Torsion - Revision Notes
Torsion is the twisting of a structural or mechanical member caused by a moment (torque) applied about its longitudinal axis. In the PRC Civil Engineer Licensure Examination, torsion problems appear consistently in the Strength of Materials section and typically test three core competencies: (1) computing maximum shear stress in circular shafts, (2) finding the angle of twist, and (3) solving power-transmission problems involving rotating shafts. Additional board items cover flanged bolt couplings and thin-walled closed tubes. Mastery of the torsion formula, careful unit management (N·mm vs N·m), and the power-to-torque conversion are the keys to scoring full marks on this topic. This chapter also provides the foundation for combined loading problems where torsion interacts with bending or axial forces.
1. Fundamentals of Torsion in Circular Shafts
Formulas
τ = Tρ / J
Example
Solid shaft d = 60 mm, T = 2 kN·m: ρ = 30 mm, J = π(60)⁴/32 = 1.272×10⁶ mm⁴, τ = (2×10⁶)(30)/(1.272×10⁶) = 47.2 MPa
Variables
τ = shear stress at radius ρ (MPa); T = internal torque (N·mm); ρ = radial distance from centroid (mm); J = polar moment of inertia (mm⁴)
Application
Finds the shear stress at ANY radius ρ within the cross-section. Use ρ = c (outer radius) for the maximum.
τ_max = Tc / J = 16T / (πd³)
Example
Solid shaft d = 50 mm, T = 1.2 kN·m = 1.2×10⁶ N·mm: τ_max = 16(1.2×10⁶)/[π(50)³] = 1.92×10⁷/3.927×10⁵ = 48.9 MPa
Variables
c = outer radius = d/2 (mm); d = outer diameter (mm); T = torque (N·mm); J = polar moment of inertia (mm⁴)
Application
Direct formula for maximum shear stress in a SOLID circular shaft. Memorize 16T/(πd³) for quick board-exam use.
J_solid = πd⁴/32 = πc⁴/2
Example
d = 80 mm: J = π(80)⁴/32 = π(40,960,000)/32 = 4.021×10⁶ mm⁴
Variables
d = diameter (mm); c = radius = d/2 (mm)
Application
Polar moment of inertia for a solid circular shaft. Choose the diameter form (πd⁴/32) to avoid confusion with radius vs diameter.
J_hollow = π(D⁴ − d⁴)/32
Example
D = 80 mm, d = 60 mm: J = π(80⁴ − 60⁴)/32 = π(40,960,000 − 12,960,000)/32 = π(28,000,000)/32 = 2.749×10⁶ mm⁴
Variables
D = outer diameter (mm); d = inner diameter (mm)
Application
Polar moment of inertia for a hollow circular shaft. Subtract fourth powers of diameters — NOT areas.
τ_max (hollow) = TD/(2J) = 16TD/[π(D⁴−d⁴)]
Example
D = 80 mm, d = 60 mm, T = 3 kN·m = 3×10⁶ N·mm: τ_max = 16(3×10⁶)(80)/[π(80⁴−60⁴)] = 3.84×10⁸/8.796×10⁷ = 43.7 MPa
Variables
D = outer diameter (mm); d = inner diameter (mm); T = torque (N·mm)
Application
Maximum shear stress in a hollow circular shaft; stress occurs at the outer surface.
Exam tips
- Memorize τ_max = 16T/(πd³) for solid shafts — it combines Tc/J into one direct formula saving time.
- For hollow shafts, derive τ_max step-by-step: compute J, then τ = T·(D/2)/J to avoid formula confusion.
- When the problem gives diameter in mm and torque in kN·m, convert torque to N·mm first: 1 kN·m = 10⁶ N·mm.
- Ratio problems (hollow vs solid, same τ_max or same torque) are common — set up the ratio of J/c for each section.
Key points
- Torsion is twisting due to a torque T applied along the longitudinal axis of a member.
- The torsion formula is valid ONLY for circular cross-sections (solid or hollow) of linearly elastic, homogeneous material.
- Shear stress τ varies linearly from zero at the centroidal axis to a maximum τ_max at the outer surface.
- The cross-section of a circular shaft remains plane (does not warp) — this is a critical assumption.
- Non-circular sections DO warp under torsion and require Saint-Venant's warping theory, not the standard formula.
- The polar moment of inertia J is the geometric property that governs torsional resistance.
- A hollow shaft is more material-efficient than a solid shaft for the same torque and allowable stress because the low-stressed core material is removed.
Definitions
Torque (T)
A moment applied about the longitudinal axis of a member, causing it to twist. Units: N·m or N·mm (use one system consistently throughout a problem).
Importance
Every torsion problem begins by identifying or computing T. In power problems, T is derived from P and N.
