Torque and Twisting in Circular Shafts – Comprehensive Lecture Notes

Setup and coordinate system

• Consider a long, slender shaft/object where the largest cross-sectional dimension defines a characteristic scale; along the length, there are many such cross-sections, so the object is seen as a stack of shorter segments.
• The x-axis lies along the direction of the object (the shaft). The cross-section lies in the y-z plane.
• In the cross-section, the axes are y and z. If the object has two planes of symmetry (one vertical, one horizontal), the geometric centroid coincides with the centroid of area; otherwise, you must compute the centroid explicitly via an exercise (review topic).
• If you don’t have two planes of symmetry, you must compute the centroid position (to find lever arms and internal stresses). This is a reviewed topic and will be revisited in class.

Review of static equilibrium (chapter context)

• We study static equilibrium with six equations (sum of forces and sum of moments in three directions). In most problems this semester, not all six are needed; often one or two equations suffice.
• In particular problems in this lecture, the sum of moments about the x-axis will be the primary, sometimes the only, moment equation used.
• The course emphasizes turning external loading into internal loading and then into stress.

Twisting moments and torque (chapter 3 focus)

• A twisting moment about the x-axis is a torque that causes the shaft to twist about its axis.
• The quantity m_x is the torque (often called a twisting moment).
• The torque about the x-axis generates shear stress (not normal stress) in the cross-section.
• Terminology: tau denotes shear stress; sigma denotes normal stress.
• The external loading is converted into an internal torque distribution through free-body diagrams of shaft segments.

Key concepts: torque, twist, and material moduli

• The twist angle theta (φ or ϕ in some texts) is computed similarly to deflection in axial problems: it is an integral of the local rotation rate, i.e. $\theta = \int \frac{T(x)}{J(x) \, G(x)} \, dx.$
• If the torque T, the polar moment J, and the modulus G are constant over a segment, the twist contribution reduces to $\Delta \theta = \frac{T \, L}{J \, G}.$
• J is the polar moment of the cross-section area; for torsion about the central axis, J measures the resistance to twisting.
• G is the shear modulus in the material (note: the lecturer sometimes refers to it in context as a bulk-like modulus; in standard torsion, the relevant modulus is the shear modulus G). E is the Young’s (axial) modulus, and ν is Poisson’s ratio; G is related to E and ν via $G = \frac{E}{2(1+\nu)}.$
• For a moment about the x-axis, if the bar is straight and homogeneous, the twist angle accumulates along the length as you move along x.

Notation and geometry for torsion

• t denotes the internal torque inside the shaft (internal torsional moment).
• L denotes the segment length along the shaft over which the torque is considered.
• j (capital J) is the polar moment of area (second polar moment) of the cross-section; for a circular cross-section, J depends on the radius.
• ρ (rho) is the radial distance from the center of the cross-section; the shear strain at radius ρ is related to the local torque and geometry.
• The outer radius is R; the inner limit is at zero radius (center) for polar calculations.
• The maximum shear stress occurs at the outer surface (ρ = R).

Stress under torsion for circular cross-sections

• Shear strain in torsion is related to tau and ρ by: $\tau(\rho) = \frac{T \, \rho}{J}.$
• Maximum shear stress occurs at the outer surface: $\tau_{\max} = \frac{T \, R}{J}.$
• For a circular cross-section with radius r, the polar moment of area is: $J = \frac{\pi}{2} \, r^{4}.$
• The law used in this course for shear stress is: $\tau = \frac{T \, \rho}{J}.$
• In this course, T is the internal torque; the external torques produce internal torque values that are piecewise-constant across segments as the cross-section or material changes.

Material properties relevant to torsion

• G is the shear modulus (in the transcript referred to as a bulk-like modulus; standard torsion uses G). It is a material property that resists shear.
• E is the Young’s modulus; ν is Poisson’s ratio. The relation between E, G, and ν is $G = \frac{E}{2(1+\nu)}.$
• When mass is considered, the polar moment of inertia can be thought of analogously to J, but in dynamics you’d consider mass distribution; here the focus is on the polar moment of area J and the material modulus G.
• The modulus g (as written in the transcript) is treated as a material property akin to bulk modulus; in the torsion context, you’ll be given G (or its relation to E and ν) for the problem. The instructor notes that G can be given and may be related to Poisson’s ratio.

Length changes, segments, and piecewise constants

• If the shaft has changes in geometry or material, J and G can change along the length. In such cases, you break the shaft into segments where T, J, and G are constant within each segment.
• The total twist is the sum of the twists in each segment: $\theta = \sum<em>i \frac{T</em>i \; L<em>i}{J</em>i \; G_i}.$
• When there is a change in cross-section, the torque may be piecewise constant (TA in one segment, then TD in another, etc.). The internal torque can change when crossing a junction where the geometry changes or external torques are applied.

