CELE Strength of Materials — TorsionExam Answer Templates

Marks

1

Topic

Torsion of Circular Shafts — Basic Formula

Difficulty

easy

Template Id

T1

Examiner Tip

For 1-mark VSA, examiners want precision in ONE sentence plus the formula — do not elaborate; extra incorrect statements can cancel the mark.

Model Answer

Torsion is the twisting of a structural member caused by a moment (torque) applied about its longitudinal axis. The maximum shear stress in a solid circular shaft of diameter d under torque T is: τ_max = 16T / (πd³).

Question Type

very_short_answer

Answer Structure

• Line 1: One-sentence definition of torsion stating 'twisting' and 'longitudinal axis' [½ mark]
• Line 2: Correct formula τ_max = 16T/πd³ with all symbols identified [½ mark]

Scoring Breakdown

Common Mark Deductions

• Writing τ = Tc/J without specifying it equals 16T/πd³ for solid shafts loses the recognition mark.
• Omitting the word 'longitudinal' in the definition — examiners distinguish torsion from bending.
• Writing the formula with d as radius instead of diameter.

Key Phrases To Include

• twisting
• longitudinal axis
• torque
• τ_max = 16T/πd³
• polar moment of inertia

Marks

1

Topic

Polar Moment of Inertia

Difficulty

easy

Template Id

T2

Examiner Tip

Examiners specifically test the hollow shaft formula because students confuse it with the area formula. Write the exponent '4' clearly — '2' is wrong.

Model Answer

(a) Solid shaft: J = πd⁴/32 (b) Hollow shaft: J = π(D⁴ − d⁴)/32 Note: In the hollow formula, the inner fourth-power is SUBTRACTED from the outer fourth-power.

Question Type

very_short_answer

Answer Structure

• Part (a): Correct J for solid shaft [½ mark]
• Part (b): Correct J for hollow shaft showing subtraction of d⁴ [½ mark]

Scoring Breakdown

Common Mark Deductions

• Writing π(D² − d²)²/32 instead of π(D⁴ − d⁴)/32 — squaring differences, not subtracting fourth powers.
• Using radius c instead of diameter d in πd⁴/32 formula (correct radius form is πc⁴/2).
• Adding instead of subtracting the inner diameter contribution.

Key Phrases To Include

• J = πd⁴/32
• J = π(D⁴ − d⁴)/32
• polar moment of inertia
• fourth power

Marks

2

Topic

Torsion of Circular Shafts — Maximum Shear Stress

Difficulty

easy

Template Id

T3

Examiner Tip

Always show the unit conversion explicitly on its own line — examiners award a mark for the conversion step alone in 2-mark numericals.

Model Answer

Given: d = 50 mm, T = 1.2 kN·m = 1.2 × 10⁶ N·mm Formula: τ_max = 16T / (πd³) Substituting: τ_max = 16(1.2 × 10⁶) / [π(50)³] τ_max = 19.2 × 10⁶ / [π × 125 000] τ_max = 19.2 × 10⁶ / 392 699 ∴ τ_max = 48.9 MPa

Question Type

numerical

Answer Structure

• Line 1: List given data with unit conversion (kN·m → N·mm) [½ mark]
• Line 2: State formula τ_max = 16T/πd³ [½ mark]
• Line 3: Substitute values and compute numerator and denominator [½ mark]
• Line 4: Final boxed answer with unit MPa [½ mark]

Scoring Breakdown

Common Mark Deductions

• Using T = 1.2 kN·m directly without converting to N·mm gives an answer 10⁶ times too small — zero credit for the computation mark.
• Using d = 25 mm (radius instead of diameter) — answer is 8 times too large.
• Forgetting to write the unit 'MPa' on the final answer.

Key Phrases To Include

• τ_max = 16T/πd³
• N·mm
• MPa
• 48.9 MPa

Marks

2

Topic

Angle of Twist

Difficulty

easy

Template Id

T4

Examiner Tip

Examiners test whether you know θ comes out in radians — stating this explicitly always earns the definition mark even if you make a later error.