Polar Moment of Inertia (J)
The second moment of area of a cross-section about its centroidal axis perpendicular to the plane of the section. J = ∫ρ² dA. Units: mm⁴.
Importance
J appears in both the torsion formula and the angle-of-twist formula. A larger J means lower shear stress and less twist for the same torque.
Torsional Rigidity (JG)
The product of the polar moment of inertia J and the shear modulus G. Units: N·mm² (or N·m²). It measures a shaft's resistance to twisting.
Importance
A high torsional rigidity JG means a small angle of twist. Used directly in θ = TL/(JG).
Shear Modulus (G)
Also called the modulus of rigidity. Relates shear stress to shear strain: G = τ/γ. For steel, G ≈ 80 GPa; for aluminum, G ≈ 28 GPa.
Importance
Required for all angle-of-twist calculations. G is always given or must be recalled for common materials.
Common mistakes
- Using DIAMETER instead of RADIUS for c in τ = Tc/J — always c = d/2.
- Subtracting areas instead of fourth-power diameters in J_hollow = π(D⁴−d⁴)/32.
- Mixing units: computing J in mm⁴ but T in N·m — either convert T to N·mm or J to m⁴ (m⁴ is awkward; use mm system).
- Applying the torsion formula τ = Tc/J to non-circular (rectangular, I-section) members — it does NOT apply there.
- Forgetting that shear stress at the centroidal axis is ZERO, not maximum.
2. Angle of Twist
Formulas
θ = TL / (JG)
Example
Solid shaft d = 40 mm, L = 2 m = 2000 mm, T = 800 N·m = 800,000 N·mm, G = 80 GPa = 80,000 MPa: J = π(40)⁴/32 = 251,327 mm⁴; θ = (800,000)(2000)/[(251,327)(80,000)] = 1.6×10⁹/2.011×10¹⁰ = 0.0796 rad = 4.56°
Variables
θ = angle of twist (radians); T = torque (N·mm); L = length (mm); J = polar moment of inertia (mm⁴); G = shear modulus (MPa = N/mm²)
Application
Single uniform shaft segment under constant torque. Result is in radians automatically if T in N·mm, L in mm, J in mm⁴, G in N/mm² (MPa).
θ_total = Σ (Tᵢ Lᵢ) / (Jᵢ Gᵢ)
Example
Two-segment shaft: Segment AB — T = 600 N·m, L = 1 m, d = 30 mm, G = 80 GPa; Segment BC — T = 400 N·m, L = 1.5 m, d = 40 mm, G = 80 GPa. Compute J_AB and J_BC separately, then sum θ_AB + θ_BC.
Variables
Subscript i denotes segment number. Each segment may have different T, L, J (diameter), or G (material).
Application
Stepped shafts or shafts with torques applied at intermediate points. Draw the torque diagram first to find Tᵢ in each segment.
γ_max = τ_max / G = c·θ / L
Example
c = 25 mm, θ = 0.0796 rad, L = 2000 mm: γ = 25(0.0796)/2000 = 9.95×10⁻⁴ (dimensionless)
Variables
γ_max = maximum shear strain (dimensionless); c = outer radius (mm); θ = angle of twist (radians); L = length (mm)
Application
Links shear stress, shear strain, and angle of twist. Useful when the problem asks for surface shear strain or as a check.
Exam tips
- Unit consistency rule: if T in N·mm, L in mm, J in mm⁴, G in N/mm², then θ is automatically in radians — no conversion factors needed.
- For a solid shaft at constant G: θ = 32TL/(πd⁴G). Derive this directly from θ = TL/(JG) by substituting J = πd⁴/32 to verify.
- A shaft twisted 'to failure' means τ_max has reached the shear yield strength — convert: τ_y ≈ 0.577 F_y (von Mises) or 0.5 F_y (Tresca).
- Statically indeterminate torsion (shaft fixed at both ends with an intermediate torque) follows the same approach as axially indeterminate bars: ΣT = 0 (equilibrium) + θ = 0 (compatibility).
Key points
- The angle of twist θ measures the relative angular rotation between two cross-sections separated by length L.
- θ = TL/(JG) gives the result in RADIANS; convert to degrees by multiplying by (180/π) only when the problem asks for degrees.
- For a stepped shaft (different diameters or materials along its length), apply θ = Σ(TᵢLᵢ)/(JᵢGᵢ) by summing contributions of each segment.
- For a shaft with continuously varying torque, integration is required, but board-exam problems almost always involve piecewise constant segments.
- The angle of twist is directly proportional to torque T and length L, and inversely proportional to torsional rigidity JG.
- A shaft that satisfies stress requirements may still fail the stiffness (twist) requirement — always check both.
Definitions
Angle of Twist (θ)
The angular rotation of one end of a shaft relative to the other end when subjected to a torque. Measured in radians (or degrees after conversion). Positive direction follows the right-hand rule with the torque vector.
