Notes
Influence lines (ILs) are indispensable tools for designing structures subjected to moving loads — bridges, crane girders, and floor beams carrying live traffic. Unlike a shear or moment diagram (which fixes the load position and varies the section), an influence line FIXES the section (or the response of interest) and MOVES a unit load across the span, plotting the resulting response at every load position. Mastering influence lines is essential for the PRC Civil Engineer Licensure Examination, appearing consistently in the Structural Theory and Analysis portion. This chapter covers: (1) construction of ILs for determinate beams, (2) Müller-Breslau principle, (3) using ILs for point loads and distributed loads, and (4) absolute maximum moment under a series of moving loads.
Sections
Formulas
Example
A 60 kN load sits at x = 3 m from A on a 10 m simple beam. If the IL ordinate for M_C at x = 3 m is 1.8 m, then M_C = 60 × 1.8 = 108 kN·m.
Formula
R = P × η_x
Variables
R = response (kN or kN·m); P = actual point load (kN); η_x = IL ordinate at load position x
Application
Computing the response at the fixed section due to a single concentrated moving load placed at position x.
Example
A UDL w = 20 kN/m covers a 6 m simple beam. IL area for midspan moment = ½(6)(1.5) = 4.5 m². M_mid = 20 × 4.5 = 90 kN·m (check: wL²/8 = 20×36/8 = 90 ✓).
Formula
R = w × A_IL
Variables
R = response; w = intensity of uniformly distributed load (kN/m); A_IL = area under the IL over the loaded length (m² for moment IL, m for shear/reaction IL)
Application
Computing the response due to a uniformly distributed load (UDL) applied over all or part of the span.
Exam Tips
- In board exams, a question will specify 'determine the maximum reaction/shear/moment at section C' — this is your cue to use influence lines.
- Quickly sketch the IL shape using straight-line properties of determinate beams before computing ordinates.
- For reactions: IL ordinate at the support itself is always 1; at the opposite support it is always 0.
- Remember the sign convention: positive shear IL ordinate means upward on the left face of the cut.
Key Points
- An influence line (IL) for a specific response R (reaction, shear V, or moment M) at a fixed section is a graph of R vs. the position x of a unit moving load (1 kN or 1 unit) as it traverses the span.
- The IL is NOT the same as a shear force diagram or bending moment diagram. A shear/moment diagram plots the distribution of the response along the entire beam for a FIXED load position; an IL plots the variation of a SINGLE response at one section as the load MOVES.
- IL ordinates are dimensionless (for reactions), in m⁻¹ units (for shears — actually dimensionless, kN/kN), or in m (for moments — kN·m/kN).
- Practical use: once the IL is drawn, the maximum (or minimum) value of the response due to any loading is quickly determined.
- Influence lines are applicable to both statically determinate and indeterminate structures, but this chapter focuses on determinate beams.
Definitions
Term
Influence Line (IL)
Definition
A graph showing the variation of a specific structural response (reaction, shear, moment) at a FIXED point as a unit load moves across the structure.
Importance
Core concept — every subsequent formula and procedure in this chapter is built on this definition.
Term
IL Ordinate (η)
Definition
The value of the response at the fixed section when the unit load is at position x. The unit depends on the response: dimensionless for reactions and shears (kN per kN), and meters (m) for moments (kN·m per kN).
Importance
The ordinate is what you multiply by P (or integrate with w) to get the actual response.
Term
Live Load vs. Dead Load
Definition
Live loads are moving or variable loads (vehicles, cranes, pedestrians). Dead loads are fixed and do not move. ILs are used exclusively to analyse live loads.
Importance
Understanding this distinction is fundamental — ILs have no application to dead loads because dead loads do not change position.
Section Title
1. Fundamental Concept — What Is an Influence Line?
Common Mistakes
- Confusing the influence line with the bending moment diagram — they answer entirely different questions.
- Forgetting that IL ordinates have units: for reactions/shears they are dimensionless (kN/kN); for moments they are in metres (kN·m/kN).
- Using the IL ordinate for the wrong section — always confirm which fixed section the IL is drawn for before reading ordinates.
- Forgetting that the IL is drawn by moving a UNIT load (1 kN), not the actual applied load.
Formulas
Example
L = 8 m, x = 6 m: η = (8−6)/8 = 0.25. A 100 kN load at x = 6 m gives R_A = 100 × 0.25 = 25 kN.
