CELE Structural Theory & Analysis — Deflections of StructuresExam Answer Templates

Marks

1

Topic

Virtual Work — Trusses

Difficulty

easy

Template Id

T1

Examiner Tip

A 1-mark VSA rewards recall precision. Write the formula exactly as it appears in standard texts and define every symbol on the same line — examiners spend 15 seconds on these answers.

Model Answer

The virtual work equation for truss deflection is: δ = Σ(nNL/AE), where N = real member axial force, n = member axial force due to a unit virtual load applied at the point and direction of the desired deflection, L = member length, A = cross-sectional area, and E = modulus of elasticity.

Question Type

very_short_answer

Answer Structure

• Line 1: Write the formula with all variables identified [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing δ = ΣNL/AE (omitting the virtual force n) — loses the mark entirely
• Correct formula but no variable definitions when the question implies identification is needed
• Using incorrect symbol (e.g., writing ΔnNL/AE with Δ undefined)

Key Phrases To Include

• δ = ΣnNL/AE
• unit virtual load
• real member force N
• virtual member force n

Marks

1

Topic

Virtual Work — Beams and Frames

Difficulty

easy

Template Id

T2

Examiner Tip

Examiners reward the pairing: real → M(x) and virtual → m(x). State both pairs explicitly even in one sentence.

Model Answer

The real system consists of the actual applied loads on the structure, producing real bending moments M(x) and real deflections. The virtual system consists of a single unit load (1 kN force or 1 kN·m moment) applied at the point and in the direction of the desired deflection, producing virtual bending moments m(x), with no real loads present.

Question Type

very_short_answer

Answer Structure

• Part A: Define real system — actual loads, produces M(x) [0.5 mark implied]
• Part B: Define virtual system — unit load only, produces m(x) [0.5 mark implied]

Scoring Breakdown

Common Mark Deductions

• Stating the virtual system uses actual loads — fundamentally wrong concept
• Failing to specify that the unit load acts at the point of desired deflection

Key Phrases To Include

• actual applied loads
• unit load
• real bending moment M(x)
• virtual bending moment m(x)

Marks

2

Topic

Castigliano's Second Theorem

Difficulty

medium

Template Id

T3

Examiner Tip

The 2-mark answer must show BOTH the general statement and the specific bending integral. Many reviewees only write ∂U/∂P and lose the second mark.

Model Answer

Castigliano's Second Theorem states: 'The deflection at the point of application of a load, in the direction of that load, equals the partial derivative of the total strain energy U with respect to that load.' Mathematically: δᵢ = ∂U/∂Pᵢ For a beam where strain energy is due to bending, U = ∫[M²/(2EI)]dx, and therefore: δ = ∫[M/(EI)](∂M/∂P) dx where M = real bending moment as a function of the load P, EI = flexural rigidity, and ∂M/∂P is analogous to the virtual moment m(x) in the unit-load method.

Question Type

short_answer

Answer Structure

• Line 1: State the theorem in words [1 mark]
• Line 2: Write the bending deflection integral δ = ∫(M/EI)(∂M/∂P)dx with variables defined [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating δ = ∂U/∂P only without defining what U is for a beam — incomplete for 2-mark level
• Writing δ = ∫M²/(EI)dx (omitting the ∂M/∂P factor — this is the strain energy, not the deflection)
• Confusing Castigliano's First Theorem (force = ∂U/∂δ) with the Second

Key Phrases To Include

• partial derivative
• strain energy U
• ∂U/∂P
• ∫(M/EI)(∂M/∂P)dx
• flexural rigidity EI

Marks

2

Topic

Standard Beam Deflection Formulas

Difficulty

easy

Template Id

T4

Examiner Tip

For 2-mark numericals, show the formula and the final numerical answer clearly. The two marks correspond to these two elements — method and result.

