MAE 3201 - Mechanics of Materials Course Notes · MAE 3201 - Mechanics of Materials Course Notes Brandon Runnels About These notes are for the personal use of students who are enrolled

MAE 3201 - Mechanics of MaterialsCourse NotesBrandon Runnels

About

These notes are for the personal use of students who are enrolled in or have taken MAE3201 at the Universityof Colorado Colorado Springs in the Fall 2017 semester. Please do not share or redistribute these notes withoutpermission.ContentsLECTURE 1 0 Introduction to Mechanics of Materials 1.1

1 Introduction and Statics Review 1.11.1 Linear algebra review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1

1.1.1 Scalars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.11.1.2 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.11.1.3 Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2

1.2 Rigid body equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.31.2.1 Equilibrium equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.31.2.2 Distributed loadings and effective loading . . . . . . . . . . . . . . . . . 1.31.2.3 Reaction forces and moments . . . . . . . . . . . . . . . . . . . . . . . 1.41.2.4 Solution strategy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4

LECTURE 2 1.3 Shear/moment diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.21.4 Centroids, centers of mass, and second moments of area in 2D . . . . . . . . . 2.4

LECTURE 3 2 Stress and Strain 3.12.1 Normal stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1

2.1.1 Engineering versus true stress . . . . . . . . . . . . . . . . . . . . . . . 3.22.1.2 Sign convention . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.22.1.3 Continuous uniaxial stress distribution . . . . . . . . . . . . . . . . . . . 3.32.1.4 Saint-Venant’s principle . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3

2.2 Normal strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.42.2.1 True strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4

2.3 Material Response in 1D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.52.3.1 Stress-strain plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.52.3.2 Hooke’s law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.62.3.3 Plastic strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7

LECTURE 4 2.3.4 Strain energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.12.3.5 Thermal strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1

2.4 Problems in uniaxial loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3LECTURE 5 2.4.1 Stress concentration factors . . . . . . . . . . . . . . . . . . . . . . . . 5.1

2.4.2 Factor of safety . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2

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2.5 Shear stress and shear strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.22.5.1 Hooke’s law for shear stress and strain . . . . . . . . . . . . . . . . . . . 5.4

2.6 Stress and strain in 2D and 3D . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.42.6.1 The stress tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4

LECTURE 6 2.6.2 Stress equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.12.6.3 The strain tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.22.6.4 Hooke’s law in 2D and 3D . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3

LECTURE 7 2.6.5 Other material constants . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.13 Transverse Loading: Mechanics of Beams 7.1

3.1 Euler-Bernoulli beam theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.13.1.1 Derivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.13.1.2 Boundary conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4

LECTURE 8 3.2 Singularity functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.23.2.1 Dirac delta distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.23.2.2 Heaviside functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.43.2.3 Ramp functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.43.2.4 Unit doublet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.53.2.5 Singularity bracket notation . . . . . . . . . . . . . . . . . . . . . . . . . 8.5

LECTURE 9 3.3 Solution strategy and examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1LECTURE 10 3.4 Composite beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1LECTURE 11 3.5 Shear stresses in beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1

3.5.1 The shear formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.13.5.2 Shearing stresses in thin-walled beams . . . . . . . . . . . . . . . . . . 11.33.5.3 Shear flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.5

3.6 Stress concentrations in beams . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.53.7 Beam failure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.53.8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.6

LECTURE 12 4 Torsional Loading: Mechanics of Shafts 12.14.1 Derivation of torsion equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.14.2 Boundary conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.5

LECTURE 13 4.3 Composite shafts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.14.4 Stress concentrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.44.5 Power transmission in shafts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.44.6 Non-axisymmetric shafts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.5

LECTURE 14 4.7 Combined loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.1LECTURE 15 4.7.1 Von Mises stress & yield criterion . . . . . . . . . . . . . . . . . . . . . . 15.1

5 Axial Loading: Mechanics of Columns 15.25.1 Derivation of buckling equations and critical buckling loads . . . . . . . . . . . . 15.2

5.1.1 Buckling loads derived from the pin-pin solution . . . . . . . . . . . . . 15.4LECTURE 16 5.1.2 Buckling load equations derived independently from the pin-pin solution 16.1

5.1.3 Effective Length Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.3LECTURE 17 6 Volumetric Loading: Thin-wall Pressure Vessels 17.1

6.1 Derivation of pressure vessel formulae . . . . . . . . . . . . . . . . . . . . . . . 17.16.1.1 Spherical pressure vessels . . . . . . . . . . . . . . . . . . . . . . . . . . 17.1

LECTURE 18 6.1.2 Cylindrical pressure vessels . . . . . . . . . . . . . . . . . . . . . . . . . 18.1


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6.2 Stress concentrations in pressure vessels . . . . . . . . . . . . . . . . . . . . . . 18.27 Transformations of Stress and Strain 18.3

7.1 Review of tensors, normal stress, and shear stress . . . . . . . . . . . . . . . . . 18.3LECTURE 19 7.2 Tensor transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.1LECTURE 20 7.3 Principal stresses and directions . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.1

7.3.1 Solution strategy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.3LECTURE 21 7.4 Mohr’s circle for plane stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.1LECTURE 22 7.5 Principal Strains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.2LECTURE 23 8 Energy Methods 23.1

8.1 Strain energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23.18.1.1 Strain energy in Euler-Bernoulli beams . . . . . . . . . . . . . . . . . . . 23.28.1.2 Strain energy in shafts under torsion . . . . . . . . . . . . . . . . . . . . 23.3

8.2 Applied work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23.48.2.1 Work done by a loading on a beam . . . . . . . . . . . . . . . . . . . . . 23.4

LECTURE 24 8.2.2 Work done by couple moments . . . . . . . . . . . . . . . . . . . . . . . 24.18.3 Principle of minimum potential energy . . . . . . . . . . . . . . . . . . . . . . . . 24.3

8.3.1 Castigliano’s first theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 24.38.3.2 Castigliano’s second theorem . . . . . . . . . . . . . . . . . . . . . . . . 24.5

A Table of parameters for common materials A.1B Selected stress concentration factor calculations B.1

B.1 Uniaxial Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B.1B.2 Beam Bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B.1B.3 Shaft Torsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B.2




ExamplesExample 1.1: Matrix-vector multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3Example 1.2: 2D Equilibrium Example [2] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1Example 1.3: Shear-moment for cantilever beam with uniform load . . . . . . . . . . . . . . . . . . . . . . . 2.3Example 1.4: Moment of inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5Example 2.1: Ultimate tensile load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2Example 2.2: Fracture strain of steel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4Example 2.3: Yield strain of steel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6Example 2.4: Shrink fit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2Example 2.5: Uniaxial loading in linkage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3Example 2.6: Thermal expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4Example 2.7: Stress concentration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1Example 2.8: Normal/shear stress using statics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3Example 2.9: Tractions in material under pure shear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5Example 2.10: Strain tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2Example 2.11: 3D stress/strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4Example 3.1: Simply supported beam with point load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6Example 3.2: Statically indeterminate beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1Example 3.3: Beam with U-shaped cross-section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1Example 3.4: Simply supported beam with singularity functions . . . . . . . . . . . . . . . . . . . . . . . . . 9.4Example 3.5: Carbon reinforced steel beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2Example 3.6: Shear stress in rectangular beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2Example 3.7: Thin-wall shearing force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3Example 4.1: Loaded shaft with bearings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.3Example 4.2: Distributed load on a statically indeterminate beam . . . . . . . . . . . . . . . . . . . . . . . . 12.5Example 4.3: Woven carbon fiber-reinforced drive shaft . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.1Example 4.4: Composite drive shaft . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2Example 4.5: Combined loads on a shaft . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.1Example 5.1: Thermal buckling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.3Example 6.1: Beach ball . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.2Example 6.2: Cylindrical pressure vessel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.1Example 7.1: Uniaxial tension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.1Example 7.2: Numerical example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.2Example 7.3: Maximum shear stress in a cylinder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.3Example 7.4: Computation of principal stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.2Example 7.5: Manual computation of eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.4Example 7.6: Uniaxial tension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2Example 7.7: Pressure vessel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3Example 7.8: Eigenvalues using Mohr’s circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.4Example 7.9: Combined torsion and bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.1Example 7.10: Strain Rosette . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.3Example 8.1: Strain energy of simply supported beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23.2Example 8.2: Shaft under constant twisting moment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23.3Example 8.3: Work done on cantilever beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23.4Example 8.4: Work done by point force on simply supported beam . . . . . . . . . . . . . . . . . . . . . . . 24.1Example 8.5: Cantilever beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24.3Example 8.6: Simply supported beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24.5Example 8.7: Cantilever beam continued . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24.6



Lecture 1 Introduction

0 Introduction to Mechanics of Materials

Welcome to Mechanics of Materials. This is class is a natural sequal to Engineering Statics, as statics forms thebasis for the problems we will be solving here.1. Solving problems that are statically indeterminate (and consequently unsolvable using themethods of statics)by accounting for the deflection of the material itself.2. Determining the material response and possible conditions for failure by solving for stresses inside the mate-

rial under a set of loading conditions.In this course we will begin with a brief statics review, and then spend some time introducing the subjects ofstress and strain in a material. From there, we will take these tools and use them to solve problems for a varietyof different loading scenarios. This includes beam theory, shaft theory, mechanics of columns, and mechanicsof pressure vessels. We will also develop the notions of principal stresses and strains, and conclude with energymethods.

1 Introduction and Statics Review

Almost all of thematerial covered in Statics is crucial to being able to solve problems in themechanics ofmaterials,so we will start with a brief review.1.1 Linear algebra review

We begin by defining the elementary mathematical structures from linear algebra that we will use in this class.1.1.1 Scalars

A quantity is said to be a scalar quantity if it requires only one number to be completely described. It can be positiveor negative but has no direction. Examples of scalar quanitites are: mass, temperature, volume, density, time, etc.1.1.2 Vectors

A vector is a quantity that has both a direction and a magnitude. In course notes we represent vectors using bold-face (e.g. x) and in handwriting/on the board we use an underline (e.g. x). Here vector quantities typically areindicated by a lowercase letter. Examples of vector quantities are: position, velocity, acceleration, force, displace-ment, etc.Let us review the notation and some of the basic operations when dealing with vectors:NOTATION As an example of vector notation, suppose that we have a vector x with components 1, 2, and 3

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in the x , y , z directions, respectively. Ways of writing this vector include:x =

[123

]= ex + 2ey + 3ez = i + 2j + 3k = 〈1, 2, 3〉 (1.1)

Here, we will stick to either column vector notation or basis vector notation. i, j, k notation androw vector notation is allowed but discouraged.VECTOR NORM The norm of a vector is its total magnitude and is computed using the Pythagorean theorem.For instance the norm of a vector x = a ex + b ey + c ez is

|x| =√a2 + b2 + c2 (1.2)

DOT PRODUCT A dot product (or inner product, or scalar product) is the product of two vectors that returns thescalar product of their magnitudes times the cosine of the area between them. The dot productof two vectors x, y is given byx =

[abc

]y =

[def

]〈x, y〉 = xTy = [a b c]

[def

]= ad + be + cf = |x| |y| cos θ (1.3)

Note that you can compute the norm of a vector by taking the square root it dotted with itself:|x| =

√〈x, x〉 (1.4)

Also note that if the dot product of two non-zero vectors is zero, then cos θ = 0, implying thatθ = 90, and the vectors are said to be orthogonal.

CROSS PRODUCT The cross product of two vectors returns another vector that is orthogonal to both. The mag-nitude of the cross product of two vectors is equal to the product of their magnitudes timesthe sine of the angle between them. The cross product is most easily computed using thedeterminant mnemonic:x× y = det

[ex ey eza b cd e f

]= ex (bf − ce) + ey (cd − af ) + ez (ae − bd) (1.5)|x× y| = |x||y| sin θ (1.6)

Note that if the cross product is zero than sin θ = 0 and θ = 0, implying that the vectors arecolinear.1.1.3 Tensors

It is likely that you have never heard of tensors before, and that’s ok. We’ll introduce a somewhat limited, but fairlyaccurate definition:Definition 1.1. A tensor is a matrix that has transformation properties.

You don’t need to be too concerned with what the “transformation properties” are right now – that will come uplater. For all current intents and purposes, a tensor is a 3x3 matrix. Here is an example of a tensor in 3D:T =

[a b cd e fg h i

](1.7)

It is possible to represent tensors using basis vectors (similarly to vectors) but we will stick with this representationhere.Why do we need tensors? Just as velocity cannot be represented with a scalar, there are physical quantities (tensorquantities) that cannot be represented with vectors. Some examples are:All content © 2017, Brandon Runnels 1.2



• Curvature• Thermal conductivity in anisotropic materials

• Moment of inertia• Stress and strain

(In mechanics of materials, we will talk about stress and strain tensors a lot.)Another way of thinking about a tensor (or a 3x3 matrix) is as a machine that transforms one vector into another.Let’s illustrate this with an example:

Example 1.1: Matrix-vector multiplication

Compute Ax, ifA =

[cos θ − sin θ 0sin θ cos θ 0

0 0 1

]x =

[100

](1.8)

To find this we use matrix-vector multiplication:Ax =

[cos θ − sin θ 0sin θ cos θ 0

0 0 1

][100

]=

[(cos θ)(1) + (− sin θ)(0) + (0)(0)(sin θ)(1) + (cos θ)(0) + (0)(0)

(0)(1) + (0)(0) + (1)(0)

]=

[cos θsin θ

0

](1.9)

which is a vector that has been rotated by an angle θ about the z axis. So in this case, A is called a rotationmatrix about the z axis.

1.2 Rigid body equilibrium

Here we review briefly some of the key concepts learned in MAE 2103 that will carry over to this course.1.2.1 Equilibrium equations

The equations for a rigid body in equilibrium are:∑fi = 0

∑i

Mi =∑

i

(ri × fi ) +∑

i

Mcouplei = 0 (1.10)

where fi are all of the different forces acting on our system and ri are the vectors pointing from the “reference point”(which can be any point in space) to the point of application of the load.In 2D, equilibrium conditions result in three equations (force balance in the x direction, force balance in the y direc-tion, moment balance in the z direction). In 3D, they result in six equations – force balance in all three directions,and moment balance in all three directions.Therefore, a rigid body in 2D must have exactly three unknowns, while a rigid body in 3D must have exactly six. Ifthe rigid body has more equations than unknowns, it is said to have insufficient constraints. On the other hand, ifit has more unknowns than equations, it is said to be statically indeterminate.1.2.2 Distributed loadings and effective loading

We often treat loadings as if they are concentrated at a point. However, in reality, loadings are always distributedover some finite distance. Additionally, there are often loads that are spread out and vary over a distance, like asack of sand.




f (x)ey feffèff

L L

If we have a distributed loading f (x) over a beam of length L we can compute the total force exerted by the forceusing integration:f =

∫ L

0

f (x) ey dx = feff ey (1.11)We can also compute the total moment generated by this distributed load about the origin:

M =

∫ L

0

(x ex )× (f (x) ey ) dx =

∫ L

0

f (x) x dxez = Meff ez (1.12)We can then use this result to compute what distance (èff ) a point load with magnitude feff would have to be togenerate the same moment, and find that

èff =Meff

feff(1.13)

It can be convenient to reduce large distributed loads to point loads when doing static analysis. However: do notreduce distributed loads to point loads when doing beam theory!.1.2.3 Reaction forces and moments

In most cases, rigid bodies are subjected both to applied loads as well as some type of constraint. Constraints fixthe displacement or rotation (or both) of the member by applying a force or moment of an unknown magnitude.Below are a couple of types of constraints in 2D:

froll = frolyey froll = frolxex + frolyey

ffix = ffixxex + ffixyey

Mfix = Mfixzez

free end - 0 unknowns roller - 1 unknown pivot - 2 unknowns fixed - 3 unknowns

Generally, the number of unknowns from the constraints must add up to three (in 2D) or six (in 3D); for instancea system can be constrained with three rollers, a roller and a pivot, or a fixed constraint, etc. Notice that eachconstraint contributes a force component in each direction inwhich force is constrained, and amoment componentfor each direction in which moment is constrained. This is a good way to determine what unknowns contribute toa system.1.2.4 Solution strategy

The following strategy is recommended for solving equilibrium problems:




Solution strategy for static equilibrium of a rigid body1. Determine how many unknowns are in the system.2. Draw a free body diagram.3. Write down the equations of equilibrium. (Make sure the number of equations matches the number ofunknowns.)4. Solve the system for the unknowns.5. Sanity check: units, limiting values6. Substitute numbersWe’ll illustrate this with an example.



Lecture 2 Equilibrium, internal forces, centroid, moments of area

Example 1.2: 2D Equilibrium Example [2]

Find the reactions at the pivot and the roller supporting the following rigid body subjected to a point loadand a graduated load as shown in the figure below

f

L

w = w0xL

H

L/2

To solve this problem, follow the general solution strategy:1. Unknowns: fpiv ,x , fpiv ,y , frol

2. Free body diagram:first, let’s reduce the distributed load: find the total force

Feff ,y =

∫ L

0

w0x

Ldx =

w0x2

2L

∣∣∣∣L0

=w0L

2(1.14)

find the total moment:Meff ,z = −

∫ L

0

w0x2

Ldx =

w0x3

3L

∣∣∣∣L0

=w0L

2

3(1.15)

and the effective location:Leff = −Meff

Feff=

w0L2/3

w0L/2=

2

3L (1.16)

and use this in our FBD:

f =

[0−f0

]L

H

fpiv =

[fpiv ,xfpiv ,y

0

]frol =

[0

frol ,y0

]2L/3 Feff =

[0

−w0L/20

]L/2



3. Write equations:force equilibrium: [

fpiv ,x

fpivy + frol ,y − f − w0L/20

]=

[000

](1.17)

moment equilibrium: select the pivot as the point, so the distance vectors arer =

[L/2H0

]reff =

[2L/3

00

]rrol =

[L00

](1.18)

and the moments areM =

[00

−fL/2

]Meff =

[00

−w0L2/3

]Mrol =

[00

Lfrol ,y

](1.19)

so the moment equilibrium is [00

−fL/2− w0L2/3 + Lfrol ,y

]=

[000

](1.20)

4. Solve: from x equilibrium we have that fpiv ,x = 0. What does this mean?moment equilibrium gives:

frol ,y =1

2f +

1

3w0L (1.21)

y force equilibrium gives:fpiv ,y = f +

1

2w0L− frol ,y = f +

1

2w0L−

1

2f − 1

3w0L =

f

2+

w0L

6(1.22)

5. Sanity check: do the units check out? Yes. If I substitute f = w0 = 0 do I get zero for all of myreactions? Yes.

1.3 Shear/moment diagrams

Consider a beam subjected to a loading w(x) as shown in the following figure:w(x)

This loading induces internal loading within the beam in the form of normal force (N(x)), shear force (V (x)), andbending moment M(x), where we adopt the following directionality convention.




M(x1)

N(x1)

V (x1)

N(x1)

V (x1)M(x1)

N(x2)

V (x2)

M(x2)

M(x2)

N(x2)

V (x2)

It is of interest to determine what the shear/normal forces, and bending moments are for all points in the beam.There are two methods of doing this:1. Method of sections: take a section of the beam, and use force/moment balance to solve forM(x),N(x),V (x).2. Integration: one can show that

dV

dx= w(x)

dM

dx= V (x)e

dN

dx= 0 (1.23)

Integrating loading once returns the shear, and integrating the shear gives the bending moment. Boundaryconditions can then be applied to solve for the integration constants.We will illustrate how this works with a simple example:

Example 1.3: Shear-moment for cantilever beam with uniform load

Consider a beam with a fixed end subjected to a uniform load −w0 as shown in the following figure:

w(x) = −w0

L

Find the equations for and plot the shear and moment diagrams for this beam.We can solve this one of two ways, we will use both the method of sections and the method of integration.method of sections: For this method we must first find the reactions at the fixed end. For this, we canreplace the distributed load with a point load magnitude Lw0 acting downwards at L/2. Consequently wehave that

V (0) = w0L M(0) = −w0L2

2(1.24)

Now, we draw a free body diagram with variable width:

x

V (0)

M(0)

V (x)

M(x)

We can now solve for V (s) by using force balance in the y direction:∑fy = w0L− w0x − V (x) = 0 =⇒ V (x) = w0(L− x) , (1.25)




and we solve for M(x) using moment balance in the z direction about the right end:∑

Mz = −w0Lx +w0x

2

2+

w0L2

2+ M(x) =

w0

2(L− x)2 + M(x) = 0 =⇒ M(x) = −w0

2(L− x)2 (1.26)

integration: For this method we simply integrate.V (x) =

∫w(x) dx = −w0x + C1 = 0 (1.27)

Using boundary conditions:V (L) = −w0L + C1 = 0 =⇒ C1 = w0L =⇒ V (x) = w0(L− x) X (1.28)

Integrating again to get moment:M(x) =

∫V (x) dx = w0

(Lx − x2

2

)+ C2 (1.29)

Using boundary conditions:M(L) =

w0L2

2+ C2 = 0 =⇒ C2 = −w0L

2

2=⇒ M(x) = −w0

2(L− x)2 X (1.30)

1.4 Centroids, centers of mass, and second moments of area in 2D

Given a region Ω, the coordinates of the centroid of the area are given byx =

1

|Ω|

∫Ω

x dA y =1

|Ω|

∫Ω

y dA (1.31)where |Ω| is the area of the region. If the region corresponds to a body with uniform density, the center of mass isthe same as the centroid. If the region does not have uniform density, then the coordinates of the center of massis given by

x =1

m

∫Ω

ρ(x , y) x dA y =1

m

∫Ω

ρ(x , y) y dA (1.32)wherem is the total mass of the body and ρ is the density (mass per unit area). If there are several bodies Ω1, Ω2, ...with masses m1,m2, ... and centers of gravity (x1, y1), (x2, y2), ..., then the coordinates of the center of mass of thecombined body is

xcombined =m1x1 + m2x2 + ...

m1 + m2 + ...ycombined =

m1y1 + m2y2 + ...

m1 + m2 + ...(1.33)

The following figure illustrates possible centers of mass/centroids for some objects.

xy

x

y(x1, y1)

(x2, y2)

(x , y)

Ω Ω

ρ(x , y)

Ω1

Ω2

centroid center of mass composite c.o.m./centroid




The second moments of area for an object Ω are defined as the following:Ixx =

∫Ω

y2 dA Iyy =

∫Ω

x2 dA Ixy =

∫Ω

x y dA Izz =

∫Ω

(x2 + y2) dA = Ixx + Iyy (1.34)where Ixx is the moment of area about the x axis, Iyy about the y axis, and Izz about the z axis. The value of themoment of area is dependent on the placement of the axes with respect to the object. Here we note the followingimportant points:

• Secondmoments of area are often referred to asmoments of inertia. This is incorrect as moments of inertiainclude the mass of the material, which is necessary when doing dynamics. This misnomer is very frequently(almost universally) used so you should be aware of it.• Here, and predominantly in beam theory, the moment of inertia about the x axis is used far more frequentlythan the others, and is typically denoted by I.• The quantity Izz is frequently called the polar moment of inertia and is often denoted as J .We recall the use and computation of second moments of area by doing a simple example:

Example 1.4: Moment of inertia

Compute the second moment of area Ixx about the two horizontal axes below for the following rectangularregion.

H

W

For the lower axis:Ixx =

∫ W

0

∫ H

0

y2 dy dx =

∫ W

0

H3

3dx =

WH3

3(1.35)

For the center axis:Ixx =

∫ W

0

∫ H/2

−H/2

y2 dy dx =

∫ W

0

y3

3

∣∣∣H/2

−H/2dx =

WH3

12(1.36)

If we already know Ixx about the axis corresponding to the centroid of the body, we can compute Ixx for thelower axis using the parallel axis theorem:I′xx = |Ω|∆y2 + Ixx (1.37)

where |Ω| is the area of the region and ∆y is the distance between the two parallel x axes. In this case wehaveI′xx = (HW )

(H2

)2

+WH3

12=

WH3

4+

WH3

12=

WH3

3X (1.38)



Lecture 3 Introduction to stress and strain in 1D

2 Stress and Strain

Let us suppose that we want to measure the properties of a material. We consider a bar with length L and cross-sectional area A. We subject it to a force and increase the load gradually until it breaks. Wemeasure that the beamelongated by a total distance of ∆u before it broke, and that the force required to break it was f .

f1 f1

u1

But suppose someone comes along and tries to repeat the experiment with a bar of the samematerial that has thesame length but a greater cross-sectional area. What happens?

f1 f2

u21 u2

2

The experimenter concludes that the force f does not break the bar, in fact, it does not even produce the sameelongation. Suppose that yet another experimenter attempts to repeat the experiment with a bar of the samematerial and cross section, but different length. What happens?

f1 f1

u11

The experimenter finds that upon loading thematerial with a force f it does indeed break, but that id did not elongatenearly as much before failure.What is the problem with these experiments? They do not tell us anything about the material itself, because itis measuring quantities that are dependent on the geometry of the specimen. To measure the properties of thematerial, we need to measure stress and strain.2.1 Normal stress

Stress is defined as force per unit area; the SI units for stress are Pascals, the same as those for pressure. Considerthe following case:ff

A1

A2



The force on each bar is the same, but the stress in the beam is, on average,:σ1 =

f

A1σ2 =

f

A2(2.1)

Because A2 > A1, σ1 > σ2; therefore, we conclude that because the material breaks under the same stress, thefirst beam would break under a lighter load.Example 2.1: Ultimate tensile load

AISI 1010 carbon steel can withstand a tensile stress of approximately 365 MPa. What is the maximumlongitudinal tensile load that a cylindrical sample with diameter 1cm can take before breaking?We know that σ = 365× 106Pa. The cross-sectional area is A = πr2 = π(0.005m)2 = 7.8540× 10−5m2. So,σ =

f

A=⇒ f = σ A = (365× 106Pa)× (7.8540× 10−5m2) = 28, 667N = 28.667kN (2.2)

2.1.1 Engineering versus true stress

Consider a sample of material with cross section Ainitial. The material is subjected to a load f that causes thecross-setional area to change in one of the following ways shown below:

Ainitial

Aload

Aload

Almost all materials shrink laterally when subjected to a tensile load (we will talk about this in a later section).Another phenomon that is generally observed is called “necking” in which the strain concentrated around a regionprior to failure. You see that we have two different areas to choose from – so which do we choose when computingstress? The multiple choices of cross-sectional area prompt two seperate definitions of stress:Definition 2.1. Uniaxial engineering stress is defined as the force per unit undeformed area.Definition 2.2. Uniaxial true stress is defined as the force per unit deformed area.

