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

Examiner tip

One clean, technically precise sentence is all that is needed. Examiners penalize vague answers like 'the moment when the beam is fixed' because 'fixed' is not qualified.

Model answer

A fixed-end moment (FEM) is the moment developed at either end of a fully fixed (built-in) beam when it is subjected to transverse loading, assuming both ends are restrained against rotation and translation.

Answer structure

• One precise sentence defining FEM: mention fully-fixed condition AND moment at the ends due to loading [1 mark]

Scoring breakdown

Common mark deductions

• Defining FEM as 'the moment at a support' without specifying fixed-end boundary conditions.
• Confusing FEM with reaction moment at a pinned support.

Key phrases to include

• fully fixed
• restrained against rotation
• transverse loading
• moment at the ends

Examiner tip

Memorize all five FEM cases in the table. The UDL case is the most tested. The denominator 12 distinguishes it from the simple-beam moment (denominator 8).

Model answer

Using the clockwise-positive convention:
FEM_AB = −wL²/12 (near end A)
FEM_BA = +wL²/12 (far end B)

Answer structure

• State both FEM expressions with correct signs and formula [1 mark]

Scoring breakdown

Common mark deductions

• Both FEMs given as positive (sign error).
• Writing wL²/8 — confusing with the simple-span maximum moment.
• Omitting the sign convention statement.

Key phrases to include

• wL²/12
• clockwise-positive
• FEM_AB
• FEM_BA
• opposite signs

Examiner tip

The examiner wants to see that you know DF is a property of the JOINT, not the member in isolation. Show the Σk denominator explicitly.

Model answer

The distribution factor for a member at a joint is the fraction of the unbalanced moment absorbed by that member:
DF = k / Σk
where k = 4EI/L is the stiffness of the member (or 3EI/L if the far end is pinned) and Σk is the sum of stiffnesses of all members meeting at the joint.

At a fixed support, the joint cannot rotate; consequently, the support absorbs no moment from the distribution process. Therefore, DF = 0 at a fixed far end.

Answer structure

• Line 1–2: Define DF with formula DF = k/Σk and define k [1 mark]
• Line 3–4: State that DF at a fixed support = 0 and justify why [1 mark]

Scoring breakdown

Common mark deductions

• Stating DF = 1 at a fixed support (common confusion with carry-over).
• Giving the formula without defining k.
• Not distinguishing between k = 4EI/L and k = 3EI/L.

Key phrases to include

• DF = k/Σk
• k = 4EI/L
• k = 3EI/L for far-end pin
• DF = 0 at fixed support
• cannot rotate

Examiner tip

The coefficient pattern (2, 2, 1, −3) is the most tested detail. Write the equation first, then define terms — never the reverse.

Model answer

The slope-deflection equation for end A of member AB is:
M_AB = (2EI/L)(2θ_A + θ_B − 3ψ) + FEM_AB

where:

• M_AB = moment at end A of member AB (kN·m)
• E = modulus of elasticity of the member (kN/m²)
• I = second moment of area of the cross-section (m⁴)
• L = length of the member (m)
• θ_A = rotation of joint A (rad), positive clockwise
• θ_B = rotation of joint B (rad), positive clockwise
• ψ = Δ/L = chord rotation due to relative settlement Δ (rad)
• FEM_AB = fixed-end moment at A due to applied loading

Answer structure

• Line 1: Write the equation in full [1 mark]
• Lines 2–8: Define each symbol with units [1 mark]

Scoring breakdown

Common mark deductions

• Writing coefficient as 4EI/L in the equation (confusion with stiffness k).
• Omitting the −3ψ term.
• Not defining ψ as the chord rotation Δ/L.

Key phrases to include

• 2EI/L
• 2θ_A + θ_B
• −3ψ
• chord rotation
• ψ = Δ/L
• FEM_AB

Examiner tip

Note that |FEM_AB| > |FEM_BA| because the load is closer to A. This asymmetry is a useful self-check. If your answers are equal, recheck a and b.

Model answer

Given: L = 6 m, P = 40 kN, a = 2 m (from A), b = L − a = 4 m
Sign convention: clockwise-positive on the member.

