Simple Stresses and Strains - Exam Answer Templates
Proper answer writing is the bridge between what you know and the marks you earn. In the PRC Civil Engineer Licensure Examination (CELE), examiners award marks not just for the correct final answer but for demonstrating a clear, logical solution path — correct formula citation, proper unit handling, and systematic computation. A student who knows the concept but writes a disorganized answer routinely loses 30–50% of available marks. These templates show you EXACTLY how a full-mark answer looks for every mark level in Simple Stresses and Strains — the most foundational topic in Strength of Materials (PSAD). Master these answer structures, and you will convert knowledge into maximum scores on exam day.
Answer templates
Define normal (axial) stress and state its SI unit.
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).
Answer structure
- Line 1: State that normal stress is force per unit area acting perpendicular to the section [1 mark]
Scoring breakdown
Criteria
Correct definition mentioning 'force per unit area' AND 'perpendicular to the cross-section' with correct SI unit (Pa or MPa)
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
Differentiate between single shear and double shear in a bolted connection.
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.
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
Criteria
Correctly identifies the number of shear planes (1 vs. 2) AND writes the correct formula for each case
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)
State Hooke's Law for normal stress and define the modulus of elasticity.
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.
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
Criteria
Correct statement of Hooke's Law with formula σ = Eε and mention of 'within the proportional limit'
Criteria
Correct definition of E as the ratio of stress to strain and its value for steel (200 GPa)
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
Explain bearing stress and write the formula for a bolt bearing on a plate.
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.
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
Criteria
Correct definition of bearing stress as compressive contact pressure on the projected area
Criteria
Correct formula σ_b = P/(d·t) with d and t correctly identified
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
A steel rod 20 mm in diameter and 2 m long carries an axial tensile load of 40 kN. Given E = 200 GPa, compute (a) the normal stress and (b) the elongation.
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
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
Criteria
Correct area calculation A = π/4 × (20)² = 314.16 mm²
Criteria
Correct stress σ = P/A with result in MPa and stress type (tensile) stated
Criteria
Correct elongation δ = PL/(AE) with result in mm
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
A 16 mm bolt connects two plates in double shear and carries a total load of 48 kN. Compute the average shear stress in the bolt.
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
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
Criteria
Correctly identifies double shear condition and computes A = 201.06 mm²
Criteria
Applies τ = P/(2A) correctly and arrives at 119.4 MPa
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
Define strain and Poisson's ratio. Derive the relationship G = E / [2(1 + ν)].
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)] ✓
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
Criteria
Correct definitions of strain (δ/L) and Poisson's ratio (−ε_lat/ε_ax) with typical values
Criteria
Logical derivation reaching G = E/[2(1+ν)] — method marks awarded for each correct step
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°
A steel bar is fixed between two rigid walls with no initial stress at 25°C. The temperature rises to 65°C. Given α = 11.7 × 10⁻⁶/°C and E = 200 GPa, determine the thermal stress developed in the bar.
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)
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
Criteria
Correctly identifies ΔT = 40°C and states that the member is fully restrained
Criteria
Writes the correct formula σ_T = EαΔT with brief justification
Criteria
Correct numerical answer 93.6 MPa with nature of stress stated as compressive
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
Explain the stress–strain diagram for a ductile material (mild steel) and identify all key points.
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:
- Proportional Limit (A): The highest stress at which σ ∝ ε (the curve is perfectly straight — Hooke's Law holds here).
- Elastic Limit (B): The highest stress at which the material returns to its original length upon unloading — no permanent deformation.
- 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.
- Ultimate Strength (D): The maximum stress on the diagram — the highest load the specimen carries before necking begins.
- 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.
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
Criteria
Correct identification of proportional limit and elastic limit with meaningful distinction between them
Criteria
Correct description of yield point and its engineering significance (plastic flow begins)
Criteria
Correct distinction between ultimate strength (max load) and rupture strength (failure point)
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
A stepped steel bar (E = 200 GPa) has two segments: Segment 1 is 30 mm diameter and 500 mm long; Segment 2 is 20 mm diameter and 800 mm long. An axial tensile load of 50 kN is applied throughout. Determine the total elongation of the bar.