Importance
Service-limit-state criterion in shaft design alongside the stress check. Some specifications limit twist to 1°/m or similar.
Torsional Stiffness (k_T)
k_T = JG/L (N·mm/rad). The torque required to produce unit angle of twist. Analogous to axial stiffness k = AE/L.
Importance
Used in torsional spring analogy and in statically indeterminate torsion problems where torque distribution requires compatibility equations.
Common mistakes
- Leaving θ in radians when the problem asks for DEGREES — always check the required unit.
- Using L in meters but J and G in mm-based units, causing a factor-of-10³ error in θ.
- For stepped shafts, using the SAME J for all segments — must compute J for each diameter.
- Forgetting to draw the internal torque diagram before applying Σ(TᵢLᵢ/JᵢGᵢ); the torque in each segment is found by equilibrium from either end.
- Using G = E (Young's modulus) instead of G (shear modulus) — common slip especially under exam pressure.
3. Power Transmission by Rotating Shafts
Formulas
P = Tω = T(2πN/60)
Example
P = 75 kW = 75,000 W, N = 1200 rpm: T = (75,000 × 60)/(2π × 1200) = 4,500,000/7539.8 = 596.8 N·m
Variables
P = power (W = N·m/s); T = torque (N·m); ω = angular velocity (rad/s); N = rotational speed (rpm)
Application
Fundamental power-torque-speed relationship for a rotating shaft. Rearrange to T = P×60/(2πN) to find torque from power and rpm.
T = 60P / (2πN)
Example
P = 50 kW = 50,000 W, N = 600 rpm: T = 60(50,000)/[2π(600)] = 3,000,000/3769.9 = 795.8 N·m
Variables
T = torque (N·m); P = power (W); N = speed (rpm)
Application
Direct rearrangement of the power equation to isolate torque. This is the form used in every power-transmission board problem.
P = 2πfT
Example
f = 20 rev/s, T = 200 N·m: P = 2π(20)(200) = 25,133 W ≈ 25.1 kW
Variables
P = power (W); f = frequency (rev/s = Hz); T = torque (N·m)
Application
Use when speed is given in rev/s rather than rpm. Equivalent to P = Tω since ω = 2πf.
d = [16T / (π τ_allow)]^(1/3)
Example
T = 596.8 N·m = 596,800 N·mm, τ_allow = 60 MPa: d = [16(596,800)/(π×60)]^(1/3) = [50,650]^(1/3) = 37.0 mm → round up to next standard size
Variables
d = required shaft diameter (mm); T = torque (N·mm); τ_allow = allowable shear stress (MPa)
Application
Design equation for minimum solid shaft diameter given torque and allowable shear stress. Derived from τ_max = 16T/(πd³) solved for d.
Exam tips
- Write out T = 60P/(2πN) explicitly with numbers before evaluating — this prevents the common rpm/rps error.
- If power is in kW and you want T in N·mm directly: T = (60 × P_kW × 10⁶)/(2πN) — the 10⁶ converts kN·m to N·mm.
- The formula d = [16T/(πτ)]^(1/3) requires a CUBE ROOT — use the x^(1/3) key or trial-and-check for board efficiency.
- After computing d, always round UP to the next standard or whole-number size to ensure the design is safe (conservative).
Key points
- The most common board pattern is: given power P and rotational speed N (rpm), find T, then find d or τ.
- Power P = Tω = 2πfT where f is frequency in rev/s (Hz). If speed is in rpm, use P = 2πNT/60.
- Always identify the unit of power: 1 kW = 1000 W = 1000 N·m/s; 1 horsepower (hp) = 745.7 W.
- After finding T from P and N, proceed with the torsion formula exactly as in any other torsion problem.
- The step sequence is always: P and N → T → stress (or diameter). Never skip the intermediate torque calculation.
Definitions
Power (P)
The rate of doing work or transferring energy. In shaft problems: P = Tω. SI unit is the watt (W = N·m/s = J/s). For large values, use kW (kilowatts).
Importance
The starting point for all power-transmission shaft design problems. Must be converted to watts (W) before use in SI formulas.
Rotational Speed (N)
Number of complete revolutions per minute (rpm). Convert to angular velocity: ω = 2πN/60 (rad/s), or to frequency: f = N/60 (rev/s).
Importance
Board problems almost always give speed in rpm. The factor 60 in the denominator is the most frequently forgotten element.
Common mistakes
- Forgetting to divide by 60 when N is in rpm: using P = 2πNT instead of P = 2πNT/60.
- Using kilowatts without converting to watts: T = P/ω with P in kW gives T in kN·m — check your units.
- Not converting T from N·m to N·mm before applying the torsion formula with dimensions in mm.
