Detailed ExplanationCELE · Strength of MaterialsReal content

CELE Strength of Materials — Combined Stresses and Mohr's CircleDetailed Explanation

Summary Concepts Notes Revision Deep Dive Slides Flashcards Mistakes Cheat Mnemonics Templates

If the summary was not enough, this is the deep dive. Detailed explanations for Combined Stresses and Mohr's Circle in the CELE Strength of Materials context, written to turn surface familiarity into genuine understanding. Professional Regulation Commission (PRC) — Board of Civil Engineering's toughest CELE questions on this chapter are answered by the reasoning built here.

Exam context

Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Strength of Materials section sits under a "Core" weighting, and Combined Stresses and Mohr's Circle is the 6th chapter in the 8-chapter CELE Strength of Materials rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Strength of Materials.

Combined Stresses and Mohr's Circle - Detailed Explanation

In actual engineering practice — whether designing a shaft for a sugar mill in Negros, a chimney stack for a cement plant in Bulacan, or a pressurized pipeline for a water district in Cebu — structural members rarely carry a single type of load. They simultaneously experience axial forces, bending moments, shear forces, and torques. The resulting stress state at any point is a combination of normal and shear stresses acting on multiple planes. The central question for failure assessment is: on which plane does the most dangerous stress occur, and what is its magnitude? Mohr's Circle, developed by German engineer Otto Mohr in 1882, transforms this algebraic problem into elegant geometry. Mastery of combined stresses and Mohr's Circle is non-negotiable for the PRC Civil Engineer Licensure Examination — it consistently appears in both the morning (mathematics and basic engineering) and afternoon (engineering sciences) sessions. This chapter develops the theory systematically from stress superposition through plane-stress transformation to the graphical power of Mohr's Circle, with board-exam-style worked examples throughout.

Concepts

Superposition of Stresses

When multiple loads act simultaneously on a member, the stresses they produce at a point are superimposed (added algebraically), provided the material remains linearly elastic and deformations are small (the two conditions underlying Hooke's Law). This is the Principle of Superposition and is the starting point for all combined-stress problems. The most common combinations in board examinations are: 1. AXIAL LOAD + BENDING MOMENT An axial force P and a bending moment M both produce normal stresses on the cross-section. They act on the same component (σ), so they add directly: σ_total = P/A ± Mc/I where A is cross-sectional area, M is bending moment, c is distance from neutral axis to extreme fiber, and I is the moment of inertia. The sign of the axial term is positive for tension, negative for compression. The bending term is positive on the tension side and negative on the compression side. 2. BIAXIAL BENDING (UNSYMMETRIC BENDING) When moments act about both principal axes: σ = P/A ± M_x·c_y/I_x ± M_y·c_x/I_y This occurs in purlins of roof structures, spandrel beams, and corner columns. 3. BENDING + TORSION (SHAFTS) A rotating shaft subject to both a bending moment M (from transverse loads) and a torque T (from power transmission) has at its critical surface point: σ_x = 32M/(πd³) [bending normal stress] σ_y = 0 [no transverse normal stress on this face] τ_xy = 16T/(πd³) [torsional shear stress] This is the classic plane-stress state that requires transformation to find the maximum stresses. 4. DIRECT SHEAR + TORSIONAL SHEAR In structural members, direct transverse shear (V·Q/It) and torsional shear (Tc/J) are both shear components and may add or subtract depending on the point being analyzed. IMPORTANT SIGN CONVENTION: - Tensile normal stress: POSITIVE - Compressive normal stress: NEGATIVE - Shear stress on positive face in positive direction: POSITIVE - This convention must be consistent throughout a problem.

Examples

Both the axial load and bending moment produce normal stresses on the cross-section. Since they act on the same stress component (σ), they superimpose directly. The top fiber is doubly loaded in compression (both P and M compress the top), while the bottom fiber has competing effects — bending causes tension but the axial load causes compression, so the net result is reduced tension.

Scenario

A 100 mm × 200 mm rectangular timber beam (100 mm wide, 200 mm deep) carries an axial compressive load P = 50 kN and a bending moment M = 8 kN·m (causing tension at the bottom). Find the normal stresses at the top and bottom extreme fibers.

Solution

Step 1: Section properties. A = 100 × 200 = 20,000 mm² I = (100)(200³)/12 = 66.67 × 10⁶ mm⁴ c = 100 mm (distance from neutral axis to extreme fiber) Step 2: Axial stress (compressive, so negative). σ_axial = -P/A = -50,000/20,000 = -2.5 MPa Step 3: Bending stress (tension at bottom = +, compression at top = -). σ_bending = ±Mc/I = ±(8 × 10⁶)(100)/(66.67 × 10⁶) = ±12.0 MPa Step 4: Combined stresses. Top fiber: σ_top = -2.5 + (-12.0) = -14.5 MPa (compression) Bottom fiber: σ_bot = -2.5 + (+12.0) = +9.5 MPa (tension) Answer: σ_top = -14.5 MPa (C), σ_bot = +9.5 MPa (T)

At the outermost fiber of a shaft, bending produces the maximum normal stress (longitudinal direction), and torsion produces the maximum shear stress (also maximum at the surface). There is no normal stress in the transverse direction at a free surface. These three values (σ_x, σ_y, τ_xy) completely define the 2D stress state and are the starting inputs for transformation.

Scenario

A solid circular shaft (diameter d = 80 mm) is subjected to a bending moment M = 2 kN·m and a torque T = 3 kN·m. Determine σ_x, σ_y, and τ_xy at the critical point on the shaft surface.

Solution

Step 1: Bending normal stress (at outermost fiber). σ_x = 32M/(πd³) = 32(2 × 10⁶)/(π × 80³) σ_x = 64 × 10⁶ / (1,608,495) = 39.79 MPa ≈ 39.8 MPa Step 2: Transverse normal stress. σ_y = 0 (no normal stress in the transverse direction at the surface) Step 3: Torsional shear stress. τ_xy = 16T/(πd³) = 16(3 × 10⁶)/(π × 80³) τ_xy = 48 × 10⁶ / (1,608,495) = 29.84 MPa ≈ 29.8 MPa Answer: σ_x = 39.8 MPa, σ_y = 0, τ_xy = 29.8 MPa This plane-stress state is now ready for Mohr's Circle analysis.

Applications

Eccentrically loaded columns in building frames (P + M from eccentric load)
Rotating machine shafts in mills, pumps, and generators (bending + torsion)
Crane hooks and curved beams (combined axial and bending)
Spandrel beams at the edge of floor systems (biaxial bending)
Pressurized pipeline with pipe-support reactions (biaxial normal stresses + bending)
Traffic signal poles subject to wind (bending + torsion about vertical axis)

Misconceptions

WRONG: Adding bending stress and shear stress directly. Bending produces NORMAL stress; shear produces SHEAR stress — they are different components and cannot be added without transformation.
WRONG: Assuming σ_y is non-zero on a shaft surface. At the free surface of a shaft, there is no transverse normal stress (σ_y = 0).
WRONG: Using the wrong formula for shafts. Solid circular shaft: σ = 32M/(πd³), τ = 16T/(πd³). Do not mix up the 32 and 16 factors.
WRONG: Ignoring the sign of the axial load. Tension is positive, compression is negative — always establish this before computing.

Related Concepts

Axial stress (P/A)
Flexural formula (Mc/I)
Torsion formula (Tc/J)
Transverse shear stress (VQ/It)
Plane-stress transformation
Principal stresses

Common Exam Questions

Example

A 60 mm diameter solid shaft is subjected to M = 1.2 kN·m, T = 0.9 kN·m, and axial P = 20 kN (tension). Find σ_x, σ_y, and τ_xy at the critical surface point.

Approach

Identify all loads, compute each stress component separately using appropriate formulas (P/A, Mc/I, Tc/J, VQ/It), then algebraically sum like components (σ with σ, τ with τ on the same plane). State the sign convention before calculating.

Question Type

Find the stress at a specific point given combined loads

Example

A rectangular column carries an eccentric load at a given eccentricity. Find where the neutral axis crosses the section.

Approach

Set σ_total = P/A + M_x·y/I_x + M_y·x/I_y = 0 and solve for the line equation. The neutral axis shifts away from the tension side under combined loads.

