CELE Strength of Materials — Simple Stresses and StrainsExam Answer Templates

Marks

1

Topic

Normal Stress

Difficulty

easy

Template Id

T1

Examiner Tip

The two keywords that earn this mark are 'perpendicular' and 'force per unit area.' Include both in one concise sentence.

Model Answer

Normal stress (σ) is the internal resisting force per unit area acting perpendicular to a cross-section. Its SI unit is the Pascal (Pa) or, for structural calculations, the megapascal (MPa = N/mm²).

Question Type

very_short_answer

Answer Structure

• Line 1: State that normal stress is force per unit area acting perpendicular to the section [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing 'parallel to the cross-section' — that describes shear stress, not normal stress
• Omitting the SI unit entirely
• Writing 'kPa' without context — MPa is the standard engineering unit for stress

Key Phrases To Include

• force per unit area
• perpendicular to the cross-section
• MPa or N/mm²
• σ = P/A

Marks

1

Topic

Shear Stress

Difficulty

easy

Template Id

T2

Examiner Tip

The distinction is purely the denominator — make it obvious by writing both formulas side by side.

Model Answer

In single shear, the bolt resists on ONE shear plane, so τ = P/A. In double shear, the bolt resists on TWO shear planes, so τ = P/(2A), giving half the shear stress for the same load.

Question Type

very_short_answer

Answer Structure

• Line 1: Define single shear — one plane, τ = P/A [0.5 mark]
• Line 2: Define double shear — two planes, τ = P/(2A) [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Writing τ = P/2A for single shear — reverses the formulas
• Mentioning planes but not the formula — loses half mark

Key Phrases To Include

• one shear plane
• two shear planes
• τ = P/A
• τ = P/(2A)

Marks

2

Topic

Hooke's Law and Elastic Constants

Difficulty

easy

Template Id

T3

Examiner Tip

Examiners want to see the phrase 'within the proportional limit' — it shows you know Hooke's Law has a validity boundary.

Model Answer

Hooke's Law states that, within the proportional limit, stress is directly proportional to strain: σ = Eε where E is the modulus of elasticity (Young's modulus) — the ratio of normal stress to the corresponding normal strain. It represents the material's stiffness. For structural steel, E ≈ 200 GPa.

Question Type

short_answer

Answer Structure

• Line 1: State proportionality and write σ = Eε [1 mark]
• Line 2: Define E as stress-to-strain ratio (stiffness) and give steel value [1 mark]

Scoring Breakdown

Common Mark Deductions

• Omitting 'within the proportional limit' — Hooke's Law is not valid beyond this point
• Confusing E with G (shear modulus) — they are different constants
• Not specifying units of E (GPa or kN/mm²)

Key Phrases To Include

• proportional limit
• σ = Eε
• modulus of elasticity
• stiffness
• 200 GPa for steel

Marks

2

Topic

Bearing Stress

Difficulty

easy

Template Id

T4

Examiner Tip

The word 'projected' is the critical differentiator — without it, you look like you're guessing which area to use.

Model Answer

Bearing stress (σ_b) is the contact pressure between two bodies in compression — for example, a bolt shank pressing against the wall of a plate hole. It is computed on the PROJECTED (rectangular) area, not the curved contact surface: σ_b = P / (d × t) where d = bolt diameter (mm) and t = plate thickness (mm). This stress governs plate crushing and hole tear-out in connection design.

Question Type

short_answer

Answer Structure

• Line 1: Define bearing stress as contact pressure; mention projected area [1 mark]
• Line 2: Write σ_b = P/(d·t) and identify each variable [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using the bolt's circular cross-sectional area A = πd²/4 instead of the projected area d·t
• Confusing bearing stress with shear stress in the bolt
• Omitting the word 'projected' when describing the area

Key Phrases To Include

• contact pressure
• projected area
• σ_b = P/(d·t)
• bolt diameter
• plate thickness

Marks

3

Topic

Axial Deformation

Difficulty

medium

Template Id

T5

Examiner Tip

Always convert to N and mm at the 'Given' stage — never mix unit systems mid-solution. Write the formula, substitute, then compute.

