CELE Structural Theory & Analysis — Influence Lines and Moving LoadsExam Answer Templates

Marks

1

Topic

Definition of Influence Lines

Difficulty

easy

Template Id

T1

Examiner Tip

The key contrast that earns the mark is: SECTION FIXED, LOAD MOVES — the opposite of a shear/moment diagram where the LOAD IS FIXED and the SECTION VARIES. Include this contrast for extra clarity.

Model Answer

An influence line is a graph that shows the variation of a specific structural response (reaction, shear, or moment at a fixed section) as a unit load moves across the span of the structure.

Question Type

very_short_answer

Answer Structure

• One complete sentence: define IL as a graph/plot of a specific response at a FIXED section as a UNIT LOAD MOVES across the span [1 mark]

Scoring Breakdown

Common Mark Deductions

• Describing it as a 'shear or moment diagram' — this is the most common conceptual error and earns zero marks
• Omitting 'unit load' — saying just 'a load moves' is incomplete
• Not specifying that the section is fixed while the load position varies

Key Phrases To Include

• fixed section
• unit load moves
• variation of a response
• reaction, shear, or moment

Marks

1

Topic

Müller-Breslau Principle

Difficulty

easy

Template Id

T2

Examiner Tip

Name the principle explicitly — 'Müller-Breslau Principle' — because the examiner awards the mark partly for citing the correct theorem by name.

Model Answer

The Müller-Breslau Principle states that the influence line for any force response has the same shape as the deflected form of the structure when the restraint corresponding to that response is removed and a unit displacement is imposed in the direction of that response.

Question Type

very_short_answer

Answer Structure

• One sentence: state IL shape = deflected shape after releasing the response constraint and imposing unit displacement [1 mark]

Scoring Breakdown

Common Mark Deductions

• Saying 'apply a unit load at the section' instead of 'impose a unit displacement at the released constraint'
• Not mentioning 'release the restraint' — just saying 'deform the structure' is insufficient

Key Phrases To Include

• Müller-Breslau Principle
• deflected form
• release the restraint
• unit displacement
• same shape as

Marks

2

Topic

Influence Line for Moment at a Section

Difficulty

easy

Template Id

T3

Examiner Tip

A small labeled sketch of the triangular IL showing the peak ab/L at C and zeros at A and B will earn the first mark even if the formula derivation has a minor error.

Model Answer

The influence line for moment at section C is a triangle with the following ordinates: • At support A (x = 0): η = 0 • At section C (x = a): η_peak = ab/L [peak ordinate] • At support B (x = L): η = 0 For a unit load at position x (0 ≤ x ≤ a): η = (b/L)·x For a unit load at position x (a ≤ x ≤ L): η = (a/L)·(L − x) The peak ordinate at C is: η_max = ab/L

Question Type

short_answer

Answer Structure

• Line 1–2: Identify IL shape as a triangle; state boundary conditions (zero at both supports) [1 mark]
• Line 3–4: State the peak ordinate formula η = ab/L at section C, with correct derivation or labeling [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing η = PaB/L instead of ab/L (including P — the IL is for a unit load, P = 1)
• Not defining a and b in terms of the beam geometry
• Drawing the IL with a negative region — the moment IL for a simple beam is always positive (above the baseline)

Key Phrases To Include

• triangular influence line
• peak ordinate = ab/L
• zero at supports
• a = distance from A to C
• b = distance from C to B

Marks

2

Topic

Influence Lines vs. Bending Moment Diagram

Difficulty

easy

Template Id

T4

Examiner Tip

Use a two-column comparison format in your answer: left column for BMD, right column for IL. Examiners scan for structured comparisons and reward clarity.

Model Answer

Bending Moment Diagram (BMD): The load position is FIXED; the bending moment is plotted at EVERY SECTION along the beam for that fixed load configuration. Influence Line (IL) for Moment: The SECTION is FIXED; the moment at that section is plotted as a UNIT LOAD MOVES to every position along the beam. Key distinction: A BMD answers 'what is the moment everywhere for this load?' while an IL answers 'how does the moment at this point change as the load moves?'