Model Answer

Given: P = 24 kN = 24,000 N; L = 6 m = 6,000 mm; EI = 1.44 × 10¹³ N·mm² Applicable formula (simply supported beam, central point load): δ_mid = PL³/(48EI) Substituting: δ_mid = (24,000)(6,000)³ / [48 × (1.44 × 10¹³)] = (24,000)(2.16 × 10¹¹) / (6.912 × 10¹⁴) = 5.184 × 10¹⁵ / 6.912 × 10¹⁴ δ_mid = 7.50 mm (downward) ✓

Question Type

numerical

Answer Structure

• Step 1: List given data with unit conversions to mm and N [0.5 mark]
• Step 2: Identify and state the formula δ = PL³/48EI [0.5 mark]
• Step 3: Substitute values and compute numerically [0.5 mark]
• Step 4: State final answer with unit and direction [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using L = 6 m instead of converting to mm when EI is in N·mm² — unit inconsistency gives wrong answer
• Using wrong formula (e.g., PL³/3EI for cantilever instead of 48EI for simply supported)
• Omitting direction of deflection from final answer

Key Phrases To Include

• δ = PL³/48EI
• midspan deflection
• downward
• N·mm² units consistent

Marks

3

Topic

Virtual Work — Beams and Frames

Difficulty

medium

Template Id

T5

Examiner Tip

For 3-mark virtual-work problems, the three marks map directly to: real moments, virtual moments, and the integral result. Show each step on a new line.

Model Answer

Given: P = 10 kN = 10,000 N; L = 3 m = 3,000 mm; EI = 2 × 10¹³ N·mm² Step 1 — Real system (measuring x from the free end A toward fixed support B): M(x) = −Px = −10,000x [N·mm, hogging negative] Step 2 — Virtual system (unit downward load at A): m(x) = −(1)x = −x [N·mm per unit load] Step 3 — Apply virtual work integral: δ_A = ∫₀ᴸ [m(x)·M(x)/EI] dx = ∫₀ᴸ [(−x)(−10,000x)/EI] dx = ∫₀ᴸ [10,000x²/EI] dx = (10,000/EI) · [x³/3]₀ᴸ = (10,000 · L³) / (3EI) = (10,000)(3,000)³ / [3 × 2 × 10¹³] = (10,000)(2.7 × 10¹⁰) / (6 × 10¹³) = 2.7 × 10¹⁴ / 6 × 10¹³ δ_A = 4.50 mm (downward) ✓ Note: This confirms the standard formula δ = PL³/3EI.

Question Type

numerical

Answer Structure

• Step 1: Define real bending moment M(x) with sign convention [1 mark]
• Step 2: Define virtual bending moment m(x) for unit load [0.5 mark]
• Step 3: Set up and evaluate the integral ∫mM/EI dx [1 mark]
• Step 4: Substitute, compute, and state δ with unit and direction [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Not stating where x is measured from — ambiguous coordinate system loses method mark
• Sign error: writing m(x) = +x instead of −x, then getting negative product and negative deflection
• Computing (3)³ = 9 instead of 27 — arithmetic error loses the result mark but not method marks
• Units mixed: P in kN but L not converted to m for EI in N·mm²

Key Phrases To Include

• real system M(x)
• virtual system m(x)
• unit load at free end
• ∫mM/EI dx
• PL³/3EI confirmed

Marks

3

Topic

Castigliano's Second Theorem — Dummy Load

Difficulty

medium

Template Id

T6

Examiner Tip

The phrase 'set Q = 0 AFTER differentiating, not before' is the key examiner discriminator for this topic. Write it explicitly.

Model Answer

Concept of Dummy Load: Castigliano's Second Theorem requires a real load P to exist at the point where deflection is sought, because δ = ∂U/∂P requires a real load to differentiate with respect to. When no real load acts at the desired deflection point, an artificial load called a dummy load Q is introduced at that point in the direction of the desired deflection. Procedure: 1. Apply the dummy load Q (= 0 in reality) at the target point alongside all real loads. 2. Express the bending moment M in terms of all loads including Q. 3. Compute ∂M/∂Q — this is the influence of the dummy load on moments. 4. Evaluate δ = ∫[M/EI](∂M/∂Q) dx with all real loads AND Q. 5. After integration, set Q = 0 to obtain the actual deflection. Illustrative Scenario: A simply supported beam carries a UDL w over its full span. If the midspan deflection is required but no concentrated load acts at midspan, apply dummy load Q ↓ at midspan. After differentiating and integrating, set Q = 0. The result δ_mid = 5wL⁴/384EI is recovered. Key rule: Q is set to zero AFTER differentiation, not before.