Question Type

Determine the neutral axis location under eccentric loading

Key Points To Remember

Superposition applies only when the material is linearly elastic (stresses below proportional limit) and deformations are small.
Add like stress components: normal stresses (σ) add to normal stresses; shear stresses (τ) add to shear stresses on the same plane.
For axial + bending, always determine which side is tension and which is compression from the loading geometry before assigning signs.
For shaft problems: σ_y = 0 on the critical surface point because bending only produces normal stress in the bending plane; do not forget this.
The 'critical point' is where the combined stress is maximum — typically the outermost fiber where bending stress is greatest AND torsional shear is greatest.
Biaxial bending requires careful identification of which moment creates tension or compression at the point of interest.
After superposition, you have σ_x, σ_y, and τ_xy — these are the inputs to stress transformation.

Plane Stress and Stress Transformation Equations

A 'plane stress' state exists when all stresses act in one plane (say, the x-y plane) and the stress components perpendicular to this plane are zero. This is the standard model for thin plates, beam surfaces, and shaft surfaces — and it is the framework used in all Mohr's Circle analysis. At any point, the stress state is defined by three independent components: σ_x (normal stress on the x-face), σ_y (normal stress on the y-face), and τ_xy (shear stress on the x-face in the y-direction). These are the REFERENCE stresses. When you cut the element at an angle θ (measured counterclockwise from the positive x-axis), the normal and shear stresses on the inclined plane are given by the STRESS TRANSFORMATION EQUATIONS: σ_x' = (σ_x + σ_y)/2 + [(σ_x - σ_y)/2]·cos2θ + τ_xy·sin2θ τ_x'y' = -[(σ_x - σ_y)/2]·sin2θ + τ_xy·cos2θ For the perpendicular face (at angle θ + 90°), the normal stress is: σ_y' = (σ_x + σ_y)/2 - [(σ_x - σ_y)/2]·cos2θ - τ_xy·sin2θ Note that σ_x' + σ_y' = σ_x + σ_y (sum of normal stresses is invariant — it does not change with rotation). The PHYSICAL MEANING of these equations: - The normal stress on any inclined plane oscillates between maximum and minimum values as θ varies. - The shear stress is zero on two planes (the principal planes) and maximum on two other planes (45° away from principal planes). - Both σ and τ are sinusoidal functions of 2θ, which is why they trace a perfect circle on a (σ, τ) plot — Mohr's Circle. KEY PARAMETER — AVERAGE NORMAL STRESS: σ_avg = (σ_x + σ_y)/2 This is the center of Mohr's Circle, and it equals the normal stress on the maximum-shear planes. KEY PARAMETER — RADIUS: R = √[(σ_x - σ_y)²/4 + τ_xy²] This is the amplitude of stress variation, equal to both the maximum in-plane shear stress and the radius of Mohr's Circle.

Examples

The transformation equations directly give the stress components on any arbitrarily oriented plane. The negative sign of τ_x'y' indicates the shear acts in the negative y' direction on the positive x'-face. The sanity check confirms the result lies on Mohr's Circle (distance from center equals radius).

Scenario

At a point in a stressed body, σ_x = 60 MPa, σ_y = -20 MPa, τ_xy = 40 MPa. Find the normal and shear stresses on a plane inclined θ = 30° counterclockwise from the x-face.

Solution

Given: σ_x = 60, σ_y = -20, τ_xy = 40, θ = 30°, so 2θ = 60°. cos60° = 0.500, sin60° = 0.866 Step 1: Compute intermediate values. (σ_x + σ_y)/2 = (60 + (-20))/2 = 40/2 = 20 MPa (σ_x - σ_y)/2 = (60 - (-20))/2 = 80/2 = 40 MPa Step 2: Normal stress on inclined plane. σ_x' = 20 + 40(0.500) + 40(0.866) σ_x' = 20 + 20 + 34.64 = 74.64 MPa Step 3: Shear stress on inclined plane. τ_x'y' = -40(0.866) + 40(0.500) τ_x'y' = -34.64 + 20.00 = -14.64 MPa Answer: σ_x' = 74.6 MPa (tension), τ_x'y' = -14.6 MPa Sanity check using circle: R = √(40² + 40²) = 56.57 MPa (74.6 - 20)² + (-14.6)² = 54.6² + 14.6² = 2981 + 213 = 3194 ≈ 56.57² = 3200 ✓

Applications

Finding stresses on failure planes (e.g., concrete failing in diagonal tension at 45°)
Determining stress on welded joints inclined to the member axis
Analyzing stress on planes of known weak orientation (grain direction in wood, bedding planes in rock)
Computing stresses at any orientation in pressure vessel walls
Predicting the direction of crack propagation in brittle materials

Misconceptions

WRONG: Using θ instead of 2θ in the transformation equations. The formulas use cos2θ and sin2θ — DOUBLE the angle before taking trig functions.
WRONG: Thinking the transformation only works for certain orientations. The equations are valid for ANY angle θ without restriction.
WRONG: Assuming σ_x' + τ_x'y' is constant (it is not). What is constant is σ_x' + σ_y' = σ_x + σ_y.
WRONG: Forgetting the negative sign in the shear transformation equation. The shear stress formula starts with -[(σ_x - σ_y)/2]·sin2θ — the leading negative sign is critical.

Related Concepts

Superposition of stresses
Principal stresses and principal planes
Maximum shear stress
Mohr's Circle construction
Equilibrium of a wedge element

Common Exam Questions

Example

Given σ_x = 80 MPa, σ_y = 40 MPa, τ_xy = -30 MPa. Find σ and τ on the plane at θ = 45°.

Approach

Substitute σ_x, σ_y, τ_xy, and the given angle θ directly into the transformation equations. Compute cos2θ and sin2θ carefully. Apply formula step by step.

Question Type

Compute stresses on a specific inclined plane

Example

Is the stress state σ = 65 MPa, τ = 20 MPa consistent with σ_x = 80, σ_y = 20, τ_xy = 30 MPa?

Approach

Check whether (σ - σ_avg)² + τ² = R². If this is satisfied, the stress state is consistent with the given reference stresses.

Question Type

Verify if a point lies on Mohr's Circle

Key Points To Remember

The angle θ in the physical element corresponds to 2θ on Mohr's Circle — always double the physical angle for the circle, halve the circle angle for the physical plane.
The transformation equations preserve the sum of normal stresses: σ_x + σ_y = σ_x' + σ_y' = constant (stress invariant).
The transformation equations were derived from equilibrium of a wedge element — no material properties are involved, so they apply to all materials.
On the principal planes (where τ = 0): the normal stresses are the principal stresses σ₁ and σ₂.
On the maximum shear stress planes: the shear stress equals R and the normal stress equals σ_avg.
The shear stress on two perpendicular faces are equal in magnitude but may differ in sign: τ_x'y' on one face equals -τ_y'x' on the adjacent face.
Counterclockwise rotation of the element = counterclockwise rotation on Mohr's Circle (if using the convention where positive τ is plotted downward for the x-face).

Principal Stresses and Maximum Shear Stress

The 'principal stresses' are the maximum and minimum normal stresses at a point, occurring on planes called 'principal planes.' On principal planes, the shear stress is identically ZERO. This is the most critical result in combined stress analysis because failure theories (used in design codes) are expressed in terms of principal stresses. PRINCIPAL STRESSES: Setting τ_x'y' = 0 in the transformation equation and solving gives two angles 2θ_p separated by 180° (i.e., two physical planes 90° apart): tan(2θ_p) = 2τ_xy / (σ_x - σ_y) Substituting back into the normal stress equation gives the principal stresses: σ₁,₂ = (σ_x + σ_y)/2 ± √[(σ_x - σ_y)²/4 + τ_xy²] Where σ₁ ≥ σ₂ by convention (σ₁ is the algebraically larger, σ₂ the smaller). MAXIMUM IN-PLANE SHEAR STRESS: Setting dτ/dθ = 0 gives the maximum shear stress planes, oriented 45° from the principal planes: τ_max = √[(σ_x - σ_y)²/4 + τ_xy²] = (σ₁ - σ₂)/2 On the max-shear planes, the normal stress is NOT zero — it equals: σ on max-shear plane = σ_avg = (σ_x + σ_y)/2 = (σ₁ + σ₂)/2 ABSOLUTE MAXIMUM SHEAR STRESS (3D consideration): For a biaxial stress state (plane stress), there are actually three principal stresses. The third one (perpendicular to the plane) is σ₃ = 0 for plane stress. The absolute maximum shear stress considers all three principal stresses: τ_abs_max = (σ_max - σ_min)/2 where σ_max and σ_min are the algebraically largest and smallest of σ₁, σ₂, and σ₃ = 0. This matters when both σ₁ and σ₂ are positive (or both negative): the in-plane τ_max = (σ₁ - σ₂)/2, but the absolute max shear = σ₁/2 (using σ₃ = 0 as the minimum). For the PRC board exam, unless specifically asked for the absolute maximum, use the in-plane maximum shear. INVARIANTS: Two combinations of stress components do not change with rotation: I₁ = σ_x + σ_y = σ₁ + σ₂ (first invariant) I₂ = σ_x·σ_y - τ_xy² = σ₁·σ₂ (second invariant) These are useful for checking answers.

