CELE Steel & Timber Design — Steel Compression MembersExam Answer Templates

Marks

1

Topic

Elastic Buckling Stress

Difficulty

easy

Template Id

T1

Examiner Tip

One complete correct formula with variables defined earns full marks. No derivation needed at 1-mark level.

Model Answer

The elastic buckling stress is Fe = π²E / (KL/r)², where E is the modulus of elasticity of steel (200,000 MPa), K is the effective-length factor, L is the unbraced length, and r is the least radius of gyration. It represents the Euler critical stress at which a perfectly straight, elastic column becomes unstable.

Question Type

very_short_answer

Answer Structure

• Line 1: State the formula Fe = π²E / (KL/r)² and identify all variables [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing Fe = π²E / L² (omitting K and r) — loses the mark
• Confusing Fe with Fcr — these are different quantities

Key Phrases To Include

• Fe = π²E / (KL/r)²
• elastic buckling stress
• least radius of gyration
• effective-length factor K

Marks

1

Topic

Inelastic vs Elastic Buckling Transition

Difficulty

easy

Template Id

T2

Examiner Tip

Memorize 4.71, not 4.17. Write both the formula and the inequality direction to secure the full mark.

Model Answer

The transition slenderness ratio is KL/r = 4.71√(E/Fy). Members with KL/r ≤ 4.71√(E/Fy) undergo inelastic buckling; those with KL/r > 4.71√(E/Fy) undergo elastic buckling.

Question Type

very_short_answer

Answer Structure

• Line 1: State the transition value 4.71√(E/Fy) [0.5 mark]
• Line 2: State which region corresponds to inelastic and which to elastic [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Writing the coefficient as 4.17 instead of 4.71 (digit transposition)
• Reversing the inequality — stating elastic governs when KL/r is small

Key Phrases To Include

• 4.71√(E/Fy)
• transition slenderness
• inelastic buckling
• elastic buckling

Marks

2

Topic

Critical Stress Formulas

Difficulty

easy

Template Id

T3

Examiner Tip

Always write the condition alongside each formula. An examiner cannot give full marks if they cannot see that you know when each formula applies.

Model Answer

Under NSCP 2015 / AISC 360 Section E3: (1) Inelastic buckling — when KL/r ≤ 4.71√(E/Fy): Fcr = [0.658^(Fy/Fe)] × Fy (2) Elastic buckling — when KL/r > 4.71√(E/Fy): Fcr = 0.877 Fe where Fe = π²E / (KL/r)² is the elastic buckling stress. The factor 0.877 accounts for the effect of initial member out-of-straightness in the elastic range.

Question Type

short_answer

Answer Structure

• Line 1: Inelastic regime formula Fcr = [0.658^(Fy/Fe)]Fy with condition [1 mark]
• Line 2: Elastic regime formula Fcr = 0.877Fe with condition and explanation of the 0.877 factor [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing 0.685 instead of 0.658 as the exponent base — common typo that loses the mark
• Omitting the condition (inequality) that determines which formula to use
• Not defining Fe within the answer

Key Phrases To Include

• 0.658^(Fy/Fe)
• 0.877Fe
• inelastic buckling
• elastic buckling
• initial out-of-straightness
• 4.71√(E/Fy)

Marks

2

Topic

Design Strength

Difficulty

easy

Template Id

T4

Examiner Tip

φ_c = 0.90 for steel columns is worth at least one mark in almost every numerical problem. Never omit it.

Model Answer

The LRFD design axial compressive strength (NSCP 2015 / AISC 360 Section E1) is: φ_c Pn = φ_c × Fcr × Ag where: • φ_c = 0.90 (resistance factor for compression) • Fcr = critical stress determined by the applicable buckling formula • Ag = gross cross-sectional area of the member The required strength Pu must satisfy: Pu ≤ φ_c Pn

Question Type

short_answer

Answer Structure

• Line 1: State φ_c Pn = φ_c Fcr Ag [1 mark]
• Line 2: State φ_c = 0.90 explicitly, define Fcr and Ag [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using φ_c = 0.65 or 0.75 (RC column values from ACI 318) — this is a steel column
• Omitting Ag or substituting net area An without justification

Key Phrases To Include

• φ_c = 0.90
• Pn = Fcr Ag
• gross cross-sectional area
• LRFD
• resistance factor

Marks

3

Topic

Inelastic Buckling — Critical Stress

Difficulty

medium

Template Id

T5

Examiner Tip

In board exams, the five-step structure shown here maps cleanly to partial marks. Never jump from KL/r straight to Fcr — always show Fe.

