CELE Strength of Materials — Combined Stresses and Mohr's CircleExam Answer Templates

Marks

1

Topic

Principal Stresses

Difficulty

easy

Template Id

T1

Examiner Tip

Examiners want two ideas in one sentence: (1) they are extremes of normal stress, and (2) shear is zero on those planes. Both must appear for full credit.

Model Answer

Principal stresses are the maximum and minimum normal stresses at a point, acting on planes (called principal planes) where the shear stress is zero.

Question Type

very_short_answer

Answer Structure

• Single sentence: define principal stresses as extreme normal stresses [0.5 mark]
• State the defining characteristic: shear stress is zero on principal planes [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Defining principal stresses only as 'maximum stress' without specifying normal stress loses the mark
• Omitting the zero-shear condition on principal planes gives an incomplete definition

Key Phrases To Include

• maximum and minimum normal stress
• principal planes
• shear stress is zero

Marks

1

Topic

Maximum Shear Stress

Difficulty

easy

Template Id

T2

Examiner Tip

The '÷ 2' is the most dropped element in this formula. Always write the denominator explicitly.

Model Answer

The maximum in-plane shear stress is: τ_max = (σ₁ − σ₂) / 2 It acts on planes oriented 45° from the principal planes, and the average normal stress σ_avg = (σ₁ + σ₂)/2 acts simultaneously on those planes.

Question Type

very_short_answer

Answer Structure

• State the formula τ_max = (σ₁ − σ₂)/2 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing τ_max = (σ₁ − σ₂) without dividing by 2 — the most frequent halving error
• Confusing this with the absolute maximum shear (which may involve the third principal stress)

Key Phrases To Include

• τ_max = (σ₁ − σ₂)/2
• 45° from principal planes

Marks

1

Topic

Mohr's Circle

Difficulty

easy

Template Id

T3

Examiner Tip

Always distinguish between the angle on the physical element (θ) and the angle on Mohr's circle (2θ). State both if you are unsure which the examiner wants.

Model Answer

The planes of maximum shear stress are oriented 45° from the principal planes in the physical element, which corresponds to a 90° rotation on Mohr's circle.

Question Type

very_short_answer

Answer Structure

• State the physical angle: 45° [0.5 mark]
• State the Mohr's circle angle: 90° [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Answering '90°' as the physical angle (confusing Mohr's circle angle with actual angle)

Key Phrases To Include

• 45° from the principal planes
• 90° on Mohr's circle

Marks

2

Topic

Principal Stresses

Difficulty

easy

Template Id

T4

Examiner Tip

Even for 'simple' problems (τ = 0), show all three steps. An examiner cannot award partial credit for a correct answer with no working shown.

Model Answer

Given: σ_x = +50 MPa, σ_y = +30 MPa, τ_xy = 0 MPa Step 1 – Average stress (center of Mohr's circle): σ_avg = (σ_x + σ_y)/2 = (50 + 30)/2 = 40 MPa Step 2 – Radius: R = √[(σ_x − σ_y)²/4 + τ_xy²] = √[(50 − 30)²/4 + 0] = √[100] = 10 MPa Step 3 – Principal stresses: σ₁ = σ_avg + R = 40 + 10 = 50 MPa σ₂ = σ_avg − R = 40 − 10 = 30 MPa Check: σ₁ + σ₂ = 80 MPa = σ_x + σ_y ✓ Answer: σ₁ = 50 MPa, σ₂ = 30 MPa (no shear exists; original axes are already principal axes).

Question Type

numerical

Answer Structure

• Step 1: Compute σ_avg = (σ_x + σ_y)/2 [0.5 mark]
• Step 2: Compute R using the correct formula [0.5 mark]
• Step 3: State σ₁ = σ_avg + R and σ₂ = σ_avg − R with correct values and units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using (σ_x − σ_y)² instead of (σ_x − σ_y)²/4 inside the radical
• Omitting MPa in the final answer
• Not checking invariant sum

Key Phrases To Include

• σ_avg = (σ_x + σ_y)/2
• R = √[(σ_x − σ_y)²/4 + τ_xy²]
• σ₁ = σ_avg + R
• σ₂ = σ_avg − R

Marks

2

Topic

Maximum Shear Stress

Difficulty

medium

Template Id

T5

Examiner Tip

Many students forget that the maximum shear plane carries a non-zero normal stress equal to σ_avg. This is a frequent second-mark differentiator.

