Table of contents

The 1D case

If we take a look at a so called infinitesimal element of size Δx\Delta x then at some arbitrary point xx we have a displacement u(x)u\left(x\right) and at the other end of that small element, at the point x+Δxx+\Delta x we have the displacement u(x+Δx)u\left(x+\Delta x\right). The engineering strain is thus defined as

ε:=elongationoriginallength=u(x+Δx)u(x)Δxdudx\varepsilon :=\dfrac{\mathrm{elongation} }{\mathrm{original} \mathrm{length} }=\dfrac{u\left(x+\Delta x\right)-u\left(x\right)}{\Delta x}\to \dfrac{d u}{d x}

In 1D the relation between the strain and stress is done using the linear elasticity modulus (tensile modulus or Young's modulus) EE

σ=Eε\sigma =E\varepsilon

In two dimensions we have multiple strains to take into consideration. First there is shear strain, which can be defined using the infinitesimal element in 2D:

Given some shear stress τ\tau the element will be deformed in a way which can be described using the shear angle γ\gamma. The Hookean constitutive relationship is give by

τ=Gγ\tau = G\gamma

where GG is known as the shear modulus.

Adding tensile deformation to the mix we get the general state of deformation on a infinitesimal element.

In two dimensions we will have two tensile deformations (described by uxu_x and uyu_y) and two shear deformations (described by γ1\gamma_1 and γ2\gamma_2).

The strain in the x-direction can be derived from the infinitesimal element getting displaced

ux(x+Δx,y)ux(x,y)Δxuxx=:εx\dfrac{u_x \left(x+\Delta x,y\right)-u_x \left(x,y\right)}{\Delta x}\to \dfrac{\partial u_x }{\partial x}=:\varepsilon_x

and in the y-direction

uy(x,y+Δy)uy(x,y)Δyuyy=:εy\dfrac{u_y \left(x,y+\Delta y\right)-u_y \left(x,y\right)}{\Delta y}\to \dfrac{\partial u_y }{\partial y}=:\varepsilon_y

The shear strain is derived by

uy(x+Δx,y)uy(x,y)Δxuyx\dfrac{u_y \left(x+\Delta x,y\right)-u_y \left(x,y\right)}{\Delta x}\to \dfrac{\partial u_y }{\partial x}

Using trigonometry we can see that

tan(γ1)=uy(x+Δx,y)uy(x,y)Δxuyx\tan \left(\gamma_1 \right)=\dfrac{u_y \left(x+\Delta x,y\right)-u_y \left(x,y\right)}{\Delta x}\to \dfrac{\partial u_y }{\partial x}

We assume small deformations such that we can apply the small angle approximation

tan(γ1)γ1\tan \left(\gamma_1 \right)\approx \gamma_1

The total shear angle is given by

γxy=γ1+γ2\gamma_{x y} =\gamma_1 +\gamma_2

So we get the total shear strain

γxy=uyx+uxy\gamma_{x y} =\dfrac{\partial u_y }{\partial x}+\dfrac{\partial u_x }{\partial y}

This reasoning is extended into 3D such that we get

εz=uzz,γxz=uxz+uzx,γyz=uyz+uzy\varepsilon_z =\dfrac{\partial u_z }{\partial z},\gamma_{x z} =\dfrac{\partial u_x }{\partial z}+\dfrac{\partial u_z }{\partial x},{ \gamma }_{y z} =\dfrac{\partial u_y }{\partial z}+\dfrac{\partial u_z }{\partial y}

In total we get six strains, which can be represented in a matrix

ε=[εx12γxy12γxz12γxyεy12γyz12γxz12γyzεz] \def\arraystretch{2.0} \bm \varepsilon =\left\lbrack \begin{array}{ccc} \varepsilon_x & \dfrac{1}{2}\gamma_{x y} & \dfrac{1}{2}\gamma_{x z} \\ \dfrac{1}{2}\gamma_{x y} & \varepsilon_y & \dfrac{1}{2}\gamma_{y z} \\ \dfrac{1}{2}\gamma_{x z} & \dfrac{1}{2}\gamma_{y z} & \varepsilon_z \end{array}\right\rbrack

where 12γxy\frac{1}{2}\gamma_{x y} is known as the tensor shear strain. This is because the shear tensor can be also expressed using index notation