Model Answer

Step 1 — Compute the transition slenderness: 4.71√(E/Fy) = 4.71√(200,000/248) = 4.71 × 28.40 = 133.7 Step 2 — Compare: KL/r = 70 < 133.7 → INELASTIC BUCKLING governs. Step 3 — Elastic buckling stress: Fe = π²E / (KL/r)² = π²(200,000) / 70² = 1,973,921 / 4,900 = 402.8 MPa Step 4 — Compute exponent ratio: Fy/Fe = 248 / 402.8 = 0.616 Step 5 — Apply inelastic formula: Fcr = (0.658^0.616)(248) = 0.7733 × 248 = 191.8 MPa ∴ Fcr ≈ 191.8 MPa

Question Type

numerical

Answer Structure

• Step 1: Compute transition slenderness 4.71√(E/Fy) [0.5 mark]
• Step 2: Compare KL/r and state governing regime [0.5 mark]
• Step 3: Compute Fe = π²E/(KL/r)² [0.5 mark]
• Step 4: Compute Fy/Fe ratio [0.5 mark]
• Step 5: Apply Fcr = [0.658^(Fy/Fe)]Fy and solve [1 mark]

Scoring Breakdown

Common Mark Deductions

• Skipping the transition check — examiner cannot award regime identification mark
• Using 0.658 as a multiplier instead of as the base of an exponent: 0.658 × 0.616 × Fy is wrong
• Not showing the Fe calculation before substituting into the Fcr formula

Key Phrases To Include

• 4.71√(E/Fy) = 133.7
• inelastic buckling
• Fe = 402.8 MPa
• Fy/Fe = 0.616
• Fcr = 0.658^(Fy/Fe) × Fy
• 191.8 MPa

Marks

3

Topic

Elastic Buckling — Critical Stress

Difficulty

medium

Template Id

T6

Examiner Tip

The 0.877 factor is mandatory for elastic columns. It accounts for initial crookedness — mention this physically in long-answer versions to impress examiners.

Model Answer

Step 1 — Transition slenderness: 4.71√(200,000/248) = 133.7 Step 2 — Compare: KL/r = 150 > 133.7 → ELASTIC BUCKLING governs. Step 3 — Elastic buckling stress: Fe = π²(200,000) / 150² = 1,973,921 / 22,500 = 87.73 MPa Step 4 — Apply elastic formula: Fcr = 0.877 × Fe = 0.877 × 87.73 = 76.9 MPa ∴ Fcr ≈ 76.9 MPa

Question Type

numerical

Answer Structure

• Step 1: Compute transition slenderness [0.5 mark]
• Step 2: Identify elastic buckling regime (KL/r > 133.7) [0.5 mark]
• Step 3: Compute Fe = 87.73 MPa [1 mark]
• Step 4: Apply Fcr = 0.877Fe = 76.9 MPa [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using the inelastic formula (0.658^(Fy/Fe)Fy) for an elastic column — wrong regime, loses Fcr marks
• Forgetting the 0.877 factor and writing Fcr = Fe directly

Key Phrases To Include

• KL/r = 150 > 133.7
• elastic buckling
• Fe = 87.73 MPa
• Fcr = 0.877Fe
• 76.9 MPa

Marks

5

Topic

Full Design Strength Computation — Inelastic Column

Difficulty

medium

Template Id

T7

Examiner Tip

A clean, numbered step solution with all intermediate values shown earns full marks even if the final number is slightly off due to rounding. Show your work — every step is a potential mark.