- Using horsepower directly in SI formulas — always convert: 1 hp = 745.7 W.
- Solving for d using the formula for stress (τ = 16T/πd³) but then not cubing the result properly when solving for d — take the cube root, not the square root.
4. Flanged Bolt Couplings
Formulas
T = n × A × τ × R
Example
6 bolts, d_bolt = 20 mm, R = 150 mm, τ = 70 MPa: A = π(20)²/4 = 314.16 mm²; P = 314.16×70 = 21,991 N; T = 6×21,991×150 = 19.79×10⁶ N·mm = 19.8 kN·m
Variables
T = torque capacity (N·mm); n = number of bolts; A = cross-sectional area of one bolt (mm²) = π d_bolt²/4; τ = bolt shear stress (MPa); R = bolt-circle radius (mm)
Application
Single bolt circle with n identical bolts. Assumes all bolts are at the same radius and carry equal shear.
P₁/R₁ = P₂/R₂ (compatibility for two bolt circles)
Example
R₁ = 100 mm, R₂ = 150 mm: P₁/100 = P₂/150 → P₂ = 1.5 P₁. Then T = n₁P₁R₁ + n₂P₂R₂ = n₁P₁(100) + n₂(1.5P₁)(150).
Variables
P₁, P₂ = shear force per bolt in ring 1 and ring 2 (N); R₁, R₂ = bolt-circle radii (mm). Assumes identical bolt areas and same G.
Application
Compatibility condition when bolts are on two concentric circles. The bolt on the larger radius deforms more, hence carries more force.
T = n₁P₁R₁ + n₂P₂R₂
Example
n₁ = 4 bolts at R₁ = 100 mm, n₂ = 6 bolts at R₂ = 150 mm. If P₂ = 1.5P₁ and τ_allow = 80 MPa for 16 mm bolts (A = 201 mm²), then P₂_max = 201×80 = 16,080 N (R₂ governs). P₁ = 16,080/1.5 = 10,720 N. T = 4(10,720)(100) + 6(16,080)(150) = 4,288,000 + 14,472,000 = 18.76×10⁶ N·mm = 18.76 kN·m
Variables
n₁, n₂ = number of bolts in each ring; P₁, P₂ = shear force per bolt (N) from compatibility; R₁, R₂ = bolt-circle radii (mm)
Application
Total torque for a two-ring bolt coupling after determining P₁ and P₂ from the compatibility condition.
Exam tips
- For a single bolt circle: T = n(πd²/4)(τ)(R). Write this out in one line to avoid intermediate errors.
- For two bolt circles: identify the larger R, set P_outer = A×τ_allow (it governs), then back-calculate P_inner using P₁/R₁ = P₂/R₂.
- Board problems often ask for the torque capacity — give the answer in kN·m (divide N·mm result by 10⁶).
- Quick check: if the problem gives shaft diameter and bolt data, it may be asking for the shaft τ or θ after finding T from the coupling — read the question carefully.
Key points
- A flanged bolt coupling transfers torque from one shaft to another through bolts arranged in a circular pattern (bolt circle).
- Each bolt resists shear. The shear force in each bolt is P = Aτ, where A is the bolt's cross-sectional area and τ is the bolt shear stress.
- For n identical bolts on a single bolt circle of radius R, the total torque capacity is T = nARτ.
- When bolts are on TWO concentric bolt circles (two rings of bolts), compatibility requires that shear force is proportional to bolt-circle radius: P₁/R₁ = P₂/R₂ (assuming identical bolt areas and material).
- The total torque with two bolt circles: T = n₁P₁R₁ + n₂P₂R₂ (resolve using the proportionality condition to find P₁ and P₂).
- Design procedure: identify the critical bolt (highest stress), set its shear stress equal to τ_allow, find forces, then compute T.
Definitions
Bolt Circle Radius (R)
The distance from the shaft centerline to the centerline of the bolts in the coupling. The moment arm for each bolt's shear force.
Importance
Determines the torque capacity directly — doubling R doubles the torque capacity for the same bolt shear capacity.
Shear Flow Compatibility (Two-Ring Couplings)
In a two-ring coupling, the flanges are rigid, so both rings rotate the same angle. This means bolt deformation is proportional to R, and thus bolt force is proportional to R (since F = kδ and δ is proportional to R).
Importance
Without applying this compatibility condition, the force distribution is indeterminate. The condition P/R = constant resolves the indeterminacy.
Common mistakes
- Using bolt DIAMETER instead of bolt AREA — always compute A = πd²/4 for each bolt.
- Forgetting the number of bolts n in T = nARτ — a pure arithmetic oversight under exam pressure.
- For two-ring couplings, assuming EQUAL force in all bolts regardless of radius — the outer ring bolts carry MORE shear.
- Checking stress in the wrong bolt circle — check the OUTER ring (larger R, larger force) as the critical one.
