CELE Strength of Materials — TorsionStudy Notes

When a torque T is applied to a circular shaft, the material undergoes shear deformation. The key insight is that shear strain (and hence shear stress) varies linearly from the axis of the shaft outward to its outer surface. At the center, the strain is zero; at the maximum radius c, the strain (and stress) is maximum. For any point at radius ρ from the shaft axis, the shear stress is: τ = (T·ρ)/J where J is the polar moment of inertia of the cross-section. The maximum shear stress occurs at the outer radius c (or d/2 for a solid shaft): τ_max = (T·c)/J This linear relationship is identical to the flexure formula for bending stress (σ = M·y/I) but applied to shear rather than normal stress. The polar moment of inertia J plays the same role for torsion as the second moment of area I plays for bending. For a solid circular shaft of diameter d: J = πd⁴/32 = πc⁴/2 (where c = d/2) This can be rewritten in a convenient form for stress calculation: τ_max = 16T/(πd³) For a hollow circular shaft (outer diameter D, inner diameter d): J = π(D⁴ − d⁴)/32 Understanding the underlying assumptions is critical for exam success: (1) The material is linearly elastic and homogeneous (uniform properties throughout). (2) The shaft cross-section is circular and remains plane during twisting (no warping in circular sections). (3) Shear stress does not exceed the proportional limit of the material. (4) The applied torque acts perpendicular to the cross-section and about the longitudinal axis. Non-circular cross-sections (rectangles, I-sections) do not satisfy assumption (2)—they warp out of plane—and require more complex analysis. The PRC exam typically restricts non-circular torsion to thin-walled closed tubes, which are treated separately.

While maximum shear stress is the primary design criterion for strength, the angle of twist is crucial for vibration analysis, alignment tolerances, and dynamic performance. The relative rotation θ between two cross-sections separated by length L is: θ = TL / (JG) where: - θ is the angle of twist (in radians) - T is the internal torque (constant over length L) - L is the length of the shaft segment - J is the polar moment of inertia - G is the shear modulus (material property) For steel, G ≈ 80–82 GPa; for aluminum, G ≈ 25–27 GPa. The product JG is called the torsional rigidity; a larger value means the shaft twists less under the same torque. For a stepped shaft or one with varying torque over its length, the total angle of twist is the sum of contributions from each segment: θ_total = Σ(T_i × L_i) / (J_i × G_i) This is analogous to summing deflections in a stepped beam under varying loads. Shear strain at the outer surface is related to angle of twist by: γ_max = τ_max / G = (c × θ) / L where γ is the shear strain (dimensionless). For design purposes, limits on angle of twist are often imposed to prevent: - Excessive vibration in rotating machinery - Misalignment of coupled shafts - Performance degradation in power transmission systems Typical allowable angles of twist in industrial shafts range from 0.5° to 2° per meter of length, though this varies by application. The PRC exam often requires conversion between radians and degrees: 1 radian = 180/π degrees ≈ 57.3° 1 degree = π/180 radians ≈ 0.01745 radians

In industrial and mechanical applications, rotating shafts transmit power. The relationship between mechanical power P, torque T, and rotational speed ω is fundamental to shaft sizing. Power is defined as: P = T·ω where P is in watts (W), T is in newton-meters (N·m), and ω is in radians per second (rad/s). If rotational speed is given in revolutions per minute (rpm), denoted N, the conversion is: ω = 2πN / 60 (rad/s) Therefore: P = T·ω = T·(2πN/60) = (2πNT) / 60 where P is in watts, N is in rpm, and T is in N·m. This is the workhorse formula for power transmission problems in the PRC exam. Alternatively, if speed is given in revolutions per second (rev/s or Hz), denoted f: P = 2πfT For engineering problems, power is often given in kilowatts (kW), so: P(W) = T(N·m) × [2πN(rpm) / 60] Rearranging to solve for torque: T = [60P] / [2πN] = [9.549P] / N (with P in watts, N in rpm, T in N·m) Or, if P is in kilowatts: T = [60,000P] / [2πN] = [9549P] / N (with P in kW, N in rpm, T in N·m) The standard two-step solution approach for power transmission problems is: (1) Convert power and speed into torque using the P-ω-N relationship. (2) Use the torque to find maximum shear stress or required diameter using τ_max = 16T/(πd³). In the Philippines, small marine engines and agricultural machinery frequently operate at 1000–1500 rpm, while large industrial motors may run at 600, 900, or 1200 rpm. These are common speeds in PRC exam problems. For a given power requirement, lower speed implies higher torque; higher speed implies lower torque. This is why large-horsepower hydraulic systems (which require very high torques at low speeds) use thick, short shafts, while high-speed gear motors use thin, long shafts.