Model Answer

Given: Ag = 8,000 mm², r = 50 mm, KL = 3,500 mm, Fy = 248 MPa, E = 200,000 MPa Step 1 — Slenderness ratio: KL/r = 3,500 / 50 = 70 ✓ (< 200, acceptable) Step 2 — Transition slenderness: 4.71√(E/Fy) = 4.71√(200,000/248) = 4.71 × 28.40 = 133.7 Step 3 — Regime check: KL/r = 70 < 133.7 → INELASTIC BUCKLING governs. Step 4 — Elastic buckling stress: Fe = π²E / (KL/r)² = π²(200,000) / (70)² Fe = 1,973,921 / 4,900 = 402.8 MPa Step 5 — Exponent ratio: Fy/Fe = 248 / 402.8 = 0.616 Step 6 — Critical stress (inelastic formula): Fcr = [0.658^(Fy/Fe)] × Fy = [0.658^0.616] × 248 Fcr = 0.7733 × 248 = 191.8 MPa Step 7 — Nominal and design strength: Pn = Fcr × Ag = 191.8 × 8,000 = 1,534,400 N φ_c Pn = 0.90 × 1,534,400 = 1,380,960 N ∴ φ_c Pn ≈ 1,381 kN

Question Type

numerical

Answer Structure

• Step 1: Compute KL/r and check ≤ 200 [0.5 mark]
• Step 2: Compute transition slenderness 4.71√(E/Fy) [0.5 mark]
• Step 3: Identify inelastic regime and state it explicitly [0.5 mark]
• Step 4: Compute Fe = 402.8 MPa with formula shown [1 mark]
• Step 5–6: Compute Fy/Fe and apply Fcr = [0.658^0.616]×248 = 191.8 MPa [1.5 marks]
• Step 7: Compute φ_c Pn = 0.90 × Fcr × Ag = 1,381 kN [1 mark]

Scoring Breakdown

Common Mark Deductions

• Omitting the slenderness limit check (KL/r ≤ 200) — minor deduction but shows incomplete solution
• Not checking the transition and blindly using one formula
• Using φ_c = 0.85 or 0.65 instead of 0.90
• Arithmetic error in 0.658^0.616 — use logarithms: 0.616 × ln(0.658) = 0.616 × (−0.419) = −0.258; e^(−0.258) = 0.773

Key Phrases To Include

• KL/r = 70
• 4.71√(E/Fy) = 133.7
• inelastic buckling governs
• Fe = 402.8 MPa
• Fcr = 191.8 MPa
• φ_c = 0.90
• φ_c Pn = 1,381 kN

Marks

5

Topic

Full Design Strength Computation — Elastic Column

Difficulty

medium

Template Id

T8

Examiner Tip

After finding KL/r, immediately write: 'Since ___ > 133.7, elastic buckling governs, Fcr = 0.877Fe.' This sentence alone secures regime identification marks.

Model Answer

Given: KL/r = 150, Ag = 8,000 mm², Fy = 248 MPa, E = 200,000 MPa Step 1 — Slenderness check: KL/r = 150 < 200 ✓ Step 2 — Transition slenderness: 4.71√(200,000/248) = 133.7 Step 3 — Regime: KL/r = 150 > 133.7 → ELASTIC BUCKLING governs. Step 4 — Elastic buckling stress: Fe = π²(200,000) / 150² = 1,973,921 / 22,500 = 87.73 MPa Step 5 — Critical stress (elastic formula): Fcr = 0.877 × Fe = 0.877 × 87.73 = 76.94 MPa Step 6 — Nominal strength: Pn = Fcr × Ag = 76.94 × 8,000 = 615,520 N = 615.5 kN Step 7 — Design strength: φ_c Pn = 0.90 × 615.5 = 553.9 kN ∴ φ_c Pn ≈ 554 kN

Question Type

numerical

Answer Structure

• Step 1: Slenderness acceptability check [0.5 mark]
• Step 2–3: Transition computed and elastic regime identified [1 mark]
• Step 4: Fe = 87.73 MPa [1 mark]
• Step 5: Fcr = 0.877 × 87.73 = 76.94 MPa [1 mark]
• Steps 6–7: φ_c Pn = 554 kN with φ_c = 0.90 [1.5 marks]

Scoring Breakdown

Common Mark Deductions

• Applying the inelastic formula for this elastic-range column — single largest source of error
• Omitting the 0.877 factor and using Fcr = Fe directly

Key Phrases To Include

• KL/r = 150 > 133.7
• elastic buckling
• Fe = 87.73 MPa
• 0.877Fe
• Fcr = 76.94 MPa
• φ_c Pn = 554 kN

Marks

5

Topic

High-Strength Steel Column Design

Difficulty

hard

Template Id

T9

Examiner Tip

High-strength steel (Fy = 345 MPa or higher) has a lower transition slenderness — always recompute 4.71√(E/Fy) for the given Fy. Do not memorize 133.7 as a universal constant.