Model Answer

Given: σ_x = 80 MPa, σ_y = 20 MPa, τ_xy = 30 MPa Step 1 – Radius (= τ_max): R = √[(σ_x − σ_y)²/4 + τ_xy²] R = √[(80 − 20)²/4 + 30²] R = √[(60)²/4 + 900] R = √[900 + 900] R = √1800 = 42.43 MPa Therefore: τ_max = 42.43 MPa Step 2 – Average normal stress on the max-shear plane: σ_avg = (σ_x + σ_y)/2 = (80 + 20)/2 = 50 MPa Answer: τ_max = 42.43 MPa; the average normal stress on those planes = 50 MPa.

Question Type

numerical

Answer Structure

• Step 1: Write and substitute the R formula correctly [1 mark]
• Step 2: State σ_avg with correct value and explain it acts on the max-shear plane [1 mark]

Scoring Breakdown

Common Mark Deductions

• Forgetting to divide (σ_x − σ_y) by 2 inside the square root
• Stating zero normal stress on the max-shear plane (only true for pure shear)

Key Phrases To Include

• τ_max = R = √[(σ_x − σ_y)²/4 + τ_xy²]
• σ_avg = (σ_x + σ_y)/2
• normal stress on the maximum shear plane equals σ_avg

Marks

3

Topic

Mohr's Circle

Difficulty

medium

Template Id

T6

Examiner Tip

A small hand-drawn Mohr's circle sketch with C, R, X, Y, and σ₁,₂ labeled is worth at least 0.5 to 1 mark in many rubrics, even in a written-explanation question.

Model Answer

Construction of Mohr's Circle for Plane Stress: 1. Establish axes: Plot normal stress σ on the horizontal axis (tension positive, rightward) and shear stress τ on the vertical axis. 2. Plot the reference points: - Point X: coordinates (σ_x, +τ_xy) — represents the x-face of the element. - Point Y: coordinates (σ_y, −τ_xy) — represents the y-face. Note the sign reversal on τ. 3. Draw the circle: Connect X and Y with a straight line; its midpoint C is the center. C lies on the σ-axis at C = (σ_x + σ_y)/2. The radius is R = √[(σ_x − σ_y)²/4 + τ_xy²]. 4. Information obtained from the circle: - Principal stresses: σ₁ = C + R and σ₂ = C − R (where the circle crosses the σ-axis, τ = 0). - Maximum in-plane shear stress: τ_max = R (top and bottom of circle). - Principal-plane angle: the angle from X to the σ₁ point on the circle is 2θ_p; divide by 2 for the physical angle. - Any inclined plane rotated by θ in the element corresponds to a rotation of 2θ in the same sense on the circle.

Question Type

short_answer

Answer Structure

• State the axes convention (σ horizontal, τ vertical, tension positive) [0.5 mark]
• Correctly describe points X(σ_x, +τ_xy) and Y(σ_y, −τ_xy) [1 mark]
• Define center C and radius R with formulas [0.5 mark]
• List at least two results read from the circle (σ₁,₂ and τ_max) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Plotting Y with +τ_xy instead of −τ_xy (sign reversal is essential)
• Omitting the 2θ vs θ angle relationship
• Not mentioning that the circle crosses the σ-axis at principal stresses (where τ = 0)

Key Phrases To Include

• Point X(σ_x, +τ_xy)
• Point Y(σ_y, −τ_xy)
• center C = (σ_x + σ_y)/2
• radius R = √[(σ_x − σ_y)²/4 + τ_xy²]
• σ₁ = C + R, σ₂ = C − R
• physical rotation θ equals 2θ on Mohr's circle

Marks

3

Topic

Principal Stresses and Mohr's Circle

Difficulty

medium

Template Id

T7

Examiner Tip

Always handle the sign of σ_y carefully. The difference (σ_x − σ_y) is the most error-prone step. Write it out fully: 60 − (−20) = 80.

