CELE Strength of Materials — TorsionDetailed Explanation
Detailed explanations for CELE Strength of Materials — Torsion. This page treats you like a serious reviewer: we unpack the concepts thoroughly, show worked examples of how Professional Regulation Commission (PRC) — Board of Civil Engineering frames Torsion questions, and explain the underlying reasoning that gets you to the right answer every time.
Exam context
Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Strength of Materials section sits under a "Core" weighting, and Torsion is the 2nd chapter in the 8-chapter CELE Strength of Materials rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Strength of Materials.
Torsion - Detailed Explanation
Torsion is the internal twisting effect produced in a structural or machine member when an external moment — called a torque — is applied about the member's longitudinal axis. In practice, every rotating drive shaft (pumps, motors, turbines), every propeller shaft on a vessel, and every spandrel beam in a building frame experiences torsion. For the PRC Civil Engineer Licensure Examination, torsion problems fall into four recurring patterns: (1) find the maximum shear stress in a circular shaft, (2) find the angle of twist, (3) size a shaft for a given power and speed, and (4) analyse a flanged bolt coupling or thin-walled tube. Mastery of this chapter also builds the foundation for combined loading (bending + torsion) in later topics. All formulas in this chapter assume linearly elastic, homogeneous, isotropic material and circular cross-sections unless otherwise stated.
Concepts
The Torsion Formula for Circular Shafts
When a torque T is applied to a circular shaft, every cross-section of the shaft resists that torque through shear stresses distributed across the cut face. The key insight from the theory of elasticity (and confirmed by experiment) is that plane cross-sections remain plane and undistorted — they simply rotate. This means the shear strain — and therefore the shear stress — varies LINEARLY from zero at the centroidal axis to a maximum at the outermost fibre. The general torsion formula is: τ = Tρ / J where: τ = shear stress at radius ρ from the shaft axis (MPa or N/mm²) T = internal torque at the cross-section (N·mm or N·m) ρ = radial distance from the centroidal axis to the point of interest (mm) J = polar moment of inertia of the cross-section (mm⁴) At the outermost surface where ρ = c (the outer radius), the shear stress is maximum: τ_max = Tc / J For a SOLID circular shaft of diameter d (radius c = d/2): J = πd⁴/32 = πc⁴/2 τ_max = Tc/J = T(d/2) / (πd⁴/32) = 16T / (πd³) For a HOLLOW circular shaft (outer diameter D, inner diameter d, outer radius R, inner radius r): J = π(D⁴ − d⁴)/32 = π(R⁴ − r⁴)/2 τ_max = TD/2 / [π(D⁴−d⁴)/32] = 16TD / [π(D⁴−d⁴)] The solid shaft shortcut τ_max = 16T/(πd³) is the single most-tested formula in CE Board torsion problems. Memorise it.
Examples
The conversion of kN·m to N·mm (multiply by 10⁶) is critical. The resulting stress is well below the typical allowable shear stress for structural steel (~90–100 MPa), so this shaft is safe. In board exams, answer choices are often given to 3 significant figures, so 48.9 MPa is the expected answer.
Scenario
A solid steel shaft 50 mm in diameter transmits a torque of 1.2 kN·m. Determine the maximum shear stress.
Solution
Given: d = 50 mm, T = 1.2 kN·m = 1.2×10⁶ N·mm Using the shortcut formula: τ_max = 16T / (πd³) = 16(1.2×10⁶) / [π(50)³] = 19.2×10⁶ / [π × 125,000] = 19.2×10⁶ / 392,699 = 48.9 MPa Answer: τ_max = 48.9 MPa
Note how the hollow shaft (which has 43.75% less cross-sectional area than a solid 80 mm shaft) carries the torque at only slightly higher stress than a comparable solid shaft — this is why hollow shafts are used in weight-sensitive applications such as ship propeller shafts and aircraft components.
Scenario
A hollow shaft has outer diameter D = 80 mm and inner diameter d = 60 mm. It carries a torque of 3 kN·m. Find the maximum shear stress.
Solution
Given: D = 80 mm, d = 60 mm, T = 3 kN·m = 3×10⁶ N·mm Step 1 — Polar moment of inertia: J = π(D⁴−d⁴)/32 = π(80⁴ − 60⁴)/32 = π(40,960,000 − 12,960,000)/32 = π(28,000,000)/32 = 2,748,893 mm⁴ ≈ 2.749×10⁶ mm⁴ Step 2 — Maximum shear stress (at outer surface, ρ = D/2 = 40 mm): τ_max = T(D/2)/J = (3×10⁶)(40) / 2.749×10⁶ = 120×10⁶ / 2.749×10⁶ = 43.7 MPa Answer: τ_max = 43.7 MPa
Applications
- Design of drive shafts and propeller shafts in mechanical and marine engineering
- Checking torsional stress in pump shafts and motor shafts
- Analysis of spandrel (edge) beams in building frames subjected to torsion from one-sided slab loads
- Design of drill rods and boring tools
- Torsional stress checks in combined bending-torsion problems (circular columns, crane hooks)
Misconceptions
- Using diameter instead of radius in J = πc⁴/2 (correct formula uses c = radius, so J = π(d/2)⁴/2 = πd⁴/32 — use J = πd⁴/32 to avoid confusion).
- Forgetting that J subtracts FOURTH powers of the diameters for hollow shafts, NOT second powers (areas).
- Applying the torsion formula to non-circular sections (rectangle, I-section) — those sections warp, and the formula is invalid.
- Using the bending stress formula (σ = Mc/I) instead of the torsion formula (τ = Tc/J) — they look similar but I is the second moment of area about a bending axis, while J is the polar (second) moment.
- Confusing the shear stress from torsion with the shear stress from transverse loads — they are different; torsion gives maximum stress at the SURFACE of the shaft, while transverse shear is maximum at the NEUTRAL AXIS.
