CELE Strength of Materials — Combined Stresses and Mohr's CircleStudy Notes

When multiple loads act on a structural member simultaneously, the stress state at any point is the superposition (sum) of the individual stress components. This principle is valid as long as the material remains in the elastic range and deformations are small—assumptions fundamental to civil engineering design under NSCP 2015 and AISC 360. **Stress Superposition Principle:** At a point in a loaded member, the total stress is the algebraic sum of stresses from each load: - Axial load P produces uniform normal stress σ = P/A - Bending moment M produces linear normal stress σ = Mc/I (maximum at outer fibers) - Torsional torque T produces shear stress τ = Tc/J (maximum at outer edge) For a typical case (e.g., a shaft with bending and torsion), the combined normal stresses simply add: σ_total = σ_axial + σ_bending ± σ_secondary. Similarly, shear stresses from torsion and shear force add vectorially. **Real-World Example (Philippine Context):** Consider a reinforced concrete column in a multi-story building (designed per ACI 318 and NSCP 2015) carrying both vertical load and lateral wind moment. The stress at a corner fiber is the sum of: (1) axial compression from building weight, and (2) bending tension/compression from the overturning moment. A combined-stress analysis at the critical point determines whether the section needs larger reinforcement or thicker concrete. **Key Concept:** Once the three stress components (σ_x, σ_y, τ_xy) are known at a point, we can find: - Stresses on any inclined plane (transformation equations) - The planes with maximum and minimum normal stress (principal stresses) - The plane with maximum shear stress - The magnitude and orientation of all extremes This is the bridge from load analysis to failure-theory application.

Consider a stress state defined by σ_x, σ_y, and τ_xy on the x and y reference planes. To find the normal and shear stress on a plane inclined at angle θ (measured counterclockwise from the x-axis), we use the transformation equations. These are derived from equilibrium of a wedge element and are fundamental in strength of materials. **Transformation Equations for Normal Stress:** The normal stress on the inclined plane (call it the x' direction) is: σ_x′ = (σ_x + σ_y)/2 + (σ_x − σ_y)/2 · cos(2θ) + τ_xy · sin(2θ) **Transformation Equations for Shear Stress:** The shear stress on the inclined plane is: τ_x′y′ = −(σ_x − σ_y)/2 · sin(2θ) + τ_xy · cos(2θ) **Critical Observations:** 1. The term (σ_x + σ_y)/2 is the **average normal stress** and appears in all transformations—it is invariant (same for all θ). 2. The angle θ appears as 2θ in the trig functions; a 45° physical rotation becomes 90° on the mathematical equation. 3. At θ = 0°, we recover σ_x′ = σ_x and τ_x′y′ = τ_xy (verification). 4. These are periodic with period 180° in θ (or 360° in 2θ), reflecting the symmetry of a stress element. **Physical Meaning:** As you mentally rotate a tiny element through different orientations, the normal stress on the top face oscillates between a maximum (σ_1) and minimum (σ_2), while shear reaches extremes at ±45° from the principal planes. The transformation equations quantify this variation. **Practical Application (NSCP 2015 and AISC 360):** In beam design, we often check stress on inclined planes due to combined bending and shear. For connections and welds, stresses on skew planes (not aligned with member axes) are critical. The transformation equations provide the rigorous method to evaluate these.

