CELE Structural Theory & Analysis — Indeterminate Structures: Force MethodsExam Answer Templates

Marks

1

Topic

Degree of Indeterminacy

Difficulty

easy

Template Id

T1

Examiner Tip

Board examiners reward the word 'redundant' and the phrase 'equilibrium alone'. Including both in a single sentence almost guarantees full marks at this level.

Model Answer

A statically indeterminate structure is one in which the support reactions and/or internal forces cannot be determined by the equations of static equilibrium alone. The degree of indeterminacy (DI) is the number of redundant forces (reactions or internal forces) that must be removed to reduce the structure to a stable, determinate primary structure.

Question Type

very_short_answer

Answer Structure

• Sentence 1: Define 'statically indeterminate' — equilibrium insufficient [½ mark]
• Sentence 2: Define 'degree of indeterminacy' as the count of redundants [½ mark]

Scoring Breakdown

Common Mark Deductions

• Defining only 'indeterminate' without explaining DI — earns ½ mark only.
• Saying 'too many unknowns' without linking to the concept of redundants.

Key Phrases To Include

• equations of static equilibrium alone
• redundant forces
• degree of indeterminacy
• primary (released) structure
• stable and determinate

Marks

2

Topic

Method of Consistent Deformation

Difficulty

easy

Template Id

T2

Examiner Tip

Explicitly state 'Δ = 0 for an unyielding support' — this single phrase distinguishes a well-prepared answer from a memorised one and earns the second mark.

Model Answer

The compatibility equation is: δ₀ + R · δ₁₁ = Δ Where: • δ₀ = deflection at the redundant location in the primary structure due to applied loads (downward positive) • R = unknown redundant reaction • δ₁₁ = deflection at the redundant location per unit value of the redundant (flexibility coefficient) • Δ = prescribed displacement at the support (zero for an unyielding support; equals the settlement value if the support settles) Rearranging: R = (Δ − δ₀) / δ₁₁

Question Type

short_answer

Answer Structure

• Line 1: Write the symbolic equation δ₀ + R·δ₁₁ = Δ [1 mark]
• Lines 2–5: Define all four terms (δ₀, R, δ₁₁, Δ) with physical meaning [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing δ₀ = R·δ₁₁ (omitting Δ or setting it always to zero without justification).
• Not defining δ₁₁ as a per-unit-redundant value (confusing it with the total deflection from R).
• Omitting the rearranged formula for R.

Key Phrases To Include

• δ₀ — deflection due to applied loads on primary structure
• δ₁₁ — flexibility coefficient
• compatibility
• unyielding support (Δ = 0)
• redundant

Marks

2

Topic

Three-Moment Equation

Difficulty

medium

Template Id

T3

Examiner Tip

Board problems almost always use UDL or central point loads, so memorising 'wL³/4' and '3PL²/8' as the load terms converts a 2-mark question into a 30-second answer.

Model Answer

The three-moment equation (Clapeyron's theorem) for three consecutive supports A, B, C with spans L₁ (span AB) and L₂ (span BC) is: M_A·L₁ + 2M_B(L₁ + L₂) + M_C·L₂ = −6[(A₁x̄₁/L₁) + (A₂x̄₂/L₂)] Where: • M_A, M_B, M_C = bending moments at supports A, B, C respectively • L₁, L₂ = span lengths • A₁, A₂ = areas of the simple-beam (free) BMD of spans AB and BC • x̄₁, x̄₂ = distances from A and C respectively to the centroids of those areas For a UDL w on a span L: the load term 6Ax̄/L = wL³/4. For a central point load P on a span L: the load term 6Ax̄/L = 3PL²/8.

Question Type

short_answer

Answer Structure

• Line 1: Write the full symbolic three-moment equation [1 mark]
• Lines 2–5: Define M_A, M_B, M_C, L₁, L₂, A, x̄ with brief physical meaning [½ mark]
• Lines 6–7: State the UDL and central-point-load special cases for the load term [½ mark]

Scoring Breakdown

Common Mark Deductions

• Writing +6 instead of −6 on the right-hand side.
• Confusing x̄₁ (measured from A) and x̄₂ (measured from C).
• Not remembering the load-term formula for UDL — citing general Ax̄ without the simplified result.