- Using the bolt-circle radius R in mm but torque in kN·m — keep everything in N·mm for consistency.
5. Torsion of Thin-Walled Closed Tubes
Formulas
τ = T / (2 A_m t)
Example
Rectangular closed tube: outer 100×60 mm, wall t = 5 mm. A_m = (100−5)(60−5) = 95×55 = 5,225 mm². T = 4 kN·m = 4×10⁶ N·mm. τ = 4×10⁶/(2×5,225×5) = 4×10⁶/52,250 = 76.6 MPa
Variables
τ = shear stress in the wall (MPa); T = torque (N·mm); A_m = area enclosed by the median wall (mm²); t = wall thickness at the point considered (mm)
Application
Shear stress in a thin-walled CLOSED tube of any cross-sectional shape. For variable t, compute τ at the thinnest wall.
q = τ × t = T / (2 A_m)
Example
T = 4×10⁶ N·mm, A_m = 5,225 mm²: q = 4×10⁶/(2×5,225) = 382.8 N/mm. At t = 5 mm: τ = 382.8/5 = 76.6 MPa
Variables
q = shear flow (N/mm); τ = shear stress (MPa); t = wall thickness (mm); T = torque (N·mm); A_m = enclosed median area (mm²)
Application
Shear flow is constant around the closed cross-section. Useful when comparing stresses at different wall thicknesses.
θ = TL∮(ds/t) / (4 A_m² G)
Example
Rectangular tube (100−5)×(60−5) mm, t = 5 mm, L = 1 m = 1000 mm, T = 4×10⁶ N·mm, G = 80,000 MPa: ∮(ds/t) = [2(95+55)]/5 = 300/5 = 60; θ = (4×10⁶)(1000)(60)/[4(5225)²(80,000)] = 2.4×10¹¹/8.734×10¹³ = 0.00275 rad = 0.157°
Variables
θ = angle of twist (rad); ∮(ds/t) = line integral around median perimeter (mm/mm = dimensionless); L = length (mm); A_m = enclosed area (mm²); G = shear modulus (MPa)
Application
Angle of twist for a thin-walled closed tube. For constant t: ∮(ds/t) = Perimeter/t.
Exam tips
- The phrase 'thin-walled tube' or 'box section under torsion' in a board problem is the trigger to use τ = T/(2A_m t).
- A_m is ALWAYS the median enclosed area — for a circular tube of mean radius r_m: A_m = πr_m². For a square tube of mean side a: A_m = a².
- Shear flow q = T/(2A_m) is a convenient intermediate result — divide by t at any location to get local shear stress.
- If two walls have different thicknesses, maximum τ is at minimum t — state this explicitly in your solution.
Key points
- The thin-walled tube (Bredt-Batho) formula applies to ANY closed cross-sectional shape (box, circular tube, elliptical tube) provided the wall is thin relative to overall dimensions.
- The shear flow q = τt is CONSTANT around the entire perimeter of a thin-walled closed section under pure torsion.
- τ = T / (2 A_m t) where A_m is the area enclosed by the median (midthickness) wall centerline — NOT the gross area.
- For a variable-thickness closed section, the maximum shear stress occurs at the point of MINIMUM wall thickness.
- For a circular thin-walled tube of mean radius r and constant thickness t: A_m = πr², giving τ = T/(2πr²t) — this matches the exact thick-wall result as t→0.
- Open thin-walled sections (channel, angle, I-beam) have MUCH lower torsional stiffness than closed sections and warp significantly — the Bredt formula does NOT apply to open sections.
Definitions
Shear Flow (q)
The internal shear force per unit length along the wall of a thin-walled section: q = τt (N/mm). For a closed section under pure torsion, q is constant around the entire perimeter.
Importance
The constancy of shear flow is the key insight of thin-walled tube theory (Bredt's theorem). It allows τ to be found at any wall simply as q/t.
Enclosed Median Area (A_m)
The area bounded by the centerline (mid-thickness line) of the tube wall. For a rectangular tube of outer dimensions B×H and wall thickness t: A_m = (B−t)(H−t).
Importance
A_m appears squared in the angle-of-twist formula, so an error in computing it is amplified. Use mid-wall dimensions, not outer dimensions.
Common mistakes
- Using the OUTER cross-section area instead of the median-wall enclosed area A_m — leads to an overestimated torsional capacity.
- Applying τ = T/(2A_m t) to OPEN sections (channels, angles) — the formula is valid ONLY for closed tubes.
- For variable-thickness walls, computing τ at the THICKEST wall (lowest τ) instead of the critical THINNEST wall.
- Forgetting the factor of 2 in the denominator: τ = T/(2A_m t), not T/(A_m t).
