CELE Strength of Materials — Columns and BucklingExam Answer Templates

Marks

1

Topic

Effective Length and End Conditions

Difficulty

easy

Template Id

T1

Examiner Tip

Even for 1-mark definitions, include the formula. Examiners reward the formula symbol alongside the words.

Model Answer

The effective length (Le = KL) of a column is the equivalent pin-ended length that produces the same Euler buckling load as the actual column with its given end conditions, where K is the effective length factor and L is the actual column length.

Question Type

very_short_answer

Answer Structure

• One complete sentence: define Le = KL and state its physical meaning [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining it as simply 'the length of the column' without mentioning end conditions
• Omitting the formula Le = KL
• Confusing effective length with unsupported length

Key Phrases To Include

• effective length
• Le = KL
• K = effective length factor
• end conditions
• equivalent pin-ended length

Marks

1

Topic

Euler's Buckling Formula

Difficulty

easy

Template Id

T2

Examiner Tip

The word 'least' before moment of inertia is the most tested nuance in buckling — always write it.

Model Answer

Euler's critical buckling load for a pin-ended column is: Pcr = π²EI / L², where E = modulus of elasticity (MPa), I = least moment of inertia of the cross-section (mm⁴), and L = unsupported length (mm).

Question Type

very_short_answer

Answer Structure

• Write the formula clearly [0.5 mark]
• Identify all three variables with units [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Writing I without specifying it is the LEAST (minimum) moment of inertia
• Omitting π² from the numerator
• Not including units for the variables

Key Phrases To Include

• Pcr = π²EI / L²
• least moment of inertia
• modulus of elasticity
• pin-ended

Marks

2

Topic

Slenderness Ratio and Column Classification

Difficulty

easy

Template Id

T3

Examiner Tip

Always pair the formula with its physical meaning. Examiners look for both quantitative definition and qualitative interpretation.

Model Answer

The slenderness ratio is defined as KL/r, where KL is the effective length and r = √(I/A) is the radius of gyration (least value). It is dimensionless. A high slenderness ratio indicates that the column is long and slender relative to its cross-sectional dimensions. Such a column will fail by elastic (Euler) buckling at a stress well below the material yield stress, making it susceptible to sudden lateral instability.

Question Type

short_answer

Answer Structure

• Line 1: Define slenderness ratio as KL/r and define r [1 mark]
• Line 2: State that high KL/r → elastic buckling / Euler regime / failure below yield stress [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using L/r instead of KL/r (forgetting the effective length factor K)
• Not specifying 'least' radius of gyration
• Saying high slenderness means 'strong' or 'safe' — it means MORE susceptible to buckling

Key Phrases To Include

• slenderness ratio = KL/r
• radius of gyration r = √(I/A)
• elastic buckling
• below yield stress
• Euler regime

Marks

2

Topic

Effective Length and End Conditions

Difficulty

easy

Template Id

T4

Examiner Tip

A small table of four rows is the fastest, clearest way to present end conditions. It takes 30 seconds and earns full marks.

Model Answer

The four common end conditions and their theoretical effective length factors K are: 1. Pinned–Pinned: K = 1.0 2. Fixed–Fixed: K = 0.5 3. Fixed–Pinned: K = 0.7 4. Fixed–Free (cantilever): K = 2.0 The Fixed–Fixed condition (K = 0.5) provides the highest buckling resistance because it produces the shortest effective length (Le = 0.5L), and since Pcr ∝ 1/(KL)², the critical load is four times that of a pin-ended column of the same length.

Question Type

short_answer

Answer Structure

• List all four end conditions with correct K values in a table or numbered list [1 mark]
• Identify Fixed-Fixed as strongest and justify using Pcr ∝ 1/(KL)² [1 mark]

Scoring Breakdown

Common Mark Deductions

• Confusing Fixed-Pinned (K=0.7) with Fixed-Free (K=2.0)
• Stating Fixed-Free has highest resistance — it is the WEAKEST
• Using recommended design K values (0.65, 0.80) when theoretical values are asked

Key Phrases To Include

• K = 0.5 for Fixed-Fixed
• K = 2.0 for Fixed-Free
• Pcr ∝ 1/(KL)²
• shortest effective length
• four times

Marks

3

Topic

Euler's Buckling Formula

Difficulty

medium

Template Id

T5

Examiner Tip

Write unit conversions explicitly in the 'Given' section. Many students lose marks computing the right number in the wrong unit.