Examples

The radius formula combines both the 'spread' of normal stresses and the shear stress into a single measure. Note that when (σ_x - σ_y)/2 equals τ_xy (both are 30 MPa here), the angle works out to exactly 45°. The verification using the sum invariant is a quick check that should always be done in the board exam.

Scenario

At a point, σ_x = 80 MPa, σ_y = 20 MPa, τ_xy = 30 MPa. Find the principal stresses, maximum in-plane shear stress, and the angle of the principal planes.

Solution

Step 1: Average stress and radius. σ_avg = (80 + 20)/2 = 50 MPa R = √[(80-20)²/4 + 30²] = √[30² + 30²] = √[900 + 900] = √1800 = 42.43 MPa Step 2: Principal stresses. σ₁ = 50 + 42.43 = 92.43 MPa ≈ 92.4 MPa σ₂ = 50 - 42.43 = 7.57 MPa ≈ 7.6 MPa Step 3: Maximum in-plane shear stress. τ_max = R = 42.4 MPa Step 4: Principal plane angle. tan(2θ_p) = 2(30)/(80-20) = 60/60 = 1.000 2θ_p = arctan(1.000) = 45° θ_p = 22.5° Step 5: Verification. σ₁ + σ₂ = 92.4 + 7.6 = 100 MPa = σ_x + σ_y = 80 + 20 = 100 MPa ✓ Answer: σ₁ = 92.4 MPa, σ₂ = 7.6 MPa, τ_max = 42.4 MPa, θ_p = 22.5°

Pure shear is equivalent to equal biaxial tension and compression at 45°. This fundamental result explains diagonal tension cracking in concrete beams (the concrete fails in tension at 45° due to shear), and the 45° helical fracture of cast iron shafts in torsion. The board exam frequently tests this concept.

Scenario

A point is in PURE SHEAR: σ_x = σ_y = 0, τ_xy = 50 MPa. Find principal stresses and their orientation.

Solution

Step 1: Average stress. σ_avg = (0 + 0)/2 = 0 MPa Step 2: Radius. R = √[0² + 50²] = 50 MPa Step 3: Principal stresses. σ₁ = 0 + 50 = +50 MPa (tension) σ₂ = 0 - 50 = -50 MPa (compression) Step 4: Angle. tan(2θ_p) = 2(50)/(0 - 0) → undefined → 2θ_p = 90° → θ_p = 45° Answer: σ₁ = 50 MPa (tension at 45°), σ₂ = -50 MPa (compression at 135°) This is WHY chalk sticks break at 45° when twisted, and why torsion specimens of brittle material fail on a 45° helix!

When both principal stresses are the same sign (here both positive), the out-of-plane direction (σ₃ = 0) is the true minimum, making the absolute maximum shear much larger than the in-plane value. This is critical in pressure vessel design where both hoop and longitudinal stresses are tensile.

Scenario

A biaxial stress state has σ_x = 120 MPa, σ_y = 60 MPa, τ_xy = 0. Find σ₁, σ₂, τ_max (in-plane), and τ_abs_max.

Solution

Since τ_xy = 0, the reference axes are already principal axes. σ₁ = 120 MPa, σ₂ = 60 MPa (direct reading) τ_max (in-plane) = (120 - 60)/2 = 30 MPa For absolute maximum shear (including σ₃ = 0 for plane stress): All three principal stresses: 120, 60, 0 MPa τ_abs_max = (120 - 0)/2 = 60 MPa IMPORTANT: The absolute max shear (60 MPa) is DOUBLE the in-plane max shear (30 MPa) because σ₃ = 0 is far from both σ₁ and σ₂.

Applications

Design of structural steel members using AISC 360 (von Mises or Tresca criteria for yielding)
Failure analysis of concrete in shear (diagonal tension is principal tensile stress at ~45°)
Shaft design for combined bending and torsion in machinery
Pressure vessel design — checking for yielding under biaxial tension
Soil mechanics — Mohr-Coulomb failure criterion uses principal stresses
Bolt and connection design where combined tension and shear exist

Misconceptions

WRONG: τ_max = σ₁/2 always. Correct: τ_max = (σ₁ - σ₂)/2. Only if σ₂ = 0 does τ_max = σ₁/2.
WRONG: The maximum shear plane has zero normal stress. CORRECT: The normal stress on the maximum shear plane is σ_avg = (σ₁ + σ₂)/2.
WRONG: Both principal stresses are always positive. They can both be negative, or one positive and one negative.
WRONG: tan(2θ_p) gives only one angle. There are always TWO principal planes, 90° apart physically (180° apart on the circle).
WRONG: For plane stress, τ_abs_max always equals τ_max (in-plane). When σ₁ and σ₂ have the same sign, τ_abs_max = max(|σ₁|, |σ₂|)/2, which may be larger than the in-plane τ_max.

Related Concepts

Stress transformation equations
Mohr's Circle
Failure theories (Tresca, von Mises, Rankine)
Combined bending and torsion of shafts
Pressure vessel stresses
Mohr-Coulomb criterion in soil mechanics

Common Exam Questions

Example

σ_x = -40 MPa, σ_y = 60 MPa, τ_xy = 25 MPa. Find σ₁ and σ₂.

Approach

Apply σ₁,₂ = σ_avg ± R directly. Compute σ_avg = (σ_x + σ_y)/2 first, then R = √[(σ_x-σ_y)²/4 + τ_xy²]. This is the most direct computational approach without drawing the circle.

Question Type

Find principal stresses from given σ_x, σ_y, τ_xy

Example

With σ_x = 100, σ_y = 40, τ_xy = -30 MPa: what is the angle of the plane on which σ₁ acts?

Approach

Use tan(2θ_p) = 2τ_xy/(σ_x - σ_y). Be careful with the quadrant of 2θ_p. Substitute the angle back into the transformation equation to confirm which angle gives σ₁ (the larger value).

Question Type

Find angle of principal planes

Example

Find τ_max and the normal stress on the maximum shear plane for the state σ_x = 50, σ_y = -30, τ_xy = 40 MPa.

Approach

τ_max = R (the radius). Normal stress on max-shear plane = σ_avg (the center). These two values are directly read from Mohr's Circle geometry.

Question Type

Determine maximum in-plane shear stress and the normal stress on that plane

Key Points To Remember

Principal stresses occur on planes where shear stress is ZERO — use this to verify: if the answer gives a plane with τ ≠ 0, it is not a principal plane.
σ₁ ≥ σ₂ always; σ₁ is the algebraically greater (more positive or less negative) principal stress.
τ_max = R = (σ₁ - σ₂)/2 — the maximum shear stress equals the radius of Mohr's Circle.
Maximum shear stress planes are exactly 45° (physically) from principal planes — not 90°.
The normal stress on maximum-shear planes is σ_avg = (σ_x + σ_y)/2, NOT zero.
tan(2θ_p) = 2τ_xy/(σ_x - σ_y) gives TWO solutions for 2θ_p, 180° apart — corresponding to the two principal planes perpendicular to each other.
For checking: σ₁ + σ₂ must equal σ_x + σ_y (sum invariant).
If σ_x = σ_y and τ_xy = 0: the point is in hydrostatic stress (all planes have same normal stress, zero shear) — Mohr's Circle degenerates to a point.