Key Phrases To Include

• 2M_B(L₁ + L₂)
• simple-beam BMD area
• 6Ax̄/L
• UDL load term wL³/4
• central point load term 3PL²/8

Marks

3

Topic

Method of Consistent Deformation — Propped Cantilever (UDL)

Difficulty

medium

Template Id

T4

Examiner Tip

The formula R_B = 3wL/8 is a board standard. Write it as a named formula after the derivation — 'Using the propped-cantilever formula: R_B = 3wL/8' — this signals confident mastery and saves time.

Model Answer

Given: w = 8 kN/m, L = 5 m. Redundant: R_B (prop reaction at free end B). Primary structure: cantilever fixed at A. Step 1 — Deflection at B due to UDL on primary cantilever (downward): δ₀ = wL⁴ / (8EI) Step 2 — Deflection at B per unit of R_B on primary cantilever (upward): δ₁₁ = L³ / (3EI) Step 3 — Compatibility (unyielding support, Δ = 0): δ₀ = R_B · δ₁₁ wL⁴/(8EI) = R_B · L³/(3EI) R_B = 3wL/8 = 3(8)(5)/8 = 15 kN ← Step 4 — Vertical equilibrium: R_A = wL − R_B = 8(5) − 15 = 25 kN Step 5 — Moment equilibrium about A: M_A = wL²/2 − R_B·L [taking moments of all forces about A] = 8(5)²/2 − 15(5) = 100 − 75 = 25 kN·m (hogging) Alternatively, using the board formula: M_A = wL²/8 = 8(25)/8 = 25 kN·m ✓ Answer: R_B = 15 kN (upward); M_A = 25 kN·m (hogging).

Question Type

numerical

Answer Structure

• Line 1: Identify redundant and primary structure [setup — method mark]
• Step 1: Write δ₀ formula with substitution [1 mark]
• Step 2: Write δ₁₁ formula [included in method mark]
• Step 3: Write and solve compatibility equation → R_B [1 mark]
• Steps 4–5: Equilibrium to get R_A and M_A [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using wrong δ₀ formula (e.g., PL³/3EI for a point load instead of wL⁴/8EI for UDL).
• Forgetting to state the moment direction (hogging vs. sagging) — loses ½ mark in strict marking.
• Not writing the compatibility equation in symbolic form before substituting numbers.

Key Phrases To Include

• redundant R_B
• primary structure — cantilever fixed at A
• δ₀ = wL⁴/8EI
• δ₁₁ = L³/3EI
• compatibility: δ₀ = R_B · δ₁₁
• R_B = 3wL/8
• hogging

Marks

3

Topic

Method of Consistent Deformation — Propped Cantilever (Point Load)

Difficulty

medium

Template Id

T5

Examiner Tip

Memorise both cantilever deflection cases: tip load (PL³/3EI) and intermediate load (Pa²(3L−a)/6EI). The board regularly tests the intermediate case precisely because many candidates confuse the two.

Model Answer

Given: P = 16 kN, L = 4 m, load at a = L/2 = 2 m from A. Redundant: R_B. Primary: cantilever fixed at A. Step 1 — Free-end deflection δ₀ from P at midspan on primary cantilever: For a cantilever with a point load P at distance a from the fixed end: δ₀ = Pa²(3L − a) / (6EI) = 16(2)²[3(4) − 2] / (6EI) = 16(4)(10) / (6EI) = 640/(6EI) = 5PL³/(48EI) [substituting a = L/2] Step 2 — Flexibility coefficient: δ₁₁ = L³/(3EI) = 64/(3EI) Step 3 — Compatibility (Δ = 0): R_B = δ₀/δ₁₁ = [640/(6EI)] / [64/(3EI)] = 640 × 3 / (6 × 64) = 1920/384 = 5 kN ← Step 4 — Verification: R_B = 5P/16 = 5(16)/16 = 5 kN ✓ Answer: R_B = 5 kN (upward).

Question Type

numerical

Answer Structure

• Step 1: State and apply δ₀ = Pa²(3L−a)/6EI with a = L/2 [1 mark]
• Step 2: State δ₁₁ = L³/3EI [included in method mark]
• Step 3: Apply compatibility and solve for R_B numerically [1 mark]
• Step 4: Express as R_B = 5P/16 and verify [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using δ₀ = PL³/3EI (tip-load formula) instead of the intermediate-load formula — loses the first mark entirely.
• Arithmetic errors in substituting a = 2 m, L = 4 m into the formula.