In power transmission systems, two shafts are often connected by a flanged coupling (also called a flanged bolted coupling or plate coupling). The torque is transmitted from one shaft to the other through shear forces in a ring of bolts. Understanding the torque distribution is essential for coupling design and bolt sizing. For a simple coupling with n bolts of equal cross-sectional area A, all arranged on a single bolt circle of radius R from the shaft centerline, the shear force in each bolt is equal: P = A × τ_bolt where τ_bolt is the shear stress in the bolt material (not the shaft). Each bolt arm acts at radius R, so the torque contribution of each bolt is T_bolt = P × R. The total torque capacity of the coupling is: T = P × R × n = A × τ_bolt × R × n Rearranging: T = n × A × τ_bolt × R This is the fundamental equation for single-circle bolt coupling design. In practice, the allowable bolt shear stress τ_bolt is typically 75–100 MPa for steel bolts, depending on grade and safety factor. For dual-circle couplings (bolts on two concentric circles at radii R₁ and R₂), the situation is more complex. Both circles share the torque, but not equally. From deformation compatibility (the angle of twist must be equal for all bolts), the shear force in a bolt is proportional to its radius: P₁ / R₁ = P₂ / R₂ The total torque is: T = (P₁ × R₁ × n₁) + (P₂ × R₂ × n₂) If we denote the shear stress in bolts at radius R₁ as τ₁ and at radius R₂ as τ₂, then: P₁ = A₁ × τ₁ and P₂ = A₂ × τ₂ For design purposes, bolts are typically sized so that all bolts reach their allowable shear stress simultaneously (τ₁ = τ₂ = τ_allow), which ensures uniform utilization and prevents one circle of bolts from failing first. In industrial coupling design (as covered by AISC 360 for steel connections), the design equations account for bolt hole friction, slip resistance, and bearing stress at the flange. However, the PRC exam typically simplifies the problem to the direct shear model above.

When a closed thin-walled tube (such as a hollow rectangular or circular tube with constant small wall thickness t) is subjected to torsion, the classical solid-shaft formula τ = T·c/J does not apply because the section does not warp uniformly. Instead, the shear stress flows around the perimeter of the tube in what is called shear flow. For a thin-walled closed tube, the key concept is that the product τ·t (shear stress times wall thickness) is constant around the perimeter. This product is called the shear flow q: q = τ·t = constant The maximum shear stress in the tube occurs where the wall is thinnest (because q is constant, if t decreases, τ must increase). For a tube with uniform thickness t, the shear stress is: τ = T / (2·A_m·t) where A_m is the area enclosed by the median (centerline) of the wall. For a rectangular tube of internal width b and height h with uniform wall thickness t: A_m ≈ b·h (when t is small compared to b and h) For a circular tube of radius r_m (measured to the centerline): A_m = π·r_m² The fundamental relationship for the torque capacity of a thin-walled tube is: T = 2·A_m·t·τ_allow where τ_allow is the allowable shear stress in the tube material. This formula is particularly useful for hollow structural sections (HSS) commonly used in Filipino construction and bridge engineering. Unlike solid shafts where most material near the center is underutilized, thin-walled tubes distribute material optimally for torsional resistance. For angle of twist of a thin-walled tube: θ = T·L / (4·A_m²·G·∫(ds/t)) where the integral is taken around the perimeter (ds is an element of the centerline perimeter). For uniform thickness: ∫(ds/t) = Perimeter / t In practice, this becomes complex, and problems usually focus on stress rather than twist for thin-walled tubes.

