CELE Strength of Materials — TorsionRevision Notes

Formulas

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

Section Title

1. Fundamentals of Torsion in Circular Shafts

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.

Formulas

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

Section Title

2. Angle of Twist

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.

Formulas

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

Section Title

3. Power Transmission by Rotating Shafts

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.

Formulas

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

Section Title

4. Flanged Bolt Couplings

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.

Formulas

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

Section Title

5. Torsion of Thin-Walled Closed Tubes

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).

Formulas

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

Section Title

6. Combined Torsion and Bending (Equivalent Torque and Moment)

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.