Formula
η_{R_A}(x) = 1 − x/L = (L−x)/L
Variables
x = position of unit load from A; L = span; η_{R_A} = IL ordinate for reaction R_A
Application
Determine R_A when any load is at position x.
Example
L = 10 m, a = 4 m, b = 6 m. Just left of C: η = −4/10 = −0.4. Just right of C: η = 6/10 = +0.6.
Formula
η_{V_C}(x) = −x/L for 0 ≤ x < a; η_{V_C}(x) = (L−x)/L for a < x ≤ L
Variables
a = distance from A to section C; x = unit load position
Application
Construct the shear IL for section C.
Example
L = 10 m, a = 4 m, b = 6 m. Peak at C: η = (4)(6)/10 = 2.4 m. A 50 kN load at C gives M_C = 50 × 2.4 = 120 kN·m.
Formula
η_{M_C}(x) = bx/L for 0 ≤ x ≤ a; η_{M_C}(x) = a(L−x)/L for a ≤ x ≤ L
Variables
a = distance A to C; b = L − a; x = unit load position
Application
Construct the moment IL for section C. Peak value at C = ab/L.
Example
P = 80 kN, a = 3 m, b = 7 m, L = 10 m: M_C,max = 80 × (3×7)/10 = 80 × 2.1 = 168 kN·m.
Formula
M_{C,max} = P × (ab/L)
Variables
P = concentrated load; a = distance A to C; b = L − a
Application
Maximum moment at section C due to a single moving concentrated load — load placed AT the section.
Exam Tips
- For a midspan section (a = b = L/2): moment IL peak = L/4; maximum moment = PL/4.
- For a quarter-point section (a = L/4, b = 3L/4): moment IL peak = (L/4)(3L/4)/L = 3L/16.
- The area of the triangular moment IL = ½ × L × (ab/L) = ab/2. For a UDL: M_C,max = w × ab/2.
- The area under the positive region of the shear IL at C = ½ × b × (b/L) = b²/(2L). The negative region area = ½ × a × (a/L) = a²/(2L).
Key Points
- For a simple beam of span L, all ILs are made up of STRAIGHT LINE segments — property of statically determinate structures.
- IL for Reaction R_A: a straight line from ordinate 1.0 at A to ordinate 0 at B.
- IL for Reaction R_B: a straight line from ordinate 0 at A to ordinate 1.0 at B.
- IL for Shear V_C (at section C, distance a from A, b = L − a from B): the IL has ordinate +b/L just to the RIGHT of C, and −a/L just to the LEFT of C. The two segments are parallel (same slope), with a unit drop (from +b/L to −a/L = drop of 1) at C.
- IL for Moment M_C (at section C): a triangle with zero at both A and B, peaking at C with ordinate ab/L.
- Sign convention for shear IL: positive ordinate means the unit load causes positive shear (upward on the left face); negative ordinate means the load causes negative shear.
- To maximise positive shear at C: place the load just to the RIGHT of C (ordinate = +b/L). To maximise negative shear: place just to the LEFT (ordinate = −a/L).
- Moment IL is always positive for loads between A and B (no overhang), so the entire span is loaded for maximum positive moment.
Definitions
Term
Section C
Definition
The fixed section of interest in the beam, located at distance a from support A and b from support B, where a + b = L.
Importance
All IL ordinates are computed and read AT this section; the peak of the moment IL is always at this section.
Term
Peak Ordinate of Moment IL
Definition
The maximum ordinate of the triangular moment IL, occurring at section C itself. Its value is ab/L (in metres).
Importance
Directly used to compute the maximum moment due to any single point load: M_max = P × ab/L.
Section Title
2. Influence Lines for a Simply Supported Beam
Common Mistakes
- For the shear IL, forgetting the SIGN: the left segment (load to the LEFT of C) gives NEGATIVE shear, and the right segment gives POSITIVE shear.
- When computing the peak of the moment IL, using a+b in the denominator instead of L (they are equal, but using L avoids confusion).
- Forgetting that the two segments of the shear IL are PARALLEL (same slope), creating a unit jump at C.
- Computing M_C = Pab/L only when the load is AT the section C — for any other position, use η_{M_C}(x) × P.
Exam Tips
- Use Müller-Breslau by inspection to SKETCH the IL quickly in exam conditions, then compute key ordinates using statics.
- For any reaction: the IL is always 1 at that support and 0 at all other supports (for determinate beams).
