CELE Strength of Materials — Beam DeflectionsExam Answer Templates

Marks

1

Topic

Elastic Curve — Governing Differential Equation

Difficulty

easy

Template Id

T1

Examiner Tip

This is a recall question. Full mark requires the equation in recognisable form AND at least a minimal definition of the symbols — a bare equation with no labelling is penalised in some rubrics.

Model Answer

EI d²y/dx² = M(x), where E is the modulus of elasticity, I is the moment of inertia of the cross-section about the neutral axis, y is the transverse deflection, x is the position along the beam axis, and M(x) is the internal bending moment at section x.

Question Type

very_short_answer

Answer Structure

• Line 1: Write the equation in correct mathematical form with all symbols defined [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing EI dy/dx = M(x) — confusing slope equation with the moment equation
• Omitting EI on the left-hand side
• Writing M without the argument (x), implying a constant moment

Key Phrases To Include

• EI d²y/dx² = M(x)
• flexural rigidity
• bending moment
• elastic curve

Marks

1

Topic

Serviceability — Code Deflection Limits

Difficulty

easy

Template Id

T2

Examiner Tip

Memorise the NSCP 2015 deflection table in pairs: L/360 (LL, damage-prone finish) and L/240 (total load, damage-prone finish). Board exams frequently ask for the specific fraction.

Model Answer

Per NSCP 2015 Table 405.2, the allowable live-load deflection for beams supporting floor construction likely to be damaged by large deflections is L/360, where L is the beam span.

Question Type

very_short_answer

Answer Structure

• Line 1: State the fraction L/360 and cite the code table [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating L/240 instead of L/360 — that is the total-load limit for the same condition
• Omitting the qualifier 'live load' — L/240 applies to total load under different conditions
• No code citation when one is specifically asked for

Key Phrases To Include

• L/360
• NSCP 2015 Table 405.2
• live-load deflection
• susceptible to damage

Marks

2

Topic

Area-Moment (Moment-Area) Theorems

Difficulty

easy

Template Id

T3

Examiner Tip

The examiners award one mark per theorem. Write each theorem as a clear one-sentence statement followed by the integral or symbolic equation — do not combine them into a single sentence.

Model Answer

First Moment-Area Theorem: The change in slope between two points A and B on the elastic curve equals the area of the M/EI diagram between those two points. θ_B/A = ∫[A to B] (M/EI) dx = Area of (M/EI) diagram from A to B Second Moment-Area Theorem: The vertical deviation of point B from the tangent drawn at point A equals the first moment of the M/EI area between A and B, taken about point B. t_B/A = ∫[A to B] (M/EI) · x̄_B dx = (Area_AB) · x̄_B where x̄_B is the horizontal distance from B to the centroid of the M/EI area.

Question Type

short_answer

Answer Structure

• Theorem 1 statement — change in slope = area of M/EI diagram [1 mark]
• Theorem 2 statement — tangential deviation = first moment of M/EI area about the far point [1 mark]

Scoring Breakdown

Common Mark Deductions

• Confusing θ_B/A with absolute slope — it is a change in slope, not the slope at B
• Saying moment about A instead of about B for the second theorem
• Omitting the M/EI qualifier and just writing 'moment diagram'

Key Phrases To Include

• M/EI diagram
• change in slope
• tangential deviation
• first moment
• centroid of the M/EI area

Marks

2

Topic

Conjugate-Beam Method

Difficulty

medium

Template Id

T4

Examiner Tip

The support transformation is the most commonly tested detail of this method. Memorise the three pairs as a table and reproduce it in your answer.

Model Answer

The conjugate-beam method replaces the real beam with a fictitious 'conjugate' beam of the same length, loaded with the real beam's M/EI diagram as a distributed load. The relationship is: • Shear in the conjugate beam = Slope in the real beam • Bending moment in the conjugate beam = Deflection in the real beam Support transformations: • Real pin/roller (y = 0, y' ≠ 0) → Conjugate pin/roller (V ≠ 0, M = 0) [unchanged] • Real fixed end (y = 0, y' = 0) → Conjugate free end (V = 0, M = 0) • Real free end (y ≠ 0, y' ≠ 0) → Conjugate fixed end (V ≠ 0, M ≠ 0) Once the conjugate beam is established, standard statics gives slopes and deflections of the real beam.