Question Type

short_answer

Answer Structure

• Line 1: Define BMD — fixed load, moment varies along all sections [1 mark]
• Line 2: Define IL — fixed section, moment varies as unit load moves; state the key distinction [1 mark]

Scoring Breakdown

Common Mark Deductions

• Saying both are 'graphs of bending moment' without specifying what varies in each
• Confusing direction: stating IL varies the section (incorrect)

Key Phrases To Include

• load position fixed
• section fixed
• unit load moves
• moment at every section vs. moment at one section

Marks

3

Topic

Influence Line for Shear — Single Moving Load

Difficulty

medium

Template Id

T5

Examiner Tip

A quick labeled sketch of the IL for shear — showing the +0.75 block to the right, the −0.25 block to the left, and the unit jump at C — earns the first mark even before any computation. Always draw the IL before computing.

Model Answer

Given: P = 60 kN, L = 12 m, section C at a = 3 m from A, b = 9 m from B. Step 1 — IL for shear at C: The IL for shear at section C of a simple beam consists of two parallel line segments: • For load to the RIGHT of C (a < x ≤ L): η_V = +b/L = +9/12 = +0.75 [positive shear region] • For load to the LEFT of C (0 ≤ x < a): η_V = −a/L = −3/12 = −0.25 [negative shear region] There is a unit jump discontinuity at C. Step 2 — Maximum positive shear: Place P just to the right of C (at the peak of the positive region): V_max(+) = P × η_V(+) = 60 × 0.75 = +45 kN Step 3 — Maximum negative shear: Place P just to the left of C (at the peak of the negative region): V_max(−) = P × η_V(−) = 60 × (−0.25) = −15 kN Answer: Maximum positive shear at C = +45 kN; Maximum negative shear at C = −15 kN.

Question Type

numerical

Answer Structure

• Step 1: Draw and describe the IL for shear at C — positive ordinate (+b/L) to the right, negative ordinate (−a/L) to the left [1 mark]
• Step 2: Compute maximum positive shear by placing P at the positive IL peak [1 mark]
• Step 3: Compute maximum negative shear by placing P at the negative IL peak; state both answers with units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using a/L for the positive region and b/L for the negative region — these are swapped
• Reporting only the positive shear and ignoring the maximum negative shear
• Omitting the sign convention for shear

Key Phrases To Include

• IL for shear has two segments with unit jump at C
• positive ordinate = +b/L to the right
• negative ordinate = −a/L to the left
• place load at peak ordinate for maximum response
• kN (units)

Marks

3

Topic

Influence Line for Moment — Distributed Moving Load

Difficulty

medium

Template Id

T6

Examiner Tip

Note that M = w × A_IL = 20 × 9.375 = 187.5 kN·m. You can verify with the formula M = wab²/L + wa²b/L... but the IL method is cleaner. Always cross-check with an alternative formula if time permits — examiners reward self-checking.

Model Answer

Given: w = 20 kN/m, L = 10 m, section C at a = 2.5 m from A, b = 7.5 m from B. Step 1 — IL for moment at C: The IL is a triangle with: • Peak at C: η_peak = ab/L = (2.5)(7.5)/10 = 1.875 m • Zero at both supports A and B. Step 2 — Area of the IL (entire positive triangle): The IL for moment at C on a simple beam is entirely positive. A_IL = (1/2)(base)(height) = (1/2)(10)(1.875) = 9.375 m² Step 3 — Maximum moment (load the entire span): M_max = w × A_IL = 20 × 9.375 = 187.5 kN·m Answer: Maximum moment at C = 187.5 kN·m

Question Type

numerical

Answer Structure

• Step 1: Compute peak ordinate of triangular moment IL: η = ab/L = 1.875 m [1 mark]
• Step 2: Compute area of IL triangle: A = (1/2)(L)(η) = 9.375 m² [1 mark]
• Step 3: Multiply by w to get maximum moment; state answer with units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using M = wL²/8 directly without showing the IL method (the question asks for the IL approach)
• Computing A_IL = (1/2)(a)(η) or (1/2)(b)(η) instead of (1/2)(L)(η) for the full triangle
• Omitting units (kN·m)

Key Phrases To Include

• triangular IL for moment
• peak ordinate = ab/L
• IL area = (1/2)(base)(height)
• response = w × IL area
• load entire positive region for maximum

Marks

3

Topic

Absolute Maximum Moment — Single Moving Load

Difficulty

medium

Template Id

T7

Examiner Tip

The word 'derive' in a question means full mathematical development from first principles. Simply quoting the formula PL/4 without derivation will earn zero marks regardless of correctness. Show every algebraic step.

