CELE Strength of Materials — TorsionSummary

Torque is a moment applied about the longitudinal axis of a member, causing it to twist. Torsion is the deformation resulting from applied torque. Unlike bending, which creates longitudinal stress variation, torsion produces shear stress that varies radially from the axis. The internal torque T resists external applied torques and is constant along a member unless distributed torques act.

The polar moment of inertia is a geometric property measuring a cross-section's resistance to torsion, analogous to the moment of inertia in bending. For a solid circular shaft: J = πd⁴/32 = πc⁴/2. For a hollow shaft: J = π(D⁴ − d⁴)/32. The larger the J, the greater the torsional rigidity. Hollow shafts are far more efficient than solid shafts—removing material near the center has minimal impact on J but significantly reduces weight.

The fundamental relationship for shear stress in a circular shaft under torsion. At any radius ρ from the axis, τ = Tρ/J. Maximum shear stress occurs at the outer radius c: τ_max = Tc/J. For a solid shaft, this simplifies to τ_max = 16T/(πd³). The formula assumes linearly elastic material, circular cross-section, plane sections remaining plane (no warping), and stress below the proportional limit. Shear stress distribution is linear—zero at the center, maximum at the outer surface.

The relative rotation between two cross-sections separated by length L under torque T. θ is measured in radians and is directly proportional to applied torque and length, inversely proportional to the torsional rigidity JG. For stepped shafts or varying torque, sum contributions: θ_total = Σ(T_i L_i)/(J_i G_i). The shear modulus G is a material property (e.g., 80 GPa for steel). Converting θ to degrees requires multiplying by 180/π.

A rotating shaft transmits power through torque. The relationship P = Tω connects power (watts), torque (N·m), and angular velocity ω (rad/s). In practical terms: P = 2πfT where f is frequency in rev/s, or P = (2πN/60)T where N is speed in rpm. The factor 2π/60 ≈ 0.1047 converts rpm to rad/s. To find required shaft diameter for power transmission: (1) calculate torque from power and speed, (2) apply allowable stress to find diameter using τ_max = 16T/(πd³). This two-step pattern is standard on boards.

The shear modulus G relates shear stress to shear strain: τ = Gγ. For steel, G ≈ 80 GPa; for aluminum, G ≈ 27 GPa. The product JG is the torsional rigidity—a material-and-geometry property governing resistance to twisting. Larger JG means less twist for the same torque. In composite or layered shafts, each material contributes to the overall stiffness according to its G value and cross-sectional area.

Two shafts are often connected via bolted flange couplings. The torque is transmitted through shear in a ring of bolts. For n bolts of area A each on a bolt circle of radius R: T = P·R·n where P = A·τ is the shear force per bolt. If bolts are on two concentric circles (a common design), torque is shared based on radius proportion: P₁/R₁ = P₂/R₂ (compatibility of deformation). Total torque capacity: T = (P₁R₁n₁) + (P₂R₂n₂). The coupling must be designed so the bolts do not slip or fail in shear.

For a closed thin-walled hollow tube of any shape (rectangular, triangular, circular), the shear stress is given by τ = T/(2A_m·t), where A_m is the area enclosed by the median (centerline) of the wall and t is the wall thickness. The key insight is that shear flow q = τ·t is constant around the perimeter. This formula is valid when t is small compared to the cross-sectional dimensions (typically t < d/10). Thin-walled formulas are far simpler than solving the full stress distribution, making them practical for design.

Real shafts often carry torque, bending moment, and axial load simultaneously. Each stress component is calculated separately: bending stress σ_b = My/I, axial stress σ_a = P/A, torsional shear stress τ_t = Tc/J. These are combined using Mohr's circle or principal-stress methods to find equivalent stress. A common simplification for shafts in torsion and bending: equivalent torque T_e = √(M² + T²) and equivalent moment M_e = ½(M + √(M² + T²)). The appropriate failure criterion (von Mises or maximum shear stress) is applied to these equivalent quantities.

Shear strain γ is the distortion angle (in radians) at a material element. From τ = Gγ, the surface shear strain in a circular shaft is γ = τ_max/G = (Tc/J)/G. This relates to the angle of twist per unit length: dθ/dL = c·γ/L = T/(JG). Large shear strains indicate material near yielding or failure. The angle of twist per unit length dθ/dL is proportional to torque and inversely proportional to JG.

Applied torque T produces internal shear stress distributed linearly across the circular cross-section. The torsion formula τ = Tc/J directly links applied moment to resulting stress. Larger J (either bigger diameter or hollow design) reduces stress for the same T.

Torque causes relative rotation (twist) between sections. The angle θ = TL/(JG) increases linearly with applied torque and length, but decreases with material stiffness G and geometric stiffness J. Shafts designed for low twist (high stiffness) require both large J and stiff material.