The **principal stresses** are the extreme (maximum and minimum) normal stresses at a point. They act on planes where the shear stress is exactly zero. These planes are perpendicular to each other and are the reference orientation most naturally suited to analyzing the material's response to stress. **Principal Stress Formula (2D Case):** σ₁, σ₂ = (σ_x + σ_y)/2 ± √[((σ_x − σ_y)/2)² + τ_xy²] where σ₁ ≥ σ₂ (σ₁ is the algebraically larger principal stress). Note: If both are positive, σ₁ is the larger tension; if both are negative, σ₂ is the larger compression. If one is positive and one negative, their magnitudes determine the severity. **Maximum In-Plane Shear Stress:** τ_max = √[((σ_x − σ_y)/2)² + τ_xy²] = (σ₁ − σ₂)/2 This equals the radius of Mohr's circle. **Principal Plane Orientation:** tan(2θ_p) = 2τ_xy / (σ_x − σ_y) Solving for θ_p gives the angle of the principal planes (there are two, 90° apart). The maximum principal stress σ₁ acts on the plane labeled θ_p (measured counterclockwise from x-axis). **Planes of Maximum Shear:** Located at θ = θ_p ± 45° (i.e., 45° away from the principal planes). **Important Property:** On the plane of maximum shear, the normal stress equals the average: σ = (σ_x + σ_y)/2. **Physical Interpretation (Critical for Failure Theories):** Different failure theories use different stress measures: - **Maximum normal-stress theory (brittle materials):** failure when σ₁ reaches the tensile strength or σ₂ reaches the compressive strength in magnitude. - **Maximum shear-stress theory (ductile materials, Tresca):** failure when τ_max = σ_yield/2. - **Distortion-energy theory (most accurate for ductile materials, von Mises):** σ_von Mises = √[σ₁² − σ₁σ₂ + σ₂²] compared to yield strength. Understanding principal stresses is essential for applying these criteria correctly in design (AISC 360 for steel, ACI 318 for concrete, NSCP 2015 for both).

**What is Mohr's Circle?** Mohr's circle is a two-dimensional graphical representation of the stress state. Each point on the circle corresponds to the normal and shear stress on some plane through the material point. The circle encodes all the transformation equations and makes finding principal stresses, maximum shear, and stress on any plane a matter of geometry rather than trigonometry. **Circle Equation and Center:** For a stress state (σ_x, σ_y, τ_xy), Mohr's circle is defined in the (σ, τ) coordinate system (stress space, not physical space): - **Center:** C = ((σ_x + σ_y)/2, 0) on the horizontal σ-axis - **Radius:** R = √[((σ_x − σ_y)/2)² + τ_xy²] - **General equation:** (σ − C_σ)² + τ² = R² **Principal Stresses (Points on σ-axis):** Where the circle intersects the σ-axis (τ = 0): - σ₁ = C + R = (σ_x + σ_y)/2 + R (rightmost point) - σ₂ = C − R = (σ_x + σ_y)/2 − R (leftmost point) **Maximum Shear Stress:** The topmost and bottommost points on the circle: - τ_max = ±R - The normal stress at these points is C = (σ_x + σ_y)/2 **Construction of Mohr's Circle (Step-by-Step):** 1. **Plot point X** on the (σ, τ) plane: X = (σ_x, τ_xy) 2. **Plot point Y** on the (σ, τ) plane: Y = (σ_y, −τ_xy) — note the sign flip on shear 3. **Find the center:** The midpoint of line XY is C = ((σ_x + σ_y)/2, 0) 4. **Find the radius:** Measure the distance from C to X (or Y); this is R 5. **Draw the circle:** Center at C, radius R 6. **Read results:** - Where circle meets σ-axis: σ₁ (right), σ₂ (left) - Top of circle: τ_max (positive shear) - Bottom of circle: −τ_max (negative shear) - At max-shear points: normal stress = C **Angle Relationship (Critical):** If the stress element is rotated by angle θ counterclockwise in physical space, the corresponding point on Mohr's circle rotates by **2θ in the same direction** (counterclockwise). Conversely, if a point on the circle is at angle 2α from the positive σ-axis (measured from center), the corresponding physical plane is at angle α from the x-reference. **Example: Reading Principal Planes from the Circle:** Draw a line from center C to point X (or Y). The angle this line makes with the positive σ-axis is 