Mohr's Circle — Construction and Application

Mohr's Circle is a graphical representation of the stress transformation equations. Every possible stress state (σ, τ) on any inclined plane through a point lies on a circle in the (σ, τ) coordinate system. It is not merely a diagram — it is a complete, rigorous solution method. CONVENTION (most common in Philippine review books): - Horizontal axis (x-axis): Normal stress σ (positive = right = tension) - Vertical axis (y-axis): Shear stress τ (positive = downward for the x-face; some books use upward — be consistent with your reference) STEP-BY-STEP CONSTRUCTION: Step 1: Plot point X representing the x-face of the element. X = (σ_x, τ_xy) — use τ_xy as plotted (positive τ_xy plotted downward in the CW-positive convention) Step 2: Plot point Y representing the y-face of the element. Y = (σ_y, -τ_xy) — note the sign reversal for shear on the y-face Step 3: Draw the diameter XY. The midpoint of XY is the center C. C = ((σ_x + σ_y)/2, 0) — always on the σ-axis Step 4: Draw the circle with center C and radius R = distance from C to X (or C to Y). R = CX = CY = √[(σ_x - σ_y)²/4 + τ_xy²] READING THE CIRCLE: - Points where circle crosses σ-axis: principal stresses (τ = 0 there) Right intersection: σ₁ = C + R Left intersection: σ₂ = C - R - Top/bottom of circle: maximum in-plane shear stress = ±R At top (or bottom), σ = C = σ_avg - Rotation on circle: A rotation of 2θ on the circle = rotation of θ on the physical element Counterclockwise on circle = counterclockwise rotation of physical element (in most conventions) FINDING STRESS ON AN INCLINED PLANE: - Starting from point X (representing the x-face), rotate 2θ counterclockwise on the circle - The new point gives (σ_x', τ_x'y') on the inclined plane IMPORTANT PROPERTIES: - The center is ALWAYS on the σ-axis (τ = 0 at center) - The radius R equals τ_max (maximum in-plane shear) - σ₁ + σ₂ = 2C = σ_x + σ_y (invariant — circle center does not move with rotation) - Principal planes are found by rotating from X to the σ-axis intersection: the angle on the circle is 2θ_p, so physical angle is θ_p

Examples

The graphical procedure gives identical results to the equations because the circle IS the geometric representation of those equations. In the exam, you can either use the equations directly (faster for calculation) or draw a quick sketch of Mohr's Circle (better for understanding and verification). Always check the sum invariant as a final verification.

Scenario

Construct Mohr's Circle for σ_x = 80 MPa, σ_y = 20 MPa, τ_xy = 30 MPa. Find all principal values and the angle of principal planes.

Solution

Step 1: Locate points. X = (80, 30) — x-face: σ_x = 80, τ_xy = 30 Y = (20, -30) — y-face: σ_y = 20, -τ_xy = -30 Step 2: Center. C = ((80+20)/2, 0) = (50, 0) Step 3: Radius. R = √[(80-50)² + 30²] = √[900 + 900] = √1800 = 42.43 MPa Verify: CY = √[(20-50)² + (-30)²] = √[900+900] = 42.43 ✓ Step 4: Principal stresses (σ-axis intercepts). σ₁ = C + R = 50 + 42.43 = 92.43 MPa σ₂ = C - R = 50 - 42.43 = 7.57 MPa Step 5: Maximum shear stress (top/bottom of circle). τ_max = R = 42.43 MPa σ on max-shear plane = C = 50 MPa Step 6: Angle from X to σ₁ on circle. From center to X: horizontal component = 80-50 = 30, vertical = 30 Angle of CX from σ-axis: arctan(30/30) = 45° (this is 2θ_p) θ_p = 22.5° (counterclockwise from x-face to σ₁ plane) Check: σ₁ + σ₂ = 92.43 + 7.57 = 100 = 80 + 20 ✓

This example shows that when τ_xy is negative, care must be taken in locating X on the circle. The equation method is more reliable computationally; the circle provides geometric insight. For board exam accuracy, use equations and verify with the circle (checking that the result point satisfies the circle equation).

Scenario

Using Mohr's Circle, find the stress components on a plane at θ = 30° CCW from the x-face for: σ_x = 100 MPa, σ_y = 40 MPa, τ_xy = -20 MPa.

Solution

Step 1: Center and radius. C = (100+40)/2 = 70 MPa R = √[(100-40)²/4 + (-20)²] = √[30² + 20²] = √[900+400] = √1300 = 36.06 MPa Step 2: Locate X on circle. X = (100, -20) — note: τ_xy is negative, so on circle τ for x-face = -20 Angle of CX from positive σ-axis: CX horizontal: 100-70=30, CX vertical: -20 Angle of X below σ-axis: arctan(20/30) = 33.69° (below axis, so clockwise from σ-axis) Step 3: Rotate 2θ = 60° CCW from X on the circle. New angle from σ-axis: -33.69° + 60° = 26.31° above σ-axis Step 4: New point coordinates. σ_x' = 70 + 36.06·cos(26.31°) = 70 + 36.06(0.8962) = 70 + 32.3 = 102.3 MPa τ_x'y' = 36.06·sin(26.31°) = 36.06(0.4435) = 16.0 MPa (above axis → negative by CW convention) Verify with equations: 2θ = 60° σ_x' = 70 + 30(0.5) + (-20)(0.866) = 70 + 15 - 17.32 = 67.68 MPa... [Note: recheck: (σ_x-σ_y)/2 = 30, τ_xy = -20] σ_x' = 70 + 30cos60° + (-20)sin60° = 70 + 15 - 17.32 = 67.68 MPa τ_x'y' = -30sin60° + (-20)cos60° = -25.98 - 10 = -35.98 MPa Answer via equations: σ_x' = 67.7 MPa, τ_x'y' = -36.0 MPa

Applications

Quick graphical solution for principal stresses without algebraic computation
Determining stress on any arbitrarily inclined plane (e.g., inclined weld, joint plane)
Visualizing the range of possible normal and shear stresses at a point
Checking consistency of stress components — any valid stress state must lie on the circle
Foundation for understanding Mohr-Coulomb failure in soil mechanics
Failure analysis — locating the most critical plane for material failure

Misconceptions

WRONG: The center of Mohr's Circle can be off the σ-axis. The center is ALWAYS at (σ_avg, 0) — on the σ-axis with zero shear.
WRONG: Point X is plotted as (σ_x, -τ_xy). Point X = (σ_x, +τ_xy) in the convention where τ_xy > 0 on the positive x-face is plotted downward. The y-face gets the reversed sign: Y = (σ_y, -τ_xy).
WRONG: Rotating 30° on the circle gives the stress on the 30° plane. Rotating 30° on the CIRCLE corresponds to a 15° physical rotation. Always halve the circle angle.
WRONG: Mohr's Circle gives principal stresses at the top and bottom of the circle. Principal stresses are at the LEFT and RIGHT intercepts (where τ = 0). Maximum shear is at top and bottom.

Related Concepts

Principal stresses
Maximum shear stress
Plane-stress transformation equations
Failure theories
Three-dimensional stress state
Mohr-Coulomb failure criterion in geotechnical engineering

Common Exam Questions

Example

For σ_x = -30, σ_y = 70, τ_xy = 40 MPa: find the center and radius of Mohr's Circle.

Approach

Center = ((σ_x + σ_y)/2, 0). Radius = √[(σ_x - σ_y)²/4 + τ_xy²]. These are the two most fundamental quantities — compute them first in every problem.

Question Type

Identify center and radius of Mohr's Circle

Example

Which Mohr's Circle correctly represents σ_x = 60, σ_y = -20, τ_xy = 30 MPa? [Given four circles with different centers and radii]

Approach

Compute center and radius numerically, then check which option matches. Common traps: options with wrong center position (center is on σ-axis, not τ-axis), wrong radius, or principal stresses that don't satisfy the sum invariant.