Which stress is correct? The answer is that neither are “correct” because stress is an invention that we use tounderstand material and we can define it however we wish. Furthermore (and more importantly) both definitionsof stress are useful and significant as long as they are used consistently.True stress is a more accurate measure of the physical state of the material. However, it is very tricky to measureexperimentally, as it requires the experimenter to constantly adjust the cross sectional area during the experiment.Engineering stress is the easiest to measure and is the standard for measuring material properties. (In addition,material deformation is often so small that the difference between engineering and true stress is negligible.) So,in this course (and in most other cases as well) you can assume that “stress” refers to engineering stress unlessspecified otherwise.2.1.2 Sign convention

Here we pause to make an important point regarding the positivity/negativity of stress. In the above cases, howdo we know that the stress is positive? In fact, because stress in the material is the result of equal and oppositeforce, there is no obvious way to define the sign.All content © 2017, Brandon Runnels 3.2



tension - positive compression-negative

Therefore, we adopt the convention that uniaxial stress in a material is positive if the material is in tension, andnegative if the material is in compression.Why do we emphasize this? We stress its importance because it is actually the reverse of what is intiutive and ofthe convention used in fluid mechanics. As an example: we always assume that atmospheric pressure is positive,not negative, even though the pressure is compressive. In solid mechanics it really does make more sense fortension to be negative, but keep in mind that it is not a universal convention.2.1.3 Continuous uniaxial stress distribution

So far we have not defined uniaxial stress precisely, rather we have specified a means of approximating the stressas a uniform field. Here we introduce a more comprehensive definition of uniaxial stress:Definition 2.3. Uniaxial stress σ is defined such that, for any cross-sectional area A, the integral over the area isequal to the total force f acting on the area, i.e. ∫

A

σ dA = f (2.3)In other words, the stress field is defined as the field such that integrating σ over any region gives the total forceacting on the region:

σ(x , y) f

2.1.4 Saint-Venant’s principle

We take this opportunity to justify a simplification that we started making initially. When a material is loaded witha single point load, what is the stress at the point of application? It is infinite, because it is a finite amount of loadconcentrated at a infinitesimally small region. However, this effect dies out further along the beam, as shown inthe following figure:f

This is an effect that is a result of Saint-Venant’s principle, which states“The difference between the effects of two different but statically equivalent loads becomes very smallat sufficiently large distances from load”

Or, put another way,For a longitudinal load, at a sufficient distance from the point of application of a point load, the stress isdistributed evenly across the cross-section of the material.




2.2 Normal strain

Consider a sample of material of length L subjected to a force. The force causes the material to deform such thatit elongates to a length L + u as shown in the figure below.L L u

The engineering strain for the material is given asε =

∆L

L=

(L + u)− L

L=

u

L(2.4)

What is useful about this definition? Suppose we had a sample of material that was half as long, L/2, and the totaldisplacement is also half as long, u/2. What is the strain in this case?ε =

(u/2)

(L/2)=

u

L(2.5)

...exactly the same as the strain in the case. Engineering strain is useful because it is a measure of deformationthat is independent of the length of the material.Example 2.2: Fracture strain of steel

A 10cm piece of steel is subjected to a steadily increasing load until it breaks. Just before it breaks, it ismeasured to have a length 10.01cm. What was the strain of the material when it broke?ε =

10.01cm − 10cm

10cm=

.01cm

10cm= 0.001 = 0.1% (2.6)

Notice that strain is a unitless quantity, and we occasionally refer to strain using percentages. (Don’t forget thesuitable conversion from percent to decimal value, i.e. 0.1% = 0.001.)2.2.1 True strain

Recall that we faced a dilemma when defining stress regarding which area we used – undeformed or deformed.We face a similar issue when working with strain, although it may not be as immediately obvious. When a materialchanges length, its length changes. That may seem obvious, but it’s non-trivial when the length itself is used tocompute strain. As an example, consider the following material undergoing two successive strains:L0 L1

u1 L2u2

What is the strain?ε12 =

u1

L0ε23 =

u2

L1ε13 =

u1 + u2

L0(2.7)




But if the strain is defined correctly, then ε13 = ε12 + ε23. Does this hold?ε12 + ε23 =

u1 + L0u2/L1

L06= u1 + u2

L0(2.8)

Apparently it does not. But what if we suppose that strain is the composition of multiple small strains that use thecurrent length of the member i.e.

ε = ∆ε01 + ∆ε12 + ∆ε23 + ... =L1 − L0

L1+

L2 − L1

L2+

L3 − L2

L3+ ... (2.9)

We can express this more compactly asε =

∑i

∆Li

Li(2.10)

If we take the limit as ∆L→ 0 then we have the following definition for strain:εtrue =

∫ L+u

L

dL

L= ln(L)

∣∣∣L+u

L= ln

(L + u

L

)= ln(1 + ε) (2.11)

where εtrue is true strain (or sometimes called log strain) and ε is engineering strain. As with engineering stress,engineering strain is significantly easier to measure and therefore it is safe to assume here and subsequently thatstrain refers to engineering strain unless specifically stated otherwise.2.3 Material Response in 1D

Because stress and strain are geometry-independent quantities, they are ideal physical quantities for measuringmaterial behavior. In the following subsections we will explore one-dimensional material behavior.2.3.1 Stress-strain plots

Basic material tests are getnerally conducted using a tensile tester such as those manufactured by Instron. Theidea is simple: pull the material until it breaks. But this simple test can give us a lot of information about thematerial. By plotting the stress in the material as a function of strain, we can identify different regimes of behavior.Consider the following two stress-strain diagrams:σ

ε

σuts

σ0

E E

elastic plastic hardening necking

elas

ticload

/unloa

d

truestres

s

failure

ductile response

brittle response

ε

σ

σuts

E

tensile tester

dogb

one

These two diagrams represent the response of a ductile material (center) and a brittle material (right). Examplesof ductile materials includemetals such as copper and aluminum. Brittle materials include carbon-reinforced com-posites, concrete.Let us make a few notes about the ductile material first:All content © 2017, Brandon Runnels 3.5



1. The first regime is the linear elastic region, where stress and strain are linearly related, and the original ge-ometry is recovered on unloading.2. The second regime corresponds to plastic yield, where the material exhibits permanent deformation evenonce it has been unloaded.3. In the third regime, the material begins to neck, where strain localizes to the point of eventual failure. Whydoes the stress go down during this stage? Remember, we are using engineering stress which uses theundeformed area. But necking causes the area of the material to decrease, so the true stress actually is stillincreasing.4. The quantityσ0 is called the yield stress and is themaximumstress thematerial can undergo before exhibitingpermanent damage.5. The quantity σUTS is the maximum engineering stress that the material can exhibit before necking occurs. Itis generally referred to as the ultimate tensile stress, and sometimes abbreviated UTS.6. When used to refer to material properties, the word “strength” is sometimes used instead of stress. Forinstance, σ0 is often referred to as the “yield strength” or σuts as the “ultimate tensile strength.” For mostintents and purposes, these terms can be used interchangably.

Some examples of tensile strengths in materials are enumerated in Appendix A. You can refer to this table forvarious material properties in homework assignments.Brittle materials do not have a yield stress because there is no plastic regime. There are a number of reasons whybrittle materials are less ideal than ductile materials for many applications. In general, materials that are brittleunder tension are poor structural materials because they are very unforgiving. Whereas ductile materials “flow”without breaking upon reaching their yield strength, brittle materials generally fail catastrophically.2.3.2 Hooke’s law

Now that we have expored different regimes of material behavior, let us explore how to model this behavior mathe-matically. Suppose that we have some function σ(ε) that describes the evolution of stress with strain. Performinga first order Taylor expansion about ε = 0 we have

σ(ε) = σ|ε=0 + εdσ

dε

∣∣∣ε=0

+1

2ε2 d2σ

dε2

∣∣∣ε=0

+ ... (2.12)Here we introduce the first major assumption that we will continue to make throughout our study of the mechanicsof materials: the the small strain assumption.Definition 2.4. The small strain assumption is making the approximation that ε is so small that ε2, ε3,etc. is smallenough to be safely ignored, i.e. O(ε2) ≈ 0

If we introduce this approximation into the above expression, noting that σ(0) = 0, we haveσ = E ε E =

dσ

dε, (2.13)

i.e. we have a linear relationship between stress and strain. Because this is a direct consequence of the smallstrain assumption, the small strain assumption is sometimes refered to as the “linear strain assumption.”. It turnsout that the small strain assumption is quite good, and is sufficient for most engineering problems. Therefore it issafe to assume linearized strain in this class thatThe constant E is called “Young’s Modulus.” Because ε is unitless, E must have the same units of stress. We willillustrate this with an example:




Example 2.3: Yield strain of steel

What is the yield strain of steel?Using Hooke’s law

σ0 = E ε0 (2.14)and the fact that σ0 = 250MPa, and E = 200GPa, we have

ε0 =250× 106Pa

200× 109Pa= 0.00125 (2.15)

2.3.3 Plastic strain

In most engineering analysis applications, onset of plasticity is tantamount to failure, because materials that aredeforming plastically are generally close to breaking. So, in general, you can treat the yield stress σ0 as the maxi-mum allowable stress.Consider the following experiment: a material is subjected to an increasing load up to and beyond its yield strengthas shown in the following figure. In the initial elastic regime, the material obeys Hooke’s law:

σ = E ε (2.16)After some plastic deformation, the sample is unloaded, and the stress goes according to the following:

σ = E (ε− εp) (2.17)When the sample is unloaded so that σ = 0, the sample has a permament strain ε = εp

σ

ε

σ0

σ=

Eε

εp

tensile tester

dogb

one

σ

ε

σ′0

εp

hardening

σ=

E(ε−ε p

)

σ=

E(ε−ε p

)

If the material is loaded again, it continues to obey σ = E (ε − εp), right up until the point where it intersects theoriginal curve, and so this time it has a higher yield stress. This increase in yield stress is a great way to make amaterial stronger, so it is often done to materials to increase their strength. This process is called work hardeningor strain hardening.What causes plasticity? To understand the mechanics of plasticity we have to look at the atomistic structure ofthe material itself. There are different mechanisms of plasticity (many of which are active topics of research!) butwe will talk about the type that is most common in structural materials: dislocation-mediated plasticity.For metals with face-centered cubic structure, it is easy for dislocations to move. When subjected to a critical load,themetal attempts to reduce its stress bymoving dislocations around, as you can see in the following figure. Whenthe load is small enough, the crystal simply relaxes to its original configuration when the load is removed. But aftera dislocation has been moved through the material, the deformation is permanent.All content © 2017, Brandon Runnels 3.7



unloaded elastic dislocation dislocation permanentresponse nucleation motion deformation

In general, though, you can think of plasticity as permanent “damage” to a metal’s crystal lattice that cannot beundone by unloading. A more sophisticated example of plastic behavior of a material under nano-indentation canbe found here, https://youtu.be/x4xu89dVkHU (courtesy J. Amelang)



https://youtu.be/x4xu89dVkHU

https://youtu.be/x4xu89dVkHU

Lecture 4 Thermal strain, 1D examples

2.3.4 Strain energy

The area under a stress-strain curve corresponds to the energy put into the system in the form of strain work. Notethat it is similar to a pressure-volume diagram in thermodynamics. In the elastic regime, we compute strain energyas:W =

∫ ε

0

σ d ε =

∫ ε

0

E ε d ε =E

2ε2 =

1

2Eσ2 (2.18)

Are the units correct? We have units of:[pressure] =

[force][area] ×

length][length] =

[force][length][volume] =

[energy][volume] X (2.19)

For elastic loading and unloading, we see that we have no net strain energy – all energy that goes into the materialis recovered upon unloading. However, when the material is deformed plastically, energy is spent that cannot berecovered. This energy typically is understood to dissipate into heat. Therefore plastic deformation is said to be adissipative mechanism.2.3.5 Thermal strain

Temperature is ameasure of atomic vibration, so higher temperature causes the atoms inmaterials to have a higheramplitude of vibration. As temperature increases, to accommodate higher amplitude vibrations, most materialsexpand. We represent this using the following material properties, defined as:

αL =1

L

∆L

∆T=εthermal

∆T→ 1

[temperature] αV =1

V

∆V

∆T→ 1

[temperature] (2.20)

αL is the coefficient of linear expansion, andαV the coefficient of volumetric expansion. (Some values are includedin Appendix A.) Therefore the thermal strain can be computed as εthermal = αL∆T . Then the 1D stress induced by1D thermal strain is simplyσ = E (ε− αL∆T ) ≡ Hooke′sLaw (2.21)

In the elastic regime, temperature changes correspond to lateral shifts in the stress-strain curve as shown in thefollowing diagram.



σ

ε

heating

εthermalεthermal

σthermal < 0 (compression)

σthermal > 0

(tension)

Let us illustrate how to do analysis for a material undergoing thermal loading.Example 2.4: Shrink fit

A sample of 6061 Al at room temperature is designed to fit snugly on a rigid substrate with length L. Thesample cutout length is a length u less than the substrate. Find the temperature to which the sample to beheated to in order for it to fit on the substrate, and the residual stress after the sample cools.u

L

If L = 0.5m, what is the largest allowable difference in length u that does not cause material failure?We know that the thermal strain required is

εthermal =∆L

L0=

(L)− (L− u)

L− u=

u

L− u(2.22)

Therefore the change in temperature is∆T =

εthermal

αL=

u

αL(L− u)(2.23)

and the residual stress is justσ =

Eu

L− u(2.24)

Letting σ → σ0, the maximum value for u isu =

σ0L

E + σ0(2.25)

Substituting σ0 = 250× 106Pa, E = 200× 109, L = 0.5m,u = 0.000624m = 0.624mm = 624µm (2.26)




2.4 Problems in uniaxial loading

Let us do some examples to illustrate how accounting for stress and strain enables us to solve problems that maynot have been solvable using the methods of statics.Example 2.5: Uniaxial loading in linkage

Consider the following linkage where the three vertical bars are identical, each having an elastic modulus Eand cross-sectional area A. The horizontal bar has an elastic modulus Ehor >> E such that the bar can betreated as rigid.

u(x) = ax + bx

yW W

L

f

Determine the stresses in the bars.We start out by doing statics on the linkage. We know that each of the three vertical bars are two-forcemembers, so the statics are trivial. Doing force balance on the rigid beam:

∑f =

[0t10

]+

[0t20

]+

[0t30

]−

[0f0

]= 0 (2.27)

Doing moment balance about the leftmost pin gives:∑

M =

[00

Wt2

]−

[00

−3Wf /2

]+

[00

2Wt3

]= 0 (2.28)

Apparently we have only two equations but three unknowns, so we need to allow for deformation. We knowthe horizontal bar is “rigid” so we can assume that it remains straight, so the end deformation of each baris given as u(x) = ax + b. Now we just need to solve for a and b. Substituting in to get strains:ε1 = −b

Lε2 = −aW + b

Lε3 =

2aW + b

L(2.29)

σ1 = −Eb

Lσ2 = −E (aW + b)

Lσ3 =

E (2aW + b)

L(2.30)

t1 = −EAb

Lt2 = −EA(aW + b)

Lt3 = −EA(2aW + b)

L(2.31)

Now we have the tensions; next substitute into the force balance equation:∑fy = −EAb

L− EA(aW + b)

L− EA(2aW + b)

L− f = 0 (2.32)




aW + b = − fL

3EA(2.33)

And substituting into moment balance:∑Mz@origin = t2W −

3

2Wf + 2t3W =

(− EA(aW + b)

L

)W − 3

2Wf + 2

(− EA(2aW + b)

L

)W (2.34)

= −(WEA(5aW + 3b)

L

)− 3

2Wf = 0 (2.35)

=⇒ 5aW + 3b = − 3fL

2EA(2.36)

We now have two equations and two unkonwns that we can solve using Cramer’s rule:[W 1

5W 3

] [ab

]=

[−fL/3EA−3fL/2EA

](2.37)

Solving using Cramer’s rule:a = − fL

4WEAb = − fL

12EA(2.38)

You may have noticed that we did something a little sneaky. If all of the beams are attached to a rigidly rotating bar,then the attachment points will move not just horizontally but vertically as well.

uy

ux

Let us compute exactly the total new length of the bar in terms of both uy and ux :L =

√(L + uy )2 + u2

x (2.39)Remember, however, that we are making the small strain assumption so that O(u2) = 0. Let us write the Taylorseries expansion about ux = 0, uy = 0:

L = L(0, 0) +∂

∂uxL∣∣∣ux =uy−0

ux +∂

∂uyL∣∣∣ux =uy =0

uy +O(u2)

≈√

(L + uy )2 + u2x

∣∣∣ux =uy =0

+ux√

(L + uy )2 + u2x

∣∣∣ux =uy =0

ux +(L + uy )√

(L + uy )2 + u2x

∣∣∣ux =uy =0

uy

= L + 0 +(L)√L2

∣∣∣ux =uy =0

uy = L + uy (2.40)In other words, as long as we are making the small strain assumption (and throwing out higher powers of u) we donot have to account for the lateral deflection of the beam.




Example 2.6: Thermal expansion

Two bars with cross-sectional areas A1,A2, elastic moduli E1,E2, coefficients of linear expansion αL1,αL2,and lengths L1, L2 are placed between two rigid walls as shown in the figure below. The temperature israised by ∆T .L1 L2

u

Determine the deflection of the interface and the residual stresses in both bars.Statics: our unknowns are the tensions in both bars. Doing a simple cut shows that

t1 = t2 =⇒ A1σ1 = A2σ2 (2.41)We have two unknowns but only one equation, so we must take into account deflection. Let the interfacedeflect by u. Then the strains are

ε1 =u

L1ε2 = − u

L2(2.42)

Now to get from strain to stress we must use Hooke’s law. But remember that both beams have undergonethermal expansion, so we haveσ1 = E1

( u

L1− αL1∆T

)σ2 = E2

(− u

L2− αL2∆T

) (2.43)Substituting into the statics equation we have

E1A1

( u

L1− αL1∆T

)= E2A2

(− u

L2− αL2∆T

) (2.44)Solving gives

u =L1L2(E1A1αL1 − E2A2αL2)∆T

E1A1L2 + E2A2L1(2.45)

Sanity check: do the units work out? Yes. What happens when both beams have identical properties? ThenE1A1αL1 − E2A2αL2 → 0 and there is no deflection. Otherwise, this term determines whether the deflectionis positive or negative.



Lecture 5 Shear stress, stress/strain tensors

2.4.1 Stress concentration factors

Why do airplanes have rounded windows? This was not always the case: the first commercial jetliner, the “de Hav-illand DH 106 Comet” had square windows. In 1954 the planes started crashing because of failure at the windows.Engineers eventually realized that this was because the square, riveted corners of the windows caused a very highstress concentration. Subsequent designs included rounded corners, reducing the total stress significantly, andeliminating windows as a cause for fuselage failure.

uniform stress infinite stress decreased stress

When working with materials undergoing loading, it is important to take into account geometric causes of excessstress concentration that can cause premature failure. We account for this phenomenologically by including astress concentration factor in our analysis, defined as:

K =σmaxσnominal ≡ Stress Concentration Factor (2.46)

σnominal, the nominal stress, is the maximum estimated stress in the material computed using a simplistic analysis.It is then multiplied by K to estimate the maximum stress exhibited in the material. How do we find K? Stressconcentration factors are usually computed using experimental methods (strain gauges, photoelasticity, digitalimage correlation), computational methods (finite element analysis) or – in a couple of rare cases – using ananalytic solution. When you need a stress concentration factor, you can usually look it up online or in a book. Wewill show how this works using an example.Example 2.7: Stress concentration

A member with thickness t is subjected to a load f as shown in the following diagram:

H h

r

Given the following parameters,H = 6cm h = 3cm r = 1cm t = 1cm f = 100N (2.47)



find the maximum stress in the bar.We begin by computing the stress in the thick and thin sections:

σthick =f

A=

100N

.06m × .01m= 167kPa σthin =

f

A=

100N

.03m × .01m= 333kPa (2.48)

On first glance, it seems that the maximum tension is 333kN . But we’re not done: the fillet creates a stressconcentration that we must account for. Using www.amesweb.info, we estimate that the stress concentra-tion factor is

K = 1.72 (2.49)Therefore we multiply the maximum estimated stress (called that nominal stress) by K to get

σmax = Kσnom = 573kPa (2.50)When you need a stress concentration factor, you can use online tools like www.amesweb.info or books like Roark’sFormulas for Stress and Strain to calculate them. (Just be sure you cite your reference because not all sourcesreturn exactly the same K .)2.4.2 Factor of safety

The “Factor of Safety” (FoS, FS) is defined as the followingFactor of Safety =

failure loaddesign load (2.51)

Using a factor of safety is a fancy way of saying “we’re pretty surewe did everything right, but we’re going to reduceour maximum design load by [FoS] amount just to be on the safe side.” Engineering failures do happen (sometimesat the cost of human lives) so it is the job of the engineer to ensure that a design will never even come close to itsfailure point.2.5 Shear stress and shear strain

Consider a block of material subjected to a shear load as shown in the followingu fshear

θ

H A

The shear stress τ is defined as the shear force divided by area, i.e.τ =

fshear

A(2.52)



www.amesweb.info

www.amesweb.info


Similarly, the shear strain γ is defined (here) as the elongation in the deformation direction divided by the length inthe normal direction, i.e.:γ =

u

H= tan(θ) (2.53)

We notice that we can use the small strain assumption to linearize tan θ about θ = 0 using a Taylor series expansionas follows:tan(θ) = tan(0) +

d

dθtan(θ)

∣∣∣θ=0

θ +O(θ2) = 0 +1

cos2 θ

∣∣∣θ=0

θ +O(θ2) ≈ θ =⇒ γ ≈ θ (2.54)In other words, for small deformations, the shear strain is approximately equal to the angle of strain.Note: in some texts (e.g. Hibbeler) shear stress is defined as the angle between the undeformed line and thedeformed line. As such, you may sometimes see the units of shear strain given in radians. It is recommended thatyou avoid this: it is sometimes incorrect and often confusing – shear strain, just like normal strain, is unitless.

Example 2.8: Normal/shear stress using statics

Consider the following member of length L√

2 situated at an angle of 45 subjected to a force f as shownin the following figure:

f

A

Each of the pins has a cross sectional area A. Estimate the shear stress in each of the pins under thisapplied loading.Begin by doing static analysis to find the reaction forces:∑

f =

[fpivx − frolxfpivy − f

0

]= 0

∑M =

[00

−fL/2 + frolxL

]= 0 (2.55)

Solving shows thatfpivy = f frolx = f /2 fpivx = f /2 (2.56)

So the magnitude of the force exerted on the pin and the roller is|fpiv | =

f√

5

2|frol | =

f

2(2.57)

Therefore the shear stresses areτpiv =

f√

5

2Aτrol =

f

2A(2.58)




2.5.1 Hooke’s law for shear stress and strain

Shear stress and shear strain are generally related linearly via the shear modulus µ:τ = µγ (2.59)

(Note that the shear modulus is sometimes denoted G or S .)2.6 Stress and strain in 2D and 3D

Now that we have gotten a feel for how stress and strain work in one dimension, let us explore material responsein 2D and 3D.2.6.1 The stress tensor

Consider a square (or block in 3D) ofmaterial that is undergoing a state of stress. This is similar to drawing sectionsof a beam for shear-moment diagrams, but this time we are expressing all forces in terms of the stress tensor. Wedraw all of our possible shear and normal stresses in this way:

σyy

x

y

z

σzy

σxy

σxx

σyx

σzx

σzz

σyz

σxz

σxx

σyy

σxy

σyx

2D stresses 3D stresses

where τxy = τ , the shear stress, and σxx ,σyy are the normal stresses in the x and y directions, respectively. Withthis convention we can express all stresses compactly in tensor form:σ =

[σxx τxyτyx σyy

] (2D) σ =

[σxx τxy τxzτyx σyy τyzτzx τzy σzz

](3D) (2.60)

Intuitively, the row of the stress component represents the direction of the force/area and the column correspondsto the direction of the face on which it acts. If the face and the direction are the same (i.e. along the diagonal ofthe tensor) the component is a normal stress, otherwise it is a shear stress.The stress tensor is a way to compactly express the state of stress in a material, but it is also a handy way todetermine stress in a particular direction. Recall how we referred to tensors as things that “turn vectors into othervectors.” In the same way, the stress tensor turns normal vectors into force vectors:

σ : [normal vector to a surface]→ [traction vector (force per unit area) acting on the surface]σ : [area vector for a surface]→ [force vector on the surface]

We can use this result to compute the total shear and normal stresses on a surface:




Normal stress is given by computing the projection of thetraction vector on the normal vector:σnormal = n · σn (2.61)

The shear stress is computed by vector subtraction andtaking the magnitude of the result:τshear = |σn− σnormaln| (2.62)

n

t = σnnor

mal stres

s

traction vectorshear stress

Let us illustrate how this works with an example:Example 2.9: Tractions in material under pure shear

Consider a material undergoing pure shear in which there are no normal components of the tensor, justshear:

σ =[

0 ττ 0

] −ex

ey

−ey

ex

σex

σey

−σex

−σey

n1

σn1

n2

σn2

Determine the force per unit area acting on each of the four surfacesUsing matrix-vector multiplication:[

0 ττ 0

] [10

]=[

0τ

] [0 ττ 0

] [−10

]= −

[0τ

] [0 ττ 0

] [01

]=[τ0

] [0 ττ 0

] [0−1

]= −

[τ0

] (2.63)This is exactly what we expect, and we note that the sign change on each of the faces is a direct result of thesign change in the normal vectors themselves. Furthermore, we can also find the stress acting on internalsurfaces: [

0 ττ 0

] [1/√

21/√

2

]=

[τ/√

2τ/√

2

] [0 ττ 0

] [−1/√

21/√

2

]=

[τ/√

2τ/√

2

](2.64)

(Notice how for these particular faces, the traction vector is parallel to the normal vector. These are calledeigenvectors of the stress tensor, or in engineering parlance, principal directions. We will discuss these ingreater detail in a later section.)