Question Type

short_answer

Answer Structure

• Explain the analogy: M/EI diagram as load; shear = slope; moment = deflection [1 mark]
• State the support transformation rules, particularly fixed ↔ free end [1 mark]

Scoring Breakdown

Common Mark Deductions

• Reversing the analogy — stating conjugate moment = slope and conjugate shear = deflection
• Stating pin stays pin but not explaining why (boundary conditions must match)
• Omitting the support transformation entirely

Key Phrases To Include

• M/EI diagram as distributed load
• conjugate shear = real slope
• conjugate moment = real deflection
• fixed end becomes free end
• free end becomes fixed end

Marks

3

Topic

Superposition — Simply Supported Beam Under UDL

Difficulty

easy

Template Id

T5

Examiner Tip

This is one of the most frequently appearing board-exam question types. Practise the unit conversion drill until it is automatic: w in N/mm, L in mm, E in N/mm², I in mm⁴ → δ in mm.

Model Answer

Given: w = 10 kN/m = 10 N/mm, L = 6 m = 6 000 mm, E = 200 GPa = 200 000 N/mm², I = 80 × 10⁶ mm⁴ Formula (simply supported beam, full UDL): δ_max = 5wL⁴ / (384EI) Substitution: δ_max = [5 × 10 × (6 000)⁴] / [384 × 200 000 × 80 × 10⁶] = [5 × 10 × 1.296 × 10¹⁵] / [6.144 × 10¹⁵] = 6.48 × 10¹⁶ / 6.144 × 10¹⁵ = 10.55 mm ↓ Serviceability check (NSCP 2015 Table 405.2): Allowable δ = L/360 = 6 000/360 = 16.67 mm Actual δ = 10.55 mm < 16.67 mm ∴ COMPLIANT

Question Type

numerical

Answer Structure

• List given data with unit conversions to N and mm [½ mark]
• State the correct formula δ = 5wL⁴/384EI [1 mark]
• Substitute and compute numerically, showing intermediate steps [1 mark]
• Compute L/360 and write explicit compliance statement [½ mark]

Scoring Breakdown

Common Mark Deductions

• Using L in metres throughout without converting — gives answer in metres × 10⁻³ only if EI is also in kN·m² — risky; always use mm and N
• Citing formula as wL⁴/384EI (omitting the factor 5) — wrong formula, loses formula mark
• Computing the deflection correctly but skipping the serviceability comparison — loses ½ mark

Key Phrases To Include

• 5wL⁴/384EI
• w = 10 N/mm
• L = 6 000 mm
• L/360 = 16.67 mm
• COMPLIANT / SAFE
• NSCP 2015 Table 405.2

Marks

3

Topic

Superposition — Cantilever Under Point Load

Difficulty

easy

Template Id

T6

Examiner Tip

The four 'magic denominators' for board exams are 3 (cantilever point load), 8 (cantilever UDL), 48 (SS point load), and 384/5 (SS UDL). Write them in a small table at the top of your exam paper during reading time.

Model Answer

Given: P = 8 kN = 8 000 N, L = 2.5 m = 2 500 mm, EI = 1.2 × 10¹³ N·mm² Formula (cantilever, point load at free end): δ = PL³ / (3EI) Substitution: δ = (8 000 × 2 500³) / (3 × 1.2 × 10¹³) = (8 000 × 1.5625 × 10¹⁰) / (3.6 × 10¹³) = 1.25 × 10¹⁴ / 3.6 × 10¹³ = 3.47 mm ↓ (at the free end)

Question Type

numerical

Answer Structure

• State given values with unit conversions [½ mark]
• Cite the formula δ = PL³/3EI [1 mark]
• Substitute numerically and compute, showing L³ step [1 mark]
• State the final answer with units and direction [½ mark]

Scoring Breakdown

Common Mark Deductions

• Using the simply supported formula PL³/48EI instead of the cantilever formula — completely wrong result
• Forgetting the factor 3 in the denominator — using PL³/EI instead of PL³/3EI
• Not converting L from metres to millimetres before cubing

Key Phrases To Include

• PL³/3EI
• P = 8 000 N
• L = 2 500 mm
• 3.47 mm
• free end
• downward

Marks

5

Topic

Double-Integration Method — Simply Supported Beam, Central Point Load

Difficulty

hard

Template Id

T7

Examiner Tip

The double-integration derivation is a 'show that' question — every single step must be written out. Examiners mark the process, not just the result. If your final formula differs from PL³/48EI, show your algebra clearly so partial marks can be awarded for correct method.