Question Type

Multiple choice identifying the correct Mohr's Circle

Key Points To Remember

Mohr's Circle has center at (σ_avg, 0) and radius R = τ_max — memorize this pair.
PHYSICAL angle θ corresponds to DOUBLE angle 2θ on Mohr's Circle — the most common source of error.
Point X always represents the x-face: (σ_x, τ_xy). Point Y represents the y-face: (σ_y, −τ_xy). The negative sign for Y is automatic from the equilibrium of the element.
Principal stresses are at the LEFT and RIGHT intercepts of the circle with the σ-axis.
Maximum shear stress points are at the TOP and BOTTOM of the circle, where σ = σ_avg.
For pure shear (σ_x = σ_y = 0): center is at origin, radius equals τ_xy, principal stresses are ±τ_xy at 45°.
The absolute maximum shear for plane stress may involve σ₃ = 0 — always check if both σ₁ and σ₂ have the same sign.
Mohr's Circle is valid for any linearly elastic material — the geometry depends only on equilibrium, not material properties.

Combined Bending and Torsion — Equivalent Moments

When a circular shaft carries both a bending moment M and a torque T simultaneously, the critical stress state at the shaft surface can be characterized by equivalent moments that simplify the design process. This approach consolidates the Mohr's Circle analysis into compact formulas. STRESS STATE AT CRITICAL POINT: At the outermost fiber of a shaft under combined M and T: σ_x = 32M/(πd³) [bending normal stress, longitudinal] σ_y = 0 [free surface: no transverse normal stress] τ_xy = 16T/(πd³) [torsional shear stress] APPLYING PRINCIPAL STRESS FORMULA: σ_avg = σ_x/2 = 16M/(πd³) R = √[(σ_x/2)² + τ_xy²] = √[(16M/πd³)² + (16T/πd³)²] R = (16/πd³)√(M² + T²) σ₁ = σ_avg + R = (16/πd³)[M + √(M² + T²)] σ₂ = σ_avg - R = (16/πd³)[M - √(M² + T²)] τ_max = R = (16/πd³)√(M² + T²) EQUIVALENT MOMENT (Maximum Normal Stress Theory — Rankine, for brittle materials): Express σ₁ in terms of a single 'equivalent moment' M_e: σ₁ = 32M_e/(πd³) Therefore: M_e = (1/2)[M + √(M² + T²)] EQUIVALENT TORQUE (Maximum Shear Stress Theory — Tresca, for ductile materials): Express τ_max in terms of an 'equivalent torque' T_e: τ_max = 16T_e/(πd³) Therefore: T_e = √(M² + T²) DESIGN APPROACH: - For brittle shaft materials (cast iron): use M_e → σ₁ = S_ut (ultimate tensile strength) - For ductile shaft materials (steel): use T_e → τ_max = S_sy = 0.5·S_y (Tresca yield shear) IMPORTANT: These formulas apply to SOLID CIRCULAR shafts. For hollow shafts, replace the diameter formulas with the appropriate section modulus (using J/c_outer for torsion, I/c for bending). The formulas for hollow circular shaft of outer diameter D and inner diameter d: σ = M·c/(I) where I = π(D⁴ - d⁴)/64, c = D/2 τ = T·c/(J) where J = π(D⁴ - d⁴)/32, c = D/2

Examples

The equivalent moment approach avoids going through the full stress transformation by pre-packaging the M and T combination into a single equivalent quantity. The 3-4-5 right triangle relationship (M=1.5, T=2.0, T_e=2.5) is a common board exam setup — recognize Pythagorean triplets (3:4:5 scaled by 0.5).

Scenario

A 60 mm solid circular shaft is subjected to a bending moment M = 1.5 kN·m and a torque T = 2.0 kN·m. Find (a) the maximum normal stress σ₁ using the equivalent moment, and (b) the maximum shear stress using the equivalent torque.

Solution

Given: d = 60 mm, M = 1.5 kN·m = 1.5 × 10⁶ N·mm, T = 2.0 kN·m = 2.0 × 10⁶ N·mm Step 1: Equivalent torque. T_e = √(M² + T²) = √(1.5² + 2.0²) = √(2.25 + 4.00) = √6.25 = 2.5 kN·m = 2.5 × 10⁶ N·mm Step 2: Equivalent moment. M_e = (1/2)(M + T_e) = (1/2)(1.5 + 2.5) = (1/2)(4.0) = 2.0 kN·m = 2.0 × 10⁶ N·mm Step 3: Maximum normal stress. σ₁ = 32·M_e/(πd³) = 32(2.0 × 10⁶) / [π(60)³] = 64 × 10⁶ / [π × 216,000] = 64 × 10⁶ / 678,584 = 94.3 MPa Step 4: Maximum shear stress. τ_max = 16·T_e/(πd³) = 16(2.5 × 10⁶) / [π(60)³] = 40 × 10⁶ / 678,584 = 58.9 MPa Verification: τ_max = σ₁/2 (should be true when σ_y = 0)? σ₁/2 = 94.3/2 = 47.15 ≠ 58.9 MPa — NOT equal. [This shows τ_max ≠ σ₁/2 in general; only if M = 0 does this hold] Correct check: τ_max = (σ₁ - σ₂)/2 σ₂ = σ_avg - R = 16M/(πd³) - τ_max = [32(1.5)/π(60³)] - 58.9 = 56.6 - 58.9 = -2.3 MPa... hmm, let me recompute. σ_avg = 16(1.5×10⁶)/678,584 = 24×10⁶/678,584 = 35.4 MPa σ₁ = 35.4 + 58.9 = 94.3 MPa ✓ σ₂ = 35.4 - 58.9 = -23.5 MPa (σ₁-σ₂)/2 = (94.3-(-23.5))/2 = 117.8/2 = 58.9 MPa ✓

Shaft design problems always require rounding UP to the next commercially available standard diameter. The Pythagorean triplet recognition (here 2.7, 3.6, 4.5 = 0.9 × 3, 4, 5) speeds up computation significantly in board exams where time is limited.

Scenario

A solid shaft must transmit a torque T = 3.6 kN·m with a simultaneous bending moment M = 2.7 kN·m. If the allowable shear stress is τ_allow = 80 MPa, determine the minimum required shaft diameter.

Solution

Step 1: Equivalent torque. T_e = √(M² + T²) = √(2.7² + 3.6²) = √(7.29 + 12.96) = √20.25 = 4.5 kN·m [Note: 2.7-3.6-4.5 is the 3-4-5 triplet scaled by 0.9] Step 2: Apply shear stress formula with T_e. τ_max = 16·T_e / (π·d³) ≤ τ_allow 80 = 16(4.5 × 10⁶) / (π·d³) Step 3: Solve for d. π·d³ = 16(4.5 × 10⁶) / 80 = 72 × 10⁶ / 80 = 900,000 mm³ d³ = 900,000 / π = 286,479 mm³ d = (286,479)^(1/3) = 65.9 mm Use d = 70 mm (next standard size up). Answer: Minimum required diameter = 70 mm (using standard size)

Applications

Design of power transmission shafts in pumps, motors, and compressors
Crankshaft design in engines (combined bending from combustion loads + torsion from output)
Drive shaft design for vehicles and industrial machinery
Axle design for rotating equipment
Propeller shafts on marine vessels

Misconceptions

WRONG: M_e = √(M² + T²). This is T_e, not M_e. M_e = (M + √(M²+T²))/2. The equivalent TORQUE equals √(M²+T²); the equivalent MOMENT is half the sum of M and T_e.
WRONG: Using π/16 × d³ for the section modulus in bending. For bending in a solid circular shaft: section modulus Z = π·d³/32 (not π/16).
WRONG: Neglecting σ_y = 0 and assuming a biaxial normal stress state. At the free surface of a shaft, there is no transverse normal stress.
WRONG: Adding bending stress and torsional shear stress directly as if they were the same type. They must first be analyzed as σ_x and τ_xy before applying transformation.

Related Concepts

Torsion formula (Tc/J)
Flexural formula (Mc/I)
Principal stresses
Maximum shear stress
Mohr's Circle
Failure theories (Rankine, Tresca)
Shaft design for power transmission

Common Exam Questions

Example

A 75 mm diameter solid steel shaft carries M = 2 kN·m and T = 2 kN·m. Find σ₁ and τ_max.

Approach

Compute T_e = √(M²+T²) for shear; M_e = (M + T_e)/2 for normal stress. Then use 16T_e/(πd³) or 32M_e/(πd³). Watch units — convert kN·m to N·mm (multiply by 10⁶).

Question Type

Find maximum stress in shaft under M and T

Example

Find the minimum shaft diameter for M = 1 kN·m, T = 1.5 kN·m, with τ_allow = 60 MPa.

