Stress and Strain
When a force is applied to an elastic body, the body deforms. The way in which the body deforms depends upon the type of force applied to it.
Compression force makes the body shorter. A tensile force makes the body longer.
Tensile and compressive forces are called direct forces
Stress is the force per unit area upon which it acts, denoted by $\sigma$.
$$ \sigma = \frac{F}{A} [N/m^2] $$
We also usually use the unit Pa (Pascal) for stress, which is equivalent to $1 N/m^2$.
In each case, a force $F$ produces a deformation $x$. In engineering, we usually change this force into stress and the deformation into strain denoted by $\epsilon$,
$$ \epsilon = \frac{x}{L} [m/m] $$
Strain has no unit since it is a ratio of two lengths.
The mechanical properties of materials used in engineering are determined by experiments performed on small specimens.
These experiments are conducted in laboratories equipped with testing machines capable of loading in tension or compression.
Modulus of elasticity $(E)$
Elastic materials always spring back into shape when released. They also obey Hoke’s Law.
This is the law of springs which states that deformation is directly proportional to the force.
$$ \frac{F}{x} = \text{stiffness} [kN/m] $$
The stiffness is different for the different material and different sizes of the material. We may eliminate the size by using stress and strain instead of force and deformation.
Let $F$ be the direct stress and $x$ be the strain,
$$ F = \sigma A \quad \text{and} \quad x = \epsilon L $$
Therefore,
$$ \frac{F}{x} = \frac{\sigma A}{\epsilon L} \quad \text{and} \quad \frac{FL}{Ax} = \frac{\sigma}{\epsilon}. $$
This is what we call the Modulus of Elasticity (E),
$$ E = \frac{FL}{Ax} = \frac{\sigma}{\epsilon}. $$
A graph of stress against strain will be straight line with gradient of $E$. The units of $E$ are the same as the unit of stress.
Shear Strain
Shear stress denoted by $\tau$, is the force per unit area carrying the load. This means the cross-sectional area of the material being cut, the beam and pin.
$$ \tau = \frac{F}{A} [N/m^2] $$
The force causes the material to deform as shown. The **shear strain denoted by $\gamma$, is defined as the ratio of the distance deformed to the height. We can say that,
$$ \gamma = \frac{x}{L} $$
modulus of rigidity $(G)$
If we conduct an experiment and measure $x$ for various values of $F$, we will find that if the material is elastic, it behaves like a spring and so long as we do not damage the material by using too big force, the graph of $F$ and $x$ is a straight line.
The gradient of the graph is constant, so,
$$ \frac{F}{x} = \text{constant}. $$
This is the spring stiffness of the block in $N/m$.
If we divide $F$ by area $A$ and $x$ by the height $L$, the relationship is still a constant and we get,
$$ G = \frac{\frac{F}{A}}{\frac{x}{L}} = \frac{FL}{Ax} = \frac{\tau}{\gamma} = \text{constant}. $$
Poisson’s Ratio
For clarity, this physical fact may be restated thus.
If a solid body is subjected to an axial tensionl, it contracts laterally.
On the other hand, the material “squashes out” sideways.
This constant is a definite property of a material called Poisson’s ratio. It will be denoted by $\nu$ and is defined as, $$ \nu = \left\vert \frac{\text{lateral strain}}{\text{axial strain}} \right\vert = - \frac{\text{lateral strain}}{\text{axial strain}} $$
Homogeneous isotropic linear elastic materials (with a very basic material properties).
$$ G = \frac{E}{2(1 + \nu)} $$
Bending Deformation
- Bending moment causes beam to deform.
- $x$: longitudinal axis.
- $y$: axis of symmetry.
Assumptions:
- Neutral surface - no change in length.
- All cross-sections remain plane and perpendicular to longitudinal axis.
- Neglect deformation of cross-section within its own plane.
$$ \epsilon = \lim_{\Delta s \to 0} \frac{\Delta s^{\prime} - \Delta x}{\Delta x} = \lim_{\Delta s \to 0} \frac{(\rho_c - y) \Delta \theta - \rho_c \Delta \theta}{\rho_c \Delta \theta} = - \frac{y}{\rho_c} $$
$$ \epsilon = - \frac{y}{\rho_c} \epsilon_{\max} $$
Bending Stress
From Hooke’s Law we can expect that,
$$ \sigma = -\frac{y}{c} \sigma_{\max} $$
Moment-Curvature Relation
$$ m = \int_A y\ dF = \int_A y(-\sigma)\ dA = \int_A y \left(\frac{y}{c} \sigma_{\max}\right)\ dA = \frac{\sigma_{\max}}{c} \int_A y^2\ dA $$
The last term in our final expression is the moment of inertia denoted by $I$.
Therefore, we can obtain,
$$ \sigma = - \frac{My}{I} \rightarrow E\epsilon = - \frac{My}{I} \rightarrow M = - \frac{EI}{\rho_c} $$
Deflection Curve
The deflection $v$ is the displacement in the $y$ direction of any point on the axis of the beam.
When the beam is bent, there is not only a deflection at each point along the axis, but also a rotation.
The angle of rotation $\theta$ of the axis of the beam is the angle between the $x$-axis and the tangent to the deflection curve.
Radius of Curvature
The radius of curvature $\rho_c$ has the following relation,
$$ ds = \rho_c\ d\theta \rightarrow \frac{1}{\rho_c} = \frac{d\theta}{ds}. $$
The slope of the deflection curve is the first derivative $dv/dx$ of the deflection $v$.
$$ \frac{1}{\rho_c} = \frac{\frac{d^2 v}{dx^2}}{\left(1 + \left(\frac{dv}{dx}\right)^2\right)^{3/2}} $$
for small $dv/dx$, $$ \frac{1}{\rho_c} \approx \frac{d^2 v}{dx^2} $$
The Beam Equations
Hence, we obtain,
$$ M(x) = EI \frac{d^2 y}{dx^2}. $$
Shear force equation,
$$ V = - \frac{dM}{dx} \rightarrow V = - \frac{d}{dx} \left(EI \frac{d^2 y}{dx^2}\right) $$
The load equation, $$ w = - \frac{dV}{dx} \rightarrow w = \frac{d^2}{dx^2} \left(EI \frac{d^2 y}{dx^2}\right) $$
An Application of Micro-Beams
- Accelerometers
- With thick proof-mass
- Over-range protection by bottom wafer.
Membrane Structures
Membrane structures are thin structures that are subjected to tension.
We use SiO$_2$ as a stop layer and etch mask.
Applications of membrane structures
- Micropumps and microvalves
- KOH etched structrues
- 4 layers of structures including:
- Beams (functioning as check valves).
- A membrane (to provide volume change).