Model Answer

Setup: Place origin at left support A. By symmetry, reactions are R_A = R_B = P/2. For 0 ≤ x ≤ L/2 (left half only, using symmetry): M(x) = (P/2)x Differential Equation: EI d²y/dx² = M(x) = (P/2)x ... (1) First Integration (slope): EI dy/dx = (P/2)(x²/2) + C₁ = Px²/4 + C₁ ... (2) Second Integration (deflection): EI y = Px³/12 + C₁x + C₂ ... (3) Boundary Conditions: BC1: At x = 0 (pin support A), y = 0: → EI(0) = 0 + 0 + C₂ ∴ C₂ = 0 BC2: At x = L/2 (midspan, by symmetry), dy/dx = 0: → EI(0) = P(L/2)²/4 + C₁ → 0 = PL²/16 + C₁ ∴ C₁ = −PL²/16 Deflection Equation (0 ≤ x ≤ L/2): EI y = Px³/12 − (PL²/16)x y = P(4x³ − 3L²x) / (48EI) Maximum Deflection at x = L/2: y_max = P[4(L/2)³ − 3L²(L/2)] / (48EI) = P[4(L³/8) − 3L³/2] / (48EI) = P[L³/2 − 3L³/2] / (48EI) = P(−L³) / (48EI) |y_max| = PL³ / (48EI) ↓ (downward, as expected) Result: δ_max = PL³/48EI at midspan.

Question Type

long_answer

Answer Structure

• Free-body diagram and reaction calculation R_A = R_B = P/2 [½ mark]
• Write M(x) expression for the appropriate interval [½ mark]
• Set up and write EI y'' = M(x) clearly [½ mark]
• First integration to get slope equation with constant C₁ [½ mark]
• Second integration to get deflection equation with constant C₂ [½ mark]
• State and apply BC1 (y = 0 at support) to find C₂ [½ mark]
• State and apply BC2 (slope = 0 at midspan by symmetry) to find C₁ [½ mark]
• Write final deflection equation and compute δ_max = PL³/48EI [1 mark]

Scoring Breakdown

Common Mark Deductions

• Omitting integration constants C₁ or C₂ — automatic loss of at least 1 mark
• Not stating boundary conditions before applying them — examiner cannot award BC marks
• Using the full-span M(x) without applying Macaulay brackets — leads to an inconsistent equation for the right half
• Correctly finding C₁ and C₂ but not substituting back to derive the deflection equation explicitly
• Arithmetic error in evaluating y at x = L/2 — show substitution step clearly

Key Phrases To Include

• EI d²y/dx² = M(x)
• boundary condition
• y = 0 at x = 0
• dy/dx = 0 at x = L/2
• C₁ = −PL²/16
• C₂ = 0
• PL³/48EI

Marks

5

Topic

Superposition — Cantilever with Combined Loading

Difficulty

medium

Template Id

T8

Examiner Tip

Always write the superposition equation (δ_total = δ₁ + δ₂) as the first line after stating the principle. This earns a method mark immediately, even before any numbers appear.

Model Answer

Given: P = 5 kN = 5 000 N, w = 3 kN/m = 3 N/mm, L = 3 m = 3 000 mm, EI = 2.0 × 10¹³ N·mm² Principle of Superposition: δ_B = δ_P + δ_w Case 1 — Point load at free end: δ_P = PL³ / (3EI) = (5 000 × 3 000³) / (3 × 2.0 × 10¹³) = (5 000 × 2.7 × 10¹⁰) / (6.0 × 10¹³) = 1.35 × 10¹⁴ / 6.0 × 10¹³ = 2.25 mm Case 2 — Full UDL on cantilever: δ_w = wL⁴ / (8EI) = (3 × 3 000⁴) / (8 × 2.0 × 10¹³) = (3 × 8.1 × 10¹³) / (1.6 × 10¹⁴) = 2.43 × 10¹⁴ / 1.6 × 10¹⁴ = 1.52 mm Total Deflection at B: δ_B = δ_P + δ_w = 2.25 + 1.52 = 3.77 mm ↓

Question Type

numerical

Answer Structure

• State principle of superposition and identify the two load cases [½ mark]
• Write formula for Case 1 (PL³/3EI) and substitute with correct units [1 mark]
• Compute δ_P with clear intermediate calculation of L³ [½ mark]
• Write formula for Case 2 (wL⁴/8EI) and substitute with correct units [1 mark]
• Compute δ_w with clear intermediate calculation of L⁴ [½ mark]
• Sum the two deflections and state total δ_B with units and direction [½ mark]