In real-world engineering applications, shafts rarely carry pure torsion. A drive shaft in a vehicle transmits both torque (torsion) and vertical load (bending moment from the weight of pulleys or gears). Similarly, a propeller shaft in a ship experiences torsion along with bending from hydrodynamic forces. The ability to handle combined loading is critical for professional practice. When a shaft carries both bending moment M and torque T, the stress state at any point on the outer surface is two-dimensional: - Normal stress from bending: σ_bending = M·c / I (acting perpendicular to the cross-section) - Shear stress from torsion: τ_torsion = T·c / J (acting in the plane of the cross-section) The principal stresses (the maximum and minimum normal stresses at that point) must be found using Mohr's circle or the principal stress formulas. For a point on the outer surface of a circular shaft: σ₁, σ₂ = [σ_bending / 2] ± √[(σ_bending / 2)² + τ_torsion²] The maximum principal stress (algebraically larger) is: σ_max = [σ_bending / 2] + √[(σ_bending / 2)² + τ_torsion²] The maximum shear stress (from Mohr's circle) is: τ_max(combined) = √[(σ_bending / 2)² + τ_torsion²] For design, several equivalent-stress theories are used: 1. **Von Mises (Distortion Energy) Theory**: Most commonly applied. σ_equivalent = √[σ_bending² + 3·τ_torsion²] (for principal stresses σ_bending and 0) This is the most realistic for ductile materials like steel. 2. **Maximum Shear Stress Theory**: Sometimes used for brittle materials. τ_max(design) = √[(σ_bending / 2)² + τ_torsion²] 3. **Equivalent Torque** (a shorthand often used in practice): T_equivalent = √[M² + T²] Then use τ = T_equivalent·c / J as if pure torsion were applied to this equivalent value. However, this is an approximation and not rigorous. For a shaft carrying known values of M and T, the design approach is: (1) Calculate σ_bending = M·c / I and τ_torsion = T·c / J. (2) Compute the equivalent stress (e.g., von Mises): σ_eq = √[σ_bending² + 3·τ_torsion²]. (3) Compare σ_eq with the allowable stress (yield strength / safety factor). (4) If σ_eq exceeds the allowable, increase the shaft diameter. In the Philippines, AISC 360 (which is the basis for many Philippine design standards for steel structures) covers combined loading, and the combined stress check is mandatory for critical shafts in building frames and industrial structures.

For rapid problem-solving during the PRC examination, the following formulas should be memorized and immediately applicable: **Solid Circular Shaft (most common):** τ_max = 16T / (πd³) [T in N·m or N·mm; d in mm; result in MPa if T in N·m and d in mm] J = πd⁴ / 32 [result in mm⁴ if d in mm] **Hollow Circular Shaft:** J = π(D⁴ − d⁴) / 32 [both D and d in mm] **Angle of Twist (radians):** θ = TL / (JG) [T in N·m, L in m, J in mm⁴ → convert units carefully] **Power Transmission:** P = 2πNT / 60 [P in watts, N in rpm, T in N·m] T = 60P / (2πN) = 9.549P / N [for quick rearrangement] **Flanged Coupling (single circle):** T = n·A·τ_allow·R [n bolts, area A each, radius R, allowable shear stress τ_allow] **Thin-Walled Closed Tube:** τ = T / (2·A_m·t) [T = torque, A_m = enclosed median area, t = wall thickness] **Combined Torsion and Bending:** σ_eq (von Mises) = √[σ_bending² + 3·τ_torsion²] σ_bending = M·c / I [M = bending moment, c = outer radius, I = second moment of area] τ_torsion = T·c / J [standard torsion formula] **Useful Ratios for Circular Sections:** For a solid circular shaft: J = 2I [where I = πd⁴/64] **Unit Conversion Reminders:** 1 kN·m = 10⁶ N·mm 1 MPa = 1 N/mm² 1 rad = 180/π degrees ≈ 57.3° 1 rev/s = 60 rpm