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CELE Strength of MaterialsCombined Stresses and Mohr's CircleRevision Notes

Final-week revision notes for Combined Stresses and Mohr's Circle. If you have already studied the full chapter, this page is your go-to refresher before sitting the CELE. Compact, high-yield, and aligned with what Professional Regulation Commission (PRC) — Board of Civil Engineering tests in the Strength of Materials subtest.

Exam context

The Civil Engineer Licensure Examination is conducted by Professional Regulation Commission (PRC) — Board of Civil Engineering and is scheduled for May and November 2026. The Strength of Materials subtest is marked as "Core" in the official pattern, and Combined Stresses and Mohr's Circle appears in position 6th of 8 in the CELE Strength of Materials review rotation. Passing mark: 70% weighted average, no sub-test below 50%. Recent CELE 2026 papers have drawn roughly a meaningful share of questions from this subject.

Combined Stresses and Mohr's Circle - Revision Notes

Real structural members seldom carry a single type of load. A roof column carries axial compression and bending from wind; a drive shaft twists and bends simultaneously; a pressurized pipeline feels hoop stress, longitudinal stress, and sometimes torsion. The stress analyst must determine the worst-case normal and shear stresses at the critical point — and on which plane they act — before checking against material strength. Mohr's Circle is the graphical engine that converts any plane-stress state into its principal values and orientations with minimal algebra. This chapter is consistently tested in the PRC Civil Engineer Licensure Examination under Strength of Materials, covering superposition, transformation equations, principal stresses, maximum shear, Mohr's Circle construction, combined bending-torsion of shafts, and failure theories.

Sections

Formulas

Example

A 200×200 mm column carries P = 500 kN compression and M = 40 kN·m. At the extreme tension fiber: σ = –500,000/40,000 + 40×10⁶×100/66.67×10⁶ = –12.5 + 60.0 = +47.5 MPa (tension governs).

Formula

σ = P/A ± Mc/I

Variables

P = axial load (N); A = cross-sectional area (mm²); M = bending moment (N·mm); c = distance from neutral axis to extreme fiber (mm); I = moment of inertia (mm⁴)

Application

Combines axial and bending normal stresses. The ± depends on whether bending causes tension or compression at the point of interest.

Example

A W-section carries axial load P and moments M_x and M_y simultaneously. Check all four corners of the flange for the critical combination.

Formula

σ = P/A ± M_x·c_y/I_x ± M_y·c_x/I_y

Variables

M_x, M_y = bending moments about x and y centroidal axes; c_y, c_x = distances to the fiber from respective axes; I_x, I_y = second moments of area about respective axes

Application

Biaxial bending with or without axial load — common in asymmetrically loaded columns and crane girder flanges.

Example

At the outer surface of a shaft: τ_total = 16T/(πd³) ± 4V/(πr²) — note torsion and shear may add or subtract depending on the point location.

Formula

τ = VQ/(Ib) + Tc/J

Variables

V = transverse shear (N); Q = first moment of area above the point (mm³); b = width at the point (mm); T = torque (N·mm); J = polar moment of inertia (mm⁴)

Application

Combined transverse shear and torsional shear — critical on the outer fiber of circular shafts where both are maximum.

Exam Tips

  • Identify σ_x, σ_y, and τ_xy FIRST before touching any principal-stress formula. Write them down explicitly.
  • For rectangular sections: I = bh³/12; for circular sections: I = πd⁴/64, J = πd⁴/32.
  • At the neutral axis of a beam, σ_bending = 0, so only axial and shear stresses need combining.
  • At the extreme fiber of a beam, τ_transverse = 0 (Q = 0 at the extreme fiber), so only normal stresses and torsional shear combine.

Key Points

  • Superposition is valid only when the material is linearly elastic and deformations are small (Hooke's Law region).
  • Only stresses of the SAME TYPE (both normal, or both shear) on the SAME plane can be added directly.
  • Axial stress and bending stress are both normal stresses acting on the cross-sectional plane — they add algebraically.
  • Bending shear and torsional shear are both shear stresses on the cross-section — they add algebraically at the same point.
  • After superposition, you obtain σ_x, σ_y, and τ_xy at the critical point — the inputs to Mohr's Circle.
  • Always identify the critical point (maximum combined stress) before applying transformation equations.