Model Answer

The angle of twist for a uniform circular shaft is: θ = TL / (JG) Where: T = applied torque (N·mm) L = length of shaft (mm) J = polar moment of inertia of cross-section (mm⁴) G = shear modulus (modulus of rigidity) of the material (MPa or N/mm²) The angle θ is in RADIANS. Convert to degrees by multiplying by (180/π) if required.

Question Type

short_answer

Answer Structure

• Line 1: Correct formula θ = TL/(JG) [1 mark]
• Lines 2–5: Correct identification of all four terms T, L, J, G with units [½ mark]
• Line 6: Explicit statement that θ is in radians [½ mark]

Scoring Breakdown

Common Mark Deductions

• Writing θ in degrees directly without stating that the formula gives radians.
• Omitting G or confusing G (shear modulus) with E (Young's modulus).
• Writing J incorrectly as second moment of area (I) instead of polar moment.

Key Phrases To Include

• θ = TL/JG
• torsional rigidity JG
• radians
• shear modulus G

Marks

2

Topic

Power Transmission

Difficulty

easy

Template Id

T5

Examiner Tip

The board exam almost always specifies rpm. Memorize T = 60P/(2πN) exactly as written — it eliminates the rpm-to-rad/s conversion error.

Model Answer

Given: P = 75 kW = 75 000 W = 75 000 N·m/s, N = 1 200 rpm Formula: P = 2πNT/60 → T = 60P / (2πN) Substituting: T = 60(75 000) / [2π(1 200)] T = 4 500 000 / 7 539.82 ∴ T = 596.8 N·m ≈ 597 N·m

Question Type

numerical

Answer Structure

• Line 1: Convert P to watts (N·m/s) and state N in rpm [½ mark]
• Line 2: State formula T = 60P/(2πN) [½ mark]
• Line 3: Substitute and compute [½ mark]
• Line 4: Final answer in N·m [½ mark]

Scoring Breakdown

Common Mark Deductions

• Using P = 2πfT (rev/s formula) but substituting N in rpm without dividing by 60 — gives answer 60 times too small.
• Converting 75 kW as 75 N·m/s instead of 75 000 N·m/s.
• Leaving the answer in N·mm without stating the conversion.

Key Phrases To Include

• P = 2πNT/60
• T = 60P/(2πN)
• N·m
• rpm

Marks

3

Topic

Angle of Twist

Difficulty

medium

Template Id

T6

Examiner Tip

Show all three steps explicitly numbered — examiners award partial credit at each step. A unit-consistent numerical answer in radians earns 2 of 3 marks even if degree conversion is wrong.

Model Answer

Given: d = 40 mm, L = 2 m = 2 000 mm, T = 800 N·m = 800 × 10³ N·mm, G = 80 GPa = 80 000 MPa Step 1 — Compute J: J = πd⁴/32 = π(40)⁴/32 = π(2 560 000)/32 = 251 327 mm⁴ Step 2 — Compute θ in radians: θ = TL/(JG) = (800 × 10³)(2 000) / [(251 327)(80 000)] θ = 1.6 × 10⁹ / (2.011 × 10¹⁰) θ = 0.07954 rad Step 3 — Convert to degrees: θ = 0.07954 × (180/π) = 0.07954 × 57.296 ∴ θ = 4.56°

Question Type

numerical

Answer Structure

• Step 1: Correct unit conversions (mm, N·mm, MPa) [½ mark]
• Step 2: Correct computation of J = πd⁴/32 = 251 327 mm⁴ [1 mark]
• Step 3: Correct substitution into θ = TL/JG giving 0.0795 rad [1 mark]
• Step 4: Conversion to degrees → 4.56° [½ mark]

Scoring Breakdown

Common Mark Deductions

• Using L = 2 m directly without converting to mm — gives answer 1000 times too small.
• Using G = 80 GPa = 80 in MPa instead of 80 000 MPa.
• Skipping the radian-to-degree conversion when the question explicitly asks for degrees.
• Computing J with radius c = 20 mm using πc⁴/2 correctly but then forgetting the factor — answer is off by 2.