Power P transmitted by a rotating shaft relates to torque and speed: P = Tω. For fixed power, high-speed shafts carry less torque (and thus lower stress and smaller diameter), while low-speed shafts carry high torque (larger diameter). This speed-torque trade-off drives machine design choices.

A typical design problem: (1) Power and speed given → calculate required torque using P = (2πN/60)T, (2) apply allowable stress τ_allow to find minimum diameter using τ_max = 16T/(πd³), (3) check angle of twist using θ = TL/(JG) against allowable θ_allow, (4) if twist is excessive, increase diameter to increase J. The governing constraint (stress or twist) determines final diameter.

The product JG (geometric polar moment × shear modulus) quantifies torsional rigidity. For the same J, a stiffer material (larger G) reduces twist. For the same material, larger J reduces twist. Design optimization often involves balancing these: high-G materials may be expensive, so larger geometry (larger J) can achieve the same stiffness cost-effectively.

In flanged couplings, the torque capacity is directly proportional to bolt count (more bolts share the load), bolt area A (larger bolts carry more force), allowable shear stress τ_allow, and bolt circle radius R (larger moment arm). This relationship T = A·τ_allow·R·n shows why both bolt size and arrangement matter.

For thin-walled closed sections, the constant shear flow q = τ·t eliminates the need to integrate stress over the cross-section. The simple formula τ = T/(2A_m·t) works for any closed shape (circular, rectangular, triangular) as long as walls are thin. This demonstrates how geometric insight can simplify complex stress distributions.

When torque and bending moment coexist, principal stresses are computed from both components. The equivalent torque or moment then governs whether failure occurs. This relationship shows that neither torque nor moment alone determines failure—their combined effect on the stress state does.

Motors deliver power through rotating shafts. Given motor power (kW) and speed (rpm), engineers calculate required shaft diameter. The shaft must resist shear stress from torque and not exceed allowable twist (typically ≤ 0.3° per meter for machinery). In Philippine industrial plants, AISC 360 and local machinery standards guide shaft sizing. Example: a 100 kW motor at 1500 rpm requires substantial shaft diameter to keep stress within 60–80 MPa for steel and twist within limits.

Centrifugal pumps used in Philippine water supply systems transmit power from motor to impeller through a shaft. The coupling must securely join motor and pump shafts. Flanged bolt couplings carry torque through bolts in shear. Design checks: (1) does the coupling bolt arrangement safely transmit pump torque? (2) does the shaft itself withstand the torque without exceeding stress limits? A miscalculated coupling has caused catastrophic failures in industrial systems.

When lateral loads (wind, earthquake) are applied asymmetrically to a building, they create torsional moments about the building's vertical axis. NSCP 2015 requires engineers to account for torsional effects in design. Torsion in building columns and beams is analyzed similarly to shaft torsion: compute internal torques, apply stress and deformation limits. Reinforced concrete members are often more sensitive to torsion than steel members because concrete is weaker in shear (governed by ACI 318 torsional provisions).

Marine propeller shafts and hydroelectric turbine shafts experience extreme torques. The shaft must be large enough (high J) to limit stress and twist. Hollow shafts are standard—they reduce weight while maintaining torsional capacity. Example: a 5 MW hydroelectric turbine shaft at 300 rpm requires careful sizing to keep shaft stress below material limits while controlling twist to preserve bearing alignment.

Modern construction increasingly uses hollow steel tubes (circular, rectangular, square) to reduce weight while maintaining strength. The thin-walled tube torsion formula τ = T/(2A_m·t) is used to design these members. In transmission towers and temporary structures common in the Philippines, hollow tubes provide excellent torsional rigidity-to-weight ratios, reducing foundation costs and improving transportability.

When equipment is retrofitted with a more powerful motor, the existing shaft may be undersized. Torsion analysis reveals whether the shaft can handle the new torque or must be replaced. A larger-diameter solid shaft or a switch to a hollow shaft can solve the problem. This is common when industrial plants in the Philippines upgrade production capacity without replacing the entire drive train.

Flanged couplings, gearbox input/output shafts, and other bolted joints in machinery must transmit torque through bolt shear. Engineers verify that bolt shear capacity (τ_bolt = P/A where P is the shear force per bolt) exceeds the applied stress. The design must also account for preload, fatigue (if cyclic), and the clamping action of the bolts. This is critical for safety and reliability in Philippine industrial and construction equipment.

Lifting lugs and hooks often carry both vertical bending load and twisting moment (if the lifted load is off-center or rotating). Combined stress analysis using torsion and bending formulas determines the maximum equivalent stress. Design standards (e.g., local crane codes) specify allowable combined stresses. This is essential for safe design of lifting equipment used widely in Philippine construction.