Lecture 6 Equilibrium, strain tensor, 2d Hooke’s law

2.6.2 Stress equilibrium

Consider a piece of material in 2D with a stress tensor varying in space as shown below:

σxx (x + 12 ∆x , y)

σyy (x , y + 12 ∆y)

τxy (x , y + 12 ∆y)

τyx (x + 12 ∆x , y)

σyy (x , y − 12 ∆y)

τxy (x , y − 12 ∆y)

σxx (x − 12 ∆x , y)

τyx (x − 12 ∆x , y)

x

y

b

The material is also subjected to a “body force” (force per unit area, e.g. weight) b. The dimensions of the blockare ∆x and ∆y . What is the equation of equilibrium? We begin by taking the force balance:∑fx = ∆yσxx (x + ∆x/2, y)−∆yσxx (x −∆x/2, y) + ∆xτxy (x + ∆x/2)−∆xτxy (x + ∆x/2) + bx ∆x∆y = 0∑fy = ∆yτyx (x + ∆x/2, y)−∆yτyx (x −∆x/2, y) + ∆xσyy (x + ∆x/2)−∆xσyy (x + ∆x/2) + by ∆x∆y = 0

(2.65)Dividing both equations by ∆x∆y :

σxx (x + ∆x/2, y)− σxx (x −∆x/2, y)

∆x+τxy (x + ∆x/2)− τxy (x + ∆x/2)

∆y+ bx = 0

τyx (x + ∆x/2, y)− τyx (x −∆x/2, y)

∆x+σyy (x + ∆x/2)− σyy (x + ∆x/2)

∆y+ by = 0 (2.66)

When we take the limit as ∆x , ∆y → 0, the fractions become derivatives and we end up with∂σxx

∂x+∂τxy

∂y+ bx = 0

∂τyx

∂x+∂σyy

∂y+ by = 0 (2.67)

This is called the stress divergence equation and is a fundamental equation of solid mechanics (like the Navier-Stokes equations in fluid mechanics.) It can easily be generalized to 3D using the exact same process.The stress divergence equation is the continuous version of force balance, but what about moment balance? Letus compute moment balance in the above case:

∆x

2×∆y(τyx (x + ∆x/2, y) + τyx (x −∆x/2, y))− ∆y

2×∆x(τxy (x , y + ∆y/2) + τxy (x , y −∆y/2)) = 0 (2.68)

Dividing by ∆x∆y

1

2(τyx (x + ∆x/2, y)− τyx (x −∆x/2, y)) =

1

2(τxy (x , y + ∆y/2) + τxy (x , y −∆y/2)) (2.69)

Taking the limit as ∆x , ∆y → 0

τyx = τxy (2.70)In other words, moment balance requires that the stress tensor must be symmetric.All content © 2017, Brandon Runnels 6.1


2.6.3 The strain tensor

Similarly to the stress tensor, the strain tensor can be expressed in the following way:ε =

[εxx εxyεyx εyy

]=

[εxx

12 (γxy + γyx )

12 (γxy + γyx ) εyy

](2.71)

(and similarly for 3D). The row corresponds to the reference direction, and the column corresponds to the directionof displacement. (For instance, εxy corresponds to the displacement in the x direction along the y axis).Notice that there is something unusual going on with the shear strains. This is beacuse while γxy is not necessarilyequal to γyx , it is important that the strain tensor be symmetric (for reasons that are outside the scope of thisclass.) As you can see, this form guarantees that εxy = εyx . In general, the cases that we will deal with will besimple enough so that this should not cause any confusion.Similarly to the stress tensor, the strain tensor is also a machine that performs an action on a vector, except thistime it turns distance vectors into displacement vectors:

ε : [unit vector along a direction]→ [strain vector in that direction]ε : [length vector along a direction]→ [displacement in that direction]

Normal and shear strain are computed in the same way as with normal and shear stresses; i.e.εnormal = n · εn εshear = |εn− εnormaln| (2.72)

where n is a unit vector corresponding to the direction we are looking at, ε is the 3x3 strain tensor, and εnormal , εshearare the normal and shear strains in the direction of n. Let us illustrate how this works with an example:Example 2.10: Strain tensor

Find the strain tensor for the following deformation:

L

hy

L

r2

r1

r3

u2u3

u1hx

In the x direction we have normal strain εxx = hx/L, in the y direction we have normal strain εyy = −hy/L.There is no shear in the x or y directions so the strain tensor isε =

[hx/L 0

0 −hy/L

](2.73)

Let us use this tensor to compute the displacement along three vectors r1, r2, r3 as shown in the figure:u1 =

[hx/L 0

0 −hy/L

] [L0

]=[hx0

]u2 =

[hx/L 0

0 −hy/L

] [0L

]=[

0−hy

]u3 =

[hx/L 0

0 −hy/L

] [LL

]=[hx−hy

](2.74)

We see that the displacements u1,u2 are parallel to the direction vector, indicating that we have pure normalstrain in those directions. But what about u3? It is not strictly in the same direction as r3, indicating that




there is both normal and shear strain in this direction. We can compute this directly using projection:u3|| =

u3 · r3|r3|

=L(hx − hy )

L√

2=

hx − hy√2

(2.75)so the parallel displacement is

ε3,normal =u3||

|r3|=

hx − hy

2L(2.76)

Now we compute the orthogonal component of u:u3⊥ = u3 − u3,||

[1/√

21/√

2

]=[hx−hy

]− hx − hy

2

[11

]=

hx + hy

2

[1−1

]=

hx + hy√2

[1/√

2−1/√

2

](2.77)

indicating that the magnitude of the displacement in the orthogonal direction is (hx + hy )/√

2, so the totalshear strain isεshear =

u3⊥

|r3|=

hx + hy

2L(2.78)

What does this mean? Had we not already determined that there was no shear strain in this material? Infact, we know that there is no shear strain relative to the x and y axes, but there is shear relative to thediagonal axis. This is why we need tensors to express shear strain. We will delve into this in greater detailin the chapter on transformations of stress and strain.

2.6.4 Hooke’s law in 2D and 3D

Consider a material subjected to a uniaxial stress σxx as shown below. We know that εxx = σxx/E . But notice thatthere is strain both in the x direction and the y direction.

εxx > 0

εyy < 0

How do we capture this? We need another material parameter: this is called Poisson’s ratio. It is denoted ν and isdefined asν = −εtransverse

εaxial≡ Poisson’s Ratio (2.79)

For instance, if a material has ν = 0 there is no lateral deformation. It is possible to show theoretically (see MAE5201 notes if you are interested!) that Poisson’s ratio must be in the range−1 ≤ ν ≤ 1

2(2.80)




which indicates that it is possible for materials with a negative Poisson’s ratio to exist! Such materials are calledauxetic materials and have some very interesting applications.We can use this information to construct Hooke’s law for materials in three dimensions:

εxx =1

E(σxx − νσyy − νσzz ) εyy =

1

E(σyy − νσzz − νσxx ) εzz =

1

E(σzz − νσxx − νσyy )

εyz =1 + ν

Eτyz =

γyz

2εzx =

1 + ν

Eτxz =

γzx

2εxy =

1 + ν

Eτxy =

γxy

2(2.81)

These equations can be inverted to give stress as a function of strain:σxx =

E ((1− ν) εxx + ν εyy + ν εzz )

(1 + ν)(1− 2ν)σyy =

E (ν εxx + (1− ν) εyy + ν εzz )

(1 + ν)(1− 2ν)σzz =

E (νεxx + ν εyy + (1− ν) εzz )

(1 + ν)(1− 2ν)

τyz =E εyz

1 + ν=

E γyz

2(1 + ν)τzx =

E εzx

1 + ν=

E γzx

2(1 + ν)τxy =

E εxy

1 + ν=

E γxy

2(1 + ν)(2.82)

Example 2.11: 3D stress/strain

A stress-free beamof structural steel is constrained at either end and subjected to a uniaxial load of 100MPa.What is the strain of the material?Steel has properties E = 200GPa, ν = 0.30. Using Equations 2.81 we have:εxx =

100MPa− (0.30)(0)− (0.30)(0)

200GPa= 0.0005 = .05% (2.83)

εyy = εzz =(0)− (0.30)(0)− (0.30)(100MPa)

200GPa= −0.00015 = −0.015% (2.84)

Notice that even if all strains are in the x and y plane, there will still be stress in the z direction. Conversely, if thereall stresses are in the x and y plane, there will be strain in the z direction. This prompts the following definitions:Definition 2.5. Amaterial is said to be in plane stress if it is loaded such that all stress components in one directionare zero. (E.g. τxz = τyz = σzz = 0.) Plane stress often takes place in thin geometry.

Definition 2.6. A material is said to be in plane strain if it is loaded such that all strain components in one directionare zero. (E.g. εxz = εyz = εzz = 0.) Plane stress often takes place in a thick geometry.

The takeaway is that there is no such thing as a purely 1D or purely 2D solid mechanics problem – all solid me-chanics problems are in 3D. This does not mean that we will be solving highly complex solids problems, but it doesmean that you often need to think about what is going on in all directions in a solid mechanics problem.



Lecture 7 Introduction to beam theory

2.6.5 Other material constants

We have introduced Young’s modulus E , the shear modulus µ, and Poisson’s ratio ν. Other constants include thebulk modulus, which is defined in the linear regime as:

κ =∆p

∆V /V= V

∆P

∆V≡ Bulk modulus (2.85)

Othermoduli include the p-wave/constrainedmodulusM and the Laméparameterλ. Howmanymaterial constantsdo we need? For an isotropic material, only two material constants are required to define small-strain elasticbehavior. In other words, given two material parameters you can always find the other parameters as functions ofthe first two; for instance, the shear modulus as a function of Young’s modulus and Poisson’s ratio is:

µ =E

2(1 + ν)(2.86)

In general, for this class, we will primarily work with E , ν , and occasionally µ.

3 Transverse Loading: Mechanics of Beams

In this section we will develop the theory of flexible, linearized beams. (Note that Hibbler divides this section intotwo chapters, one on “Bending” and one on “Deflection of Beams and Shafts”). We will develop both together,because these two subjects are interdependent and it does not make much sense to split them up.3.1 Euler-Bernoulli beam theory

Formulation of Euler-Bernoulli beam theory was first attempted by Gallileo and Leonardo da Vinci, but it was notuntil after the development of calculus and Hooke’s law that it was fully formulated. The main contributors tothe original development were Leonhard Euler and Daniel Bernoulli, and it became one of the first useful tools inmechanics for solving engineering problems.3.1.1 Derivation

Let us consider a beam with properties that vary only along the x axis and which undergoes small deformationparameterized by a function h(x) as shown in the following figure:

x

z z

y

neutralaxis

h(x)

deflection cross-section

Ω(x)

The dashed line represents the neutral axis which has the following definition:All content © 2017, Brandon Runnels 7.1


Definition 3.1. The neutral axis in an Euler-Bernoulli beam is the plane along which there is no elongation or con-traction, and is (by convention) the plane in which the origin of the reference coordinate system is located.

In other words, the neutral axis is the line along which there is no normal strain in the x direction. How do we knowsuch an axis exists, and where is it in relation to the cross-sectional area? We will answer these questions in just amoment.Let us first compute the normal strain in the x direction, εxx . We begin by considering a section of length ∆x thatundergoes a deformation resulting in a constant curvature with total angle change ∆θ (note that the figure is highlyexaggerated.) What is the strain along the neutral axis? Substituting we have

εneutral axisxx =`new − òld

òld=

∆x −∆x

∆x= 0, (3.1)

as expected.

∆θ

∆x

∆x

zθ = tan−1(dz/dx)

R

What about the strain of a line segment located a distance z above the neutral axis? First, let us compute the newlength of the segment – we know the old one already. The radius of curvature isR =

∆x

∆θ(3.2)

so the arc length along the line segment is`new = (R − z)(∆θ) = ∆x − z ∆θ (3.3)

Now we can compute the strain and taking the limit:εxx (z) =

(∆x − z ∆θ)−∆x

∆x= −z ∆θ

∆xlim

∆x ,∆z→0→ εxx (z) = −z dθ

dx(3.4)

This is helpful, but it is in terms of θ – we would much prefer to have our strain in terms of the parameterizationfunction h(x). Computing θ in terms of h:θ = tan−1

(∆h

∆x

)→ θ = tan−1

(dhdx

) (3.5)Now let us compute the derivative (using the chain rule)

dθ

dx=

1

1 + (dh/dx)2

d

dx

dh

dx≈ d2h

dx2→ θ ≈ dh

dx(3.6)

Substituting this back and assuming a Young’s modulus constant along the cross-section, E (x) we haveεxx (x , z) = −z d2h

dx2→ σxx (x , z) = −E (x) z

d2h

dx2(3.7)

Note that stress varies linearly with z , and that normal stress in the x direction is zero at the neutral axis.




−σ(x , z)df = −σ(x , z) dy dz

y

z

x

z

dM = −zσ(x , z) dy dz

ΩM

In order to maintain cohesion in the beam, the total stress in the beam must sum to zero; i.e.∫Ω

σ(x , z) dA = 0 (3.8)Substituting in our form for stress, (), we have

−∫

Ω

E (x) zd2h

dx2dA = −E (x)

d2h

dx2

∫Ω

z dA = 0 =⇒∫

Ω

z dA = z = 0 (3.9)This tells us that the vertical distance between the neutral axis and the centroid, z , is zero – in other words, theneutral axis passes through the centroid of the cross-sectional area.Now, let us compute the total effective moment, Meff generated by this distributed stress about the neutral axis,which we now know is also the axis passing through the centroid of the cross section. We compute it simply byintegration:

Meff = −∫

Ω

dM =

∫Ω

σ(x , z) z dy dz =

∫Ω

E (x) zd2h

dx2z dy dz (3.10)

Notice that we can take everything that does not depend on y or z outside of the integral:= E (x)

d2h

dx2

∫Ω

z2 dy dz (3.11)Does this integral look familiar? It is nothing other than Iyy , the moment of inertia about the y axis, which is againalso the centroidal axis. Here, we will denote it simply I(x), adopting the convention that it is with reference to theneutral axis, and noting that it can change along the x axis as Ω changes. The result is

M(x) = E (x) I(x)d2h

dx2(3.12)

Substituting this into (3.1.1), we haveσxx = −M z

I≡ The General Flexure Formula (3.13)

In other words, given the bendingmomentM and cross sectionalmoment of inertia I, we can compute the stress inthe beam for any distance from the neutral axis z . If themaximumdistance from the neutral axis to the upper/loweredge of the beam is c , the formula is often reduced to the following formσmax =

M c

I≡ The Flexure Formula (3.14)

where σmax 7→ max |σxx | the maximum tensile/compressive stress, M 7→ |M| the absolute value of the moment,and c = max |z | ∈ Ω, the maximum vertical neutral axis-to-edge distance. Remember that this formula loses allinformation about signs so you need to think about that every time you use it. I discourage memorizing formulas;however, this is one of those that basically every engineer should remember, so it is worth filing it away.All content © 2017, Brandon Runnels 7.3



To recapitulate, we have derived a differential relationship between moment and deflection h. Recall that if thereis an applied loading w(x), we have the following:w(x) =

dV

dx=

d2M

dx2(3.15)

Substituting in the equation we have just derived, we have the following differential equation describing the rela-tionship between deflection h and applied load w :w(x) =

d2

dx2

(E I

d2h

dx2

)≡ Euler-Bernoulli Beam Equation (3.16)

Written in terms of other beam bending quantities:d2

dx2

(E I

d2h

dx2

)=

d2

dx2

(E I

dθ

dx

)=

d2M

dx2=

dV

dx= w(x) (3.17)

This forms the basis for Euler-Bernoulli beam theory, one of the first successful applications of the academic studyof mechanics to real-world mechanical engineering problems. Before we move to applications, let us pause tounderstand the key assumptions the theory we have developed:1. small strain: remember that this beam theory fails when dealing with large deflections. If you ever need tostudy beams undergoing large deformation, you may want to look up the theory of elastica [1].2. constant normals: when we computed strains in the beam, we assumed that lines originally normal to thesurface remained normal. However, materials can in fact undergo in-plane shear that cause the normals torotate as illustrated below:

Euler-Bernoulli Timoshenko

A theory developed by Stephen Timoshenko allows for in-plane shear by introducing an additional variableφ(x) to describe additional rotation. It is more accurate (and far more suitable for some applications) butalso harder to solve.

3. constant properties: we have assumed that there is no variation of material properties within the cross-sectional area of the material. In a subsequent section we will modify our theory to handle cases where thisis not true.Here and subsequently, we will stick to conventional (Euler-Bernoulli) beam theory, but you may encounter caseswhere more sophisticated approaches are needed, and you should always know what assumptions you are using.3.1.2 Boundary conditions

The Euler-Bernoulli beam equation is a fourth order ordinary differential equation. Therefore we know there willbe four integration constants that we must solve for using boundary conditions. Let us consider a selection ofboundary conditions:Free end Consider a beam that is free to move at the end.




We recall that because the beam is free to move and to rotate, that no shear or moment can besustained, and henceh (xfree end ) =?? θ (xfree end ) =??

V (xfree end ) = 0 M (xfree end ) = 0 (3.18a)Pivot/roller end Consider a pin constrained by a pivot or a roller as shown:

What are the boundary conditions for this end? We know that it is constrained to move but it isfree to rotate. Our boundary conditions, therefore, areh (xpinned end ) = 0 θ (xpinned end ) =??

V (xpinned end ) =?? M (xpinned end ) = 0 (3.18b)Fixed rotation Consider a beam that is free to translate but has constrained rotation:

The boundary conditions here areh (xroller ) =?? θ (xroller ) = 0

V (xroller ) = 0 M (xroller ) =?? (3.18c)It is worth noting here that the boundary conditions here are exactly the opposite of that of thepivot/roller end.

Fixed end Consider a beam that is fixed at one end, that we might draw as:

As usual a fixed end constrains all of the degrees of freedom of our system, so we know thath (xfix ) = 0 θ (xfix ) = 0

V (xfix ) =?? M(xfix ) =?? (3.18d)Appliedshear/moment

Consider a beam that is fixed at one end, that we might draw as:All content © 2017, Brandon Runnels 7.5



V0

M0

As usual a fixed end constrains all of the degrees of freedom of our system, so we know thath (xapplied ) =?? θ (xapplied ) =??

V (xapplied ) = V0 M(xapplied ) = M0 (3.18e)NB: don’t forget that a downward shear on the other end of the bar would result in a negativevalue for V at that end.

Let us illustrate how this works with a couple of examples.Example 3.1: Simply supported beam with point load

Consider the following beam of length L with a square cross-sectional area, and simply supported bound-ary conditions subjected to a load at the center with magnitude f . What is the maximum normal stressexperienced by the beam?L/2 L/2f

W

W

This beam is not statically indeterminate, so we can find the shear and moment at the end quite easily.Drawing a FBD in terms of M,V evaluated at the beginning and end, we haveL/2 L/2f

V (0) V (L)

M(L)M(0)

and we know that, because of the roller/pinned ends, M(0) = M(L) = 0. Static analysis shows us thatV (0) =

f

2V (L) = − f

2(3.19)

Then it is straightforward to find the shear-moment diagrams:f /2

−f /2

fL/4

M(x

) =fx/

2

V (x) M(x)

We are in the business of computing the maximum normal stress that the beam undergoes. We can usethe flexure formula (3.13) or (3.14) to find it. Computing the following ingredients:Mmax =

fL

4c =

W

2I =

W 4

12(3.20)




so the maximum normal stress isσ = ±M c

I= ± fL

4× W

2× 12

W 4= ± 3fL

2W 3(3.21)

Sanity check: do the units work out? we have [force][length]/[length]3 = [pressure], X



Lecture 8 Beam theory continued

Aswe saw in the previous example, all we need is to find themaximumbendingmomentM to extract themaximumtensile force. This is easy to find if the beam is statically determinate, but what if it is not? In this case, we musttake into account the beam’s deflection as well. We’ll illustrate this with the following example:Example 3.2: Statically indeterminate beam

Consider a beam with Young’s modulus E , identical to that in the previous example, except that the pinnedjoint has been replaced with a fixed joint and the distributed load has been replaced with a point load, asshown beloww(x) = −w0

W

W

Unlike the previous example, this beam is statically indeterminate. Wemust therefore solve the fourth-orderEuler-Bernoulli ODE (3.16); because E and I are constant we have the following equation:E I

d4h

dx4= −w0 (3.22)

Because this is a fourth order ODE, we know we will require four boundary conditions. Let us list what weknow about the boundaries given the supports:V (0) =?? M(0) =?? V (L) =?? M(L) = 0

h(0) = 0 θ(0) = 0 h(L) = 0 θ(L) =?? (3.23)Fortunately this is an easy ODE to solve – we just use successive integration.

E Id3h

dx3= V (x) = −

∫w0 dx = C1 − w0x (3.24)

Unfortunately, we have no boundary conditions for V (x) – whereas we have two for the deflection h. There-fore we must solve for h in order to use BCs for it. Integrating again:E I

d2h

dx2= M(x) =

∫V (x) dx = C1 x −

1

2w0x

2 + C2 (3.25)We can use M(L) = 0 to eliminate one of the integration constants:

M(L) = C1L−1

2w0L

2 + C2 = 0 =⇒ C2 =1

2w0L

2 − C1 L (3.26)substituting:

M(x) = C1 (x − L) +1

2w0(L2 − x2) (3.27)

We must integrate again to get θ(x):dh

dx= θ(x) =

1

E I

∫M(x) dx =

1

E I

[C1

(x2

2− L x

)+

1

2w0

(L2x − x3

3

)+ C3

] (3.28)



Use the boundary condition θ(0) = 0 to show that C3 = 0. Integrate once more to get h(x):h(x) =

∫θ(x) dx =

1

E I

[C1

(x3

6− L x2

2

)+

1

2w0

(L2x2

2− x4

12

)+ C4

] (3.29)h(0) = 0 implies that C4 = 0; now we use h(L) = 0 to solve at last for C1:

h(L) =1

E I

[− C1L

3

3+

5w0L4

24

]= 0 =⇒ C1 =

5w0L

8(3.30)

therefore our moment equation isM(x) =

5w0L

8(x − L) +

1

2w0(L2 − x2) = −w0

8

[L2 − 5xL + 4x2

]= −w0

8(x − L)(4x − L) (3.31)

But we want to find the maximum moment. This occurs when M ′(x) = 0; evaluating the derivativeM ′(x) = −w0

8(−5L + 8x) = 0 =⇒ x =

5

8L (3.32)

Substituting back in we getM(5L/8) = −w0

8

(− 3L

8

)(3L

2

)=

9

128w0 L

2 (3.33)Now that we have our maximummoment, we compute the maximum normal stress by substituting into theflexure formula:

σ = ±Mc

I= ±9w0L

2

128× W

2× 12

W 4= ±27w0L

2

64W 3(3.34)

3.2 Singularity functions

Up until now we have had two distinct ways of handling beams: the case where we have point loads and the casewhere we have distributed loads. The point loads are generally handled using the method of sections, while wehandle distributed loads by integration. However, as integration is typically a lot easier, we seek a way to handlepoint loads along with distributed loads so we can integrate everything. To do this, we introduce a couple of newfunctions:3.2.1 Dirac delta distributions

We have established a differential relationship between applied load, shear, and moment. But what happens if wehave a point load? How do we integrate that thing? To do this, we need Dirac delta distributions.(Dirac distributions are often incorrectly called “delta functions” by engineers. This is a misnomer–is is not, strictlyspeaking, a function.)Consider the following box function.




f (x) =

1L −L/2 < x < L/2

0 else (3.35)

L

1L

What happens if I integrate this curve?∫ ∞−∞

f (x) =

∫ L/2

−L/2

(1/L)dx =L/2− (−L/2)

L= 1 (3.36)

I will get a value of 1. What happens if I make L really, really small?limL→0

f (x) (3.37)

The value of f(x) goes to infinity at 0 and 0 elsewhere...but we still have that

limL→0

∫ ∞−∞

f (x) = 1 (3.38)This is called a Dirac distribution, and we generally write them as

δ(x) (3.39)Suppose we want to put the Dirac distribution at a different location?

x1 x2 x3

M1

M2

M3

We can express this asδ(x − x0) (3.40)

What if we want the Dirac distribution to have a greater magnitude? (That is, have an integral greater than 1). Justuse a multiplier:Mδ(x − x0) =⇒

∫ ∞−∞

Mδ(x − x0)dx = M

∫ ∞−∞

δ(x − x0)dx = M (3.41)




3.2.2 Heaviside functions

Let us suppose that we have a Dirac distribution centered at 0 withmagnitude 1. Suppose we perform the followingintegral: ∫ t

−∞δ(x)dx (3.42)

What will the value of this integral be for t < 0? It will be zero.What is the value for t > 0 It will be 1.Let’s give this thing a name:

u(t) =

∫ t

−∞δ(x)dx =

0 x < 0

1 x > 0(3.43)

x0 x0

δ(x) u(x)

This is called the Heaviside function, also called the “Step function.” From the fundamental theorem of calculuswe see that, by definition,δ(x) =

du(x)

dx(3.44)

3.2.3 Ramp functions

What happens if we integrate the Heaviside function?

x0 x0

u(x) r(x)

For x < 0 we just have 0, but for x > 0 we get x . So we have:∫ x

−∞u(x)dx =

0 x < 0

x x > 0= x u(x) = r(x) (3.45)




where r(x) is called the ramp function.We can write the following:

αδ(x − x0) =d

dx(αu(x − x0)) αu(x − x0) =

d

dx(αr(x − x0)) (3.46)

We can use this to solve beam equations much more easily.3.2.4 Unit doublet

How do we represent point moments in this framework? Suppose we have a point couple moment with magnitudeM0 acting at a location x0. Then we would expect that

M(x) = M0u(x − x0) =⇒ V (x) =dM

dx= M0δ(x − x0) (3.47)

In general, we don’t want to prescribe a loading in terms of V (x) on our beam; we would rather prescribe it in termsof the loading function w(x). If we use our differential relationship, that indicates thatw(x) =

dV

dx= M0

d

dxδ(x − x0) (3.48)

How do we take the derivative of the Dirac delta? This is one of the places where we start to run into trouble: thedelta function is not actually a function, and so we can’t exactly take derivatives of it in the usual way. However, ifwe treat it as a distribution, functional analysis allows us to define the distributional derivative of it, which is calledthe unit doublet, u1(x).We can roughly visualize the unit doublet as the limit of the following:

u1(x)

where we let the width of the function go to zero and the magnitude approach infinity. You can see that if weapplied a loading w(x) of this form, we would be applying “equal and opposite” forces that are infinite in magnitudeand infinitesimally close to each other. In other words, there would be no net force, but there would be a resultantcouple moment.3.2.5 Singularity bracket notation

It can be tricky to keep track of the notation with Dirac deltas, Heavisides, etc., so we introduce a handy notationthat generalizes all of our singularity functions:α〈x − x0〉−2 = αu1(x − x0) α〈x − x0〉−1 = αδ1(x − x0) α〈x − x0〉0 = αu(x − x0) (3.49)

(3.50)All content © 2017, Brandon Runnels 8.5



and for exponents with n if n > 0,α〈x − x0〉n = αu(x − x0)(x − x0)n =

α(x − x0)n x > x0

0 else (3.51)To help visualize how these functions work, consider the plot below:

x

x0

x

x0

x

x0

x

x0

x

x0

α〈x − x0〉−2 α〈x − x0〉−1 α〈x − x0〉0 α〈x − x0〉1 α〈x − x0〉2

α

α

α

doublet Dirac delta Heaviside (step) ramp polynomial

The real utility of these functions is the easy with which they are differentiated and – especially – integrated.Consider the following examples: (ignoring integration constants):∫α〈x − x0〉−2dx = α〈x − x0〉−1

∫α〈x − x0〉−1dx = α〈x − x0〉0

∫α〈x − x0〉0dx = α〈x − x0〉1∫

α〈x − x0〉1dx =1

2α〈x − x0〉2

∫α〈x − x0〉2dx =

1

3α〈x − x0〉3

∫α〈x − x0〉3dx =

1

4α〈x − x0〉4

In other words, you treat singularity brackets just like parentheses when integrating except when it is raised to anegative power, because of the unique way we define these brackets with a negative power. So, generally, use thefollowing rule for singularity bracket integration:∫α〈x − x0〉ndx =

1

n+1α〈x − x0〉n+1 n ≥ 0

α〈x − x0〉n+1 else (3.52)



Lecture 9 Beam theory examples

3.3 Solution strategy and examples

We have formulated Euler-Bernoulli beam theory and developed the tools necessary to robustly solve many typesof beam problems. The following outlines a general solution strategy for analyzing beams.Solution Strategy for Euler-Bernoulli beam problems1. Express the beam loading function w(x), using singularity functions if necessary2. Write down all four boundary conditions – there should always be two at each end3. Write down additional constraints (e.g. h(x0) = 0 if there is a support at x0).4. Integrate successively to find V ,M, θ, h:

V (x) =

∫w(x)dx M(x) =

∫V (x)dx θ(x) =

∫M(x)

E (x)I(x)dx h(x) =

∫θ(x)dx (3.53)

5. Apply boundary conditions and additional constraints (this is often done during the previous step)6. Determine the neutral axis of the cross-section, i.e. the centroid7. Determine the moment of inertia about the neutral axis for the cross-section8. Use the flexure formula to convert from bending moment to normal stressσ = −M(x) z

I(3.54)

9. Sanity check: units, limit analysis10. Plug in numbersWe will illustrate how this process works by working through a couple of examples.