Key Phrases To Include

• J = πd⁴/32
• θ = TL/JG
• radians
• × (180/π)
• 4.56°

Marks

3

Topic

Flanged Bolt Couplings

Difficulty

medium

Template Id

T7

Examiner Tip

The three-step structure (Area → Force → Torque) is universal for bolt coupling problems — memorize this sequence and write it even before reading the full question.

Model Answer

Given: n = 6 bolts, d_b = 20 mm, R = 150 mm, τ_allow = 70 MPa Step 1 — Bolt cross-sectional area: A = π d_b²/4 = π(20)²/4 = 314.16 mm² Step 2 — Shear force per bolt: P = A × τ_allow = 314.16 × 70 = 21 991 N Step 3 — Torque capacity: T = P × R × n = 21 991 × 150 × 6 = 19 792 000 N·mm ∴ T = 19.79 × 10⁶ N·mm ≈ 19.8 kN·m

Question Type

numerical

Answer Structure

• Step 1: Bolt area A = πd²/4 = 314.16 mm² [1 mark]
• Step 2: Force per bolt P = Aτ = 21 991 N [1 mark]
• Step 3: T = PRn = 19.8 kN·m with unit [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using diameter 150 mm instead of radius 150 mm for R — answer is halved.
• Computing bolt area using πd²/2 or πd (perimeter) instead of πd²/4.
• Forgetting to multiply by n = 6 (number of bolts).
• Not converting final answer to kN·m — examiners penalize unit omission.

Key Phrases To Include

• A = πd²/4
• P = Aτ
• T = PRn
• bolt circle radius

Marks

3

Topic

Power Transmission and Shaft Design

Difficulty

medium

Template Id

T8

Examiner Tip

The two-step Power→Torque→Diameter pattern appears in nearly every board examination on torsion. Practice it until it is automatic.

Model Answer

Given: P = 75 000 W, N = 1 200 rpm, τ_allow = 60 MPa Step 1 — Compute torque: T = 60P/(2πN) = 60(75 000)/[2π(1 200)] T = 4 500 000/7 539.82 = 596.8 N·m = 596 831 N·mm Step 2 — Apply torsion formula for diameter: τ_max = 16T/(πd³) → d³ = 16T/(π × τ_allow) d³ = 16(596 831)/[π(60)] d³ = 9 549 296/188.496 = 50 661 mm³ d = (50 661)^(1/3) = 37.0 mm ∴ Use d ≥ 37 mm (round up to next standard size)

Question Type

numerical

Answer Structure

• Step 1: Torque from power and speed: T = 597 N·m = 596 831 N·mm [1 mark]
• Step 2: Rearrange torsion formula to solve for d³ [1 mark]
• Step 3: d = 37.0 mm with recommendation to round up [1 mark]

Scoring Breakdown

Common Mark Deductions

• Forgetting to convert T from N·m to N·mm before using it with d in mm.
• Solving for d² instead of d³ (confusing torsion with bending).
• Not stating that d must be rounded UP — safety requires the next larger standard size.

Key Phrases To Include

• T = 60P/(2πN)
• d³ = 16T/(πτ)
• 37 mm
• round up to standard size

Marks

5

Topic

Torsion Formula Derivation

Difficulty

hard

Template Id

T9

Examiner Tip

Long-answer derivation questions are structured around the three pillars: GEOMETRY (kinematics) → MATERIAL LAW → EQUILIBRIUM. Write these as labeled steps — examiners mark against this structure.