Model Answer

Given: d = 20 mm, L = 2000 mm, P = 40 000 N, E = 200 000 MPa Step 1 — Cross-sectional area: A = (π/4)(20)² = 314.16 mm² Step 2 — Normal stress: σ = P/A = 40 000 / 314.16 σ = 127.3 MPa (tensile) Step 3 — Elongation: δ = PL/(AE) = (40 000 × 2000) / (314.16 × 200 000) δ = 80 000 000 / 62 832 000 δ = 1.27 mm Answers: σ = 127.3 MPa (T); δ = 1.27 mm

Question Type

numerical

Answer Structure

• Line 1: List all given data with consistent units (N, mm) [0 marks — but shows organisation]
• Step 1: Compute A = πd²/4 [1 mark]
• Step 2: Apply σ = P/A correctly [1 mark]
• Step 3: Apply δ = PL/(AE) correctly [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using L = 2 m instead of converting to 2000 mm — stress will be in wrong units
• Forgetting to convert P from kN to N (use 40 000 N, not 40)
• Not stating whether stress is tensile or compressive

Key Phrases To Include

• A = πd²/4
• σ = P/A
• δ = PL/(AE)
• tensile
• MPa
• mm

Marks

2

Topic

Shear Stress

Difficulty

medium

Template Id

T6

Examiner Tip

Write 'double shear: 2 planes' explicitly before the formula. This one phrase protects your method mark.

Model Answer

Given: d = 16 mm, P = 48 000 N, double shear (2 shear planes) Step 1 — Bolt cross-sectional area: A = (π/4)(16)² = 201.06 mm² Step 2 — Shear stress (double shear): τ = P / (2A) = 48 000 / (2 × 201.06) τ = 48 000 / 402.12 τ = 119.4 MPa Answer: τ = 119.4 MPa

Question Type

numerical

Answer Structure

• Step 1: Identify double shear and compute A = πd²/4 [1 mark]
• Step 2: Apply τ = P/(2A) and compute [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using τ = P/A (single shear formula) — halves the correct answer
• Using the projected area d·t instead of the circular cross-sectional area for shear

Key Phrases To Include

• double shear
• two shear planes
• τ = P/(2A)
• A = πd²/4

Marks

3

Topic

Poisson's Ratio and Elastic Constants

Difficulty

hard

Template Id

T7

Examiner Tip

Even if you cannot complete the full derivation, show the principal stress approach at 45° — this earns at least partial method marks.

Model Answer

Normal strain (ε) is the deformation per unit length of a member under axial load: ε = δ/L (dimensionless) Poisson's ratio (ν) is the negative ratio of lateral strain to axial strain when the member is loaded uniaxially: ν = −ε_lateral / ε_axial Typical values: steel ≈ 0.27–0.30; concrete ≈ 0.15–0.20. Derivation of G = E / [2(1 + ν)]: Consider a square element under pure shear τ at 45°. The principal stresses on the diagonal planes are σ₁ = +τ and σ₂ = −τ. Using the generalized Hooke's Law, the principal strain at 45° is: ε₁ = (σ₁ − νσ₂)/E = (τ − ν(−τ))/E = τ(1 + ν)/E But by the geometry of shear, ε₁ = γ/2 and τ = Gγ, so γ = τ/G and ε₁ = τ/(2G). Equating the two expressions for ε₁: τ(1 + ν)/E = τ/(2G) ∴ G = E / [2(1 + ν)] ✓

Question Type

short_answer

Answer Structure

• Part A: Define ε = δ/L and state it is dimensionless [0.5 mark]
• Part B: Define ν = −ε_lateral/ε_axial and give typical values [0.5 mark]
• Part C: Correctly derive G = E/[2(1+ν)] via principal stress or any valid method [2 marks]

Scoring Breakdown

Common Mark Deductions

• Simply stating the formula without any derivation — earns 0 for the derivation marks
• Confusing ε (normal strain) and γ (shear strain) in the derivation
• Not stating the negative sign in Poisson's ratio definition

Key Phrases To Include

• ε = δ/L
• dimensionless
• ν = −ε_lateral/ε_axial
• G = E/[2(1+ν)]
• principal stresses at 45°

Marks

3

Topic

Thermal Stress

Difficulty

medium

Template Id

T8

Examiner Tip

State 'FULLY RESTRAINED' loudly and clearly. Many students lose a mark by jumping to the formula without justifying why stress exists. Stress only arises because expansion is prevented.