Example 3.3: Beam with U-shaped cross-section

Consider a beam with Young’s modulus E and length L subjected to a point load centered at `, as shown inthe following figure.f

L

`W

W3W /4

W /2

Determine (1) the deflection of the beam and (2) the maximum tensile/compressive normal stresses in thebeam:Step (1): determine the loading function

w(x) = −f 〈x − `〉−1 (3.55)



Step (2): determine the boundary conditionsu(0) = 0 θ(0) = 0 u(L) = 0 M(L) = 0 (3.56)

Step (3): there are no additional constraintsStep (4-5): First, integrate to get shear,V (x) =

∫w(x)dx = −f 〈x − `〉0 + C1 (3.57)

then integrate to get moment.M(x) =

∫V (x)dx = −f 〈x − `〉1 + C1x + C2 (3.58)

Apply one of the boundary conditions – we won’t be able to solve yet, though.M(L) = −f (L− `) + C1L + C2 = 0 =⇒ C1L + C2 = f (L− `) (3.59)

Integrate again to get angle, and use one of the boundary conditions to eliminate C3

θ(x) =1

EI

∫M(x)dx =

1

EI

[− 1

2f 〈x − `〉2 +

1

2C1x

2 + C2x +>0

C3

] (3.60)Integrate again to get deflection, and use one of the boundary conditions to eliminat C4.

h(x) =

∫θ(x) =

1

EI

[− 1

6f 〈x − `〉3 +

1

6C1x

3 +1

2C2x

2 +>0

C4

] (3.61)Now we have our form for the deflection, and we still have one BC left to use.

h(L) =1

EI

[− 1

6f (L− `)3 +

1

6C1L

3 +1

2C2L

2]

= 0 (3.62)C1L

3 + 3C2L2 = f (L− `)3 (3.63)

Simultaneously solving this equation and the one we obtained earlier, we haveC1 =

f (L− `)(2L2 + 2L`− `2)

2L3C2 = − f `(2L− `)(L− `)

2L2(3.64)

Plotting the deflection and moment for a couple values of ` gives:

0.0 0.2 0.4 0.6 0.8 1.00.010

0.005

0.000

0.005

0.010

`= 0L

`= 0. 25L

`= 0. 5L

`= 0. 75L

`= 1L

Deflection

0.0 0.2 0.4 0.6 0.8 1.00.20

0.15

0.10

0.05

0.00

0.05

0.10

0.15

0.20

Bending moment




which is what we expect. Now let us determine the maximum bending moment. Because M is discontin-uous, we cannot find the maximum moment using optimization. Note that the maximum moment occurseither at x = 0 or x = `. Therefore, we can state that

Mmax = max|C2|, |C1`+ C2| (3.65)and the exact one will be determined by the value of `. Let us evaulate for ` = 0.5, so that

C1 =11

16f C2 = − 3

16f L (3.66)

ThenMmax = max

3

16fL,

5

32fL

=3

16fL (3.67)

Step (6): wemust determine the neutral axis, i.e. the centroid of the cross section. The best way to computeit for a geometry like this is to start with a box and subtract the geometry for the smaller box from it. Mea-suring all distances from the bottom of the cross-section we have that the centroids for the two sectionsarex1 =

W

2x2 =

3W

8(3.68)

Then using the formula for centroids of composite areas,x =

x1A1 − x2A2

A1 − A2=

(W /2)(W 2)− (3W /8)(3W 2/8)

W 2 − 3W 2/8=

23W 3/64

5W 2/8=

23W

40(3.69)

Step (7): we can now compute the moments of inertia for the outer box and the inner box using the parallelaxis theorem:I′1 = I1 + A1(x − x1)2 =

W 4

12+ W 2

(3W

40

)2

=427

4800W 4 (3.70)

I′2 = I2 + A2(x − x2)2 =(W /2)(3W /4)3

12+(3W 2

8

)(W5

)2

=417

12800W 4 (3.71)

and subtract the moments of inertia to getI = I′1 − I′2 =

422

7680W 4 ≈ 0.05638W 4 (3.72)

and the max distances to the top and bottom of the beam arec+ = W − 23

40W =

17

40W = 0.425W c− =

23

40W − 0 =

23

40W = 0.575W (3.73)

Step (8): we have all the ingredients we need to use the flexure formula to get normal stress, s we have:

σcompression =Mmaxc+

I= 1.4134

fLW

W 3σtension =

Mmaxc−I

= 1.912fLW

W 3(3.74)

Step (9): do the units make sense? Yes.Step (10): We have already plugged in some numbers, but if we have additional values for f , L,W we canget an exact numerical answer.




Beam problems are often tricky and can involve a great deal of algebra, regardless of whether symbolic notationor real numbers are used. You should feel free to use computer algebra systems to help solve some of theseproblems, but remember that you should always be able to solve them completely by hand when necessary.Example 3.4: Simply supported beam with singularity functions

Consider a beam with Young’s modulus E that is simply supported and subjected to the following loading:w0 f

L L L L

W

W

Find the deflection and the normal stress in the beam as a function of position.Step (1): we begin by formulating a w(x) function that captures all of the loadings on the member:

w(x) = −w0 + w0〈x − L〉0 − f 〈x − 2L〉−1 + fpiv 〈x − L〉−1 + frol〈x − 3L〉−1 (3.75)Step (2): our boundary conditions are:

V (0) = 0 M(0) = 0 V (4L) = 0 M(4L) = 0 (3.76)Step (3): We also have two auxiliary conditions

u(L) = 0 u(3L) = 0 (3.77)(Note that these auxiliary conditions are the result of the introduction of two force unknowns.)Step (4-5): To find the shear we proceed by integration:

V (x) =

∫w(x) dx = −w0x + w0〈x − L〉1 − f 〈x − 2L〉0 + fpiv 〈x − L〉0 + frol〈x − 3L〉0 + >

0C1

V (0)=0(3.78)

Try plugging in the end boundary condition:V (4L) = −4w0L + w0〈3L〉1 − f 〈2L〉0 + fpiv 〈3L〉0 + frol〈L〉0 = −w0L− f + fpiv + frol = 0 (3.79)

Rearrangingfpiv + frol = w0L + f (3.80)

To find the moment we integrate again:M(x) =

∫V (x) dx = −1

2w0x

2 +1

2w0〈x − L〉2 − f 〈x − 2L〉1 + fpiv 〈x − L〉1 + frol〈f − 3L〉1 + >

0C2

M(0)=0(3.81)

Try plugging in the end boundary condition:M(4L) = −1

2w0(4L)2 +

1

2w0〈3L〉2 − f 〈2L〉1 + fpiv 〈3L〉1 + frol〈L〉1 (3.82)

= −7

2w0L

2 − 2fL + 3fpivL + frolL = 0 (3.83)




Rearranging3fpivL + frolL =

7

2w0L

2 + 2fL (3.84)Simultaneously solving (3.80) and (3.84) we have

fpiv =1

2f +

5

4w0L frol =

1

2f − 1

4w0L (3.85)

To find the angle we integrate again, remembering to divide by (EI):θ(x) =

1

E I

∫M(x) dx =

1

E I

[− 1

6w0x

3 +1

6w0〈x − L〉3 − f

2〈x − 2L〉2 +

fpiv

2〈x − L〉2 +

frol

2〈f − 3L〉2 + C3

]and then integrate one final time to get deflection:h(x) =

∫θ(x) dx =

1

E I

[− 1

24w0x

4 +1

24w0〈x − L〉4 − f

6〈x − 2L〉3 +

fpiv

6〈x − L〉3 +

frol

6〈f − 3L〉3 + C3x + C4

](3.86)

Now we implement our two auxiliary constraints to solve for C3,C4: first,h(L) =

1

E I

[− 1

24w0L

4 + C3L + C4

]= 0 =⇒ C3L + C4 =

1

24w0L

4 (3.87)and then

h(3L) =1

E I

[− 27

9w0L

4 +2

3w0L

4 − f

6L3 +

4fpiv

3L3 + 3C3L + C4

]= 0 (3.88)

=⇒ 3C3L + C4 =7

3w0L

4 +f

6L3 +

4fpiv

3L3 (3.89)

Solving these two equations givesC3 =

1

2w0L

3 − 1

4fL2 C4 = −11

24w0L

4 +1

4L3 (3.90)

Plotting the results for a couple values of w0 and f gives

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.00.5

0.4

0.3

0.2

0.1

0.0

0.1

0.2

0.3

undeformedw0=0Lw0=ff=0

Deflection

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.00.6

0.4

0.2

0.0

0.2

0.4

0.6

w0=0Lw0=ff=0

Bending moment




Because the moment is discontinuous, we must again look at the plot to determine where the maximummoment occurs. Here, the location is dependent on the loading type, but we can find the stress using theflexure formula.Step (6): The beam is a simple square cross-section so the neutral axis passes through the center.Step (7): Since the beam is a simple square cross-section, we haveI =

W 4

12c =

W

2(3.91)

Step (8): Using the flexure formula, we have the maximum tensile/compressive stress as a function of x :σ =

6

W 3M(x) (3.92)

where M(x) is what we defined earlier.Step (9): In this case, the best way to conduct a sanity check (as we have already done) is to plot the shear,moment, and deflection to ensure that it makes sense and that all boundary conditions are satisfied.Step (10): Plug in numbers if needed.



Lecture 10 Composite beams

3.4 Composite beams

Composite beams are used in many structures to increase their strength without a significant increase in weight orcost. Previously we have assumed that material properties are constant throughout the cross-section of the beam;now, we will allow for two or more materials to exist in the cross section.

ε(x , z)

y

z

x

z

ΩM

Ω1.E1

Ω2,E2

Recall thatε = −(z − zna)

d2h

dx2σ1 = −E1 (z − zna)

d2h

dx2σ2 = −E2 (z − zna)

d2h

dx2(3.93)

where zna is the location of the neutral axis. (Note that we assumed before that the coordinate system coincidedwith the neutral axis, so we let zna = 0. Here we treat it slightly more generally.) The implications of this are:1. Strain is continuous throughout the cross-section (i.e. no large jumps in strain), but2. Stress is discontinuous if E is discontinuous (i.e. there may exist jumps in stress)

as can be seen in the following figure.ε(x , z) σ(x , z)

This implies that all of the assumptions we made about strain were valid up to this point; however, we must re-doour calculations that involve stress.We begin by finding the location of the neutral axis; we use the fact that the integral over the area of the normalstress must be zero:

0 =

∫Ω

σdA = −∫

Ω

E zd2h

dx2dA = −d2h

dx2

∫Ω

E (z − zna) dy dz =⇒∫

Ω

E (z − zna) dy dz = 0. (3.94)We can no longer pull out E because it varies over the integral. However it is constant within subsections of Ω, solet us assume that E = E1 in Ω1 and E = E2 in Ω2. Then we have:

E1

∫Ω1

z dA︸︷︷︸A1 z1

+E2

∫Ω2

z dA︸︷︷︸A2 z2

−(E1

∫Ω1

dA︸︷︷︸A1

+E2

∫Ω2

dA︸︷︷︸A2

)zna = E1A1z1 + E2A2z2 − (E1A1 + E2A2)zna = 0 (3.95)

Rearranging our equation we arrive at:zna =

E1A1z1 + E2A2z2

E1A1 + E2A2≡ Location of neutral axis for bimaterial composite beam (3.96)



(A similar form holds for 3 or more material types.)We must now find a modified version of the flexure formula for a composite cross-section; we do this by findingthe relationship between the stress distribution and the effective moment. Recall that

Meff =

∫Ω

σ (z − zna) dA =

∫Ω

E (z − zna)d2h

dx2(z − zna) dA =

d2h

dx2

∫Ω

E (z − zna)2 dA (3.97)=

d2h

dx2

[E1

∫Ω1

(z − zna)2 dA︸︷︷︸I1

+E2

∫Ω2

(z − zna)2 dA︸︷︷︸I2

]=

d2h

dx2(E1I1 + E2I2) (3.98)

where I1 and I2 are the moments of inertia of regions 1 and 2, respectively, about the neutral axis.The main implication is that, since stress differs between materials, we need two flexure formulae, one for eachmaterial (or more if working with more materials.As such, we must keep track of the individual stresses and elastic moduli when substituting:

M = − σ1

E1 z(E1I1 + E2I2) = − σ2

E2 z(E1I1 + E2I2) (3.99)

Rearranging these formulae we have:σ1 = − M z E1

E1I1 + E2I2σ2 = − M z E2

E1I1 + E2I2≡ Flexure formulae for bimaterial composite beam (3.100)

Finally, recalling thatd2M

dx2= w(x) (3.101)

we can write the Euler-Bernoulli beam equation for a composite beam:w(x) =

d2

dx2

[(E1I1 + E2I2)

d2h

dx2

]≡ Euler-Bernoulli Beam Equation for bimaterial composite beam (3.102)

Example 3.5: Carbon reinforced steel beam

Carbon fiber exhibits exceptional strength under tension and, because it is lightweight and inexpensive, isa possible way to increase the load capacity of structures. Consider the following beam that has carbonreinforcement applied to the bottom.

h

h

hb

b

t

w(x) = −w0

Determine the maximum stress with and without the carbon fiber.




Let us begin by computing the maximum moment using integrating.V (x) =

∫w(x) dx = −w0x + C1 (3.103)

M(x) =

∫V (x) dx = −1

2w0x

2 + C1x +>0

C2 (3.104)M(L) = −w0L

2

2+ C1L = 0 =⇒ C1 =

w0L

2=⇒ M(x) =

w0(xL− x2)

2(3.105)

(The maximum moment occurs whendM

dx=

w0(L− 2x)

2= 0 =⇒ x =

L

2(3.106)

as expected.) SoMmax =

w0(L2/2− L2/4)

2=

w0L2

8(3.107)

The moduli of elasticity for steel and carbon fiber areE1 = 200GPa E2 = 1000GPa (3.108)

and the dimensions of the I-beam are specified to beb = 15cm h = 2cm t = 0.5cm (3.109)

Moments of inertia for steel I-beam I1 and carbon fiber section I2 about their centroids:I1 =

hb3

12+ 2×

(bh3

12+ (bh)

(b + h

2

)2)=

7

12b3h +

2

3bh3 + b2h2 = 4917.5cm4 I2 =

bt3

12= 0.1562cm4

(3.110)Neutral axis of the composite beam (measured from bottom)

zna =E1(3bh)(t + h + b/2) + E2(tb)(t/2)

E1(3bh) + E2(tb)= 7.13cm (3.111)

Modified moments of inertia (parallel axis theorem)I′1 = I1 + (3bh)(t + h + b/2− zna)2 = 5657cm4 I′2 = I2 + (tb)(t/2− zna)2 = 335.4cm4 (3.112)

Finally, the maximum vertical distances c from theneutral axis to the edge of the beam:c+

1 = t + 2h + b − zna = 0.1237m (3.113)c−1 = t − zna = −0.0663m (3.114)c2 = −zna = −0.0713m (3.115)

c+1

z1

z2

zna c−1 c2

Using the flexure formulae:σ+

1 = − Mc+1 E1

E1I′1 + E2I

′2

= (−207.9m−3)w0L2 σ−1 = − Mc−1 E1

E1I′1 + E2I

′2

= (111.5m−3)w0L2 (3.116)




σ2 = − Mc2 E2

E1I′1 + E2I

′2

= (119.9m−3)w0L2 (3.117)

Comparing this stress to the amount carried by the beam if there was no carbon fiber:σ± = ±M (b + h/2)

I1= ±(407m−3)w0L

2 (3.118)Suppose that (with a factor of safety) themaximum allowable stress for the steel in tension or compressionis 100MPa, and suppose that the beam is required to have a length L = 10m. Without the carbon reinforcedbeams we have :

w0max =100MPa

407m−3(10m)2= 2.5kN (3.119)

but with the carbon reinforcement the maximum load per length isw0max =

100MPa

207.9m−3(10m)= 4.8

kN

m(3.120)

a double in the beam’s loading capacity.



Lecture 11 Beam theory concluded

3.5 Shear stresses in beams

Recall that we relate the normal stress, σxx , to the bending moment of the beam, and we also require that the totaleffective force exerted by the normal stress must be zero. Pictorally,

σxx

M =∫σxxz dA

0 =∫σxxdA=

But we have neglected some of the other components of stress. Recall that we are looking at a face that is inthe direction of x ; therefore all of the stresses acting on this face are σxx , τxy , τxz . What effect do we expect thesestresses to have? Consider τxz first:

σxz V =∫σxzdA

=

We see that the total shear force, V , is nothing other than the effective force resulting from integration over theentire face. Until now we considered V (x) to be constant throughout the cross-section of the beam, but in fact, itis the total force resulting from an uneven distribution of shear stress throughout the beam. It is important for usto know what the shear distribution is, because we have to allow for the possibility that the material may fail dueto exessive shear stress just like it might fail due to excessive normal stress.3.5.1 The shear formula

Consider a region Ω′ ⊂ Ω that is the part of the cross-sectional region above z as shown in the figure below.Let us make the following assumptions:1. There is no shear stress on the boundaryof Ω.2. The shear stress is constant along thelength of the cut at z .3. We ignore the vertical stress εzz

z ′

t(z)

Ω′

zσxx

σxz

σzz

We begin by using the stress divergence equation (2.67) from Section 2, where we consider the x direction only:Then the equation reduces to∂σxx

∂x+∂τxy

∂y+∂τxz

∂z= 0→ ∂τxy

∂y+∂τxz

∂z= −∂σxx

∂x(3.121)



We substitute the general flexure formula (3.13) to obtain∂τxy

∂y+∂τxz

∂z= − ∂

∂x

(− Mz

I

)=

z

I

∂M

∂x(3.122)

and then substituting V = dM/dx we obtain∂τxy

∂y+

dτxz

dz=

zV

I(3.123)

Now, let us integrate both sides over Ω′:∫Ω′

(∂τxy

∂y+

dτxz

dz

)dA =

∫Ω′

zV

IdA =

V

I

∫Ω′

z dA =V z ′ A′

I(3.124)

where z ′ is the centroid of Ω′ andA′ is the area of Ω′. (Note that I is still themoment of inertia of the entire section.)To evaluate the integral on the other side we make use of the divergence theorem (a full discussion of which isoutside the scope of this class, but which is well worth looking into) and assumptions (1-2) to find∫Ω′

(∂τxy

∂y+

dτxz

dz

)dA =

∫∂Ω′

τxz z d` = t(x) τxz (3.125)where t(x) is the width of the cross-section at z . Putting both sides together we arrive at

τxz =V z ′ A′

I t=

V Q

I t≡ The Shear Formula for Horizontal Cuts (3.126)

where Q = z ′A′. Although the derivation of this formula is somewhat tricky, its usage is quite straightforward. Wewill illustrate how this works with an example.Example 3.6: Shear stress in rectangular beam

Consider a beam with rectangular cross section having width b and height h, subjected to a shear V . Com-pute the distribution of shear within the cross-section.We begin by computing the various quantities for the shear formula:

z ′ =1

2

(h2

+ z)

A′ = b(h

2− z)

Q = z ′A′ =b

2

(h2

4− z2

)t = b I =

bh3

12(3.127)

so the shear stress isτxz =

VQ

It=

6V

bh3

(h2

4− z2

) (3.128)We see that τxz (±h/2) = 0, as expected. Furthermore we see that the maximum stress occurs at z = 0,giving us a maximum shear stress of

τxz,max =6V

bh3(h/2)2 =

3V

2bh(3.129)




3.5.2 Shearing stresses in thin-walled beams

We have considered shearing stresses on horizontal sections of thebeam, allowing us to solve for τxz (the shearing force in the x directionon a face with a z-facing normal).Consider a beam in which all walls are sufficiently thin so that it canbe assumed that the shearing force is constant throughout the widthof the section. Let us now do the same thing for a vertical sectionof the beam, solving for τxy . The analysis is the same as before, thedifference is in our analysis of the integral:∫ (∂τxy

∂y+∂τxz

∂z

)dA = t τxz (3.130)

y

x

z

σxy

σxz

Substituting this into the integral expression, we haveτxy =

V z ′A′

It=

V Q

I t≡ The Shear Formula for Vertical Cuts (3.131)

where V is the shear force, z ′ is the centroid of the cut region only (measured with respect to the neutral axis), A′the area of the cut region, I the moment of inertia of the entire section, and t the thickness of the cut. The usageis exactly the same; the only difference is that we are working with the xy shear stress and the cut is vertical. Letus illustrate how this works with an example:Example 3.7: Thin-wall shearing force

Compute the shearing stresses in the beam in the figure below.

h

h

hb

b2fL L

The first thing that we need is the total shearing force in the beam as a function of x . It should be fairlyobvious by this point that the shearing force isV (x) =

f 0 < x < L

−f L < x < 2L(3.132)

(If you are still struggling with shear moment diagrams, it is highly recommended that you do some addi-tional review and get some extra practice.)So, we see that the shear is always V = ±f , which will make our analysis easier. We will simply take V = f ,and change the sign for when V = −f . Now, we will apply both shear formulae for τxy and τyz , and we willlook at two particular cuts:




h

h

hb

b

h

z ′

y

h

h

b

b

z ′z

Next we find the ingredients that we need for the shear formula: first, we need the moment of inertia I ofthe entire cross-section, which isI =

7

12b3h +

2

3bh3 + b2h2 (3.133)

Beginning with the upper cross-section we have:z ′ =

b + h

2A′ = h (b/2− y) t = h (3.134)

Substituting these values into the shear formula we have:τxy =

V z ′A′

It=

f ((b + h)/2) (h(b/2− y))

Ih=

f (b + h) (b/2− y)

2I(3.135)

Note that the form is linear in y , so the extreme values take place at y = 0 and y = b/2: Computing themaximum and minimum shear we find thatτxy (y = b/2) = 0 τxy (y = 0) =

fb(b + h)

4I(3.136)

which matches expectations.Now let us apply the shear formula to the middle section. Everything is the same except the centroid andcross-sectional area of the cross-section, so we havez ′A′ =

A1z1 + A2z2

A1 + A2× (A1 + A2) = A1z1 + A2z2 (3.137)

= (bh)((b + h)/2) + (h(b/2− z))((b/2 + z)/2) =h

8(5b2 + 4bh − 4z2) (3.138)

t = h (3.139)Substituting into the shear formula we find

τyz =V z ′A′

It=

f

8I(5b2 + 4bh − 4z2) (3.140)

The maximum shear stress is parabolic, and (since there is no term that is linear in z) we see that themaximum shear stress takes place at the center of the beam. Computing the maximum shear stress gives:τyz (z = 0) =

f (5b2 + 4bh)

8I(3.141)

Also the minimum stress must take place at the end points, giving usτyz (z = b/2) =

fb(b + h)

2I(3.142)

which we note is twice twice themaximum τxy shear stress. Sketching the distribution of stress in our beamwe have the figure on the left below:




σxy

y

σxz

z

y

−σxy

fb(b+h)4I

fb(b+h)4I

fb(b+h)2I

fb(5b+4h)8I

Recall that τxy = τyx , i.e. the stress in the x direction on the y face is equal to the stress in the y direction onthe x face; also τxz = τzx , the stress in the x direction on the z face equals the stress in the z direction on thex face. In other words, we can sketch all of the shear stresses on the x face (i.e. the cross-section) to getthe above figure on the right.

3.5.3 Shear flow

A slight modification to the shear formula is to multiply both sides by the thickness of the cut t , givingq =

V Q

I≡ Shear Flow Formula (3.143)

where q is the force per unit length of the beam. The notion of “shear flow” is that you can think of the distribution ofshear stress throughout the cross-section as a type of nonconservative flow. (Note that this is not to be confusedwith shear flow in fluid mechanics.) q can then be plotted to give an idea of how the shear is directed within thebeam as a result of its geometry.3.6 Stress concentrations in beams

Irregularities in beams can cause additional stress distributions that are considerably more complex than thatwhich can be described using simple beam theory. For this it is generally necessary to use a stress concentrationfactor K whereσmax = K σnominal (3.144)

where σnominal is the maximum stress calculated using simple beam theory. To estimate the stress concentrationfactor, you can use online tools or references, or consult Appendix B.3.7 Beam failure

Beam failure is never simple and can be the result of many mechanisms that we have not yet considered heresuch as buckling, plasticity, fatigue, etc. However we can imagine, simplistically, two possible types of failuremechanisms that can occur when a beam is subjected to a loading that causes its normal or shear failure stressesto exceed maximum. The following figure outlines some of these cases.

shear failurenormal stress failure




Notice that normal failure occurs on the outer part of the beam (where normal stress is at a maximum), whereasshear failure occurs at the center (where shear stress is maximum). It is useful to keep these mechanisms in mindwhen designing beams.3.8 Summary

In this section we have derived the theory for simple beam bending, deflection, and stress analysis. To summarize:1. For a statically determinate beam use the shear-moment formulae or the method of sections to find M(x)

V (x) =

∫w(x) dx M(x) =

∫V (x) dx (3.145)

2. For a statically indeterminate beam use the 4th order Euler-Bernoulli beam equation to find the deflectionh(x), and use this to get the bending moment M(x).

w(x) =d2

dx2

(E I

d2h

dx2

) (3.146)Singularity functions are a convenient way to solve this ODE if there are discontinuous loadings present.