Model Answer

DERIVATION OF THE TORSION FORMULA Assumptions: 1. The material is linearly elastic, homogeneous, and isotropic. 2. The cross-section is circular (solid or hollow) and remains plane after twisting (no warping). 3. Plane cross-sections remain plane and rotate as rigid bodies. 4. Shear stress is proportional to shear strain (Hooke's Law in shear: τ = Gγ). 5. The angle of twist is small. Derivation: Consider a circular shaft of length L and radius c subject to torque T. Step 1 — Geometry (kinematics): For a fiber at radius ρ from the axis, the shear strain γ is: γ = ρ × (θ/L) ... (i) where θ is the angle of twist over length L. This shows γ is proportional to ρ. Step 2 — Material law: From Hooke's Law in shear: τ = Gγ Substituting (i): τ = G(ρθ/L) = (Gθ/L)ρ ... (ii) This confirms shear stress τ is LINEAR in ρ: zero at the center, maximum at r = c. Step 3 — Equilibrium (moment resultant): The torque T is the resultant of all shear-stress couples over the cross-section: T = ∫ ρ × τ × dA = ∫ ρ × (Gθ/L)ρ dA = (Gθ/L) ∫ ρ² dA Since J = ∫ ρ² dA (polar moment of inertia): T = (Gθ/L) × J → θ = TL/(JG) ... (iii) ← Angle of twist formula Step 4 — Stress formula: From (ii): τ = (Gθ/L)ρ. Substituting Gθ/L = T/J from (iii): τ = Tρ/J At the outer surface ρ = c: τ_max = Tc/J ← TORSION FORMULA For a solid shaft, c = d/2 and J = πd⁴/32: τ_max = T(d/2)/(πd⁴/32) = 16T/(πd³) ← convenient solid-shaft form

Question Type

long_answer

Answer Structure

• Part 1: State all five assumptions — 1 mark
• Part 2: Geometric/kinematic step showing γ = ρθ/L — 1 mark
• Part 3: Material law application giving τ = (Gθ/L)ρ — 1 mark
• Part 4: Equilibrium integration defining J = ∫ρ²dA and deriving θ = TL/JG — 1 mark
• Part 5: Final substitution giving τ = Tc/J and solid-shaft form τ = 16T/πd³ — 1 mark

Scoring Breakdown

Common Mark Deductions

• Skipping the assumptions — this section alone is worth 1 full mark.
• Jumping straight to τ = Tc/J without showing the derivation steps — examiners require the integral.
• Confusing J (polar moment) with I (second moment of area about a bending axis) — using I instead of J.
• Not explaining WHY stress is linear (skipping the geometric step) — loses the kinematic mark.
• Not writing the final solid-shaft form 16T/πd³ — common boards-level expectation.

Key Phrases To Include

• plane sections remain plane
• linear elastic
• γ = ρθ/L
• τ = Gγ
• J = ∫ρ²dA
• τ = Tc/J
• τ_max = 16T/πd³

Marks

3

Topic

Torsion of Circular Shafts — Hollow Section

Difficulty

medium

Template Id

T10

Examiner Tip

Always pause and verify: for τ_MAX, use the OUTER radius c = D/2 — maximum stress is at the outermost fiber, not the inner.

Model Answer

Given: D = 80 mm, d = 60 mm, T = 3 kN·m = 3 × 10⁶ N·mm Part (a) — Polar moment of inertia: J = π(D⁴ − d⁴)/32 J = π[(80)⁴ − (60)⁴]/32 J = π[40 960 000 − 12 960 000]/32 J = π(28 000 000)/32 J = 2 748 893 mm⁴ ≈ 2.749 × 10⁶ mm⁴ Part (b) — Maximum shear stress: τ_max = T × c / J, where c = D/2 = 40 mm τ_max = (3 × 10⁶ × 40) / (2 748 893) τ_max = 120 × 10⁶ / 2 748 893 ∴ τ_max = 43.7 MPa