Related Concepts
- Polar moment of inertia (J)
- Shear modulus G and its relationship to E and ν (G = E/[2(1+ν)])
- Angle of twist θ = TL/(JG)
- Combined bending and torsion — equivalent torque method
- Principal stresses from torsional shear — Mohr's circle
Common Exam Questions
Example
A 60 mm solid shaft carries 2.5 kN·m. Find τ_max. → T = 2.5×10⁶ N·mm; τ = 16(2.5×10⁶)/[π(60³)] = 58.9 MPa
Approach
Convert torque to N·mm, apply τ_max = 16T/(πd³) directly.
Question Type
Direct stress calculation — solid shaft
Example
T = 800 N·m, τ_allow = 55 MPa → d = ∛(16×800,000/[π×55]) = ∛(74,074) = 42.0 mm → use d = 45 mm
Approach
Rearrange τ_max = 16T/(πd³) → d = ∛(16T/[π·τ_allow]), then round UP to the next standard size.
Question Type
Find required diameter given allowable stress
Example
If τ_max must equal 50 MPa under T = 1 kN·m, the hollow shaft (D/d ratio = 2) will have a smaller outer diameter than a solid shaft.
Approach
Set τ_max equal for both cross-sections; use J formulas to relate diameters.
Question Type
Compare solid vs hollow shaft of equal τ_max
Key Points To Remember
- Shear stress is LINEAR across the radius — zero at the centre, maximum at the surface.
- Solid shaft: τ_max = 16T/(πd³); J = πd⁴/32.
- Hollow shaft: J = π(D⁴−d⁴)/32; use D (not d) in the numerator for τ_max.
- J uses the FOURTH power of diameter; errors in this exponent are the most common mistake.
- The torsion formula is valid ONLY for circular cross-sections (solid or hollow). Non-circular sections warp and require different formulas.
- Units must be consistent: if d is in mm, T must be in N·mm to get τ in MPa (N/mm²).
- 1 kN·m = 1×10⁶ N·mm — always convert before substituting.
Angle of Twist
The angle of twist θ is the angular displacement between two cross-sections of a shaft separated by distance L, caused by the applied torque T. It is the rotational analogue of axial deformation (δ = PL/AE). For a uniform shaft under constant torque: θ = TL / (JG) (result in RADIANS) where: T = torque (N·mm) L = length between the two sections (mm) J = polar moment of inertia (mm⁴) G = shear modulus of the material (MPa or N/mm²) For steel: G ≈ 80,000 MPa (80 GPa) For aluminum: G ≈ 26,000–28,000 MPa For brass: G ≈ 37,000–40,000 MPa The product JG is called TORSIONAL RIGIDITY. A stiff shaft has large JG. For a STEPPED SHAFT (different segments with different T, L, J, or G), apply the principle of superposition: θ_total = Σ (T_i × L_i) / (J_i × G_i) Be careful with sign conventions when torques are applied in multiple locations — use the method of sections to find internal torque T_i in each segment. Important conversion: 1 radian = 180°/π ≈ 57.296° To convert from radians to degrees: θ° = θ_rad × (180/π) The board exam ALWAYS asks for the angle in DEGREES — compute in radians, then convert at the very last step.
Examples
Note how L and d must both be in mm when T is in N·mm. Using mixed units (m for L but mm for d) is the single most frequent arithmetic error in angle-of-twist problems. When in doubt, convert EVERYTHING to N and mm first.
Scenario
A 40 mm solid steel shaft 2 m long carries a torque of 800 N·m. With G = 80 GPa, find the angle of twist in degrees.
Solution
Given: d = 40 mm, L = 2 m = 2,000 mm, T = 800 N·m = 800,000 N·mm, G = 80,000 MPa Step 1 — Polar moment of inertia: J = πd⁴/32 = π(40)⁴/32 = π(2,560,000)/32 = 251,327 mm⁴ Step 2 — Angle of twist: θ = TL/(JG) = (800,000)(2,000) / [(251,327)(80,000)] = 1.6×10⁹ / 2.011×10¹⁰ = 0.07956 radians Step 3 — Convert to degrees: θ = 0.07956 × (180/π) = 0.07956 × 57.296 = 4.56° Answer: θ = 4.56°
In a stepped shaft problem, the internal torque must be found in each segment by the method of sections before applying θ = TL/(JG). In this problem, since no intermediate torque is applied between A and C, the internal torque is the same (1,500 N·m) throughout. If a torque were applied at B, the internal torques in AB and BC would differ.
Scenario
A stepped steel shaft (G = 80 GPa) consists of two segments: Segment AB has d = 50 mm, L = 1.0 m, T = 1,500 N·m; Segment BC has d = 40 mm, L = 0.8 m, T = 1,500 N·m. Find the total angle of twist from A to C.
Solution
Convert: T = 1,500 N·m = 1.5×10⁶ N·mm; G = 80,000 MPa Segment AB (d = 50 mm, L = 1,000 mm): J_AB = π(50)⁴/32 = 613,592 mm⁴ θ_AB = (1.5×10⁶)(1,000) / [(613,592)(80,000)] = 1.5×10⁹ / 4.909×10¹⁰ = 0.03056 rad Segment BC (d = 40 mm, L = 800 mm): J_BC = π(40)⁴/32 = 251,327 mm⁴ θ_BC = (1.5×10⁶)(800) / [(251,327)(80,000)] = 1.2×10⁹ / 2.011×10¹⁰ = 0.05967 rad Total: θ_total = 0.03056 + 0.05967 = 0.09023 rad θ_total = 0.09023 × (180/π) = 5.17° Answer: θ = 5.17°
Applications
- Checking that shaft twist does not cause misalignment in precision machinery
- Verifying serviceability (deformation) limits for shafts — analogous to deflection limits for beams
- Statically indeterminate torsion problems where compatibility (equal twist) establishes additional equations
- Calculating the power angle (phase angle) between input and output shafts in power transmission systems
Misconceptions
- Leaving θ in radians when the problem asks for degrees — always check the units required.
- Forgetting to convert L from metres to millimetres when using N·mm for torque.