Key Phrases To Include

• δ₀ = Pa²(3L − a)/6EI
• a = L/2 = 2 m
• 5PL³/48EI
• compatibility equation
• R_B = 5P/16

Marks

5

Topic

Three-Moment Equation — Two-Span Beam

Difficulty

medium

Template Id

T6

Examiner Tip

In a 5-mark problem, the examiner expects all five distinct steps shown. An equilibrium check (Step 4) is a dedicated mark in board exam rubrics — never skip it. Writing 'Check: ΣFy = 120 = 2wL ✓' takes 10 seconds and earns a full mark.

Model Answer

Given: L₁ = L₂ = L = 6 m, w = 10 kN/m, M_A = M_C = 0 (simple ends). Step 1 — Load terms (UDL, both spans): 6A₁x̄₁/L₁ = wL₁³/4 = 10(6)³/4 = 10(216)/4 = 540 kN·m² 6A₂x̄₂/L₂ = wL₂³/4 = 540 kN·m² (same by symmetry) Step 2 — Three-moment equation at B: M_A·L₁ + 2M_B(L₁ + L₂) + M_C·L₂ = −(540 + 540) 0 + 2M_B(6 + 6) + 0 = −1080 24M_B = −1080 M_B = −45 kN·m (hogging) ← Step 3 — Reactions (use superposition: simple-beam reaction ± correction from M_B): For span AB (M_A = 0, M_B = −45 kN·m): Simple-beam reaction at A = wL/2 = 10(6)/2 = 30 kN Correction from M_B: −M_B/L = −(−45)/6 = +7.5 kN (A gets negative, B gets positive) R_A = 30 − 7.5 = 22.5 kN ← R_B(left) = 30 + 7.5 = 37.5 kN For span BC (M_B = −45 kN·m, M_C = 0): by symmetry R_C = 22.5 kN ← R_B(right) = 37.5 kN R_B(total) = 37.5 + 37.5 = 75 kN ← Step 4 — Equilibrium check: ΣFy = 22.5 + 75 + 22.5 = 120 kN = w(2L) = 10(12) = 120 kN ✓ Answer: M_B = 45 kN·m (hogging) R_A = R_C = 22.5 kN; R_B = 75 kN

Question Type

numerical

Answer Structure

• Step 1: Compute 6Ax̄/L load terms for both spans [1 mark]
• Step 2: Write and apply three-moment equation; solve for M_B [1 mark]
• Step 3a: Derive R_A using superposition for span AB [1 mark]
• Step 3b: State R_C by symmetry and compute total R_B [1 mark]
• Step 4: Equilibrium check ΣFy = 0 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using 6Ax̄/L = wL³/12 (wrong formula — that is the area of the parabola, not the load term).
• Not applying the negative sign: M_B should come out negative (hogging) and students who omit the sign lose the moment-calculation mark.
• Computing reactions without accounting for the moment correction from M_B.
• Skipping the equilibrium check — this is a free mark that many candidates leave on the table.

Key Phrases To Include

• M_A = M_C = 0 (simple ends)
• 6Ax̄/L = wL³/4
• three-moment equation at B
• M_B = −45 kN·m (hogging)
• superposition of simple-beam reaction and moment correction
• ΣFy check

Marks

2

Topic

Method of Consistent Deformation — Flexibility Coefficients

Difficulty

easy

Template Id

T7

Examiner Tip

The phrase 'per unit value of the redundant applied to the primary structure' is the exact phrase examiners mark for. The word 'unit' must appear in your definition.

Model Answer

The flexibility coefficient δ₁₁ is the displacement (deflection or rotation) at the location and in the direction of redundant 1, caused by a unit value of redundant 1 applied to the primary (released) structure. Physical meaning: if the redundant R has a magnitude of R kN, then the displacement it produces at its own point of application on the primary structure is R·δ₁₁. Sketch description: [Primary structure (e.g., cantilever fixed at A, free at B)] Apply unit upward load at B → measure downward deflection at B = δ₁₁ = L³/3EI For a propped cantilever of span L and stiffness EI: δ₁₁ = L³/(3EI) [standard cantilever tip deflection per unit load]

Question Type

short_answer

Answer Structure

• Line 1: Define δ₁₁ formally — displacement at redundant location per unit redundant [1 mark]
• Line 2: Sketch showing primary structure + unit load → displacement [½ mark]
• Line 3: Give formula for at least one standard case [½ mark]

Scoring Breakdown

Common Mark Deductions

• Defining δ₁₁ as 'the deflection due to the redundant force' without specifying 'per unit' — loses ½ mark.
• Applying the unit load to the original (indeterminate) structure instead of the primary structure.