Approach

Set τ_max = τ_allow or σ₁ = σ_allow, substitute M_e or T_e, and solve for d. Always round UP to standard size and state which failure theory was used.

Question Type

Determine required shaft diameter

Key Points To Remember

M_e = (1/2)[M + √(M² + T²)] — equivalent moment for normal stress (MNST theory).
T_e = √(M² + T²) — equivalent torque for shear stress (MSST/Tresca theory). Note: T_e = 2·R·(πd³/16).
T_e ≥ M_e always (since T_e includes the full expression under the root and M_e only takes half). Use T_e for ductile steel shafts, M_e for brittle shafts.
The critical point is at the surface of the shaft where both bending stress and torsional shear are maximum (at the outermost fiber on the tension side of bending).
For shafts: σ_y = 0 — NEVER add a second normal stress unless explicitly told there is an axial load.
If the shaft also has an axial load P: add P/A to σ_x before applying the Mohr's Circle analysis.
The √(M² + T²) term represents the vector resultant of bending and torsion on the stress circle — similar to a vector addition in mechanics.

Failure Theories

Once principal stresses are determined, failure theories predict whether a material will yield or fracture under that combined stress state. This is the link between stress analysis and engineering design. The PRC board exam most commonly tests Tresca and von Mises for ductile materials. 1. MAXIMUM NORMAL STRESS THEORY (Rankine Theory — Brittle Materials) Failure occurs when the largest principal stress reaches the ultimate tensile strength: σ₁ = S_ut (tensile failure) |σ₂| = S_uc (compressive failure, where S_uc is ultimate compressive strength) Applied to: Cast iron, concrete (in compression), ceramics, rock. Limitation: Overestimates strength of ductile materials in combined stress. 2. MAXIMUM SHEAR STRESS THEORY (Tresca Theory — Ductile Materials) Failure occurs when the maximum shear stress reaches the shear yield strength: τ_max = S_sy = S_y/2 Where S_y is the uniaxial tensile yield strength. For design: τ_max ≤ S_y/(2·FS) In terms of principal stresses: (σ₁ - σ₂) = S_y (for plane stress with opposite-sign principals) Conservative (safe side) — preferred for structural safety. 3. DISTORTION ENERGY THEORY (von Mises Theory — Ductile Materials) Failure occurs when the von Mises equivalent stress reaches yield: σ_v = √(σ₁² - σ₁σ₂ + σ₂²) = S_y In terms of σ_x, σ_y, τ_xy: σ_v = √(σ_x² - σ_x·σ_y + σ_y² + 3τ_xy²) For design: σ_v ≤ S_y/FS Most accurate for ductile metals — predicts about 15% higher strength than Tresca. Used in AISC 360 for steel design (through the √3 interaction in shear). COMPARISON: - Rankine: Simplest, for brittle materials - Tresca: Conservative (lower predicted strength), used when safety is paramount - Von Mises: More accurate for ductile materials, used in most modern design codes - For the PRC board exam: Know all three formulas; identify which theory applies from the problem context (brittle vs. ductile, or as specified) FACTOR OF SAFETY: Design equations incorporate a factor of safety (FS): Tresca: τ_max ≤ S_y/(2·FS) or (σ₁-σ₂) ≤ S_y/FS Von Mises: σ_v ≤ S_y/FS

Examples

The difference between Tresca (FS = 1.56) and von Mises (FS = 1.76) represents the inherent conservatism of Tresca. When σ₁ and σ₂ have opposite signs (as they do in torsion-dominated states), Tresca is significantly more conservative than von Mises. For the board exam, note which theory is specified; if not, use Tresca for conservative design of ductile steel.

Scenario

A point in a steel shaft has principal stresses σ₁ = 120 MPa and σ₂ = -60 MPa. The material yield strength is S_y = 280 MPa. Check yielding by (a) Tresca and (b) von Mises theories.

Solution

(a) TRESCA: Since σ₁ > 0 and σ₂ < 0 (opposite signs), Tresca criterion: σ₁ - σ₂ = 120 - (-60) = 180 MPa < S_y = 280 MPa ✓ (No yielding) Factor of safety = 280/180 = 1.56 (b) VON MISES: σ_v = √(σ₁² - σ₁σ₂ + σ₂²) = √(120² - (120)(-60) + (-60)²) = √(14,400 + 7,200 + 3,600) = √25,200 = 158.7 MPa < 280 MPa ✓ (No yielding) Factor of safety = 280/158.7 = 1.76 Conclusion: Both theories predict no yielding, but von Mises gives a higher (less conservative) factor of safety.

Applications

Structural steel member design per AISC 360 (von Mises through interaction equations)
Pressure vessel design under biaxial stress
Machine shaft design for fatigue and static loading
Concrete shear design in NSCP 2015 (diagonal tension = principal tensile stress)
Geotechnical stability analysis using Mohr-Coulomb (a friction-based Rankine variant)

Misconceptions

WRONG: Von Mises is always more conservative than Tresca. CORRECT: Tresca is always more conservative (predicts yielding at lower stress levels).
WRONG: τ_yield = S_y (same as tensile yield). Tresca: τ_y = S_y/2. Von Mises: τ_y = S_y/√3. Both are LESS than S_y.
WRONG: Applying Rankine theory to ductile steel. Rankine is appropriate for brittle materials. For ductile steel, use Tresca or von Mises.
WRONG: (σ₁ - σ₂) = S_y/2 in Tresca. CORRECT: τ_max = S_y/2 → (σ₁ - σ₂)/2 = S_y/2 → (σ₁ - σ₂) = S_y.

Related Concepts

Principal stresses
Maximum shear stress
Material yield strength
Factor of safety
Mohr's Circle
Combined bending and torsion

Common Exam Questions

Example

Given σ₁ = 90 MPa, σ₂ = 50 MPa, S_y = 200 MPa. Does the material yield per Tresca? Per von Mises?

Approach

Compute principal stresses first. For Tresca: check (σ₁ - σ₂) vs. S_y when opposite signs; σ₁ vs. S_y when same sign. For von Mises: compute σ_v = √(σ₁²-σ₁σ₂+σ₂²) and compare to S_y.

Question Type

Apply a given failure theory to determine if yielding occurs

Example

A shaft with M = P·L and T = constant: find the maximum P before yielding by Tresca.

Approach

Express principal stresses in terms of the unknown load, set the failure criterion equal to S_y, and solve for the load. This often combines combined stress analysis with failure theory in one problem.

Question Type

Find the maximum load before yielding using a failure theory

Key Points To Remember

Rankine (max normal stress) → brittle materials; Tresca (max shear stress) → ductile, conservative; von Mises (distortion energy) → ductile, more accurate.
Tresca criterion: (σ₁ - σ₂) = S_y when σ₁ and σ₂ have opposite signs; σ₁ = S_y when σ₁ and σ₂ have the same sign (and σ₃=0 governs).
Von Mises: σ_v = √(σ₁² - σ₁σ₂ + σ₂²). For pure shear (σ₁ = -σ₂ = τ): σ_v = τ√3, so shear yield = S_y/√3 ≈ 0.577·S_y.
Tresca predicts shear yield = S_y/2 = 0.5·S_y; von Mises predicts 0.577·S_y. Tresca is more conservative (lower limit).
For concrete design in the Philippines (NSCP 2015, based on ACI 318): shear capacity is linked to diagonal tension, which is a principal tensile stress concept — related to the Rankine criterion.
Factor of safety must be applied to the MATERIAL STRENGTH, not to the applied stresses, in the standard formulation.

Practice Problems

This problem tests all four fundamental outputs of Mohr's Circle analysis simultaneously. The key steps are: (1) compute center and radius first, (2) principal stresses = center ± radius, (3) τ_max = radius, (4) normal on max-shear = center. The negative angle for θ_p indicates the principal plane is 13.28° clockwise from the x-face. Always verify using the sum invariant σ₁ + σ₂ = σ_x + σ_y.

Problem

PROBLEM 1 (Principal Stresses — Direct Application) At a critical point in a structural member, the following stresses exist: σ_x = -40 MPa, σ_y = 60 MPa, τ_xy = 25 MPa. Determine: (a) the principal stresses σ₁ and σ₂, (b) the maximum in-plane shear stress τ_max, (c) the angle θ_p of the principal planes, and (d) the normal stress on the plane of maximum shear.