2θ_p (approximately). The principal plane corresponding to σ₁ is at θ_p counterclockwise from the x-axis. **Three-Dimensional Extension (Important for Out-of-Plane Stresses):** For a 3D stress state, there are three principal stresses σ₁ ≥ σ₂ ≥ σ₃, and three principal planes. The 2D Mohr's circle (shown for plane 1–2) is one of three possible circles: - Circle 1–2 (radius = (σ₁ − σ₂)/2) - Circle 2–3 (radius = (σ₂ − σ₃)/2) - Circle 1–3 (radius = (σ₁ − σ₃)/2, typically the largest) For a surface (where σ₃ = 0), the largest circle is 1–3, with radius σ₁/2. This is crucial for surface-failure analysis (e.g., fatigue). **Advantages of Mohr's Circle:** - Rapid graphical solution (useful in exams and field work) - Immediate visualization of principal stresses, max shear, and any inclined plane - Easy to check reasonableness (e.g., are σ₁ and σ₂ on opposite sides of the center?) - Elegant geometric insight into why max shear occurs at 45° from principals

**Practical Context:** Rotating machinery—pumps, compressors, turbines, and power-transmission shafts—simultaneously carry bending moment (from self-weight, gear teeth forces, or misalignment) and torque (from the power transmission). This combined loading is one of the most common design scenarios in mechanical and civil (industrial) engineering. **Stress State at the Shaft Surface:** For a circular shaft of diameter d (radius r = d/2) with bending moment M and torque T: **Bending stress (at the outer fiber):** σ_x = Mc/I = (M · r) / I = (32M) / (πd³) where I = πd⁴/64 for a solid circular section. **Torsional shear stress (at the outer fiber):** τ_xy = Tc/J = (T · r) / J = (16T) / (πd³) where J = πd⁴/32 for a solid circular shaft. **Notes:** - σ_y = 0 (no stress perpendicular to the shaft surface at the outer fiber) - Bending produces normal stress (σ_x) on the cross-section - Torsion produces shear stress (τ_xy) tangent to the circumference - The two stresses are perpendicular—a classic plane-stress state **Equivalent Stress for Design (Using Maximum Normal Stress Theory):** For ductile shafts (steel, aluminum), design often uses the equivalent (or combined) stress. The principal stresses are: σ₁, σ₂ = σ_x/2 ± √[(σ_x/2)² + τ_xy²] With σ_x = 32M/(πd³) and τ_xy = 16T/(πd³): σ₁ = 16/(πd³) · [M + √(M² + T²)] ← **Equivalent moment approach** Defining the equivalent moment: **M_e = (1/2)[M + √(M² + T²)]** (sometimes written as M_e = √(M² + T²) for maximum-shear theory; different conventions exist) Then: σ_max = (32 M_e) / (πd³) **Equivalent Torque (Maximum Shear Theory):** Alternatively, using τ_max = √[(σ_x/2)² + τ_xy²]: **T_e = √(M² + T²)** Then: τ_max = (16 T_e) / (πd³) = (16√(M² + T²)) / (πd³) **Design Criterion:** Choose the equivalent moment or equivalent torque method based on the material and failure criterion: - **Normal-stress theory:** Design for σ_max ≤ σ_allowable, using M_e - **Shear-stress theory (Tresca):** Design for τ_max ≤ τ_allowable = σ_yield/2, using T_e - **Distortion-energy theory (von Mises):** More refined; gives results between the other two For ductile metals (steel), the maximum-shear and von-Mises theories are most accurate. AISC 360 and various machinery design codes often use one of these methods. **Physical Insight:** Imagine the shaft rotating in bending (the stress reverses with each rotation). Meanwhile, torque is applied continuously. The combined effect produces principal stresses that are more severe than bending or torsion alone. The equivalent moment or torque "accounts" for this combined effect in a single number, which is then plugged into familiar design equations (e.g., σ = M/W). **Step-by-Step Design Procedure for a Shaft Under M and T:** 1. Determine M and T from the application (loads, geometry, kinematics). 2. Calculate M_e = (1/2)[M + √(M² + T²)] or T_e = √(M² + T²) 3. Choose allowable stress (from material data, design codes) 4. Solve for required diameter: d_required = (32 M_e / (π σ_allow))^(1/3) [normal-stress approach] 5. Round up to a standard size; verify other criteria (deflection, critical speed, etc.)

