CELE Strength of Materials — TorsionCheat Sheet
Torsion cheat sheet for CELE aspirants. If you could only take one sheet of paper into your review session, this is what it would look like. Professional Regulation Commission (PRC) — Board of Civil Engineering's most-tested concepts, all in one place.
Exam context
On the CELE 2026, the Strength of Materials subtest carries a "Core" weight in Professional Regulation Commission (PRC) — Board of Civil Engineering's pattern. Torsion lands at position 2nd out of 8 in the standard review order. Target score is 70% weighted average, no sub-test below 50%, and roughly a meaningful share of items come from Strength of Materials on a typical CELE paper.
Torsion - Cheat Sheet
Your last-minute revision companion for the Strength of Materials chapter on Torsion. This quick reference covers circular shaft stress and angle formulas, power transmission, flanged couplings, thin-walled tubes, and the most common board-exam pitfalls. Review the must-remember section first, then drill the formulas.
Sections
Formulas
Formula
τ_max = (T × c) / J = (16T) / (π d³)
Meaning
τ_max = maximum shear stress (Pa or MPa); T = applied torque (N⋅m); c = outer radius (m); J = polar moment of inertia (m⁴); d = shaft diameter (m)
Watch Out
Unit inconsistency kills this: keep T in N⋅mm and d in mm for d³ formula, OR use T in N⋅m and d in m. Mixing units inflates d³ errors by a factor of 1000+.
When To Use
Find max shear stress in a solid or hollow circular shaft under torsion.
Formula
J_solid = π d⁴ / 32 = π c⁴ / 2
Meaning
J = polar moment of inertia (m⁴); d = diameter (m); c = radius (m)
Watch Out
Never confuse with area moment of inertia I (used for bending). J is for torsion and always uses fourth power of a linear dimension.
When To Use
Calculate J for solid circular cross-section.
Formula
J_hollow = π (D⁴ − d⁴) / 32
Meaning
D = outer diameter; d = inner diameter; both in same units (m or mm).
Watch Out
Subtract fourth powers of diameters, NOT areas. Common error: J = π(D² − d²)/32 is WRONG.
When To Use
Hollow (tubular) shaft torsion problem.
Formula
θ = (T × L) / (J × G)
Meaning
θ = angle of twist (radians); T = torque (N⋅m); L = length (m); J = polar moment (m⁴); G = shear modulus (Pa)
Watch Out
Result is in RADIANS. Convert to degrees at the end: θ_deg = θ_rad × (180/π). Forgetting this is a classic board slip.
When To Use
Find relative twist angle between two sections, or total rotation of shaft.
Formula
γ_max = τ_max / G = (c × θ) / L
Meaning
γ = shear strain (unitless); relates max shear stress and strain to angle of twist.
Watch Out
Strain must be in radians; if using degrees, convert first.
When To Use
Relate shear strain at outer fiber to angle of twist (strain-displacement check).
Formula
τ = (T × ρ) / J
Meaning
τ = shear stress at distance ρ from centerline; stress is LINEAR across the radius.
Watch Out
This is a linear profile: zero at center (ρ = 0), maximum at ρ = c (outer edge).
When To Use
Find shear stress at any radius (not just the outer surface).
Common Values
Value
80 GPa
Symbol
G
Quantity
Shear modulus — Steel
Value
27 GPa
Symbol
G
Quantity
Shear modulus — Aluminum
Value
42 GPa
Symbol
G
Quantity
Shear modulus — Bronze
Section Title
Fundamental Torsion Formulas — Circular Shafts
Important Facts
- Shear stress in a circular shaft is LINEAR: zero at center, max at outer surface.
- A hollow shaft is much more efficient than solid at same diameter (higher J-to-weight ratio).
- Angle of twist formula θ = TL / JG is valid only within elastic limit (linear material).
- Circular sections remain plane during torsion (no warping); non-circular sections warp.
- The torsion formula assumes stress ≤ proportional limit and homogeneous material.
- Shear strain at outer fiber: γ_max = τ_max / G = c θ / L.
Key Definitions
Term
Torque (T)
Example
A drive shaft twisted by a motor applies 500 N⋅m of torque.
Definition
A moment (rotational force) applied about the longitudinal axis of a member; measured in N⋅m.