Model Answer

Given: KL/r = 100, Ag = 6,000 mm², Fy = 345 MPa, E = 200,000 MPa Step 1 — Slenderness check: KL/r = 100 < 200 ✓ Step 2 — Transition slenderness: 4.71√(200,000/345) = 4.71√(579.71) = 4.71 × 24.08 = 113.4 Step 3 — Regime: KL/r = 100 < 113.4 → INELASTIC BUCKLING governs. Step 4 — Elastic buckling stress: Fe = π²(200,000) / 100² = 1,973,921 / 10,000 = 197.4 MPa Step 5 — Exponent ratio: Fy/Fe = 345 / 197.4 = 1.748 Step 6 — Critical stress: Fcr = [0.658^1.748] × 345 Using logarithms: 1.748 × ln(0.658) = 1.748 × (−0.4193) = −0.733 e^(−0.733) = 0.4807 Fcr = 0.4807 × 345 = 165.8 MPa Step 7 — Design strength: φ_c Pn = 0.90 × 165.8 × 6,000 = 895,320 N ≈ 895 kN ∴ φ_c Pn ≈ 895 kN

Question Type

numerical

Answer Structure

• Step 1: KL/r < 200 acceptable [0.5 mark]
• Step 2–3: Transition = 113.4, inelastic regime identified [1 mark]
• Step 4: Fe = 197.4 MPa [1 mark]
• Step 5–6: Fy/Fe = 1.748, Fcr = 165.8 MPa using 0.658^1.748 [1.5 marks]
• Step 7: φ_c Pn = 895 kN [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using the same transition slenderness for Fy = 248 (133.7) instead of recomputing for Fy = 345 (113.4)
• Computational error in the exponent 0.658^1.748 — show the logarithm calculation step explicitly

Key Phrases To Include

• Fy = 345 MPa
• transition = 113.4
• inelastic buckling
• Fy/Fe = 1.748
• 0.658^1.748
• Fcr = 165.8 MPa
• φ_c Pn ≈ 895 kN

Marks

2

Topic

Slenderness Limits

Difficulty

easy

Template Id

T10

Examiner Tip

Distinguishing between 'recommended' and 'mandatory' limits shows code literacy — a quality examiners reward at the board level.

Model Answer

Per NSCP 2015 / AISC 360, the recommended maximum slenderness ratio for compression members is KL/r ≤ 200. This is not a hard failure criterion but a serviceability-based limit. When KL/r > 200, the column becomes excessively slender — it is highly susceptible to accidental lateral loads, vibrations, handling stresses during construction, and large lateral deformations. The design compressive strength also drops dramatically as Fe ∝ 1/(KL/r)², making the member structurally inefficient.

Question Type

short_answer

Answer Structure

• Line 1: State KL/r ≤ 200 as the recommended limit [1 mark]
• Line 2: Explain the physical consequences (serviceability, vulnerability to lateral loads, dramatically reduced strength) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Calling KL/r = 200 a 'mandatory code limit' — it is a recommendation, not a prohibition
• Failing to provide any physical explanation and only quoting the number

Key Phrases To Include

• KL/r ≤ 200
• recommended limit
• serviceability
• lateral loads
• Fe ∝ 1/(KL/r)²

Marks

2

Topic

Governing Axis and Slenderness

Difficulty

medium

Template Id

T11

Examiner Tip

Say 'lowest Fcr' and 'weakest axis' — these are the keywords that connect the physics to the design requirement and earn concept marks.