FEM_AB = −Pab²/L² = −(40)(2)(4²)/(6²) = −(40 × 2 × 16)/36 = −1280/36 = −35.6 kN·m

FEM_BA = +Pa²b/L² = +(40)(2²)(4)/(6²) = +(40 × 4 × 4)/36 = +640/36 = +17.8 kN·m

∴ FEM_AB = −35.6 kN·m, FEM_BA = +17.8 kN·m

Answer structure

• Step 1: Identify given data and compute a and b [0.5 mark implicit — sets up formula correctly]
• Step 2: Apply FEM_AB = −Pab²/L² and substitute [1 mark]
• Step 3: Apply FEM_BA = +Pa²b/L² and substitute [1 mark]

Scoring breakdown

Common mark deductions

• Swapping a and b (computing −Pa²b/L² for FEM_AB).
• Using the central-load formula −PL/8 instead of the off-center formula.
• Omitting the negative sign on FEM_AB.

Key phrases to include

• FEM_AB = −Pab²/L²
• FEM_BA = +Pa²b/L²
• a = 2 m
• b = 4 m
• clockwise-positive

Examiner tip

Linking the carry-over factor back to the slope-deflection equation demonstrates deeper understanding and usually earns the bonus reasoning mark.

Model answer

Carry-Over Factor (COF):
When a moment is balanced at a joint and distributed to the near end of a member, exactly half of that distributed moment is automatically induced at the far end of the same member — this fraction is the carry-over factor.

COF = 1/2 (for a prismatic member with a fixed far end)

Physical basis: Distributing moment M at the near end of a fixed–fixed prismatic member produces, from the slope-deflection equation, a moment of M/2 at the far end (the cross-stiffness term θ_B = 0, θ_A ≠ 0 yields M_BA = (2EI/L)(θ_A) = ½ × M_AB).

For a pinned far end: The pin cannot sustain a moment (M = 0 at pin). If the reduced stiffness k = 3EI/L is used for that member at the outset, the carry-over to the pinned end is zero — there is nothing to carry over to a free-moment boundary. This eliminates one full iteration cycle and simplifies the distribution table.

Answer structure

• Paragraph 1: Define carry-over factor [1 mark]
• Paragraph 2: State COF = 1/2 for fixed far end and derive or explain its basis [1 mark]
• Paragraph 3: Explain why COF = 0 (or k = 3EI/L used) for pinned far end [1 mark]

Scoring breakdown

Common mark deductions

• Stating COF = 1 instead of 1/2.
• Not explaining the physical or mathematical reason for 1/2.
• Not mentioning the modified stiffness 3EI/L for pinned far end.

Key phrases to include

• carry-over factor = 1/2
• fixed far end
• pinned far end
• k = 3EI/L
• zero moment at pin
• cross-stiffness

Examiner tip

The answer −wL²/8 = −25 kN·m is the well-known propped cantilever result. If you get this, you have a strong cross-check. Stating it explicitly shows exam savvy.

Model answer

Sign convention: clockwise-positive. No settlement, so ψ = 0.

Step 1 — Identify unknowns.
Joint A: fixed → θ_A = 0.
Joint B: pin → M_BA = 0 (boundary condition); θ_B = unknown.

Step 2 — Fixed-end moments.
FEM_AB = −wL²/12 = −8(5²)/12 = −16.67 kN·m
FEM_BA = +wL²/12 = +8(5²)/12 = +16.67 kN·m

Step 3 — Slope-deflection equations.
M_AB = (2EI/L)(2θ_A + θ_B − 0) + FEM_AB
= (2EI/5)(0 + θ_B) − 16.67
= (2EI/5)θ_B − 16.67

M_BA = (2EI/L)(2θ_B + θ_A − 0) + FEM_BA
= (2EI/5)(2θ_B) + 16.67
= (4EI/5)θ_B + 16.67

Step 4 — Apply boundary condition at B: M_BA = 0.
(4EI/5)θ_B + 16.67 = 0
θ_B = −16.67 × 5/(4EI) = −20.83/EI

Step 5 — Back-substitute for M_AB.
M_AB = (2EI/5)(−20.83/EI) − 16.67
= −8.33 − 16.67
= −25.0 kN·m

∴ M_A = 25.0 kN·m (hogging / negative = tension on top)

Answer structure

• Step 1: Identify boundary conditions (θ_A = 0, M_BA = 0) [0.5 mark]
• Step 2: Correct FEMs [0.5 mark]
• Step 3: Write both slope-deflection equations [1 mark]
• Step 4–5: Apply M_BA = 0, solve θ_B, back-substitute [1 mark]

Scoring breakdown

Common mark deductions

• Using θ_A as an unknown instead of setting it to zero (fixed end).
• Incorrect FEM formula (e.g., wL²/8).
• Forgetting to add FEM when back-substituting.
• Final answer has correct magnitude but wrong sign/direction.