Model Answer

Given: σ_x = +60 MPa, σ_y = −20 MPa, τ_xy = +40 MPa Sign convention: tension = positive, shear as given. Step 1 – Center of Mohr's circle: C = (σ_x + σ_y)/2 = (60 + (−20))/2 = 40/2 = 20 MPa Step 2 – Radius: R = √[((σ_x − σ_y)/2)² + τ_xy²] R = √[((60 − (−20))/2)² + 40²] R = √[(40)² + (40)²] R = √[1600 + 1600] = √3200 = 56.57 MPa Step 3 – Principal stresses: σ₁ = C + R = 20 + 56.57 = 76.57 MPa ≈ 76.6 MPa σ₂ = C − R = 20 − 56.57 = −36.57 MPa ≈ −36.6 MPa Check: σ₁ + σ₂ = 76.6 − 36.6 = 40 MPa = σ_x + σ_y ✓ Step 4 – Maximum in-plane shear stress: τ_max = R = 56.57 MPa ≈ 56.6 MPa Step 5 – Principal-plane angle: tan 2θ_p = 2τ_xy/(σ_x − σ_y) = 2(40)/(60 − (−20)) = 80/80 = 1.0 2θ_p = arctan(1.0) = 45° → θ_p = 22.5° Answer: σ₁ = 76.6 MPa, σ₂ = −36.6 MPa, τ_max = 56.6 MPa, θ_p = 22.5° (counterclockwise from x-face).

Question Type

numerical

Answer Structure

• State sign convention [implicit, no dedicated mark but penalized if absent]
• Compute center C = 20 MPa [0.5 mark]
• Compute radius R = 56.57 MPa [1 mark]
• State σ₁ and σ₂ with correct values and units [0.5 mark]
• State τ_max = R [0.5 mark]
• Compute θ_p using tan 2θ_p formula [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Forgetting the negative sign on σ_y when computing (σ_x − σ_y): (60 − (−20)) = 80, not 40
• Reporting θ_p = 45° instead of 22.5° (angle doubling error)
• Omitting the invariant check which could flag arithmetic errors

Key Phrases To Include

• C = (σ_x + σ_y)/2
• R = √[((σ_x − σ_y)/2)² + τ_xy²]
• σ₁ = C + R
• σ₂ = C − R
• τ_max = R
• tan 2θ_p = 2τ_xy/(σ_x − σ_y)

Marks

3

Topic

Mohr's Circle — Pure Shear

Difficulty

medium

Template Id

T8

Examiner Tip

Board questions on pure shear almost always ask for the physical explanation. Prepare a one-sentence answer about the 45° helical fracture and brittle materials — it earns the conceptual mark every time.

Model Answer

Given: σ_x = 0, σ_y = 0, τ_xy = 60 MPa (pure shear state) Step 1 – Center and Radius: C = (0 + 0)/2 = 0 MPa (center at the origin) R = √[(0)² + (60)²] = 60 MPa Step 2 – Principal stresses: σ₁ = 0 + 60 = +60 MPa (tension) σ₂ = 0 − 60 = −60 MPa (compression) The principal planes are oriented at θ_p = 45° from the x-face (since C = 0, the Mohr's circle is centered at the origin and the principal axes are exactly 45° from the shear-stressed faces). Physical significance for shafts in torsion: A shaft subjected to torque has pure shear on its cross-section. The equivalent state creates equal tensile and compressive stresses at 45° to the shaft axis. Brittle materials (e.g., cast iron or concrete cylinders) fail by cracking along the 45° helix — in tension along the σ₁ direction — rather than by direct shear. This is why a chalk stick in torsion shows a 45° spiral fracture surface.