Model Answer

A doubly symmetric section (e.g., W-shape) has two principal axes with potentially different radii of gyration: r_x (strong axis) and r_y (weak axis), and potentially different effective lengths K_x L_x and K_y L_y due to different bracing conditions. The column will buckle about the axis that produces the LOWEST resistance — that is, the axis with the LARGEST slenderness ratio KL/r. Using the largest KL/r ensures the design is governed by the critical (weakest) mode of buckling, resulting in the lowest Fcr and lowest φ_c Pn. Using the smaller KL/r would unconservatively overestimate the column's capacity.

Question Type

short_answer

Answer Structure

• Line 1: Explain that different axes have different r and potentially different K and L [1 mark]
• Line 2: State that the largest KL/r gives the smallest Fcr and must govern for a conservative, safe design [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not explaining why the largest value is used — just stating 'use the largest' without justification
• Confusing the largest r (strong axis) with the governing axis — it is the largest KL/r ratio (smallest r usually governs)

Key Phrases To Include

• weak axis
• strong axis
• largest KL/r
• critical buckling mode
• unconservative
• r_x and r_y

Marks

2

Topic

Local Buckling and Slender Elements

Difficulty

medium

Template Id

T12

Examiner Tip

Mention Q = 1.0 and Q < 1.0 explicitly — the Q factor is the code mechanism and its mention signals familiarity with NSCP 2015 Section E7.

Model Answer

Local buckling is the premature buckling of individual thin plate elements (flanges, webs) of a cross-section before the overall (global) flexural buckling of the member occurs. It reduces the effective area of the cross-section. NSCP 2015 (AISC 360 Section E7) addresses local buckling through the Q factor (slenderness reduction factor) for columns with slender cross-sectional elements. For compact and non-compact sections (λ ≤ λr), Q = 1.0 and full Ag is used. For sections with slender elements (λ > λr), Q < 1.0, reducing the effective Fcr and Pn to account for the loss of plate stability.

Question Type

short_answer

Answer Structure

• Line 1: Define local buckling as buckling of individual plate elements, distinct from global flexural buckling [1 mark]
• Line 2: Explain the Q factor approach in NSCP 2015/AISC 360 E7 — Q = 1.0 for compact, Q < 1.0 for slender elements [1 mark]

Scoring Breakdown

Common Mark Deductions

• Confusing local buckling with overall flexural buckling
• Not mentioning the Q factor or the width-to-thickness ratio limit λr

Key Phrases To Include

• local buckling
• slender elements
• Q factor
• AISC 360 Section E7
• effective area
• compact section

Marks

2

Topic

Transition Slenderness for Different Steel Grades

Difficulty

easy

Template Id

T13

Examiner Tip

These two transition values (133.7 and 113.4) appear frequently in PRC board problems. Compute them cleanly and state the physical implication for bonus recognition.

Model Answer

(a) Fy = 248 MPa: (KL/r)_transition = 4.71√(E/Fy) = 4.71√(200,000/248) = 4.71 × 28.40 = 133.7 (b) Fy = 345 MPa: (KL/r)_transition = 4.71√(200,000/345) = 4.71 × 24.08 = 113.4 Note: Higher-strength steel (larger Fy) has a LOWER transition slenderness — the elastic range starts at a smaller KL/r.

Question Type

numerical

Answer Structure

• Part (a): Correct computation of 4.71√(E/248) = 133.7 [1 mark]
• Part (b): Correct computation of 4.71√(E/345) = 113.4 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using the same value 133.7 for both — not recomputing for Fy = 345
• Arithmetic error in the square root

Key Phrases To Include

• 4.71√(E/Fy)
• 133.7 for Fy = 248
• 113.4 for Fy = 345
• higher Fy → lower transition slenderness

Marks

5

Topic

Complete Column Design — Board-Exam Style

Difficulty

hard

Template Id

T14

Examiner Tip

The problem asks for 'maximum factored load Pu' — this is the design strength φ_c Pn. Make sure to state this equivalence explicitly: 'Maximum Pu = φ_c Pn = ...'