Definitions

Term

Plane Stress

Definition

A stress state where all stresses act in a single plane (the xy-plane), with σ_z = τ_xz = τ_yz = 0. This is the standard assumption for thin members and surface points of thick members.

Importance

Almost all board exam problems are plane-stress problems. Recognizing this allows direct application of the transformation equations and Mohr's Circle.

Term

Superposition Principle

Definition

The total stress at a point due to several simultaneous loads equals the algebraic sum of the stresses produced by each load acting alone, provided the material is linearly elastic.

Importance

Allows complex combined loading to be broken into simple cases, solved individually, then summed — the foundation of all combined-stress analysis.

Section Title

Superposition of Stresses

Common Mistakes

  • Adding a normal stress (σ) and a shear stress (τ) directly — they act on different faces and CANNOT be summed; they must be combined via the transformation equations.
  • Forgetting to check all critical points (top fiber, bottom fiber, neutral axis) when bending and axial stresses combine.
  • Using the wrong sign for axial stress: compressive P gives negative σ = –P/A.
  • Assuming the neutral axis location does not shift when axial load is combined with bending.

Formulas

Example

For σ_x = 60, σ_y = –20, τ_xy = 40 MPa, at θ = 30°: σ_x' = 20 + 40·cos60° + 40·sin60° = 20 + 20 + 34.64 = 74.64 MPa

Formula

σ_x' = (σ_x + σ_y)/2 + (σ_x – σ_y)/2 · cos2θ + τ_xy · sin2θ

Variables

σ_x, σ_y = normal stresses on the reference x and y faces (MPa); τ_xy = shear stress on the x-face in the y-direction (MPa); θ = angle of rotation from x-axis to the new x'-axis (degrees or radians, CCW positive)

Application

Finds the normal stress on any inclined plane. Used when the board asks for stress on a specific oblique plane.

Example

Continuing the example above at θ = 30°: τ_x'y' = –40·sin60° + 40·cos60° = –34.64 + 20 = –14.64 MPa

Formula

τ_x'y' = –(σ_x – σ_y)/2 · sin2θ + τ_xy · cos2θ

Variables

Same variables as above. τ_x'y' is the shear stress on the inclined plane; positive value = CCW sense on the positive x'-face.

Application

Finds the shear stress on any inclined plane. Combine with σ_x' to fully define stress on the oblique plane.

Exam Tips

  • Memorize: normal stress equation has cosine of 2θ; shear stress equation has sine of 2θ — opposite from what many expect.
  • When asked for stress on a plane at θ = 45°, use 2θ = 90°: cos90° = 0, sin90° = 1. This simplifies the equations greatly.
  • The transformation equations are rarely needed in full for board problems — Mohr's Circle is faster and less error-prone.
  • Check: at θ = 0, you should recover σ_x and τ_xy from the transformation equation — verify your formula.

Key Points

  • Stress is NOT a scalar — its value depends on the orientation of the plane on which it acts.
  • The transformation equations give the normal stress σ_x' and shear stress τ_x'y' on a plane whose normal is rotated θ counterclockwise from the x-axis.
  • The sum (σ_x + σ_y) is INVARIANT under rotation — it does not change with plane orientation.
  • As θ varies from 0° to 180°, the stresses trace one full circle on the Mohr's Circle diagram.
  • A rotation of θ in the physical element corresponds to a rotation of 2θ on Mohr's Circle.
  • The transformation equations are derived from equilibrium of a wedge element — no material property is involved.

Definitions

Term

Sign Convention for τ_xy

Definition

τ_xy is positive when it acts in the +y direction on the +x face (and simultaneously in the –x direction on the +y face). This convention must be consistent when plotting Mohr's Circle.

Importance

The most common source of sign errors in board problems. Establish the convention at the start and never switch within a problem.

Term

Stress Invariant

Definition

Quantities that remain constant regardless of the coordinate rotation. The first invariant is I₁ = σ_x + σ_y = σ_1 + σ_2. The second invariant is I₂ = σ_x·σ_y – τ_xy² = σ_1·σ_2.

Importance

Use I₁ as a quick check: after computing principal stresses, verify that σ_1 + σ_2 = σ_x + σ_y.