Model Answer

Given: End condition: Pinned–Pinned → K = 1.0 L = 5 m = 5 000 mm I_min = 6 × 10⁶ mm⁴ E = 200 GPa = 200 000 MPa = 200 000 N/mm² Formula (Euler, general form): Pcr = π²EI / (KL)² Substituting: Pcr = π²(200 000)(6 × 10⁶) / (1.0 × 5 000)² Pcr = (9.8696)(200 000)(6 × 10⁶) / (25 × 10⁶) Pcr = 1.1844 × 10¹³ / 25 × 10⁶ Pcr = 4.737 × 10⁵ N ∴ Pcr = 474 kN

Question Type

numerical

Answer Structure

• Step 1: List all given data with unit conversions (mm, N/mm²) [0.5 mark]
• Step 2: Write the Euler formula Pcr = π²EI/(KL)² [0.5 mark]
• Step 3: Substitute values correctly [1 mark]
• Step 4: Compute numerator and denominator separately, then divide [0.5 mark]
• Step 5: State final answer in kN with correct rounding [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using L in meters while E and I are in N and mm — unit inconsistency
• Squaring only L and not KL (though K=1 here, showing the step matters for other K values)
• Using the larger moment of inertia instead of the least
• Forgetting to convert the final answer to kN

Key Phrases To Include

• K = 1.0 (pinned-pinned)
• Pcr = π²EI/(KL)²
• I_min
• E = 200 000 N/mm²
• L converted to mm

Marks

3

Topic

Effective Length and End Conditions

Difficulty

medium

Template Id

T6

Examiner Tip

The question specifies 'theoretical K' — always check whether theoretical (0.5, 0.7, 2.0) or recommended design values (0.65, 0.80, 2.10) are required.

Model Answer

Given: End condition: Fixed–Free → K = 2.0 (theoretical) L = 4 m = 4 000 mm I_min = 3.5 × 10⁶ mm⁴ E = 200 000 N/mm² Effective length: Le = KL = 2.0 × 4 000 = 8 000 mm Formula: Pcr = π²EI / (KL)² Substituting: Pcr = π²(200 000)(3.5 × 10⁶) / (8 000)² Pcr = (9.8696)(200 000)(3.5 × 10⁶) / (64 × 10⁶) Pcr = 6.909 × 10¹² / 6.4 × 10⁷ Pcr = 1.080 × 10⁵ N ∴ Pcr ≈ 108 kN Note: This is only 1/4 of the pin-ended value (since K² = 4.0 relative to K = 1.0), confirming Fixed-Free is the weakest configuration.

Question Type

numerical

Answer Structure

• Step 1: Identify end condition and K = 2.0 (theoretical) explicitly [0.5 mark]
• Step 2: Compute Le = KL = 8 000 mm [0.5 mark]
• Step 3: State and apply the Euler formula [1 mark]
• Step 4: Correct numerical answer ≈ 108 kN [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using K = 2.10 (recommended design value) when theoretical K is asked
• Forgetting to square the effective length in the denominator
• Using L = 4 000 mm (not applying K)

Key Phrases To Include

• Fixed-Free
• K = 2.0 theoretical
• Le = KL = 8 000 mm
• weakest end condition

Marks

3

Topic

Slenderness Ratio and Column Classification

Difficulty

medium

Template Id

T7

Examiner Tip

Use a structured numbered list. Examiners check for three distinct categories with three distinct failure modes.

Model Answer

Columns are classified based on slenderness ratio KL/r: 1. Short Columns (low KL/r): Failure mode: Material crushing/yielding when σ = σy. Design: P = σy × A (direct compressive strength). 2. Intermediate Columns (KL/r between short and long limits): Failure mode: Inelastic buckling — combined yielding and lateral instability. Design: Empirical formulas are required, e.g., NSCP/AISC parabolic formula or the Rankine-Gordon formula. Pure Euler is unconservative here. 3. Long (Slender) Columns (KL/r > Cc, where Cc = √(2π²E/σy)): Failure mode: Elastic (Euler) buckling at stress below the proportional limit. Design: Euler formula Pcr = π²EI/(KL)² is valid and governs. Key point: As slenderness increases, the critical stress decreases rapidly. Most real structural columns fall in the intermediate range.