εij=12(uixj+ujxi)\varepsilon_{i j} =\dfrac{1}{2}\left(\dfrac{\partial u_i }{\partial x_j }+\dfrac{\partial u_j }{\partial x_i }\right)

so for e.g., εx=ε11=12(u1x1+u1x1)=u1x1=uxx\varepsilon_{x } =\varepsilon_{11} =\frac{1}{2}\left(\frac{\partial u_1 }{\partial x_1 }+\frac{\partial u_1 }{\partial x_1 }\right)=\frac{\partial u_1 }{\partial x_1 }=\frac{\partial u_x }{\partial x}

εxy=ε12=12(u1x2+u2x1)=uxy+uyx\varepsilon_{x y} =\varepsilon_{12} =\dfrac{1}{2}\left(\dfrac{\partial u_1 }{\partial x_2 }+\dfrac{\partial u_2 }{\partial x_1 }\right)=\dfrac{\partial u_x }{\partial y}+\dfrac{\partial u_y }{\partial x}

The stress tensor

The stresses can be expressed in the same way

σ=[σxτxyτxzτxyσyτyzτxzτyzσz]\bm \sigma =\left\lbrack \begin{array}{ccc} \sigma_{x} & \tau_{x y} & \tau_{x z} \\ \tau_{x y} & \sigma_{y} & \tau_{y z} \\ \tau_{x z} & \tau_{y z} & \sigma_{z} \end{array}\right\rbrack

Deriving the force equilibrium

On an infinitesimal element, we can assume σy(x,y)\sigma_y \left(x,y\right) and σy(x,y+Δy)\sigma_y \left(x,y+\Delta y\right) are constant in the xx-direction, and do the corresponding assumption for the other two directions. This helps simplify the following force equilibrium equations

Fx=0ΔσxΔy+ΔτyxΔx+fxΔxΔy=0\sum F_x =0\Rightarrow {\Delta \sigma }_x \Delta y+{\Delta \tau }_{y x} \Delta x+f_x \Delta x\Delta y=0 :(σx(x+Δx,y)σx(x,y))Δy+(τyx(x,y+Δy)τyx(x,y))Δx+fxΔxΔy=0\longrightarrow :\left(\sigma_x \left(x+\Delta x,y\right)-\sigma_x \left(x,y\right)\right)\Delta y+\left(\tau_{y x} \left(x,y+\Delta y\right)-\tau_{y x} \left(x,y\right)\right)\Delta x+f_x \Delta x\Delta y=0

and similarly for the yy-direction.

This leads to the system

σxxτyxy=fxσyyτxyx=fy \def\arraystretch{2.5} \begin{array}{l} -\dfrac{\partial \sigma_x }{\partial x}-\dfrac{\partial \tau_{y x} }{\partial y}=f_x \\ -\dfrac{\partial \sigma_y }{\partial y}-\dfrac{\partial \tau_{x y} }{\partial x}=f_y \end{array}

This can be expressed using matrix notation

x[σxτxy]y[τxyσx]=[fxfy]-\dfrac{\partial }{\partial x}\left\lbrack \begin{array}{c} \sigma_x \\ \tau_{x y} \end{array}\right\rbrack -\dfrac{\partial }{\partial y}\left\lbrack \begin{array}{c} \tau_{x y} \\ \sigma_x \end{array}\right\rbrack =\left\lbrack \begin{array}{c} f_x \\ f_y \end{array}\right\rbrack

Recall the gradient operator \nabla

=[xy] \def\arraystretch{2.5} \nabla =\left\lbrack \begin{array}{c} \dfrac{\partial }{\partial x}\\ \dfrac{\partial }{\partial y} \end{array}\right\rbrack

The divergence of the stress matrix is

σ=f=x[σxτxy]y[τxyσx]=[fxfy]-\nabla \cdot \sigma =f=-\dfrac{\partial }{\partial x}\left\lbrack \begin{array}{c} \sigma_x \\ \tau_{x y} \end{array}\right\rbrack -\dfrac{\partial }{\partial y}\left\lbrack \begin{array}{c} \tau_{x y} \\ \sigma_x \end{array}\right\rbrack =\left\lbrack \begin{array}{c} f_x \\ f_y \end{array}\right\rbrack

