Multiaxial Stress and Combined Loading

Introduction

Real mechanical components rarely experience one idealized load. Shafts transmit torque while carrying bending moments; pressure-vessel walls experience membrane tension in two directions; brackets may be subjected simultaneously to axial force, shear, bending, and torsion. Strength-of-materials analysis therefore requires a method for combining stress components and determining the critical normal and shear stresses on any plane.

This chapter develops the stress-transformation equations for plane stress, introduces principal stresses and maximum shear stress, and explains Mohr’s circle as a graphical interpretation of the transformation equations. The same ideas are extended to plane strain, combined axial–bending loading, circular shafts under combined bending and torsion, and thin-walled pressure vessels.

Learning Objectives

By the end of this chapter, the reader should be able to:

1. Represent a two-dimensional stress state using normal and shear components.
2. Transform stresses to an inclined plane.
3. Determine principal stresses, principal directions, and maximum in-plane shear stress.
4. Construct and interpret Mohr’s circle for plane stress and plane strain.
5. Analyze members subjected to combined axial force, bending moment, torsion, and transverse shear.
6. Apply thin-walled pressure-vessel equations and identify their assumptions and limitations.
7. Select a suitable failure criterion for ductile or brittle materials.

1. Stress at a Point and the Sign Convention

Consider an element with x- and y-faces. The stress state in plane stress is described by the normal stresses σₓ and σᵧ and the shear stress τₓᵧ. Plane stress means that the out-of-plane components are negligible:

σ_z = τ_xz = τ_yz = 0

This condition is often appropriate for a thin plate loaded in its own plane. It is not generally valid near concentrated loads, holes, thick sections, or constrained three-dimensional regions.

A positive normal stress is tensile. A common engineering convention defines τₓᵧ as positive when it acts in the positive y-direction on the positive x-face, with complementary shear stresses maintaining moment equilibrium. Other conventions are valid, but the selected convention must be used consistently in equations and Mohr’s circle.

For an infinitesimal element in static equilibrium, complementary shear stresses are equal:

The units of all stress components are pressure units, normally Pa, MPa, or ksi.

2. Stress Transformation in Plane Stress

Let an inclined plane have an outward normal making an angle θ counterclockwise from the x-axis. Resolving forces on the inclined plane gives the transformed normal stress σₙ and shear stress τₙ:

Here θ is measured from the x-axis to the normal of the plane, not to the plane itself. The equations assume a small element, static equilibrium, and a two-dimensional stress state. The signs of τₙ depend on the adopted face and positive-direction convention.

The transformed normal stress perpendicular to the plane and the transformed shear stress parallel to it are obtained simultaneously. The factor 2θ is fundamental: a physical rotation of θ corresponds to a rotation of 2θ on Mohr’s circle.

Principal stresses

Principal planes are planes on which shear stress is zero. Setting τₙ = 0 gives the principal directions. The principal stresses are

where σ₁ is conventionally the larger algebraic value. The principal angle satisfies

The quadrant must be selected using the signs of both numerator and denominator; using an ordinary single-value arctangent can give the wrong orientation. The two principal planes are perpendicular.

Maximum in-plane shear stress

The maximum in-plane shear stress is the radius of Mohr’s circle:

It occurs on planes 45° from the principal planes. The corresponding normal stress is the average normal stress:

For a plane-stress state, σ₃ = 0 is the third principal stress. Consequently, the absolute maximum shear stress in three dimensions is

where σ_max and σ_min are the largest and smallest of σ₁, σ₂, and 0. The in-plane value is not always the absolute three-dimensional maximum.

3. Mohr’s Circle for Plane Stress

Mohr’s circle provides a graphical representation of all transformed stress states. On axes of σ horizontally and τ vertically, plot the points

The center and radius are

The circle equation is

Its intersections with the σ-axis are σ₁ and σ₂. The top and bottom points give the maximum shear stress in magnitude. Depending on convention, the positive τ direction on a textbook Mohr diagram may be drawn downward rather than upward. The numerical results are unchanged if the convention is used consistently.

Mohr’s circle is especially useful for visualizing the relationship between physical-plane rotation and stress transformation, checking calculations, and identifying critical orientations. It should not replace equilibrium-based reasoning when sign conventions are uncertain.

4. Worked Example: Plane-Stress Transformation

A material point is subjected to σₓ = 80 MPa, σᵧ = −20 MPa, and τₓᵧ = 30 MPa. Determine the principal stresses and maximum in-plane shear stress.

The average stress is

The radius is

Therefore,

and

The principal angle is found from

Thus 2θₚ ≈ 30.96°, giving θₚ ≈ 15.48° for the appropriate quadrant. The perpendicular principal plane is at approximately 105.48°. Since σ₃ = 0, the absolute maximum shear is

in this particular case, because zero lies between the two in-plane principal stresses.