Term
Polar Moment of Inertia (J)
Example
A larger J means the shaft resists twisting better; hollow shafts have high J-to-weight ratio.
Definition
A measure of resistance to torsion; the second moment of area with respect to the polar axis. For circular sections, J = π d⁴/32.
Term
Shear Modulus (G)
Example
Steel with G = 80 GPa is stiffer in shear than aluminum with G = 27 GPa.
Definition
Material property relating shear stress to shear strain; τ = G × γ. For steel, G ≈ 80 GPa; for aluminum, G ≈ 27 GPa.
Term
Angle of Twist (θ)
Example
A 2 m shaft twists 0.05 rad ≈ 2.9° under 1 kN⋅m.
Definition
The relative rotation (in radians) between two sections of a shaft separated by length L under applied torque.
Term
Torsional Rigidity (JG)
Example
A hollow steel shaft may have JG = 200 × 10⁹ N⋅mm², resisting twist more than a smaller solid shaft.
Definition
The product of polar moment and shear modulus; higher JG means less twist for the same torque.
Diagrams To Know
- Cross-section of solid circular shaft with shear stress distribution (linear, zero at center).
- Cross-section of hollow (tubular) shaft; outer and inner diameters marked.
- Longitudinal view of shaft showing internal torque diagram (constant torque between applied couples).
- Deformation diagram: shaft twisting; angle θ marked between two sections.
Formulas
Formula
P = T × ω = 2π f × T
Meaning
P = power (W or kW); T = torque (N⋅m); ω = angular velocity (rad/s); f = rotational speed (rev/s or Hz)
Watch Out
ω must be in rad/s; if given in rpm, convert first or use the rpm formula below.
When To Use
Relate power, torque, and rotational speed. Most common form in board exams.
Formula
P = (2π N T) / 60
Meaning
N = rotational speed (rpm); T = torque (N⋅m); P = power (W). The 60 converts minutes to seconds.
Watch Out
Forgetting the factor of 60 is a CLASSIC SLIP. Result of 2πNT without dividing by 60 is 60× too large.
When To Use
When speed is given in rpm (revolutions per minute) — the most common form in Philippine exams.
Formula
T = (P × 60) / (2π N)
Meaning
Rearranged form: solve for torque from power and rpm.
Watch Out
This is the FIRST STEP in most board sizing problems: Power → Torque → Diameter.
When To Use
Power and speed are given; find the torque before computing shaft diameter or stress.
Common Values
Value
60–80 MPa
Symbol
τ_allow
Quantity
Typical allowable shear stress (steel shaft)
Section Title
Power Transmission and Speed Relationships
Important Facts
- Board problems almost always give speed in rpm; use P = (2π N T) / 60 directly.
- Power is proportional to both torque AND speed: doubling rpm doubles power at constant torque.
- In design, solve Power → Torque → Diameter in three steps; never skip the middle step.
- Typical steel shaft allowable shear stress is 60–80 MPa; use this after computing torque.
Key Definitions
Term
Power (P)
Example
A motor delivering 50 kW at 1200 rpm transmits torque T = (50,000 × 60) / (2π × 1200) ≈ 398 N⋅m.
Definition
Rate of energy transfer; in torsion, P = T ω. Units: watts (W) or kilowatts (kW).
Term
Angular Velocity (ω)
Example
1200 rpm = (2π × 1200) / 60 ≈ 125.7 rad/s.
Definition
Rotational speed in radians per second; ω = 2π N / 60 for N in rpm.
Diagrams To Know
- Power transmission diagram: motor delivering P (kW) at N (rpm); shaft carries torque T.
Formulas
Formula
d = ∛[16T / (π τ_allow)]
Meaning
d = required shaft diameter (m); T = torque (N⋅m); τ_allow = allowable shear stress (Pa).
Watch Out
Always round UP to next standard size. If calc gives 37.2 mm, use 38 or 40 mm (never round down).
When To Use
Given torque and allowable stress, find minimum solid shaft diameter.
Formula
τ_actual = (16T) / (π d³)
Meaning
Check: compute actual stress in the sized shaft.
Watch Out
Keep units consistent: N⋅mm with mm, or N⋅m with m.