Scoring Breakdown

Common Mark Deductions

• Using δ_w = wL⁴/384EI (the simply-supported formula) instead of the cantilever formula wL⁴/8EI
• Using w in kN/m instead of N/mm — gives L⁴ term in mm⁴ × kN/m which is dimensionally inconsistent
• Forgetting to add the two deflections and reporting only one case
• Not stating the direction of deflection (downward)

Key Phrases To Include

• superposition
• PL³/3EI
• wL⁴/8EI
• δ_P = 2.25 mm
• δ_w = 1.52 mm
• δ_B = 3.77 mm
• downward

Marks

3

Topic

Serviceability Design — Required Moment of Inertia

Difficulty

medium

Template Id

T9

Examiner Tip

Always make the 'required I' problem a two-step: first find δ_allow from the code ratio, then rearrange the standard formula. Setting up I ≥ (numerator)/(denominator) earns a method mark before any arithmetic.

Model Answer

Given: w = 12 kN/m = 12 N/mm, L = 5 m = 5 000 mm, E = 200 GPa = 200 000 N/mm² Serviceability limit (NSCP 2015 Table 405.2): δ_allow = L/360 = 5 000/360 = 13.89 mm Set actual deflection ≤ allowable: 5wL⁴/(384EI) ≤ δ_allow I ≥ 5wL⁴ / (384 × E × δ_allow) Substitution: I ≥ [5 × 12 × (5 000)⁴] / [384 × 200 000 × 13.89] = [5 × 12 × 6.25 × 10¹⁴] / [1.067 × 10⁹] = 3.75 × 10¹⁶ / 1.067 × 10⁹ = 3.514 × 10⁷ mm⁴ ∴ I_min = 35.14 × 10⁶ mm⁴

Question Type

numerical

Answer Structure

• Convert units; compute allowable deflection δ = L/360 numerically [½ mark]
• Write deflection formula and rearrange for I [1 mark]
• Substitute values and compute I_min [1 mark]
• State I_min with units and code reference [½ mark]

Scoring Breakdown

Common Mark Deductions

• Computing δ_allow in metres (0.01389 m) and mixing with mm values in the formula
• Solving for EI instead of I alone — forgetting to divide by E
• Rounding δ_allow to 14 mm without showing L/360 calculation

Key Phrases To Include

• δ_allow = L/360 = 13.89 mm
• I ≥ 5wL⁴/(384Eδ)
• I_min = 35.14 × 10⁶ mm⁴
• NSCP 2015

Marks

3

Topic

Superposition — Simply Supported Beam, Central Point Load

Difficulty

easy

Template Id

T10

Examiner Tip

The L³ step is where arithmetic errors occur most. Write L³ = (8 000)³ = 5.12 × 10¹¹ mm³ explicitly before proceeding — it costs no time and prevents a common 10× error.

Model Answer

Given: P = 30 kN = 30 000 N, L = 8 m = 8 000 mm, E = 200 000 N/mm², I = 120 × 10⁶ mm⁴ Formula (simply supported, central point load): δ_max = PL³ / (48EI) Substitution: δ_max = (30 000 × 8 000³) / (48 × 200 000 × 120 × 10⁶) = (30 000 × 5.12 × 10¹¹) / (1.152 × 10¹⁵) = 1.536 × 10¹⁶ / 1.152 × 10¹⁵ = 13.33 mm ↓ Serviceability check: δ_allow = L/360 = 8 000/360 = 22.22 mm 13.33 mm < 22.22 mm ∴ SAFE — deflection is within the NSCP 2015 limit.

Question Type

numerical

Answer Structure

• Convert units to N and mm [½ mark]
• Cite PL³/48EI and substitute [1 mark]
• Compute δ_max (correct value ≈ 13.33 mm) [½ mark]
• Compute L/360 and state compliance [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using 5PL³/384EI (the UDL formula) instead of PL³/48EI
• Not converting kN to N — result is off by factor of 1 000
• Failing to compare δ_actual with δ_allow explicitly

Key Phrases To Include

• PL³/48EI
• 30 000 N
• 8 000 mm
• 13.33 mm
• L/360 = 22.22 mm
• SAFE

Marks

2

Topic

Flexural Rigidity and Section Properties

Difficulty

easy

Template Id

T11

Examiner Tip

Concept questions at 2 marks require one 'what' and one 'why' or one 'how much'. State the formula (what), then provide the numerical comparison (how much).