Question Type

short_answer

Answer Structure

• Part 1: Define dummy load and state why it is needed (no real load at target point) [1 mark]
• Part 2: Describe the 5-step procedure including 'set Q = 0 after differentiation' [1 mark]
• Part 3: Provide a specific scenario/example with correct formula result [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating Q = 0 before differentiation — this eliminates Q from M and gives a trivially zero ∂M/∂Q
• Confusing dummy load with the unit load in virtual work — they are analogous but in different methods
• Scenario given with no conclusion or formula result — incomplete example

Key Phrases To Include

• dummy load Q
• set Q = 0 after differentiation
• ∂M/∂Q
• no real load at target point
• ∫(M/EI)(∂M/∂Q)dx

Marks

3

Topic

Virtual Work — Trusses

Difficulty

medium

Template Id

T7

Examiner Tip

A tidy tabulation (Member | n | N | L | nNL/AE) shows the examiner exactly where every mark is earned. Use it for all truss deflection problems.

Model Answer

Step 1 — Virtual system: Apply unit downward load (1 kN) at C. By proportion (load scales linearly), virtual forces are: n_AC = −15.63/20 = −0.7815 kN n_BC = −15.63/20 = −0.7815 kN n_AB = 12/20 = 0.60 kN Step 2 — Tabulate nNL/AE for each member: Member AC: nNL/AE = (−0.7815)(−15.63)(3.905) / (2×10⁵) = (47.73) / (2×10⁵) = 2.387 × 10⁻⁴ m Member BC: nNL/AE = (−0.7815)(−15.63)(3.905) / (2×10⁵) = 47.73 / (2×10⁵) = 2.387 × 10⁻⁴ m [same as AC by symmetry] Member AB: nNL/AE = (0.60)(12)(6) / (2×10⁵) = 43.2 / (2×10⁵) = 2.160 × 10⁻⁴ m Step 3 — Sum: δ_C = 2.387 × 10⁻⁴ + 2.387 × 10⁻⁴ + 2.160 × 10⁻⁴ = 6.934 × 10⁻⁴ m δ_C = 0.693 mm (downward) ✓

Question Type

numerical

Answer Structure

• Step 1: Compute virtual (unit-load) member forces n by proportion or independent analysis [1 mark]
• Step 2: Tabulate nNL/AE for each member, showing sign of each product [1 mark]
• Step 3: Sum all contributions and state final δ with unit (mm or m) and direction [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using n = N (real forces) instead of computing separate unit-load forces
• Forgetting to divide by AE — computing ΣnNL only
• Sign error: treating (−)(−) as negative, getting negative contribution
• Unit error: mixing AE in N while N is in kN, giving answer 1000× too small

Key Phrases To Include

• δ = ΣnNL/AE
• virtual forces by proportion
• sign: tension positive
• symmetry of AC and BC
• downward deflection

Marks

2

Topic

Castigliano's Second Theorem

Difficulty

medium

Template Id

T8

Examiner Tip

The phrase '∂M/∂P plays the role of m(x)' is the precise equivalence statement examiners expect. Write it explicitly.

Model Answer

The two methods are mathematically equivalent for linear elastic beams under bending. Virtual work method: δ = ∫[m(x)·M(x)/EI] dx, where m(x) = bending moment due to a unit virtual load. Castigliano's method: δ = ∂U/∂P = ∫[M(x)/EI·(∂M/∂P)] dx. The equivalence arises because ∂M/∂P — the rate of change of real bending moment with respect to the load P — is exactly equal to m(x), the bending moment produced by a unit load at the same point. Both methods integrate the product of a 'moment influence function' and the real bending moment over EI. The only practical difference is computational: Castigliano uses differentiation of the real moment expression, while virtual work requires an independent unit-load analysis.