- Assuming G for steel is 200 GPa (that is E, not G). G_steel ≈ 80 GPa.
- In stepped shafts: adding angles without checking if the internal torque is constant throughout — use free body diagrams in each segment.
- Confusing torsional rigidity JG with flexural rigidity EI — they are analogous but involve different material constants and cross-section properties.
Related Concepts
- Axial deformation δ = PL/(AE) — torsional analogue
- Shear modulus G and its relation to E: G = E/[2(1+ν)]
- Statically indeterminate torsion (compatibility equations)
- Torsional stiffness k_T = JG/L
- Power transmission (torque computed from power and speed)
Common Exam Questions
Example
30 mm shaft, 1.5 m long, G = 80 GPa, T = 400 N·m → θ = (400×10³)(1,500)/[(πd⁴/32)(80,000)] → compute J first, then θ.
Approach
Apply θ = TL/(JG); substitute in consistent units (N·mm and mm); convert radians to degrees at the end.
Question Type
Find angle of twist of a uniform shaft
Example
A 30 mm shaft 1.5 m long twists 3°. G = 80 GPa. Find T. → θ = 3×π/180 = 0.05236 rad; J = π(30)⁴/32 = 79,522 mm⁴; T = (0.05236)(79,522)(80,000)/1,500 = 222 N·m
Approach
Rearrange: T = θ × JG / L; substitute θ in radians.
Question Type
Find unknown torque from a given angle of twist
Example
Two-segment shaft with different diameters and same torque applied at the ends.
Approach
Isolate each segment using free body diagrams; find T in each segment; sum θ = Σ T_i L_i/(J_i G_i).
Question Type
Stepped shaft total twist
Key Points To Remember
- θ = TL/(JG) gives radians — convert to degrees only at the final step.
- For stepped shafts: θ_total = Σ T_i L_i / (J_i G_i).
- G_steel ≈ 80 GPa = 80,000 MPa; G_aluminum ≈ 27 GPa.
- Torsional rigidity = JG (larger = stiffer = less twist per unit torque).
- Sign convention: positive torque follows right-hand rule (thumb points in +x direction, fingers curl in direction of positive twist).
- If the problem gives shaft stiffness k_T = JG/L, then θ = T/k_T.
Power Transmission
Rotating shafts transmit mechanical power. The relationship between power P, torque T, and angular velocity is: P = T × ω = T × (2πf) where ω = angular velocity in rad/s and f = rotational frequency in rev/s (Hz). If speed is given in REVOLUTIONS PER MINUTE (rpm = N): ω = 2πN/60 (rad/s) P = T × 2πN/60 → Solve for torque: T = 60P / (2πN) = P / (2πN/60) Unit summary: P in Watts (W) = N·m/s T in N·m N in rpm → T (N·m) = [P(W) × 60] / (2π × N_rpm) → T (N·m) = P(W) / (2π × f_Hz) Conversions to remember: 1 kW = 1,000 W 1 horsepower (hp) = 746 W Board exam strategy: STEP 1: Convert power to Watts. STEP 2: Convert speed to rpm (if given in another unit). STEP 3: Solve T = 60P/(2πN). STEP 4: Apply the torsion formula or angle-of-twist formula as required. This two-step (power → torque → stress) approach handles 90% of power transmission questions on the board exam.
Examples
This is the exact prototype of the most common CE Board power transmission problem. Note: ALWAYS round UP the computed diameter to ensure the actual stress ≤ allowable stress. Rounding down would make τ_max > τ_allow.
Scenario
A shaft transmits 75 kW at 1,200 rpm. If the allowable shear stress is 60 MPa, find the required minimum solid shaft diameter.
Solution
Step 1 — Torque: T = 60P/(2πN) = 60 × 75,000 / (2π × 1,200) = 4,500,000 / 7,539.82 = 596.8 N·m = 596,800 N·mm Step 2 — Required diameter (rearranging τ_max = 16T/[πd³]): d³ = 16T / (π × τ_allow) = 16 × 596,800 / (π × 60) = 9,548,800 / 188.50 = 50,659 mm³ d = ∛50,659 = 37.0 mm Answer: Use d ≥ 37 mm (round up to next standard size, e.g., 40 mm)
Since 44.5 MPa < typical allowable shear stress (~60–80 MPa for shaft steel), the 45 mm diameter is adequate. In a design problem, if the computed stress exceeds allowable, you must use the larger computed diameter.
Scenario
A solid steel shaft is to transmit 50 kW at 600 rpm. If d = 45 mm, find the torque and the maximum shear stress.
Solution
Step 1 — Torque: T = 60P/(2πN) = 60 × 50,000 / (2π × 600) = 3,000,000 / 3,769.91 = 795.8 N·m = 795,800 N·mm Step 2 — Maximum shear stress: τ_max = 16T/(πd³) = 16 × 795,800 / [π × (45)³] = 12,732,800 / [π × 91,125] = 12,732,800 / 286,279 = 44.5 MPa Answer: T = 795.8 N·m; τ_max = 44.5 MPa
Applications
- Sizing motor shaft diameters for electric motors in pumping stations and water treatment facilities (relevant to Philippine infrastructure projects)
- Design of transmission shafts in industrial plants
- Shaft design for wind turbines and hydroelectric generators
- Checking torsional stress in automotive drive shafts
Misconceptions
- Using N in rpm directly in P = Tω without dividing by 60 (ω must be in rad/s).
- Forgetting to convert kW to W before computing torque.
- Using 1 hp = 750 W instead of the more precise 746 W — check if the exam specifies the conversion.
- Rounding the computed shaft diameter DOWN instead of UP — always round up to ensure safety.
- Mixing N·m and N·mm inconsistently within the same calculation.