Question Type

short_answer

Answer Structure

• Identify σ_x = 0, σ_y = 0 and set up pure shear state [0.5 mark]
• Compute C = 0 and R = 60 MPa [0.5 mark]
• State σ₁ = +60 MPa and σ₂ = −60 MPa [1 mark]
• Explain 45° principal planes and physical significance (brittle failure on 45° helix) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Failing to explain the physical significance (the third mark is entirely for conceptual explanation)
• Stating principal planes are at 90° instead of 45°

Key Phrases To Include

• pure shear: σ_x = σ_y = 0
• center at origin
• σ₁ = +τ, σ₂ = −τ
• 45° principal planes
• brittle materials fail at 45° helix
• tensile fracture along σ₁ direction

Marks

3

Topic

Stress Transformation

Difficulty

hard

Template Id

T9

Examiner Tip

In theory-derivation questions, each logical step earns a mark. Never skip from formula to result; show differentiation explicitly.

Model Answer

For a plane-stress element with σ_x, σ_y, and τ_xy, the normal stress on a plane whose outward normal makes angle θ (measured counterclockwise) with the x-axis is obtained by equilibrium of a wedge element: σ_θ = (σ_x + σ_y)/2 + (σ_x − σ_y)/2 · cos 2θ + τ_xy · sin 2θ This can be rewritten as: σ_θ = σ_avg + (Δσ/2)·cos 2θ + τ_xy·sin 2θ where σ_avg = (σ_x + σ_y)/2 and Δσ/2 = (σ_x − σ_y)/2. To find the maximum (principal) stress, differentiate with respect to θ and set dσ_θ/dθ = 0: dσ_θ/dθ = −(σ_x − σ_y)·sin 2θ + 2τ_xy·cos 2θ = 0 Dividing: tan 2θ_p = 2τ_xy/(σ_x − σ_y) Substituting back yields: σ₁,₂ = (σ_x + σ_y)/2 ± √[((σ_x − σ_y)/2)² + τ_xy²] The ± corresponds to two roots separated by 90° that give the maximum (σ₁) and minimum (σ₂) normal stresses. At these orientations, inspection of the shear-stress transformation formula shows τ = 0, confirming that principal planes carry no shear.

Question Type

short_answer

Answer Structure

• Write the stress-transformation formula for σ_θ [1 mark]
• Differentiate and set dσ/dθ = 0 to obtain tan 2θ_p [1 mark]
• Substitute back to obtain the principal stress formula and note τ = 0 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Omitting the sin 2θ term (the τ_xy contribution) in the transformation formula
• Not connecting the derivation result to τ = 0 on principal planes

Key Phrases To Include

• σ_θ = (σ_x + σ_y)/2 + (σ_x − σ_y)/2 · cos 2θ + τ_xy · sin 2θ
• dσ_θ/dθ = 0
• tan 2θ_p = 2τ_xy/(σ_x − σ_y)
• σ₁,₂ = σ_avg ± R
• shear stress is zero on principal planes

Marks

5

Topic

Combined Bending and Torsion

Difficulty

hard

Template Id

T10

Examiner Tip

State the criterion (maximum-shear-stress = Tresca) explicitly in your first line. Then compute T_e. Examiners see students switch criteria mid-solution and lose a mark for inconsistency.