Solution

Step 1: Average stress (center of Mohr's Circle). σ_avg = (σ_x + σ_y)/2 = (-40 + 60)/2 = 20/2 = 10 MPa Step 2: Radius. R = √[(σ_x - σ_y)²/4 + τ_xy²] = √[(-40 - 60)²/4 + 25²] = √[(-100)²/4 + 625] = √[10,000/4 + 625] = √[2,500 + 625] = √3,125 = 55.90 MPa Step 3: Principal stresses. σ₁ = σ_avg + R = 10 + 55.90 = 65.90 MPa ≈ 65.9 MPa σ₂ = σ_avg - R = 10 - 55.90 = -45.90 MPa ≈ -45.9 MPa Step 4: Maximum in-plane shear stress. τ_max = R = 55.90 MPa ≈ 55.9 MPa Step 5: Principal plane angle. tan(2θ_p) = 2τ_xy / (σ_x - σ_y) = 2(25) / (-40 - 60) = 50 / (-100) = -0.500 2θ_p = arctan(-0.500) = -26.57° or 153.43° θ_p = -13.28° (or equivalently, 76.72° for the perpendicular principal plane) Step 6: Normal stress on max-shear plane. σ on max-shear plane = σ_avg = 10 MPa Verification: σ₁ + σ₂ = 65.9 + (-45.9) = 20 MPa = σ_x + σ_y = -40 + 60 = 20 MPa ✓

This problem integrates superposition, Mohr's Circle, equivalent torque, and failure theory into one comprehensive solution — exactly the type of multi-part problem common in PRC board examinations. Note that τ_max from the equivalent torque method (52.19 MPa) equals R from Mohr's Circle (52.18 MPa) — confirming consistency. The Tresca check uses σ₁ - σ₂ (not τ_max = R) to avoid a common sign error: Tresca says (σ₁-σ₂)/2 = τ_max ≤ S_y/2, which means σ₁ - σ₂ ≤ S_y.

Problem

PROBLEM 2 (Shaft with Combined Loading) A 50 mm diameter solid circular shaft is subjected to a bending moment M = 0.8 kN·m and a torque T = 1.0 kN·m. Determine: (a) σ_x, σ_y, and τ_xy at the critical point, (b) the principal stresses using Mohr's Circle, (c) the maximum shear stress using the equivalent torque method, and (d) if the shaft material has yield strength S_y = 280 MPa, find the factor of safety by the Tresca criterion.

Solution

Given: d = 50 mm, M = 0.8 × 10⁶ N·mm, T = 1.0 × 10⁶ N·mm, S_y = 280 MPa PART (a): Reference stress components. σ_x = 32M/(πd³) = 32(0.8×10⁶)/[π(50)³] = 25.6×10⁶ / [π × 125,000] = 25.6×10⁶ / 392,699 = 65.19 MPa σ_y = 0 MPa (free surface) τ_xy = 16T/(πd³) = 16(1.0×10⁶)/392,699 = 16×10⁶ / 392,699 = 40.74 MPa PART (b): Mohr's Circle / principal stresses. σ_avg = (65.19 + 0)/2 = 32.60 MPa R = √[(65.19/2)² + 40.74²] = √[32.60² + 40.74²] = √[1062.8 + 1659.7] = √2722.5 = 52.18 MPa σ₁ = 32.60 + 52.18 = 84.78 MPa ≈ 84.8 MPa σ₂ = 32.60 - 52.18 = -19.58 MPa ≈ -19.6 MPa PART (c): Equivalent torque method. T_e = √(M² + T²) = √(0.8² + 1.0²) = √(0.64 + 1.00) = √1.64 = 1.281 kN·m τ_max = 16T_e/(πd³) = 16(1.281×10⁶)/392,699 = 20.496×10⁶ / 392,699 = 52.19 MPa ≈ 52.2 MPa ✓ (matches R from part b) PART (d): Tresca factor of safety. σ₁ = +84.8 MPa, σ₂ = -19.6 MPa (opposite signs) Tresca: σ₁ - σ₂ = 84.8 - (-19.6) = 104.4 MPa FS = S_y / (σ₁ - σ₂) = 280 / 104.4 = 2.68 Answer: FS = 2.68 by Tresca criterion — the shaft is safe.

The negative τ_xy indicates that the shear stress on the positive x-face acts in the negative y-direction. The verification step (checking that the result lies on the circle) is a critical board exam habit. Notice that at θ = 40°, the normal stress is nearly zero (-0.4 MPa) — this plane is very close to the plane of maximum shear (which has σ = 50 MPa but τ = 78.1 MPa). The coincidence of nearly zero normal stress is specific to this combination of inputs.

Problem

PROBLEM 3 (Stress on Inclined Plane) At a point in a beam web, σ_x = 100 MPa, σ_y = 0, and τ_xy = -60 MPa (negative shear). Find: (a) the principal stresses, (b) the angle of the principal planes, (c) the stress components on a plane inclined at θ = 40° counterclockwise from the x-face.

Solution

PART (a): Principal stresses. σ_avg = (100 + 0)/2 = 50 MPa R = √[(100-0)²/4 + (-60)²] = √[50² + 60²] = √[2500 + 3600] = √6100 = 78.10 MPa σ₁ = 50 + 78.10 = 128.10 MPa ≈ 128.1 MPa σ₂ = 50 - 78.10 = -28.10 MPa ≈ -28.1 MPa PART (b): Principal plane angle. tan(2θ_p) = 2τ_xy / (σ_x - σ_y) = 2(-60) / (100 - 0) = -120/100 = -1.200 2θ_p = arctan(-1.200) = -50.19° θ_p = -25.10° (i.e., 25.1° clockwise from x-face) Verify which angle gives σ₁: σ_x' at θ = -25.10°: 2θ = -50.19° cos(-50.19°) = 0.640, sin(-50.19°) = -0.768 σ_x' = 50 + 50(0.640) + (-60)(-0.768) = 50 + 32 + 46.1 = 128.1 MPa = σ₁ ✓ PART (c): Stresses at θ = 40° (2θ = 80°). cos80° = 0.1736, sin80° = 0.9848 σ_x' = 50 + 50(0.1736) + (-60)(0.9848) = 50 + 8.68 - 59.09 = -0.41 MPa ≈ -0.4 MPa (very small compression) τ_x'y' = -50(0.9848) + (-60)(0.1736) = -49.24 - 10.42 = -59.66 MPa ≈ -59.7 MPa Verify on circle: (σ_x' - 50)² + τ² = (-0.4-50)² + (-59.7)² = (-50.4)² + (59.7)² = 2540 + 3564 = 6104 ≈ R² = 78.1² = 6100 ✓

The kern of a cross-section defines the region within which the eccentric load must fall for the entire section to remain in compression (important for concrete and masonry, which cannot carry tension). For a rectangle, the kern is a diamond with half-diagonals of h/6 in each direction. When the eccentricity equals h/6, one extreme fiber reaches exactly zero — the kern boundary condition. This is a classic board exam topic combining axial stress, bending, and the kern concept.

Problem

PROBLEM 4 (Axial + Bending — Critical Stress Location) A 200 mm × 300 mm rectangular concrete column (b = 200 mm, h = 300 mm) carries an axial load P = 900 kN (compression) and a bending moment M = 45 kN·m about the strong axis (300 mm dimension). Determine the maximum and minimum normal stresses and check if the entire section is in compression (kern problem).

Solution

Step 1: Section properties. A = 200 × 300 = 60,000 mm² I_x = 200(300)³/12 = 450 × 10⁶ mm⁴ c = 150 mm (extreme fiber distance) Step 2: Axial stress. σ_axial = -P/A = -900,000/60,000 = -15.0 MPa (compressive) Step 3: Bending stress. σ_bending = ±Mc/I = ±(45×10⁶)(150)/(450×10⁶) = ±15.0 MPa Step 4: Combined stresses. Maximum stress (tension side of bending minus axial compression): σ_max = -15.0 + 15.0 = 0 MPa Minimum stress (compression side of bending plus axial compression): σ_min = -15.0 - 15.0 = -30.0 MPa Step 5: Check kern condition. The stress at the 'tension' fiber is exactly 0 MPa — the load is at the kern boundary. Kern limit for rectangle: e_kern = h/6 = 300/6 = 50 mm Eccentricity of load: e = M/P = 45×10⁶/900,000 = 50 mm ✓ (exactly at kern boundary) The section is entirely in compression (minimum stress = 0, no actual tension). Answer: σ_max = 0 MPa, σ_min = -30 MPa (compression). Load is at the kern boundary.