Question Type

short_answer

Answer Structure

• Define and describe short column: failure mode + design approach [1 mark]
• Define and describe intermediate column: inelastic buckling + empirical formulas [1 mark]
• Define and describe long column: elastic Euler buckling + Cc comparison [1 mark]

Scoring Breakdown

Common Mark Deductions

• Saying Euler applies to ALL columns — it only applies to long/slender ones
• Not mentioning Cc or a slenderness boundary between intermediate and long
• Describing only two categories instead of three

Key Phrases To Include

• crushing/yielding
• inelastic buckling
• elastic (Euler) buckling
• Cc = √(2π²E/σy)
• empirical formulas
• Rankine-Gordon
• unconservative

Marks

5

Topic

Slenderness Ratio and Column Classification

Difficulty

hard

Template Id

T8

Examiner Tip

The three-step sequence — compute KL/r, compute Cc, compare — must be shown explicitly every time. It earns marks even if arithmetic is slightly off.

Model Answer

Given: K = 1.0 (pin-pin), L = 6 000 mm, r_min = 42 mm A = 7 200 mm², E = 200 000 MPa, σy = 250 MPa (a) Slenderness Ratio: KL/r = (1.0 × 6 000) / 42 = 142.9 (b) Limiting Slenderness Cc: Cc = √(2π²E / σy) Cc = √(2 × π² × 200 000 / 250) Cc = √(2 × 9.8696 × 200 000 / 250) Cc = √(15 791.4) Cc = 125.7 (c) Classification and Critical Stress: Since KL/r = 142.9 > Cc = 125.7, the column is a LONG (SLENDER) column → Euler formula applies. σcr = π²E / (KL/r)² σcr = π²(200 000) / (142.9)² σcr = 1 973 921 / 20 420.4 σcr = 96.7 MPa Check: σcr = 96.7 MPa < σy = 250 MPa ✓ (Euler is valid — stress is below yield) Critical load: Pcr = σcr × A = 96.7 × 7 200 = 696 240 N ≈ 696 kN ∴ The column fails by elastic buckling at σcr = 96.7 MPa; Pcr ≈ 696 kN.

Question Type

numerical

Answer Structure

• Part (a): Compute KL/r = 142.9 [1 mark]
• Part (b): Set up and solve for Cc = 125.7 [1 mark]
• Part (c): Compare KL/r vs Cc and correctly classify as long/slender [1 mark]
• Part (c): Apply Euler σcr formula and compute 96.7 MPa [1 mark]
• Part (c): Verify σcr < σy and compute Pcr in kN [1 mark]

Scoring Breakdown

Common Mark Deductions

• Computing KL/r but skipping the Cc comparison before applying Euler
• Applying intermediate-column formula when Euler is correct, or vice versa
• Not verifying that σcr < σy (omitting the validity check)
• Not converting L to mm before computing KL/r

Key Phrases To Include

• KL/r = 142.9
• Cc = √(2π²E/σy) = 125.7
• KL/r > Cc → long column
• Euler governs
• σcr = π²E/(KL/r)²
• σcr < σy (validity check)

Marks

3

Topic

Intermediate-Column Formulas (Rankine-Gordon)

Difficulty

medium

Template Id

T9

Examiner Tip

Show the denominator computation as a separate numbered step. Students who rush this step make the most arithmetic errors.

Model Answer

Given: A = 5 000 mm², Le/r = 90, σy = 250 MPa, a = 1/7 500 Rankine-Gordon Formula: Pcr = (σy × A) / [1 + a(Le/r)²] Computing numerator: σy × A = 250 × 5 000 = 1 250 000 N Computing denominator: a(Le/r)² = (1/7 500)(90)² = (1/7 500)(8 100) = 8 100 / 7 500 = 1.08 1 + a(Le/r)² = 1 + 1.08 = 2.08 Critical load: Pcr = 1 250 000 / 2.08 = 600 962 N ∴ Pcr ≈ 601 kN

Question Type

numerical

Answer Structure

• Step 1: Write the Rankine-Gordon formula [0.5 mark]
• Step 2: Compute numerator = σy × A [0.5 mark]
• Step 3: Compute denominator step-by-step: a(Le/r)², then add 1 [1 mark]
• Step 4: Divide and state final answer in kN [1 mark]