Question Type

numerical

Answer Structure

• Line 1: Unit conversion of T to N·mm [½ mark]
• Part (a): Correct J formula with subtraction of fourth powers, J = 2.749 × 10⁶ mm⁴ [1 mark]
• Part (b): Use c = D/2 = 40 mm, correct τ = Tc/J calculation [1 mark]
• Final: τ_max = 43.7 MPa with unit [½ mark]

Scoring Breakdown

Common Mark Deductions

• Computing J = π(D² − d²)²/32 — squaring the difference rather than subtracting fourth powers (most common error).
• Using c = d/2 = 30 mm (inner radius) instead of c = D/2 = 40 mm for τ_max.
• Using T = 3 kN·m directly without converting to N·mm.

Key Phrases To Include

• J = π(D⁴ − d⁴)/32
• c = D/2 = 40 mm (outer radius)
• τ_max = Tc/J
• 43.7 MPa

Marks

3

Topic

Thin-Walled Closed Tubes

Difficulty

medium

Template Id

T11

Examiner Tip

The key phrase to write is 'A_m is the area enclosed by the median (centerline) of the wall' — this shows you understand Bredt's formula and earns the formula mark.

Model Answer

Given: t = 5 mm, T = 2 kN·m = 2 × 10⁶ N·mm Outer dimensions: 100 mm × 60 mm Formula for thin-walled closed tube: τ = T / (2 A_m t) Step 1 — Find A_m (area enclosed by median line): Median line dimensions: Width = 100 − 2(t/2) = 100 − 5 = 95 mm [subtract half-thickness on each side] Height = 60 − 2(t/2) = 60 − 5 = 55 mm A_m = 95 × 55 = 5 225 mm² Step 2 — Compute shear stress: τ = T / (2 A_m t) τ = (2 × 10⁶) / (2 × 5 225 × 5) τ = 2 × 10⁶ / 52 250 ∴ τ = 38.3 MPa

Question Type

numerical

Answer Structure

• Line 1: State formula τ = T/(2A_m t) [½ mark]
• Step 1: Compute median-line dimensions (subtract t/2 each side) and A_m = 5 225 mm² [1 mark]
• Step 2: Substitute into formula and compute τ [1 mark]
• Final: 38.3 MPa with unit [½ mark]

Scoring Breakdown

Common Mark Deductions

• Using the outer area (100 × 60 = 6 000 mm²) instead of the median-line area — common and costly error.
• Forgetting the factor of 2 in the denominator 2A_m t.
• Not converting torque to N·mm before substituting.

Key Phrases To Include

• τ = T/(2A_m t)
• median line
• enclosed area A_m
• thin-walled
• shear flow q = τt

Marks

5

Topic

Flanged Bolt Couplings — Two Concentric Circles

Difficulty

hard

Template Id

T12

Examiner Tip

Compatibility for bolt couplings is often worth 2 of 5 marks — the phrase 'bolt deformation is proportional to its distance from the center' and the ratio τ₁/R₁ = τ₂/R₂ are the mark-earning statements.

Model Answer

Given: R₁ = 100 mm, n₁ = 4 bolts; R₂ = 160 mm, n₂ = 6 bolts; d_b = 16 mm; τ_allow = 60 MPa Step 1 — Bolt area (same for all bolts): A = π(16)²/4 = 201.06 mm² Step 2 — Deformation compatibility (shear stress proportional to radius): For a rigid flange, bolt deformation ∝ radius: τ₁/R₁ = τ₂/R₂ → τ₁ = τ₂(R₁/R₂) Step 3 — Identify critical (maximum-stress) bolt circle: Outer bolts are farther from center, so they deform more and attract higher stress. Assume outer bolts reach the limit first: τ₂ = 60 MPa Inner bolt stress: τ₁ = 60 × (100/160) = 37.5 MPa Step 4 — Forces in each bolt: Outer bolt: P₂ = A × τ₂ = 201.06 × 60 = 12 063.7 N Inner bolt: P₁ = A × τ₁ = 201.06 × 37.5 = 7 539.8 N Step 5 — Total torque: T = (P₁ × R₁ × n₁) + (P₂ × R₂ × n₂) T = (7 539.8 × 100 × 4) + (12 063.7 × 160 × 6) T = 3 015 920 + 11 581 152 T = 14 597 072 N·mm ∴ T = 14.60 kN·m