**Why Failure Theories?** At a point in a loaded member, we have principal stresses σ₁, σ₂ (and σ₃ = 0 for plane stress). The question is: **what combination of these stresses causes the material to fail (yield for ductile, fracture for brittle)?** Different materials fail under different stress conditions, so multiple theories exist, each suited to different materials and loading modes. **1. Maximum Normal Stress Theory (Rankine)** **Criterion:** Failure occurs when either principal stress reaches the ultimate (or yield) strength. **For ductile materials:** σ₁ ≤ σ_yield and |σ₂| ≤ σ_yield **For brittle materials:** σ₁ ≤ σ_tensile and |σ₂| ≤ σ_compressive **Advantage:** Simple, intuitive; used for brittle materials (concrete, cast iron, stone). **Disadvantage:** Ignores shear stress entirely; often unconservative for ductile metals under shear-dominated loading. **Example:** If σ₁ = 100 MPa, σ₂ = 0, material yields at 200 MPa, failure load is when σ₁ reaches 200 MPa (straightforward). But in pure shear (σ₁ = 100, σ₂ = −100, σ₃ = 0), the normal-stress theory predicts failure at 200 MPa, which experiments show is wrong for ductile metals—they fail at ~115 MPa (half the yield) due to shear. **2. Maximum Shear Stress Theory (Tresca)** **Criterion:** Failure occurs when the maximum shear stress reaches half the yield strength (for ductile metals). **Formula:** τ_max = (σ₁ − σ₃)/2 ≤ σ_yield/2 In 2D plane stress (σ₃ = 0): τ_max = σ₁/2 ≤ σ_yield/2 → σ₁ ≤ σ_yield But if both σ₁ and σ₂ have the same sign (both tension or both compression), then the absolute maximum shear is (σ₁ − σ₃)/2 = σ₁/2 (if σ₃ = 0). **Advantage:** Accounts for shear; matches experimental data for many ductile metals better than normal-stress theory. **Disadvantage:** Ignores the intermediate principal stress σ₂; predicts the same failure stress for (σ₁ = 100, σ₂ = −100) and (σ₁ = 100, σ₂ = 0), which is not always accurate. **Example:** Pure shear τ = 50 MPa → σ₁ = 50, σ₂ = −50 → τ_max = 50 MPa. Failure at τ_max = σ_yield/2 → material yields at σ_yield = 100 MPa. Experiments confirm this is better than the normal-stress prediction (200 MPa). **3. Distortion-Energy Theory (von Mises)** **Concept:** The energy stored in the material due to shear (distortion), not volumetric change (hydrostatic stress), causes yielding. **Criterion:** The equivalent von Mises stress **σ_v = √[σ₁² − σ₁σ₂ + σ₂²]** should not exceed the yield strength: σ_v ≤ σ_yield **For plane stress (σ₃ = 0):** σ_v = √[σ₁² − σ₁σ₂ + σ₂²] **In terms of original components (σ_x, σ_y, τ_xy):** σ_v = √[σ_x² − σ_x σ_y + σ_y² + 3τ_xy²] **Advantage:** Most accurate for ductile metals over a wide range of stress states; accounts for all three principal stresses; predictions closely match experimental failure data. **Disadvantage:** More complex; requires calculation of principal stresses (or use of the direct formula above). **Example:** σ₁ = 100 MPa, σ₂ = 0 → σ_v = √(10000 − 0 + 0) = 100 MPa. Failure at 100 MPa (normal stress alone, no shear). But σ₁ = 100, σ₂ = −100 → σ_v = √(10000 + 10000 + 10000) = √30000 = 173.2 MPa. Higher equivalent stress because shear is larger; failure expected later (higher load) than in the first case—this matches experiments. **4. Comparison and Selection** | Theory | Best For | Formula (Plane Stress) | Difficulty | |--------|----------|--------|-------------| | Normal