Section Title

Stress Transformation Equations

Common Mistakes

  • Using θ instead of 2θ in the trigonometric terms — the equations use cos2θ and sin2θ, not cosθ and sinθ.
  • Mixing up which sign of τ_xy to use when setting up the equations — always establish and follow one convention.
  • Forgetting that σ_x' + σ_y' = σ_x + σ_y (invariant) — use this to find σ_y' after computing σ_x' without re-solving.
  • Not checking the answer by confirming the computed (σ_x', τ_x'y') lies on Mohr's Circle.

Formulas

Example

σ_x = 80, σ_y = 20, τ_xy = 30 MPa. Center = 50, R = √(30² + 30²) = 42.43 MPa. σ_1 = 92.43 MPa, σ_2 = 7.57 MPa.

Formula

σ_1,2 = (σ_x + σ_y)/2 ± √[(σ_x – σ_y)²/4 + τ_xy²]

Variables

σ_1 = maximum principal stress (MPa); σ_2 = minimum principal stress (MPa); all other variables as previously defined. The ± gives both values from a single expression.

Application

THE master formula for principal stresses. Works for all plane-stress states.

Example

From the example above: τ_max = 42.43 MPa. Check: (92.43 – 7.57)/2 = 84.86/2 = 42.43 MPa ✔

Formula

τ_max = √[(σ_x – σ_y)²/4 + τ_xy²] = (σ_1 – σ_2)/2

Variables

τ_max = maximum in-plane shear stress (MPa). This equals the radius R of Mohr's Circle.

Application

Critical for shear-based failure checks (Tresca criterion) and shaft design. Also equals R in Mohr's Circle.

Example

σ_x = 80, σ_y = 20, τ_xy = 30 MPa. tan2θ_p = 2(30)/(80–20) = 60/60 = 1.0. 2θ_p = 45°. θ_p = 22.5°.

Formula

tan2θ_p = 2τ_xy / (σ_x – σ_y)

Variables

θ_p = angle from the x-axis to the principal plane carrying σ_1 (degrees). The result of arctan gives one value; the other principal plane is at θ_p + 90°.

Application

Determines the ORIENTATION of the principal planes — required when the board asks 'at what angle do the principal stresses act?'

Example

For the example above: σ_avg = (80 + 20)/2 = 50 MPa. The max-shear planes carry both τ_max = 42.43 MPa AND σ = 50 MPa simultaneously.

Formula

σ_avg = (σ_x + σ_y)/2

Variables

σ_avg = average normal stress (MPa) — equals the center of Mohr's Circle. Acts on the planes of maximum shear stress.

Application

The normal stress that accompanies the maximum shear stress. Often forgotten by examinees, leading to incorrect specification of the stress state on the max-shear planes.

Exam Tips

  • QUICK CHECK: σ_1 + σ_2 must equal σ_x + σ_y. If your computed sum differs, you made an arithmetic error.
  • For PURE SHEAR (σ_x = σ_y = 0): σ_1 = +τ_xy, σ_2 = –τ_xy, θ_p = 45°. Memorize this — it is a classic board case.
  • For UNIAXIAL STRESS (σ_y = 0, τ_xy = 0): σ_1 = σ_x, σ_2 = 0, τ_max = σ_x/2 at 45°. Also memorize.
  • When the board gives you principal stresses and asks for σ_x, σ_y: use the invariant σ_x + σ_y = σ_1 + σ_2 and the R formula.

Key Points

  • Principal stresses σ_1 and σ_2 are the MAXIMUM and MINIMUM normal stresses at the point, acting on planes where shear stress is ZERO.
  • The algebraic maximum is σ_1 (could be tension or compression); the algebraic minimum is σ_2.
  • Maximum in-plane shear stress τ_max equals the RADIUS of Mohr's Circle, R.
  • The principal planes are oriented at θ_p from the reference x-face; the max-shear planes are at θ_p ± 45°.
  • The normal stress on the planes of maximum shear equals the average stress σ_avg = (σ_x + σ_y)/2 — it is NOT zero unless σ_x + σ_y = 0.
  • For a complete 3D analysis, the absolute maximum shear may be larger than the in-plane maximum if both principal stresses have the same sign.