Model Answer

Given: Diameter d = 60 mm, M = 1.5 kN·m = 1.5 × 10⁶ N·mm, T = 2.0 kN·m = 2.0 × 10⁶ N·mm Step 1 – Identify the stress state at the critical point (top or bottom surface, outermost fiber): The bending moment produces a normal stress on the x-face: σ_x = 32M/(πd³) = 32(1.5 × 10⁶)/(π × 60³) σ_x = 48.0 × 10⁶/(π × 216,000) σ_x = 48.0 × 10⁶/678,584 = 70.74 MPa The y-face (circumferential) carries no bending stress: σ_y = 0 MPa The torque produces torsional shear stress: τ_xy = 16T/(πd³) = 16(2.0 × 10⁶)/(π × 60³) τ_xy = 32.0 × 10⁶/678,584 = 47.16 MPa Step 2 – Compute the equivalent torque (maximum-shear criterion): T_e = √(M² + T²) = √(1.5² + 2.0²) = √(2.25 + 4.00) = √6.25 = 2.5 kN·m T_e = 2.5 × 10⁶ N·mm Step 3 – Maximum shear stress: τ_max = 16T_e/(πd³) = 16(2.5 × 10⁶)/(π × 216,000) τ_max = 40.0 × 10⁶/678,584 τ_max = 58.95 MPa ≈ 58.9 MPa Verification using Mohr's circle formula: Center C = (70.74 + 0)/2 = 35.37 MPa R = √[(35.37)² + (47.16)²] = √[1251.0 + 2224.1] = √3475.1 = 58.95 MPa τ_max = R = 58.95 MPa ✓ Answer: τ_max = 58.9 MPa (This value must not exceed the allowable shear stress of the shaft material; for structural steel, τ_allow ≈ 0.6F_y per AISC 360.)

Question Type

numerical

Answer Structure

• Convert M and T to N·mm; state d = 60 mm [0.5 mark]
• Compute σ_x = 32M/(πd³) = 70.74 MPa [1 mark]
• Compute τ_xy = 16T/(πd³) = 47.16 MPa [1 mark]
• Compute T_e = √(M² + T²) = 2.5 kN·m [1 mark]
• Compute τ_max = 16T_e/(πd³) = 58.9 MPa [1 mark]
• Verify using Mohr's circle R formula [0.5 mark, cross-check mark]

Scoring Breakdown

Common Mark Deductions

• Using d in meters without converting — produces wildly wrong MPa values
• Forgetting σ_y = 0 (bending stress acts only on one face)
• Confusing M_e (used for σ₁) with T_e (used for τ_max)
• Not identifying which failure criterion is being used before computing

Key Phrases To Include

• σ_x = 32M/(πd³)
• τ_xy = 16T/(πd³)
• T_e = √(M² + T²)
• τ_max = 16T_e/(πd³)
• maximum-shear-stress criterion
• verification using Mohr's circle R

Marks

5

Topic

Combined Bending and Torsion — Shaft Design

Difficulty

hard

Template Id

T11

Examiner Tip

Board problems almost always ask you to size a shaft, not just find stress. Always conclude with a design statement ('use d = X mm') — examiners reward engineering judgment beyond arithmetic.

Model Answer

Given: d = 50 mm, M = 0.8 kN·m = 0.8 × 10⁶ N·mm, T = 1.0 kN·m = 1.0 × 10⁶ N·mm Criterion: Maximum-normal-stress (Rankine) Step 1 – Equivalent moment: M_e = (1/2)[M + √(M² + T²)] M_e = (1/2)[0.8 + √(0.8² + 1.0²)] × 10⁶ M_e = (1/2)[0.8 + √(0.64 + 1.00)] × 10⁶ M_e = (1/2)[0.8 + √1.64] × 10⁶ M_e = (1/2)[0.8 + 1.281] × 10⁶ M_e = (1/2)(2.081) × 10⁶ = 1.040 × 10⁶ N·mm Step 2 – Maximum normal stress on the 50 mm shaft: σ_max = σ₁ = 32M_e/(πd³) σ_max = 32(1.040 × 10⁶)/(π × 50³) σ_max = 33.28 × 10⁶/(π × 125,000) σ_max = 33.28 × 10⁶/392,699 σ_max = 84.7 MPa Step 3 – Check against allowable (σ_allow = 80 MPa): 84.7 MPa > 80 MPa → the 50 mm shaft is INADEQUATE. Step 4 – Required diameter: From σ_allow = 32M_e/(πd³): d³ = 32M_e/(π × σ_allow) d³ = 32(1.040 × 10⁶)/(π × 80) d³ = 33.28 × 10⁶/251.33 d³ = 132,400 mm³ d = (132,400)^(1/3) = 50.94 mm Use next standard size: d_required ≥ 50.94 mm → use d = 55 mm (or 51 mm if non-standard). Answer: σ_max = 84.7 MPa on the 50 mm shaft; required minimum diameter = 50.94 mm ≈ 51 mm.