Scoring Breakdown

Common Mark Deductions

• Forgetting to add 1 to a(Le/r)² in the denominator (getting denominator = 1.08 instead of 2.08)
• Squaring the entire Le/r fraction incorrectly — must square the numerical value of Le/r first
• Not converting the answer from N to kN

Key Phrases To Include

• Pcr = σyA / [1 + a(Le/r)²]
• a = 1/7500
• denominator = 1 + a(Le/r)²
• numerator = σy × A

Marks

3

Topic

Intermediate-Column Formulas

Difficulty

hard

Template Id

T10

Examiner Tip

The word 'unconservative' (meaning the formula predicts a HIGHER load than actual) is the key engineering word here. Use it explicitly.

Model Answer

Euler's formula assumes the column material remains perfectly elastic up to the point of buckling. For intermediate columns (KL/r between zero and Cc), the actual compressive stress at buckling can equal or exceed the proportional limit, meaning the material yields before or simultaneously with buckling. This is called inelastic buckling. Additionally, real steel columns contain residual stresses from manufacturing (rolling, welding, cooling). These stresses effectively lower the proportional limit to approximately σy/2, meaning yielding begins at a lower applied load than a residually stress-free column would suggest. AISC/NSCP account for this by setting the elastic-inelastic boundary at Cc = √(2π²E/σy) (based on σy/2 rather than σy). Euler predicts a higher critical stress than the column can actually sustain in the intermediate range — it is therefore unconservative (unsafe). The Rankine-Gordon formula (or the NSCP inelastic buckling equation) provides a continuous empirical curve that matches test data better by blending the crushing limit (short columns) and the Euler limit (long columns), giving conservative results across the intermediate range.

Question Type

short_answer

Answer Structure

• Explain Euler's elastic assumption and its violation in intermediate range — inelastic buckling [1 mark]
• Mention residual stresses in steel and the σy/2 basis for Cc [1 mark]
• State that Rankine/NSCP formula bridges the short-column and Euler limits — conservative [1 mark]

Scoring Breakdown

Common Mark Deductions

• Saying Euler is conservative (it is actually unconservative for intermediate columns)
• Not mentioning residual stresses as a practical reason for the correction
• Treating Rankine as only a historical formula without relating it to code philosophy

Key Phrases To Include

• inelastic buckling
• proportional limit
• residual stresses
• Cc = √(2π²E/σy)
• unconservative
• Rankine-Gordon
• blends crushing and Euler limits

Marks

5

Topic

Column Classification and Formula Comparison

Difficulty

hard

Template Id

T11

Examiner Tip

A comparison table at the end (three rows, three columns: Method | σcr | P) earns presentation marks and clearly demonstrates engineering judgment.

Model Answer

Given: A = 4 000 mm², Le/r = 100, σy = 248 MPa, E = 200 000 MPa --- Classification --- Cc = √(2π²E / σy) = √(2 × 9.8696 × 200 000 / 248) = √(15 894) = 126.1 Since KL/r = 100 < Cc = 126.1 → INTERMEDIATE column (a) Pure Crushing Load: P_crush = σy × A = 248 × 4 000 = 992 000 N = 992 kN (Upper bound — ignores buckling entirely; highly unconservative for slender columns.) (b) Euler Critical Stress and Load: σcr,Euler = π²E / (Le/r)² = π²(200 000) / (100)² = 1 973 921 / 10 000 = 197.4 MPa P_Euler = σcr × A = 197.4 × 4 000 = 789 600 N ≈ 790 kN (Still exceeds σy/2 = 124 MPa, so inelastic — Euler is also unconservative.) (c) Rankine-Gordon Load: Pcr,Rankine = σy × A / [1 + a(Le/r)²] = (248 × 4 000) / [1 + (1/7 500)(100)²] = 992 000 / [1 + 1.333] = 992 000 / 2.333 = 425 300 N ≈ 425 kN Summary: P_crush = 992 kN (unconservative — ignores buckling) P_Euler = 790 kN (unconservative — inelastic range, elastic assumption violated) P_Rankine = 425 kN (most appropriate for intermediate column) Conclusion: For an intermediate column (KL/r < Cc), the Rankine-Gordon (or NSCP inelastic formula) governs. Euler over-predicts by ~86% compared to Rankine, confirming it is unsafe to use Euler here.