Model Answer

Let the load P be at distance x from A. By statics: • R_A = P(L − x)/L • Moment under load: M(x) = R_A · x = P(L − x)x/L = P(Lx − x²)/L To maximize, differentiate M with respect to x and set to zero: dM/dx = P(L − 2x)/L = 0 → L − 2x = 0 → x = L/2 The maximum moment occurs at MIDSPAN (x = L/2): M_abs_max = P(L − L/2)(L/2)/L = P(L/2)(L/2)/L = PL/4 Conclusion: For a single moving load, the absolute maximum moment = PL/4, occurring at midspan.

Question Type

short_answer

Answer Structure

• Step 1: Write R_A and moment M(x) as functions of load position x [1 mark]
• Step 2: Differentiate M(x) with respect to x, set dM/dx = 0, solve for x = L/2 [1 mark]
• Step 3: Substitute x = L/2 to get M_abs_max = PL/4; state the conclusion [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not deriving the result — just stating PL/4 without derivation earns 0 for a 'derive' question
• Differentiating with respect to P instead of x
• Not equating dM/dx = 0 or not solving for x explicitly

Key Phrases To Include

• R_A = P(L−x)/L
• M(x) = Px(L−x)/L
• dM/dx = 0
• x = L/2 (midspan)
• M_abs_max = PL/4

Marks

2

Topic

Absolute Maximum Moment — Series of Moving Loads

Difficulty

medium

Template Id

T8

Examiner Tip

Memorize the exact phrasing: 'the centerline of the beam bisects the distance between the resultant of all loads on the span and the load under which the maximum moment is sought.' Boards have asked this verbatim.

Model Answer

The absolute maximum bending moment in a simply supported beam under a series of moving loads occurs UNDER ONE OF THE CONCENTRATED LOADS. The critical load position satisfies the following criterion: The beam's centerline (midspan point) must bisect the distance between the resultant of ALL loads on the beam and the specific concentrated load under which the maximum moment is sought. In practice: compute the resultant R and its location. For each candidate load P_i, position the load group so that the midspan is midway between P_i and R. Then compute the moment under P_i and select the largest value.

Question Type

short_answer

Answer Structure

• Sentence 1: State that abs max moment occurs under one of the loads [0.5 mark implied]
• Sentence 2: State the bisection criterion — midspan bisects the distance between the critical load and the resultant [1 mark]
• Sentence 3: Describe practical procedure — check each load as candidate, pick the maximum [0.5 mark implied]

Scoring Breakdown

Common Mark Deductions

• Stating 'place the resultant at midspan' — this is the rule for a single load, not a series
• Not mentioning that each load must be checked as a candidate
• Confusing 'resultant on the span' with the resultant of all loads in the train (only loads on the span are considered)

Key Phrases To Include

• absolute maximum moment occurs under one of the loads
• midspan bisects the distance
• resultant of all loads on the span
• critical load and resultant straddle the midspan equally

Marks

2

Topic

Influence Line for Moment — Single Concentrated Load

Difficulty

easy

Template Id

T9

Examiner Tip

For single load problems, always state: 'The maximum moment at C is obtained by placing P directly at C.' This one sentence of reasoning can be the difference between 1 and 2 marks.

Model Answer

Given: L = 8 m, a = 2 m (from A), b = L − a = 8 − 2 = 6 m, P = 50 kN. Influence line for M at C: triangular, peak ordinate at C: η_peak = ab/L = (2)(6)/8 = 1.5 m Maximum moment (place P at section C, i.e., at x = 2 m from A): M_max = P × η_peak = 50 × 1.5 = 75 kN·m Answer: M_max at C = 75 kN·m

Question Type

numerical

Answer Structure

• Step 1: Identify a, b, and compute peak IL ordinate η = ab/L = 1.5 m [1 mark]
• Step 2: Compute M_max = P × η = 50 × 1.5 = 75 kN·m; state load position and include units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using M = Pab/L correctly but not mentioning that the load is placed at C — the positioning justification earns a mark
• Arithmetic error in computing b = L − a
• No units on the final answer

Key Phrases To Include

• peak ordinate η = ab/L
• place load at section C for maximum moment
• M_max = P × η
• 75 kN·m

Marks

3

Topic

Influence Line for Reaction — Distributed Moving Load

Difficulty

medium

Template Id

T10

Examiner Tip

The IL for any reaction at a simple support is a TRIANGLE (not a rectangle). This is the most frequently confused shape. The peak ordinate is 1 (dimensionless) at the support itself, and zero at the opposite support.