5. Plane Strain and the Stress–Strain Relationship

Stress transformation concerns forces per unit area. Strain transformation concerns deformation. For small strains in a two-dimensional state, the normal strains εₓ and εᵧ and engineering shear strain γₓᵧ transform as

The factor γₓᵧ/2 appears because engineering shear strain is twice the tensor shear-strain component.

For isotropic, linear elastic material under plane stress,

where E is Young’s modulus, ν is Poisson’s ratio, and G is the shear modulus. They are related by

For plane strain, ε_z = 0, which is common in long bodies where deformation in the z-direction is constrained. The out-of-plane stress is generally not zero:

for isotropic linear elasticity under plane strain. The complete three-dimensional constitutive equations should be used when substantial out-of-plane effects or nonisotropic materials are present.

Principal strains are found in the same geometric manner as principal stresses, using εₓ, εᵧ, and γₓᵧ/2. The maximum in-plane engineering shear strain is

6. Combined Axial Force and Bending

For a prismatic beam subjected to an axial force P and bending moments Mₓ and Mᵧ, the normal stress at a point in the cross-section can be written, with a consistent coordinate convention, as

Here A is cross-sectional area, x and y locate the point, and Iₓ and Iᵧ are second moments of area about the relevant centroidal axes. The signs of the bending terms depend on the axis directions and moment convention.

For bending about one centroidal principal axis,

where M is the bending moment, c is the distance from the neutral axis, and I is the second moment of area. The combined stress must be evaluated at the extreme points of the section because the axial stress is uniform while bending stress changes linearly across the section.

If transverse shear V is also present, the average shear stress V/A is usually inadequate except for simple approximate estimates. For beams, the distribution is obtained from

where Q is the first moment of the area outside the chosen level about the neutral axis, I is the section’s second moment of area, and t is the local width or thickness. This equation assumes beam theory, a prismatic member, and a stress distribution adequately represented by the cross-sectional formulation.

Worked Example: Axial Force and Bending

A rectangular section is 40 mm wide and 80 mm deep. It carries P = 20 kN and a bending moment M = 1.2 kN·m about its centroidal horizontal axis. Find the extreme normal stresses.

The area is

The axial stress is

For bending about the horizontal centroidal axis, with b = 40 mm and h = 80 mm,

The extreme-fiber distance is c = 40 mm. Converting M = 1.2 kN·m = 1.2 × 10⁶ N·mm,

Thus the extreme stresses are

One side is in tension and the opposite side is in compression. If shear or torsion is also present, these normal stresses must be combined with the corresponding shear stresses and transformed or assessed using a failure criterion.

7. Combined Bending and Torsion in Circular Shafts

A circular shaft carrying bending moment M and torque T has, at its outer surface, bending normal stress and torsional shear stress:

For a solid circular shaft of diameter d,

so that

At a surface point where the bending stress is tensile or compressive, the local state is plane stress with σₓ = σ_b, σᵧ ≈ 0, and τₓᵧ = τ_t. The principal stresses are therefore

For a ductile material, the maximum-distortion-energy (von Mises) stress for this state is

This criterion is appropriate for many homogeneous, isotropic, ductile metals under static or slowly varying loading. It does not directly account for fatigue, stress concentrations, residual stress, plastic anisotropy, or brittle fracture.

The maximum-shear-stress (Tresca) criterion uses

and compares it with the material’s allowable shear stress or yield condition. Tresca is generally more conservative than von Mises for ductile yielding.

8. Thin-Walled Pressure Vessels

A vessel is considered thin-walled when wall thickness t is small relative to radius r; a common engineering guideline is t/r ≤ 0.1, although the acceptable limit depends on design code and geometry. Thin-wall equations assume uniform membrane stress through the thickness, negligible bending stress in the wall, and a sufficiently large vessel away from discontinuities.

For a cylindrical vessel of internal pressure p, radius r, and wall thickness t, the hoop or circumferential stress is

The longitudinal stress is

Thus the hoop stress is twice the longitudinal stress. These equations follow by cutting the vessel through a longitudinal plane and through a transverse plane and balancing pressure-resultant forces against wall tension.

For a thin-walled spherical vessel,

in every tangential direction. The radial stress is approximately −p at the inner surface and nearly zero at the outer surface relative to the membrane stresses when the vessel is thin. Near nozzles, supports, welds, end closures, and abrupt changes in thickness, local stresses can be much higher than membrane values.