Model Answer

Given: Ag = 9,500 mm², r = 65 mm, K = 1.0, L = 5,000 mm, Fy = 248 MPa, E = 200,000 MPa Step 1 — Slenderness ratio: KL/r = (1.0 × 5,000) / 65 = 76.92 ✓ (< 200) Step 2 — Transition slenderness: 4.71√(200,000/248) = 133.7 Step 3 — Regime: KL/r = 76.92 < 133.7 → INELASTIC BUCKLING governs. Step 4 — Elastic buckling stress: Fe = π²(200,000) / (76.92)² = 1,973,921 / 5,916.7 = 333.6 MPa Step 5 — Exponent ratio: Fy/Fe = 248 / 333.6 = 0.7435 Step 6 — Critical stress: Fcr = [0.658^0.7435] × 248 ln(0.658) = −0.4193; 0.7435 × (−0.4193) = −0.3117; e^(−0.3117) = 0.7322 Fcr = 0.7322 × 248 = 181.6 MPa Step 7 — Design strength: φ_c Pn = 0.90 × 181.6 × 9,500 = 1,551,672 N ≈ 1,552 kN ∴ Maximum Pu = φ_c Pn ≈ 1,552 kN

Question Type

numerical

Answer Structure

• Step 1: KL/r = 76.92 and slenderness check [0.5 mark]
• Step 2–3: Transition 133.7 and inelastic regime [0.5 mark]
• Step 4: Fe = 333.6 MPa [1 mark]
• Steps 5–6: Fy/Fe = 0.7435, Fcr = 181.6 MPa [1.5 marks]
• Step 7: φ_c Pn = 1,552 kN [1.5 marks]

Scoring Breakdown

Common Mark Deductions

• Not converting L to mm before computing KL/r (L = 5 m = 5,000 mm)
• Rounding KL/r to 77 and using 77² instead of 76.92² — causes a small but cascading error

Key Phrases To Include

• KL/r = 76.92
• inelastic buckling
• Fe = 333.6 MPa
• Fcr = 181.6 MPa
• φ_c Pn = 1,552 kN
• maximum factored load

Marks

5

Topic

Pipe Column Design — Pinned-Pinned

Difficulty

medium

Template Id

T15

Examiner Tip

When a problem says 'pinned at both ends,' immediately write K = 1.0. This earns a mark and sets up the correct KL/r calculation. Boundary condition identification is tested independently.

Model Answer

Given: K = 1.0 (pinned-pinned), L = 4,000 mm, r = 60 mm, Ag = 5,000 mm², Fy = 248 MPa Step 1 — Slenderness: KL/r = (1.0 × 4,000) / 60 = 66.67 ✓ (< 200) Step 2 — Transition: 4.71√(200,000/248) = 133.7 Step 3 — Regime: 66.67 < 133.7 → INELASTIC BUCKLING governs. Step 4 — Elastic buckling stress: Fe = π²(200,000) / (66.67)² = 1,973,921 / 4,444.9 = 443.9 MPa Step 5 — Exponent: Fy/Fe = 248 / 443.9 = 0.5587 Fcr = [0.658^0.5587] × 248 0.5587 × ln(0.658) = 0.5587 × (−0.4193) = −0.2342 e^(−0.2342) = 0.7912 Fcr = 0.7912 × 248 = 196.2 MPa Step 6 — Design strength: φ_c Pn = 0.90 × 196.2 × 5,000 = 882,900 N ≈ 883 kN ∴ φ_c Pn ≈ 883 kN

Question Type

numerical

Answer Structure

• Step 1: K = 1.0 stated for pinned-pinned; KL/r = 66.67 [0.5 mark]
• Step 2–3: Transition 133.7 and inelastic regime [0.5 mark]
• Step 4: Fe = 443.9 MPa [1 mark]
• Step 5: Fcr = 196.2 MPa [1.5 marks]
• Step 6: φ_c Pn = 883 kN [1.5 marks]

Scoring Breakdown

Common Mark Deductions

• Not stating K = 1.0 and its basis (pinned-pinned boundary condition) — loses the K justification mark
• Forgetting to convert L = 4 m to 4,000 mm before dividing by r in mm

Key Phrases To Include

• K = 1.0 pinned-pinned
• KL/r = 66.67
• Fe = 443.9 MPa
• inelastic buckling
• Fcr = 196.2 MPa
• φ_c Pn = 883 kN