Related Concepts
- Torsion formula τ_max = 16T/(πd³) for solid shaft
- Angular velocity ω = 2πN/60
- Work and energy in rotating systems
- Efficiency of power transmission (η = P_out/P_in)
- Flanged bolt couplings (how torque is transferred between shafts)
Common Exam Questions
Example
100 kW at 900 rpm in a 60 mm solid shaft → T = 60(100,000)/(2π×900) = 1,061.0 N·m → τ = 16(1,061,000)/[π(60³)] = 24.9 MPa
Approach
T = 60P/(2πN); then τ = 16T/(πd³). Three-step solution.
Question Type
Find torque from given power and speed, then find stress
Example
150 kW, 720 rpm, τ_allow = 70 MPa → T = 1,989 N·m → d = ∛[16T/(π×70)] = 52.6 mm → use 55 mm
Approach
Find T from power; rearrange τ_max = 16T/(πd³) to solve for d; round up.
Question Type
Find minimum diameter for given power, speed, and allowable stress
Example
200 hp at 1,800 rpm → P = 200×746 = 149,200 W → T = 60(149,200)/(2π×1,800) = 791.8 N·m
Approach
Convert hp to W first (×746), then use T = 60P/(2πN).
Question Type
Power converted from hp instead of kW
Key Points To Remember
- T = 60P/(2πN) when P is in Watts and N is in rpm.
- T = P/(2πf) when P is in Watts and f is in Hz (rev/s).
- 1 kW = 1,000 W; 1 hp = 746 W.
- The result T is in N·m; convert to N·mm (×10³) before using in mm-based stress formulas.
- Forgetting the factor 60 when converting rpm is the most common board exam error in this topic.
- After finding T, the same torsion and angle-of-twist formulas from the previous sections apply.
Flanged Bolt Couplings
A flanged bolt coupling is a device that connects two shafts end-to-end by bolting their flanges together. The torque is transmitted entirely through the shear in the bolts. For a SINGLE BOLT CIRCLE with n bolts, each of cross-sectional area A_b, on a bolt circle of radius R: Shear force in one bolt: V = A_b × τ_bolt Torque capacity: T = n × V × R = n × A_b × τ_bolt × R For CONCENTRIC (two or more) BOLT CIRCLES, deformation compatibility requires that each bolt deforms in proportion to its distance from the centre. Since all flanges are rigid (assumed), the shear deformation of a bolt is proportional to R. For bolts with the same shear modulus G: γ ∝ R → τ ∝ R V ∝ A × τ ∝ A × R Compatibility condition: V₁/R₁ = V₂/R₂ (for equal bolt sizes; the ratio of shear STRESS is R₁/R₂) For two bolt circles: T = n₁ × V₁ × R₁ + n₂ × V₂ × R₂ with the constraint: τ₁/R₁ = τ₂/R₂ (if bolt sizes are equal, determine which bolt is critical first) Design approach: 1. Identify the CRITICAL bolt — the one on the LARGEST radius has the highest stress. 2. Set τ_critical = τ_allow. 3. Find stresses in all other bolts by proportion: τ_other = τ_allow × (R_other/R_max). 4. Compute T = Σ n_i × A_i × τ_i × R_i.
Examples
The setup is straightforward. All six bolts are at equal radius, so all carry equal shear force. The torque is the moment arm (R) times the total shear force (n × V).
Scenario
A coupling has six 20 mm bolts on a 150 mm-radius bolt circle. Allowable bolt shear stress = 70 MPa. Find the torque capacity.
Solution
Given: n = 6 bolts, d_bolt = 20 mm, R = 150 mm, τ_allow = 70 MPa Bolt area: A = π(20)²/4 = 314.16 mm² Shear force per bolt: V = A × τ = 314.16 × 70 = 21,991 N Torque capacity: T = n × V × R = 6 × 21,991 × 150 = 19,792,000 N·mm = 19.79 kN·m Answer: T = 19.8 kN·m
The compatibility condition τ₁/R₁ = τ₂/R₂ is the key governing equation for multi-circle couplings. The outer bolts always govern (highest stress), so set their stress equal to allowable; then inner bolt stress is always lower. Compute torque contributions separately and add.
Scenario
A coupling has two concentric bolt circles. Inner circle: 4 bolts, 16 mm dia, R₁ = 80 mm. Outer circle: 8 bolts, 16 mm dia, R₂ = 140 mm. Allowable shear stress = 80 MPa. Find the torque capacity.
Solution
Step 1 — Identify critical bolt (outer circle at R₂ = 140 mm): τ₂ = τ_allow = 80 MPa Step 2 — Stress in inner bolts by compatibility: τ₁/R₁ = τ₂/R₂ → τ₁ = τ₂(R₁/R₂) = 80(80/140) = 45.71 MPa Step 3 — Bolt area (same for all): A = π(16)²/4 = 201.06 mm² Step 4 — Torque from each circle: T₁ = n₁ × A × τ₁ × R₁ = 4 × 201.06 × 45.71 × 80 = 2,934,600 N·mm T₂ = n₂ × A × τ₂ × R₂ = 8 × 201.06 × 80 × 140 = 18,015,400 N·mm Step 5 — Total torque: T = T₁ + T₂ = 2,934,600 + 18,015,400 = 20,950,000 N·mm = 20.95 kN·m Answer: T = 20.95 kN·m
Applications
- Connecting motor shafts to pump impeller shafts in waterworks facilities
- Joining sections of long rotating shafts in industrial conveyors
- Field connections between shaft segments where welding is impractical
- Transmission couplings in sugar mill and rice mill machinery (common in Philippine agricultural industry)
Misconceptions
- Applying the same τ to all bolt circles regardless of radius — compatibility requires τ ∝ R.
- Using the outer diameter of the flange instead of the bolt circle radius R.
- Forgetting to account for the NUMBER of bolts (n) — the total resisting torque scales with n.
- Using bolt diameter directly instead of bolt AREA A = πd²/4.
- Assuming the inner bolt circle governs — it never does (outer is always critical for same bolt size).