Question Type

numerical

Answer Structure

• State criterion: maximum-normal-stress (Rankine); identify M_e formula [0.5 mark]
• Compute √(M² + T²) = 1.281 kN·m correctly [1 mark]
• Compute M_e = 1.040 kN·m [0.5 mark]
• Compute σ_max = 32M_e/(πd³) = 84.7 MPa and compare to allowable [1 mark]
• State shaft is inadequate [0.5 mark]
• Solve for required d = 50.94 mm [1.5 marks]

Scoring Breakdown

Common Mark Deductions

• Using T_e instead of M_e for the normal-stress criterion (criterion confusion = 2-mark loss)
• Using σ = Mc/I instead of the compact 32M_e/(πd³) formula without showing the equivalence
• Not concluding whether the shaft is adequate or not (a required design decision)

Key Phrases To Include

• M_e = (1/2)[M + √(M² + T²)]
• σ₁ = 32M_e/(πd³)
• maximum-normal-stress criterion (Rankine)
• d³ = 32M_e/(π · σ_allow)
• inadequate — exceeds allowable stress

Marks

5

Topic

Mohr's Circle — Full Analysis

Difficulty

hard

Template Id

T12

Examiner Tip

In 5-mark diagram questions, the diagram itself is worth 1 full mark. Spend 1–2 minutes sketching a neat circle with all 5 labeled elements: C, R or τ_max, σ₁, σ₂, and points X and Y.

Model Answer

Given: σ_x = −40 MPa, σ_y = +60 MPa, τ_xy = +25 MPa Sign convention: tension positive; τ_xy plotted downward on x-face (per standard Mohr's circle convention). (a) Center and Radius: C = (σ_x + σ_y)/2 = (−40 + 60)/2 = 20/2 = 10 MPa R = √[((σ_x − σ_y)/2)² + τ_xy²] R = √[((−40 − 60)/2)² + 25²] R = √[(−50)² + 625] R = √[2500 + 625] = √3125 = 55.9 MPa (a) Principal Stresses: σ₁ = C + R = 10 + 55.9 = 65.9 MPa (tension) σ₂ = C − R = 10 − 55.9 = −45.9 MPa (compression) Invariant check: σ₁ + σ₂ = 65.9 − 45.9 = 20 MPa = σ_x + σ_y ✓ (b) Maximum in-plane shear stress: τ_max = R = 55.9 MPa (with σ_avg = 10 MPa acting on the same plane) (c) Principal-plane angle: tan 2θ_p = 2τ_xy/(σ_x − σ_y) = 2(25)/(−40 − 60) = 50/(−100) = −0.5 2θ_p = arctan(−0.5) = −26.57° → θ_p = −13.28° (Negative: principal plane rotates clockwise from x-face by 13.3°) (d) Mohr's Circle Description: - Horizontal axis: σ (MPa); Vertical axis: τ (MPa) - Center C at (10, 0) - Radius R = 55.9 MPa - Point X: (−40, +25) — x-face - Point Y: (+60, −25) — y-face - Right intercept (σ-axis): σ₁ = 65.9 MPa - Left intercept (σ-axis): σ₂ = −45.9 MPa - Top of circle: τ_max = 55.9 MPa at σ = 10 MPa

Question Type

diagram_based

Answer Structure

• Compute C = 10 MPa and R = 55.9 MPa [1 mark]
• State σ₁ = 65.9 MPa and σ₂ = −45.9 MPa [1 mark]
• State τ_max = R = 55.9 MPa with σ_avg = 10 MPa [1 mark]
• Compute θ_p = −13.3° with correct direction stated [1 mark]
• Draw or describe Mohr's circle with center, radius, X, Y, and intercepts labeled [1 mark]

