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General State of Stress
Sign Convention
Plane Stress
Plane Stress Transformation
Principal Stresses
- Alternative Approach: Eigenvalues
Maximum Shear Stress

Stress Transformation

General State of Stress

The general state of stress at a point is characterized by three independent normal stress components and three independent shear stress components, and is represented by the stress tensor. The combination of the state of stress for every point in the domain is called the stress field.

Stress tensor.

T = [\begin{matrix} σ_{x} & τ_{x y} & τ_{x z} \\ τ_{x y} & σ_{y} & τ_{y z} \\ τ_{x z} & τ_{y z} & σ_{z} \end{matrix}]

Warning: Stress is a physical quantity and as such, it is independent of the chosen coordinate system. #str-ind

Sign Convention

Sign conventions on 2D elements

Positive normal stress acts outward from all faces.
Positive shear stress points towards the positive axis direction in a positive face.
Positive shear stress points towards the negative axis direction in a negative face.

Plane Stress

Plane stress occurs when two faces of the cube element are stress free.

Plane stress.

σ_{z} = τ_{z x} = τ_{z y} = 0

Example: Thin plates subject to forces acting in the mid-plane of the plate. #thn-plt

Plane Stress Transformation

We think of stresses acting on faces, so we often associate the state of stress with a coordinate system. However, the selection of a coordinate system is arbitrary (materials don't know about coordinates - it's a mathematical construct!) and we could choose to express the stress state acting on any set of faces aligned with any coordinate system axes. Furthermore, we can relate the states of stress in each coordinate system to one another through stress transformation equations.

Heads up!

Cauchys stress theorem builds on this content in later solid mechanics courses.

For any surface that divides the body ( imaginary or real surface), the action of one part of the body on the other is equivalent to the system of distributed internal forces and moments and it is represented by the stress vector $t^{n}$ (also called traction), defined on the surface with normal unit vector $n$ .

The state of stress at a point in the body is defined by all the stress vectors $t^{n}$ associated with all planes (infinite in number) that pass through that point.

Cauchys stress theorem.

t^{n} = T n

Cauchys stress theorem states that there exists a stress tensor

T

(which is independent of

n

), such that

t^{n}

is a linear function of

n

Transformation Sign Convention

Both the $x - y$ (original) and $x^{'} - y^{'}$ (transformed) systems follow the right-hand rule.
The orientation of an inclined plane (on which the normal and shear stress components are to be determined) will be defined using the angle $θ$ . The angle $θ$ is measured from the positive $x$ to the positive $x^{'}$ -axis. It is positive if it follows the curl of the right-hand fingers.

Stress transformation equations. #str-trn

\begin{aligned} σ_{x}^{'} & = \frac{σ_{x} + σ_{y}}{2} + \frac{σ_{x} - σ_{y}}{2} \cos (2 θ) + τ_{x y} \sin (2 θ) \\ σ_{y}^{'} & = \frac{σ_{x} + σ_{y}}{2} - \frac{σ_{x} - σ_{y}}{2} \cos (2 θ) - τ_{x y} \sin (2 θ) \\ τ_{x^{'} y^{'}} & = - \frac{σ_{x} - σ_{y}}{2} \sin (2 θ) + τ_{x y} \cos (2 θ) \end{aligned}

Use the equations from stress in inclined planes.

\begin{aligned} σ_{x}^{'} & = σ_{x} \cos^{2} (θ) + 2 τ_{x y} \sin (θ) \cos (θ) + σ_{y} \sin^{2} (θ) \\ τ_{x^{'} y^{'}} & = (σ_{y} - σ_{x}) \sin (θ) \cos (θ) + τ_{x y} (\cos^{2} (θ) - \sin^{2} (θ)) \end{aligned}

Trigonometric relations.

\begin{matrix} \cos^{2} θ = \frac{1 + \cos (2 θ)}{2} & \sin (2 θ) = 2 \sin θ \cos θ \\ \sin^{2} θ = \frac{1 - \cos (2 θ)}{2} & \cos (2 θ) = \cos^{2} θ - \sin^{2} θ \end{matrix}

Plugging them in.

\begin{aligned} σ_{x}^{'} & = σ_{x} \frac{1 + \cos (2 θ)}{2} + τ_{x y} \sin (2 θ) + σ_{y} \frac{1 - \cos (2 θ)}{2} \\ τ_{x^{'} y^{'}} & = (σ_{y} - σ_{x}) \frac{\sin (2 θ)}{2} + τ_{x y} (\frac{1 + \cos (2 θ)}{2} - \frac{1 - \cos (2 θ)}{2}) \end{aligned}