Key phrases to include

• θ_A = 0
• M_BA = 0
• FEM = ±wL²/12
• θ_B = −20.83/EI
• M_A = −25.0 kN·m
• hogging

Examiner tip

A clear comparison table in 3-mark answers almost always earns full marks because it organizes information exactly as the examiner's marking scheme is laid out.

Model answer

                 | indeterminacy                 | many displacement DOF
Basis of matrix  | Flexibility matrix [F]        | Stiffness matrix [K]; forms basis
analysis         |                               | of all modern FEM software
Example methods  | Three-moment equation,        | Slope-deflection, moment
                 | virtual work                  | distribution

Key distinction: In the force method, the structure is released and compatibility conditions restore continuity. In the displacement method, all joints are first locked (restraint), loads are applied, then joints are released sequentially while equilibrium is maintained — this scales efficiently for complex frames.

Answer structure

• Comparison table or paired paragraph covering unknowns [1 mark]
• Coverage of applicability / number of equations [1 mark]
• Computational advantage / basis for matrix analysis [1 mark]

Scoring breakdown

Common mark deductions

• Describing both methods using the same unknowns.
• Not linking displacement method to the stiffness/FEM matrix formulation.
• Vague statements without reference to degree of indeterminacy.

Key phrases to include

• redundant forces
• joint rotations and displacements
• static indeterminacy
• kinematic indeterminacy
• stiffness matrix
• FEM software
• equilibrium

Examiner tip

The moment distribution table is your work shown for partial marks. Even if you make an arithmetic error, a clearly laid-out table with correct methodology can earn 4 out of 5 marks. Never skip the table.

Model answer

Sign convention: clockwise-positive on member ends. No settlement.

--- Step 1: Fixed-End Moments ---
FEM_AB = −12(6²)/12 = −36.0 kN·m
FEM_BA = +12(6²)/12 = +36.0 kN·m
FEM_BC = 0 (no load on BC)
FEM_CB = 0

--- Step 2: Stiffness and Distribution Factors ---
k_BA = 4EI/6 = 0.667EI (A is fixed → use 4EI/L)
k_BC = 4EI/6 = 0.667EI (C is fixed → use 4EI/L)
Σk at B = 1.333EI
DF_BA = 0.667EI / 1.333EI = 0.50
DF_BC = 0.667EI / 1.333EI = 0.50
DF at A = 0 (fixed), DF at C = 0 (fixed)

--- Step 3: Moment Distribution Table ---
Joint: A BA BC C
FEM: −36.0 +36.0 0.0 0.0
Balance B: −18.0 −18.0
Carry-over: −9.0 −9.0 (CO = ½)

FINAL: −45.0 +18.0 −18.0 −9.0 (kN·m)

--- Step 4: Joint Equilibrium Check ---
At B: M_BA + M_BC = 18.0 + (−18.0) = 0 ✓

--- Results ---
M_A = −45.0 kN·m (hogging at A)
M_BA = +18.0 kN·m
M_BC = −18.0 kN·m
M_C = −9.0 kN·m (hogging at C)

Answer structure

• Step 1: Correct FEMs for both spans [1 mark]
• Step 2: Correct stiffness values and DF = 0.50 for both members at B; DF = 0 at fixed ends [1 mark]
• Step 3: Distribution table — balancing round with correct distributed moments [1 mark]
• Step 3 (cont.): Carry-over to fixed ends correctly (−9 kN·m to A and C) [1 mark]
• Step 4: Final moments correct AND joint equilibrium check shown [1 mark]

Scoring breakdown

Common mark deductions

• Distributing the full unbalanced moment (−36) to one side only.
• Carrying over 1/2 of the FEM instead of 1/2 of the balancing moment.
• Not showing the joint equilibrium check (costs the verification mark).
• Sign errors in carry-over (applying +9 instead of −9 to A).
• Using k = 3EI/L for fixed far ends (only used when far end is pinned).

Key phrases to include

• DF = k/Σk
• DF_BA = DF_BC = 0.50
• carry-over = 1/2
• Balance at B = −36 kN·m
• M_A = −45 kN·m
• M_C = −9 kN·m
• joint equilibrium check

Examiner tip

Recognizing and invoking symmetry to set θ_B = 0 and θ_C = −θ_A reduces the system from 3 equations to 1. Explicitly stating this symmetry argument earns a methodology mark.