6. Combined Torsion and Bending (Equivalent Torque and Moment)
Formulas
σ = Mc/I = 32M/(πd³) and τ = Tc/J = 16T/(πd³)
Example
d = 60 mm, M = 1.5 kN·m = 1.5×10⁶ N·mm, T = 2 kN·m = 2×10⁶ N·mm: σ = 32(1.5×10⁶)/[π(60)³] = 4.8×10⁷/678,584 = 70.7 MPa; τ = 16(2×10⁶)/[π(60)³] = 3.2×10⁷/678,584 = 47.2 MPa
Variables
σ = bending stress (MPa); M = bending moment (N·mm); τ = torsional shear stress (MPa); T = torque (N·mm); c = d/2; I = πd⁴/64; J = πd⁴/32
Application
Individual bending and torsion stresses at the outer fiber of a circular shaft. Note J = 2I for circular sections.
T_e = √(M² + T²)
Example
M = 1.5 kN·m, T = 2 kN·m: T_e = √(1.5² + 2²) = √(2.25 + 4) = √6.25 = 2.5 kN·m. τ_max = 16(2.5×10⁶)/[π(60)³] = 59.0 MPa
Variables
T_e = equivalent torque (N·mm); M = bending moment (N·mm); T = torque (N·mm)
Application
Use with τ_max = 16T_e/(πd³) to find the maximum shear stress (or design d) under combined loading. Based on maximum shear stress theory (Tresca).
M_e = ½[M + √(M² + T²)] = ½(M + T_e)
Example
M = 1.5 kN·m, T_e = 2.5 kN·m: M_e = ½(1.5 + 2.5) = 2.0 kN·m. σ_max = 32(2.0×10⁶)/[π(60)³] = 94.3 MPa
Variables
M_e = equivalent moment (N·mm); M = bending moment (N·mm); T_e = equivalent torque (N·mm)
Application
Use with σ_max = 32M_e/(πd³) to find the maximum normal stress under combined loading. Based on maximum normal stress theory.
Exam tips
- In combined loading problems: if asked for 'maximum shear stress,' compute T_e = √(M² + T²) and use τ_max = 16T_e/(πd³).
- If asked for 'maximum normal stress,' compute M_e = ½(M + T_e) and use σ_max = 32M_e/(πd³).
- Both M and T must be in the same units (both N·mm or both N·m) before computing T_e and M_e.
- Some board problems give a shaft with a horizontal force (creating both shear and bending) plus a torque — identify M at the critical section by statics before applying equivalent formulas.
Key points
- Shafts in practice carry BOTH bending moments M and torques T simultaneously (e.g., a shaft with a gear or pulley at mid-span).
- The critical point on the cross-section has both a bending stress σ = Mc/I and a torsional shear stress τ = Tc/J acting simultaneously.
- Principal stress analysis (or Mohr's circle) is applied at the critical point to find the maximum normal stress σ₁ and maximum shear stress τ_max.
- Equivalent Torque T_e = √(M² + T²) — gives the torque that would produce the same MAXIMUM SHEAR STRESS as the combined loading.
- Equivalent Moment M_e = ½[M + √(M² + T²)] — gives the moment that would produce the same MAXIMUM NORMAL STRESS as the combined loading.
- These equivalent values are used with the standard torsion and bending formulas to design or check circular shafts under combined loading.
Definitions
Equivalent Torque (T_e)
The single torque that, acting alone, would produce the same maximum shear stress at the surface of a circular shaft as the actual combination of bending moment M and torque T. T_e = √(M² + T²).
Importance
Provides a one-formula shortcut for checking shear stress under combined loading without going through the full Mohr's circle construction.
Equivalent Moment (M_e)
The single bending moment that, acting alone, would produce the same maximum normal stress at the surface as the combined M and T. M_e = ½[M + √(M² + T²)].
Importance
Provides a one-formula shortcut for checking normal (bending) stress under combined loading. Both T_e and M_e should be checked in design.
Common mistakes
- Using T_e for normal stress check — T_e applies to maximum SHEAR stress; use M_e for maximum NORMAL stress.
- Forgetting the ½ factor in M_e = ½(M + T_e).
- Applying equivalent torque/moment to non-circular shafts — these formulas are derived assuming circular cross-sections.
- Not identifying the CRITICAL POINT on the cross-section — combined bending and torsion is worst at the surface point where bending stress is maximum.
Connections
- STRESS TRANSFORMATION (Chapter 6): Shear stresses from torsion appear as τ on the stress element. Combined with bending stress σ, Mohr's circle gives principal stresses and maximum shear stress — directly uses the equivalent torque/moment formulas.
- SHEAR STRESS IN BEAMS (Chapter 5): Both beam shear and torsion produce shear stress τ on cross-sections. Beam shear varies parabolically (max at neutral axis); torsional shear varies linearly (max at surface). They can be superimposed at a point but act on different faces.