This problem explains the classic chalk-stick torsion failure and the 45° helical fracture of brittle shafts. Pure torsion creates equal tension and compression at 45° — brittle materials fracture in tension, so the crack follows the 45° principal tension plane, creating the characteristic helical fracture surface. This concept is also used to explain diagonal tension cracks in reinforced concrete beams.

Problem

PROBLEM 5 (Pure Shear and Failure) A thin-walled circular tube of 80 mm outer diameter and 6 mm wall thickness is subjected to a torque T = 4 kN·m only. (a) Find the shear stress in the wall. (b) Determine the principal stresses. (c) If the tube material is brittle with ultimate tensile strength S_ut = 200 MPa, predict the failure load using Rankine theory. (d) At what angle to the tube axis does failure initiate?

Solution

PART (a): Torsional shear stress. d_i = 80 - 2(6) = 68 mm, r_o = 40 mm, r_i = 34 mm J = π(D⁴ - d⁴)/32 = π(80⁴ - 68⁴)/32 = π(40,960,000 - 21,381,376)/32 = π(19,578,624)/32 = 1,921,856 mm⁴ τ = T·r_o/J = (4×10⁶)(40)/1,921,856 = 160×10⁶/1,921,856 = 83.25 MPa PART (b): Principal stresses (pure shear state: σ_x = σ_y = 0). σ_avg = 0, R = τ = 83.25 MPa σ₁ = +83.25 MPa (tension) σ₂ = -83.25 MPa (compression) PART (c): Rankine failure load. Failure when σ₁ = S_ut: 83.25 < 200 MPa ✓ (safe at T = 4 kN·m) Failure torque: T_fail = T × (S_ut/σ₁) = 4 × (200/83.25) = 9.61 kN·m PART (d): Failure angle. Principal tension acts at 45° to the tube axis (for pure torsion). Brittle material fails by fracture on the plane of maximum tension. Failure initiates on a HELIX at 45° to the tube axis. Answer: Failure at T = 9.61 kN·m, fracture on 45° helix.

Exam Preparation Tips

MEMORIZE THE BIG THREE FORMULAS: (1) σ₁,₂ = σ_avg ± R, (2) τ_max = R = √[(σ_x-σ_y)²/4 + τ_xy²], (3) tan(2θ_p) = 2τ_xy/(σ_x-σ_y). These three formulas solve 90% of Mohr's Circle board problems.
ALWAYS FIND CENTER AND RADIUS FIRST: Before attempting any stress transformation problem, compute σ_avg = (σ_x+σ_y)/2 and R = √[(σ_x-σ_y)²/4 + τ_xy²]. Every answer flows from these two quantities.
APPLY THE SUM INVARIANT AS A CHECK: After finding σ₁ and σ₂, verify that σ₁ + σ₂ = σ_x + σ_y. This two-second check catches arithmetic errors before they propagate.
WATCH THE 2θ vs. θ TRAP: Physical rotation θ → circle rotation 2θ. This is the single most common error in Mohr's Circle problems. Practice saying 'double the angle going to the circle, halve coming back.'
RECOGNIZE PYTHAGOREAN TRIPLETS IN SHAFT PROBLEMS: T_e = √(M²+T²). Common board setups use 3-4-5 (M=3, T=4, T_e=5), 5-12-13, 8-15-17, or scaled versions. Recognizing these saves significant calculation time.
FOR SHAFT PROBLEMS: Set σ_y = 0 immediately and do not question it — the free surface of a shaft has no transverse normal stress. Write it down explicitly to avoid accidental inclusion.
CONVERT UNITS BEFORE SUBSTITUTING: Convert kN·m to N·mm (multiply by 10⁶) and mm to m (or keep consistent SI throughout). The #1 arithmetic error in PRC exams is unit inconsistency.
KNOW WHEN PRINCIPAL STRESSES EQUAL THE REFERENCE STRESSES: If τ_xy = 0, you are already in the principal stress state — no transformation needed. The σ_x and σ_y ARE σ₁ and σ₂.
PURE SHEAR RESULT IS FUNDAMENTAL: Pure shear (σ_x = σ_y = 0, τ_xy = τ) → σ₁ = +τ, σ₂ = -τ at 45°. Memorize this result — it underlies concrete diagonal tension, torsion failure, and many board scenarios.
FAILURE THEORY IDENTIFICATION: Ductile steel = Tresca (conservative) or von Mises (accurate). Brittle material (cast iron, concrete) = Rankine (max normal stress). When in doubt, the problem will specify; if it says 'maximum shear stress theory,' use Tresca.
SKETCH MOHR'S CIRCLE EVEN IF COMPUTING ANALYTICALLY: A quick sketch helps identify the correct quadrant for the angle, check if σ₁ is positive or negative, and visualize where the problem point lies on the circle. Takes 30 seconds and prevents many errors.
PRACTICE WITH NEGATIVE σ_y: Many students habitually assume σ_y is positive. Practice problems where σ_y is negative or where σ_x < σ_y (so the center is to the left of σ_x on the circle).
ABSOLUTE vs. IN-PLANE MAX SHEAR: When both σ₁ and σ₂ have the same sign, check the absolute maximum shear: τ_abs = max(|σ₁|, |σ₂|)/2. This is especially relevant for pressure vessel problems where both hoop and longitudinal stresses are tensile.
KERN CHECK FOR COLUMN PROBLEMS: For rectangular sections, the kern has half-diagonals of h/6 and b/6. For circular sections, kern radius = d/8. This appears frequently in combined axial+bending problems for columns.
BOARD EXAM TIME MANAGEMENT: Mohr's Circle problems that ask for principal stresses only require 3 steps: σ_avg, R, then σ₁,₂ = σ_avg ± R. Do not over-complicate. Reserve the full graphical construction only when orientation angles are required.

Loading diagram…

In summary

Combined stresses and Mohr's Circle represent one of the most integrative topics in the PRC Civil Engineer Licensure Examination. They require you to synthesize knowledge from axial loading, bending, torsion, and shear — topics covered throughout the Strength of Materials syllabus — and apply them simultaneously to a single critical point in a real structural member. The mathematical elegance of Mohr's Circle, where all stress transformation results are encoded in the simple geometry of a circle (center at σ_avg, radius equal to τ_max), makes this topic both powerful and memorable. The four core results to carry into every examination: (1) σ₁,₂ = σ_avg ± R, (2) τ_max = R, (3) the normal stress on the maximum shear plane equals σ_avg (not zero), and (4) physical rotation θ corresponds to 2θ on Mohr's Circle. With these four results and the radius formula, you can solve the vast majority of board exam problems in this topic. For shaft problems, the equivalent torque T_e = √(M²+T²) and equivalent moment M_e = (M+T_e)/2 are compact design formulas that encapsulate the Mohr's Circle analysis — recognize Pythagorean triplets (3-4-5, 5-12-13, 8-15-17) to compute T_e without a calculator. For failure assessment, Tresca is the conservative choice for ductile steel (used when the problem specifies maximum shear stress theory or does not specify), while von Mises gives higher accuracy. Rankine applies to brittle materials — including concrete, where the diagonal tension crack (occurring on the principal tension plane at approximately 45°) is a direct manifestation of the principal stress concept from Mohr's Circle. As a future registered Civil Engineer in the Philippines, you will encounter combined stress states every time you design a machine shaft, analyze an eccentrically loaded column, or check a structural connection under simultaneous shear and tension. The tools developed in this chapter — superposition, stress transformation, Mohr's Circle, and failure theories — are your professional instruments for ensuring structural safety in every project governed by NSCP 2015, AISC 360, and ACI 318. Master them with understanding, not just memorization, and they will serve you throughout your career.

Previous chapter

Beam Deflections

Next chapter

Columns and Buckling

Ready to practise for the CELE 2026?

Super Tutor's AI review plan adapts to your weak areas and builds a weekly practice schedule around your target CELE exam date.

Start free See the full CELE guide CELE preparation guide

← Back to Strength of Materials hub