Composite shafts and indeterminate problems (chapter 3 context)

• Composite shafts may have different materials and radii in different segments, so J and G differ by segment.
• Problems can be statically determinate or statically indeterminate depending on the number of unknown internal torques and the available equations.
• Example scenario described: a shaft with two external torques and a wall support at one end, leading to one unknown internal torque if there is a single wall reaction (static determinacy). If there are walls (supports) at both ends, you typically have two unknowns and a statically indeterminate problem.
• When statically indeterminate (redundant supports), use compatibility conditions (e.g., the total twist between ends must satisfy boundary conditions, such as zero rotation at a fixed end) to obtain additional equations and solve for unknown torques.
• A practical note from the lecturer: for a problem with two end walls, the end twist must satisfy the boundary condition that the end does not rotate (total twist is zero). This provides an additional equation to solve for unknown torques.
• In the described problems, the total twist is set to zero at the fixed end, leading to equations like: $\phi_{total} = 0.$ This, combined with segmental twist expressions, allows solving for the unknowns TA, TD, etc.

Worked problem structure (described in class)

• First redacted problem: two externally applied torques with a wall at one end; only one external unknown torque (the wall’s torque) makes the problem statically determinate if there is only one wall reaction. If two unknowns exist (e.g., another wall or another applied torque), you cannot solve with equilibrium alone.
• Second redacted problem: a shaft with a wall at each end and multiple segments with changes in J and G. This problem is statically indeterminate (two unknown torques across ends). Solve by combining equilibrium with compatibility (total twist condition) and segmental twist expressions to determine TA, TD, etc.
• In the solutions, sometimes the problem is reformulated with the changes in torque across segments and the changes in J and G across the junctions. The total twist is then set equal to the boundary condition (e.g., zero twist at the fixed end) to determine the unknowns.

Practical notes and problem-solving strategy emphasized by the instructor

• Free-body diagrams of parts of the shaft are essential to determine internal torques in each segment.
• When geometry changes, you must account for the different J and G in each segment in the twist equation.
• The goal is to transform external loading into internal loading and then into the stress distribution, focusing on the maximum stress since it drives failure.
• Be prepared for statically indeterminate problems; you will see such problems on quizzes and homework; compatibility conditions are used to close the system.
• Homework problems from chapter 3 are designed to reinforce these concepts and mirror the redacted problems used in class.

Historical note and example context (brief anecdote)

• For non-circular cross-sections, determining the exact stress distribution under torque is historically complex. A hand calculation by Guran (with multiple graduate students) on a square cross-section under torsion is mentioned as an early finite-element-style approach before FE was common.
• The anecdote emphasizes the complexity of non-circular torsion and why, in this introductory course, we focus on circular cross sections where the formulas are tractable.

Summary of key equations to memorize (as introduced in this lecture)

• Twist angle (general): $\theta = \int \frac{T(x)}{J(x) \; G(x)} \, dx.$
• If constants over a segment: $\Delta \theta = \frac{T \, L}{J \, G}.$
• Circular cross-section polar moment: $J = \frac{\pi}{2} \; r^{4}.$
• Shear stress under torsion: $\tau(\rho) = \frac{T \, \rho}{J}.$
• Maximum shear stress: $\tau_{\max} = \frac{T \, R}{J}.$
• Stress definitions: $\tau = \text{shear stress}, \quad \sigma = \text{normal stress}.$
• Material modulus relation (standard): $G = \frac{E}{2(1+\nu)}.$
• Axial normal stress (centroidal axial loading): $\sigma = \frac{P}{A}.$
• For circular cross-section, the specific J value: $J = \frac{\pi}{2} \, r^{4}.$

Practical takeaways for exam preparation

• Be able to set up internal torque distributions for a shaft with piecewise geometry/materials using free-body diagrams.
• Be able to write the segmental twist sum and solve for unknown torques, especially in statically indeterminate cases using the total twist compatibility condition.
• Recognize that the maximum stress due to torsion occurs at the outer surface and use the formula for tau_max accordingly.
• Be comfortable with moving between global loading, internal loading, and stress results, with emphasis on the endpoint behavior (boundary conditions) and compatibility.
• Remember that problems can involve multiple materials and radii, requiring separate J and G terms per segment and summing twists accordingly.

Instructor tips and classroom logistics mentioned

• Encourage students to physically move forward in the classroom to facilitate questions and discussion during problem-solving time.
• There will be a quiz on Chapter 2 (static equilibrium context) on Monday; be prepared for statically indeterminate problems as part of Chapter 3 material.
• The class will transition to new material after the review week; the current session emphasizes tying Chapter 2 concepts to Chapter 3 torsion.