Model Answer

Given: w = 12 kN/m, L = 10 m. Step 1 — IL for reaction R_A: The IL for R_A is a straight line: • At x = 0 (support A): η = 1 • At x = L (support B): η = 0 This is a right triangle with base L = 10 m and peak ordinate of 1 (dimensionless). Step 2 — Area of the positive IL region: To maximize R_A, load the ENTIRE span (since the IL is entirely positive): A_IL = (1/2)(L)(1) = (1/2)(10)(1) = 5 m Step 3 — Maximum reaction: R_A(max) = w × A_IL = 12 × 5 = 60 kN Verification: For UDL over full span, R_A = wL/2 = 12(10)/2 = 60 kN ✓ Answer: Maximum R_A = 60 kN

Question Type

numerical

Answer Structure

• Step 1: Describe the IL for R_A — triangular, peak = 1 at A, zero at B [1 mark]
• Step 2: Compute IL area = (1/2)(L)(1) = 5 m; note that full span is loaded [1 mark]
• Step 3: R_A(max) = w × A_IL = 60 kN with verification [1 mark]

Scoring Breakdown

Common Mark Deductions

• Drawing the IL as a rectangle with ordinate 1 over the full span (incorrect shape)
• Not explaining why the full span is loaded
• Computing A_IL = L × 1 = 10 m instead of the triangular area 5 m

Key Phrases To Include

• IL for R_A is a triangle from 1 at A to 0 at B
• load the entire span for maximum reaction
• R_A = w × IL area
• A_IL = (1/2)(L)(1)
• 60 kN

Marks

5

Topic

Absolute Maximum Moment — Series of Moving Loads

Difficulty

hard

Template Id

T11

Examiner Tip

For 5-mark moving load problems, organize your solution into clearly numbered steps and label each candidate check (Candidate A, Candidate B). Examiners award partial marks for each correctly computed step, so a structured layout protects your score even if you make an arithmetic error in one step.

Model Answer

Given: P₁ = 40 kN, P₂ = 60 kN, spacing d = 3 m, span L = 12 m. Step 1 — Resultant and its location: R = P₁ + P₂ = 40 + 60 = 100 kN Taking moments from P₁: x_R = (60 × 3)/100 = 1.8 m from P₁ (i.e., the resultant is 1.8 m from P₁ toward P₂). Step 2 — Check each load as the critical load: Candidate A: Maximum moment under P₁ Bisection criterion: midspan (x = 6 m from A) bisects the distance between P₁ and R. Distance from P₁ to R = 1.8 m; half = 0.9 m. Place P₁ at 6 − 0.9 = 5.1 m from A → R is at 5.1 + 1.8 = 6.9 m from A. P₂ is at 5.1 + 3 = 8.1 m from A. (Both loads on span — valid.) R_A = R × (L − x_R)/L = 100 × (12 − 6.9)/12 = 100 × 5.1/12 = 42.5 kN M under P₁ = R_A × 5.1 = 42.5 × 5.1 = 216.75 kN·m Candidate B: Maximum moment under P₂ Bisection criterion: midspan bisects the distance between P₂ and R. Distance from P₂ to R = 3 − 1.8 = 1.2 m; half = 0.6 m. Place P₂ at 6 − 0.6 = 5.4 m from A → P₁ is at 5.4 − 3 = 2.4 m from A. R is at 5.4 + (−1.2) ... corrected: R at 2.4 + 1.8 = 4.2 m from A. ✗ Let P₂ be candidate: R is at 5.4 − 1.2 = 4.2 m from A → R_A = 100 × (12 − 4.2)/12 = 100 × 7.8/12 = 65 kN M under P₂ = R_A × 5.4 − P₁ × (5.4 − 2.4) = 65 × 5.4 − 40 × 3 = 351 − 120 = 231 kN·m Step 3 — Select the absolute maximum: M under P₁ = 216.75 kN·m M under P₂ = 231 kN·m Absolute Maximum Moment = 231 kN·m (occurring under P₂ = 60 kN).