When To Use
Verify that selected shaft diameter gives stress ≤ τ_allow.
Section Title
Shaft Design and Sizing
Important Facts
- Design procedure: Power (kW) → Torque (N⋅m) → Diameter (m).
- Hollow shafts reduce weight; outer diameter D is larger but wall thickness t is small.
- For stepped shafts with different diameters or materials, sum angle of twist over each segment.
- Never accept a shaft diameter smaller than the calculated value; always round up.
Key Definitions
Term
Allowable Shear Stress (τ_allow)
Example
At τ_allow = 70 MPa, a 50 mm solid shaft can safely carry τ_max = 16(1.2×10⁶) / [π(50)³] ≈ 48.9 MPa < 70 MPa. ✓
Definition
Maximum safe shear stress for the material and design code (NSCP 2015, AISC 360). Typically 60–80 MPa for steel shafts.
Diagrams To Know
- Typical step-by-step design flow: Power → rpm → Torque formula → Diameter cube-root formula → Select standard size.
Formulas
Formula
T = A τ R n
Meaning
T = torque capacity (N⋅m); A = bolt cross-sectional area (m²); τ = allowable shear stress in bolt (Pa); R = radius of bolt circle (m); n = number of bolts
Watch Out
Each bolt shears on one plane; if bolts are in double shear, use A τ for each shear plane.
When To Use
Find torque capacity of a bolted flange coupling with bolts on one circle.
Formula
P = A τ (force in one bolt)
Meaning
P = shear force per bolt; total torque T = P R n.
Watch Out
Ensure τ is allowable shear stress for the bolt material (usually lower than shaft).
When To Use
Intermediate step: find per-bolt force, then sum torque.
Formula
T = (P₁ R₁ n₁) + (P₂ R₂ n₂) (two concentric circles)
Meaning
Two rings of bolts at different radii R₁, R₂; force ratio by compatibility: P₁/R₁ = P₂/R₂.
Watch Out
Solve compatibility equations first to find P₁ and P₂; do not assume equal loads.
When To Use
Coupling with bolts on two different bolt circles.
Common Values
Value
70–100 MPa (usually less than shaft)
Symbol
τ_bolt
Quantity
Typical allowable bolt shear stress
Section Title
Flanged Bolt Couplings
Important Facts
- Bolt shear stress is typically lower allowable than shaft shear stress (bolts are weaker).
- For concentric bolt circles, deformation compatibility requires P₁/R₁ = P₂/R₂.
- Always check that τ_bolt ≤ τ_allow, and that bolt diameter is adequate.
- Coupling torque capacity is limited by either shaft or bolts (whichever is weaker).
Key Definitions
Term
Bolt Circle
Example
Six bolts on a 100 mm bolt circle; each bolt has moment arm R = 100 mm = 0.1 m.
Definition
The radius R (measured from shaft centerline to bolt centerline) on which bolts are arranged; torque arm for each bolt.
Term
Shear Plane
Example
A bolt in single shear has area A; in double shear, effective shear area is 2A (two planes resist).
Definition
The cross-sectional area where shear force acts in the bolt. Single shear: one plane; double shear: two planes (bolt passes through two plates).
Diagrams To Know
- End view of coupling: flange plates with n bolts arranged on bolt circle of radius R.
- Flange coupling with two concentric bolt circles at R₁ and R₂; force vectors in each bolt.
Formulas
Formula
τ = T / (2 A_m t)
Meaning
τ = average shear stress in wall (Pa); T = torque (N⋅m); A_m = area enclosed by median (centerline) of wall (m²); t = wall thickness (m)
Watch Out
A_m is NOT the outer area; it is the area bounded by the CENTERLINE of the wall thickness. For a rectangular box b × h with thickness t: A_m ≈ b × h (use outer dims minus t).
When To Use
Shear stress in thin-walled closed hollow section (box, tube, any closed shape with small t).
Formula
q = τ × t = T / (2 A_m) (shear flow)
Meaning
q = shear flow (units: force per unit length, N/m); constant around the perimeter of a closed tube.
Watch Out
Shear flow is constant AROUND the perimeter for closed section; stress varies inversely with t (thinner wall → higher stress).
When To Use
Verify constant shear flow in thin-walled closed section; alternative to stress formula.