Scoring Breakdown

Common Mark Deductions

• Arithmetic error in (σ_x − σ_y)/2: (−40 − 60)/2 = −50, not −5 (decimal place)
• Missing the negative sign on σ_x when computing center
• Drawing Mohr's circle without labeling X and Y points loses the diagram mark
• Not stating direction (clockwise/counterclockwise) for θ_p

Key Phrases To Include

• C = (σ_x + σ_y)/2 = 10 MPa
• R = 55.9 MPa
• σ₁ = 65.9 MPa (tension)
• σ₂ = −45.9 MPa (compression)
• τ_max = R
• tan 2θ_p = 2τ_xy/(σ_x − σ_y)
• Point X(−40, +25), Point Y(60, −25)

Marks

3

Topic

Superposition — Axial Plus Bending

Difficulty

medium

Template Id

T13

Examiner Tip

Always write the general formula σ = ±P/A ± Mc/I (or ±M/S) before substituting. This earns a setup mark even if arithmetic is wrong downstream.

Model Answer

Given: P = 200 kN = 200,000 N (compressive → negative) M = 10 kN·m = 10 × 10⁶ N·mm (bending) A = 5,000 mm², S = 250,000 mm³ Principle: Superposition of axial and bending stresses. Step 1 – Axial stress (uniform compression): σ_axial = −P/A = −200,000/5,000 = −40 MPa (Negative because compressive.) Step 2 – Bending stress at extreme fibers: σ_bending = ±M/S = ±10 × 10⁶/250,000 = ±40 MPa (Plus on tension fiber, minus on compression fiber.) Step 3 – Combined stresses by superposition: Maximum (least compressive or tensile) fiber: σ_max = σ_axial + σ_bending(+) = −40 + 40 = 0 MPa Minimum (most compressive) fiber: σ_min = σ_axial + σ_bending(−) = −40 + (−40) = −80 MPa Answer: σ_max = 0 MPa (neutral — tension fiber exactly cancels axial compression) σ_min = −80 MPa (compression — governs design) Note: This is a combined axial-bending (P-M interaction) case. Under NSCP 2015 Section 505 for steel columns, both axial and flexural capacity must be checked simultaneously using the interaction equation.

Question Type

numerical

Answer Structure

• Compute axial stress: σ_axial = −P/A = −40 MPa [1 mark]
• Compute bending stress: σ_bending = ±M/S = ±40 MPa [1 mark]
• Apply superposition to get σ_max = 0 MPa and σ_min = −80 MPa [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using P as positive for a compressive load without adjusting sign
• Adding magnitudes instead of algebraic superposition
• Confusing section modulus S with moment of inertia I

Key Phrases To Include

• σ = −P/A ± M/S
• superposition
• tension fiber: −P/A + M/S
• compression fiber: −P/A − M/S
• compressive stress negative

Marks

2

Topic

Failure Theories

Difficulty

medium

Template Id

T14

Examiner Tip

Remember the hierarchy by conservatism for ductile materials: Tresca (most conservative) > von Mises > test results. This sequence appears directly on board exam MCQs.

Model Answer

Tresca Criterion (Maximum Shear Stress): Failure occurs when τ_max = S_y/2, i.e., when (σ₁ − σ₂)/2 = S_y/2. Equivalently: σ₁ − σ₂ = S_y. Von Mises Criterion (Distortion Energy): Failure occurs when the von Mises (effective) stress reaches yield: σ_v = √(σ₁² − σ₁σ₂ + σ₂²) = S_y Comparison: Tresca is more conservative because it predicts failure at a lower load than von Mises. Geometrically, the Tresca envelope is a hexagon inscribed within the von Mises ellipse in principal-stress space — the Tresca boundary is always inside or touching the von Mises boundary. In practice, von Mises is more accurate for ductile metals (steel, aluminum), while Tresca is simpler and used in conservative design codes. For the special case of pure shear (σ₁ = −σ₂ = τ), Tresca predicts failure at τ = 0.5 S_y versus von Mises at τ = 0.577 S_y — a 15% difference.