Question Type

numerical

Answer Structure

• Step 1: Bolt area A = πd²/4 = 201.06 mm² [½ mark]
• Step 2: Compatibility condition τ₁/R₁ = τ₂/R₂ stated [1 mark]
• Step 3: Identify outer bolts govern; compute τ₁ = 37.5 MPa [1 mark]
• Step 4: Forces P₁ and P₂ computed correctly [1 mark]
• Step 5: T = ΣPRn = 14.60 kN·m with unit [½ mark + ½ mark for each circle correct]

Scoring Breakdown

Common Mark Deductions

• Assuming τ₁ = τ₂ = 60 MPa for all bolts — ignores compatibility, overestimates torque.
• Applying τ_allow to the inner bolts instead of the outer bolts as the critical circle.
• Forgetting to multiply by the number of bolts in each ring.
• Not summing both torque contributions — computing only one ring's contribution.

Key Phrases To Include

• deformation compatibility
• τ₁/R₁ = τ₂/R₂
• outer bolts control
• T = ΣPRn
• rigid flange assumption

Marks

2

Topic

Solid vs. Hollow Shaft Comparison

Difficulty

easy

Template Id

T13

Examiner Tip

Conceptual questions require engineering reasoning, not just memorized facts. Open with the physical principle (linear stress distribution), then derive the conclusion — examiners reward this logical flow.

Model Answer

In torsion, shear stress varies linearly from ZERO at the shaft center to a MAXIMUM at the outer surface (τ = Tρ/J). The core material (near the axis) carries negligible stress and thus contributes little to torque resistance. A hollow shaft removes this low-stress core material, reducing weight significantly while retaining most of the torsional strength. For the same outer diameter and torque, a hollow shaft has lower J (due to the removed core) but the reduction in J is small compared to the weight saved when d_i/d_o is moderate (e.g., 0.6). Conclusion: Hollow shafts provide higher torque-per-unit-mass ratio, making them structurally and economically superior for power transmission applications such as propeller shafts and turbine rotors.

Question Type

short_answer

Answer Structure

• Sentence 1: State that stress is zero at center and maximum at surface — explains why core is inefficient [1 mark]
• Sentence 2–3: State that hollow shaft removes low-stress core → same strength, less weight [½ mark]
• Conclusion: Higher torque-per-unit-mass (material efficiency) justifies hollow shafts [½ mark]

Scoring Breakdown

Common Mark Deductions

• Stating hollow is stronger than solid without qualification — hollow has lower J and lower τ capacity for same outer diameter; the advantage is efficiency, not raw strength.
• Not linking the explanation back to the physics (linear stress distribution).

Key Phrases To Include

• linear stress distribution
• zero at center
• maximum at outer surface
• low-stress core
• material efficiency
• torque-per-unit-mass

Marks

5

Topic

Angle of Twist — Back-calculation of Torque

Difficulty

hard

Template Id

T14

Examiner Tip

When the question says 'find torque AND shear stress', plan for both steps before writing — managing your solution path reduces errors. Mark your sub-answers clearly with brackets.