Model answer

Sign convention: clockwise-positive. EI = constant, ψ = 0 (no settlement).

Step 1 — Degree of kinematic indeterminacy.
Unknown rotations: θ_A, θ_B, θ_C (with M_A = M_C = 0 at simple ends).
By symmetry of loading: θ_C = −θ_A, θ_B = 0.

Step 2 — Fixed-end moments.
FEM_AB = −wL²/12 = −10(6²)/12 = −30.0 kN·m
FEM_BA = +wL²/12 = +30.0 kN·m
FEM_BC = −30.0 kN·m (symmetry)
FEM_CB = +30.0 kN·m

Step 3 — Slope-deflection equations.
M_AB = (2EI/6)(2θ_A + θ_B) + FEM_AB
= (EI/3)(2θ_A) − 30.0 [since θ_B = 0]

M_BA = (2EI/6)(2θ_B + θ_A) + FEM_BA
= (EI/3)(θ_A) + 30.0 [since θ_B = 0]

Step 4 — Boundary condition at A: M_AB = 0.
(EI/3)(2θ_A) − 30.0 = 0
(2EI/3)θ_A = 30.0
θ_A = 45.0/EI = 45/EI

Step 5 — Compute M_BA.
M_BA = (EI/3)(45/EI) + 30.0 = 15.0 + 30.0 = 45.0 kN·m

Step 6 — By symmetry, M_BC = −45.0 kN·m.
Check joint B: M_BA + M_BC = 45.0 − 45.0 = 0 ✓

∴ M_B = 45.0 kN·m (interior support moment, hogging)

Answer structure

• Step 1: Identify unknowns, apply symmetry (θ_B = 0, θ_C = −θ_A) [0.5 mark]
• Step 2: Correct FEMs ±30.0 kN·m for both spans [1 mark]
• Step 3: Write slope-deflection equations for M_AB and M_BA [1 mark]
• Step 4: Apply M_AB = 0, solve θ_A = 45/EI [1 mark]
• Step 5–6: Compute M_BA = 45 kN·m, invoke symmetry, check joint B [1.5 marks]

Scoring breakdown

Common mark deductions

• Not invoking symmetry, leading to unnecessarily complex system of equations.
• Using FEM = ±wL²/8 (simple beam formula) instead of ±wL²/12.
• Not applying M_AB = 0 as the boundary condition for the simple end.
• Skipping the joint equilibrium check.

Key phrases to include

• FEM = ±wL²/12
• θ_B = 0 by symmetry
• M_AB = 0 at simple end
• θ_A = 45/EI
• M_B = 45.0 kN·m
• joint equilibrium check

Examiner tip

Always scan the structure for exterior simple supports or pinned ends BEFORE computing DFs. Using 3EI/L from the start is cleaner and faster — and examiners know whether you used it.

Model answer

When the far end of a member is a pin (or roller), the moment there is permanently zero. In this case, the effective rotational stiffness of the member as seen from the near joint is:
k = 3EI/L (modified stiffness for far-end pin)
instead of the standard 4EI/L for a fixed far end.

When to use: Apply k = 3EI/L in the distribution factor calculation at any joint connected to a member whose far end is pinned or has a free-moment boundary (e.g., an overhanging end, an exterior simple support).

Advantage: No carry-over moment is sent to the pinned far end (COF = 0 for that member), eliminating the need to balance that end in subsequent iterations. This shortens the distribution table and reduces arithmetic — a significant time-saver in board exams.

Answer structure

• Line 1–2: State formula k = 3EI/L and when it applies [1 mark]
• Line 3–4: State the advantage — zero carry-over, fewer iterations [1 mark]

Scoring breakdown

Common mark deductions

• Applying 3EI/L to both ends rather than only to members with a pinned far end.
• Not linking 3EI/L to the elimination of carry-over.
• Confusing modified stiffness with the distribution factor itself.

Key phrases to include

• k = 3EI/L
• far end pin or roller
• carry-over factor = 0
• fewer iterations
• zero moment boundary

Examiner tip

The memory trick: the larger FEM (denominator 20) goes to the end with MORE load (end B). Denominator 30 → lighter-load end A.

Model answer

FEM_AB = −wL²/30 (at the zero-load end A)
FEM_BA = +wL²/20 (at the peak-load end B)
(Clockwise-positive convention; the larger moment is at the loaded end B.)