Question Type

numerical

Answer Structure

• Step 1: Compute resultant R = 100 kN and locate it (1.8 m from P₁) [1 mark]
• Step 2a: Apply bisection criterion for P₁ as candidate; find positions of all loads and R_A [1 mark]
• Step 2b: Compute M under P₁ using equilibrium [1 mark]
• Step 3a: Apply bisection criterion for P₂ as candidate; find positions and R_A [1 mark]
• Step 3b: Compute M under P₂; compare and select the absolute maximum [1 mark]

Scoring Breakdown

Common Mark Deductions

• Checking only one load as the candidate and missing the other — always check ALL loads
• Incorrectly computing the location of the resultant (moments from wrong reference point)
• Applying bisection criterion: placing the RESULTANT at midspan instead of bisecting the midspan between load and resultant
• Computing R_A without correctly accounting for loads to the left of the candidate load in the moment calculation
• Not selecting and comparing both candidates — the examiner expects a conclusion

Key Phrases To Include

• resultant R = ΣP
• location of resultant by moments
• bisection criterion: midspan bisects distance between candidate load and resultant
• check each load as candidate
• moment under candidate load by equilibrium
• absolute maximum is the largest among all candidates

Marks

5

Topic

Influence Line for Shear — Distributed Moving Load

Difficulty

hard

Template Id

T12

Examiner Tip

For UDL problems with shear IL, remember: 'positive shear → load the positive IL region only (right of C for a simple beam); negative shear → load the negative IL region only (left of C).' Loading both regions simultaneously gives neither the maximum positive nor maximum negative shear.

Model Answer

Given: w = 25 kN/m, L = 16 m, a = 4 m from A, b = 12 m from B, section C at 4 m. Step 1 — IL for shear at C: • For loads to the RIGHT of C: η_V = +b/L = +12/16 = +0.75 (positive region) • For loads to the LEFT of C: η_V = −a/L = −4/16 = −0.25 (negative region) • Jump at C from −0.25 to +0.75 (unit = 1). Step 2 — Area of positive IL region (right of C): Positive region spans from C to B: length = b = 12 m, peak = +0.75 A+ = (1/2)(12)(0.75) = 4.5 m Step 3 — Maximum positive shear (load the positive region only): V_max(+) = w × A+ = 25 × 4.5 = 112.5 kN Step 4 — Area of negative IL region (left of C): Negative region spans from A to C: length = a = 4 m, peak = 0.25 A− = (1/2)(4)(0.25) = 0.5 m Step 5 — Maximum negative shear (load the negative region only): V_max(−) = w × A− = 25 × 0.5 = 12.5 kN (in magnitude) V_max(−) = −12.5 kN (by sign convention) Answers: (a) Maximum positive shear at C = +112.5 kN (b) Maximum negative shear at C = −12.5 kN

Question Type

numerical

Answer Structure

• Step 1: Draw and describe IL for shear at C — state both ordinate values (+0.75 and −0.25) [1 mark]
• Step 2: Compute area of positive region A+ = (1/2)(b)(+b/L) = 4.5 m [1 mark]
• Step 3: V_max(+) = w × A+ = 112.5 kN [1 mark]
• Step 4: Compute area of negative region A− = (1/2)(a)(a/L) = 0.5 m [1 mark]
• Step 5: V_max(−) = −w × A− = −12.5 kN; state both answers with units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Loading the ENTIRE span for both maximum positive and negative shear — you must load only the corresponding sign region
• Using rectangular areas instead of triangular areas for the IL regions
• Ignoring the negative shear calculation (answering only part a)

Key Phrases To Include

• load only the positive region for maximum positive shear
• load only the negative region for maximum negative shear
• positive ordinate = +b/L
• negative ordinate = −a/L
• triangular IL areas
• V = w × IL area

Marks

2

Topic

Müller-Breslau Principle — Application

Difficulty

easy

Template Id

T13

Examiner Tip

Always follow the three-step Müller-Breslau procedure: (1) identify and release the constraint, (2) impose a unit deformation, (3) the deformed shape IS the IL. Naming this procedure explicitly earns the methodology mark.