Model Answer

Given: T₁ = 25°C, T₂ = 65°C → ΔT = 40°C α = 11.7 × 10⁻⁶ /°C, E = 200 000 MPa Condition: FULLY RESTRAINED (rigid walls → zero net deformation) Concept: The bar would expand freely by δ_T = αLΔT, but the walls prevent this movement. The wall therefore exerts a compressive reaction force, inducing compressive thermal stress. Formula for fully restrained bar: σ_T = E × α × ΔT Substituting: σ_T = 200 000 × (11.7 × 10⁻⁶) × 40 σ_T = 200 000 × 4.68 × 10⁻⁴ σ_T = 93.6 MPa Answer: σ_T = 93.6 MPa (COMPRESSIVE)

Question Type

numerical

Answer Structure

• Step 1: Compute ΔT and state the boundary condition (fully restrained) [1 mark]
• Step 2: Write formula σ_T = EαΔT and justify it (restraint prevents free expansion) [1 mark]
• Step 3: Substitute and compute; state compressive nature [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not stating that the bar is compressive — temperature rise in a restrained bar always produces compression
• Using the free thermal deformation formula δ_T = αLΔT and stopping there — that gives deformation, not stress
• Not converting α to consistent units (11.7 × 10⁻⁶ /°C is already correct for MPa calculation)

Key Phrases To Include

• fully restrained
• ΔT = 40°C
• σ_T = EαΔT
• 93.6 MPa
• compressive

Marks

3

Topic

Stress-Strain Diagram

Difficulty

medium

Template Id

T9

Examiner Tip

Use the labels A, B, C, D, E consistently. If you sketch the σ-ε diagram, label all five points — a labelled sketch earns marks on its own.

Model Answer

A tensile test of mild steel produces the following stress–strain diagram with five key landmarks: 1. Proportional Limit (A): The highest stress at which σ ∝ ε (the curve is perfectly straight — Hooke's Law holds here). 2. Elastic Limit (B): The highest stress at which the material returns to its original length upon unloading — no permanent deformation. 3. Yield Point (C): The stress at which strain increases significantly with little or no increase in stress (the material 'flows'). For structural steel, σ_y ≈ 248–250 MPa. 4. Ultimate Strength (D): The maximum stress on the diagram — the highest load the specimen carries before necking begins. 5. Rupture/Fracture Point (E): The actual stress at failure. The apparent drop from D to E on an engineering diagram is due to the use of original area — true stress continues to rise. The region from O to A is the ELASTIC region (Hooke's Law applies). Beyond C is the PLASTIC region.

Question Type

short_answer

Answer Structure

• Identify and describe Point A (Proportional Limit) [0.5 mark]
• Identify and describe Point B (Elastic Limit) [0.5 mark]
• Identify and describe Point C (Yield Point) with typical value [1 mark]
• Identify Points D (Ultimate) and E (Rupture) with distinction [1 mark]

Scoring Breakdown

Common Mark Deductions

• Mixing up proportional limit and elastic limit — they are different (elastic limit is always ≥ proportional limit)
• Saying ultimate strength is the failure point — fracture/rupture is the failure point
• Not mentioning necking in connection with the drop from ultimate to rupture

Key Phrases To Include

• proportional limit
• elastic limit
• yield point
• ultimate strength
• rupture strength
• Hooke's Law
• plastic region

Marks

5

Topic

Axial Deformation — Variable Cross-Section

Difficulty

medium

Template Id

T10

Examiner Tip

The key insight is Σ(PL/AE) — each segment gets its own AE. Organize your solution in a clear table-like format to avoid missing a segment.