- Positive IL regions → load that region for maximum POSITIVE response; negative IL regions → load that region for maximum NEGATIVE (or minimum) response.
- In indeterminate structures (not the focus of this chapter), the shapes become curves — important to note for future chapters.
Key Points
- The Müller-Breslau principle states: the influence line for any linear response function (reaction, shear, moment) has the same shape as the deflected form of the structure obtained by REMOVING the restraint corresponding to that response and imposing a UNIT DISPLACEMENT (or rotation) in the direction of the response.
- For the reaction R_A: remove the vertical support at A, apply a unit upward displacement at A — the resulting elastic curve is the IL for R_A.
- For the shear V_C: introduce a shear release (internal hinge that allows relative sliding but not rotation) at C, impose a unit relative vertical displacement — the resulting shape (two straight-line segments) is the IL for V_C.
- For the moment M_C: introduce a moment release (internal hinge at C), impose a unit relative rotation — the resulting triangle shape is the IL for M_C.
- For determinate structures, the Müller-Breslau shapes are straight lines (rigid body motion of segments); for indeterminate structures, the shapes are curves (elastic deflections).
- Practical use: the principle allows you to sketch the IL shape by INSPECTION without statics calculations — critical for quickly identifying which regions to load for maximum effects.
- This principle is also referred to as the principle of virtual work or the reciprocal theorem in its foundation.
Definitions
Term
Müller-Breslau Principle
Definition
The IL for a response R has the same shape as the deflected structure obtained by releasing the restraint R and applying a unit generalized displacement corresponding to R.
Importance
Allows rapid IL shape identification by inspection — a major time-saver in examinations.
Term
Shear Release
Definition
A device that allows relative vertical displacement between two adjacent beam segments but prevents relative rotation. Used conceptually in the Müller-Breslau principle for shear IL construction.
Importance
Understanding this release mechanism clarifies why the shear IL has two parallel segments with a unit jump.
Term
Moment Release (Internal Hinge)
Definition
A connection that allows relative rotation between two segments but carries no moment. Used conceptually in the Müller-Breslau principle for moment IL construction.
Importance
Explains why the moment IL is triangular, peaking at C.
Section Title
3. Müller-Breslau Principle
Common Mistakes
- Applying a unit LOAD instead of a unit DISPLACEMENT in the Müller-Breslau principle — the displacement (or rotation) is what generates the IL shape.
- Forgetting that for DETERMINATE structures, all segments remain straight in the Müller-Breslau deflected shape.
- Not recognising that the IL ordinates must be scaled correctly — the total imposed displacement must equal 1.0 (unit displacement).
Formulas
Example
Two loads: P_1 = 30 kN at η_1 = 2.0 m, P_2 = 50 kN at η_2 = 1.5 m for moment IL. M_C = 30(2.0) + 50(1.5) = 60 + 75 = 135 kN·m.
Formula
R_total = Σ P_i × η_i
Variables
P_i = individual point loads; η_i = IL ordinate at the position of load P_i
Application
Superposition for multiple moving point loads at known positions.
Example
w = 25 kN/m, L = 10 m, a = 4 m, b = 6 m: V_C,max+ = 25 × 36/(20) = 25 × 1.8 = 45 kN.
Formula
V_{C,max+} = w × b²/(2L)
Variables
w = UDL intensity; b = L − a (distance from C to B); L = span
Application
Maximum positive shear at C due to a long UDL — load the positive (right) region only.
Example
w = 25 kN/m, L = 10 m, a = 4 m: V_C,max− = −25 × 16/20 = −20 kN.
Formula
V_{C,max−} = −w × a²/(2L)
Variables
w = UDL intensity; a = distance from A to C; L = span
Application
Maximum negative shear at C — load the negative (left) region only.
Example
w = 15 kN/m, a = 4 m, b = 6 m: M_C,max = 15(4)(6)/2 = 180 kN·m. Check: compare with a fixed UDL giving M_max differently (IL approach is for MOVING loads).
Formula
M_{C,max} = w × (ab/2) = wab/2
Variables
w = UDL; a = distance A to C; b = L − a
Application
Maximum moment at C when a UDL longer than the span covers the entire beam.
Example
w = 20 kN/m, L = 8 m: M_mid = 20 × 64/8 = 160 kN·m. (IL area = ½ × 8 × 2 = 8 m²; M = 20 × 8 = 160 ✓)
Formula
M_{mid,max} (UDL) = w × L²/8
Variables
w = UDL; L = span
Application
For midspan section (a = b = L/2), IL area = L²/8, giving the familiar formula.