Pressure-vessel example

A cylindrical tank has internal pressure p = 2 MPa, internal radius r = 500 mm, and wall thickness t = 8 mm. Neglecting end effects and local discontinuities,

The membrane state is biaxial with negligible in-plane shear. The principal stresses are therefore approximately 125 MPa and 62.5 MPa, with a third stress close to zero under the thin-wall approximation. The design must still include corrosion allowance, weld efficiency, pressure transients, temperature, fatigue, and the applicable pressure-vessel code.

9. Failure Criteria and Design Interpretation

Stress transformation provides a local stress description; a design decision requires a material and loading model. Common criteria include:

For a factor of safety n based on yield stress Sᵧ, a simple von Mises elastic design requirement is

This is not a complete design standard. Real design must account for fatigue, stress concentration factors, imperfections, buckling, fracture toughness, manufacturing variability, environmental degradation, and code requirements.

10. Applied Case: Pump-Drive Shaft

Background

Consider an illustrative steel pump-drive shaft transmitting torque from a motor while a belt load produces a transverse force. The shaft has a solid circular diameter of 40 mm at a critical section. At that section, the bending moment is 600 N·m and the transmitted torque is 900 N·m.

Problem and analysis

The outer-fiber bending stress is

The torsional shear stress is

The von Mises stress is

If the shaft steel has a yield strength of 310 MPa and a required factor of safety of 2, the allowable von Mises stress is 155 MPa. The calculated value is approximately equal to, but slightly above, the allowable value. The shaft is therefore unacceptable without refinement, even though each individual stress may appear moderate.

Solution and engineering decision

Possible design changes include increasing diameter, reducing the overhung load, shortening the moment arm, selecting a stronger material, or modifying the geometry to reduce stress concentration. Because shaft loading is often cyclic, a fatigue analysis must follow the static check. Keyways, shoulders, and splines require concentration factors; using nominal stresses alone can be unsafe.

Learning points

Combined loading can produce a critical equivalent stress even when no individual load is extreme. The governing location is the point where the local combination of normal and shear stress is most severe. A valid design also requires consideration of fatigue, manufacturing details, lubrication, alignment, and applicable machinery standards.

11. Best Practices, Limitations, and Contemporary Tools

A reliable workflow is:

1. Draw the free-body diagram and determine internal forces and moments.
2. Identify the critical section and critical points on that section.
3. Calculate normal and shear stress components using consistent units.
4. State the stress sign convention.
5. Transform the stress state or calculate principal/equivalent stresses.
6. Compare results with a criterion appropriate to the material and loading.
7. Apply factors of safety and check code requirements.
8. Investigate fatigue, stress concentrations, instability, thermal stress, and local three-dimensional effects.

Finite-element analysis is valuable for complex geometry, but its results depend on mesh quality, boundary conditions, material models, and interpretation. Singular stresses at sharp idealized corners should not be accepted uncritically as physical infinite stresses. Experimental strain gauges, digital image correlation, and photoelastic methods can validate models. AI-assisted design and automated optimization can explore geometry rapidly, but engineers remain responsible for load definitions, verification, traceability, safety margins, and compliance with regulations.

Chapter Summary

Multiaxial stress analysis describes the normal and shear stresses acting on arbitrary planes through a material point. The plane-stress transformation equations determine the stresses on an inclined plane, while principal stresses occur where transformed shear stress is zero. Mohr’s circle gives the same results graphically: its center is the average normal stress and its radius is the maximum in-plane shear stress.

Plane strain requires strain transformation and a constitutive relation; unlike plane stress, it can produce a nonzero out-of-plane stress. Combined axial force and bending produce uniform plus linearly varying normal stress, while transverse shear follows the VQ/(It) relation in beam theory. Circular shafts under bending and torsion have simultaneous normal and shear stress, commonly assessed using von Mises or Tresca criteria. Thin-walled pressure vessels develop hoop and longitudinal membrane stresses, with hoop stress twice longitudinal stress in a cylinder. All formulas depend on assumptions, and sound design must additionally consider fatigue, stress concentrations, discontinuities, materials, codes, and uncertainty.

Key Terms

Stress transformation: Calculation of stress components on an inclined plane.

Plane stress: A state in which out-of-plane stress components are negligible.

Plane strain: A state in which one strain component, commonly ε_z, is constrained to zero.

Principal stress: A normal stress acting on a plane where shear stress is zero.

Principal plane: A plane carrying a principal stress and no shear stress.

Maximum in-plane shear stress: The largest shear stress obtained within the plane-stress transformation.

Absolute maximum shear stress: The largest shear stress considering all three principal stresses.

Mohr’s circle: A graphical construction representing transformed normal and shear stresses.

Normal stress: Force per unit area acting perpendicular to a surface.

Shear stress: Force per unit area acting parallel to a surface.

Axial stress: Normal stress produced by an axial force.

Bending stress: Normal stress produced by a bending moment.

Torsional stress: Shear stress produced by torque.