Model Answer

Given: P = 50 000 N throughout (same load — no intermediate loads) E = 200 000 MPa Segment 1: d₁ = 30 mm, L₁ = 500 mm Segment 2: d₂ = 20 mm, L₂ = 800 mm Step 1 — Cross-sectional areas: A₁ = (π/4)(30)² = 706.86 mm² A₂ = (π/4)(20)² = 314.16 mm² Step 2 — Stress in each segment (to verify and understand load path): σ₁ = 50 000 / 706.86 = 70.7 MPa σ₂ = 50 000 / 314.16 = 159.2 MPa Step 3 — Elongation of each segment using δ = PL/(AE): δ₁ = (50 000 × 500) / (706.86 × 200 000) = 25 000 000 / 141 372 000 = 0.177 mm δ₂ = (50 000 × 800) / (314.16 × 200 000) = 40 000 000 / 62 832 000 = 0.637 mm Step 4 — Total elongation: δ_total = δ₁ + δ₂ = 0.177 + 0.637 δ_total = 0.814 mm Answer: Total elongation = 0.814 mm (tensile elongation)

Question Type

numerical

Answer Structure

• Step 1: Compute A₁ and A₂ for both segments [1 mark]
• Step 2: Recognize same P throughout and set up δ = PL/(AE) for each [1 mark]
• Step 3: Compute δ₁ correctly [1 mark]
• Step 4: Compute δ₂ correctly [1 mark]
• Step 5: Sum δ_total = δ₁ + δ₂ with correct final answer [1 mark]

Scoring Breakdown

Common Mark Deductions

• Adding the two areas instead of computing deformation for each segment separately
• Using average diameter — areas must be computed individually per segment
• Not converting P from kN to N before substituting
• Forgetting to sum both deformations at the end

Key Phrases To Include

• δ = PL/(AE)
• sum over segments
• A = πd²/4
• same axial load throughout
• total elongation

Marks

5

Topic

Statically Indeterminate Axial Members

Difficulty

hard

Template Id

T11

Examiner Tip

Label your two equations clearly as 'Equilibrium:' and 'Compatibility:'. This two-equation label structure is the standard expected format for indeterminate problems and earns method marks even if arithmetic slips.

Model Answer

Given: P_total = 800 000 N Steel: Aₛ = 4 000 mm², Eₛ = 200 000 MPa Concrete: A_c = 80 000 mm², E_c = 25 000 MPa This is a statically INDETERMINATE composite column — need Equilibrium + Compatibility. Equation 1 — Equilibrium: Pₛ + P_c = 800 000 N ...(1) Equation 2 — Compatibility (both materials shorten equally, same L): δₛ = δ_c Pₛ L / (Aₛ Eₛ) = P_c L / (A_c E_c) [L cancels] Pₛ / (4 000 × 200 000) = P_c / (80 000 × 25 000) Pₛ / (8.0 × 10⁸) = P_c / (2.0 × 10⁹) 2.0 × 10⁹ × Pₛ = 8.0 × 10⁸ × P_c Pₛ = 0.40 P_c ...(2) Solving (1) and (2): 0.40 P_c + P_c = 800 000 1.40 P_c = 800 000 P_c = 571 429 N ≈ 571.4 kN Pₛ = 800 000 − 571 429 = 228 571 N ≈ 228.6 kN Stresses: σₛ = Pₛ / Aₛ = 228 571 / 4 000 = 57.1 MPa (compressive) σ_c = P_c / A_c = 571 429 / 80 000 = 7.14 MPa (compressive) Answers: Steel carries 228.6 kN at σₛ = 57.1 MPa (C) Concrete carries 571.4 kN at σ_c = 7.14 MPa (C)

Question Type

numerical

Answer Structure

• Step 1: Identify problem as statically indeterminate; write Equilibrium equation Pₛ + P_c = P [1 mark]
• Step 2: Write Compatibility equation δₛ = δ_c → Pₛ/(AₛEₛ) = P_c/(A_cE_c) [1 mark]
• Step 3: Simplify compatibility to get ratio Pₛ = 0.40P_c [1 mark]
• Step 4: Solve simultaneously for Pₛ and P_c [1 mark]
• Step 5: Compute stresses in each material [1 mark]