Model Answer

Applying the Müller-Breslau Principle to R_B: 1. Remove the vertical constraint at B (conceptually release support B as a roller). 2. Impose a unit upward displacement at B in the direction of R_B. 3. The resulting deflected shape of the beam is: the beam rotates rigidly about A (since A remains fixed as a pin), producing a straight line that rises from 0 at A to +1 at B. Conclusion: The IL for R_B is a straight line with ordinate 0 at A and ordinate +1 at B — identical to what statics gives.

Question Type

short_answer

Answer Structure

• Step 1: State the Müller-Breslau step — release constraint at B, impose unit upward displacement [1 mark]
• Step 2: Describe the resulting deflected shape — straight line from 0 at A to +1 at B; state conclusion [1 mark]

Scoring Breakdown

Common Mark Deductions

• Describing the deflected shape as a curve (it is a straight line for a simple beam reaction)
• Not stating that the displacement imposed equals unity (unit displacement)

Key Phrases To Include

• Müller-Breslau Principle
• release constraint at B
• unit upward displacement at B
• deflected shape is the IL
• straight line from 0 at A to 1 at B

Marks

5

Topic

Absolute Maximum Moment — Three Moving Loads

Difficulty

hard

Template Id

T14

Examiner Tip

In a 5-mark problem with three loads, the examiner expects ALL three candidates to be checked. Awarding marks is typically: 1 mark per candidate check + 1 mark for resultant + 1 mark for correct final selection. Even if you get the wrong numerical answer for one candidate, you earn partial marks for the correct procedure.

Model Answer

Given: L = 15 m, P₁ = 30 kN at x₁, P₂ = 50 kN at x₁+2, P₃ = 40 kN at x₁+4; spacing d = 2 m. Step 1 — Resultant: R = 30 + 50 + 40 = 120 kN Location of R from P₁: x_R = [50(2) + 40(4)]/120 = [100 + 160]/120 = 260/120 = 2.167 m from P₁ Step 2 — Check P₂ as candidate (often governs for heaviest central load): Distance from P₂ to R: x_R − 2 = 2.167 − 2 = 0.167 m (R is to the RIGHT of P₂) Bisection: midspan (7.5 m from A) must bisect 0.167 m → P₂ at 7.5 − 0.167/2 = 7.5 − 0.083 = 7.417 m from A R at 7.5 + 0.083 = 7.583 m from A Load positions: P₁ at 7.417−2 = 5.417 m, P₂ at 7.417 m, P₃ at 7.417+2 = 9.417 m (all within span ✓) R_A = 120(15 − 7.583)/15 = 120(7.417)/15 = 59.33 kN M under P₂ = R_A(7.417) − P₁(7.417 − 5.417) = 59.33(7.417) − 30(2) = 440.1 − 60 = 380.1 kN·m Step 3 — Check P₁ as candidate: Bisection: midspan bisects distance between P₁ (at x₁) and R (at x₁+2.167): offset = 2.167/2 = 1.083 m P₁ at 7.5 − 1.083 = 6.417 m, R at 7.5 + 1.083 = 8.583 m Positions: P₁ at 6.417 m, P₂ at 8.417 m, P₃ at 10.417 m (all on span ✓) R_A = 120(15 − 8.583)/15 = 120(6.417)/15 = 51.33 kN M under P₁ = R_A(6.417) = 51.33(6.417) = 329.4 kN·m Step 4 — Check P₃ as candidate: Distance from P₃ to R: 4 − 2.167 = 1.833 m (R is LEFT of P₃) Bisection: P₃ at 7.5 − 1.833/2 = 7.5 − 0.917 = 6.583 m; R at 6.583 − 1.833 = 4.75 m Positions: P₁ at 6.583−4 = 2.583 m, P₂ at 4.583 m, P₃ at 6.583 m (all on span ✓) R_A = 120(15 − 4.75)/15 = 120(10.25)/15 = 82 kN M under P₃ = R_A(6.583) − P₁(4) − P₂(2) = 82(6.583) − 30(4) − 50(2) = 539.8 − 120 − 100 = 319.8 kN·m Step 5 — Select absolute maximum: M under P₁ = 329.4 kN·m M under P₂ = 380.1 kN·m ← MAXIMUM M under P₃ = 319.8 kN·m Absolute Maximum Moment ≈ 380 kN·m, occurring under P₂ with P₂ at approximately 7.42 m from A.