Exam Tips
- Memorise the IL areas: for moment at C, area = ab/2; for positive shear region, area = b²/(2L); for negative shear region, area = a²/(2L).
- Cross-check moment IL results using statics (e.g., M_mid = wL²/8 for a full-span UDL at midspan).
- In PRC exams, the question often gives a 'lane load' (UDL) plus a 'truck load' (concentrated loads) — compute both contributions and add.
- For the maximum positive moment at ANY section (UDL, full span): M_max = w × ab/2. For midspan: a = b = L/2, so M_max = wL²/8.
Key Points
- POINT LOAD: Response = P × (IL ordinate under P). To maximise, place P at the IL peak (maximum ordinate).
- SERIES OF POINT LOADS: Response = Σ(P_i × η_i), where η_i is the IL ordinate under load P_i. To maximise, position the group so the dominant load is near the IL peak.
- UDL (longer than span): Load the entire span. Response = w × (total area under IL with the sign of interest).
- UDL (shorter than span, or partially distributed): For maximum POSITIVE response, load only the POSITIVE region of the IL. For maximum NEGATIVE response, load only the NEGATIVE region.
- COMBINED LOADING: Superpose contributions from each load component.
- Maximum shear at C — positive: load only the right segment (length b, peak b/L, area = b²/(2L)). V_C,max+ = w × b²/(2L).
- Maximum shear at C — negative (most negative): load only the left segment (length a, depth a/L, area = a²/(2L)). V_C,max− = −w × a²/(2L).
- Maximum moment at C (UDL covers full span): M_C,max = w × (ab/2).
Definitions
Term
Critical Load Position
Definition
The position of the moving load (or load group) that produces the maximum (or minimum) value of the response at the fixed section.
Importance
Identifying this position quickly is the practical goal of influence line analysis.
Section Title
4. Using Influence Lines — Point Loads and Distributed Loads
Common Mistakes
- For a UDL SHORTER than the span, loading the full span instead of only the positive or negative IL region — always cover only the region with the desired sign.
- Forgetting to include the negative IL area when computing the MAXIMUM NEGATIVE shear, or accidentally adding positive and negative contributions.
- When multiple point loads cross the beam, forgetting to check multiple load arrangements — the largest response may not always occur with the dominant load at the IL peak.
- Using the wrong area formula for the shear IL triangles (they are right triangles, not full triangles of the span).
Formulas
Example
P = 120 kN, L = 12 m: M_abs,max = 120 × 12/4 = 360 kN·m at midspan.
Formula
M_{abs,max} = PL/4 (single moving load)
Variables
P = single concentrated load; L = beam span
Application
Absolute maximum moment for a single moving load on a simply supported beam, occurring at midspan.
Example
Two 50 kN loads spaced 3 m apart, L = 12 m. R = 100 kN at 1.5 m from left load. d = 1.5 m. x_k = 6 − 1.5/2 = 5.25 m. Left load at 5.25 m from A, right load at 8.25 m from A.
Formula
Position of P_k from A: x_k = L/2 − d/2
Variables
L = span; d = distance from P_k to the resultant R of all loads on span; x_k = position of P_k from A
Application
Bisection rule — position the load group so midspan bisects the gap between P_k and R.
Example
R = 100 kN at x_R = 5.25 + 1.5 = 6.75 m from A, L = 12 m: R_A = 100(12 − 6.75)/12 = 100(5.25)/12 = 43.75 kN.
Formula
R_A = R × (L − x_R)/L
Variables
R = resultant of all loads on span; x_R = position of resultant from A; L = span
Application
Compute left reaction after positioning the load group by the bisection rule.
Example
No loads to left of left load. M = R_A × 5.25 = 43.75 × 5.25 = 229.7 kN·m.
Formula
M_{abs,max} = R_A × x_k − (moments of loads to left of P_k)
Variables
R_A = left reaction; x_k = position of critical load P_k from A; subtract all loads to the left of P_k × their distances from P_k
Application
Compute the moment at P_k by summing moments from left side.
Exam Tips
- For TWO equal loads separated by d: the absolute maximum moment occurs under either load at position L/2 − d/4 from the nearer support. M_abs,max = R_A × (L/2 − d/4).
- Always verify the answer is larger than the moment at midspan (with the resultant at midspan) — the bisection-rule position always gives a larger or equal moment.