Thus the short hand notation for the equilibrium equation using tensor notation is

σ=f \boxed{ -\nabla \cdot \bm \sigma = f }

Symmetry of the stress matrix

A moment equilibrium around the midpoint gives:

τxy(x+Δx,y)Δy12Δx+τxy(x,y)Δy12Δxτyx(x,y)Δx12Δyτyx(x,y+Δy)Δx12Δy=0\tau_{x y} \left(x+\Delta x,y\right)\Delta y \dfrac{1}{2} \Delta x+\tau_{x y} \left(x,y\right)\Delta y \dfrac{1}{2} \Delta x-\tau_{y x} \left(x,y\right)\Delta x\dfrac{1}{2}\Delta y-\tau_{y x} \left(x,y+\Delta y\right)\Delta x \dfrac{1}{2} \Delta y=0

This simplifies to

τxy(x+Δx,y)+τxy(x,y)=τyx(x,y+Δy)+τyx(x,y)\Rightarrow \tau_{x y} \left(x+\Delta x,y\right)+\tau_{x y} \left(x,y\right)=\tau_{y x} \left(x,y+\Delta y\right)+\tau_{y x} \left(x,y\right)

with Δx0\Delta x\to 0 and Δy0\Delta y\to 0

we have τxy=τyx\tau_{xy} =\tau_{yx} and similarly τyz=τzy\tau_{y z} =\tau_{z y} and τxz=τzx\tau_{x z} =\tau_{z x}. Thus the stress matrix is symmetric!

Boundary traction

On a deformable body Ω\Omega we apply an external force tt somewhere on the boundary Ω\partial \Omega.

The external force tt is called a traction force, and has the unit force per length (2D) or force per area unit (3D).

On an infinitesimal boundary element the boundary can be said to be a small straight line, Δs\Delta s, with an angle to the horizontal axis given by α\alpha. We have a surface normal defined positive outwards from the body.

From the figure we can establish the following kinematic relations

Δs=Δx2+Δy2Δx=cosαΔsΔy=sinαΔs\begin{array}{l} \Delta s=\sqrt{\Delta x^2 +\Delta y^2 }\\ \Delta x=\cos \alpha \Delta s\\ \Delta y=\sin \alpha \Delta s \end{array} n=[sinα,cosα]n=\left\lbrack \sin \alpha ,\cos \alpha \right\rbrack

Equilibrium gives:

:σxΔy+τxyΔx=txΔs=σxsinα+τxycosα=tx:σxΔx+τxyΔy=tyΔs=σxcosα+τxysinα=ty\begin{array}{l} \longleftarrow :\sigma_x \Delta y+\tau_{x y} \Delta x=t_x \Delta s=\sigma_x \sin \alpha +\tau_{x y} \cos \alpha =t_x \\ \downarrow :\sigma_x \Delta x+\tau_{x y} \Delta y=t_y \Delta s=\sigma_x \cos \alpha +\tau_{x y} \sin \alpha =t_y \end{array} σxnx+τxyny=txσxny+τxyny=tn}[σxτxyτxyσy][nxny]=[txty]\left.\Rightarrow \begin{array}{c} \sigma_x n_x +\tau_{x y} n_y =t_x \\ \sigma_x n_y +\tau_{x y} n_y =t_n \end{array}\right\rbrace \left\lbrack \begin{array}{cc} \sigma_x & \tau_{x y} \\ \tau_{x y} & \sigma_y \end{array}\right\rbrack \left\lbrack \begin{array}{c} n_x \\ n_y \end{array}\right\rbrack =\left\lbrack \begin{array}{c} t_x \\ t_y \end{array}\right\rbrack

or

σn=t \boxed{ \bm \sigma \cdot \bm n = \bm t }

Principal stress and strain

A 2D geometry in pure tension along the x-axis experiences mostly stress in the x-direction and some amount of transversal contraction stress in the y-direction, but no shear stress.

σ=[σxτxyτxyσy]=[σx00σy]\bm \sigma =\left\lbrack \begin{array}{cc} \sigma_x & \tau_{x y} \\ \tau_{x y} & \sigma_y \end{array}\right\rbrack =\left\lbrack \begin{array}{cc} \sigma_x & 0\\ 0 & \sigma_y \end{array}\right\rbrack

If we apply the same load case to the geometry but rotate the frame of reference than we will get shear stress since our matrix is expressed in terms of the x and y coordinates and not the local body coordinates.