Formula
θ = (T × L) / (J_w × G) [thin-walled tube angle of twist]
Meaning
J_w = torsional constant for thin-walled section; different from solid J.
Watch Out
J_w ≠ J (polar moment of solid). For closed thin-walled: J_w ≈ 4 A_m² t / ∮ (ds/t), where ∮ is perimeter integral.
When To Use
For thin-walled open (e.g., C-channel) or closed tubes, use specialized J_w (given in problem).
Section Title
Thin-Walled Closed Tubes (Shear Flow Formula)
Important Facts
- Thin-walled CLOSED tubes: use τ = T / (2 A_m t) directly; very quick and accurate.
- Shear flow q is constant; so where wall is thicker, stress is lower.
- Thin-walled OPEN sections (e.g., C-channel, I-beam): stress is non-uniform; use torsional constant J_w (usually given).
- For closed rectangular or circular tubes, membrane (shear flow) analogy applies.
Key Definitions
Term
Median Line (Centerline) of Wall
Example
A rectangular box 100 × 80 mm with wall thickness 5 mm has A_m ≈ (100 − 5) × (80 − 5) ≈ 95 × 75 ≈ 7125 mm².
Definition
The line halfway through the thickness of a thin wall; used to define A_m in shear stress and shear flow formulas.
Term
Shear Flow (q)
Example
If τ = 50 MPa and t = 2 mm, then q = 50 × 2 = 100 N/mm (constant around tube).
Definition
Product of shear stress and thickness: q = τ t; constant around the perimeter of a closed thin-walled tube.
Diagrams To Know
- Cross-section of thin-walled closed tube (rectangular or circular); centerline (median line) marked; wall thickness t shown.
- Shear stress distribution around closed tube: constant stress τ in each segment (can differ between segments if thickness varies).
Formulas
Formula
θ_total = Σ (T_i L_i) / (J_i G_i)
Meaning
θ_total = total angle of twist over multi-segment shaft; sum angle contributions from each segment.
Watch Out
Each segment must use its own J and G; forget this and your result is completely wrong.
When To Use
Shaft has different diameters, materials, or applied torques in different sections.
Formula
J_1 d_1 / T_1 L_1 = J_2 d_2 / T_2 L_2 (compatibility — same twist angle per segment)
Meaning
If all segments twist the same angle, the ratio of torque to segment stiffness must be equal.
Watch Out
Rarely asked in board exams; more common in advanced courses.
When To Use
Redundant (statically indeterminate) torque problem; use deformation compatibility to solve for internal torques.
Section Title
Stepped Shafts and Variable Loading
Important Facts
- Always sum angles of twist segment-by-segment for complex shafts.
- Torque is constant along a shaft between applied couples (draw a torque diagram first).
- Each segment contributes to total angle; ignore any segment and your answer is wrong.
Key Definitions
Term
Torsional Rigidity (J_i G_i per unit length)
Example
Segment 1: d = 50 mm, J = π(50)⁴/32 ≈ 6.14×10⁵ mm⁴, G = 80 GPa; JG ≈ 4.9×10¹⁶ N⋅mm².
Definition
The resistance of a shaft segment to twist per unit length; higher JG means less twist angle for same T.
Diagrams To Know
- Stepped shaft: diameter d₁ on length L₁, diameter d₂ on length L₂; applied torques marked.
- Torque diagram for stepped or loaded shaft: constant torque T between couples.
Formulas
Formula
σ_max = (M y) / I (bending stress)
Meaning
σ = bending stress; M = bending moment; y = distance from neutral axis; I = area moment of inertia.
Watch Out
Bending stress is NORMAL (σ), not shear; must combine with shear from torsion using Mohr or principal stress formulas.
When To Use
Compute bending stress component before combining with torsion.
Formula
τ = (T c) / J (torsional shear stress)
Meaning
Shear stress from torque.
Watch Out
This is SHEAR (τ), perpendicular to bending normal stress (σ); they are on orthogonal planes at the same point.
When To Use
Compute torsional shear before combining with bending.
Formula
σ_1, σ_2 = (σ/2) ± √[(σ/2)² + τ²] (principal stresses)
Meaning
σ₁, σ₂ = max and min normal stress; σ = bending stress; τ = shear stress from torsion.