Rearranging terms.

\begin{aligned} σ_{x}^{'} & = \frac{σ_{x}}{2} + \frac{σ_{x} \cos (2 θ)}{2} + \frac{σ_{y}}{2} - \frac{σ_{y} \cos (2 θ)}{2} + τ_{x y} \sin (2 θ) \\ τ_{x^{'} y^{'}} & = - \frac{(σ_{x} - σ_{y})}{2} \sin (2 θ) + (\frac{τ_{x y}}{2} + \frac{τ_{x y} \cos (2 θ)}{2} - \frac{τ_{x y}}{2} + \frac{τ_{x y} \cos (2 θ)}{2}) \end{aligned}

We can then simplify to the equations above. Note that to derive

σ_{y}^{'}

, we use the following.

σ_{x}^{'} + σ_{y}^{'} = σ_{x} + σ_{y}

Normal and shear stresses can be shown graphically as a function of

θ

to see how they overlap.

Stresses based on the transformation angle chosen.

2-D Mohr's Circle

Mohr's circle is a graphical representation of stress transformations. The equations for stress transformations actually describe a circle if we consider the normal stress

σ

to be the x-coordinate and the shear stress

π

to be the y-coordinate.

Circle centroid.

C = σ_{a v g} = \frac{σ_{x} + σ_{y}}{2} = \frac{σ_{1} + σ_{2}}{2}

Cirlce radius.

R = \sqrt{(\frac{σ_{x} - σ_{y}}{2})^{2} + τ_{x y}^{2}}

Two points of the circle.

\begin{aligned} P o i n t X & : (σ_{x}, τ_{x y}) \\ P o i n t Y & : (σ_{y}, - τ_{x y}) \end{aligned}

The principal stresses are where the circle crosses the x-axis, and the maximum shear stress is the highest y-coordinate of the circle.

Interactive Mohr's circle

Example Problem: Interactive Mohr's circle #int-mrh

$σ_{x}$ : -80

$σ_{y}$ : 50

$τ_{x y}$ : -25

$σ_{1}$ : 54.64

$σ_{2}$ : -84.64

$τ_{m a x}$ : 69.64

$σ_{x^{'}}$ : -80

$σ_{y^{'}}$ : 50

$τ_{x^{'} y^{'}}$ : -25

Angle:

θ =

^{\circ}

3-D Mohr's Circle

Circle centroid.

C = σ_{a v g} = \frac{σ_{1} + σ_{3}}{2}

Cirlce radius.

R = \frac{σ_{1} - σ_{3}}{2}

Points of the circle are plotted using each face.

(σ_{p l a n e}, τ_{p l a n e})

The principal stresses are where the circle crosses the x-axis, and the maximum shear stress is the highest y-coordinate of the circle.

Stresses on Inclined Planes

Stresses on inclined planes. #str-inc

\begin{aligned} σ_{x^{'}} & = n \cdot t^{n} = n \cdot T n = σ_{x} \cos^{2} (θ) + 2 τ_{x y} \sin (θ) \cos (θ) + σ_{y} \sin^{2} (θ) \\ τ_{x^{'} y^{'}} & = s \cdot t^{n} = s \cdot T n = (σ_{y} - σ_{x}) \sin (θ) \cos (θ) + τ_{x y} (\cos^{2} (θ) - \sin^{2} (θ)) \end{aligned}

Equilibrium equations.

\sum F_{x} : - σ_{x} (A \cos θ) - τ_{x y} (A \sin θ) + (σ_{x}^{'} \cos θ) A - (τ_{x^{'} y^{'}} \sin θ) A = 0

\sum F_{y} : - σ_{y} (A \sin θ) - τ_{x y} (A \cos θ) + (σ_{x}^{'} \sin θ) A - (τ_{x^{'} y^{'}} \cos θ) A = 0

Rearrange terms.

A [σ_{x}^{'} \cos θ - τ_{x^{'} y^{'}} \sin θ] = A [σ_{x} \cos θ + τ_{x y} \sin θ]

A [σ_{x}^{'} \sin θ - τ_{x^{'} y^{'}} \cos θ] = A [σ_{x} \sin θ + τ_{x y} \cos θ]

Combine into a matrix.

[\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}] [\begin{matrix} σ_{x}^{'} \\ τ_{x^{'} y^{'}} \end{matrix}] = [\begin{matrix} σ_{x} \cos θ & τ_{x y} \sin θ \\ σ_{y} \sin θ & τ_{x y} \cos θ \end{matrix}]

Multiply by the inverse.