Definitions

Term

Principal Stress

Definition

The maximum or minimum normal stress at a point, acting on a plane where the shear stress is identically zero. Every stress state has exactly two principal stresses for plane stress (σ_1 ≥ σ_2).

Importance

Principal stresses are the values compared against material strength (yield stress, ultimate stress) in failure theories. They are the most frequently asked values in board exams.

Term

Maximum In-Plane Shear Stress

Definition

The largest shear stress considering only rotations within the xy-plane. Equals the radius R of Mohr's Circle and (σ_1 – σ_2)/2.

Importance

Used in the Tresca failure criterion and in shaft design. Note: the ABSOLUTE maximum shear (considering 3D) may be larger.

Term

Absolute Maximum Shear Stress

Definition

For plane stress with the third principal stress σ_3 = 0, the absolute maximum shear is max(|σ_1–σ_2|/2, |σ_1|/2, |σ_2|/2). If σ_1 and σ_2 have opposite signs, τ_abs = (σ_1–σ_2)/2 = τ_max,in-plane. If same sign, τ_abs = σ_1/2 (for σ_1 > σ_2 > 0).

Importance

Board problems sometimes specify 'absolute maximum shear stress' — do not assume it equals τ_max,in-plane when both principal stresses are same sign.

Section Title

Principal Stresses and Maximum In-Plane Shear Stress

Common Mistakes

  • Halving error: writing τ_max = √[(σ_x–σ_y)² + τ_xy²] — missing the division by 2 inside the radical. The correct term is (σ_x–σ_y)/2, not (σ_x–σ_y).
  • Assuming zero normal stress on the maximum shear plane — it equals σ_avg, not zero.
  • Forgetting the second principal plane is at θ_p + 90° — and that the sign of τ reverses.
  • Using τ_max = (σ_1 + σ_2)/2 instead of (σ_1 – σ_2)/2.
  • Not checking whether the absolute maximum shear exceeds the in-plane maximum when both principal stresses are of the same sign.

Formulas

Example

σ_x = 80 MPa, σ_y = 20 MPa → C = (50, 0)

Formula

Center C = ((σ_x + σ_y)/2, 0)

Variables

C = center of Mohr's Circle, located on the σ-axis at the average normal stress. The τ-coordinate of the center is always zero.

Application

First step in constructing Mohr's Circle. Locates the center before computing the radius.

Example

σ_x = 80, σ_y = 20, τ_xy = 30 MPa → R = √(30² + 30²) = 42.43 MPa

Formula

R = √[(σ_x – σ_y)²/4 + τ_xy²]

Variables

R = radius of Mohr's Circle (MPa). Also equals τ_max and (σ_1 – σ_2)/2.

Application

Second step: computing the radius. Once C and R are known, the circle — and all transformed stresses — are completely determined.

Example

C = 50, R = 42.43 → σ_1 = 92.43 MPa, σ_2 = 7.57 MPa

Formula

σ_1 = C + R ; σ_2 = C – R

Variables

C = center value (MPa); R = radius (MPa). σ_1 = maximum principal stress; σ_2 = minimum principal stress.

Application

Read off principal stresses directly from the circle — the two x-intercepts.

Exam Tips

  • FASTEST METHOD for board exams: Write C and R, then σ_1 = C + R, σ_2 = C – R, τ_max = R. Four lines of work.
  • To find stress on a specific plane at angle θ from the x-face: move 2θ along the circle from point X.
  • The top and bottom of the circle (maximum and minimum τ) are always at 90° from the x-intercepts on the circle = 45° in physical space.
  • Draw a rough sketch of Mohr's Circle even for calculation problems — it confirms your answers and prevents sign errors.

Key Points

  • Mohr's Circle is a graphical representation of the stress transformation equations — every point on the circle represents the stress state on a specific plane through the point.
  • The horizontal axis is the normal stress σ (positive = tension, rightward); the vertical axis is shear stress τ (sign convention: see below).
  • CONVENTION: Plot τ downward as positive (some texts use upward) — the key is CONSISTENCY within the problem.
  • The center C is always on the σ-axis at coordinates (σ_avg, 0).
  • The radius R of the circle equals τ_max (in-plane).
  • Principal stresses are the rightmost and leftmost intercepts of the circle on the σ-axis.
  • A physical rotation of angle θ corresponds to a circle rotation of 2θ in the SAME direction.