3. The flexure formula converts the bending moment M(x) to normal stress σxx

M = −M z

I(3.147)

4. Composite beams are treated similarly to regular beams but different constants for E , I are used.5. The shear formula is used to find the distribution of shear stresses and shear flow in beams.6. Irregularities in beams that cause increased stress are taken into account using stress concentration factors.



Lecture 12 Shaft theory

4 Torsional Loading: Mechanics of Shafts

We have considered bending in the x-z plane, and it is fairly straightforward to generalize to bending in the x-yplane as well. Moments about the y and z axes are called bending moments. Now we consider moments that arenormal to the direction of the beam. In this coordinate system this type of moment is in the x direction and is calleda twisting moment.

x

yz

x

yz

x

yz

bending (xz) bending (xy) torsion (x)

My

Vz

Mz

Vy

Mx

Note that we will now start keeping track of the directionality of moments. Whereas in the previous section wereferred to all bending moments as M , we will now refer to bending moments as My ,Mz and the twisting momentas Mx . Let us now consider the response of the beam to this type of loading.4.1 Derivation of torsion equations

Here is it useful to switch to polar coordinates (r , θ, x) from (y , z , x) according to the following mappingr =

√y2 + z2 θ = tan−1

( zy

) (4.1)y = r cos θ z = r sin θ (4.2)

Suppose that the shaft twists by an angle φ(x) as shown in the following figure.

dx

rφ(x + ∆x)

φx

r dr

rdθτθzgθ

rφ(x)

Then the total shear strain between the θ and x coordinates isεθx = lim

∆x→0

rφ(x + ∆x)− rφ(x)

∆x= r

dφ

dx(4.3)

All other components of the strain are assumed to be zero. If we only have shear strain, then the shear stress isτθx = µεθx = µ r

dφ

dx(4.4)



Now, we wish to link the strain to the total moment generated by the strain on the cross-sectional surface. We canget this by integration: the magnitude of the differential force given by a differential element of area isdf = (τθxgθ)× dA = µ r

dφ

dxdA gθ (4.5)

In cylindrical coordinates, dA = r dr dθ, sodf = µ r2 dφ

dxdr dθ gθ (4.6)

Recall that the differential moment is given by crossing the distance vector with the force vector. What is thedistance vector for this differential area vector? It is just r = r gr , so the differential moment isdM = r × df = µ r2 dφ

dxdA gr × gθ = µ r2 dφ

dxdA gx (4.7)

Then we get the total moment of the stress by integrating over the surface:M =

∫A

dM =

∫A

µ r2 dφ

dxdA gx =

dφ

dx

∫A

µ r2dA gx (4.8)If µ is constant throughout the cross-section of the shaft, then

M = µdφ

dx

∫A

µ r2dA︸︷︷︸J

gx = µdφ

dxJ gx = Mxgx (4.9)

where J = Ixx , the second moment of area about the x axis. This quantity is frequently referred to as the polarmoment of inertia, but (as described earlier) this is a misnomer unless density is included. Note thatJ can be computed for all kinds of cross-sectional geometries, but we most frequently consider a solid cylinderwith radius R , for which we have

J =

∫A

r2dA =

∫ 2π

0

∫ R

0

r3 dr dθ = 2π × R4

4=

1

2πR4 (4.10)

(We also know that J = Ixx = Iyy + Izz in the planar case, so it is easy to compute for any beam’s cross-section.)Because the direction of the moment is always going to be along the axis (for the torsion case, at least), we cansimplify to consider the magnitude of the moment only, giving us the expression:

M(x) = µ Jdφ

dx(4.11)

Combining (4.4) and (4.11) to eliminate dφ/dx , we haveτθx =

M r

Jτmax =

M c

J≡ Torsion Formula (4.12)

This connects the angle of twist to the connecting moment, but how do we relate this to the applied distributedmoment? We follow a procedure similar to that for shear moment diagrams: consider a section of a shaft withlength ∆x subjected to a distributed moment m as shown in the following figure:




M(x) M(x + ∆x)m(x)

∆x

Let us do a moment balance about the x axis: assuming m = m(x) =constant along the length, we have∑Mx = M(x)−M(x + ∆x) + m(x)∆x = 0 (4.13)

Rearranging and letting ∆x → 0, we havelim

∆x→0m(x) = lim

∆x→0

M(x + ∆x)−M(x)

∆x+

dM

dx= 0 =⇒ dM

dx= −m(x) (4.14)

Substituting (4.11), we have the result for a beam (with constant µ across the cross-section)m(x) = − d

dx

(µ J

dφ

dx

)≡ Relationship Between Twist Angle and Moment (4.15)

If µ and J are constant along the length of the shaft thenm(x) = −µ J d

2φ

dx2(4.16)

If the moment is constant along the length of the beam with M(x) = M0, and if µ, J are constant as well then,substituting into (4.11) ∫ L

0

M0dx =

∫ L

0

µJdφ

dxdx (4.17)

LM0 = µJ(φ(L)− φ(0)) = µ J ∆Φ (4.18)Therefore if a beam of length L is subjected to a constant moment M0, then the angular deflection is

∆Φ =LM0

µ J≡ Twist Angle Formula (4.19)

Example 4.1: Loaded shaft with bearings

Consider a shaft made of a material with shear modulus µ, having radius R supported by two ideal bearingsas shown in the following figure. The shaft is subjected to twisting moments −M, 3M, 2M,−4M .




L

L

L

−M0

3M0

2M0

−4M0

Determine the internal twisting moment M(x) and change in twist angle ∆φ as functions of x , and themaximum shear stress in the shaft.(Sketching the moment/twist angle plots based on the solution given below)

M(x)φ(x)

By moment balance we find that, for the moment,

Mx (x) =

M0 0 < x < L

−2M0 L < x < 2L

−4M0 2L < x < 3L

0 else(4.20)

We know that for a circular cross-section,J =

πR4

2(4.21)

so the maximum shear stress isτmax =

(4M)(R)

πR4/2=

8M

πR3(4.22)

Because the moment is constant in each of the three segments we can determine ∆φ for each:∆φ1 =

(M0)(L)

µ J∆φ2 =

(−2M0)(L)

µ J∆φ3 =

(−4M0)(L)

µ J(4.23)

Then the total change in φ is simply given as∆φtotal = ∆φ1 + ∆φ2 + ∆φ3 = −5M L

µ J(4.24)

Note that we can alternatively find φ(x) using integration with singularity functions.m(x) = −M0〈x〉−1 + 3M0〈x − L〉−1 + 2M0〈x − 2L〉−1 − 4M0〈x − 3L〉−1 (4.25)




Integrating to get the moment we haveM(x) = −

∫m(x) dx = M0〈x〉0 − 3M0〈x − L〉0 − 2M0〈x − 2L〉0 + 4M0〈x − 3L〉0 +>

0C1 (4.26)

For x < 0 M(x) = 0, implying that C1 = 0. Integrating again we obtain the angle:φ(x) =

1

µJ

∫M(x) dx =

1

µJ

[M0〈x〉1 − 3M0〈x − L〉1 − 2M0〈x − 2L〉1 + 4M0〈x − 3L〉1

]+ = φ0

C2

(4.27)Note that we don’t actually know what φ0 is – it is not specified. However we know that ∆φ(x) = φ(x)− φ0,so we can write as

∆φ =1

µJ

[M0〈x〉1 − 3M0〈x − L〉1 − 2M0〈x − 2L〉1 + 4M0〈x − 3L〉1

] (4.28)Let us confirm that we obtain the correct twist angle:

∆φ(3L) =1

µJ

[M0〈3L〉1 − 3M0〈2L〉1 − 2M0〈L〉1 + 4M0〈0〉1

]=

1

µJ

[M0(3L)− 3M0(2L)− 2M0(L)

]= −5M0L

µJ(4.29)exactly as we had before.

4.2 Boundary conditions

The fundamental equation for shaft theory is a second order ODE rather than a fourth order. Consequently thereare only two boundary conditions necessary; each end has exactly one BC.

fixed end free endφ(0) = 0 M(0) = 0

Let us consider another example:Example 4.2: Distributed load on a statically indeterminate beam

Consider a shaft that is fixed on both ends and subjected to a distributed load w0 as shown in the figurebelow. Determine the twisting moments and and angles as functions of x , and the location and magnitudeof the maximum shear stress.




L

L

L

w0

2R

What is the distributed moment function? We have a loading per unit length – we turn this into a momentper unit length by multiplying by the radius:m(x) = w0r L < x < 2L (4.30)

Using singularity functions:m(x) = w0r〈x − L〉0 − w0r〈x − 2L〉0 (4.31)

Integrate to get moment:M(x) = −

∫m(x) = −w0r〈x − L〉1 + w0r〈x − 2L〉1 + C1 (4.32)

What is C1? We don’t know–all our boundary conditions are on the angle. So we integrate again to get φ(x):φ(x) =

1

µJ

∫M(x) dx =

1

µJ

[− w0

2r〈x − L〉2 +

w0

2r〈x − 2L〉2 + C1x + C2

] (4.33)Use the first BC to find C2:

φ(0) =C2

µJ=⇒ C2 = 0 (4.34)

Now use the second BC to find C1

φ(3L) =1

µJ

[− w0

2r〈2L〉2 +

w0

2r〈L〉2 + C1(3L)

]= 0 (4.35)

−2w0rL +w0rL

2+ 3C1 = 0 =⇒ C1 =

w0rL

2(4.36)

so our expressions for the deflection and moment areφ(x) =

w0rL2

2µJ

[xL− 〈x − L〉2

L2+〈x − 2L〉2

L2

]M(x) = w0rL

[1

2− 〈x − L〉1

L+〈x − 2L〉1

L

] (4.37)Key values:

φ(L) = φ(2L) =w0rL

2

2µJφ(3L/2) =

5w0rL2

8µJ(4.38)

Plotting:




M(x)φ(x)

Apparently the greatest deflection is in the center, and the greatest moment is located at either end in theregions 0 < x < L, 2L < x < 3L. Using the torsion formula,τmax =

M(0)r

(πr4/2)=

2r

πr4

(w0rL

2

)=

w0L

πr2(4.39)



Lecture 13 Shaft theory continued

4.3 Composite shafts

As before, our assumptions hold about strain – strains are linear and continuouslydistributed over the radius of the shaft. However our stresses are defined piecewise:σθx ,1 = µ1 r

dφ

dxσθx ,2 = µ2 r

dφ

dx(4.40)

ΩΩ1

Ω2

We must also reconsider the relationship between shear stress and applied moment as the stress distribution hasnow changed. Integrating over the surface yieldsMx =

∫Ω

σ r dA =

∫Ω1

σ1 r dA +

∫Ω2

σ2 r dA = µ1dφ

dx

∫Ω1

r2 dA + µ2dφ

dx

∫Ω2

r2 dA = µ1dφ

dxJ1 + µ2

dφ

dxJ1

=dφ

dx

[µ1J1 + µ2J2

] (4.41)Combining this equation with the one above, we find

σθx ,1 =Mxµ1r

µ1J1 + µ2J2σθx ,2 =

Mxµ2r

µ1J1 + µ2J2≡ Torsion formula for a bimaterial composite shaft (4.42)

which is similar in form to the flexure formula for a composite beam. Let us show how these equations are usedby looking at an example.Example 4.3: Woven carbon fiber-reinforced drive shaft

Consider a hollow steel shaft with minor and major diameters a, b, that is reinforced with a woven carbonfiber composite as shown below:

2b

2a

t

If the steel and carbon fiber have shear moduli µ1,µ2, determine the maximum twisting moment Mmax theshaft can withstand if the maximum allowalbe shear stress in the steel is τ0, and compare this with themaximum allowable twisting moment without the carbon reinforcement.First we need the polar moments of inertia:

J1 =πa4

2− πb4

2=π(a4 − b4)

2J2 =

π((a + t)4 − a4)

2(4.43)

Then with the carbon fiber:σθx ,1 =

2Mµ1a

π(µ1(a4 − b4) + µ2((a + t)4 − a4))=⇒ Mmax =

τmaxπ(µ1(a4 − b4) + µ2((a + t)4 − a4))

2aµ1(4.44)



and without the carbon fiber:σθx ,1 =

2Ma

π(a4 − b4)=⇒ Mmax =

τmaxπ(a4 − b4)

2a(4.45)

Let a = 10cm, b = 8cm, t = 1mm. For steel, τ0 = σ0/2 = 125MPa. We know that µ1 = 79.3GPa, but whatabout µ2?Sidenote: Woven carbon fiber is highly anisotropic and does not withstand compression. Let usconsider the shear strain of a single carbon fiber with length L

√2 subjected to shear stress σθx asshown below:

4545A

A√

2

σθx

L

L u

The actual shear stresses and strains areσθx =

fθ

A√

2εθx =

u

L(4.46)

The strain of the carbon fiber isε =

√(L + u)2 + L2 − L

√2

L√

2≈ L√

2 + u/√

2− L√

2

L√

2=

u

2L(4.47)

so the stress of the carbon fiber and traction forces areσ =

Eu

2Lf =

EAu

2L(4.48)

Finding the component of the force in the direction of the stress, and total stress on the face,fθ =

1√2× EAu

2L=

EAu

2√

2Lσθx =

1

A√

2× EAu

2√

2L=

Eu

4L=

E

4︸︷︷︸µ

× u

L︸︷︷︸εθx

(4.49)

so therefore we conclude that µ2 = E/4 = 1000GPa/4 = 250GPa.Substituting these numbers, we find that with carbon reinforcement,

Mmax = 14.1kN ·m (4.50)and without reinforcement

Mmax = 11.6kN ·m (4.51)showing that the carbon fiber increases the load by more than > 25%.

Let us consider another type of composite shaft with another example:




Example 4.4: Composite drive shaft

L

L

R

r

µ1

µ2

b

t1

t2

a

f

First, we must do statics to find the relationship between the force and the differences in the tensions in thebelt. ∑Mx = f (b + R) + (t2 − t1)(r + a) = 0 =⇒ f =

(t1 − t2)(r + a)

R + b(4.52)

Now, let us determine the angle of twist and twisting moments within the shaft. The moment function ism(x) = f (b + R)〈x〉−1 + (t2 − t1)(r + a)〈x − 2L〉−1 (4.53)

and the moment is given by integration:M(x) = −

∫m(x) = f (b + R)〈x〉0 + (t2 − t1)(r + a)〈x − 2L〉0 +>

0C1 (4.54)

where the boundary condition vanishes by the conidition that M(x) = 0,∀x < 0.Mx = −f (b + R) (4.55)

everywhere in the beam. BecauseMx is constant we can use the twist angle formula to determine the twistangle for both sections:∆φ1 =

LM0

µJ= −L f (b + R)

µ1(πR4/2)= −2L f (b + R)

µ1πR4(4.56)

∆φ2 = −L f (b + r)

µ1(πr4/2)= −2L f (b + R)

µ1πr4(4.57)

Finally we determine the shear stresses in both shafts using the torsion formula:σ1 =

Mc

J=

f (b + R)(R)

πR4/2=

2f (b + R)

πR3σ2 =

Mc

J=

f (b + R)(r)

πr4/2=

2f (b + R)

πr3(4.58)




4.4 Stress concentrations

As with uniaxial loading and beam bending, discontinuities in shafts induce stress concentrations. Let us see howthis works by continuing the above example:Example 4.4 Continued

Let L = 1m, R = 30cm, r = 20cm, b = 3cm, a = 2cm, t1 = 100kN , t2 = 1kN , and let both materials be madeof steel. Suppose that there is a fillet with radius 0.5cm Determine the maximum stress induced in the jointbetween the shafts.The stress concentration is given by

σmax = Ktσ2 (4.59)whereKt is the stress concentration factor. Kt can be estimated using a graph (Figure B.1). To use the graphwe compute

h

d=

(R − r)

2R= 0.16

h

r=

R − r

rfillet= 3.33 (4.60)

Looking at the chart we estimate that, for this geometry, Kt = 1.5. The stress in the second shaft isσ2 =

2f (b + R)

πr3= 5.25MPa (4.61)

so the maximum stress in the fillet isσmax = Ktσ2 = 7.88MPa (4.62)

4.5 Power transmission in shafts

For shafts that are free to rotate, static analysis works for rotating shafts as long as they are not accelerating.Shafts are frequently used to transmit power. Let us consider how power is being transmitted in the previousexample:Example 4.4 Continued

Suppose that the shaft in the above example is rotating at a constant angular speed of ω. Howmuch poweris being transmitted?We know that power = work × velocity, so let us consider the power exerted by the force f .

P = f × [speed] = f × ω(b + R) = f (b + R)× ω = Mω (4.63)where M is the twisting moment in the shaft.

We thus determine a relationship between power and twisting moment:P = Mω = 2πMf ≡ Power transmission (4.64)

where f is frequency. This allows us to determine a relationship between power transmission and frequency interms of the maximum allowable stress in the shaft. Let us consider how this works by again continuing theAll content © 2017, Brandon Runnels 13.4



previous example:Example 4.4 Continued

Consider the shaft in the previous examples, but assume that it is supported with two bearings as shownbelow:

Suppose that the bearings are not frictionless and each exert a couple momentMfriction = −κω (4.65)

where κ is a viscous constant and ω is the angular velocity.Let us again do a moment balance to determine the relationship between f and t2 − t1:∑M = f (b + R) + (t2 − t1)(r + a)− 2κω = 0 (4.66)

so the force f is dependent on the difference in tensions as well as the angular velocity.f =

(t1 − t2)(r + a) + 2κω

b + R(4.67)

And the twisting moment is given by moment balance:∑M = Mx + f (b + R)− 2κω = 0 (4.68)

Mx = −f (b + R) + 2κω = (t2 − t1)(r + a) (4.69)If the beam turns at a speed ω, what is the power transmitted in the beam?

P = ω(t2 − t1)(r + a) (4.70)What is the loss of the system? We determine the total power exerted by the force f :

Pforce = (f )(ω(b + R)) = (t1 − t2)(r + a) + 2κω (4.71)The difference in power is

P − Pforce = 2κω (4.72)as we expect.

4.6 Non-axisymmetric shafts

We close this discussion of shafts by noting one important assumption: all of the theory developed here is foraxisymmetric (circular) shafts only. This theory can be extended to beams with arbitrary cross-section, howeverthe development of this is outside the scope of this class.



Lecture 14 Combined shaft loading analysis

4.7 Combined loading

Now that we know how to treat beams subjected to bending and shafts subjected to twisting, we can combine thetwo to obtain a comprehensive analysis of shafts subjected to loads and moments. To do this, we introduce theimportant concept of superposition:Definition 4.1. The principle of superposition states (1) that the sum of individual elastic responses to a set ofindividual loadings is the same as the response of the system to the sum of the individual loadings and (2) that if aset of loadings is multiplied by a scalar value, the response is multiplied by the same scalar value.

This is a direct consequence of the linearity that results when we make the small strain linearization assumption.Let us illustrate how this works by analyzing a shaft.Example 4.5: Combined loads on a shaft

Consider the following shaft supportedwith two frictionless, idealized bearings and subjected to three loadsas shown in the following figure:

b

b

2b

2af

f

f

x

y

z

L

L

L

L

Analyze the shaft and determine all shear and normal stresses, and determine themost likely point of failure.Torsion in the x direction

Let us begin by doing a classic torsion analysis on the shaft. Our moment function is:m(x) = f (b)〈x − L〉−1 + f (b)〈x − 2L〉−1 − f (2b)〈x − 4L〉−1 (4.73)

and so our moment is given byMx (x) = −

∫m(x) dx = −f (b)〈x − L〉0 − f (b)〈x − 2L〉0 + f (2b)〈x − 4L〉0 +

C1

0 (4.74)noting that M(0)

!= 0, giving us C1 = 0. Furthermore our angle of twist is

∆φ(x) = − 2

µπa4(f (b)〈x − L〉1 + f (b)〈x − 2L〉1 − f (2b)〈x − 4L〉1) (4.75)



The moment function Mx (x) gives us our shear stress at the radius as a function of r and x :σθx (r , x) =

Mx (x)r

J=

2r f b

πa4

[〈x − L〉0 + 〈x − 2L〉0 − 2〈x − 4L〉0

] (4.76)Plotting the shear stresses as a function of position within the beam, we have

x

05

1015

2025

3035

40

y

2.01.5

1.00.5

0.00.5

1.01.5

2.0

z

2.0

1.5

1.0

0.5

0.0

0.5

1.0

1.5

2.0

Plot of σθx resulting from torsionThis is good, but we are not done finding all of the stresses in the beam. We still must determine the stressesthat result from beam bending. We have two loading directions so we must examine the bending behaviorin both directions. We begin by considering the bending in the x-z plane:Bending in the x-z plane

Let us consider the shaft as a beam with loadings in the z direction:

f f

x

z

Let us compute our loading function in the x-z plane:w(x) = −f 〈x − L〉−1 − f 〈x − 2L〉−1 + fj2z〈x − 3L〉−1 (4.77)

What do we do about the journal bearings? Let us assume that they are “well-designed” – i.e., they do notexert a moment on the beam. Therefore we will treat one end as pinned, and the other as free, i.e.hz (0) = 0 Mx (0) = 0 Vz (4L) = 0 Mx (4L) = 0 (4.78)

but note that we also have an auxiliary constraint induced by the second journal bearinghz (3L) = 0 (4.79)

Now let us integrate:Vz (x) =

∫w(x) dx = −f 〈x − L〉0 − f 〈x − 2L〉0 + fj2z〈x − 3L〉0 + C1 (4.80)

Using one of our boundary conditions we find C1:Vz (4L) = −f 〈3L〉0 − f 〈2L〉0 + fj2z〈L〉0 + C1 = −2f + fj2z + C1 =⇒ C1 = 2f − fj2z (4.81)




Integrating again:My (x) =

∫Vz (x) dx = −f 〈x − L〉1 − f 〈x − 2L〉1 + fj2z〈x − 3L〉1 + (2f − fj2z ) x +>

0C2 (4.82)

where we use a BC to eliminate C2. Now, let us use the other moment boundary condition to find fj2z :My (4L) = −f (3L)− f (2L) + fj2zL + (2f − fj2z )(4L) = 3fL− 3fj2zL = 0 =⇒ fj2z = f (4.83)

So our moment function isMy (x) = f [−〈x − L〉1 − 〈x − 2L〉1 + 〈x − 3L〉1 + x ] (4.84)

and our shear function isVz (x) = −f 〈x − L〉0 − f 〈x − 2L〉0 + f 〈x − 3L〉0 + f (4.85)

Sketching these functions we haveV (x) M(x)

With our moment of inertia Iyy = πa4/4, our normal stress comes toσxx (x , z) = −M(x) z

Iyy=

4fz

πa4[〈x − L〉1 + 〈x − 2L〉1 − 〈x − 3L〉1 − x ] (4.86)

But we can also approximate the shear stress using the shear formulaτxz (x , z) =

V (x) z ′ A′

Iyy t(4.87)

From Wikipedia (with minor alterations) we haveQ(z) = z ′A′ =

2

3(a2 − z2)3/2 (4.88)

and furthermore we have that t =√a2 − z2, so the shear formula gives

τxz (x , z) =23 (a2 − z2)3/2

(πa4/4)(√a2 − z2)

V (x) =8f (a2 − z2)

3πa4

[− 〈x − L〉0 − 〈x − 2L〉0 + 〈x − 3L〉0 + 1

] (4.89)For the vertical shear formula we see that z ′ = 0 for all cuts, so it is clearly zero. Plotting the stresses:

x

05

1015

2025

3035

40

y

2.01.5

1.00.5

0.00.5

1.01.5

2.0

z

2.0

1.5

1.0

0.5

0.0

0.5

1.0

1.5

2.0

Plot of σxx resulting from x-z bending

x

05

1015

2025

3035

40

y

2.01.5

1.00.5

0.00.5

1.01.5

2.0

z

2.0

1.5

1.0

0.5

0.0

0.5

1.0

1.5

2.0

Plot of τzx resulting from x-z bending




We’ve done everything we can for bending in this direction, now let us consider the bending in the x-y plane:Bending in the x-y plane

Now, let us consider the loadings in the y direction:f

x

y

The first step is to define our loading function:wy (x) = fj2y 〈x − 3L〉−1 (4.90)

where we account for the other two forces using boundary conditions:hy (0) = 0 Mz (0) = 0 Vy (4L) = −f Mz (4L) = 0 (4.91)

and we have the auxiliary constrainthy (3L) = 0 (4.92)

Integrating to find shear:Vy (x) =

∫wy (x) dx = fj2y 〈x − 3L〉0 + C1 (4.93)

Using a boundary condition:Vy (4L) = fj2y + C1 = −f =⇒ C1 = −(f + fj2y ) (4.94)

Integrating to find moment:Mz (x) =

∫Vy (x) dx = fj2y 〈x − 3L〉1 − (f + fj2y )x + >

0C2

M(0)=0(4.95)

Using the other moment boundary condition we findMz (4L) = fj2yL− 4(f + fj2y )L = 0 =⇒ fj2y = −4

3f (4.96)

so we haveVy (x) =

f

3

[1− 4〈x − 3L〉0

]Mz (x) =

f

3

[x − 4〈x − 3L〉1

] (4.97)Sketching the plots of shear and moment:

V (x) M(x)




From V (x),M(x) we find normal stressσxx (x , y) = −Mz (x)y

Izz= − 4fy

3πa4

[x − 4〈x − 3L〉1

] (4.98)and shear stress

τxy (x , y) =V (x) y ′ A′

Iyy t=

8(a2 − y2)

3πa4V (x) =

8f (a2 − y2)

9πa4

[1− 4〈x − 3L〉0

] (4.99)Plotting the results for normal and shear stresses:

x

05

1015

2025

3035

40

y

2.01.5

1.00.5

0.00.5

1.01.5

2.0

z

2.0

1.5

1.0

0.5

0.0

0.5

1.0

1.5

2.0

Plot of σxx resulting from x-y bending

x

05

1015

2025

3035

40

y

2.01.5

1.00.5

0.00.5

1.01.5

2.0

z

2.0

1.5

1.0

0.5

0.0

0.5

1.0

1.5

2.0

Plot of τxz resulting from x-y bendingNow that we have computed the reactions from all of the loading modes, we must consolidate. Using theprinciple of superpositionwe know that the sum of the stress distributions from the distinct loadings is thesame as the stress distribution from the sum of the distinct loadings.Both beam loadings produce normal stresses in the xx direction, so our total normal stress is

σxx (x , y , z) =4f

πa4

[z(〈x − L〉1 + 〈x − 2L〉1 − 〈x − 3L〉1 − x

)− 1

3y(x − 4〈x − 3L〉1

)] (4.100)For torsion, we found σxθ. In Cartesian coordinates, we can split this up as

τxy = −σxθ sin θ τxz = σxθ cos θ (4.101)Substituting components for the xy shear stress gives:

τxy (x , y) =2f

πa4

[4(a2 − y2)