Answer structure

• Both FEM formulas with correct denominators (30 and 20) and consistent signs [1 mark]

Scoring breakdown

Common mark deductions

• Using the same denominator for both ends (e.g., both wL²/12).
• Assigning the larger magnitude to the wrong end (FEM_AB = wL²/20 and FEM_BA = wL²/30 is reversed).
• Omitting the sign.

Key phrases to include

• wL²/30
• wL²/20
• zero-load end
• peak-load end
• larger moment at loaded end

Examiner tip

Board exam problems on frames almost always involve sidesway, making ψ essential. If a problem has a horizontal load or mentions an unbraced frame, always include ψ in your setup.

Model answer

Chord Rotation ψ:
The chord rotation of a member AB is defined as the relative transverse displacement Δ of one end with respect to the other, divided by the member length L:
ψ = Δ/L (radians)
It represents the rotation of the straight line (chord) joining the two deformed joint positions, NOT the rotation of the tangent to the elastic curve at the joint. ψ is positive when the chord rotates clockwise.

Effect on the slope-deflection equation:
The term −3ψ (equivalently −3Δ/L) is added inside the parenthesis of the slope-deflection equation:
M_AB = (2EI/L)(2θ_A + θ_B − 3ψ) + FEM_AB
This term subtracts from the end moments when the member sways — physically, sway reduces the restoring moment capacity of the member.

When must it be included:

• Whenever a support settles by an amount Δ relative to adjacent supports (e.g., differential settlement in continuous beams).
• In portal frames and multi-story frames where lateral sidesway (horizontal displacement Δ) occurs under horizontal loads.

If no settlement and no sidesway, ψ = 0 and the term is dropped.

Answer structure

• Paragraph 1: Define ψ = Δ/L — chord vs. tangent distinction [1 mark]
• Paragraph 2: Show how ψ enters the slope-deflection equation (−3ψ term) and its physical meaning [1 mark]
• Paragraph 3: Conditions requiring its inclusion: settlement, sidesway [1 mark]

Scoring breakdown

Common mark deductions

• Defining ψ as the tangent rotation at a joint (confusing with θ).
• Not mentioning the sign convention for ψ.
• Omitting sidesway as a case requiring ψ.

Key phrases to include

• ψ = Δ/L
• chord rotation
• −3ψ term
• support settlement
• sidesway
• clockwise positive

Examiner tip

Always verify that distribution factors at a free joint sum to exactly 1.0. This one-second check prevents a chain of errors in the distribution table.

Model answer

Since both far ends A and C are FIXED, use standard stiffness k = 4EI/L for both members.

k_BA = 4EI/6 = (2/3)EI
k_BC = 4EI/6 = (2/3)EI
Σk at B = (2/3)EI + (2/3)EI = (4/3)EI

DF_BA = k_BA / Σk = (2/3)EI / (4/3)EI = 0.50
DF_BC = k_BC / Σk = (2/3)EI / (4/3)EI = 0.50

Check: DF_BA + DF_BC = 0.50 + 0.50 = 1.00 ✓
(Note: DF at A = DF at C = 0 because fixed ends cannot rotate.)

Answer structure

• Step 1: Justify k = 4EI/L because both far ends are fixed [0.5 mark]
• Step 2: Compute k_BA, k_BC, Σk [0.5 mark]
• Step 3: Compute DF_BA = DF_BC = 0.50 with verification that DFs sum to 1.0 [1 mark]

Scoring breakdown

Common mark deductions

• Using k = 3EI/L for members with fixed far ends.
• Not checking that DF_BA + DF_BC = 1.0.
• Omitting the note that DF at A and C = 0.

Key phrases to include

• k = 4EI/L
• fixed far end
• Σk at B
• DF = 0.50
• sum of DFs = 1.0
• DF at fixed end = 0

Examiner tip

This type of question tests conceptual understanding as much as computation. The key insight — FEM is what the slope-deflection equation gives when all rotations are zero — is worth spelling out explicitly. It earns the reasoning mark.

Model answer

Sign convention: clockwise-positive. No settlement, ψ = 0.

--- Part 1: Fixed-End Moments ---
For a central load P at a = b = L/2 = 4 m:
FEM_AB = −PL/8 = −(60)(8)/8 = −60.0 kN·m
FEM_BA = +PL/8 = +60.0 kN·m

--- Part 2: Slope-Deflection Verification ---
Boundary conditions: Both ends fully fixed → θ_A = 0 and θ_B = 0.