Definitions

Term

Reference Point X

Definition

The point on Mohr's Circle representing the stress state on the x-face of the element: coordinates (σ_x, τ_xy). When τ_xy is positive (CCW on x-face), X is plotted BELOW the σ-axis in the conventional downward-positive-shear plot.

Importance

X and Y are the two endpoints of the initial diameter; drawing the circle from these two points is the standard construction method.

Term

Reference Point Y

Definition

The point on Mohr's Circle representing the stress state on the y-face: coordinates (σ_y, –τ_xy). Always diametrically opposite to X.

Importance

The line XY is always a diameter of Mohr's Circle. Its midpoint is the center C. This diameter makes an angle of 2θ_p with the σ-axis.

Section Title

Mohr's Circle — Construction and Application

Common Mistakes

  • Plotting both X and Y with the SAME sign of τ — they must be plotted with OPPOSITE signs of τ (one above, one below the σ-axis).
  • Measuring physical angles θ directly on the circle diagram — remember to divide by 2 to get the actual plane orientation.
  • Confusing the direction of rotation: CCW rotation in the physical element corresponds to CCW rotation on the circle (if using the standard convention).
  • Misidentifying which intercept is σ_1 (maximum) and which is σ_2 — always σ_1 is to the right (larger algebraic value).

Formulas

Example

d = 60 mm, M = 1.5 kN·m = 1.5×10⁶ N·mm: σ_x = 32(1.5×10⁶)/(π×60³) = 48×10⁶/678,584 = 70.74 MPa

Formula

σ_x = 32M/(πd³) ; τ_xy = 16T/(πd³)

Variables

M = bending moment (N·mm); T = torque (N·mm); d = shaft diameter (mm). These are the stresses at the outermost fiber of a solid circular shaft.

Application

Sets up the plane-stress state before applying principal stress formulas or Mohr's Circle for shaft problems.

Example

M = 1.5 kN·m, T = 2.0 kN·m. √(1.5² + 2.0²) = √(2.25+4) = √6.25 = 2.5 kN·m. M_e = (1/2)(1.5 + 2.5) = 2.0 kN·m

Formula

M_e = (1/2)[M + √(M² + T²)]

Variables

M_e = equivalent bending moment (N·mm); M = applied bending moment; T = applied torque. Used with σ_1 = 32M_e/(πd³) under maximum-normal-stress theory.

Application

Shaft design for BRITTLE materials (cast iron) or when the maximum tensile stress governs.

Example

M = 1.5 kN·m, T = 2.0 kN·m. T_e = √(1.5² + 2.0²) = 2.5 kN·m. τ_max = 16(2.5×10⁶)/(π×60³) = 58.9 MPa

Formula

T_e = √(M² + T²)

Variables

T_e = equivalent torque (N·mm). Used with τ_max = 16T_e/(πd³) under maximum-shear-stress theory (Tresca).

Application

Shaft design for DUCTILE materials (steel) when maximum shear stress governs — the most common shaft design scenario.

Exam Tips

  • Memorize T_e = √(M² + T²) — it appears on nearly every shaft-related board problem involving combined loads.
  • The formula τ_max = 16T_e/(πd³) is the single most-tested shaft formula in combined stress problems.
  • If the problem says 'factor of safety = 2 against yielding': S_sy(Tresca) = S_y/2, so τ_allow = S_y/4 after applying F.S.
  • For hollow shafts: replace πd³/16 with 16T_e·D/(π(D⁴–d⁴)) where D = outer diameter, d = inner diameter.

Key Points

  • A circular shaft under simultaneous bending moment M and torque T has a plane-stress state at the surface: σ_x = 32M/(πd³), σ_y = 0, τ_xy = 16T/(πd³).
  • The critical point is at the top or bottom of the cross-section (maximum bending stress) OR at the side (where bending stress is zero but torsional shear is maximum) — always check both.
  • Equivalent moment M_e is used with the MAXIMUM NORMAL STRESS theory (Rankine/brittle).
  • Equivalent torque T_e is used with the MAXIMUM SHEAR STRESS theory (Tresca/ductile).
  • These equivalent quantities allow shaft sizing using standard bending and torsion formulas without explicitly running through Mohr's Circle each time.
  • The NSCP 2015 and machine design references use these equivalent moments extensively for power transmission shaft checks.