[\begin{matrix} σ_{x}^{'} \\ τ_{x^{'} y^{'}} \end{matrix}] = [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}] [\begin{matrix} σ_{x} \cos θ & τ_{x y} \sin θ \\ σ_{y} \sin θ & τ_{x y} \cos θ \end{matrix}]

Pure Shear

A circular shaft under torsion develops pure shear on cross-sections between longitudinal planes (the faces of element

a

are parallel and perpendicular to the axis of the shaft).

Torsion on inclined planes. #tor-inc

\begin{array}{r} σ_{x^{'}} = 2 τ_{m a x} \sin θ \cos θ = τ_{m a x} \sin (2 θ) \\ τ_{x^{'} y^{'}} = τ_{m a x} (\cos^{2} θ - \sin^{2} θ) = τ_{m a x} \cos (2 θ) \end{array}

Recall projected forces.

\begin{array}{r} σ_{x^{'}} = \frac{P}{A_{o}} (\cos^{2} (θ)) \\ τ_{x^{'} y^{'}} = - \frac{P}{A_{o}} \sin (θ) \cos (θ) \end{array}

Maximum normal stress at 90 degrees.

\begin{array}{r} σ_{x^{'}} = τ_{m a x} = \frac{T c}{J} \\ σ_{y^{'}} = - τ_{m a x} = - \frac{T c}{J} \\ τ_{x^{'} y^{'}} = 0 \end{array}

Maximum shear stress at 45 degrees.

\begin{array}{r} σ_{x^{'}} = 0 \\ σ_{y^{'}} = 0 \\ τ_{x^{'} y^{'}} = τ_{m a x} = \frac{T c}{J} \end{array}

Principal Stresses

σ_{x}^{'}

can be maximized to find the principal stresses.

Angle for maximum normal stress. #prn-str

\begin{array}{r} \tan (2 θ_{p 1}) = \frac{2 τ_{x y}}{σ_{x} - σ_{y}} \\ θ_{p 2} = θ_{p 1} + 90^{o} \end{array}

Recall.

σ_{x}^{'} = \frac{σ_{x} + σ_{y}}{2} + \frac{σ_{x} - σ_{y}}{2} \cos (2 θ) + τ_{x y} \sin (2 θ)

To maximize an equation, we take the derivative and set it equal to zero.

\frac{d σ_{x}^{'}}{d θ} = - 2 \sin (2 θ) [\frac{σ_{x} - σ_{y}}{2}] + 2 \cos (2 θ) τ_{x y} = 0

(σ_{x} - σ_{y}) \sin (2 θ) = 2 τ_{x y} \cos (2 θ)

\tan (2 θ) = \frac{2 τ_{x y}}{σ_{x} - σ_{y}}

The maximum/minimum normal stress values (the principal stresses) are associated with

θ_{p 1}

and

θ_{p 2}

Principal stresses

σ_{1, 2} = \frac{σ_{x} + σ_{y}}{2} \pm \sqrt{(\frac{σ_{x} - σ_{y}}{2})^{2} + τ_{x y}^{2}}

We use the convention that

σ_{1} > σ_{2}

Alternative Approach: Eigenvalues

The eigenvalues of the stress tensor are called the principal stresses, and the eigenvectors define the principal direction vectors.

Because of symmetry, the stress tensor ( $T$ ) has real eigenvalues ( $λ$ ) and mutually perpendicular eigenvectors ( $v$ ).

Eigenvalues.

T v = λ v \to (T - λ I) v = 0

From linear algebra, we know that a system of linear equations

A v = 0

has a non-zero solution

v

if, and only if, the determinant of the matrix

T

is zero.

Eigenvalues relate to stress. #egn-str

λ_{1, 2} = σ_{1, 2}

Start with.

det (T - λ I) = 0

Expand the equation.

det ([\begin{matrix} σ_{x} & τ_{x y} \\ τ_{x y} & σ_{y} \end{matrix}] - [\begin{matrix} λ & 0 \\ 0 & λ \end{matrix}]) = 0

det ([\begin{matrix} σ_{x} - λ & τ_{x y} \\ τ_{x y} & σ_{y} - λ \end{matrix}]) = 0

Evaluate the determinate.

(σ_{x} - λ) (σ_{y} - λ) - τ_{x y}^{2} = 0

σ_{x} σ_{y} - λ σ_{y} - λ σ_{x} + λ^{2} - τ_{x y}^{2} = 0

Rearrange.