Question Type

numerical

Answer Structure

• Classify column: compute Cc and compare KL/r — intermediate [1 mark]
• Compute P_crush = 992 kN [0.5 mark]
• Compute P_Euler = 790 kN using σcr = π²E/(KL/r)² [1 mark]
• Compute P_Rankine = 425 kN using full denominator [1.5 marks]
• Present comparison table/summary and state Rankine is most appropriate with justification [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not classifying the column before applying formulas
• Applying Euler and concluding it is correct without checking against Cc
• Arithmetic error in Rankine denominator (forgetting to add 1)
• Omitting the summary/comparison — just giving three numbers without comment

Key Phrases To Include

• Cc = 126.1
• intermediate column
• P_crush = 992 kN
• P_Euler = 790 kN
• P_Rankine = 425 kN
• Euler unconservative
• Rankine most appropriate

Marks

2

Topic

Euler's Buckling Formula — Radius of Gyration

Difficulty

easy

Template Id

T12

Examiner Tip

Always use the chain of logic: least I → least r → greatest KL/r → lowest Pcr. Examiners want to see this reasoning chain, not just the formula.

Model Answer

The radius of gyration (r) of a cross-section is defined as: r = √(I/A) where I is the moment of inertia and A is the cross-sectional area. It represents the distribution of area relative to the axis of bending. In column buckling, the LEAST radius of gyration (r_min = √(I_min/A)) is used because buckling occurs about the axis of LEAST stiffness — i.e., the axis with the smallest I (weakest axis). Using the least r gives the largest slenderness ratio KL/r and the lowest critical stress σcr, which is the governing (most critical) condition. Using any other r would overestimate the column's buckling resistance.

Question Type

short_answer

Answer Structure

• Define r = √(I/A) and state its physical meaning [1 mark]
• Explain that buckling occurs about weakest axis → least I → least r → largest KL/r → lowest σcr [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining r correctly but not explaining WHY the least value is used
• Saying the greatest r is used — this is wrong; least r gives the critical case

Key Phrases To Include

• r = √(I/A)
• least moment of inertia
• weakest axis
• largest slenderness ratio
• lowest critical stress
• governing condition

Marks

3

Topic

Euler's Buckling Formula

Difficulty

medium

Template Id

T13

Examiner Tip

Derivation questions require every algebraic step shown. Skipping even one intermediate step can cost a mark.

Model Answer

Starting from Euler's critical buckling load: Pcr = π²EI / (KL)² ... (1) Divide both sides by the cross-sectional area A to get the critical stress: σcr = Pcr / A = π²EI / [A(KL)²] ... (2) Substitute I = Ar² (where r = √(I/A) is the radius of gyration): σcr = π²E(Ar²) / [A(KL)²] σcr = π²Er² / (KL)² σcr = π²E / (KL/r)² ... (3) Result: σcr = π²E / (KL/r)² Significance of KL/r (slenderness ratio): σcr is inversely proportional to the SQUARE of the slenderness ratio. This means: • Doubling KL/r reduces σcr by a factor of 4. • High slenderness → very low critical stress → column buckles far below yield. • This makes KL/r the single most important parameter in column design: a small change in effective length or cross-sectional shape has a powerful effect on buckling resistance.

Question Type

short_answer

Answer Structure

• Start from Pcr = π²EI/(KL)² and divide by A [0.5 mark]
• Substitute I = Ar² and simplify to σcr = π²E/(KL/r)² [1 mark]
• State the σcr ∝ 1/(KL/r)² relationship and physical significance [1.5 marks]

Scoring Breakdown

Common Mark Deductions

• Skipping the substitution I = Ar² — just quoting the final formula without derivation
• Not completing the simplification to the (KL/r)² form
• Providing the formula without discussing the significance of the slenderness ratio

Key Phrases To Include

• divide by A
• I = Ar²
• σcr = π²E/(KL/r)²
• inversely proportional to square of KL/r
• doubling KL/r reduces σcr by factor of 4

Marks

5

Topic

Combined: End Conditions + Slenderness + Classification + Euler

Difficulty

hard

Template Id

T14

Examiner Tip

When KL/r is close to Cc, always state both values and explicitly compare. The comparison statement itself earns a classification mark.