Question Type

short_answer

Answer Structure

• State Tresca criterion: τ_max = S_y/2 or σ₁ − σ₂ = S_y [0.5 mark]
• State von Mises criterion: σ_v = √(σ₁² − σ₁σ₂ + σ₂²) = S_y [0.5 mark]
• Explain which is more conservative and why (Tresca is inside von Mises envelope) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Reversing which criterion is more conservative (von Mises is NOT more conservative)
• Confusing Rankine (maximum normal stress — for brittle) with Tresca (for ductile)

Key Phrases To Include

• τ_max = S_y/2 (Tresca)
• σ_v = √(σ₁² − σ₁σ₂ + σ₂²) (von Mises)
• Tresca is more conservative
• Tresca hexagon inscribed within von Mises ellipse
• ductile materials

Marks

5

Topic

Thin-Walled Pressure Vessel — Combined State

Difficulty

hard

Template Id

T15

Examiner Tip

Pressure vessel problems almost always have a 'trap' in the shear stress: when both principals are the same sign, the absolute maximum shear involves the zero third principal stress and is LARGER than the in-plane maximum shear. Always compute and compare both.

Model Answer

Given: p = 3 MPa, r = 500 mm, t = 10 mm Check: r/t = 500/10 = 50 > 10 → thin-wall assumption is valid. Step 1 – Hoop (circumferential) stress: σ_h = pr/t = 3(500)/10 = 150 MPa (tension) Step 2 – Longitudinal (axial) stress: σ_L = pr/(2t) = 3(500)/20 = 75 MPa (tension) Step 3 – Principal stresses (no shear on these surfaces for a closed-ended vessel with no torsion): σ₁ = σ_h = 150 MPa (hoop — larger) σ₂ = σ_L = 75 MPa (longitudinal) σ₃ = 0 MPa (through-thickness, outer surface) Step 4 – Maximum in-plane shear stress: τ_max(in-plane) = (σ₁ − σ₂)/2 = (150 − 75)/2 = 37.5 MPa Step 5 – Absolute maximum shear stress (considering the through-thickness direction): Since both principal stresses are positive (same sign) and σ₃ = 0: τ_abs = (σ₁ − σ₃)/2 = (150 − 0)/2 = 75 MPa This τ_abs governs over τ_max(in-plane) and must be used for failure assessment. Note: NSCP 2015 Chapter 4 and ASME pressure vessel codes require that the design pressure account for weld efficiency and temperature derating factors, but stress formulas remain σ_h = pr/t. Answer: σ₁ = 150 MPa, σ₂ = 75 MPa, τ_max(in-plane) = 37.5 MPa, τ_abs_max = 75 MPa.

Question Type

numerical

Answer Structure

• Verify thin-wall: r/t = 50 > 10 ✓ [0.5 mark]
• Compute σ_h = pr/t = 150 MPa [1 mark]
• Compute σ_L = pr/(2t) = 75 MPa [1 mark]
• State principal stresses σ₁, σ₂, σ₃ = 0 [0.5 mark]
• Compute τ_max(in-plane) = 37.5 MPa [1 mark]
• Compute τ_abs = 75 MPa using σ₃ = 0 and explain it governs [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using σ_L = pr/t (forgetting the factor of 2 in the denominator)
• Computing only τ_max(in-plane) and missing τ_abs — losing the 'out-of-plane' mark
• Not checking whether the vessel qualifies as thin-walled before applying the formula

Key Phrases To Include

• σ_h = pr/t (hoop stress)
• σ_L = pr/(2t) (longitudinal stress)
• thin-wall valid: r/t > 10
• σ₃ = 0 on outer surface
• τ_abs = (σ₁ − σ₃)/2 governs when both in-plane principals are same sign