- STATICALLY INDETERMINATE MEMBERS (Chapter 4): A shaft fixed at both ends with an intermediate applied torque is statically indeterminate — solved by the same equilibrium + compatibility (force method) approach used for axially indeterminate bars, with TL/(JG) replacing PL/(AE).
- DEFLECTION AND STIFFNESS (Chapter 9): Torsional stiffness k_T = JG/L parallels axial stiffness k = AE/L. Both are used in matrix stiffness methods and spring-analogy problems for deformation calculations.
- MATERIAL PROPERTIES (Chapter 1-2): G = E/[2(1+ν)] connects shear modulus to Young's modulus and Poisson's ratio. For steel: E = 200 GPa, ν = 0.3, so G = 200/[2(1.3)] = 76.9 ≈ 80 GPa. Understanding this eliminates having to memorize G separately.
- THIN-WALLED PRESSURE VESSELS (Chapter 8): Both thin-walled tubes and pressure vessels use the concept of membrane (in-plane) stress — shear flow in torsion parallels hoop-stress uniformity in cylindrical pressure vessels. The median-area concept reappears in both topics.
- POWER AND ENERGY (Engineering Sciences): P = Tω is the rotational analog of P = Fv (linear power). This connection to physics reinforces the formula and aids in dimensional analysis during board exams.
- NSCP 2015 / AISC 360 APPLICATION: In building frames, beams subjected to torsion are designed per AISC 360 Chapter H (Combined Forces) which accounts for warping torsion in open sections (W-shapes). Circular hollow sections (CHS) follow the pure St. Venant torsion covered in this chapter. Understanding the distinction is important for professional practice.
Quick review questions
A solid circular shaft 70 mm in diameter carries a torque of 2.5 kN·m. What is the maximum shear stress?
Answer
τ_max = 16T/(πd³) = 16(2.5×10⁶)/[π(70)³] = 4.0×10⁷/1,077,566 = 37.1 MPa
Convert T to N·mm: 2.5 kN·m = 2.5×10⁶ N·mm. Apply the compact formula τ_max = 16T/(πd³) directly. Note d³ = 70³ = 343,000 mm³ and π×343,000 = 1,077,566 mm³.
A hollow shaft has outer diameter D = 100 mm and inner diameter d = 70 mm. Torque = 5 kN·m. Find τ_max.
Answer
J = π(100⁴ − 70⁴)/32 = π(100,000,000 − 24,010,000)/32 = π(75,990,000)/32 = 7,460,390 mm⁴. τ_max = T·(D/2)/J = (5×10⁶)(50)/7,460,390 = 33.5 MPa
Step 1: Compute J using hollow formula — subtract FOURTH POWERS of diameters. Step 2: The maximum stress is at the outer surface, so use c = D/2 = 50 mm. Step 3: τ = Tc/J.
A shaft transmits 90 kW at 1500 rpm. What is the torque transmitted?
Answer
T = 60P/(2πN) = 60(90,000)/[2π(1500)] = 5,400,000/9,424.8 = 573.0 N·m
Convert P to watts: 90 kW = 90,000 W. Apply T = 60P/(2πN). The factor 60 converts rpm to rev/s; 2π converts revolutions to radians. Always write out the formula with units to avoid the common factor-of-60 error.
A 50 mm diameter solid shaft 3 m long is made of steel (G = 80 GPa) and carries T = 1.5 kN·m. Find the angle of twist in degrees.
Answer
J = π(50)⁴/32 = 613,592 mm⁴. θ = TL/(JG) = (1.5×10⁶)(3000)/[(613,592)(80,000)] = 4.5×10⁹/4.909×10¹⁰ = 0.09167 rad × (180/π) = 5.25°
Consistent units: T in N·mm, L in mm, J in mm⁴, G in N/mm² (MPa). Result is in radians; convert to degrees by multiplying by 180/π. G = 80 GPa = 80,000 MPa = 80,000 N/mm².
A flanged coupling has 8 bolts of 18 mm diameter on a 140 mm radius bolt circle. Allowable bolt shear stress = 65 MPa. Find the torque capacity.
Answer
A = π(18)²/4 = 254.47 mm². P = Aτ = 254.47×65 = 16,540 N. T = nPR = 8×16,540×140 = 18.52×10⁶ N·mm = 18.52 kN·m
Step 1: Area of one bolt. Step 2: Shear force per bolt = A×τ_allow. Step 3: Total torque = n × Force × Radius. All n bolts carry equal force since they are at the same radius.
A square thin-walled tube has mean side dimension 80 mm and wall thickness 4 mm. Find τ if T = 1.8 kN·m.