Stress | Brittle materials (concrete, stone, ceramics) | σ₁ ≤ f_t | Simple | | Maximum Shear (Tresca) | Ductile metals, conservative | τ_max = (σ₁ − σ₃)/2 ≤ σ_y/2 | Moderate | | von Mises (Distortion Energy) | Ductile metals, most accurate | σ_v = √[σ₁² − σ₁σ₂ + σ₂²] ≤ σ_y | Moderate | **Design Practice (AISC 360, ACI 318, NSCP 2015):** - **Structural steel (AISC 360):** Uses a combination of the theories, with safety factors and interaction equations. For simple combined bending and torsion, equivalent moment/torque methods (which approximate the shear-stress or von-Mises theory) are common. - **Reinforced concrete (ACI 318, NSCP 2015):** Principal-stress analysis is rarely done; instead, empirical interaction equations for columns, beams, and combined stress are prescribed. - **Machinery and rotating shafts:** Often use maximum-shear or von-Mises; AGMA (gearing) and API (pressure vessels) standards specify which. **Critical Note for Exam Preparation:** The PRC CE Licensure Exam often asks to: 1. **Calculate principal stresses** from combined loads 2. **Apply a failure criterion** (usually maximum normal or maximum shear) 3. **Determine a safety factor** or **verify adequacy** of a section Read the question carefully to identify which theory is implied (or explicitly stated). If not stated, **maximum-shear theory is the safest default for ductile metals**, and **maximum normal-stress for brittle materials** (concrete, typical in civil structures).

**Common Errors in Mohr's Circle and Transformation Problems:** 1. **Factor-of-2 errors:** - The radius of Mohr's circle is √[((σ_x − σ_y)/2)² + τ_xy²], not √[(σ_x − σ_y)² + τ_xy²] - Physical angle θ becomes 2θ on the circle; always divide by 2 when reading the angle back - tan(2θ_p) = 2τ_xy / (σ_x − σ_y), not tan(θ_p) = τ_xy / (σ_x − σ_y) 2. **Sign convention for τ_xy:** - Establish a consistent convention (e.g., τ_xy > 0 means clockwise shear on the right face) - When plotting Y on Mohr's circle, use −τ_xy (the opposite of the original) - If a problem statement is ambiguous, clarify the convention in your solution 3. **Forgetting σ_y = 0 in shaft problems:** - For bending on a shaft, only the outer fiber carries normal stress; σ_y = 0, not σ_y = −M/(something) - Torsion produces pure shear: τ_xy ≠ 0, but σ_x and σ_y may both be zero (pure shear) or one is present (combined) 4. **Neglecting the third principal stress (σ₃):** - For a point on the **surface** of a loaded member (e.g., outside of a beam, outer edge of a shaft), the stress perpendicular to the surface is zero: σ₃ = 0 - The absolute maximum shear stress in 3D is (σ₁ − σ₃)/2, not (σ₁ − σ₂)/2 - If σ₁ = 100 MPa, σ₂ = 80 MPa, σ₃ = 0 (surface), then τ_abs_max = 50 MPa, not 10 MPa 5. **Confusing equivalent moment and equivalent torque:** - Equivalent moment M_e = (1/2)[M + √(M² + T²)] is used with the normal-stress formula σ = 32M_e/(πd³) - Equivalent torque T_e = √(M² + T²) is used with the shear formula τ = 16T_e/(πd³) - They give slightly different design diameters; check which theory (normal-stress or shear-stress) the problem asks for 6. **Reading angles from Mohr's circle incorrectly:** - The angle from the center of the circle to a point is 2θ_physical (double the angle) - If you measure an angle of 60° on the circle (from positive σ-axis to the radius), the physical plane is at 30° - Always divide the circle