9

(1− 4〈x − 3L〉0

)−r b

(〈x − L〉0 + 〈x − 2L〉0 − 2〈x − 4L〉0

)sin θ

] (4.102)and for the xz shear stress:

τxz (x , z) =2f

πa4

[4(a2 − z2)

3

(1− 〈x − L〉0 − 〈x − 2L〉0 + 〈x − 3L〉0

)+rb

(〈x − L〉0 + 〈x − 2L〉0 − 2〈x − 4L〉0

)cos θ

] (4.103)There are no σyy ,σzz , τyz stresses. Plotting, we have




x

05

1015

2025

3035

40

y

2.01.5

1.00.5

0.00.5

1.01.5

2.0

z

2.0

1.5

1.0

0.5

0.0

0.5

1.0

1.5

2.0

Plot of total σxx

x

05

1015

2025

3035

40

y

2.01.5

1.00.5

0.00.5

1.01.5

2.0

z

2.0

1.5

1.0

0.5

0.0

0.5

1.0

1.5

2.0

Plot of total τxy

x

05

1015

2025

3035

40

y

2.01.5

1.00.5

0.00.5

1.01.5

2.0

z

2.0

1.5

1.0

0.5

0.0

0.5

1.0

1.5

2.0

Plot of total τxz



Lecture 15 Shaft theory concluded, intro to column buckling

4.7.1 Von Mises stress & yield criterion

In the previous example we never really answered the question of where the beam is most likely to fail. We havethree different types of stresses (σxx , τxy , τxz ), all of which have maxima at different locations.This motivates the need for a single stress measure, the maximum of which defines the location of greatest netstress that leads to yield. This prompts the definition of von Mises stress:Definition 4.2. Given a stress tensor σ that has normal stress components σxx ,σyy ,σzz and shear stress componentsτyz , τzx , τxy , the von Mises Stress σvm is given by

σvm =

√1

2

[(σyy − σzz )2 + (σzz − σxx )2 + (σxx − σyy )2 + 6 (τ 2

yz + τ 2zx + τ 2

xy )] (4.104)

The von Mises stress provides a means for defining a 3D yield criterionDefinition 4.3. If a material with yield stress σ0 yields when

σvm = σ0 (4.105)it is said to obey the von Mises yield criterion

Example 4.5 Continued

For the previous example, we plot the von Mises stress:

x

05

1015

2025

3035

40

y

2.01.5

1.00.5

0.00.5

1.01.5

2.0

z

2.0

1.5

1.0

0.5

0.0

0.5

1.0

1.5

2.0

x

05

1015

2025

3035

40

y

2.01.5

1.00.5

0.00.5

1.01.5

2.0

z

2.0

1.5

1.0

0.5

0.0

0.5

1.0

1.5

2.0

It appears the maximum stress occurs whenx = 3L y = a z = 0 (r = a, θ = 0) (4.106)

So let us compute the individual stress components: beginning with the normal stress,σxx (3L, a, 0) =

4f

πa4

[(0)(〈3L− L〉1 + 〈3L− 2L〉1 − 〈3L− 3L〉1 − 3L

)︸︷︷︸

=0

−1

3a(

3L− 4〈3L− 3L〉1︸︷︷︸=0

)]= − 4fL

πa3

(4.107)



the first shear stress,τxy (3L, a, 0) =

2f

πa4

[4(a2 − a2)

9

(1− 4〈3L− 3L〉0

)−ab

(〈3L− L〉0 + 〈3L− 2L〉0 − 2〈3L− 4L〉0

)sin(0)

]= 0 (4.108)

and the second shear stress,τxz (3L, a, 0) =

2f

πa4

[4(a2 − 02)

3

(1− 〈3L− L〉0 − 〈3L− 2L〉0 + 〈3L− 3L〉0

)+ab

(〈3L− L〉0 + 〈3L− 2L〉0 − 2〈3L− 4L〉0

)cos(0)

]=

4f

πa2(4.109)

Computing the von Mises stress givesσvm =

√1

2

[(σyy − σzz )2 + (σzz − σxx )2 + (σxx − σyy )2 + 6 (τ 2

yz + τ 2zx + τxy )

]=√σ2

xx + 3 τ 2xz =

√16f 2L2

π2a6+ 3

16f 2

π2a4

=4f

πa3

√L2 + 3a2 (4.110)

5 Axial Loading: Mechanics of Columns

Consider the following beam that is subjected to an axial load f as shown below:f f

If the beam is relatively slender, what happens as the magnitude of the force is increased? Intuition tells us thatthe beam will buckle, deflecting away from the loading axis, as shown in the following figure

h(x)

f f

As engineers, we are interested in understanding what load is required to cause buckling, as this plays an importantrole in the design of beams undergoing compression.5.1 Derivation of buckling equations and critical buckling loads

Let us consider a section of the beam as shown in the figure below:




ff

M(x)

h(x)

x

Notice that because the beam has now deflected by h(x), the axial force now generates a moment that must bebalanced by M(x). Put formally, ∑M = M(x) + f h(x) = 0 (5.1)

But recalling (3.12) from beam theory for a beam with constant properties,M(x) = E I

d2h

dx2(5.2)

Substituting, we now have a differential eqaution for h(x):d2h

dx2+

f

E Ih(x) = 0 (5.3)

which has the general solutionh(x) = C1 cos(λx) + C2 sin(λx) λ =

√f

E I(5.4)

If the beam has length L, then we know that h(0) = h(L) = 0. The first boundary condition impliesh(0) = C1 = 0 (5.5)

and the second boundary condition impliesh(L) = C2 sin(λL)

!= 0 =⇒ λ =

√f

E I=

nπ

L=⇒ fc =

n2π2E I

L2n ∈ Z (5.6)

(that is, n = 0, 1, 2, etc.) where λ are the eigenvalues and sin(λL) the corresponding eigenfunctions of the differen-tial equation.We see that the boundary conditions themselves impose restrictions on f . But what is n? Let us plot the shape ofthe buckled beam for several values of n:

n=0

n=1

n=2

n=3

fc = 0

fc = π2EIL2

fc = 4π2EIL2

fc = 9π2EIL2

These are called buckling modes. We observe intuitively that the first buckling mode is the most likely to occurphysically; therefore we conclude thatfc =

π2EI

L2≡ Critical buckling load for pinned-end column (5.7)




5.1.1 Buckling loads derived from the pin-pin solution

We have derived the solution for two pinned ends, but what about cases with different boundary conditions? Letus consider a beam that is fixed on one end and free on the other subjected to the axial load f as shown below.

f

L

Note that there is now a couple moment exerted at the fixed end, which we did not account for in our originaldifferential equation. So we now have to solveEI

d2h

dx2+ f h(x) = Mfix (5.8)

which involves more work. However, we can actually pull a little trick to get the critical buckling load with almostno effort. Let us imagine that we reflect the beam horizontally so that the fixed end becomes the midpoint of apinned-pinned beam with twice the length. There is functionally no difference between the two:

f

2L

f

We can therefore re-use the same critical buckling load that we had before, except that we plug in twice the length.The resulting buckling load isfc =

π2EI

(2L)2≡ Critical buckling load for fixed-free column (5.9)

Note that the factor of two reduces the buckling load by 50%. Thismatches our intuition that buckling of a fixed-freecolumn requires less force.Now, what if both ends are fixed? Consider the beam shown below:

fL

f

Let us attempt to pull the same trick as before: imagine that we are looking at a section of a longer beam withmode three buckling:f

3L/2

f

The critical buckling force for this beam (with length 3L/2) isfc =

9π2EI

(3L/2)2=

π2EI

(L/2)2≡ Critical buckling load for fixed-fixed column (5.10)




Note that the factor of 0.5 in the denominator increases the critical load by 400% from pin-pin, which matches ourintuition that it is much harder to buckle a fixed-end column.



Lecture 16 Buckling continued

5.1.2 Buckling load equations derived independently from the pin-pin solution

Let us consider onemore buckling scenario that cannot be handled by reducing to a pin-pin buckling case. Considera beam that is fixed on one end by a beam, and on the other by a pivot that does not allow lateral deflection:

f

L

Let us revise our buckling equation: take a cut of the beam and draw a free body diagram:

x

f

M(x)

h(x)

V0M0

f

Doing a moment balance around the cut we haveM(x) + f h(x)−M0 − V0x = 0 (5.11)

Rearranging and substituting (3.12) we haveEI

d2h

dx2+ f h(x) = M0 + V0 x (5.12)

a nonhomogeneous differential equation. The general solution for this ODE ish(x) = C1 cos(λx) + C2 sin(λx) +

M0 + fx

fλ =

√f

EI(5.13)

To solve for the constants we need boundary conditions. We recall from beam theory that we haveh(0) = 0 θ(0) = 0 h(L) = 0 M(L) = 0 (5.14)

We begin by applying the boundary conditions at x = 0:h(0) = C1 +

M0

f= 0 =⇒ C1 = −M0

f(5.15)

θ(0) = h′(0) = λC2 +V0

f= 0 =⇒ C2 = −V0

f λ(5.16)

Now let us apply the other two boundary conditions:h(L) =

M0 + V0L

f− M0

fcos(λL)− V0

f λsin(λL) = 0 (5.17)

M(x) = EId2h

dx2= EI

[λ2 M0

fcos(λL) + λ2 V0

f λsin(λL)

]= 0 (5.18)



Simplifying these two equations:(L− 1

λsin(λL)

)V0 +

(1− cos(λL)

)M0 = 0 (5.19)(

sin(λL))V0 +

(λ cos(λL)

)M0 = 0 (5.20)

Writing these equations in matrix form:[L− 1

λ sin(λL) 1− cos(λL)sin(λL) λ cos(λL)

] [V0M0

]= 0 (5.21)

One solution to this equation is, obviously, the trivial V0 = 0,M0 = 0. But we seek a solution where the V0 6= 0,M0 6=0. Because the right hand side is zero, this means that the matrix must be rank-deficient; that is, it must have anullspace that contains [V0 M0]. We know from linear algebra that this can be true if and only if the determinant iszero. Therefore, we can set the determinant to zero to solve for λ via:

det =(L− 1

λsin(λL)

)(λ cos(λL)

)−(

1− cos(λL))(

sin(λL))

= Lλ cos(λL)− sin(λL) cos(λL)− sin(λL) + sin(λL) cos(λL)

= Lλ cos(λL)− sin(λL) = 0 (5.22)Rearranging we have

λL = tan(λL) (5.23)

Plotting x = λL and x = tan(λL) we see that there are aninfinite number of intersection points. The transcedentalequation is not possible to solve analytically, but using acomputer to find the roots numerically we obtainλL =0,

(1.430296653)π,

(2.459024033)π,

(3.470889724)π

...

λL = 0 is the trivial solution so we select the first nontriv-ial solution λ = (1.430296653)π. 0 2 4 6 8 10 120

2

4

6

8

10

12 tan(x)

x

x= tan(x)

Expanding out we have √fcE I

L = (1.430296653)π (5.24)Solving:

fc =E Iπ2

(L/1.430296653)2=

E Iπ2

(0.6991556596 L)2≈ E Iπ2

(0.7 L)2(5.25)




5.1.3 Effective Length Factor

All of the formulae that we have developed for critical buckling have had the formfc =

E Iπ2

(KL)2≡ General formula for critical buckling load (5.26)

whereK = 1 for a pin-pin end,K = 0.7 for fix-pin, etc. For a selection of columnswith different boundary conditiosnwe havepin-pin

(free-free)fix-fix

(free translation)fix-pin fix-fix fix-free

K = 1.0 K = 1.0 K = 0.7 K = 0.5 K = 2.0

To determine the critical buckling load, the proper boundary conditionsmust be identified to find the effective lengthfactor, and then the general formula for critical buckling load can be used to find fc . We will show how this workswith a couple of examples:Example 5.1: Thermal buckling

A pipe with outer and inner radii a and b and is supported by two bearings spaced a distance L from eachother as shown in the following diagram.a

b

L

The pipe is made of a material with Young’s modulus E and coefficient of thermal expansion α. What is themaximum increase in temperature ∆T before the pipe buckles?Suppose that the temperature increases by ∆T . Stress is given byσ = E (ε− α∆T ) = −E α∆T (5.27)

We convert this to force by multiplying by area:f = −E α∆T × π(a2 − b2) = −E α∆Tπ(a2 − b2) (5.28)

We are working with a fix-fix boundary condition so the buckling formula isfc =

EIπ2

(L/2)2=

4EIπ2

L2(5.29)

The bending moment of inertia isI =

π(a4 − b4)

4(5.30)




Substituting in and equating the (negative) force with the critical load:E α∆Tπ(a2 − b2) =

4E (π(a4 − b4)/4)π2

L2(5.31)

and solving for ∆T we find∆T =

(a4 − b4)π2

α L2(a2 − b2)(5.32)



Lecture 17 Thin-wall pressure vessels

6 Volumetric Loading: Thin-wall Pressure Vessels

Pressure vessels are vessels designed to hold a gas or fluid that is maintained at a hydrostatic pressure.6.1 Derivation of pressure vessel formulae

Wewill consider two types of pressure vessels here: spherical and cylindrical. In all cases, wewillmake the thin-wallassumption, defined as follows:Definition 6.1. The thin-wall assumption for pressure vessels states that the thickness of the pressure vessel wall(t) is much less than the radius and/or length of the vessel (R), i.e. t << R . Implications of the thin-wall assumptioninclude

1. O(t2) ≈ 0 and (R + t)1 ≈ R1.2. There is no radial compressive stress, σrr = 0.3. Stresses are uniform throughout the thickness of the wall.

We show how this works in the case of spherical and cylindrical pressure vessels.6.1.1 Spherical pressure vessels

Consider a spherical vessel with interior radius R and wall thickness t subjected to an internal gauge pressure(pressure difference) of p. We seek to determine the magnitude of the stress developed inside the walls of thevessel.

p

σθθ

2R

t

t

σθθ

σφφ

σθθ

σφφ

The analysis turns out to be surprisingly easy. Taking a cut down through the center of the vessel, we compute theforces resulting from the internal pressure and the normal stress σθθ:∑f = −p × (πR2) + σθθ × (π((R + t)2 − R2)) (6.1)

Using the thin-wall assumption:(R + t)2 − R2) = R2 + 2Rt +7

0

t2 − R2 = 2Rt (6.2)All content © 2017, Brandon Runnels 17.1


Substituting, and noting that σθθ must equal σφφ (since the analysis would be identcal for both) we haveσθθ = σφφ =

pR

2t≡ Normal stress in a spherical thin-wall pressure vessel (6.3)

Example 6.1: Beach ball

A beach ball has a 12cm radius and a thickness of1.25mm. It is made of rubber, with material properties

E = 3.7MPa ν = 0.48 (6.4)If the ball is subjected to a pressure of 65kPa, deter-mine the stress and strain in the ball.Also, if the maximum allowable strain is 0.425, deter-mine the minimum allowable thickness of the wall.This is a very direct application of the pressure vessel formula for a spherical pressure vessel. We find thatthe stress is given by

σθθ =pR

2t=

(65E3Pa)(0.12m)

2(0.00125m)= 3.120MPa. (6.5)

Therefore we know that the stresses areσθθ = σφφ = 3.120MPa σrr = 0 σrθ = σθφ = σφr = 0 (6.6)

Now we use eqs. (2.81) to convert the stress to strain.εθθ =

1

E(σθθ − νσφφ − νσrr ) =

(1− ν)σθθE

=(1− ν)pR

2tE= 0.43 (6.7)

εφφ =1

E(σφφ − νσθθ − νσrr ) =

(1− ν)pR

2tE= 0.43 (6.8)

εrr =1

E(σrr − νσφφ − νσrr ) = −2νσφφ

E= −νpR

tE= 0.81 (6.9)

Apparently the out-of-plane strain is the greatest. Therefore if the maximum allowable strain is 0.425, thewe can solve for t:t =

νpR

νmaxE=

(.48)(65e3Pa)(.12m)

(.425)(3.7e6)= 2.38mm (6.10)



Lecture 18 Pressure vessels concluded

6.1.2 Cylindrical pressure vessels

Cylindrical pressure vessels are another common type of vessels. Railroad tank cars are an example of cylindricalpressure vessels that are used extensively. Unlike with spherical vessels, cylindrical pressure vessels have differentstresses in different directions.It is simple to show that the stress along the length of the cylinder (σzz ) is exactly the same as the stress in aspherical pressure vessel. This is called the longitudinal stress.

2R

t

t

p

σzzσθθ

p

longitudinal circumferential

L

With spherical we were able to show that all of the stresses were the same in all directions, but that trick does notwork here. Let us take a section of the vessel (length L) and cut it in half as shown above. Force balance gives us∑f = −2pLR + 2σθθLt = 0 (6.11)

Solving for σθθ and noticing that the Ls cancel, we haveσθθ =

pR

t(6.12)

This stress is called the circumferential stress, and is often referred to as the hoop stress. To summarize, we havethe following stresses in a cylindrical pressure vessel:σzz =

pR

2t≡ Longitudinal stress σθθ =

pR

t≡ Circumferential/Hoop stress (6.13)

Example 6.2: Cylindrical pressure vessel

Find the von Mises stress in a cylindrical pressure vessel with the following parametersE = 200GPa ν = 0.3 p = 800kPa t = 20mm R = 1.8m (6.14)

First, determine longitudinal and circumferential stress.σzz =

pR

2t= 36MPa σθθ =

pR

t= 72MPa (6.15)

Substitute into the von Mises equation (4.104):σvM =

√1

2

[(σrr − σθθ)2 + (σθθ − σzz )2 + (σzz − σrr )2 + 6(σ2

rθ + σ2θz + σ2

zr )] (6.16)



we haveσvM =

√1

2

[(0− pR

t

)2

+(pR

t− pR

2t

)2

+(pR

2t− 0)2

+ 6(0 + 0 + 0)]

=pR√

3

2t= 62.4MPa (6.17)

6.2 Stress concentrations in pressure vessels

We have presented a highly simplified method for pressure vessel analysis. However, given the assumptionswe made in the derivations, we must keep in mind that there are a number of things that will violate our originalassumptions. Examples include:• Corners or sharp edges of any kind• Openings (such as valves) within a load-bearing section of the pressure vessel• Discontinuous material properties

We illustrate this by looking at a finite element analysis simulation of a cylindrical pressure vessel with a cylindricalend cap. (Note that the simulation only takes into account 1/8 of the vessel because of symmetry.) We observethat the there is a smooth transition from the cylindrical portion of the cylinder to the spherical portion, and the totalstress is reduced in the cap region. Furthermore we also notice that the assumption of uniform stress distributionthroughout the thickness is a good one.

von Mises stress hoop stress longitudinal stress

Now, let us consider an identical cylinder, except with a flat-capped end. For a sufficient distance from the end cap,there is uniform hoop and longitudinal stress. However, we see that the stress significantly increases at the corner.Furthermore we observe a very nonuniform stress distribution within the end cap – because there is no curvature,the load must be carried via plate bending (similar to beam bending) which is why we see a linear stress profile.

von Mises stress hoop stress longitudinal stress

Consequently, we conclude that simplified pressure vessel analysis must be used with care, and should only beused for back-of-the-envelope estimation.




7 Transformations of Stress and Strain

Here we shift gears. In sections 3-6 we considered several different loading scenarios for different problem ge-ometries, and developed simplified solutions for their deflection and twist angle as well as stress and strain. Inthis section, we return to looking at the interaction between shear and normal stresses and strains, as well as thestress and strain tensors.7.1 Review of tensors, normal stress, and shear stress

note: in the first version of these notes there was some inconsistent notation. The following have beenupdated and will be used consistently:• The stress tensor will be denoted using bold face (σ) instead of normal face (σ)• The strain tensor will be denoted using bold face (ε) instead of normal face (ε).• Shear stresses will be denoted using tau instead of sigma. For instance, shear stress in the xy directionwill be represented as τxy instead of τxy .Consider a piece of material with cross-sectional area A pulled at either end by a force f. What is the stress in thematerial? Let us begin by proceeding in the same way with which we are familiar: by making a cut and doing forcebalance.

∑f =

[−f00

]+

[σnormalAτsheearA

0

]= 0 =⇒ σxx =

f

A, τxy = 0 (7.1)

which is as we expect. Now, suppose we take a cut that is inclined at an angle θ. Notice that the area of the tiltedcut is equal to Anew = A/ cos θ. For our previous cut, we found σxx (the normal stress) and τxy (the shear stress).Let us compute the normal and shear stresses for the cut by using force balance:∑

f =

[−f00

]+σnormalA

cos θ

[cos θsin θ

0

]+τshearA

cos θ

[− sin θcos θ

0

]= 0 (7.2)

Simplifying we haveσnormal

[cos θsin θ

0

]+ τshear

[− sin θcos θ

0

]=

[f cos θ/A

00

](7.3)

and writing in matrix form [cos θ − sin θsin θ cos θ

] [σnormalτshear

]=[f cos θ/A

0

] (7.4)If you have taken linear algebra, you know that this particular coefficient matrix is called an orthogonal matrixwhich means you can invert it by taking the transpose. (If you haven’t taken linear algebra, you can solve for thestresses simply by using Cramer’s rule.) We have[

σnormalτshear

]=[

cos θ sin θ− sin θ cos θ

] [f cos θ/A

0

]=

f cos θ

A

[cos θ− sin θ

] (7.5)so we have

σnormal =f cos2 θ

Aτshear = − f cos θ sin θ

A(7.6)




Now, you may recall that there is an easier way to find shear and normal stresses for a cut with a normal vector n.If we can find the stress tensor σ, we have the following formulae for the normal and shear stresses:σnormal = n · σn ≡ Normal stress τshear = |σn− σnn| ≡ Shear stress (7.7)

Let’s apply these formulae to this example. First, we find the stress tensor σ and the normal vector n for our cutregion:σ =

[f /A 0

0 0

]n =

[cos θsin θ

] (7.8)Plugging into our formula we get:

σnormal = [cos θ sin θ][f /A 0

0 0

] [cos θsin θ

]= [cos θ sin θ]

[f cos θ/A

0

]=

f cos2 θ

AX (7.9)

as expected. Similarly for shear,σn− σnn =

[f /A 0

0 0

] [cos θsin θ

]− f cos2 θ

A

[cos θsin θ

]=

f

A

[cos θ − cos3 θ− cos2 θ sin θ

]=

f

A

[cos sin2 θ− cos2 θ sin θ

]=

f cos θ sin θ

A

[sin θ− cos θ

](7.10)

Computing the magnitude:τshear = |σn− σnn| =

∣∣∣ f cos θ sin θ

A

[sin θ− cos θ

] ∣∣∣ =f cos θ sin θ

AX (7.11)

giving us the same result as before.



Lecture 19 Tensor transformation

7.2 Tensor transformation

Wenowpresent a key new fundamental concept in stress analysis: transformation of tensors (i.e. stress and strain)in rotated frames. To illustrate the general idea, let us consider the following illustration. Suppose we consider abody that is subjected to a normal stress σxx in the x direction, σyy in the y direction, and a shear stress of τxy = τyx .We graphically express the components of this tensor in the figure on the left.

σxx

σyy

σyy

σxx

τxy

τxy

τyx

τyx

x

y

x ′y ′

θ

σ′y ′y ′

σ′y ′y ′

σ′x′x′

σ′x′x′τ ′x′y ′

τ ′y ′x′

τ ′x′y ′

We would express our stress tensor in the following way:σ =

[σxx τxyτxy σyy

] (7.12)and we could then use it to compute normal and shear stresses as needed.But what if we chose a different set of coordinates (x ′, y ′) such that the new coordinate axes are rotated by anangle θ? (This is illustrated in the figure to the right.) What is the relationship between the stress components inthe rotated frame (σ′x′x′ ,σ′y ′y ′ , τ ′x′y ′) and the stress components in the original frame?If you have taken linear algebra, you may recall that the formula for tensor rotation in 2D is given by

σ′ = RσRT (7.13)where R is the rotation matrix corresponding to angle θ and RT is its transpose, with

R =[cos θ − sin θsin θ cos θ

] (7.14)so that the formula is[

σ′x′x′ τ ′x′y ′

τ ′y ′x′ σ′y ′y ′

]=[cos θ − sin θsin θ cos θ

] [σxx τxyτyx σyy

] [cos θ sin θ− sin θ cos θ

]≡ Components of stress in a rotated frame (7.15)

Let us show how this works using a couple of examples.



Example 7.1: Uniaxial tension

Consider a bar with cross-sectional area A subjected to a uniaxial force f as shown below. Find the compo-nents of stress for the two rotations shown below:ff ff θ

Let us start by finding the components of our tensor in the unrotated frame:σ =

[f /A 0

0 0

] (7.16)To find the components of σ′, we simply apply the formula and do the matrix-matrix multiplication:[


τ ′y ′x′ σ′y ′y ′

]=[cos θ − sin θsin θ cos θ

] [f /A 0

0 0


] (7.17)=[cos θ − sin θsin θ cos θ

] [f cos θ/A f sin θ/A

0 0

] (7.18)=

[f cos2 θ/A f sin θ cos θ/A

f sin θ cos θ/A f sin2 θ/A

](7.19)

g This tells us that the normal stress in the x ′ direction (σ′x′x′ , and the normal stress in the y ′ direction (σ′y ′y ′)and the shear stress τ ′x′y ′ are given byσ′x′x′ =

f cos2 θ

Aσ′y ′y ′ =

f sin2 θ

Aτ ′x′y ′ =

f sin θ cos θ

A(7.20)

This should seem familiar – you’ll notice that σ′x′x′ and τ ′x′y ′ are exactly the same as the normal and shearstresses that we computed in the earlier example.

θ

f cos2 θ/Af sin2 θ/A

f sin θ cos θ/A

f /A

Example 7.2: Numerical example

A material is subjected to vertial normal stress of 100kPa and a shear stress of 50kPa as shown below.100kPa

50kPa 30

? ?

?

Determine the normal and shear stresses in a frame rotated by 30.




Begin by writing down the stress tensor in the unrotated frame:σ =

[0 50kPa

50kPa 100kPa

] (7.21)Substituting into our formula we find[


τ ′y ′x′ σ′y ′y ′

]=[cos 30 − sin 30

sin 30 cos 30] [

0 50kPa50kPa 100kPa

] [cos 30 sin 30

− sin 30 cos 30]

=[cos 30 − sin 30

sin 30 cos 30] [ −50kPa sin 30 50kPa cos 30

50kPa cos 30 − 100kPa sin 30 50 sin 30kPa + 100 cos 30kPa

]=[cos 30 − sin 30

sin 30 cos 30] [ −25kPa 43.3kPa−6.70kPa 111.6kPa

]=

[(−25kPa) cos 30 − (−6.7kPa) sin 30 (43.3kPa) cos 30 − (111.6kPa) sin 30

(−25kPa) sin 30 + (−6.7kPa) cos 30 (43.3kPa) sin 30 + (111.6kPa) cos 30

]=[−18.3kPa −18.3kPa−18.3kPa 118kPa

] (7.22)

Example 7.3: Maximum shear stress in a cylinder

Consider a pressure vessel with radius R and thickness t subjected to an internal pressure p, as shown inthe following figure.