σ=[σxτxyτxyσy]=[σx0τxy0τxy0σy0]\bm \sigma =\left\lbrack \begin{array}{cc} \sigma_x & \tau_{x y} \\ \tau_{x y} & \sigma_y \end{array}\right\rbrack =\left\lbrack \begin{array}{cc} \sigma_x \not= 0 & \tau_{x y} \not= 0\\ \tau_{x y} \not= 0 & \sigma_y \not= 0 \end{array}\right\rbrack

Computing these "local coordinate stresses" is important since it gives us an understanding if the stress is in tension or compression. This change of reference system such that we get rid of the shear stress is called Principal stress.

If the traction is normal to the surface, then we can write

t=λn\bm t = \lambda \bm n

where λ\lambda denotes the principal stress as a scalar value. In general 2D we can write this

σn=tσn=λn\bm \sigma \cdot \bm n = t \Rightarrow \bm \sigma \cdot \bm n = \lambda \bm n

or

(σλI)n=0\left(\bm \sigma - \lambda \bm I\right)\cdot \bm n = \bm 0

This is an eigenvalue problem

We can find non-zero solutions by solving

det(σλI)=0\det \left( \bm \sigma -\lambda \bm I\right) = 0 σxλσxyσxyσyλ=(σxλ)(σyλ)σxy2=0\left|\begin{array}{cc} \sigma_x -\lambda & \sigma_{x y} \\ \sigma_{x y} & \sigma_y -\lambda \end{array}\right|=\left(\sigma_x -\lambda \right)\left(\sigma_y -\lambda \right)-{\sigma_{x y} }^2 =0 λ2(σx+σy)trace(σ)λ+σxσyσxy2det(σ)=0\lambda^2 -\underset{\mathrm{trace}\left(\sigma \right)}{\underbrace{\left(\sigma_x +\sigma_y \right)} } \lambda +\underset{\det \left(\sigma \right)}{\underbrace{\sigma_x \sigma_y -{\sigma_{x y} }^2 } } =0

The roots to this equation, λI\lambda_I and λII\lambda_{\mathrm{II} } are the principal stresses in this context. Usually we denote the principal stress with a numbered subindex using either roman or arabic numbering. The usual definition for the first principal stress is

σ1:=max(λ1,λ2) \sigma_1 := \max(\lambda_1, \lambda_2)

and

σ2:=min(λ1,λ2) \sigma_2 := \min(\lambda_1, \lambda_2)

von Mises stress

Using the principal stresses in a general 3D stress state, σ1,σ2,σ3\sigma_1 ,\sigma_2 ,\sigma_3 we can define the von Mises stress

σvM=12[(σ1σ2)2+(σ2σ3)2+(σ3σ1)2]\sigma_{\mathrm{vM} } =\sqrt{\dfrac{1}{2}\left\lbrack {\left(\sigma_1 -\sigma_2 \right)}^2 +{\left(\sigma_2 -\sigma_3 \right)}^2 +{\left(\sigma_3 -\sigma_1 \right)}^2 \right\rbrack }

or using the general 3D stress components

σvM=12[(σxσy)2+(σyσz)2+(σzσx)2+6(σxy2+σyz2+σzx2)]\sigma_{\mathrm{vM} } =\sqrt{\dfrac{1}{2}\left\lbrack {\left(\sigma_x -\sigma_y \right)}^2 +{\left(\sigma_y -\sigma_z \right)}^2 +{\left(\sigma_z -\sigma_x \right)}^2 +6\left({\sigma_{x y} }^2 +{\sigma_{y z} }^2 +{\sigma_{z x} }^2 \right)\right\rbrack }

For general plane stress (2D) this simplifies since σz=σyz=σzx=0\sigma_z =\sigma_{y z} =\sigma_{z x} =0

σvM=σx2σxσy+σy2+3σxy2\sigma_{\mathrm{vM} } =\sqrt{ {\sigma_x }^2 -\sigma_x \sigma_y +{\sigma_y }^2 +3{\sigma_{x y} }^2 }

More info on the failure theories is given by the efficient engineer.