Key Phrases To Include

• unit value of the redundant
• primary (released) structure
• displacement at the redundant location
• flexibility coefficient
• L³/3EI for cantilever

Marks

1

Topic

Method of Consistent Deformation — Primary Structure

Difficulty

easy

Template Id

T8

Examiner Tip

Add the word 'stable' to your definition — examiners expect you to distinguish between a valid primary structure and a mechanism (which would result from removing a non-redundant reaction).

Model Answer

The primary structure is the stable, statically determinate structure obtained by removing the redundant reactions or internal forces from the original indeterminate structure. It is chosen by releasing exactly as many constraints as the degree of indeterminacy, ensuring the resulting structure remains geometrically stable (no mechanism). Common choices include: removing an interior support (releasing a reaction) or releasing an internal hinge (releasing a moment).

Question Type

very_short_answer

Answer Structure

• Sentence 1: Define primary structure as the determinate structure after removing redundants [½ mark]
• Sentence 2: State the stability requirement and give one example of a valid release [½ mark]

Scoring Breakdown

Common Mark Deductions

• Saying 'any part of the structure removed' without specifying redundants or the stability requirement.

Key Phrases To Include

• removing redundants
• statically determinate
• geometrically stable
• degree of indeterminacy

Marks

3

Topic

Three-Moment Equation — Unequal Spans

Difficulty

medium

Template Id

T9

Examiner Tip

Write the load term formula 'wL³/4' explicitly before substituting numbers. If you mis-calculate but the formula is correct, you still earn the method mark.

Model Answer

Given: L₁ = 5 m, L₂ = 7 m, w = 12 kN/m, M_A = M_C = 0. Step 1 — Load terms: 6A₁x̄₁/L₁ = wL₁³/4 = 12(5)³/4 = 12(125)/4 = 375 kN·m² 6A₂x̄₂/L₂ = wL₂³/4 = 12(7)³/4 = 12(343)/4 = 1029 kN·m² Step 2 — Three-moment equation at B: M_A·L₁ + 2M_B(L₁ + L₂) + M_C·L₂ = −(375 + 1029) 0 + 2M_B(5 + 7) + 0 = −1404 24M_B = −1404 M_B = −58.5 kN·m Answer: M_B = 58.5 kN·m (hogging) ←

Question Type

numerical

Answer Structure

• Step 1a: Compute 6Ax̄/L for span AB using wL₁³/4 [1 mark]
• Step 1b: Compute 6Ax̄/L for span BC using wL₂³/4 [included above]
• Step 2: Assemble and solve three-moment equation for M_B [2 marks]

Scoring Breakdown

Common Mark Deductions

• Computing 5³ = 125 and 7³ = 343 incorrectly — purely arithmetic, but costs the load-term mark.
• Using L₁ + L₂ = 12 but writing 2(12) = 24 as 48 — arithmetic slip.

Key Phrases To Include

• wL³/4 for each span
• M_A = M_C = 0 (simple supports)
• 2M_B(L₁ + L₂)
• M_B = 58.5 kN·m hogging

Marks

5

Topic

Method of Consistent Deformation — Fixed-Fixed Beam

Difficulty

hard

Template Id

T10

Examiner Tip

For fixed-fixed beams, the board formula is M_f = PL/8 (central load) and M_A = wL²/12 (UDL). Even if you derive it by consistent deformation, confirming with the board formula in the last line shows mastery and is a free verification mark.