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)
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
Criteria
Correct areas A₁ = 706.86 mm² and A₂ = 314.16 mm²
Criteria
Correct formula δ = PL/(AE) identified and applied for each segment
Criteria
Correct δ₁ = 0.177 mm
Criteria
Correct δ₂ = 0.637 mm
Criteria
Correct total δ = 0.814 mm with tensile direction stated
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
A concrete-filled steel pipe short column carries a total axial load P = 800 kN. The steel pipe has area Aₛ = 4 000 mm² (Eₛ = 200 GPa) and the concrete core has area A_c = 80 000 mm² (E_c = 25 GPa). Assuming the column is short and the materials deform together, find the load carried by each material and the stress in each.
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)
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
Criteria
Correct equilibrium equation Pₛ + P_c = 800 000 N
Criteria
Correct compatibility condition δₛ = δ_c written as Pₛ/(AₛEₛ) = P_c/(A_cE_c)
Criteria
Correct algebraic ratio Pₛ = 0.40 P_c (or equivalent)
Criteria
Correct P_c ≈ 571.4 kN and Pₛ ≈ 228.6 kN
Criteria
Correct stresses σₛ = 57.1 MPa and σ_c = 7.14 MPa with compressive nature stated
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
A 25 mm diameter bolt passes through a 15 mm thick plate and carries P = 60 kN in single shear. Determine: (a) shear stress in the bolt, (b) bearing stress on the plate.
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
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
Criteria
Correct shear stress 122.2 MPa using single-shear formula τ = P/A
Criteria
Correct projected bearing area A_b = 25 × 15 = 375 mm²
Criteria
Correct bearing stress 160.0 MPa
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)
Explain St. Venant's principle and state its practical significance in stress analysis.
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.
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
Criteria
Correct statement of St. Venant's principle with the distance criterion (≈ one cross-section dimension)
Criteria
Correct practical implication — validity of σ = P/A and mention of stress concentrations near holes/notches
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
Write the generalized Hooke's Law for a biaxial stress state (σ_x and σ_y acting simultaneously) and use it to find ε_x and ε_y if σ_x = 120 MPa (T), σ_y = 60 MPa (T), ν = 0.30, E = 200 GPa.
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)
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
Criteria
Correct generalized Hooke's Law equations for biaxial state written before substitution
Criteria
Correct treatment of signs (tensile positive) and consistent MPa units
Criteria
Correct ε_x = 5.10 × 10⁻⁴
Criteria
Correct ε_y = 1.20 × 10⁻⁴
Criteria
Correct ε_z = −2.70 × 10⁻⁴ or correct explanation of Poisson contraction in z-direction
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
A steel bar (α = 11.7 × 10⁻⁶/°C, E = 200 GPa) is 3 m long, fixed at one end, and has a gap of 1.5 mm at the other end. If the temperature rises by 80°C, determine the stress in the bar.
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.
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
Criteria
Correct δ_T = 2.808 mm using δ_T = αLΔT
Criteria
Correct comparison: δ_T > gap, so gap is closed and stress develops
Criteria
Correct compatibility equation δ_T − δ_P = g, giving δ_P = 1.308 mm
Criteria
Correct stress formula σ = δ_P·E/L applied correctly
Criteria
Correct final answer 87.2 MPa compressive; states that if gap not closed, σ = 0
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
Mark wise strategy
State the definition, law, or formula directly and precisely. No derivation needed. Include the SI unit. Use the exact engineering term (e.g., 'perpendicular to the cross-section', 'projected area').
Do
- Write one clear, complete sentence containing the key concept
- Include the formula (e.g., σ = P/A) as part of the answer
- State the SI unit
- Use precise engineering vocabulary (e.g., 'tensile', 'compressive', 'projected area')
Don't
- Do not write a paragraph — one strong sentence is enough
- Do not derive the formula unless asked to
- Do not use approximate or colloquial definitions
For conceptual questions: define + formula + one example or significance. For numerical: given/required list + formula + one computation step. Each distinct point earns 1 mark — structure your answer so both marks are clearly visible.