Scoring Breakdown

Common Mark Deductions

• Assuming equal stress in both materials — the compatibility condition gives equal STRAIN (deformation), not equal stress
• Not cancelling L in the compatibility equation — L must cancel for a short column
• Forgetting to verify equilibrium check (Pₛ + P_c should equal 800 kN)
• Using MPa for E in one material and GPa in another without converting

Key Phrases To Include

• statically indeterminate
• equilibrium
• compatibility
• δₛ = δ_c
• Pₛ/(AₛEₛ) = P_c/(A_cE_c)
• stiffer material carries more load

Marks

3

Topic

Shear Stress and Bearing Stress

Difficulty

medium

Template Id

T12

Examiner Tip

Two separate areas — circular area for shear, rectangular d·t for bearing. Keep them distinct and label each clearly.

Model Answer

Given: d = 25 mm, t = 15 mm, P = 60 000 N, single shear Step 1 — Bolt cross-sectional area: A_bolt = (π/4)(25)² = 490.87 mm² (a) Shear stress (single shear — ONE plane): τ = P / A = 60 000 / 490.87 τ = 122.2 MPa (b) Bearing stress (projected area): A_b = d × t = 25 × 15 = 375 mm² σ_b = P / A_b = 60 000 / 375 σ_b = 160.0 MPa Answers: τ = 122.2 MPa; σ_b = 160.0 MPa

Question Type

numerical

Answer Structure

• Step 1: Compute bolt area A = πd²/4 [0.5 mark]
• Step 2: Apply τ = P/A (single shear) [1 mark]
• Step 3: Compute bearing area A_b = d×t and apply σ_b = P/A_b [1.5 marks]

Scoring Breakdown

Common Mark Deductions

• Using the bolt's full circular area for bearing — bearing uses projected area d·t
• Using double shear formula by mistake

Key Phrases To Include

• single shear
• τ = P/A
• bearing area = d × t
• projected area
• σ_b = P/(d·t)

Marks

2

Topic

Normal Stress

Difficulty

medium

Template Id

T13

Examiner Tip

Use the phrase 'one cross-section dimension' as the distance criterion — this is the specific quantitative statement examiners want to see.

Model Answer

St. Venant's Principle states that the stress distribution due to a concentrated (or localized) load becomes essentially uniform at a distance approximately equal to the largest dimension of the cross-section away from the point of load application. Practical significance: The formula σ = P/A (average normal stress) is valid only at cross-sections remote from load application points, holes, notches, or abrupt changes in geometry. Near these 'disturbance zones,' stress concentrations exist and the actual stress can be several times higher than the average. In design, stress concentration factors (K) are applied at these locations.

Question Type

short_answer

Answer Structure

• Line 1: State the principle — stress becomes uniform at a distance ≈ one cross-section width from the load [1 mark]
• Line 2: State the practical implication — σ = P/A applies away from load points; near disturbances use stress concentration factors [1 mark]

Scoring Breakdown

Common Mark Deductions

• Vague statement without the distance criterion
• Not connecting the principle to the validity of the basic stress formula

Key Phrases To Include

• uniform stress distribution
• one cross-section dimension
• stress concentration
• σ = P/A valid remotely from load
• notches, holes, abrupt changes

Marks

5

Topic

Generalized Hooke's Law

Difficulty

hard

Template Id

T14

Examiner Tip

The hallmark of generalized Hooke's Law is the Poisson term subtracting the perpendicular stress. Show both terms clearly before computing — this earns method marks.