angle by 2 to get the physical angle 7. **Misidentifying the average stress:** - The center of Mohr's circle is C = (σ_x + σ_y)/2, plotted on the σ-axis at τ = 0 - On the planes of maximum shear, the normal stress is always C (the average), never σ₁ or σ₂ - This is a quick sanity check: if you found a plane with maximum shear, verify that σ = C **Exam Strategy (PRC Civil Engineer Licensure Exam):** 1. **Read the problem carefully:** - Identify what is given: σ_x, σ_y, τ_xy, or loads that generate them (M, T, P) - Identify what is asked: principal stresses, max shear, stress on a specific plane, adequacy check, diameter/section sizing - Determine which failure theory (or code) applies 2. **Organize your solution:** - **Step 1:** Calculate stress components (σ_x, σ_y, τ_xy) from the given loads - **Step 2:** Calculate the center and radius of Mohr's circle - **Step 3:** Find principal stresses and max shear (read from circle or use formulas) - **Step 4:** If asked for stress on a specific plane, use transformation equations or rotate on the circle - **Step 5:** Apply failure criterion; calculate safety factor or required size - **Step 6:** State conclusions clearly 3. **Use the graphical method (Mohr's circle) as a check:** - Sketch the circle (even if rough) to verify that principal stresses are correct - Check that max shear = radius = (σ₁ − σ₂)/2 - Confirm that inclined-plane stresses lie on the circle 4. **Present calculations with units and clear labels:** - Write σ_x = 80 MPa, not σ_x = 80 (students lose points for missing units) - Label intermediate results: "Center = 50 MPa", "Radius = 42.43 MPa", etc. - Show the final answer prominently: "Principal stresses: σ₁ = 92.4 MPa, σ₂ = 7.6 MPa" 5. **For multiple-choice questions:** - If options include "principal stresses," verify that you found σ₁ and σ₂ (not σ_x and σ_y) - If options differ by a factor of 2 or √2, suspect an angle error (2θ vs. θ) or radius calculation - Use dimensional analysis: if you calculated d for a diameter, verify that d has units of length 6. **Time management:** - Mohr's circle problems are often fast if you have the formulas memorized - Transformation equations take longer; if short on time, use the circle method - Combined bending-torsion shaft problems are standard; practice them until you can solve in < 2 minutes **Quick Reference: Key Formulas** **Stress Components (from loads):** - Axial: σ = P/A - Bending: σ = Mc/I = M/W (section modulus) - Torsion: τ = Tc/J (circular); τ = T/Ip (rectangular, polar second moment) **Transformation (Plane Stress):** - σ_x′ = (σ_x + σ_y)/2 + (σ_x − σ_y)/2 cos(2θ) + τ_xy sin(2θ) - τ_x′y′ = −(σ_x − σ_y)/2 sin(2θ) + τ_xy cos(2θ) **Principal Stresses:** - σ₁, σ₂ = (σ_x + σ_y)/2 ± √[((σ_x − σ_y)/2)² + τ_xy²] - tan(2θ_p) = 2τ_xy / (σ_x − σ_y) **Mohr's Circle:** - Center: ((σ_x + σ_y)/2, 0) - Radius: √[((σ_x − σ_y)/2)² + τ_xy²] - Principal stresses: Center ± Radius - Max shear: ± Radius **Shafts (Combined M and T):** - Bending stress: σ = 32M / (πd³) - Torsional shear: τ = 16T / (πd³) - Equivalent moment (normal-stress theory): M_e = (1/2)[M + √(M² + T²)] - Equivalent torque (shear-stress theory): T_e = √(M² + T²) **Failure Criteria (Plane Stress, σ₃ = 0):** - Maximum normal stress: max(|σ₁|, |σ₂|) ≤ σ_allowable - Maximum shear (Tresca): (σ₁ − σ₂)/2 ≤ σ_yield / 2 - von Mises: √[σ₁² − σ₁σ₂ + σ₂²] ≤ σ_yield