- If the board exam gives three or more axle loads with spacings, compute the resultant position, apply bisection rule for the heaviest axle, and verify the system fits on the span.
- Quick check: for a single load, M = PL/4. For multiple loads, expect M to be somewhat less than (ΣP)L/4 because the loads are spread out.
- Memorise: 'midspan bisects P_k and R' — this one sentence contains the entire bisection rule.
Key Points
- The ABSOLUTE MAXIMUM MOMENT is the largest bending moment that can occur ANYWHERE in the beam for any position of the moving load system.
- For a SINGLE moving concentrated load P on a simple span L: the absolute maximum moment occurs at MIDSPAN and equals PL/4.
- For a SERIES of moving concentrated loads, the absolute maximum moment occurs UNDER ONE OF THE LOADS — not necessarily the heaviest, and not necessarily at midspan.
- The critical load P_k is the load under which the absolute maximum moment occurs. To identify P_k, check each load in turn.
- Positioning rule (bisection rule): Place the load group such that the beam's CENTERLINE bisects the distance between load P_k and the RESULTANT R of all loads on the span. In other words, the midpoint of the beam lies at the midpoint of the line segment between P_k and R.
- If d = distance from P_k to R, then P_k is positioned at L/2 − d/2 from the nearest support (i.e., at distance L/2 − d/2 from A if R is to the right of P_k).
- Once the system is positioned, compute reactions and moments by standard statics.
- It may be necessary to check 2–3 candidate loads and compare moments to find the true absolute maximum.
- For two equal loads spaced d apart: d/4 offset from midspan; the absolute maximum = R_A × x (where x is the position of the critical load from A).
Definitions
Term
Absolute Maximum Moment
Definition
The largest bending moment magnitude that can occur at any section of the beam under any position of the moving load system. It is a global maximum over both position along the beam AND load group position.
Importance
This is the design moment used for beam sizing — a critical exam and design quantity.
Term
Bisection Rule
Definition
The rule for positioning the moving load group to produce the absolute maximum moment: the beam midpoint must lie exactly halfway between the critical load P_k and the resultant R of all loads on the span.
Importance
The single most important formula/rule for the absolute maximum moment with multiple loads — appears frequently in PRC board exams.
Term
Resultant of Moving Loads
Definition
The single equivalent force equal to the sum of all point loads in the system, located at their centroid. Its position from any reference is found by taking moments of all loads about that reference.
Importance
Must be computed before applying the bisection rule.
Section Title
5. Absolute Maximum Moment Under Moving Loads
Common Mistakes
- Assuming the absolute maximum moment always occurs at MIDSPAN — this is only true for a single moving load. For multiple loads, it is near but not exactly at midspan.
- Forgetting to check whether all loads in the group are actually ON the span after applying the bisection rule. If a load falls off the span, exclude it and recalculate the resultant.
- Not checking multiple candidate critical loads — always check the 1–2 heaviest loads as potential P_k and compare.
- Computing the resultant position from the wrong reference point — always state clearly from which support distances are measured.
- Using the bisection rule offset in the WRONG direction — always position P_k on the SAME side of midspan as the resultant is from the overall midpoint of the load group.
Formulas
Example
V_C,max+ = 60 × (9/12) = 60 × 0.75 = 45 kN. Maximum negative shear = 60 × (−3/12) = −15 kN.
Formula
PROBLEM 1 — V_{C,max+} at quarter-point (a = L/4)
Variables
P = 60 kN, L = 12 m, a = 3 m, b = 9 m
Application
IL for shear at C: just right of C = b/L = 9/12 = 0.75; just left = −a/L = −3/12 = −0.25. Maximum positive shear = P × (b/L).
Example
M_{C,max} = 20 × 6 = 120 kN·m.
Formula
PROBLEM 2 — Max moment at quarter-point, UDL w = 20 kN/m, L = 8 m
Variables
a = 2 m (L/4), b = 6 m (3L/4), w = 20 kN/m
Application
IL peak = ab/L = (2)(6)/8 = 1.5 m. IL area = ab/2 = (2)(6)/2 = 6 m². M_max = w × IL area.