Answer
A_m = 80² = 6,400 mm². τ = T/(2A_m t) = (1.8×10⁶)/(2×6,400×4) = 1.8×10⁶/51,200 = 35.2 MPa
Key: A_m is the ENCLOSED MEDIAN AREA. Since mean side is already given as 80 mm, A_m = 80² = 6,400 mm². Apply τ = T/(2A_m t) directly. Remember the factor of 2.
A shaft carries bending moment M = 3 kN·m and torque T = 4 kN·m. Find the equivalent torque and the maximum shear stress if d = 80 mm.
Answer
T_e = √(M² + T²) = √(3² + 4²) = √(9+16) = √25 = 5 kN·m = 5×10⁶ N·mm. τ_max = 16T_e/(πd³) = 16(5×10⁶)/[π(80)³] = 8×10⁷/1,608,495 = 49.7 MPa
For combined bending and torsion: T_e = √(M² + T²) applies the Pythagorean theorem to moments. Then use T_e exactly like a pure torque in the standard torsion formula for maximum shear stress.
Two segments: AB has d = 30 mm, L = 500 mm, T = 200 N·m; BC has d = 50 mm, L = 800 mm, T = 200 N·m. Both steel, G = 80 GPa. Find total angle of twist.
Answer
J_AB = π(30)⁴/32 = 79,522 mm⁴. J_BC = π(50)⁴/32 = 613,592 mm⁴. θ_AB = (200×10³)(500)/[(79,522)(80,000)] = 10⁸/6.362×10⁹ = 0.01572 rad. θ_BC = (200×10³)(800)/[(613,592)(80,000)] = 1.6×10⁸/4.909×10¹⁰ = 0.00326 rad. θ_total = 0.01572 + 0.00326 = 0.01898 rad = 1.087°
For stepped shafts: compute J for each diameter separately, compute θ for each segment, then add. The torque is the same in both segments (no intermediate torques applied). Note T in N·mm = 200 N·m × 1000 = 200,000 N·mm.
Diagrams
Torsion — Chapter Overview Mind Map
A hierarchical mind map showing all major subtopics of the Torsion chapter, their key formulas, and how they relate to the central concept. Use this for quick orientation before studying each section.
Step-by-Step Problem-Solving Flowchart for Torsion Board Problems
A decision-tree flowchart guiding reviewees through the complete problem-solving process for any torsion board problem. Start at the top, follow the decision branches based on the given information, and arrive at the correct formula and unit check.
Power Transmission: From Power to Shaft Diameter
A linear flowchart showing the two-step power-transmission design procedure: (1) convert power and speed to torque, (2) use torque to find diameter or check stress. Includes the critical unit conversion steps and the design decision.
Flanged Bolt Coupling: Single vs Two-Ring Analysis
A flowchart distinguishing single-ring and two-ring bolt coupling analysis. For two rings, it highlights the compatibility condition (force proportional to radius) and the correct identification of the critical (outer) bolt ring.
Shear Stress Distribution States Across a Circular Shaft
A state diagram representing the progression of shear stress from zero at the centroid to maximum at the outer surface of a circular shaft under torsion. Reinforces the linear variation assumption and highlights the critical design point.
Distribution of PRC CE Board Torsion Problem Types
An approximate distribution of torsion question types appearing in recent PRC Civil Engineer Licensure Examinations. Maximum shear stress and power transmission problems together account for roughly 65% of torsion items — these should receive the most review time.
Exam strategy
Torsion problems in the PRC CE Board Exam follow three predictable patterns: (1) STRESS PATTERN — given shaft geometry and torque, find τ_max using τ = Tc/J or 16T/(πd³); (2) POWER PATTERN — given P and N, find T first, then proceed as in pattern 1; (3) ANGLE-OF-TWIST PATTERN — given shaft data, find θ = TL/(JG) and convert to degrees. For each problem: STEP 1 — Identify which pattern applies. STEP 2 — Draw the cross-section and label D, d, or both; compute J. STEP 3 — Identify T (from the problem or from P = 2πNT/60). STEP 4 — Apply the appropriate formula. STEP 5 — Check UNITS throughout — use mm and N·mm for everything, or m and N·m for everything; never mix. The most time-efficient approach is to memorize four key formulas: τ_max = 16T/(πd³) for solid shafts; T = 60P/(2πN) for power problems; θ = TL/(JG) for twist; and τ = T/(2A_m t) for thin-walled tubes. Hollow shaft and coupling problems always require an intermediate computation of J or A·τ·R. For combined loading, remember T_e = √(M²+T²) for shear and M_e = ½(M+T_e) for bending — these are the quickest paths to the answer. Always round shaft diameters UP to the nearest standard size in design problems. Avoid the three fatal errors: forgetting the factor of 60 in the power formula, subtracting areas instead of fourth-power diameters in hollow J, and leaving θ in radians when degrees are asked.