σzz

σθθ

σxx

σyy

σ′x′x′

σ′y ′y ′

τ ′x′y ′θ =?2R

In the cylindrical region, (not at the end caps) determine themaximum shear stress (and the correspondingrotation) in terms of R, t, p.Let us begin by finding σzz and σθθ which are given directly from the formulae for a pressure vessel:

σxx = σzz =pR

2tσyy = σθθ =

pR

t(7.23)

so our stress tensor isσ =

pR

2t

[1 00 2

] (7.24)Now let us apply the tensor transformation formula (noting that we factor pR/2t out to the front):[


τ ′y ′x′ σ′y ′y ′

]=

pR

2t

[cos θ − sin θsin θ cos θ

] [1 00 2


] (7.25)=

pR

2t

[cos θ − sin θsin θ cos θ

] [cos θ sin θ−2 sin θ 2 cos θ

] (7.26)=

pR

2t

[cos2 θ + 2 sin2 θ sin θ cos θ − 2 sin θ cos θ

sin θ cos θ − 2 sin θ cos θ sin2 θ + 2 cos2 theta

](7.27)




=pR

2t

[1 + sin2 θ − sin θ cos θ− sin θ cos θ 1 + cos2 theta

](7.28)

Noting the useful trigonometric identitysin θ cos θ =

1

2sin(2θ) (7.29)

we see that the shear stress is given byτ ′x′y ′ = −pR sin(2θ)

4t(7.30)

For what θ does this value equal its maximum? We know that sin(θ) is extremized at θ = (n + 1/2)π.Therefore the maximum value occurs where2θ =

(2n + 1)π

2=⇒ θ =

(2n + 1)π

4=π

4,

3π

4,

5π

4(7.31)

Therefore we have that the maximum shear stress ispR

4t(7.32)

and it occurs at θ = 45.



Lecture 20 Principal stresses

7.3 Principal stresses and directions

We have talked about finding shear and normal stresses in different directions, and we just completed an examplein which we found the orientation corresponding to the maximum shear stress.Now let us ask whether or not we can find a rotation in which there is no shear stress. Does such a case exist, andif so, under what conditions? Let us begin by looking again at the formula for shear stress:

τshear = |σn− σnn| (7.33)If τshear = 0, then we know (by the definition of the norm) that

σn− σnn = 0 (7.34)Recall that the identity matrix I is defined so that In = n, so we can modify the above to be

σn− σnIn = 0 (7.35)and factoring out n we have

(σ − σnI)n = 0 (7.36)Or, writing out in components, [

σxx − σn τxyτxy σyy − σn

] [n1n2

]=[00

] (7.37)Now, we must solve for n1 and n2 ... but if we attempt to use Cramer’s rule, we just get that n = 0. This is where wepull a linear algebra trick: we recall that for a matrix A and vector x, the only way a nontrivial solution can exist forthe equation Ax = 0 is if x is in the nullspace/kernel of A. But in order for a matrix to have a nonzero nullspace, itmust have a zero determinant.Therefore, we know that for a nontrivial solution to exist for the above, we must have

det[σxx − σn τxyτxy σyy − σn

]= (σxx − σn)(σyy − σn)− τ 2

xy = 0 (7.38)This is an equation that can be solved for σn , but it is a second order polynomial that can be solve, which meansthat there will be two solutions, which we will call σ1,σ2. These solution are very special, and we will define themas follows:Definition 7.1. Given a stress tensor σ, the solutions of the equation

det(σ − σnI) = 0 (7.39)for σn are called principal stresses, and are also known as the eigenvalues of the stress tensor. They are the normalstresses of the tensor in the rotated frame that corresponds to the shear-free rotation.

Once we have solved for σ1,σ2, we plug back into Equation (7.37). But now that we have two solutions for σ1,σ2,we know that there are two possible values for n. We will call those n1,n2, and will give them their own definition:



Definition 7.2. Given a stress tensor σ with principal stresses σ1,σ2, there are two corresponding vectors n1 and n2

that satisfy

σn1 = σ1n1 σn2 = σ2n2 (7.40)These vectors are called the principal directions, and are the eigenvectors of the stress tensor. They correspond tothe normal vectors for the planes on which there is no shear stress.

σxx

σyy

σ1

σ2

τxy

n1

n2

original stresses/directions principal stresses/directions

If you have taken a course in linear algebra, the process of finding principal stresses and directions will be familiar,as it is exactly the same process used to find eigenvalues and eigenvectors of a matrix.Example 7.4: Computation of principal stresses

Suppose that amaterial is subjected to a normal loading of 30kN in the x direction and 10kN in the y direction,and a shear loading of 20kN . Find the principal stresses and directions for this loading.First we write down our stress tensor:

σ =[30kN 20kN20kN 10kN

] (7.41)Now, we solve for the principal stresses using Equation 7.38:det[30kN − σn 20kN

20kN 10kN − σn

]= (30kN − σn)(10kN − σn)− 400kN = 300kN2 − 40kNσn + σ2

n − 400kN2 = 0

(7.42)Rearranging and putting into quadratic form we have

σ2n − 40kNσn − 100kN2 = 0 (7.43)

This is a job for the quadratic formula: with a = 1, b = −40kN, c = −100, we haveσn =

40kN ±√

1600kN2 + 400kN2

2= 20kN ± 1

2

√2000kN2 = (20± 10

√5)kN = 42.36kN,−2.3kN (7.44)

So we have σ1 = 42.36kN and σ2 = −2.3kN . We’ve completed part of the job, but we still have to find theprincipal directions corresponding to these principal stresses. We do this simply by plugging in and solving.[30kN − 42.36kN 20kN

20kN 10kN − 42.36kN

] [n1xn1y

]=

[(−12.36kN)n1x + (20kN)n1y

(20kN)n1x − (32.36kN)n1y

]= 0 (7.45)

Solving the top equation for n1y in terms of n1x we getn1y =

(12.36kN)n1x

20kN= 0.618n1x (7.46)




Substituting this into the second equation:(20kN)n1x − (32.36kN)

(0.618n1x

)= (20kN)n1x − (20kN)n1y = 0 (7.47)

So the second equation is useless. (If you’ve taken linear algebra, this should make sense.) It turns out thatthe best we are ever going to when solving for n1 is to getn1 = (n1x )

[1

0.618

] (7.48)But wait: remember that n1 is a normal vector. This gives us a way to solve for n1x :

|n1| =√n2

1x + (0.618)2n21x = 1 =⇒ n1x =

1√1 + (0.618)2

= 0.85 (7.49)Therefore we have the following result for n1:

n1 =[0.850.52

] (7.50)Now, let us do the same process for n2 by substituting in the other eigenvalue, −2.3kN :[

30kN + 2.3kN 20kN20kN 10kN + 2.3kN

] [n2xn2y

]=

[(32.3kN)n2x + (20kN)n2y

(20kN)n2x + (12.3kN)n2y

]= 0 (7.51)

Solving (either the top or the bottom equation) for n2y in terms of n2x we get n2y = −1.61n2x son2 = n2x

[1

−1.61

] (7.52)Finding the n2x that normalizes the vector gives n2x = 0.528, and

n2 =[

0.528−0.85

] (7.53)Sketching the results, we have

30kN

10kN

42.36kNn220kN

n1

−2.3kN

We note thatn1 =

[cos θsin θ

]n2 =

[− sin θcos θ

] (7.54)which we can solve to find that θ = 31.8

7.3.1 Solution strategy

Finding eigenvalues and eigenvectors can be a bit tedious if you’ve never done it before. You can easily use Matlabor Maple to find them for you, but you should also know how to do it yourself. The following steps enumerate howAll content © 2017, Brandon Runnels 20.3



to find principal stresses (eigenvalues) and principal directions (eigenvectors) in 2D:Solution strategy for finding eigenvalues and eigenvectors1. Write the stress tensor σ2. Solve the polynomial equation det(σ − σnI) = 0 to get principal stresses σn = σ1,σ23. Substitute σ1,σ2 back into

(σ − σ1I)[n1xn1y

]= 0 (σ − σ2I)

[n2xn2y

]= 0 (7.55)

to solve for n1y in terms of n1x , and n2y in terms of n2x .4. Solve for n1x , n2x using |n1| = 1, |n2| = 1.5. Solve for the rotation angle θ using any of the following:n1x = cos θ n1y = sin θ n2x = − sin θ n2y = cos θ (7.56)

Let us demonstrate this process using an example:Example 7.5: Manual computation of eigenvalues

Let us do another example to cement the process. Let the stress tensor be defined asσ =

[1kN 2kN2kN 1kN

] (7.57)The first thing we do is to compute the principal stresses / eigenvectors by solving the characteristic poly-nomial:

det[1kN − σn 2kN

2kN 1kN − σn

]= (1kN − σn)2 − 4kN2 = 0 =⇒ (1kN − σn) = ±2kN =⇒ σn = 1kN ± 2kN

(7.58)So the principal stresses are

σ1 = 3kN σ2 = −1kN (7.59)Now let us compute the principal directions / eigenvectors n1,n2:[

1kN − 3kN 2kN2kN 1kN − 3kN

] [n1xn1y

]=

[(−2kN)n1x + (2kN)n1y

(2kN)n1x + (−2kN)n1y

]= 0 (7.60)

implying that n1x = n1y . We see then that the solution must ben1 = n1x

[11

] (7.61)Solving for n1x by normalizing gives

n1 =1√2

[11

] (7.62)Following the same process for the second eigenvalue:[

1kN − (−1kN) 2kN2kN 1kN − (−1kN)

] [n2xn2y

]=

[(2kN)n2x + (2kN)n2y

(2kN)n2x + (2kN)n2y

](7.63)

Solving either equation gives n2x = −n2y . Following the same process we see that then2 =

1√2

[1−1

] (7.64)




If you have taken linear algebra before, you may be wondering what happens if the eigenvalues of the stress tensorare imaginary. It turns out that this will never happen, and the reason is because σ is symmetric. The proof of thisis quite easy to do, but requires some knowledge of complex analysis, which is out of the scope of this class.However, we can use the symmetry of σ to prove another interesting fact about principal directions:

Sidenote: notice that the two eigenvectors n1 and n2 are orthogonal to each other. In fact, it turns out thatwe can prove that they will always be orthogonal to each other based only on the fact that σ is symmetric.The proof of this is quick so we will present it here.Let n1,n2 be the eigenvectors/principal directions of σ corresponding to eigenvectors/principal stressesσ1,σ2, and assume that σ1 6= σ2 By symmetry of σ, we know that

n1 · σn2 = n2 · σn1 (7.65)We know that σn1 = σ1n1 and σn2 = σ2n2, and substituting we have

σ1n1 · n2 = σ2n2 · n1 (7.66)Rearranging

(σ1 − σ2)(n1 · n2) = 0 (7.67)Because we have enforced that σ1 6= σ2, the only way for this equation to hold is if

n1 · n2 = 0 (7.68)implying orthogonality and concluding the proof.You may be wondering about the case where σ1 = σ2. In that case, because any vector can be representedas a linear combination of n1 and n2, and because they have the same eigenvalue, then every vector in thespace spanned by n1 and n2 is an eigenvector of σ. Therefore we can pick whatever two vectors we want,and we generally pick two that orthogonal.



Lecture 21 Mohr’s circle

7.4 Mohr’s circle for plane stress

It turns out that there is, in fact, a much easier method for tensor transformation and for finding principal stressesand directions. Now that we have done it a few times the hard way, we’ll introduce a trick that helps us visualizeprincipal stresses and directions more easily.Let us consider the stress state where σxx = σ1,σyy = σ2, τxy = 0 (in other words, we’re in the state in which thestresses are equal to the principal stresses. Now, let us rotate our reference frame clockwise by an angle θ anddetermine the stress state in that rotated frame: from the previous section, we have[

σxx τxyτxy σyy

]=[

cos θ sin θ− sin θ cos θ

] [σ1 00 σ2

] [cos θ − sin θsin θ cos θ

]=

[σ1 cos2 θ + σ2 sin2 θ (σ2 − σ1) sin θ cos θ(σ2 − σ1) sin θ cos θ σ1 sin2 θ + σ2 cos2 θ

](7.69)

These expressions are a bit nasty; however, we can actually reduce them considerably by using the followingtrigonometric identities:sin θ cos θ =

1

2sin 2θ cos2 θ =

1 + cos 2θ

2sin2 θ =

1− cos 2θ

2(7.70)

Substituting in:σxx =

1

2(σ1 + σ2)− 1

2(σ2 − σ1) cos 2θ σyy =

1

2(σ1 + σ2) +

1

2(σ2 − σ1) cos 2θ τxy =

1

2(σ2 − σ1) sin 2θ (7.71)

Actually, this does not look more simple, in fact it seems considerably more complicated! But this is where wenotice something interesting: suppose we have a set of axes in which the abcissa (horizontal axis) corresponds tonormal stress and the ordinate (vertical axis) corresponds to the shear stress. Then, let us plot our stress state bymarking the following two points: (σxx , τxy ) and (σyy ,−τxy ):

2θ

σ

τ

12 (σ1 + σ2)

σ2σ1

σyy

σxx

τxy

−τxy

σ1

σ2

ττ

σ2

σ1

θ12 (σ1 − σ2)

Notice that if we plot all possible stress states in this way, we end up plotting a circle. This is called Mohr’s circleand is a very useful way to visualize at once all of the possible stress states obtainable in two dimensions. Definedformally:Definition 7.3. Mohr’s circle is the locus of points formed by plotting (σxx , τxy ) (σyy ,−τxy ) for all possible valuesof σxx ,σyy , τxy obtained by 2D rotation. The angle between the diameter connecting (σxx , τxy ) (σyy ,−τxy ) and thehorizontal axis is equal to twice the angle of rotation.

Exactly how does Mohr’s circle help us? To begin, you may have noticed that Equations 7.71 are not particularlyuseful, because the require that you already know the principal stresses.All content © 2017, Brandon Runnels 21.1


Suppose we have stresses σxx ,σyy , τxy and we wish to find principal stresses. Geometrically, looking at the circlewe know that the principal stresses can be found by computing the location of the center of the circle as well asthe circle’s radius. Let us compute these things:σaverage =

σ1 + σ2

2=σxx + σyy

2(7.72)

We can find the radius of the circle using the Pythagorean theorem:r =

√(σyy − σxx

2

)2

+ τ 2xy (7.73)

and from this we obtain the principal stresses:σ1,2 =

σxx + σyy

2±√(σyy − σxx

2

)2

+ τ 2xy ≡ Principal stresses (7.74)

We can also easily compute the maximum shear stress, which is nothing other than the radius of the circle:τmax =

∣∣∣σ1 − σ2

2

∣∣∣ =

√(σyy − σxx

2

)2

+ τ 2xy ≡ Maximum shear stress (7.75)

We can also find the angle of rotation between the current state and the principal rotation using trigonometry:tan(2θ) =

τxy

(σxx − σyy )/2(7.76)

Rearranging:θ =

1

2tan−1

( 2τxy

σxx − σyy

)≡ Rotation angle from principal direcation (7.77)

Let us revisit some of the previous examples to show how this works.Example 7.6: Uniaxial tension

Analyze the stresses in the following beam under uniaxial tensionff ff θ

Our stress state is σxx = f /A, σyy = 0, and τxy = 0. Because there is no shear stress, we are already in ourprincipal stress state and σ1 = f /A,σ2 = 0. Let us sketch how this looks on Mohr’s circle:




2θ

σ

τ

σ1 = f /A

σ2 = 0

We instantly obtain a lot of interesting information just from this plot. First, we see geometrically thatτmax =

f

2A(7.78)

and the angle of rotation corresponding to maximum shear is at 2θ = 90 or θ = 45. We can also computeσxx ,σyy , τxy using geometry:

σxx =f

2A+

f

2Acos 2θ σyy =

f

2A− f

2Acos 2θ τxy =

f

2Asin 2θ (7.79)

Example 7.7: Pressure vessel

Analyze the stresses in a pressure vessel.For a cylindrical pressure vessel the stresses in the natural coordinate system are

σxx =pR

2t= σ2 σyy =

pR

t= σ1 τxy = 0 (7.80)

where again we see that we are already in the principal frame as a result of the lack of shear forces. Let usvisualize this stress state using Mohr’s circle:

σ

τ

2θ

σ1 = pRtσ2 = pR

2t

We can easily compute the maximum shear stress as it is the radius of the circle:τmax =

pR/t − pR/2t

2=

pR/2t

2=

pR

4t(7.81)

Furthermore we know that the normal stresses in the remaining two directions are equal and have magni-




tudeσxx = σyy =

σ1 + σ2

2=

pR/t + pR/2t

2=

3pR/2t

2=

3pR

4t(7.82)

Example 7.8: Eigenvalues using Mohr’s circle

Consider a stress tensor iwth the following components:σ =

[30kPa 20kPa20kPa 10kPa

] (7.83)Analyze the stress state.This is definitely not the principal direction configuration as there are significant shear stresses present.Let us begin by sketching Mohr’s circle. From thestress tensor we gather that the following two pointsmust be located on the circle:

(30kPa, 20kPa) (10kPa,−20kPa) (7.84)Furthermore we know that the line between these twopoints is a diameter of the circle, and this allows us tosketch the circle more precisely.

σ

τ (30kN, 20kN)

(10kN,−20kN)

The average stress is given byσaverage =

10kPa + 30kPa

2= 20kPa (7.85)

and the radius of the circle is given byr =

√(30kPa− 10kPa

2

)2

+ (20kPa)2 =√

100kPa2 + 400kPa2 = 10√

5kPa (7.86)so the resulting principal values are

σ1,σ2 = 20kPa± 10√

5kPa = 42.36kPa,−2.36kPa X (7.87)We also know the maximum shear stress, which happens to be the radius of the circle:

τmax = 10√

5kPa = 22.36kPa (7.88)...and we can get the rotation required to get there from the given state. First, compute the angle from theprincipal rotation:

tan 2θ =20kPa

(30kPa− 10kPa)/2= 2.0 =⇒ θ = 31.71 X (7.89)

The maximum shear state is 45 from the principal stress state.



Lecture 22 Principal stresses concluded, energy methods

Example 7.9: Combined torsion and bending

Consider the following apparatus constructed of a A36 structural steel column and a lever arm subjectedto a force with magnitude f as shown in the following figure. Determine the stress state and the principalstresses/directions along the center of the top of the cylinder in terms of the material properties/geometryand the applied force.

LR

Wf

f

fW

This is a combined loading problem, first we must determine the stresses from beam bending and thestresses from torsion. We begin by reducing the offset loading to a center loading with moment M = fW .Let us consider the torsion first: we know that the moment throughout the center of the beam must beM(x) = f W (7.90)

Using the torsion formula,τzθ =

Mc

J=

(fW )(R)

(πR4/2)=

2fW

πR3(7.91)

Considering beam bending next: we have a loading functionw(x) = 0 (7.92)

and boundary conditionsh(0) = 0 θ(0) = 0 V (L) = f M(L) = 0 (7.93)

Integrating:V (x) = C1 =⇒ C1 = f =⇒ V (x) = f (7.94)

Integrating again:M(x) = fx + C2 (7.95)



Boundary condition:M(L) = fL + C2 = 0 =⇒ C2 = −fL =⇒ M(x) = f (x − L) (7.96)

We could go on to find angle and deflection butwe don’t need it. We nowuse the flexure formula to determinethe normal stress:σzz = −M(L/2)c

I= − (f (L/2− L))(R)

(πR4/4)=

2fL

πR3(7.97)

Sketching the stress state (left) and the corresponding Mohr’s circle (right):

2fWπR3

− 2fWπR3

2fLπR3

2fLπR3

2fLπR3

− 2fWπR3

The average normal stress is:σnormal =

1

2

(0 +

2fL

πR3

)=

fL

πR3(7.98)

and the radius of the circle isr = τmax =

√( fL

πR3

)2

+(2fW

πR3

)2

=f√L2 + 4W 2

πR3(7.99)

We also get principal stresses:σ1,2 =

fL

πR3± f√L2 + 4W 2

πR3=

f (L±√L2 + 4W 2)

πR3(7.100)

and the rotation between the original axes and the principal axes istan 2θ =

2fW /πR3

fL/πR3=

2W

L=⇒ θ =

1

2tan−1

(2W

L

) (7.101)

7.5 Principal Strains

Recall the strain tensor which, in 2D, isε =

[εxx εxyεxy εyy

] (7.102)where we recall that εxy = 1

2 (γxy + γyx ). Just as with the stress tensor, we can find principal strains – the frame inwhich the strain is purely longidinal, and principal directions – the directions along which the purely normalstrainsoccur. Furthermore, because ε is symmetric, we know that principal strains always exist and that the eigenvaluesare orthogonal.All content © 2017, Brandon Runnels 22.2



Theorem 7.1. For a linear elastic isotropic material, the principal directions for stress and strain are the same.

Proof. The proof of this requires very little work. Let us suppose that no shear stress is present – i.e. that ourframe is rotated to coincide with the principal directions for the stress tensor. From (2.81) we haveεyz =

1 + ν

Eτyz εzx =

1 + ν

Eτzx εxy =

1 + ν

Eτxy (7.103)

indicating that εyz = εzx = εxy = 0, indicating that shear strains are zero as well. Therefore we are also in therotation for the direcitons of the principal strains, meaning that both stress and strain tensors have the sameprincipal basis.What does this look like? Let us illustrate this with the case of pure shear. Consider the pure shear transformationbelow:

e1

e2

u1

u2

n1n2

u1

u2

original basis eigen/principal basis

In the original basis, the displacements corresponding to e1, e2 are not parallel to e1, e2. Therefore the box spannedby them dilates.On the other hand, looking at the box defined by the two eigenvectors n1,n2, we see that the displacements areparallel to the original vectors. As such the box deforms longitudinally, and right angles remain right.Because of Theorem 7.1, all of the same relationships hold that can be proven with Mohr’s circle. For principalstrains:

ε1,2 =εxx + εyy

2±√(εxx − εyy

2

)2

+ ε2xy ≡ Principal strains (7.104)

For the maximum in-plane shear strain1

εshear ,max =∣∣∣ε1 − ε2

2

∣∣∣ =

√(εxx − εyy

2

)2

+ ε2xy ≡ Maximum shear strain (for plane strain) (7.105)

and the angle isθ =

1

2tan−1

( 2εxy

εxx − εyy

)≡ Rotation angle from principal direction (7.106)

Let us determine how this works in practice.1This is only the maximum shear strain if working in plane strain




Example 7.10: Strain Rosette

Strain gauges are used to measure the stress state of a material undergoing loading, as shown in the fol-lowing figure.

xA

B

C

y

120

120

Each strain gauge measures a normal strain of εA, εB , εC . Determine the stress state in terms of the mea-sured strains, determine the principal strains, principal stresses, and the rotation of the principal frame fromthe original frame.We assume that we have a strain tensor:

ε =[εxx εxyεxy εyy

] (7.107)We recall that normal strain is given by the formula

εnormal = n · εn (7.108)so we just need the normal vectors corresponding to the directions of each of the three strain gauges. Doingsome trigonometry,

na =[10

]nb =

[cos 120

sin 120]

=1

2

[−1√

3

]nc =

[cos 120

− sin 120]

=1

2

[−1−√

3

](7.109)

Now, substitute into the formulae:na · εna = [1 0]

[εxx εxyεxy εyy

] [10

]= εxx = εa =⇒ εxx = εa (7.110)

nb · εnb =1

4

[−1

√3] [ εa εxyεxy εyy

] [−1√3

]=

1

4

[−1

√3] [−εa +

√3εxy

−εxy +√

3εyy

]=εa − 2

√3εxy + 3εyy

4= εb

(7.111)nc · εnb =

1

4

[1√

3] [ εa εxyεxy εyy

] [ 1√3

]=

1

4

[1√

3] [ εa +

√3εxy

εxy +√

3εyy

]=εa + 2

√3εxy + 3εyy

4= εc (7.112)

Adding the last two givesεa + 3εyy

2= εb + εc =⇒ εyy =

2(εb + εc )− εa

3(7.113)

and substituting givesεxy =

εc − εb√3

(7.114)




Sanity check: what happens if all of the strains are the same, i.e. εa = εb = εc = p? Thenεxx = p εyy =

2(p + p)− p

3= p εxy = 0 ε =

[p 00 p

] (7.115)which corresponds to hydrostatic pressure. Now let us compute principal strains:

ε1,2 =1

2

(εa +


3

)±

√(εa −


3

)2

+(εc − εb√

3

)2 (7.116)=εa + εb + εc

3±√(2(εb + εc )− 4εa

3

)2

+(εc − εb)2

3(7.117)(7.118)

and the maximum in-plane shear strain isεxy =

√(2(εb + εc )− 4εa

3

)2

+(εc − εb)2

3(7.119)

so the maximum principal stress isτxy =

1 + ν

E

√(2(εb + εc )− 4εa

3

)2

+(εc − εb)2

3(7.120)



Lecture 23 Introduction to energy methods

8 Energy Methods

In physics, andmechanics in particular, it is frequently the case that it is possible to re-derive fundamental equationsusing energy methods in such a way as to make the analysis easier. In dynamics we use Lagrangian mechanics, inthermodynamics we use the first and second laws, and in statics we use the principle of minimum potential energy(also called virtual work).Here we will extend the notion of the principal of minimum potential energy to beam bending problems.8.1 Strain energy

Strain energy density (that is, strain energy per unit volume) is defined for a 3d linear elastic material asW =

1

2ε : σ =

σxxεxx + σyyεyy + σzzεzz + τyzεyz + τzxεzx + τxyεxy

2≡ 3D strain energy density (8.1)

where the “:” is the Frobenius inner product for tensors. For a linear elastic material undergoing uniaxial stress,strain energy density reduces to:W =

1

2ε σ =

E

2ε2 =

1

2Eσ2 ≡ Strain energy density for uniaxial tension (8.2)

Alternatively, for a material undergoing pure shear with a loading τxy , the strain energy density isW =

1

2τxyεxy =

µ

2ε2

xy =1

2µτ 2

xy ≡ Strain energy density for pure shear (8.3)

The strain energy U is the total strain energy of the material, i.e.U =

∫Ω

W dV (8.4)For example, consider a beam with cross-sectional area A and length L that is elongated by length u. Then thestrain energy density is given by

W =E

2

(L + u − L

L

)2

=Eu2

2L2(8.5)

Then the total strain energy is given by integration:U =

∫Ω

Eu2

2L2dV = (AL)× Eu2

2L2=

EAu2

2L(8.6)

Sanity check: we have [force/length2]x [length]4/[length] = [force][length] = [energy ] as we expect.