Model Answer

Given: L = 6 m, P = 60 kN at midspan, fixed at both A and B (DI = 2 — both end moments are redundants). Symmetry: M_A = M_B = M_f (both fixed-end moments equal by symmetry). Step 1 — Choose redundants: M_A and M_B (end moments). Primary structure: simply-supported beam A to B. Step 2 — Midspan deflection due to P on the primary (SS) beam: δ₀ = PL³/(48EI) = 60(6)³/(48EI) = 60(216)/(48EI) = 270/EI [downward] Step 3 — Midspan deflection due to unit end moments M = 1 kN·m applied equally at both ends on the primary (SS) beam (this is equivalent to an end-moment loading): For equal end moments M₀ on a SS beam, the midspan deflection is: δ₁₁ = M₀L²/(8EI) per unit M₀ So with M₀ = 1: δ₁₁ = L²/(8EI) = 36/(8EI) = 4.5/EI [upward, opposing δ₀] Step 4 — Compatibility: midspan deflection of the actual fixed-fixed beam = 0 is automatically satisfied, but the END ROTATION must equal zero. Use slope compatibility at A: For P at midspan on SS beam: θ_A = PL²/(16EI) = 60(36)/(16EI) = 135/EI For equal end moments M_f on SS beam: θ_A = M_f·L/(6EI) + M_f·L/(3EI) = M_f·L/(2EI) [each end] Wait — use the standard rotation formula: for near end moment M_f: θ_A (from M_A alone) = M_f·L/(3EI) θ_A (from M_B alone) = M_f·L/(6EI) Total θ_A = M_f·L/(3EI) + M_f·L/(6EI) = M_f·L(2+1)/(6EI) = M_f·L/(2EI) Compatibility: θ_A(from loads) + θ_A(from M_f) = 0 [fixed end, no rotation] PL²/(16EI) − M_f·L/(2EI) = 0 [M_f opposes the rotation] M_f = PL/(8) × 2 / L × L = PL/8 M_f = PL/8 = 60(6)/8 = 45 kN·m Step 5 — Answer and check: M_A = M_B = PL/8 = 45 kN·m (hogging at both ends) Midspan moment = PL/4 − PL/8 = PL/8 = 45 kN·m (sagging) Check: reactions R_A = R_B = P/2 = 30 kN (by symmetry and ΣFy) ✓ Board formula: Fixed-end moment for central load = PL/8 = 60(6)/8 = 45 kN·m ✓ Answer: M_A = M_B = 45 kN·m (hogging); R_A = R_B = 30 kN.

Question Type

numerical

Answer Structure

• Step 1: Identify DI = 2, invoke symmetry to reduce to one unknown M_f [1 mark]
• Step 2: Compute the primary-structure slope/deflection due to P [1 mark]
• Step 3: Compute flexibility coefficient for end moments [1 mark]
• Step 4: Write and solve compatibility equation → M_f = PL/8 [1 mark]
• Step 5: State final answer with directions, verify with board formula and equilibrium [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not identifying DI = 2 and not explaining the use of symmetry — loses the first mark.
• Using deflection compatibility instead of slope compatibility (a fixed end has zero slope, not zero deflection at midspan).
• Forgetting the contribution of M_B to the slope at A when computing the flexibility coefficient.

Key Phrases To Include

• DI = 2 (fixed at both ends)
• symmetry: M_A = M_B
• primary structure: simply-supported beam
• slope compatibility at fixed end
• M_f = PL/8
• hogging at both ends

Marks

5

Topic

Three-Moment Equation — Support Settlement

Difficulty

hard

Template Id

T11

Examiner Tip

Convert mm to m before any substitution (write '10 mm = 0.01 m' explicitly on your paper). Board examiners regularly include a settlement problem precisely to check unit discipline. One forgotten unit conversion fails the entire numerical chain.

Model Answer

Given: L₁ = L₂ = L = 5 m, w = 12 kN/m, Δ_B = 10 mm = 0.01 m (downward), EI = 15,000 kN·m². M_A = M_C = 0 (simple ends). Step 1 — Load terms (UDL, equal spans): 6A₁x̄₁/L₁ = wL³/4 = 12(5)³/4 = 12(125)/4 = 375 kN·m² 6A₂x̄₂/L₂ = 375 kN·m² Step 2 — Modified three-moment equation with support settlement: The standard equation gains settlement terms on the right side: M_A·L₁ + 2M_B(L₁+L₂) + M_C·L₂ = −(ΣLoad terms) + 6EI[(δ_A − δ_B)/L₁ + (δ_C − δ_B)/L₂] Let δ_A = δ_C = 0 (no settlement at ends), δ_B = −0.01 m (downward = negative upward): Settlement terms = 6EI[(0 − (−0.01))/5 + (0 − (−0.01))/5] = 6(15000)[(0.01/5) + (0.01/5)] = 90,000 × [0.002 + 0.002] = 90,000 × 0.004 = 360 kN·m² Step 3 — Substitute into three-moment equation: 0 + 2M_B(5+5) + 0 = −(375 + 375) + 360 20M_B = −750 + 360 20M_B = −390 M_B = −19.5 kN·m Answer: M_B = 19.5 kN·m (hogging) Note: Settlement at B reduces the hogging moment compared to the no-settlement case (M_B = −wL²/8 = −37.5 kN·m). This is physically correct: settlement at B tends to relieve hogging.