Do
- Use a two-part structure: (1) concept/formula, (2) application or example
- For numerical problems, always list given data first
- Write the formula before substituting numbers
- Show units in intermediate steps
Don't
- Do not mix the two expected parts — keep them visually separate
- Do not skip to the final number without showing the formula
- Do not omit the unit in the final answer
Think of 3-mark questions as having three distinct earning points. For theory: definition + principle + significance/application. For numerical: formula + intermediate working + final answer. Organize with clear step labels (Step 1, Step 2, Step 3).
Do
- Number your steps explicitly: Step 1, Step 2, Step 3
- For numerical: list given → write formula → substitute → compute → state answer
- Box or underline the final numerical answer with unit
- For theory: use bullet points for each landmark/definition if listing multiple items
Don't
- Do not write a block of unseparated text — examiners need to see each earning point
- Do not skip intermediate calculations — method marks are awarded even if final answer is wrong
- Do not present a formula that applies to a different case (e.g., double shear formula when single shear is given)
5-mark questions reward completeness and systematic method. For numerical: given data → FBD if needed → all formulas → step-by-step computation → final answers with units and stress type. For composite/indeterminate: explicitly label EQUILIBRIUM equation and COMPATIBILITY equation before solving. Show all arithmetic.
Do
- Start with a 'Given' section listing all known values with units
- Draw a free-body diagram for force-system questions
- For indeterminate problems, write 'Equilibrium:' and 'Compatibility:' as bold labels
- Show all intermediate arithmetic — do not skip to the answer
- State the physical interpretation of each answer (e.g., 'compressive', 'steel carries more load because it is stiffer')
- Verify your answer if time allows (e.g., check Pₛ + P_c = P_total)
Don't
- Do not assume equal stress in composite members — compatibility gives equal deformation, not equal stress
- Do not mix N/mm with kN/m mid-solution
- Do not skip the gap-check step in thermal problems with gap
- Do not leave the answer without a unit and stress type (T or C)
General answer writing tips
- Always write the governing formula first before substituting numbers — examiners award a method mark for the correct formula even if arithmetic goes wrong.
- Keep units consistent throughout: use N and mm so that stress automatically comes out in MPa (N/mm²). Mixing kN with m is the single most common source of unit errors.
- For numerical problems, box or underline the final answer with the correct unit and sign (tensile/compressive, positive/negative) — examiners scan for this.
- In shear and bearing problems, explicitly state whether the configuration is single shear or double shear before writing the formula; this one phrase earns a process mark.
- For indeterminate (composite) member problems, label your two equations clearly as 'Equilibrium:' and 'Compatibility:' — this two-equation structure is what examiners expect and reward.
- Draw a free-body diagram (FBD) whenever force direction is ambiguous; even a quick sketch earns a diagram mark and prevents sign errors.
- For thermal stress questions, always state the condition — 'member is fully restrained' or 'member is free to expand' — before writing the stress formula, because the formula changes depending on the boundary condition.
- State St. Venant's principle if asked about stress distribution away from load points — this signals examiner-level understanding and can earn bonus interpretation marks.
Diagrams
Simple Stresses and Strains — Topic Mind Map
A hierarchical mind map showing all major topics in Simple Stresses and Strains. Use this as a chapter navigator to identify which template covers which topic before studying.
How to Solve Any Axial Stress Problem — Decision Flowchart
A decision tree guiding students through the correct solution path for any axial stress/deformation problem. Shows the fork between determinate and indeterminate cases, and between pure mechanical and thermal loading.
Thermal Stress Problem — Step-by-Step Procedure
Step-by-step procedure for solving thermal stress problems. The critical decision node is whether the member is fully restrained or has a gap — the most common source of errors in board exams.
Stress-Strain Diagram — States of a Ductile Material Under Tension
State diagram showing the progression of a ductile steel specimen from initial loading through all landmark points to fracture. Each transition corresponds to a key point on the stress-strain curve that board exams frequently test.
Distribution of Simple Stress Topics in PRC Board Exam Questions
Approximate distribution of Simple Stresses and Strains topics appearing in PRC Civil Engineer Licensure Examinations based on historical board exam analysis. Axial deformation and shear/bearing stress questions appear most frequently.