Definitions

Term

Equivalent Bending Moment (M_e)

Definition

A fictitious bending moment that, acting alone on the same shaft cross-section, would produce the same maximum principal (normal) stress as the combined M and T loading.

Importance

Allows direct use of the bending stress formula for principal stress evaluation — simplifies shaft design by avoiding repeated Mohr's Circle construction.

Term

Equivalent Torque (T_e)

Definition

A fictitious torque that, acting alone, produces the same maximum shear stress as the combined M and T loading. T_e = √(M² + T²).

Importance

The standard formula for ductile shaft design under combined loads — directly used to size shafts against shear stress limits.

Section Title

Combined Bending and Torsion of Shafts

Common Mistakes

  • Using M_e instead of T_e for ductile shaft design — the Tresca criterion requires T_e (maximum shear), not M_e (maximum normal).
  • Forgetting σ_y = 0 on a circular shaft surface — bending stress acts only in one direction at the extreme fiber.
  • Not converting units: M in kN·m must become N·mm (×10⁶) before substituting into the formula with d in mm.
  • Using d³ when the formula requires d³ — double-check: the section modulus for bending is πd³/32 and for torsion is πd³/16.

Formulas

Example

σ_1 = 92.43, σ_2 = 7.57 MPa: σ_v = √(92.43² – 92.43×7.57 + 7.57²) = √(8543 – 699 + 57) = √7901 = 88.9 MPa

Formula

σ_v = √(σ_1² – σ_1·σ_2 + σ_2²) [von Mises, principal stress form]

Variables

σ_v = von Mises equivalent stress (MPa); σ_1, σ_2 = principal stresses (MPa). Failure when σ_v ≥ S_y.

Application

Most accurate failure criterion for ductile steel members under combined loading. Used in AISC 360 provisions for combined axial-shear elements.

Example

σ_x = 80, σ_y = 20, τ_xy = 30 MPa: σ_v = √(6400 – 1600 + 400 + 2700) = √7900 = 88.9 MPa ✔

Formula

σ_v = √(σ_x² – σ_x·σ_y + σ_y² + 3τ_xy²) [von Mises, component form]

Variables

Same as above but expressed directly in terms of the stress components — no need to compute principal stresses first.

Application

More direct when σ_x, σ_y, τ_xy are known. Equivalent to the principal-stress form.

Exam Tips

  • For board exam: if material is STEEL → use Tresca or von Mises. If CAST IRON → use Rankine.
  • Tresca yield surface is a hexagon inscribed in the von Mises ellipse — Tresca is always ≤ von Mises limit.
  • F.S. = S_y / σ_v (von Mises) or S_y / (2τ_max) (Tresca). Know both forms for factor-of-safety problems.
  • Quick comparison: For pure shear (σ_1 = –σ_2 = τ), Tresca gives yield at τ = S_y/2; von Mises gives τ = S_y/√3 ≈ 0.577 S_y — von Mises is less conservative in pure shear.

Key Points

  • Failure theories predict when a material under combined stress will yield or fracture by comparing the multi-axial stress state to a uniaxial strength value.
  • Maximum Normal Stress Theory (Rankine): Failure when σ_1 = S_u (ultimate strength). Best for BRITTLE materials like cast iron.
  • Maximum Shear Stress Theory (Tresca): Failure when τ_max = S_y/2. Conservative (safe) estimate for DUCTILE materials.
  • Distortion Energy Theory (von Mises): Failure when von Mises stress σ_v = S_y. Most ACCURATE for ductile materials; used in AISC 360 and modern codes.
  • For exam purposes: Tresca is conservative (gives larger required cross-section), von Mises is less conservative (smaller cross-section) — both are tested.
  • For plane stress with σ_3 = 0: Tresca gives 2τ_max = σ_1 – σ_2 (if opposite signs) or max(σ_1, σ_2) (if same sign) ≤ S_y.

Definitions

Term

Tresca Criterion

Definition

Also called the Maximum Shear Stress criterion. Predicts yielding when the maximum shear stress τ_max reaches half the uniaxial yield stress S_y/2. Expressed as: σ_1 – σ_2 ≤ S_y (for opposite-sign principal stresses in plane stress).