Example
R_A = 100(10 − 5.4)/10 = 100(4.6)/10 = 46 kN. M under P_2 = 46(4.6) − 40(4.6 − 2.6) = 211.6 − 80 = 131.6 kN·m. Check under P_1: M = 46(2.6) = 119.6 kN·m. Absolute max = 131.6 kN·m (under P_2). ✓
Formula
PROBLEM 3 — Absolute max moment, two loads 40 kN and 60 kN, spacing 2 m, L = 10 m
Variables
P_1 = 40 kN (left), P_2 = 60 kN (right), spacing = 2 m
Application
R = 100 kN. x_R from P_1: (60 × 2)/100 = 1.2 m. Critical load: P_2 (heavier). d = distance from P_2 to R = 2 − 1.2 = 0.8 m. Position: x_{P2} = L/2 − d/2 = 5 − 0.4 = 4.6 m from A. x_{P1} = 4.6 − 2 = 2.6 m from A. x_R = 4.6 + 0.8 = 5.4 m from A.
Example
Note: P_1 AT A gives η = 1 and P_2 at 3 m gives η = (10−3)/10 = 0.7. R_A,max = 85 kN.
Formula
PROBLEM 4 — Max R_A, two loads P = 50 kN each, spacing 3 m, L = 10 m
Variables
IL for R_A is maximum when loads are as close to A as possible. Both loads on span: P_1 at x=0 (η=1), P_2 at x=3 m (η=7/10=0.7).
Application
R_A = P_1 × η_1 + P_2 × η_2 = 50(1) + 50(0.7) = 50 + 35 = 85 kN.
Exam Tips
- In PRC boards, these four problem types recur frequently. Practice each type until you can solve it in under 5 minutes.
- For PROBLEM 3 type questions, always write out: (1) find R and its position, (2) identify P_k, (3) apply bisection rule, (4) compute R_A, (5) compute M under P_k.
- Double-check the moment calculation by also computing from the RIGHT side (using R_B) and verify you get the same answer.
- For maximum reactions, place the larger loads closest to the support in question — this is the IL approach applied intuitively.
Key Points
- PROBLEM 1: Max shear at quarter-point, single moving load. Draw IL, read ordinate, multiply by load.
- PROBLEM 2: Max moment at a section, UDL. Use IL area formula M = wab/2.
- PROBLEM 3: Absolute maximum moment, two loads. Apply bisection rule, compute reactions, then moment under critical load.
- PROBLEM 4: Max reaction at A, series of loads.
Section Title
6. Worked Board-Style Problems
Common Mistakes
- In PROBLEM 3, not verifying both loads are on the span after positioning. x_{P1} = 2.6 m > 0 ✓ and x_{P2} = 4.6 m < 10 m ✓.
- In PROBLEM 1, confusing which ordinate (just left or just right of C) applies — the unit load position determines the response, not the section side.
- In PROBLEM 4, placing the loads too far to the left so P_1 falls off the span — always keep all loads within 0 ≤ x ≤ L.
Connections
- PREREQUISITE — Statics and Reactions: You must be able to compute support reactions for any load position before drawing IL ordinates. If you struggle with reaction calculations, review Chapter 2 (Equilibrium of Structures) first.
- LINK TO Shear and Moment Diagrams (Chapter 4): IL and BMD use the same sign conventions for shear and moment, but are constructed for entirely different purposes. Understanding BMDs makes IL interpretation easier.
- LINK TO Moving Load Design Codes: In AASHTO LRFD (used in Philippine bridge design via DPWH guidelines), truck loads and lane loads are treated as moving loads. IL analysis is the theoretical basis for load distribution factors in bridge codes.
- LINK TO Indeterminate Structures (Chapter 9): The Müller-Breslau principle extends to indeterminate beams and frames, where IL shapes become curves rather than straight lines. This chapter's straight-line ILs are the building block for those more complex cases.
- LINK TO Structural Design (NSCP 2015 Section 2): NSCP 2015 (National Structural Code of the Philippines) defines live load patterns for floors and beams. The concept of loading only the critical regions of the IL directly corresponds to NSCP provisions for pattern loading in continuous beams.
- LINK TO Truss Analysis: ILs can also be constructed for forces in individual truss members. The method (unit panel load approach) is an extension of the beam IL concepts covered here.
- LINK TO Deflection and Virtual Work: The theoretical foundation of Müller-Breslau is the principle of virtual work / Maxwell's reciprocal theorem — concepts covered in the energy methods chapter.
- RA 544 (Civil Engineering Law): As a registered civil engineer, you are responsible for correctly analysing structures under moving loads. The absolute maximum moment from IL analysis directly informs the design moment used in member sizing per NSCP 2015.