Question Type

numerical

Answer Structure

• Step 1: Compute both load terms = 375 kN·m² each [1 mark]
• Step 2: Write modified three-moment equation with settlement terms [1 mark]
• Step 2b: Compute settlement term = 360 kN·m² [1 mark]
• Step 3: Substitute and solve → M_B = −19.5 kN·m [1 mark]
• Final: State result with sign, add physical interpretation [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not converting Δ from mm to metres — produces an answer 1000× too large.
• Using Δ_B as positive in the settlement term when the beam settles downward (sign error).
• Omitting the settlement terms entirely and using the standard three-moment equation — earns load-term marks but loses settlement marks.

Key Phrases To Include

• δ_B = 0.01 m downward
• 6EI(δ/L) settlement term
• modified three-moment equation
• settlement reduces hogging moment
• EI in kN·m² units

Marks

2

Topic

Force Method vs. Stiffness Method

Difficulty

easy

Template Id

T12

Examiner Tip

Frame advantages in terms of efficiency and physical insight; frame the disadvantage in terms of scalability. Avoid vague words like 'complex' — be specific about the mechanism that causes the difficulty.

Model Answer

Advantages of the force method: 1. The number of equations equals the degree of indeterminacy (DI), which is small for structures with few redundants (e.g., propped cantilever, two-span beam) — making hand calculation efficient. 2. It provides direct physical insight into the behaviour of each redundant and how the structure releases forces, aiding intuitive understanding. Disadvantage: The number of equations grows rapidly for highly indeterminate structures (DI >> 3), making the stiffness method more computationally efficient for computer-based analysis of complex frames and multi-storey buildings.

Question Type

short_answer

Answer Structure

• Advantage 1: Small equation system for low-DI structures [½ mark]
• Advantage 2: Physical insight / intuitive interpretation [½ mark]
• Disadvantage: Impractical for high DI (where stiffness method excels) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Listing the same advantage twice in different words (counts as one point).
• Saying 'disadvantage is it is difficult' without explaining WHY (high DI = many simultaneous equations).

Key Phrases To Include

• degree of indeterminacy
• number of equations = DI
• efficient for low DI
• stiffness method preferred for high DI / computer analysis

Marks

3

Topic

Propped Cantilever — Span Moment Location

Difficulty

medium

Template Id

T13

Examiner Tip

Taking moments from support B (the simpler end) avoids carrying the fixed-end moment M_A into the computation — a common source of sign errors. Always choose the free-moment end when computing span moments.

Model Answer

From the propped-cantilever solution: R_B = 3wL/8 (upward), R_A = 5wL/8 (upward), M_A = wL²/8 (hogging) The shear force at distance x from A: V(x) = R_A − wx = 5wL/8 − wx Maximum positive moment occurs where V = 0: 5wL/8 − wx₀ = 0 x₀ = 5L/8 Maximum positive (sagging) moment at x = x₀ = 5L/8: M_max = R_A·x₀ − w·x₀²/2 − M_A [Note: M_A acts at A only, not across the span] Alternatively: take moments from B (simpler): M(x₀ from B) = R_B(L − x₀) − w(L−x₀)²/2 L − x₀ = L − 5L/8 = 3L/8 M_max = (3wL/8)(3L/8) − w(3L/8)²/2 = 9wL²/64 − 9wL²/128 = 18wL²/128 − 9wL²/128 = 9wL²/128 Answer: Maximum span moment at x = 5L/8 from A (or 3L/8 from B), magnitude = 9wL²/128 (sagging).