Question Type

short_answer

Answer Structure

• Line 1: State they are equivalent [0.5 mark]
• Line 2: Show ∂M/∂P = m(x) — the mathematical bridge [1 mark]
• Line 3: Note practical difference in computation [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Stating they give different results — fundamentally incorrect
• Correct assertion of equivalence with no mathematical justification — loses second mark

Key Phrases To Include

• mathematically equivalent
• ∂M/∂P = m(x)
• moment influence function
• unit virtual load

Marks

5

Topic

Castigliano's Second Theorem — Dummy Load

Difficulty

hard

Template Id

T9

Examiner Tip

A 5-mark derivation must show every algebraic step. The examiner checks: (1) Q in M(x), (2) ∂M/∂Q, (3) Q set to 0 after, (4) integration, (5) correct final result. Each corresponds to 1 mark.

Model Answer

Given: Simply supported beam, span L, UDL = w N/m. No concentrated load at midspan. Step 1 — Introduce dummy load: Apply a downward dummy load Q at midspan (x = L/2). Step 2 — Reactions with both w and Q: By symmetry: R_A = R_B = wL/2 + Q/2 Step 3 — Bending moment expressions (x from left support A): For 0 ≤ x ≤ L/2 (left half, no concentrated load yet): M(x) = (wL/2 + Q/2)x − wx²/2 For L/2 ≤ x ≤ L (right half, dummy load passed): By symmetry, use left half only (factor of 2). Step 4 — Partial derivative ∂M/∂Q: ∂M/∂Q = x/2 for 0 ≤ x ≤ L/2 Step 5 — Apply Castigliano: δ = ∫[M/EI](∂M/∂Q)dx, set Q = 0 after: With Q = 0: M(x) = wLx/2 − wx²/2 = (w/2)(Lx − x²) δ_mid = (2/EI)∫₀^(L/2) M(x)·(x/2) dx [factor 2 for symmetry] = (2/EI)∫₀^(L/2) [(w/2)(Lx − x²)](x/2) dx = (w/2EI)∫₀^(L/2) (Lx² − x³) dx = (w/2EI)[Lx³/3 − x⁴/4]₀^(L/2) = (w/2EI)[L(L/2)³/3 − (L/2)⁴/4] = (w/2EI)[L⁴/(24) − L⁴/(64)] = (w/2EI)·L⁴·[1/24 − 1/64] = (w/2EI)·L⁴·[8/192 − 3/192] = (w/2EI)·L⁴·(5/192) = 5wL⁴/(384EI) δ_mid = 5wL⁴/384EI (downward) ✓ This confirms the standard formula for UDL midspan deflection of a simply supported beam.

Question Type

long_answer

Answer Structure

• Step 1: Introduce and justify the dummy load Q at midspan [0.5 mark]
• Step 2: Write reactions with Q included [0.5 mark]
• Step 3: Write correct M(x) expression including Q [1 mark]
• Step 4: Compute ∂M/∂Q and set Q = 0 correctly (Q=0 AFTER differentiating) [1 mark]
• Step 5: Set up and evaluate the Castigliano integral correctly [1 mark]
• Step 6: Correct final result 5wL⁴/384EI with unit and direction [1 mark]

Scoring Breakdown

Common Mark Deductions

• Setting Q = 0 before computing ∂M/∂Q — eliminates Q term, gives wrong ∂M/∂Q = 0 for the Q part
• Not using symmetry (integrating full span with discontinuous M(x) but missing the Q term in right half)
• Arithmetic error in [L(L/2)³/3 − (L/2)⁴/4] — check: L⁴/24 − L⁴/64
• Combining fractions incorrectly: 1/24 − 1/64 ≠ 5/384 directly (correct: common denominator 192 → 5/192, then divide by 2 → 5/384)
• Forgetting the factor of 2 for symmetry when integrating only left half

Key Phrases To Include

• dummy load Q at midspan
• ∂M/∂Q = x/2
• set Q = 0 after differentiation
• symmetry factor of 2
• 5wL⁴/384EI

Marks

3

Topic

Standard Beam Deflection Formulas

Difficulty

medium

Template Id

T10

Examiner Tip

Remember the four key cantilever formulas: PL³/3EI (point load), wL⁴/8EI (UDL), ML²/2EI (end moment), PL³/48EI is for SS beams. Formula identification is worth a mark on its own.