Related Concepts
- Compatibility (deformation) conditions in structural mechanics
- Shear of rivets and bolts in eccentrically loaded connections (same principle, different geometry)
- Torque from torsion formula — the coupling must be able to transmit the shaft torque
- Friction-type vs. bearing-type bolt connections (AISC 360 Section J3, for reference only in design context)
Common Exam Questions
Example
8 bolts, 18 mm dia, R = 120 mm, τ_allow = 60 MPa → A = 254.5 mm²; T = 8(254.5)(60)(120) = 14.66 kN·m
Approach
T = n × (πd²/4) × τ_allow × R.
Question Type
Single bolt circle — find torque capacity
Example
T = 15 kN·m, n = 6, R = 130 mm, τ = 70 MPa → d = √(4×15×10⁶/[6×70×130×π]) = 21.5 mm → use 22 mm
Approach
Rearrange T = n × (πd²/4) × τ × R for d = √(4T/[n × τ × R × π]).
Question Type
Single bolt circle — find required bolt diameter
Example
As in Example 2 above.
Approach
Set outer bolt at τ_allow; find inner bolt stress by τ₁ = τ_allow × (R₁/R₂); compute T = Σ n_i A_i τ_i R_i.
Question Type
Two concentric circles — compatibility required
Key Points To Remember
- T = n × A_b × τ × R for a single bolt circle.
- For multiple bolt circles, deformation compatibility gives τ proportional to R (same bolt size).
- Critical bolt is always the one at the LARGEST radius.
- Always check whether equal bolt areas are specified; if different sizes are used, each shear force must be computed individually.
- A_b = π(d_bolt)²/4 — use the bolt shank (nominal) diameter unless otherwise specified.
- The coupling must satisfy both τ ≤ τ_allow for bolts AND the connection must be checked for bearing and friction in real design, though the board exam focuses on shear only.
Thin-Walled Closed Tubes (Bredt's Formula)
When the cross-section is not circular but is a closed thin-walled tube of ANY shape (rectangular, elliptical, triangular, etc.), the torsion formula τ = Tc/J does NOT apply. Instead, the Bredt formula (thin-walled tube theory) governs. The key concept is SHEAR FLOW, q: q = τ × t = constant around the entire perimeter This means that wherever the wall is thinner, the stress is higher — and wherever it is thicker, the stress is lower. The total torque is resisted by this shear flow acting along the closed boundary: T = 2 × A_m × q = 2 × A_m × τ × t Rearranged for shear stress: τ = T / (2 × A_m × t) where: A_m = area enclosed by the MEDIAN (mid-thickness) line of the wall (mm²) t = wall thickness at the point of interest (mm) τ = shear stress at the point with thickness t (MPa) If the wall thickness is VARIABLE, the MAXIMUM stress occurs at the THINNEST section: τ_max = T / (2 × A_m × t_min) For a closed rectangular tube of outside dimensions B × H and uniform thickness t: A_m ≈ (B − t)(H − t) ≈ B × H for thin walls IMPORTANT: This formula applies ONLY to CLOSED thin-walled sections. An open section (like a C-channel) has drastically lower torsional stiffness and must be treated differently (St. Venant torsion theory for open sections).
Examples
The median line subtraction (t from each dimension) is important for accuracy but for thin walls where t << B or H, the error from using overall dimensions is small. Board exam problems typically state 'uniform thin wall' and you can use either approach — state your assumption.
Scenario
A closed rectangular steel tube has outer dimensions 100 mm × 60 mm and uniform wall thickness t = 4 mm. It carries a torque T = 5 kN·m. Find the shear stress.
Solution
Given: B = 100 mm, H = 60 mm, t = 4 mm, T = 5 kN·m = 5×10⁶ N·mm Median-line dimensions: B_m = 100 − 4 = 96 mm (subtract one full t because median line is at t/2 from each face → (100−t) is an approximation; more precisely B_m = 100 − 2(t/2) = 100 − t) H_m = 60 − 4 = 56 mm Enclosed area: A_m = B_m × H_m = 96 × 56 = 5,376 mm² Shear stress (uniform t): τ = T / (2 × A_m × t) = 5×10⁶ / (2 × 5,376 × 4) = 5×10⁶ / 43,008 = 116.3 MPa Answer: τ = 116.3 MPa
The key skill is identifying A_m correctly. For standard shapes, use the formula for the area of the shape (rectangle: L×W; triangle: (√3/4)s²; circle: πr²) evaluated at the median dimensions.
Scenario
A closed triangular tube has equal sides of 80 mm and uniform wall thickness of 3 mm. Torque T = 2 kN·m. Find τ.
Solution
Given: equilateral triangle, side s = 80 mm (using median-line side ≈ 77 mm for thin wall), t = 3 mm, T = 2×10⁶ N·mm Median-line area (equilateral triangle with side a ≈ 77 mm): A_m = (√3/4) × a² = (√3/4)(77)² = 0.433 × 5,929 = 2,567 mm² (For simplicity in board exam: use s = 80 mm → A_m = (√3/4)(80²) = 2,771 mm²) Shear stress: τ = T / (2 × A_m × t) = 2×10⁶ / (2 × 2,771 × 3) = 2×10⁶ / 16,626 = 120.3 MPa Answer: τ ≈ 120 MPa
Applications
- Torsion in hollow box girders in bridges (common in Philippine prestressed concrete bridge design)
- Torsion in rectangular and box steel sections used as spandrel beams in building frames
- Torsion in aircraft fuselage and wing box structures
- Thin-walled pressure vessel ends under combined loading
- Analysis of closed-section reinforced concrete beams under torsion (ACI 318 Chapter 22 references thin-tube analogy)
Misconceptions
- Using the OUTSIDE area instead of the median-line enclosed area A_m — for thin walls the error is small but for thicker walls it becomes significant.
- Applying the circular shaft torsion formula to non-circular sections — wrong for any non-circular shape.
- Assuming shear flow q is constant in an OPEN section — q is constant only in closed sections.
- Thinking the thicker wall has higher stress — it is the THINNER wall (τ = T/2A_m·t; smaller t → larger τ).
- Confusing the shear stress from torsion with the shear stress from bending in thin-walled beams.