Question Type

numerical

Answer Structure

• Line 1: State R_A = 5wL/8 from standard propped-cantilever result [1 mark]
• Lines 2–3: Set V(x) = 0 to find x₀ = 5L/8 [1 mark]
• Lines 4–7: Compute M at x₀ by taking moments from the simpler end (B) → 9wL²/128 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Confusing x₀ = 3L/8 from A (that is from B — must be consistent with measurement convention).
• Not stating 'sagging' for the span moment (the fixed-end moment is hogging — confusing the two).

Key Phrases To Include

• shear force V(x) = 0 for maximum moment
• R_A = 5wL/8
• x₀ = 5L/8 from A
• M_max = 9wL²/128
• sagging

Marks

1

Topic

Support Settlement — Effect on Indeterminate Structure

Difficulty

easy

Template Id

T14

Examiner Tip

Settlement questions test conceptual understanding, not just formula recall. State the physical direction of change ('decreases', 'relieves') — examiners at the PRC board reward physical reasoning in one-mark answers.

Model Answer

If the prop at B settles downward by Δ, the compatibility condition becomes δ₀ = R_B·δ₁₁ + Δ (the total downward displacement at B now equals Δ, not zero). Solving: R_B = (δ₀ − Δ)/δ₁₁. Since Δ reduces δ₀ effectively, R_B decreases, and consequently the fixed-end moment M_A = wL²/2 − R_B·L also decreases (less hogging). Settlement at the prop relieves both the prop reaction and the fixed-end moment.

Question Type

very_short_answer

Answer Structure

• Sentence 1: State modified compatibility condition (right-hand side = Δ, not 0) [½ mark]
• Sentence 2: Conclude that R_B decreases and therefore M_A decreases [½ mark]

Scoring Breakdown

Common Mark Deductions

• Saying 'M_A increases' — physically incorrect; settlement always reduces the redundant in a propped cantilever.
• Not linking the change in R_B to the consequent change in M_A.

Key Phrases To Include

• compatibility condition: Δ (not zero)
• R_B decreases
• fixed-end moment decreases
• settlement relieves hogging

Marks

3

Topic

Three-Moment Equation — Three-Span Beam

Difficulty

hard

Template Id

T15

Examiner Tip

The key rule: apply the three-moment equation once at each interior support. For a three-span beam, write two equations. Stating 'one equation per interior support' in your first line signals to the examiner that you understand the method.

Model Answer

Given: L₁ = L₂ = L₃ = L = 4 m, w = 15 kN/m, M_A = M_D = 0. Load term for each span: 6Ax̄/L = wL³/4 = 15(4)³/4 = 15(64)/4 = 240 kN·m² Equation 1 — Apply three-moment equation at interior support B (spans AB and BC): M_A·L + 2M_B(L + L) + M_C·L = −(240 + 240) 0 + 4L·M_B + L·M_C = −480 4(4)M_B + 4M_C = −480 16M_B + 4M_C = −480 … (i) Equation 2 — Apply three-moment equation at interior support C (spans BC and CD): M_B·L + 2M_C(L + L) + M_D·L = −(240 + 240) L·M_B + 4L·M_C + 0 = −480 4M_B + 16M_C = −480 … (ii) System of equations: 16M_B + 4M_C = −480 (i) 4M_B + 16M_C = −480 (ii) (By symmetry, M_B = M_C. Substitute: 20M_B = −480 → M_B = M_C = −24 kN·m) Answer: The two simultaneous equations are equations (i) and (ii) above. By symmetry, M_B = M_C = 24 kN·m (hogging).

Question Type

numerical

Answer Structure

• Compute load term = 240 kN·m² for each span [½ mark]
• Write three-moment equation at B (Equation i) with M_A = 0 [1 mark]
• Write three-moment equation at C (Equation ii) with M_D = 0 [1 mark]
• Invoke symmetry M_B = M_C and solve → 24 kN·m [½ mark]

Scoring Breakdown

Common Mark Deductions

• Writing only one equation (at B or C) for a three-span beam with two interior supports — only one equation is insufficient.
• Not substituting M_A = 0 or M_D = 0, leaving them as unknowns in the equations.

Key Phrases To Include

• one equation per interior support
• wL³/4 = 240 kN·m² per span
• M_A = M_D = 0
• number of unknowns = number of interior supports
• symmetry: M_B = M_C