Model Answer

Flexural rigidity EI controls deflection — larger EI gives smaller deflection for the same load and span. For a rectangular section of width b and depth d: I = bd³/12 Depth appears to the third power, while width appears to the first power. Doubling the width doubles I, halving the deflection. Doubling the depth multiplies I by 2³ = 8, reducing deflection by a factor of 8. Therefore, increasing depth is far more efficient per unit of added material than increasing width when the objective is to minimise deflection.

Question Type

short_answer

Answer Structure

• State the moment of inertia formula I = bd³/12 and note the exponents [1 mark]
• Quantify the effect: width is linear, depth is cubic — with a numerical comparison (e.g., doubling d → 8× increase in I) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating 'depth is more important' without quantitative justification — gains only ½ mark
• Citing I = bh³/12 but not connecting it to the deflection formula

Key Phrases To Include

• I = bd³/12
• depth cubed
• EI flexural rigidity
• doubling depth multiplies I by 8
• deflection inversely proportional to EI

Marks

2

Topic

Double-Integration — Macaulay Bracket Method

Difficulty

medium

Template Id

T12

Examiner Tip

Use a quick example like two point loads to illustrate why a single expression is advantageous — it shows the examiner you understand the practical benefit, not just the definition.

Model Answer

The Macaulay bracket method introduces singularity functions ⟨x − a⟩ⁿ, defined as: ⟨x − a⟩ⁿ = 0 for x < a ⟨x − a⟩ⁿ = (x − a)ⁿ for x ≥ a This allows the bending-moment equation M(x) for the entire beam span to be written as a single expression, regardless of the number of load discontinuities. Each load change at position a introduces one bracket term. The expression is then integrated twice in the usual way; the brackets are integrated as ⟨x−a⟩ⁿ → ⟨x−a⟩ⁿ⁺¹/(n+1), and only two boundary conditions are needed to find the two constants C₁ and C₂. Macaulay's method is preferred when multiple point loads, moments, or partial UDLs act at different positions along the span — the alternative of writing separate M(x) expressions for each interval and matching slope/deflection continuity conditions between intervals is far more tedious.

Question Type

short_answer

Answer Structure

• Define the bracket function ⟨x − a⟩ⁿ correctly with both cases (on and off) [1 mark]
• State when/why it is preferred: single expression for multi-load beams, only two BCs needed [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining the bracket but not specifying the integration rule ⟨x−a⟩ⁿ → ⟨x−a⟩ⁿ⁺¹/(n+1)
• Not explaining the advantage over the segment-by-segment approach

Key Phrases To Include

• ⟨x − a⟩ⁿ
• singularity function
• zero for x < a
• two boundary conditions only
• discontinuous loading
• single expression

Marks

5

Topic

Indeterminate Beams — Compatibility / Consistent Deformations

Difficulty

hard

Template Id

T13

Examiner Tip

This problem is the gateway between Strength of Materials and Structural Theory. The key phrase that earns the method mark is the compatibility statement: 'net deflection at B = 0, therefore δ due to load = δ due to redundant.' Write it in words before writing equations.

Model Answer

Degree of indeterminacy: 1 (one redundant reaction R_B) Released structure: cantilever fixed at A, free at B (primary structure) Step 1 — Deflection at B due to UDL w alone (R_B released, treated as cantilever): δ_w = wL⁴ / (8EI) ↓ (downward at free end B) Step 2 — Deflection at B due to redundant R_B acting upward on the cantilever: δ_R = R_B × L³ / (3EI) ↑ (upward, opposing δ_w) Step 3 — Compatibility condition (roller at B allows no deflection): δ_w = δ_R wL⁴/(8EI) = R_B × L³/(3EI) Step 4 — Solve for R_B: R_B = wL⁴/(8EI) × (3EI/L³) = 3wL/8 Result: R_B = 3wL/8 (upward) Verification note: This is the classical result for a propped cantilever under full UDL. The fixed-end reaction R_A = wL − R_B = 5wL/8 and the fixed-end moment M_A = wL²/2 − R_B·L = wL²/8 can be computed from equilibrium as a check.