Importance

The standard design criterion for ductile metals in Philippine practice and board exams. Conservative and simple to apply.

Term

Von Mises Criterion

Definition

Also called the Distortion Energy or Maximum Octahedral Shear Stress criterion. Predicts yielding when the von Mises equivalent stress σ_v reaches S_y. Agrees better with experimental data for ductile metals.

Importance

Referenced in AISC 360 and modern structural codes. The ratio of Tresca to von Mises limits is 2/√3 ≈ 1.155 — Tresca is about 15% more conservative.

Section Title

Failure Theories

Common Mistakes

  • Applying Rankine (maximum normal stress) to ductile steel — it is only appropriate for brittle materials.
  • Using 2τ_max = S_y when both principal stresses are the same sign — the Tresca criterion in this case is max(σ_1, σ_2) = S_y, not (σ_1 – σ_2) = S_y.
  • Forgetting the factor of 3 in the von Mises component form: the shear term is 3τ_xy², not τ_xy².
  • Confusing allowable stress (S_y / F.S.) with the failure stress S_y when setting up failure equations.

Connections

  • MECHANICS OF MATERIALS FOUNDATION: Superposition builds directly on axial stress (P/A), bending stress (Mc/I), and shear stress (VQ/Ib and Tc/J) from earlier chapters — master those formulas first before tackling combined stresses.
  • STRUCTURAL STEEL DESIGN (AISC 360 / NSCP 2015 Steel): Section H of AISC 360 covers combined axial and bending with interaction equations that derive conceptually from principal stress checks. Understanding Mohr's Circle clarifies why interaction checks are necessary.
  • REINFORCED CONCRETE DESIGN (ACI 318 / NSCP 2015 Concrete): Shear-torsion interaction in beams (ACI 318 Section 22.7) is a design manifestation of combined shear and torsion — conceptually a combined stress problem.
  • GEOTECHNICAL ENGINEERING: Mohr's Circle appears again in soil mechanics for stress analysis of soil elements, Mohr-Coulomb failure criterion, and pole construction. The circle is identical; only the sign convention and failure line differ.
  • MACHINE DESIGN / SHAFT DESIGN: The equivalent moment M_e and equivalent torque T_e formulas are directly used in machine element design and appear in PRC board exam problems on rotating shafts and power transmission.
  • PRESSURE VESSEL DESIGN: Thin-walled pressure vessel stresses (hoop and longitudinal) form a biaxial stress state — a direct application of principal stress analysis and Mohr's Circle without shear on the reference planes.
  • FAILURE THEORIES: Tresca and von Mises criteria connect combined stress analysis to material strength properties tested in the Materials Science and Testing portion of the board exam.
  • RA 544 (Civil Engineering Law): Competent stress analysis, including combined loads, is a fundamental professional responsibility of licensed civil engineers under the Code of Ethics for engineering practice in the Philippines.

Exam Strategy

For PRC CE board exam problems on combined stresses and Mohr's Circle, follow this five-step protocol: (1) IDENTIFY the load type and the critical point — determine which location on the member carries the maximum combination of loads. (2) COMPUTE σ_x, σ_y, and τ_xy at that point using superposition of individual load effects (P/A, Mc/I, VQ/Ib, Tc/J). (3) CALCULATE the center C = (σ_x + σ_y)/2 and radius R = √[(σ_x–σ_y)²/4 + τ_xy²]. These two numbers give you everything: principal stresses = C ± R, max shear = R, average normal on shear plane = C. (4) FIND the angle if asked: tan2θ_p = 2τ_xy/(σ_x–σ_y), then divide by 2. Remember: max-shear planes are 45° from principal planes. (5) CHECK with invariants: σ_1 + σ_2 must equal σ_x + σ_y; the computed (σ', τ') for any angle must satisfy (σ'–C)² + τ'² = R². For SHAFT PROBLEMS: go directly to T_e = √(M²+T²) and τ_max = 16T_e/(πd³) — this bypasses the full Mohr's Circle construction and saves time. For FAILURE THEORY problems: identify material (brittle → Rankine; ductile → Tresca or von Mises) before applying the criterion. Allocate no more than 4 minutes per problem in the board exam; if you exceed this, skip and return — these problems reward practice-built pattern recognition.