Model Answer

Given: w = 15 kN/m = 15 N/mm; L = 4 m = 4,000 mm; EI = 3.6 × 10¹³ N·mm² Applicable formula (cantilever, UDL full span, free-end deflection): δ_max = wL⁴/(8EI) Note: This formula can be derived by virtual work or Castigliano; it is a standard board result. Substituting: δ_max = (15)(4,000)⁴ / [8 × 3.6 × 10¹³] = (15)(2.56 × 10¹⁴) / (2.88 × 10¹⁴) = 3.84 × 10¹⁵ / 2.88 × 10¹⁴ = 13.33 mm δ_max = 13.33 mm (downward at free end) ✓

Question Type

numerical

Answer Structure

• Step 1: Convert w and L to consistent mm/N units [0.5 mark]
• Step 2: State formula δ = wL⁴/8EI with justification [1 mark]
• Step 3: Substitute and compute — show intermediate power calculation (4000)⁴ [1 mark]
• Step 4: Final answer in mm with direction [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using δ = wL⁴/384EI (simply supported beam formula) instead of cantilever formula
• (4000)⁴ computed incorrectly — show as (4 × 10³)⁴ = 256 × 10¹² = 2.56 × 10¹⁴
• w = 15 kN/m not converted to 15 N/mm (1 kN/m = 1 N/mm)

Key Phrases To Include

• δ = wL⁴/8EI
• cantilever with UDL
• free-end deflection
• consistent units N and mm

Marks

2

Topic

Energy Methods — Strain Energy

Difficulty

easy

Template Id

T11

Examiner Tip

The factor of ½ in both formulas is critical — it comes from the linear load-deflection relationship (work = ½ × force × displacement). Never omit it.

Model Answer

Strain Energy for Axially Loaded Member: When a prismatic member of length L, cross-sectional area A, and modulus E carries a constant axial force N, the internal strain energy stored is: U_axial = N²L/(2AE) This represents the work done by N as the member elongates or shortens by δ = NL/AE. Strain Energy for Bending Member: For a beam carrying variable bending moment M(x) along its length, the strain energy stored in a differential element dx is dU = M²dx/(2EI). The total strain energy is: U_bending = ∫ M²/(2EI) dx These two expressions form the basis of Castigliano's Second Theorem applied to trusses and beams respectively.

Question Type

short_answer

Answer Structure

• Part A: Formula U = N²L/2AE with variables defined [1 mark]
• Part B: Formula U = ∫M²/(2EI)dx with variables defined [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing U = NL/AE (missing the N² and the factor of 2) — loses axial mark
• Writing U = ∫M/EI dx (missing M² and factor 2) — loses bending mark

Key Phrases To Include

• U = N²L/2AE
• U = ∫M²/2EI dx
• strain energy stored
• axial deformation
• bending

Marks

2

Topic

Energy Methods — Strain Energy

Difficulty

medium

Template Id

T12

Examiner Tip

For linearly varying M, the shortcut formula U = M_max²L/6EI saves time — derive it once in your notes and memorize it for board exam speed.

Model Answer

Given: L = 5 m = 5,000 mm; M varies linearly from 0 to M_max = 50 kN·m = 50 × 10⁶ N·mm; EI = 8 × 10¹² N·mm² Express M(x) linearly (x from zero-moment end): M(x) = (M_max/L)·x = (50 × 10⁶/5,000)·x = 10,000x N·mm Apply bending strain energy formula: U = ∫₀ᴸ M²/(2EI) dx = ∫₀^5000 (10,000x)²/[2 × 8 × 10¹²] dx = ∫₀^5000 10⁸ x² / (1.6 × 10¹³) dx = (10⁸/1.6 × 10¹³) · [x³/3]₀^5000 = (6.25 × 10⁻⁶) · (5000)³/3 = (6.25 × 10⁻⁶) · (1.25 × 10¹¹/3) = (6.25 × 10⁻⁶) · (4.167 × 10¹⁰) U = 260,417 N·mm = 260.4 J ≈ 260 N·m Alternatively: U = M_max² L / (6EI) = (50×10⁶)²(5000) / [6 × 8×10¹²] = (2.5×10¹⁵)(5000) / (4.8×10¹³) = 260.4 N·m ✓