Watch Out
This formula assumes plane stress (σ_3 = 0). For a circular shaft, use σ = (M c)/I from bending.
When To Use
Find principal stresses when shaft carries both bending moment M and torque T.
Formula
T_equiv = √(M² + T²) [equivalent torque, sometimes used]
Meaning
An alternative combined-loading rule: treat combined loading as pure torsion with T_equiv.
Watch Out
Not universal; always use principal stress method unless specifically instructed otherwise.
When To Use
Some design codes use this; check the problem statement or code cited (AISC, NSCP 2015).
Section Title
Combined Loading (Bending + Torsion)
Important Facts
- A shaft under bending + torsion experiences combined normal and shear stress.
- Maximum principal stress σ₁ governs failure (ductile or brittle); compare to allowable.
- For a circular shaft under bending moment M and torque T, max bending stress is σ = Mc/I (outer fiber).
- If problem asks for 'safe diameter,' compute combined stresses and size against σ_1 ≤ σ_allow (usually 0.6 σ_yield for ductile).
Key Definitions
Term
Principal Stress
Example
At a point on a shaft: σ = 100 MPa (bending), τ = 60 MPa (torsion) → σ₁ ≈ 130 MPa, σ₂ ≈ −30 MPa.
Definition
The maximum and minimum normal stresses (σ₁, σ₂) at a point; shear on principal planes is zero.
Term
Plane Stress
Example
A shaft surface: normal stress σ from bending, shear τ from torsion, but σ perpendicular to surface (out of plane) = 0.
Definition
A stress state where one principal stress is zero (σ₃ = 0); typical for thin-walled members and surfaces.
Diagrams To Know
- Mohr's circle: horizontal axis = normal stress σ; vertical axis = shear stress τ. Plot (σ, τ) and (σ, −τ) points; circle diameter gives 2 × radius from center to principal stress.
Must Remember
- 1. TORSION FORMULA BASICS: τ_max = Tc/J = 16T/(π d³). Never confuse polar J (torsion) with area moment I (bending). For circular solid: J = π d⁴/32.
- 2. UNIT CONSISTENCY IS CRITICAL: Use N⋅mm with mm dimensions, or N⋅m with m. The d³ term amplifies unit errors by factor of 1000 or more. Check every calculation.
- 3. HOLLOW SHAFT FORMULA: J = π(D⁴ − d⁴)/32. Subtract FOURTH POWERS, not areas. This is the #1 computational mistake on boards.
- 4. ANGLE OF TWIST IN RADIANS: θ = TL/(JG) always gives radians. Convert to degrees at the END: θ_deg = θ_rad × (180/π). Forgetting the conversion is exam suicide.
- 5. POWER → TORQUE → DIAMETER (Three-Step Design): (1) P (kW) and N (rpm) → T = (P × 60)/(2π N); (2) T and τ_allow → d = ∛[16T/(π τ)]; (3) Round up to standard size. Most board shaft problems follow this exact pattern.
- 6. RPM FORMULA — THE CLASSIC TRAP: P = (2π N T)/60 for N in rpm. Forgetting the 60 makes your answer 60× too large. This is the MOST COMMON SLIP on torsion exams.
- 7. THIN-WALLED CLOSED TUBES: τ = T/(2 A_m t) where A_m is area enclosed by the CENTERLINE (median) of the wall, not the outer area. This formula is fast and accurate for closed hollow sections.
- 8. SHEAR FLOW IN CLOSED TUBES: q = τ t = T/(2 A_m) is CONSTANT around the perimeter. If thickness varies, stress varies inversely; thinner sections have higher stress.
- 9. FLANGED BOLT COUPLINGS: T = A τ R n for single bolt circle. Each bolt shears with force P = A τ at radius R. Two circles require compatibility: P₁/R₁ = P₂/R₂. Bolt shear allowable is usually LOWER than shaft allowable.
- 10. COMBINED BENDING + TORSION: Always compute principal stresses first: σ₁,₂ = (σ/2) ± √[(σ/2)² + τ²]. Compare σ₁ to allowable (not the individual σ or τ alone). Mohr's circle is your friend here.