Exam Strategy
For PRC board exam questions on influence lines and moving loads, follow this systematic approach: (1) IDENTIFY the response asked — reaction, shear at C, moment at C, or absolute maximum moment. (2) SKETCH the IL — draw the beam, mark the fixed section, and sketch the IL shape using the key ordinates (0, 1, ab/L, b/L, −a/L). Label all key ordinates. (3) APPLY the load — for a point load, multiply P by the IL ordinate at the load position; for a UDL, multiply w by the appropriate IL area. (4) MAXIMISE — for max positive response, load the positive IL region; for max negative, load the negative region; for a single point load, place it at the IL peak. (5) ABSOLUTE MAX MOMENT with multiple loads — compute the resultant, identify the critical load, apply the bisection rule, compute R_A, then compute M under the critical load. TIME MANAGEMENT: IL questions are moderately difficult but very systematic. Allocate about 4–6 minutes per IL problem in the exam. Memorise the key formulas (ab/L for moment IL peak; b²/2L and a²/2L for shear IL areas; PL/4 for single-load absolute max; bisection rule for series of loads). Sketch ILs neatly — a clear sketch prevents sign errors. If the numbers seem messy, re-check which segment of the IL you are using. Always verify by checking if the response you computed makes physical sense (e.g., maximum positive shear at C cannot exceed R_B).
Quick Review Questions
A unit load moves across a 12 m simple beam. What is the IL ordinate for the reaction at A when the unit load is 4 m from A?
The IL for R_A is a straight line from 1.0 at A to 0 at B. At x = 4 m: η = (L − x)/L = (12 − 4)/12 = 2/3 ≈ 0.667.
For a 10 m simple beam with section C at 4 m from A, what is the peak ordinate of the moment IL at C?
a = 4 m, b = L − a = 6 m. Peak of moment IL = ab/L = 24/10 = 2.4 m. This is located AT section C.
What is the maximum moment at midspan of a 6 m simple beam due to a single moving load of 90 kN?
For midspan, a = b = 3 m. Peak ordinate = ab/L = 9/6 = 1.5 m = L/4. M_max = P × (L/4) = 90 × 1.5 = 135 kN·m. The load must be placed AT midspan.
A long UDL of 18 kN/m crosses a 10 m simple beam. What is the maximum positive shear at a section 3 m from support A?
a = 3 m, b = 7 m. Positive shear IL area (right triangle to the right of C) = b²/(2L) = 49/20 = 2.45 m. V_C,max+ = 18 × 2.45 = 44.1 kN. Load the right segment (b = 7 m) only.
What is the maximum moment at the third-point of a 9 m beam (section at 3 m from A) due to w = 12 kN/m covering the full span?
a = 3 m, b = 6 m. IL area = ab/2 = 18/2 = 9 m². M = 12 × 9 = 108 kN·m.
State the bisection rule for absolute maximum moment with a series of moving loads.
P_k is the load under which the absolute maximum moment is expected (usually the heaviest load near the centre of gravity of the group). Once positioned by this rule, compute R_A and then M under P_k by statics.
For a single moving load P on a simple beam of span L, where does the absolute maximum moment occur, and what is its magnitude?
The moment IL for midspan has the highest peak (L/4) of all sections. Placing the single load at midspan maximises the moment, giving M = P × (L/4) = PL/4.
Two 70 kN loads spaced 4 m apart move across a 14 m beam. Where is the absolute maximum moment, and how far is the critical load from the nearer support?
R = 140 kN at midpoint of the two loads (2 m from each). d = 2 m (distance from either load to resultant). x_k = L/2 − d/2 = 7 − 1 = 6 m from A. The absolute max moment is under the load at 6 m. R_A = 140(14 − 8)/14 = 60 kN. M = 60 × 6 = 360 kN·m.
What is the key difference between an influence line and a bending moment diagram?
This distinction is the most fundamental conceptual point in this chapter. Always ask: 'Am I fixing the section (IL) or fixing the load (BMD)?'
For a section C at a = 2 m from A on a 10 m beam, a UDL w = 30 kN/m shorter than the span must be placed for maximum negative shear. Where do you place it and what is the value?
The negative region of the shear IL at C is from A to just left of C (length a = 2 m), with max ordinate −a/L = −0.2 at A. Area = ½(2)(0.2) = 0.2 m. V_C,max− = −30 × 0.2 = −6 kN.
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