8.1.1 Strain energy in Euler-Bernoulli beams

It is possible to compute the strain energy in an Euler-Bernoulli beam by assuming that we can treat the normalstress as significantly higher than the shear stress, thereby treating it as uniaxial tension.Let an Euler-Bernoulli beamwith Young’smodulus E and cross-sectionAwithmoment of inertia I have a deflectioncurve h(x). We recall from Euler-Bernoulli beam theory the moment-deflection and the flexure formulae:

M(x) = E Id2h

dx2σ = −M(x) z

I= −E z

d2h

dx2(8.7)

We can then use the formula for strain energy density to compute the total strain energy density of the beam as afunction of x , z :W =

1

2Eσ2 =

1

2E

(− E z

d2h

dx2

)2

=E z2

2

(d2h

dx2

)2 (8.8)Now, let us compute the total strain energy of the entire beam, assuming constant material properties:

U =

∫Ω

W dV =

∫ L

0

∫A

E z2

2

(d2h

dx2

)2

dA dx =E

2

∫ L

0

(d2h

dx2

)2

dx

∫A

z2dA︸︷︷︸=I

(8.9)

where we see that the second integral is just the moment of inertia. Then the strain energy isU =

EI

2

∫ L

0

(d2h

dx2

)2

dx =1

2EI

∫ L

0

M2(x)dx ≡ Strain energy in Euler-Bernoulli beam (8.10)Let us see how this works with an example:

Example 8.1: Strain energy of simply supported beam

Determine the strain energy of the following following beam with Young’s modulus E .L/2 L/2f

W

W

First we must do beam theory:w(x) = −f 〈x − L/2〉−1 (8.11)V (x) = −f 〈x − L/2〉0 + C1 (8.12)M(x) = −f 〈x − L/2〉1 + C1x +>

0C2 (8.13)

M(L) = −f (L/2) + C1L = 0 =⇒ C1 =f

2(8.14)

M(x) =f

2x − f 〈x − L/2〉1 (8.15)

Using the second form we compute the total strain energy byU =

1

2EI

∫ L

0

M2(x) dx =1

2EI

∫ L

0

[ f2x − f 〈x − L/2〉1

]2

dx (8.16)




Howdowecompute integrals that contain singularity functions? It’s actually quite easy: just split the integralup into two:U =

1

2EI

∫ L/2

0

[ f2x]2

dx +1

2EI

∫ L

L/2

[ f2x − f (x − L/2)

]2

dx

=f 2

8EI

∫ L/2

0

x2dx +f 2

8EI

∫ L

L/2

(x − L)2dx

=f 2

8EI

x3

3

∣∣∣L/2

0+

f 2

8EI

(x − L)3

3

∣∣∣LL/2

=f 2

8EI

(L/2)3

3− f 2

8EI

(L/2− L)3

3

=f 2

4EI

(L/2)3

3=

f 2L3

96EI(8.17)

Do the units work out? We have[force]2[length]3

([force]/[length]2)[length]4 =[force]2[length]3[force][length]2 = [force][length] X (8.18)

8.1.2 Strain energy in shafts under torsion

Recall the moment-twist angle and torsion formula for a shaft undergoing twisting:M = µ J

dφ

dxτθx =

Mr

J= µ r

dφ

dx(8.19)

The strain energy density for a shaft under pure torsion is, thenW =

1

2µ

M2r2

J2(8.20)

and the total strain energy isU =

∫Ω

W dV =

∫A

∫ L

0

1

2µ

M2(x)r2

J2dx dA =

1

2µJ2

∫ L

0

M2(x)dx

∫A

r2 dA︸︷︷︸=J

(8.21)

Therefore we have the following expression for shaft strain energyU =

µJ

2

∫ L

0

(dφdx

)2

dx =1

2µJ

∫ L

0

M2(x)dx ≡ Strain energy for shafts under torsion (8.22)

Example 8.2: Shaft under constant twisting moment

Consider a shaft with length L and radius R subjected to a twistingmomentM0. Determine the strain energyfor the shaft.We know that the moment is constant. Substituting, we find that the strain energy for the shaft is

U =M2

0L

2µJ(8.23)




8.2 Applied work

Let us now look at another type of energy, that done by applied loadings. You may recall that the definition ofmechanical work is “force times displacement” ordV = f dx (8.24)

And in three dimensions it isdV = f · dx (8.25)

Howdoesmechanical work apply to our problems inmechanics ofmaterials? In the cases of Euler-Bernoulli beams,we have applied loading functions w(x) that induce a deflection on the beam, h(x). On the other hand, for shaftsin torsion, we have applied moment functions m(x) that induce a twist angle on the shaft, φ(x).Let us look at both of these cases in greater detail.8.2.1 Work done by a loading on a beam

Consider an Euler-Bernoulli beam subjected to a loading w(x) such that it exhibits a deflection h(x) as shown inthe following figure:

h(x)

w(x)

The work on the beam is given simply by the integralV =

∫ L

0

w(x) h(x) dx ≡ Applied work from a distributed load (8.26)

Let us show how this works by doing an example:Example 8.3: Work done on cantilever beam

Consider a cantilever beam with Young’s modulus E and moment of inertia I about the neutral axis.w(x) = −w0

Determine the work done by the external load on the beam.First we must do the beam theory.

w(x) = −w0

V (x) = −w0x + C1




V (L) = −w0L + C1 = 0 =⇒ C1 = w0L

V (x) = w0L− w0x

M(x) = w0Lx −1

2w0x

2 + C2

M(L) = w0L2 − 1

2w0L

2 + C2 = 0 =⇒ C2 =1

2w0L

2

M(x) = w0Lx −1

2w0x

2 − 1

2w0L

2

θ(x) =1

EI

[1

2w0Lx

2 − 1

6w0x

3 − 1

2w0L

2x +>0

C3

]h(x) =

1

EI

[1

6w0Lx

3 − 1

24w0x

4 − 1

4w0L

2x2 +>0

C4

] (8.27)Now that we have our deflection curve, we can compute the total work done on the beam by the load usingthe integral:

V =

∫ L

0

h(x)w(x) dx = −w0

EI

∫ L

0

[1

6w0Lx

3 − 1

24w0x

4 − 1

4w0L

2x2]dx

= −w0

EI

[ 1

24w0Lx

4 − 1

120w0x

5 − 1

12w0L

2x3]L

0

=w2

0L5

20EI(8.28)

The above process works very well for distributed loads, but what about point loads? As an example, consider thefollowing case:

h(x)

fafb

ab

We recall that we can express the loading function using Dirac delta functions:w(x) = fa〈x − a〉−1 − fb〈x − b〉−1 (8.29)

Now let us substituting in:V =

∫ L

0

(fa〈x − a〉−1 − fb〈x − b〉−1

)h(x) dx (8.30)

For discontinuity functions, we showed earlier that integrals containing singularity functions can be computedsimply by breaking the integral up into multiple parts. But that will not work here, because w(x) = 0 except at theexact points when x = a and x = b.It turns out that there is actually a very easy solution to this problem, and it comes from the functional definition ofthe Dirac delta function.Theorem 8.1 (Sifting property of the Dirac delta). Let 〈x−a〉−1 be the unit Dirac delta function centered at 0 < a < L.Then for all continuous functions h(x) defined on [0, L], the following holds:∫ L

0

h(x) 〈x − a〉−1 dx = h(a) (8.31)




In other words, the Delta function pulls out the value of the function that it is multiplied at the location where it iscentered. This is called the “sifting property” and has many interesting applications. Let us apply it to the loadingfunction above:V =

∫ L

0

(fa〈x − a〉−1 − fb〈x − b〉−1

)h(x) dx = fa h(a)− fb h(b) (8.32)

This is what we expect! The result is that the work done by point loads is exactly equal to the magnitude of thepoint loads times the displacement at that point. In general formula form,V = f h(a) ≡ Applied work from force f at x = a (8.33)



Lecture 24 Principal of minimum potential energy

Let us see how this works by means of an example, continuing an example that was started earlier.Example 8.4: Work done by point force on simply supported beam

Determine the work done by the load f on the following following beam with Young’s modulus E .L/2 L/2f

W

W

We recall that the moment was given byM(x) =

1

2f x − f 〈x − L/2〉1 (8.34)

Continuing on to find the angle and displacement:θ(x) =

1

EI

[1

4f x2 − 1

2f 〈x − L/2〉2 + C3

] (8.35)h(x) =

1

EI

[ 1

12f x3 − 1

6f 〈x − L/2〉3 + C3x +>

0C4

] (8.36)h(L) =

1

EI

[ 1

12f L3 − 1

6f (L/2)3 + C3L

]= 0 (8.37)

=⇒ C3 = − 1

16f L2 (8.38)

h(x) =f

2EI

[1

6x3 − 1

3〈x − L/2〉3 − 1

8L2x] (8.39)

The energy of the applied load is given byV =

∫ L

0

(− f 〈x − L/2〉−1

)h(x) dx = −f h(L/2) (8.40)

Substituting the expression for the deflection at that point:h(L/2) = − f L3

48EI(8.41)

Gives the resultV =

f 2 L3

48EI(8.42)

8.2.2 Work done by couple moments

Just like with applied loads, couple moments contribute applied work that must be considered when computingenergy. Let us consider how this works for an Euler-Bernoulli beam: consider the following example of a pointAll content © 2017, Brandon Runnels 24.1


moment M applied at x = a to a beam as shown. What is the applied work?θ(a)

a

M

The loading function corresponding to an applied moment is expressed using a unit doublet:w(x) = −M 〈x − a〉−2 (8.43)

(Why is it negative? Recall that a doublet is like the derivative of a Dirac delta, which would mean you can think ofit being infinitely positive just before of 0, and infinitely negative just after 0. If it were a couple moment, it wouldbe negative.) Now, let us substitute this into the expression for applied work:V = −

∫ L

0

M 〈x − a〉−2 h(x) dx (8.44)Let us simplify this by using integration by parts: noting that w(x) = dV /dx ,∫ L

0

w(x) h(x) dx = V (x) h(x)∣∣∣L0−∫ L

0

V (x) θ(x) dx (8.45)Now we must evaluate V (x)h(x) at the boundaries. This is where an interesting fact about boundary conditionscomes into play: remember that you can never specify both V (x) and h(x), and that exactly one of them must bezero? This means that at both ends, either V or h must be zero, and we can therefore write off the boundary term,leaving us with ∫ L

0

w(x) h(x) dx = −∫ L

0

V (x) θ(x) dx (8.46)Given our applied moment, our shear function is

V (x) =

∫w(x) dx = −M〈x − a〉−1 (8.47)

(Note that we can neglect the constant using a similar argument to the above.) Now we see that−∫ L

0

V (x) θ(x) dx =

∫ L

0

M〈x − a〉−1θ(x) dx = M θ(a) (8.48)In other words the work done by a couple moment is equal to the magnitude of the moment times the deflectionangle at that location. Put generally

V = M θ(a) ≡ Applied work from couple moment M at x = a (8.49)Similarly for a shaft with length L, the work done by a loading function m(x) is given by

V =

∫ L

0

m(x)φ(x) dx ≡ Applied work from distributed shaft moment (8.50)(Side note: recall that moments have units of force times distance, which is the same as energy. We noted that itmakes more sense to think of them are force times distance per radian – now when we multiply by an angle weget energy.)All content © 2017, Brandon Runnels 24.2



8.3 Principle of minimum potential energy

Why do we care about computing energy? Both strain energy and energy from applied work are key to the principalof minimum potential energy. Let us begin by defining “potential energy”Π = U − V ≡ Potential energy (8.51)

where U is the strain energy and V is the applied work. Put generally, the principal of minimum potential energystates that Π is always minimized. In the context of Euler-Bernoulli beam theory, it can be stated ash(x) = arg min

h(x)

Π [h(x)] (8.52)In other words, the deflection curve of a beam h(x) is the one for which Π is minimized. How do you minimizesomething with respect to a function? There is a whole branch of mathematics called the calculus of variationsthat is devoted exclusively to solving problems like these. However, to delve into this subject is well beyond thescope of this course. Instead, we will simplify the principle of minimum potential energy for a few select cases.8.3.1 Castigliano’s first theorem

Consider an Euler-Bernoulli beam subjected to a series of nonzero displacements such that h(x1) = h1, h(x2) =h2, .... Then we know that the corresponding applied load must be

w(x) = f1〈x − x1〉−1 + f2〈x − x2〉−1 + ... (8.53)and so on. Let us suppose that we are able to compute the strain energy as a function of these prescribed dis-placements, so we have

U = U(h1, h2, ...) (8.54)Then our potential energy function is

Π = U(h1, h2, ...)− f1h1 − f2h2 ... (8.55)Recall that theMPEstates that the deflection h(x) is the one thatminimizesΠ. Sincewehave reduced our functionalso that it only depends on certain select values of h(x), we can say that h1, h2, ... minimize Π as well. In other words,

h1, h2 ... = arg minh1,h2,...

Π(h1, h2, ...) (8.56)We recall from calculus that if h1, h2, ... minimize Π, then the partial derivatives of Π with respect to h1, h2, ... mustbe zero. Using that fact:

∂Π

∂h1=∂U

∂h1− f1 = 0

∂Π

∂h2=∂U

∂h2− f2 = 0 ... (8.57)

Solving, we havefi =

∂U(h1, ... , hn)

∂hi≡ Castigliano’s First Theorem (8.58)

where U is the strain energy density, hi is the displacement at a point and fi is the force corresponding to thatdisplacement.




Example 8.5: Cantilever beam

Consider the following cantilever beam subjected to an unknown load fL resulting in a known vertical de-flection hL at the end.

hL

fL

L

Determine the force fL required to cause the deflection hL.As usual we begin with beam theory. The loading function is

w(x) = 0 (8.59)and the beam has boundary conditions

h(0) = 0 θ(0) = 0 h(L) = hL M(L) = 0 (8.60)Integrating:

V (x) = C1 (8.61)M(x) = C1x + C2 (8.62)

M(L) = C1L + C2 = 0 =⇒ C2 = −C1L (8.63)M(x) = C1x − C1L (8.64)θ(x) =

1

EI

[1

2C1x

2 − C1Lx +>0

C2

] (8.65)h(x) =

1

EI

[1

6C1x

3 − 1

2C1Lx

2 +>0

C3

] (8.66)h(L) =

C1

EI

[1

6L3 − 1

2L3]

= −C1L3

3EI= hL =⇒ C1 = −3EIhL

L3(8.67)

M(x) =3EIhL(L− x)

L3(8.68)

Use the moment function to compute the strain energy:U =

1

2EI

∫ L

0

M2(x) dx =1

2EI

∫ L

0

(3EIhL(L− x)

L3

)2

dx =9EIh2

L

2L6

∫ L

0

(x − L)2 dx (8.69)=

9EIh2L

2L6

(x − L)3

3

∣∣∣L0

=9EIh2

L

2L6

L3

3(8.70)

=3EIh2

L

2L3(8.71)

Now, all we have to do is apply Castigliano’s theorem:fL =

∂U

∂hL=

3EIhL

L3(8.72)

Let’s check our work by evaluating the shear at the end:V (L) = C1 = −3EIhL

L3(8.73)

which is an exact match, recalling that by sign convention, an upwards force on the left end is negative.




We have proved that Castigliano’s first theorem works in the above example. But...it’s a bit cumbersome, becausefinding the energy as a function of displacements means we always have to integrate all the way down to findingh(x), which is inconvenient if we are working with a statically determinate beam.This is where Castigliano’s second theorem proves quite useful.8.3.2 Castigliano’s second theorem

Consider a beam with a single point load f0 corresponding to displacement h0, where the potential energy is Π(h0).Suppose we wish to express U in terms of f0 instead of h0. We can do this by taking the Legendre transform:U∗(f0) = max

h0

[f0h0 − U(h0)

] (8.74)where U∗(f0) is called the complementary strain energy, also known as the dual strain energy. For now you can justthink of it as the strain energy expressed in terms of f0. Note that if we solve the above problem we have

d

dh0

[f0h0 − U(h0)

]= f0 −

dU

dh0= 0 =⇒ f0 =

dU

dh0(8.75)

which gives us Castigliano’s first theorem. As you may know from thermodynamics, the Legendre transform isinvertible – that is, the Legendre transform of the Legendre transform of a function is the original function. Thismeans that we can find the original strain energy in terms of the dual strain energyU(h0) = max

f0

[f0h0 − U∗(f0)

] (8.76)where solving the problem for the optimal force yields

u0 =dU∗

df0(8.77)

Generalizing to multiple point loads with multiple displacements, we haveui =

∂U∗(f1, ... , fn)

∂fi≡ Castigliano’s second theorem (8.78)

Why is this more useful than the first theorem? It’s useful because it allows us to learn about the displacement ata point without having solved for it. Equation 8.78 is for the general case, but we can reduce it further for Euler-Bernoulli beams (with constant properties) by taking the derivative inside the integral and using the product rule:ui =

∂U∗(f1, ... , fn)

∂fi=

∂

∂ui

[ 1

2EI

∫ L

0

M2(x) dx]

=[ 1

2EI

∫ L

0

∂

∂ui(M2(x)) dx

]=

1

2EI

∫ L

0

2M∂M

∂uidx (8.79)

The result is a reduced form of Castigliano’s second theorem for Euler-Bernoulli beams that have constant E , Ithoughout the length:ui =

1

EI

∫ L

0

M∂M

∂fidx ≡ Castigliano’s second theorem for a uniform Euler-Bernoulli beam (8.80)

This form is often easier to use because pre-differentiatingM with respect to f can signficantly reduce the algebra,making the integral much easier to do. Let us show how this works by doing an example.




Example 8.6: Simply supported beam

Consider Example 8.1, the case of a simply supported beam with a point load in the center. Without solvingfor h(x), determine the deflection at the point of application of the loading.It was shown that the strain energy was

U∗(f ) =f 2L3

96EI(8.81)

To find the displacement at this point, all we have to do is apply Castigliano’s second theorem:u =

dU∗

df=

d

df

( f 2L3

96EI

)=

fL3

48EI(8.82)

As you can see, the strain energy function encodes a great deal of information about the relationship between pointloads and deflections. Let us consider the cantilever example again:Example 8.7: Cantilever beam continued

Consider the cantilever beam introduced in Example 8.6, but this time suppose that the force fL is knownand the displacement uL is not.This information means that we drop the BC h(L) = hL, and instead introduce V (L) = −fL. Re-doing thebeam theory:

w(x) = 0 (8.83)V (x) = C1 (8.84)

V (L) = −fL =⇒ V (x) = −fL (8.85)M(x) = −fLx + C2 (8.86)

M(L) = −fLL + C2 = 0 =⇒ C2 = fLL =⇒ M(x) = fL(L− x) (8.87)Now we can apply Equation 8.80:

hL =1

EI

∫ L

0

M∂M

∂fLdx =

1

EI

∫ L

0

fL(x − L)2 dx =fLL

3

3EI(8.88)

The advantage of energy methods from these examples may or may not be obvious to you at this point. How-ever, hopefully you will see in future classes that energy methods provide an extremely convenient mechanism forrelating the analyses we have done here to other much more complicated systems.

Conclusion

In this course we have covered the fundamentals of basic mechanics of materials. Hopefully you now have abetter understanding of stress and strain in 1-3 dimensions, analysis of Euler Bernoulli beams, analysis of shafts,pressure vessels, and columns under buckling. But even more importantly, you hopefully have an understanding ofhow complex solid mechanics behavior is, and how important it is to understand the assumptions that are madewhen doing hand calculations.What we have covered here is only the tip of the iceberg. You will apply much of what we have learned here in MAE




3501 - Machine Design I. But if you are interested in learning more, you may want to consider taking some of thefollowing courses:• MAE5201 - Solid mechanics: The sequel to this class. Learn about the governing equations of solid mechan-ics and how to apply them to a large number of problems like fracture and wave propagation.• MAE5160 - Finite element analysis: Learn how FEA software like Abaqus works, and how to write your ownFEA code.• MAE5100 - Continuum mechanics: Derive and apply the fundamental governing equations for materials un-der large deformation.• MAE9510 - Constitutive modeling: Learn how to model the behavior of very complex materials like humanbone or tissue.




A Table of parameters for common materials

The following is an incomplete compilation of some material properties that are relevant to the problems for thisclass and may be used for homework. Note: these values should be used for academic purposes only, they arenot to be used for design.

σ0 σuts ρ E ν µ αL

Material (MPa) (MPa) (kg/m3) (GPa) (GPa) (10−6K−1)A36 Structural Steel 250 400-550 7800 200 0.30 79.3 11.76061 Al Alloy 241 300 2700 68.9 0.32 26.0 23.614 yr old male bone 104-121 130 1600 14 0.14Carbon Nanotubes 11000-630,000 37-1340 1000Cork 1.38 710 0.0Copper (Annealed) 33.3 210 8930 110 0.343 16.4Titanium carbide 258 4940 448 0.18 7.70Lightweight concrete 30,000compressive only 950 30

All content © 2017, Brandon Runnels A.1



B Selected stress concentration factor calculations

Note: these values should be used for academic purposes only, they are not to be used for design.B.1 Uniaxial Loading

[In progress]B.2 Beam Bending

Beam with two symmetric notches

0.0 0.1 0.2 0.3 0.4 0.5hd

0.51.01.52.02.53.03.54.0

Kt

hr = 0.25hr = 0.5hr = 1hr = 1.5hr = 2

2r

h

d

Beam with fillets

0.0 0.1 0.2 0.3 0.4 0.5hd

0.81.01.21.41.61.82.02.22.4

Kt

hr = 0.25hr = 0.5hr = 1hr = 1.5hr = 2

h

d

r

All content © 2017, Brandon Runnels B.1



B.3 Shaft Torsion

Shaft with fillets

0.0 0.1 0.2 0.3 0.4 0.5hd

0.81.01.21.41.61.82.02.2

Kt

hr = 0.25hr = 0.5hr = 1hr = 2hr = 3hr = 4

Shaft with fillets

h

d

r

M

σmax = Kt16M

π(d − 2h)3(B.1)




References

[1] Raph Levien. The elastica: a mathematical history. Technical Report UCB/EECS-2008-103, EECS Department,University of California, Berkeley, Aug 2008.[2] Brandon Runnels. MAE 2103 - Engineering Mechanics I Course Notes, 2016.




Index

applied work, 23.4of a point load on an Euler-Bernoulli beam, 23.6of loading on Euler-Bernoulli beam, 23.4of moments on Euler-Bernoulli beam, 24.1of twisting moments on shafts, 24.2auxetic materials, 6.4beam theory, see Euler-Bernoulli beam theoryboundary conditionsfor Euler-Bernoulli beam theory, 7.4for shaft theory, 12.5buckling theory, 15.2critical buckling load formula, 15.2effective length, 16.3bulk modulus, κ, 7.1calculus of variations, 24.3Castigliano’s first theorem, 24.3, 24.5centroid, 2.4definition, 2.4of a composite body, 2.4circumferential stress (pressure vessel), 18.1coefficient of thermal expansion, 4.1de Haviland Comet, 5.1Dirac delta functions, 8.2dislocation-mediated plasticity, 3.7dissipative mechanism, 4.1distributed loading, 1.3dot product, 1.2doublet, 8.5, 24.2eigenvaluesof the buckling equation, 15.3of the strain tensor, see principal strainsof the stress tensor, see principal stressesenergy methods, 23.1applied work, see applied workCastigliano’s first theorem, 24.3, 24.5principle of minimum potential energy, seeprinciple of minimum potential energystrain energy, see strain energyequilibrium3D elasticity, 6.1of a rigid body, 1.3Euler-Bernoulli beam theory, 7.1assumptions and simplifications, 7.4combined with shaft theory, 14.1derivation of governing equation, 7.1for composite beams, 10.1

flexure formula, 10.2governing equation, 10.2neutral axis, 10.2governing equation, 7.4shear stresses, 11.1shear formula for horizontal cuts, 11.1shear formula for vertical cuts, 11.3solution strategy, 9.1factor of safety, 5.2flexure formula, 7.3, 11.6Heaviside functions, 8.4Hooke’s law, 3.6for 3D, 6.3for pure shear stress, 5.4for temperature change, 4.1for uniaxial normal stress/strain, 3.6hoop stress, see circumferential stressinsufficient constraints, 1.3Lamé parameter, λ, 7.1linearized strain assumption, see small strainassumptionlog strain, see true strainlongitudinal stress (pressure vessel), 18.1modulussee norm, 1.2Mohr’s circle, 21.1maximum shear stress, 21.2principal stresses, 21.2moment of inertia, 2.5about the x axis, Ixx , 12.2necking, 3.2neutral axis, 7.2nominal stress, 5.2norm, 1.2normal stress equation, 5.5, 18.4p-wave modulus, M , 7.1parallel axis theorem, 2.5plane strain, 6.4plane stress, 6.4plasticity, 3.7Poisson’s ratio, ν , 6.3polar moment of inerta, 12.2potential energy, 24.3pressure vessels, 17.1cylindrical, 18.1




spherical, 17.1principal strains, 22.2principal stresses, 20.1finding using eigenvalues, 20.3finding using Mohr’s circle, 21.2principle of minimum potential energy, 24.3Ramp functions, 8.4safety factor, see factor of safetySaint-Venant’s principle, 3.3second moment of area, see moment of inertiashaft theory, 12.1boundary conditions, 12.5combined with Euler-Bernoulli beam theory, 14.1composite shafts, 13.1torsion formula, 13.1relationship between twist angle and moment,12.3stress concentrations, 13.4torsion formula, 12.2twist angle formula, 12.3shear formula, 11.6shear modulus, µ, 5.4shear stress equation, 5.5, 18.4shear/moment diagramsmethod of integration, 2.2method of sections, 2.2sign convention, 2.2sign convention, 3.3singularity brackets, 8.5singularity functions, 8.2small strain assumption, 3.6, 4.4static indeterminacy, 1.3, 11.6strain

engineering vs true strain, 3.5normal uniaxial strain, definition, 3.4shear strain, definition, 5.2strain energy, 4.1, 23.1in Euler-Bernoulli beam, 23.2in shafts, 23.3in uniaxial loading, 23.1strain hardening, see work hardeningstrain tensor, ε, 6.2strength vs stress, 3.6stressengineering vs true, 3.2normal uniaxial stress, definition, 3.1shear stress, definition, 5.2sign convention, 3.2stress concentration factor, K , 5.1, 11.6stress tensor, σ, 5.4stress-strain plots, 3.5superposition, 14.1tensor transformation, 19.1formula using rotation matrices, 19.1thermal expansion coefficient, 4.1thin-wall assumption, 17.1Timoshenko beam theory, 7.4two-force members, 4.3ultimate tensile stress, 3.6vector notation, 1.2von Mises stress, 15.1work hardening, 3.7yield stress, σ0, 3.6



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MAE 3201 - Mechanics of Materials Course Notes · MAE 3201 - Mechanics of Materials Course Notes Brandon Runnels About These notes are for the personal use of students who are enrolled