Question Type

numerical

Answer Structure

• Step 1: Express M(x) as a linear function of x [0.5 mark]
• Step 2: Set up U = ∫M²/2EI dx and integrate [1 mark]
• Step 3: Compute numerical result with unit (N·mm or J) [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using M = 50 kN·m as constant (rectangular, not triangular) — gives U = M²L/2EI = 781 N·m (wrong by factor 3)
• Forgetting the factor of 2 in denominator: writing ∫M²/EI instead of ∫M²/2EI

Key Phrases To Include

• M(x) = linear function
• U = ∫M²/2EI dx
• U = M_max²L/6EI for linear M
• N·mm or Joules

Marks

3

Topic

Overview — Purpose of Deflection Analysis

Difficulty

medium

Template Id

T13

Examiner Tip

Board exam essay questions on deflections almost always expect the word 'compatibility.' Write it prominently — it signals to the examiner that you understand the deeper purpose.

Model Answer

Deflection calculations serve two critical roles in structural analysis: 1. Serviceability Verification: Structures must satisfy deflection limits to prevent damage to non-structural elements (partitions, cladding) and to avoid uncomfortable vibrations. The NSCP 2015 (Section 409) prescribes allowable deflections (e.g., L/360 for live load on beams supporting brittle finishes). 2. Compatibility Analysis of Indeterminate Structures: This is the more fundamental role in structural theory. A statically indeterminate structure has more unknown reactions than equilibrium equations. The additional equations needed are compatibility conditions — geometric constraints that ensure the structure deforms continuously without gaps or discontinuities. These compatibility equations are written in terms of deflections at the redundant supports or members. For example, in the force method (method of consistent deformations), the redundant reaction R is found by requiring that the deflection at the redundant's location equals zero (for a fixed support) or a prescribed value. This deflection equation is: δ_0 + f_RR · R = 0, where δ_0 = deflection from real loads and f_RR = deflection from a unit redundant — both computed using virtual work or Castigliano. Conclusion: Deflections are the bridge between equilibrium and compatibility — the two pillars of structural analysis.

Question Type

short_answer

Answer Structure

• Part 1: Serviceability role with NSCP 2015 code reference [1 mark]
• Part 2: Compatibility role — explain how δ equations replace missing equilibrium equations [1 mark]
• Part 3: Specific example of compatibility equation (force method setup) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Discussing only serviceability without the indeterminate analysis role — loses 2 marks
• Vague statement 'used in indeterminate analysis' without explaining HOW (compatibility equation) — loses last mark
• No code reference for serviceability limits

Key Phrases To Include

• compatibility conditions
• statically indeterminate
• force method
• NSCP 2015
• redundant reaction
• δ_0 + f_RR·R = 0

Marks

2

Topic

Virtual Work — Trusses

Difficulty

easy

Template Id

T14

Examiner Tip

Always write out the sign explicitly: (negative n)(negative N) = positive contribution to downward deflection. One sign error invalidates both marks.

Model Answer

Given data tabulated: Member 1: n₁ = +0.80; N₁ = +30 kN; L₁ = 4 m; A₁E = 1.5 × 10⁵ kN n₁N₁L₁/A₁E = (+0.80)(+30)(4) / (1.5 × 10⁵) = +96 / (1.5 × 10⁵) = +6.40 × 10⁻⁴ m Member 2: n₂ = −0.60; N₂ = −20 kN; L₂ = 3 m; A₂E = 1.5 × 10⁵ kN n₂N₂L₂/A₂E = (−0.60)(−20)(3) / (1.5 × 10⁵) = +36 / (1.5 × 10⁵) = +2.40 × 10⁻⁴ m Vertical deflection: δ = Σ nNL/AE = 6.40 × 10⁻⁴ + 2.40 × 10⁻⁴ δ = 8.80 × 10⁻⁴ m = 0.88 mm (downward) ✓