Related Concepts
- Shear flow in beams q = VQ/I
- St. Venant torsion for open sections (warping torsion)
- ACI 318 Section 22 — torsion in concrete members (uses thin-tube analogy)
- Bredt's formula as the basis for ACI 318 torsion design of reinforced concrete
- Torsional constant J for various cross-sections in AISC Steel Construction Manual
Common Exam Questions
Example
120×80 mm tube, t = 5 mm, T = 8 kN·m → A_m = 115×75 = 8,625 mm²; τ = 8×10⁶/(2×8,625×5) = 92.8 MPa
Approach
Compute A_m = (B−t)(H−t); apply τ = T/(2 A_m t).
Question Type
Find shear stress in a closed rectangular tube
Example
Box section with t_top = 6 mm, t_web = 4 mm → critical wall is the 4 mm web; set t = 4 mm in formula.
Approach
τ_max occurs at t_min; use τ = T/(2 A_m t_min).
Question Type
Variable wall thickness — find max stress
Key Points To Remember
- τ = T / (2 × A_m × t) — memorise this as Bredt's formula.
- A_m is the area enclosed by the MEDIAN LINE of the wall, NOT the overall outside area.
- Shear flow q = τt = constant — so stress is highest where t is smallest.
- For variable thickness: τ_max at t_min.
- Valid ONLY for CLOSED sections. Open sections (channels, angles) have much lower torsional resistance.
- For a circular tube: A_m = π(d_mean)²/4; the formula reduces correctly to the standard torsion formula.
Combined Torsion and Bending — Equivalent Torque and Moment
Many real shafts carry BOTH a bending moment M (from transverse loads or gear forces) and a torque T simultaneously. At the critical point on the shaft surface, the element experiences: - Normal stress from bending: σ = 32M/(πd³) (at the outermost bending fibre) - Shear stress from torsion: τ = 16T/(πd³) To find the MAXIMUM PRINCIPAL STRESS and MAXIMUM SHEAR STRESS at that critical point, use the principal stress equations (or Mohr's circle): σ₁,₂ = σ/2 ± √[(σ/2)² + τ²] τ_max = √[(σ/2)² + τ²] For design purposes, two equivalent quantities are defined: Equivalent Torque: T_e = √(M² + T²) ← governs maximum shear stress Equivalent Moment: M_e = (1/2)[M + √(M² + T²)] ← governs maximum normal stress These allow the design to proceed as if the shaft carried a pure torque T_e (for shear failure mode) or a pure moment M_e (for normal stress failure mode). Design equations: Based on T_e: τ_max = 16T_e/(πd³) Based on M_e: σ_max = 32M_e/(πd³) The controlling criterion depends on the material and failure theory. For ductile metals (steel), the maximum shear stress theory (T_e) is commonly used on the board exam.
Examples
This single equivalent-torque approach is much faster than separately computing σ and τ and then applying principal stress equations. It gives the same result and is the preferred method on the CE Board exam. Check: if τ_allow = 90 MPa, this shaft would be slightly overstressed and a larger diameter would be needed.
Scenario
A 60 mm solid shaft carries a bending moment M = 2.5 kN·m and a torque T = 3 kN·m simultaneously. Find the maximum shear stress using the equivalent torque method.
Solution
Given: d = 60 mm, M = 2.5 kN·m = 2.5×10⁶ N·mm, T = 3 kN·m = 3×10⁶ N·mm Equivalent torque: T_e = √(M² + T²) = √[(2.5×10⁶)² + (3×10⁶)²] = √[6.25×10¹² + 9.0×10¹²] = √(15.25×10¹²) = 3.905×10⁶ N·mm Maximum shear stress: τ_max = 16T_e/(πd³) = 16 × 3.905×10⁶ / [π × (60)³] = 62.48×10⁶ / 678,584 = 92.1 MPa Answer: τ_max = 92.1 MPa
Applications
- Design of cranked shafts in pumps where the crank arm generates both bending and torsion
- Analysis of car axles (bending from wheel load + torsion from driving torque)
- Checking combined stresses in structural members of offshore platforms
- Design of door hinges and latching mechanisms that experience combined twisting and bending
Misconceptions
- Adding bending stress σ and torsional shear stress τ algebraically — these are different stress components and must be combined using principal stress formulas, not simple addition.
- Applying T_e when the cross-section is not circular — the equivalent torque formula is derived from the circular shaft formulas.
- Forgetting that M_e gives maximum NORMAL stress while T_e gives maximum SHEAR STRESS — use the appropriate one for the failure mode being checked.
- Using the section modulus Z = πd³/32 for torsion (that is the POLAR section modulus; Z for bending is πd³/32 — coincidentally equal for circular sections but conceptually different).
Related Concepts
- Mohr's circle for stress transformation
- Principal stresses and maximum shear stress
- Failure theories: Maximum shear stress (Tresca) and distortion energy (von Mises)
- Pure bending stress σ = Mc/I = 32M/(πd³) for solid circular shafts
- Torsional shear stress τ = Tc/J = 16T/(πd³) for solid circular shafts
Common Exam Questions
Example
M = 1.8 kN·m, T = 2.4 kN·m, d = 55 mm → T_e = √(1.8²+2.4²)×10⁶ = 3.0×10⁶ N·mm → τ = 16(3.0×10⁶)/[π(55³)] = 91.6 MPa
Approach
Compute T_e = √(M²+T²); then τ = 16T_e/(πd³). Alternatively compute M_e for σ_max check.
Question Type
Given M and T, find τ_max or σ_max using equivalent quantities
Example
M = 2 kN·m, T = 2 kN·m, τ_allow = 60 MPa → T_e = 2√2 kN·m = 2.828×10⁶ N·mm → d = ∛(16×2.828×10⁶/[π×60]) = 62.5 mm
Approach
Compute T_e; rearrange τ = 16T_e/(πd³) for d = ∛(16T_e/[π·τ_allow]).