Last Minute Tips
- TIP 1 — THE POWER TRAP: 90% of board shaft-sizing questions start with power in kW and speed in rpm. Your immediate next step must be T = (P × 60)/(2π N). If the problem does not give power and rpm together, re-read for hidden info (e.g., 'motor delivers 50 kW at 1200 rpm').
- TIP 2 — ALWAYS ROUND UP SHAFT DIAMETER: Calculated d = 37.2 mm? Use 40 mm or 38 mm (next standard size up). Never round down; doing so will fail safety checks and costs you the full question. Add 1–2 mm as a safety margin if in doubt.
- TIP 3 — HOLLOW vs SOLID SPEED: A hollow shaft with outer diameter D and inner d is lighter and just as strong as a solid shaft of diameter ≈ 0.9 D. If the problem asks 'compare weight of solid and hollow carrying same torque,' hollow wins every time. Know this for short-answer/conceptual questions.
- TIP 4 — UNIT CONSISTENCY CHECKPOINT: Before you compute, write out one unit-check: e.g., 'T = 1200 N⋅m = 1.2 × 10⁶ N⋅mm; d in mm → d³ in mm³; stress in MPa.' This 30-second habit prevents 80% of arithmetic disasters.
- TIP 5 — ANGLE OF TWIST IN DEGREES: If asked 'Find the twist angle in degrees,' compute θ in radians from θ = TL/(JG), then multiply by (180/π) ≈ 57.3. Leaving the answer in radians costs full points on the problem. Conversely, if asked in radians and you give degrees, you also lose full credit.
Comparison Tables
Rows
Values
- J = π d⁴ / 32
- J = π (D⁴ − d⁴) / 32
Property
Polar Moment Formula
Values
- Higher (smaller J)
- Lower (larger J for same OD)
Property
Max Stress at Same T
Values
- Heavier (solid material throughout)
- Lighter (only wall, hollow center)
Property
Weight
Values
- Poor (core material contributes little to J)
- Excellent (most material far from center)
Property
Material Efficiency
Values
- Small shafts, low torque
- Drive shafts, pump shafts, propeller shafts
Property
Practical Use
Columns
- Property
- Solid Shaft
- Hollow Shaft
Table Title
Solid vs Hollow Shaft Comparison
Rows
Values
- Circular cross-sections (solid or hollow ring)
- Closed hollow sections (box, rectangular, polygonal, any thin wall)
Property
Applies to
Values
- One formula; J must be computed
- Direct formula; A_m is area enclosed by centerline
Property
Formula Simplicity
Values
- Cross-section remains plane; stress linear in radius
- Wall is thin (t ≪ dimensions); shear flow q = constant around perimeter
Property
Key Assumption
Values
- Linear: zero at center, max at outer edge
- Uniform (or varies inversely with thickness if thickness varies)
Property
Stress Distribution
Values
- Very common; classic question
- Common for hollow boxes and structural tubes
Property
Board Exam Frequency
Columns
- Aspect
- Solid/Hollow Circular (τ = Tc/J)
- Thin-Walled Closed (τ = T/2A_m t)
Table Title
Torsion Formula (Circular) vs Thin-Walled Closed Tube
Rows
Values
- T (N⋅m)
- T = P / ω
Property
P (W) and ω (rad/s)
Values
- T (N⋅m)
- T = (P × 60) / (2π N)
Property
P (W or kW) and N (rpm)
Values
- P (W)
- P = T × ω
Property
T (N⋅m) and ω (rad/s)
Values
- P (W)
- P = (2π N × T) / 60
Property
T (N⋅m) and N (rpm)
Columns
- Given
- Find
- Formula
Table Title
Power ↔ Torque Conversion Formulas
Rows
Values
- θ = TL / (JG)
- One material, one diameter, one torque
- Radians; convert to degrees if asked
Property
Single Segment
Values
- θ_total = Σ (T_i L_i) / (J_i G_i)
- Stepped shaft or variable torque
- Radians (sum all radians)
Property
Multiple Segments
Values
- γ_max = c θ / L → θ = γ_max × L / c
- If given strain; work backwards to angle
- Radians if γ is consistent
Property
From Shear Strain
Columns
- Form
- Formula
- When to Use
- Units for Result
Table Title
Angle of Twist: Three Common Forms
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