Model Answer

Given: d = 30 mm, L = 1.5 m = 1 500 mm, θ = 3° = 3π/180 = 0.05236 rad, G = 80 GPa = 80 000 MPa Step 1 — Polar moment of inertia: J = πd⁴/32 = π(30)⁴/32 = π(810 000)/32 = 79 522 mm⁴ Step 2 — Find torque from angle of twist formula: θ = TL/(JG) → T = θJG/L T = (0.05236)(79 522)(80 000) / 1 500 T = 0.05236 × 6 361 760 000 / 1 500 T = 333 267 840 / 1 500 T = 222 178 N·mm = 222.2 N·m Step 3 — Maximum shear stress: τ_max = 16T/(πd³) = 16(222 178)/[π(30)³] τ_max = 3 554 848 / [π × 27 000] τ_max = 3 554 848 / 84 823 ∴ T = 222.2 N·m ∴ τ_max = 41.9 MPa

Question Type

numerical

Answer Structure

• Step 1: Convert θ to radians (0.05236 rad) and L to mm [½ mark]
• Step 2: Compute J = πd⁴/32 = 79 522 mm⁴ [1 mark]
• Step 3: Rearrange and compute T = θJG/L = 222.2 N·m [1½ marks]
• Step 4: Compute τ_max = 16T/πd³ = 41.9 MPa [1½ marks]
• Step 5: State both answers with units clearly [½ mark]

Scoring Breakdown

Common Mark Deductions

• Substituting θ = 3 (degrees) directly into the formula without converting to radians — answer is 57.3 times too large.
• Forgetting to convert L from m to mm.
• Using G = 80 GPa = 80 (not 80 000 MPa) in N/mm² system.
• Not proceeding to compute τ_max — the question asks for BOTH.

Key Phrases To Include

• θ = 3° = 0.05236 rad
• J = πd⁴/32
• T = θJG/L
• τ_max = 16T/πd³
• 222 N·m
• 41.9 MPa

Marks

5

Topic

Angle of Twist — Stepped Shafts

Difficulty

hard

Template Id

T15

Examiner Tip

For stepped shafts, set up a table with columns [Segment | T | L | J | θ] — this organized layout prevents the most common errors and earns systematic method marks.

Model Answer

Given: G = 80 000 MPa for both segments Segment AB: d_AB = 60 mm, L_AB = 1 000 mm, T_AB = 1 500 N·m = 1.5 × 10⁶ N·mm Segment BC: d_BC = 40 mm, L_BC = 600 mm, T_BC = 800 N·m = 0.8 × 10⁶ N·mm Step 1 — J for each segment: J_AB = π(60)⁴/32 = π(12 960 000)/32 = 1 272 345 mm⁴ J_BC = π(40)⁴/32 = π(2 560 000)/32 = 251 327 mm⁴ Step 2 — Angle of twist for each segment: θ_AB = T_AB × L_AB / (J_AB × G) θ_AB = (1.5 × 10⁶)(1 000) / [(1 272 345)(80 000)] θ_AB = 1.5 × 10⁹ / 1.018 × 10¹¹ θ_AB = 0.014736 rad θ_BC = T_BC × L_BC / (J_BC × G) θ_BC = (0.8 × 10⁶)(600) / [(251 327)(80 000)] θ_BC = 4.8 × 10⁸ / 2.011 × 10¹⁰ θ_BC = 0.023869 rad Step 3 — Total angle of twist: θ_total = θ_AB + θ_BC = 0.014736 + 0.023869 = 0.038605 rad Convert to degrees: θ_total = 0.038605 × (180/π) = 2.21° ∴ Total angle of twist = 0.03861 rad = 2.21°

Question Type

numerical

Answer Structure

• Step 1: Correctly compute J_AB and J_BC [1 mark]
• Step 2: Correct θ_AB = 0.01474 rad with proper unit consistency [1½ marks]
• Step 3: Correct θ_BC = 0.02387 rad [1½ marks]
• Step 4: θ_total = sum of both = 2.21° [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using the same J for both segments — failing to recognize the stepped (variable) cross-section.
• Summing torques instead of summing angles of twist.
• Not converting the final angle to degrees when the question (or examiner convention) requires it.
• Unit inconsistency: mixing meters and millimeters within one segment calculation.

Key Phrases To Include

• θ_total = Σ(TL/JG)
• stepped shaft
• each segment computed separately
• angles add