Question Type
Find required diameter under combined loading
Key Points To Remember
- T_e = √(M² + T²) — equivalent torque for maximum shear stress.
- M_e = (1/2)[M + √(M² + T²)] — equivalent moment for maximum normal stress.
- Both M and T must be in the SAME units (N·mm or N·m) before combining.
- The critical location is at the surface point where both bending stress and torsional shear are maximum and additive.
- For pure torsion (M = 0): T_e = T (reduces correctly).
- For pure bending (T = 0): M_e = (1/2)[M + M] = M (reduces correctly).
- This approach assumes solid or hollow circular cross-sections.
Practice Problems
The linear variation of shear stress across the hollow cross-section (zero at axis, max at outer surface) is confirmed by the ratio. At the inner bore, the stress drops to 70% of the maximum. For this reason, hollow shafts are efficient — the low-stress material near the axis is removed without significantly reducing torque capacity.
Problem
A hollow steel shaft has outer diameter D = 100 mm and inner diameter d = 70 mm. It is subjected to a torque T = 6 kN·m. Determine: (a) the maximum shear stress, and (b) the shear stress at the inner surface.
Solution
(a) Polar moment of inertia: J = π(D⁴−d⁴)/32 = π(100⁴ − 70⁴)/32 = π(100,000,000 − 24,010,000)/32 = π(75,990,000)/32 = 7,459,400 mm⁴ T = 6 kN·m = 6×10⁶ N·mm τ_max at outer surface (ρ = D/2 = 50 mm): τ_max = Tc/J = 6×10⁶ × 50 / 7,459,400 = 300×10⁶ / 7,459,400 = 40.2 MPa (b) Shear stress at inner surface (ρ = d/2 = 35 mm): τ_inner = Tρ/J = 6×10⁶ × 35 / 7,459,400 = 210×10⁶ / 7,459,400 = 28.2 MPa Note: τ_inner/τ_max = 35/50 = 0.70 — confirming linear variation.
This is a reverse angle-of-twist problem. The shaft angle is given; torque is the unknown. Rearranging θ = TL/(JG) is straightforward algebraically. After finding T, substituting back into the torsion formula gives the stress — a routine two-step process.
Problem
A 30 mm solid steel shaft (G = 80 GPa) is 1.5 m long. It twists through 3° under an applied torque T. Find the value of T and the maximum shear stress.
Solution
Given: d = 30 mm, L = 1,500 mm, θ = 3° = 3×π/180 = 0.05236 rad, G = 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 — Solve for T: θ = TL/(JG) → T = θ × JG / L T = 0.05236 × 79,522 × 80,000 / 1,500 = 0.05236 × 6.362×10⁹ / 1,500 = 3.332×10⁸ / 1,500 = 222,100 N·mm = 222.1 N·m Step 3 — Maximum shear stress: τ_max = 16T/(πd³) = 16 × 222,100 / [π × (30)³] = 3,553,600 / 84,823 = 41.9 MPa Answer: T = 222.1 N·m; τ_max = 41.9 MPa
This three-part problem combines the three main torsion skills: power→torque, stress→diameter sizing, and angle-of-twist. Note that the diameter used for the twist calculation is the ACTUAL selected diameter (50 mm), not the theoretical minimum (49.1 mm). Using 49.1 mm would give a slightly larger angle; in a strict check, use the actual diameter.
Problem
A shaft transmits 120 kW at 900 rpm. Using an allowable shear stress of 55 MPa, find the minimum required solid shaft diameter. Then find the angle of twist over a 2.5 m length if G = 80 GPa.
Solution
Step 1 — Torque: T = 60P/(2πN) = 60 × 120,000 / (2π × 900) = 7,200,000 / 5,654.87 = 1,272.9 N·m = 1,272,900 N·mm Step 2 — Minimum diameter: d = [16T/(π × τ_allow)]^(1/3) = [16 × 1,272,900 / (π × 55)]^(1/3) = [20,366,400 / 172.79]^(1/3) = [117,866]^(1/3) = 49.1 mm → use d = 50 mm (next standard size) Step 3 — Angle of twist (using d = 50 mm): J = π(50)⁴/32 = 613,592 mm⁴ θ = TL/(JG) = 1,272,900 × 2,500 / (613,592 × 80,000) = 3.182×10⁹ / 4.909×10¹⁰ = 0.06484 rad = 0.06484 × (180/π) = 3.71° Answer: d_min = 50 mm; θ = 3.71° over 2.5 m
This classic comparison problem demonstrates WHY hollow shafts are preferred in weight-sensitive design. The hollow shaft achieves the same torque capacity and same maximum stress with approximately 30% less material. In large structures and rotating machinery, this translates to significant weight and cost savings.
Problem
Compare the efficiency of a hollow shaft (D_o = 100 mm, D_i/D_o = 0.6) versus a solid shaft that carries the same torque at the same maximum shear stress. Calculate the ratio of their cross-sectional areas (hollow/solid).
Solution
Given: hollow shaft D_o = 100 mm, D_i = 0.6 × 100 = 60 mm For equal τ_max and equal T: Solid shaft: τ = 16T/(πd_s³) → d_s³ = 16T/(π × τ) Hollow shaft: τ = 16T × D_o / [π(D_o⁴ − D_i⁴)] → T = τ × π(D_o⁴ − D_i⁴)/(16 × D_o) Equate torques to find d_s: For hollow: J_h = π(100⁴−60⁴)/32 = π(100,000,000−12,960,000)/32 = π(87,040,000)/32 = 8,545,133 mm⁴ τ_max(hollow) = T(50)/J_h → T = τ × J_h / 50 = τ × 8,545,133/50 = 170,903τ For solid: T = τ × J_s/c_s = τ × (πd_s⁴/32)/(d_s/2) = τ × πd_s³/16 Set equal: πd_s³/16 = 170,903 d_s³ = 170,903 × 16/π = 869,908 d_s = 95.5 mm Cross-sectional areas: A_hollow = π(D_o² − D_i²)/4 = π(100²−60²)/4 = π(6,400)/4 = 5,027 mm² A_solid = π(d_s)²/4 = π(95.5)²/4 = 7,163 mm² Area ratio: A_hollow/A_solid = 5,027/7,163 = 0.702 Answer: The hollow shaft uses only 70.2% of the material of the equivalent solid shaft — a 29.8% material saving.