Quick Review Questions

At a point in a structural member, σ_x = 100 MPa, σ_y = 40 MPa, and τ_xy = 30 MPa. What is the radius R of Mohr's Circle?

R = √[(σ_x – σ_y)²/4 + τ_xy²] = √[(100 – 40)²/4 + 30²] = √[30² + 30²] = √(900 + 900) = √1800 = 42.43 MPa. Note: (σ_x – σ_y)/2 = (100 – 40)/2 = 30, NOT 60 — the halving is critical.

For the stress state in Question 1, what are σ_1 and σ_2?

Center C = (100 + 40)/2 = 70 MPa. σ_1 = 70 + 42.43 = 112.43 MPa. σ_2 = 70 – 42.43 = 27.57 MPa. Quick check: σ_1 + σ_2 = 140 = σ_x + σ_y = 100 + 40 ✔

A point is in pure shear: σ_x = σ_y = 0, τ_xy = 60 MPa. What are the principal stresses and at what angle do they act?

Center = 0, R = √[0 + 60²] = 60 MPa. σ_1 = 60 MPa (tension), σ_2 = –60 MPa (compression). Angle: tan2θ_p = 2(60)/(0–0) → undefined → 2θ_p = 90° → θ_p = 45°. This explains why torsion shafts crack on a 45° helix.

A 50 mm diameter solid steel shaft carries a torque T = 1.2 kN·m and a bending moment M = 0.9 kN·m. Find τ_max using the maximum shear stress theory.

T_e = √(M² + T²) = √(0.9² + 1.2²) = √(0.81 + 1.44) = √2.25 = 1.5 kN·m = 1.5×10⁶ N·mm. τ_max = 16T_e/(πd³) = 16(1.5×10⁶)/(π×50³) = 24×10⁶/392,699 = 61.1 MPa.

On the plane of maximum shear stress, what is the normal stress for the state σ_x = 80, σ_y = 20, τ_xy = 30 MPa?

The normal stress on the maximum shear plane is ALWAYS σ_avg = (σ_x + σ_y)/2 = (80 + 20)/2 = 50 MPa. It is NOT zero. The full stress state on the max-shear plane is (50 MPa normal + 42.43 MPa shear).

For a uniaxial tension state σ_x = 120 MPa, σ_y = 0, τ_xy = 0, what is the absolute maximum shear stress?

σ_1 = 120 MPa, σ_2 = 0. In-plane τ_max = (σ_1 – σ_2)/2 = 60 MPa. Since σ_3 = 0 (plane stress), checking all three: |σ_1–σ_2|/2 = 60, |σ_1–σ_3|/2 = 60, |σ_2–σ_3|/2 = 0. Absolute max = 60 MPa.

If σ_x = 0, σ_y = 0, and τ_xy = 50 MPa, what does Mohr's Circle look like and what is the center?

Center = (σ_x + σ_y)/2 = 0. R = √[0 + 50²] = 50 MPa. The circle is centered at the origin — both positive and negative normal stresses of equal magnitude appear at 45° planes.

At a point, σ_1 = 80 MPa and σ_2 = –40 MPa. Using the Tresca criterion, at what value of uniaxial yield stress does the material just yield?

Since σ_1 and σ_2 have OPPOSITE signs, Tresca criterion gives: σ_1 – σ_2 = S_y. Therefore: 80 – (–40) = 120 MPa = S_y. With the same-sign case, Tresca would give max(|σ_1|, |σ_2|) = S_y.

A physical element is rotated 30° CCW from the reference axes. By how many degrees is the corresponding point moved on Mohr's Circle, and in which direction?

Physical rotation θ = 30° CCW corresponds to 2θ = 60° CCW on Mohr's Circle. The same-direction correspondence is maintained: CCW in physics → CCW on the circle.

The center of Mohr's Circle is at (–10, 0) MPa and the radius is 40 MPa. What are σ_1 and σ_2?

σ_1 = Center + R = –10 + 40 = +30 MPa. σ_2 = Center – R = –10 – 40 = –50 MPa. Note σ_1 > σ_2 algebraically. Quick invariant check: σ_1 + σ_2 = –20 = 2 × Center ✔

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