Question Type

numerical

Answer Structure

• Step 1: Compute R = 120 kN and locate resultant at 2.167 m from P₁ [1 mark]
• Step 2: Apply bisection criterion for P₂; compute load positions and R_A; compute M under P₂ = 380.1 kN·m [1.5 marks]
• Step 3: Apply bisection criterion for P₁; compute M under P₁ = 329.4 kN·m [1 mark]
• Step 4: Apply bisection criterion for P₃; compute M under P₃ = 319.8 kN·m [1 mark]
• Step 5: Compare all three; state absolute maximum M = 380.1 kN·m under P₂ [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Checking only the heaviest load (P₂) and not verifying the others — may miss the governing load
• Locating the resultant incorrectly (moments from wrong reference)
• Not verifying that all loads remain on the span after applying the bisection criterion
• Computing moment by summing R_A × distance but forgetting to subtract contributions of loads to the LEFT of the candidate

Key Phrases To Include

• R = ΣP = 120 kN
• resultant location by moments
• bisection criterion for each candidate load
• check all three loads
• moment by equilibrium under candidate load
• absolute maximum under P₂

Marks

3

Topic

Influence Line — Short UDL (Partial Loading)

Difficulty

hard

Template Id

T15

Examiner Tip

When the moving UDL is SHORTER than the span, position it to cover the peak of the IL (center it at the section for moment). The covered area is a TRAPEZOID, not the full triangle. This is the most commonly missed concept for UDL-shorter-than-span problems on the board exam.

Model Answer

Given: w = 18 kN/m, load length = 4 m (< span), L = 10 m, section at midspan (a = b = 5 m). Step 1 — IL for moment at midspan: Triangular IL with peak at midspan: η_peak = ab/L = (5)(5)/10 = 2.5 m Base = 10 m (from A to B). Step 2 — Portion of IL to load: Since the load is SHORTER than the span, load it over the PEAK of the IL to capture the maximum area under 4 m of span. The peak is at midspan (x = 5 m). Load symmetrically: from x = 3 m to x = 7 m (2 m each side of midspan). Step 3 — Area under the IL for the 4 m loaded segment: The IL is linear on each side of midspan. Left side: from x=3 to x=5 (2 m), ordinates at x=3: η = (3)(7)/10 = 2.1 m; at x=5: η = 2.5 m. Trapezoid area (left): A_left = (1/2)(2.1 + 2.5)(2) = 4.6 m Right side: by symmetry, η at x=7: η = (7)(3)/10 = 2.1 m; at x=5: η = 2.5 m. Trapezoid area (right): A_right = (1/2)(2.1 + 2.5)(2) = 4.6 m Total IL area over loaded region: A_IL = 4.6 + 4.6 = 9.2 m² Step 4 — Maximum moment: M_max = w × A_IL = 18 × 9.2 = 165.6 kN·m Answer: Maximum moment at midspan = 165.6 kN·m

Question Type

numerical

Answer Structure

• Step 1: Compute IL peak at midspan: η = ab/L = 2.5 m [1 mark]
• Step 2: Identify the critical position of the 4 m load — centered at midspan to capture maximum IL area [0.5 mark]
• Step 3: Compute trapezoidal IL area over 4 m loaded segment = 9.2 m² [1 mark]
• Step 4: M_max = w × A_IL = 165.6 kN·m with units [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using the full-span IL area (12.5 m²) as if the load covers the whole span — the load is only 4 m long
• Placing the 4 m load starting from A instead of centering at the peak
• Using triangular areas instead of trapezoidal areas for the loaded segment (only triangular if loaded from the support to the peak)

Key Phrases To Include

• load shorter than span — position for maximum IL area
• center the load over the IL peak for maximum moment
• trapezoidal IL area over the loaded segment
• M = w × IL area