Straightforward single-circle coupling problem. All eight bolts carry equal shear. This is a typical last-part exam question where the coupling must also handle the torque computed from a power transmission calculation in the earlier parts.
Problem
A flanged coupling uses eight 16 mm bolts on a 120 mm-radius bolt circle. Find the torque capacity if τ_allow = 75 MPa.
Solution
Given: n = 8, d_bolt = 16 mm, R = 120 mm, τ_allow = 75 MPa Bolt area: A = π(16)²/4 = 201.06 mm² Shear force per bolt: V = A × τ = 201.06 × 75 = 15,080 N Torque capacity: T = n × V × R = 8 × 15,080 × 120 = 14,476,800 N·mm = 14.48 kN·m Answer: T = 14.48 kN·m
The Bredt formula gives a uniform shear stress around all four walls (since t is constant). If the wall thickness were different on top/bottom vs sides, the thinner walls would have higher stress. The median-line dimension correction (subtracting t from each side) is the proper approach.
Problem
A closed square-section steel tube with outside dimensions 80 mm × 80 mm and wall thickness t = 5 mm carries a torque T = 4 kN·m. Find the shear stress using Bredt's formula.
Solution
Given: outer side = 80 mm, t = 5 mm, T = 4 kN·m = 4×10⁶ N·mm Median-line side length: a_m = 80 − 5 = 75 mm (median line is at mid-thickness: 80 − 2×(5/2) = 75 mm) Enclosed median area: A_m = 75 × 75 = 5,625 mm² Bredt shear stress: τ = T / (2 × A_m × t) = 4×10⁶ / (2 × 5,625 × 5) = 4×10⁶ / 56,250 = 71.1 MPa Answer: τ = 71.1 MPa
Exam Preparation Tips
- MASTER THE BIG THREE FORMULAS first: τ_max = 16T/(πd³) for solid shafts; θ = TL/(JG) in radians; and T = 60P/(2πN) for power in W, N in rpm. These three alone cover 70–80% of torsion board problems.
- UNIT CONSISTENCY is the most common source of errors. The safest approach: convert ALL quantities to N and mm at the start. Then torque is in N·mm, stress is in N/mm² = MPa, and length is in mm.
- POWER UNIT TRAPS: 1 kW = 1,000 W. If speed is in rpm (N), use T = 60P/(2πN). If speed is in rev/s (f), use T = P/(2πf). The 60 is the conversion factor from rev/min to rev/sec, and forgetting it is the #1 error in power problems.
- HOLLOW SHAFT J: Always write J = π(D⁴−d⁴)/32. The ⁴ exponent applies to EACH diameter. A common trap is computing π(D−d)⁴/32 — that is completely wrong.
- ANGLE OF TWIST UNITS: Compute θ in radians. Convert to degrees at the VERY LAST STEP using θ° = θ_rad × (180/π). The board exam answer choices are always in degrees, so leaving the answer in radians loses the point.
- SHAFT SIZING: After computing the theoretical minimum diameter, ALWAYS round UP to the next standard size. Rounding down means the actual stress exceeds allowable.
- FLANGED COUPLINGS — MULTI-CIRCLE: The outer bolt circle is always critical (highest stress). Set it at τ_allow. Find inner bolt stresses by the proportion τ_inner = τ_allow × (R_inner/R_outer). Then sum torques from all circles.
- BREDT'S FORMULA for closed thin-walled tubes: τ = T/(2A_m·t). Remember A_m is the ENCLOSED median-line area, not the outer profile area. For variable thickness, τ_max is at t_min.
- COMBINED LOADING SHORTCUT: For a shaft with bending moment M and torque T, compute T_e = √(M²+T²) first, then apply the regular torsion formula with T_e replacing T.
- G vs E: Shear modulus G ≈ 80 GPa for steel, NOT 200 GPa (that is E). Confusing E and G in the angle-of-twist formula is a very common mistake.
- STUDY THE REFERENCE MATERIAL PATTERNS: PRC board exam torsion problems always follow predictable patterns — solid shaft stress, power transmission sizing, angle of twist, and sometimes coupling design. Practise each pattern type until the solution is reflexive.
- CHECK REASONABLENESS: Maximum shear stress for steel shafts in power transmission is typically 40–80 MPa. Angles of twist for shafts under normal conditions are typically 0.5° to 5° per metre. If your answer is far outside these ranges, recheck your calculation.
In summary
Torsion is a focused, formula-intensive topic in Strength of Materials that rewards careful preparation. The core of the subject reduces to five key formulas: τ_max = 16T/(πd³) for solid circular shaft stress; J = πd⁴/32 and J = π(D⁴−d⁴)/32 for solid and hollow sections respectively; θ = TL/(JG) for angle of twist; T = 60P/(2πN) for power transmission; and τ = T/(2A_m·t) for closed thin-walled tubes. Mastery of these five formulas — together with strict unit discipline (N and mm throughout) and the rules for flanged bolt couplings (τ proportional to R for concentric circles) — is sufficient to answer any torsion question that has appeared on the PRC CE Licensure Examination. The combined bending-and-torsion concept (equivalent torque T_e = √(M²+T²)) extends these skills to more complex shaft design scenarios. For the board exam, the three most important habits are: (1) convert ALL quantities to N and mm before substituting into any formula; (2) compute angles of twist in radians and convert to degrees only at the final step; and (3) always round shaft diameters UP to the next standard size. Reviewees who internalise these habits and practise the solution flowcharts presented in this chapter will approach torsion problems with confidence and speed — two qualities essential for success in the time-pressured environment of the PRC Civil Engineer Licensure Examination.
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