Question Type

long_answer

Answer Structure

• Identify the redundant (R_B) and state the released (primary) structure [½ mark]
• Write and compute δ_w = wL⁴/8EI for the cantilever under UDL alone [1 mark]
• Write and express δ_R = R_B L³/3EI for the redundant force alone [1 mark]
• State the compatibility condition δ_w = δ_R (net deflection at B = 0) [1 mark]
• Solve the equation and derive R_B = 3wL/8 [1 mark]
• Verify using equilibrium (optional but earns examiner favour) [½ mark]

Scoring Breakdown

Common Mark Deductions

• Using the simply-supported deflection formula instead of the cantilever formula for the primary structure
• Setting up the equation as δ_w + δ_R = 0 without explaining sign convention — creates sign errors
• Forgetting that the released structure is a cantilever, not a simply supported beam — this is the most common conceptual error
• Solving for R_B but omitting the direction (upward)

Key Phrases To Include

• redundant
• released structure
• compatibility condition
• δ_w = δ_R
• wL⁴/8EI
• R_B L³/3EI
• R_B = 3wL/8
• consistent deformations

Marks

2

Topic

Exam Technique — Common Errors

Difficulty

easy

Template Id

T14

Examiner Tip

When answering a 'list' question, number each point clearly. Four numbered points are quicker to mark than a paragraph and make it obvious you have addressed all four items.

Model Answer

1. Unit inconsistency — mixing metres and millimetres in L³ or L⁴ terms. Avoid: Convert all quantities to N and mm before substituting. 2. Wrong standard formula — using PL³/48EI (simply supported) for a cantilever. Avoid: Sketch the beam type first; write the formula and label the denominator explicitly. 3. Omitting boundary conditions in double integration — guessing or skipping C₁, C₂. Avoid: State each BC in words ('At x = 0, y = 0') before applying it algebraically. 4. No serviceability check — computing δ but not comparing it with L/360. Avoid: Always write the comparison line 'δ_actual vs δ_allow = L/360' as the final step.

Question Type

short_answer

Answer Structure

• Any two well-identified errors with their corresponding avoidance strategies [1 mark]
• Two additional distinct errors with avoidance strategies [1 mark]

Scoring Breakdown

Common Mark Deductions

• Listing vague errors like 'arithmetic mistakes' without specificity — too general to earn marks
• Repeating the same error in different words

Key Phrases To Include

• unit conversion
• correct formula
• boundary conditions
• serviceability check
• L/360

Marks

5

Topic

Superposition — Simply Supported Beam, Symmetric Third-Point Loads

Difficulty

hard

Template Id

T15

Examiner Tip

If you cannot recall the off-centre formula, use the Macaulay method or area-moment to derive the midspan deflection. A correct derivation earns full formula marks even if the standard formula is not recalled verbatim.

Model Answer

Given: P = 20 kN = 20 000 N each, a = L/3 = 10 000/3 mm, L = 10 000 mm, E = 200 000 N/mm², I = 200 × 10⁶ mm⁴ For a simply supported beam with a single off-centre point load P at distance a from left support, the midspan deflection (when a ≤ L/2) is: δ_mid = Pa(3L² − 4a²) / (48EI) For a = L/3: 3L² − 4a² = 3L² − 4(L/3)² = 3L² − 4L²/9 = (27L² − 4L²)/9 = 23L²/9 Deflection due to one load P at a = L/3: δ₁ = P(L/3)(23L²/9) / (48EI) = 23PL³ / (1 296EI) By symmetry, the load at x = 2L/3 (a = L/3 measured from right) produces the same midspan deflection δ₂ = δ₁. Total midspan deflection: δ_mid = δ₁ + δ₂ = 2 × 23PL³/(1 296EI) = 46PL³/(1 296EI) = 23PL³/(648EI) Substitution: δ_mid = [23 × 20 000 × (10 000)³] / [648 × 200 000 × 200 × 10⁶] = [23 × 20 000 × 10¹²] / [648 × 4 × 10¹³] = [4.6 × 10¹⁷] / [2.592 × 10¹⁶] = 17.75 mm ↓ Alternative check using superposition of two equal symmetric loads: Standard result for two equal third-point loads: δ = 23PL³/648EI ✓

Question Type

numerical

Answer Structure

• Identify that two loads act and state superposition approach [½ mark]
• Recall or derive the off-centre point-load midspan formula δ = Pa(3L² − 4a²)/48EI [1 mark]
• Compute 3L² − 4a² with a = L/3 [1 mark]
• Apply symmetry to get total δ = 2 × δ₁ [½ mark]
• Substitute numbers correctly (N, mm) and compute δ_mid [1 mark]
• State final answer with correct value, units, and direction [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using the central point-load formula PL³/48EI for an off-centre load — wrong formula
• Not applying symmetry and computing only one load's effect
• Arithmetic error in simplifying 3L² − 4(L/3)² — show every algebra step

Key Phrases To Include

• Pa(3L² − 4a²)/48EI
• a = L/3
• symmetry
• 23PL³/648EI
• 17.75 mm