Model Answer

Given: Fixed-Pinned → K = 0.7 (theoretical) L = 6 000 mm, I_min = 12 × 10⁶ mm⁴, A = 9 600 mm² E = 200 000 MPa, σy = 250 MPa (a) Radius of Gyration and Slenderness Ratio: r_min = √(I/A) = √(12 × 10⁶ / 9 600) = √1 250 = 35.36 mm KL = 0.7 × 6 000 = 4 200 mm KL/r = 4 200 / 35.36 = 118.8 (b) Limiting Slenderness Cc: Cc = √(2π²E / σy) = √(2 × 9.8696 × 200 000 / 250) = √(15 791) = 125.7 (c) Classification: KL/r = 118.8 < Cc = 125.7 ∴ The column is an INTERMEDIATE column → Euler alone is unconservative. However, since KL/r = 118.8 is very close to Cc = 125.7 (within ~6%), the board exam commonly accepts using the Euler formula here with a note about conservatism. For full code compliance, NSCP/AISC inelastic formula should be used. Using Euler for illustration (common in board-exam context): σcr = π²E / (KL/r)² = π²(200 000) / (118.8)² = 1 973 921 / 14 113.4 = 139.9 MPa (d) Critical Load: Pcr = σcr × A = 139.9 × 9 600 = 1 342 900 N ≈ 1 343 kN Validity check: σcr = 139.9 MPa < σy = 250 MPa ✓ Note: Since this is an intermediate column, the actual design load using NSCP would be slightly lower. For board-exam purposes, Euler is commonly applied when KL/r ≈ Cc.

Question Type

numerical

Answer Structure

• Part (a): Compute r_min = 35.36 mm and KL/r = 118.8 [1 mark]
• Part (b): Compute Cc = 125.7 [1 mark]
• Part (c): Compare and correctly classify as intermediate [1 mark]
• Part (d): Apply Euler formula and compute σcr = 139.9 MPa [1 mark]
• Part (d): Compute Pcr = 1 343 kN with validity check [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using K = 0.80 (recommended design) when theoretical K is asked
• Forgetting to compute r from I and A — directly using r without derivation
• Classifying as 'long' because KL/r is close to Cc without careful comparison

Key Phrases To Include

• K = 0.7 (Fixed-Pinned theoretical)
• r_min = √(I/A)
• KL/r = 118.8
• Cc = 125.7
• intermediate column
• Euler unconservative for strict analysis

Marks

3

Topic

Eccentrically Loaded Columns

Difficulty

medium

Template Id

T15

Examiner Tip

Always draw a small sketch showing the eccentricity direction and which fiber is in maximum compression. It takes 20 seconds and prevents sign errors.

Model Answer

Given: P = 500 kN = 500 000 N, e = 30 mm, c = 80 mm A = 8 000 mm², I = 16 × 10⁶ mm⁴ Method: Combined direct stress + bending stress (P/A + Mc/I) Moment due to eccentricity: M = P × e = 500 000 × 30 = 15 × 10⁶ N·mm Direct compressive stress: σ_direct = P/A = 500 000 / 8 000 = 62.5 MPa (compression) Bending stress at extreme fiber: σ_bending = Mc/I = (15 × 10⁶ × 80) / (16 × 10⁶) = 1 200 × 10⁶ / 16 × 10⁶ = 75 MPa Maximum compressive stress (both stresses add on the compression side): σ_max = σ_direct + σ_bending = 62.5 + 75 = 137.5 MPa (compression) ∴ σ_max = 137.5 MPa (compressive)

Question Type

numerical

Answer Structure

• Step 1: Compute M = Pe = 15 × 10⁶ N·mm [0.5 mark]
• Step 2: Compute direct stress σ = P/A = 62.5 MPa [0.5 mark]
• Step 3: Compute bending stress σ = Mc/I = 75 MPa [1 mark]
• Step 4: Add both stresses and state σ_max = 137.5 MPa compressive [1 mark]

Scoring Breakdown

Common Mark Deductions

• Subtracting bending from direct stress — they ADD on the compression side
• Using the wrong c value — must use distance to the extreme fiber on the same side as the eccentricity
• Not converting P from kN to N before dividing by A

Key Phrases To Include

• M = Pe
• σ = P/A + Mc/I
• direct stress = 62.5 MPa
• bending stress = 75 MPa
• compression side