Slope-deflection equations (ψ = 0, θ_A = 0, θ_B = 0):
M_AB = (2EI/L)(2θ_A + θ_B − 3ψ) + FEM_AB
= (2EI/8)(0 + 0 − 0) + (−60.0)
= −60.0 kN·m ✓

M_BA = (2EI/L)(2θ_B + θ_A − 0) + FEM_BA
= (2EI/8)(0 + 0) + (60.0)
= +60.0 kN·m ✓

The slope-deflection equations reduce to M = FEM when both joints are fully fixed (zero rotations), confirming that FEMs are the exact member-end moments for a fully fixed-fixed beam under load.

--- Results ---
M_AB = −60.0 kN·m (hogging at A)
M_BA = +60.0 kN·m (hogging at B)
These are equal in magnitude, consistent with the symmetric loading.

Equilibrium check — sum of moments on beam:
M_AB + M_BA = −60.0 + 60.0 = 0 (internal moments balance, consistent with symmetry) ✓

Answer structure

• Part 1: FEM_AB = −PL/8 = −60 kN·m, FEM_BA = +60 kN·m [1 mark]
• Part 2 setup: State θ_A = θ_B = 0 (fully fixed condition) [1 mark]
• Part 2 calculation: Substitute into slope-deflection equation for M_AB [1 mark]
• Part 2 calculation: Substitute for M_BA [1 mark]
• Conclusion: Show that M = FEM when θ = 0; present equilibrium/symmetry check [1 mark]

Scoring breakdown

Common mark deductions

• Using the off-center formula Pab²/L² with a ≠ b (central load has a = b = L/2).
• Not setting θ_A = θ_B = 0 explicitly for the fully fixed case.
• Skipping the conclusion/conceptual statement linking FEM to the θ = 0 result.
• Sign error: writing FEM_AB = +PL/8 (should be negative).

Key phrases to include

• FEM = PL/8
• central point load
• θ_A = θ_B = 0
• M = FEM when θ = 0
• −60 kN·m
• +60 kN·m
• symmetry

Do

• Write the formula or definition in one clean sentence.
• Include units if the question involves a quantity (e.g., kN·m).
• State sign convention if the answer involves a signed quantity.
• Use standard notation: M_AB, FEM_AB, DF, k, ψ — as expected in structural analysis.

Don't

• Do not derive the formula unless asked.
• Do not write a lengthy paragraph for a one-mark item.
• Do not skip the sign — a correct magnitude with wrong sign may receive zero marks.

Do

• State the formula first, then substitute values — never jump straight to the answer.
• If comparing two concepts, use a two-column layout or paired statements.
• Show the check (e.g., DF sums to 1.0) to capture the second mark.
• Label every intermediate result (e.g., k_BA = 0.667EI).

Don't

• Do not present two marks of content in a single run-on sentence.
• Do not omit the formula — substitution without a stated formula loses a mark.
• Do not round intermediate values excessively; carry at least 3 significant figures.

Do

• Write step numbers or headings (Step 1, Step 2, Step 3) — this signals structure to the examiner.
• Include a verification step (joint check, unit check, or cross-reference to a known result).
• Use a table for comparisons — it earns marks faster than prose.
• Cite the governing formula at the top of the solution block.

Don't

• Do not write a wall of text — examiners scan for mark-earning points, not narrative.
• Do not skip the derivation of θ (unknown rotation) — it is usually the second mark.
• Do not forget to back-substitute θ into the moment equation — back-substitution is the third mark.

Do

• Begin with: 'Given:', 'Find:', 'Sign convention:' — this professional format scores methodology marks.
• Draw and complete the full moment distribution table, labeling every row (FEM, Balance, CO, Final).
• For slope-deflection: write the complete equation for every member end, even those that simplify to zero.
• Box or underline the final answer with correct units and direction (hogging/sagging).
• Always conclude with the joint equilibrium check: M_near + M_far = 0 at each balanced joint.

Don't

• Do not start computing without writing the FEM step — jumping to distribution without FEMs is a common mistake that costs 1–2 marks.
• Do not omit the distribution factor derivation — it is always a separate scoring criterion.
• Do not confuse the carry-over direction: carry-over goes FROM the balanced joint TO the far end, NOT the reverse.
• Do not leave units off the final answer.
• Do not cross out a complete solution and restart unless you have time — partial-credit scoring means a flawed but complete attempt beats a blank page.