Question Type

numerical

Answer Structure

• Step 1: Compute nNL/AE for Member 1, showing sign arithmetic [1 mark]
• Step 2: Compute nNL/AE for Member 2, showing (−)(−) = positive [0.5 mark]
• Step 3: Sum and state final answer with unit [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Computing (−0.60)(−20) = −12 (wrong sign) instead of +12 — gives negative contribution and wrong total
• Forgetting to divide by AE — treating result as kN·m instead of meters

Key Phrases To Include

• nNL/AE for each member
• (−)(−) = positive contribution
• ΣnNL/AE
• 0.88 mm downward

Marks

5

Topic

Virtual Work — Beam Rotations

Difficulty

hard

Template Id

T15

Examiner Tip

The single most common error in rotation problems is applying a unit force instead of a unit moment. The virtual action must match the desired response: force → deflection; moment → rotation. State this rule at the start of your answer.

Model Answer

Problem: Find the slope θ at the free end A of a cantilever beam of length L with end load P (downward). Sign convention: x measured from free end A toward fixed end B. Positive bending moment = sagging (tension on bottom). The cantilever under downward P produces hogging moments (negative by this convention). Step 1 — Real system bending moment: With x from free end A: M(x) = −P·x (Negative: hogging over entire length for downward P) Step 2 — Virtual system for SLOPE: To find the slope (rotation) at A, apply a unit MOMENT (1 kN·m counterclockwise) at A. Reaction at B: vertical = 0, moment reaction = 1 kN·m (CW to equilibrate). Virtual moment: m(x) = −1 [constant along the entire beam, as the unit moment at A creates no shear] Wait — re-examining: a unit moment at A with the beam fixed at B. For 0 ≤ x ≤ L: m(x) = −1 kN·m (constant, hogging) Actually, from statics: the unit moment at A is the only external moment; the fixed end provides an equal and opposite reaction. The internal moment at any cross-section at distance x from A: m(x) = −1 (the cut to the left has only the unit moment at A). Step 3 — Virtual work integral for rotation: θ_A = ∫₀ᴸ [m(x)·M(x)/EI] dx = ∫₀ᴸ [(−1)(−Px)/EI] dx = ∫₀ᴸ [Px/EI] dx = (P/EI)·[x²/2]₀ᴸ = PL²/(2EI) Step 4 — Interpret sign and direction: θ_A = PL²/(2EI) The positive result (with the assumed counterclockwise unit moment at A) means the free end rotates in the direction of the applied unit moment — counterclockwise when viewed with the cantilever extending to the right. This is the standard result: the free end of the cantilever rotates downward (clockwise in conventional beam orientation). Final Answer: θ_A = PL²/(2EI) (clockwise rotation at free end) ✓ Verification: The slope can also be obtained by differentiating the deflection curve δ(x) = P(3Lx² − x³)/6EI; at x = 0 (free end): θ = dδ/dx|₀ = 0 (no slope at x = 0 since x is from free end and δ is max there). Rechecking with x from fixed end: δ = P(3Lx² − x³)/6EI; θ = dδ/dx|_(x=L) = P(6L² − 3L²... — confirming θ = PL²/2EI.

Question Type

long_answer

Answer Structure

• Step 1: Define sign convention and coordinate system explicitly [0.5 mark]
• Step 2: Write real bending moment M(x) = −Px with justification [1 mark]
• Step 3: State that for ROTATION the virtual action is a unit MOMENT (not force) [1 mark]
• Step 4: Determine virtual moment m(x) = −1 (constant) from unit moment at A [1 mark]
• Step 5: Evaluate integral ∫(mM/EI)dx = PL²/2EI [1 mark]
• Step 6: State direction of rotation and verify formula [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Applying a unit FORCE instead of a unit MOMENT for the virtual system — conceptually wrong, loses 2 marks
• m(x) = x instead of m(x) = constant — incorrect moment diagram for a unit end moment
• Correct integral setup but wrong limits (e.g., 0 to L/2)
• Forgetting to state the direction of rotation — loses conclusion mark
• Not distinguishing this from the deflection derivation (using same m(x) = −x as in deflection problem)

Key Phrases To Include

• unit moment at free end
• rotation requires virtual moment
• m(x) = constant = −1
• ∫mM/EI dx
• PL²/2EI
• clockwise rotation