Model Answer

Generalized Hooke's Law for biaxial stress (σ_z = 0): ε_x = (1/E)[σ_x − ν·σ_y] ε_y = (1/E)[σ_y − ν·σ_x] ε_z = (1/E)[−ν(σ_x + σ_y)] [lateral contraction in z-direction] Given: σ_x = +120 MPa, σ_y = +60 MPa, ν = 0.30, E = 200 000 MPa Calculating ε_x: ε_x = (1/200 000)[120 − 0.30(60)] = (1/200 000)[120 − 18] = (1/200 000)(102) = 5.10 × 10⁻⁴ Calculating ε_y: ε_y = (1/200 000)[60 − 0.30(120)] = (1/200 000)[60 − 36] = (1/200 000)(24) = 1.20 × 10⁻⁴ Calculating ε_z (for completeness): ε_z = (1/200 000)[−0.30(120 + 60)] = (1/200 000)(−54) = −2.70 × 10⁻⁴ (contraction) Answers: ε_x = 5.10 × 10⁻⁴ (tensile) ε_y = 1.20 × 10⁻⁴ (tensile) ε_z = −2.70 × 10⁻⁴ (compressive — Poisson contraction)

Question Type

numerical

Answer Structure

• Step 1: Write generalized Hooke's Law for biaxial state (both ε_x and ε_y equations) [1 mark]
• Step 2: Correctly identify all given values with consistent units [1 mark]
• Step 3: Compute ε_x = 5.10 × 10⁻⁴ [1 mark]
• Step 4: Compute ε_y = 1.20 × 10⁻⁴ [1 mark]
• Step 5: Compute ε_z and explain physical meaning [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using ε_x = σ_x/E only (ignoring the Poisson effect from σ_y) — this is uniaxial Hooke's Law, not biaxial
• Wrong sign convention — compressive stresses must be negative
• Not stating that strains are dimensionless

Key Phrases To Include

• ε_x = (1/E)[σ_x − ν·σ_y]
• ε_y = (1/E)[σ_y − ν·σ_x]
• biaxial stress
• Poisson contraction
• dimensionless

Marks

5

Topic

Thermal Stress with Gap

Difficulty

hard

Template Id

T15

Examiner Tip

The gap problem is a two-step compatibility problem. ALWAYS check δ_T vs. gap first. If δ_T < gap → σ = 0 (free expansion fills gap). If δ_T > gap → use δ_T − δ_P = gap. Missing the gap check is the most common error in this problem type.

Model Answer

Given: L = 3000 mm, α = 11.7 × 10⁻⁶/°C, E = 200 000 MPa ΔT = 80°C, gap g = 1.5 mm Step 1 — Free thermal elongation: δ_T = α × L × ΔT = (11.7 × 10⁻⁶)(3000)(80) = 2.808 mm Step 2 — Check if gap is closed: δ_T = 2.808 mm > gap g = 1.5 mm ∴ The gap is fully closed and a compressive reaction stress develops. Step 3 — Compatibility equation (net deformation = gap): The bar expands by δ_T but is compressed back by δ_P until the net expansion equals the gap: δ_T − δ_P = g δ_P = δ_T − g = 2.808 − 1.5 = 1.308 mm Step 4 — Stress from elastic compression: δ_P = σL/E (mechanical compression) σ = δ_P × E / L = 1.308 × 200 000 / 3000 σ = 87.2 MPa Answer: σ = 87.2 MPa (COMPRESSIVE) Note: If δ_T < g (gap not closed), stress = 0 — always check this condition first.

Question Type

numerical

Answer Structure

• Step 1: Compute free thermal elongation δ_T = αLΔT [1 mark]
• Step 2: Compare δ_T with gap; determine gap is closed [1 mark]
• Step 3: Write compatibility equation δ_T − δ_P = g; solve for δ_P [1 mark]
• Step 4: Compute stress σ = δ_P·E/L [1 mark]
• Step 5: State correct answer (87.2 MPa compressive) and note the zero-stress condition if gap not closed [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not checking if the gap is closed before computing stress
• Using σ_T = EαΔT directly (fully restrained formula) — this ignores the gap
• Forgetting the gap closes first before any stress can build

Key Phrases To Include

• δ_T = αLΔT
• gap check
• compatibility: δ_T − δ_P = g
• compressive stress
• stress = 0 if gap not closed