λ^{2} - λ (σ_{y} + σ_{x}) + σ_{x} σ_{y} - τ_{x y}^{2} = 0

Solve for the eigenvalues using the quadratic equation where

a = 1

b = - (σ_{y} + σ_{x})

, and

c = σ_{x} σ_{y} - τ_{x y}^{2}

\begin{aligned} λ & = \frac{(σ_{y} + σ_{x}) \pm \sqrt{(σ_{y} + σ_{x})^{2} - 4 (σ_{x} σ_{y} - τ_{x y}^{2})}}{2} \\ = \frac{(σ_{x} + σ_{y})}{2} \pm \frac{\sqrt{σ_{y}^{2} + 2 σ_{y} σ_{x} + σ_{x}^{2} - 4 σ_{x} σ_{y} + 4 τ_{x y}^{2}}}{2} \\ = \frac{(σ_{x} + σ_{y})}{2} \pm \frac{\sqrt{σ_{y}^{2} - 2 σ_{y} σ_{x} + σ_{x}^{2} + 4 τ_{x y}^{2}}}{2} \\ = \frac{(σ_{x} + σ_{y})}{2} \pm \sqrt{\frac{(σ_{y} - σ_{x})^{2} + 4 τ_{x y}^{2}}{4}} \\ = \frac{(σ_{x} + σ_{y})}{2} \pm \sqrt{(\frac{σ_{y} - σ_{x}}{2})^{2} + τ_{x y}^{2}} \end{aligned}

This is the same result as the geometric derivation above.

To find the eigenvectors, we plug our eigenvalues back into the equation

(T - λ I) v = 0

First eigenvector angle. #fst-egn

θ_{p 1} = \tan^{- 1} (\frac{σ_{1} - σ_{x}}{τ_{x y}}) = \tan^{- 1} (\frac{τ_{x y}}{σ_{1} - σ_{y}})

λ_{1} = σ_{1}

[\begin{matrix} σ_{x} - σ_{1} & τ_{x y} \\ τ_{x y} & σ_{y} - σ_{1} \end{matrix}] [\begin{matrix} v_{11} \\ v_{12} \end{matrix}] = 0

Multiplying out gives two equations.

(σ_{x} - σ_{1}) v_{11} + τ_{x y} v_{12} = 0

τ_{x y} v_{12} + (σ_{x} - σ_{1}) v_{12} = 0

The angle of the eigenvector.

θ_{p 1} = \tan^{- 1} (\frac{v_{12}}{v_{11}})

We can repeat this procedure for the second eigenvalue,

λ_{2} = σ_{2}

Second eigenvector angle.

θ_{p 2} = \tan^{- 1} (\frac{σ_{2} - σ_{x}}{τ_{x y}}) = \tan^{- 1} (\frac{τ_{x y}}{σ_{2} - σ_{y}})

Maximum Shear Stress

The orientations for the principal stress element and max shear stress element are

45^{o}

apart.

τ_{x^{'} y^{'}}

can also be maximized.

Angle for maximum shear stress. #max-shr

\begin{array}{r} \tan (2 θ_{s 1}) = \frac{- (σ_{x} - σ_{y})}{2 τ_{x y}} \\ θ_{s 2} = θ_{s 1} + 90^{o} \end{array}

Recall.

τ_{x^{'} y^{'}} = - \frac{σ_{x} - σ_{y}}{2} \sin (2 θ) + τ_{x y} \cos (2 θ)

To maximize an equation, take the derivative and set it equal to zero.

\frac{d τ_{x^{'} y^{'}}}{d θ} = - \frac{σ_{x} - σ_{y}}{2} 2 \cos (2 θ) - τ_{x y} 2 \sin (2 θ) = 0

- 2 τ_{x y} \sin (2 θ) = (σ_{x} - σ_{y}) \cos (2 θ)

\tan (2 θ) = \frac{- (σ_{x} - σ_{y})}{2 τ_{x y}}

The maximum/minimum in plane shear stress values are associated with

θ_{s 1}

and

θ_{s 2}

Maximum shear stress.

| τ_{m a x} | = \sqrt{(\frac{σ_{x} - σ_{y}}{2})^{2} + τ_{x y}^{2}}

A maximum shear stress element has an average normal stress.

Average normal stress.

σ_{x}^{'} = σ_{y}^{'} = σ_{a v g} = \frac{σ_{x} + σ_{y}}{2}

Warning: Unlike with the principal stress element, the normal stresses are not zero. #max-shr