Second moment of area: A geometric property governing resistance to bending.

Polar moment of area: A geometric property governing resistance to circular-shaft torsion.

Hoop stress: Circumferential membrane stress in a pressure vessel.

Longitudinal stress: Axial membrane stress in a cylindrical pressure vessel.

Von Mises stress: An equivalent stress commonly used to predict ductile yielding.

Tresca criterion: A yielding criterion based on maximum shear stress.

Stress concentration: Local stress amplification caused by geometric discontinuity.

Review Questions

Short Answer Questions

1. What is meant by plane stress, and when is it a reasonable approximation?
2. Define a principal plane and a principal stress.
3. What do the center and radius of Mohr’s circle represent?
4. Why is hoop stress twice longitudinal stress in a thin cylindrical pressure vessel?
5. State two assumptions underlying thin-walled pressure-vessel equations.

Long Answer Questions

1. Derive the plane-stress transformation equations by force equilibrium on an inclined element.
2. Explain how principal stresses and maximum shear stress are obtained from Mohr’s circle.
3. Discuss the differences between plane stress and plane strain, including their constitutive relationships.
4. Explain how axial force, bending moment, and transverse shear combine in a beam cross-section.
5. Compare the von Mises and Tresca criteria for ductile-material design.

Analytical / Application Questions

1. A point has σₓ = 60 MPa, σᵧ = 20 MPa, and τₓᵧ = 40 MPa. Determine the principal stresses and maximum in-plane shear stress.
2. A solid shaft of diameter 50 mm carries M = 800 N·m and T = 1.2 kN·m. Determine the surface bending stress, torsional shear stress, and von Mises stress.
3. A thin cylindrical vessel has p = 1.5 MPa, radius 400 mm, and thickness 6 mm. Determine the hoop and longitudinal stresses and discuss whether a local discontinuity could invalidate the membrane estimate.

Multiple Choice Questions with Answers

Question 1

For a plane-stress state, the principal stresses are found from which expression?

A. σ₁,2 = (σ_x + σ_y)/2 ± √[((σ_x − σ_y)/2)² + τ_xy²]

B. σ₁,2 = σ_x + σ_y ± τ_xy

C. σ₁,2 = (σ_x − σ_y)/2 ± τ_xy²

D. σ₁,2 = σ_xσ_y/τ_xy

Correct Answer: A. σ₁,2 = (σ_x + σ_y)/2 ± √[((σ_x − σ_y)/2)² + τ_xy²]

Question 2

What does the radius of Mohr’s circle represent for plane stress?

A. Average normal stress

B. Maximum in-plane shear stress

C. Longitudinal pressure-vessel stress

D. The axial force

Correct Answer: B. Maximum in-plane shear stress

Question 3

A cylindrical thin-walled pressure vessel has p = 2 MPa, r = 500 mm, and t = 8 mm. Its hoop stress is:

A. 31.25 MPa

B. 62.5 MPa

C. 125 MPa

D. 250 MPa

Correct Answer: C. 125 MPa

Question 4

For a solid circular shaft, the torsional shear stress at the outer surface is:

A. 8T/(πd³)

B. 16T/(πd³)

C. 32T/(πd³)

D. 64T/(πd³)

Correct Answer: B. 16T/(πd³)

Question 5

In a thin cylindrical pressure vessel, the hoop stress is ideally how many times the longitudinal stress?

A. One-half

B. Equal

C. Two

D. Four

Correct Answer: C. Two

Question 6

Which criterion is commonly associated with distortional energy and ductile yielding?

A. Maximum principal stress

B. Von Mises

C. Maximum principal strain only

D. Euler buckling

Correct Answer: B. Von Mises

Question 7

For a rectangular beam section, the beam-theory transverse shear formula is:

A. τ = VQ/(It)

B. τ = Mc/I

C. τ = pr/t

D. τ = P/A

Correct Answer: A. τ = VQ/(It)

Question 8

If σₓ = 80 MPa, σᵧ = −20 MPa, and τₓᵧ = 30 MPa, the average normal stress is:

A. −50 MPa

B. 30 MPa

C. 50 MPa

D. 100 MPa

Correct Answer: B. 30 MPa

Question 9

Under plane strain in an isotropic material, ε_z = 0 generally implies that:

A. σ_z must also be zero

B. σ_z may be nonzero

C. All shear stresses vanish

D. The material is necessarily plastic

Correct Answer: B. σ_z may be nonzero

Question 10

For combined bending and torsion at a shaft surface, the von Mises stress is:

A. σ_b + τ_t

B. √(σ_b² + 3τ_t²)

C. σ_bτ_t/3

D. σ_b² − τ_t²

Correct Answer: B. √(σ_b² + 3τ_t²)