1) Stress and Strain - Stress

-1. Stress and Strain

Stress means the force acting on the unit area, such as $\sigma\,=\,\frac{F}{A} [\frac{N}{m}]\,=\,[Pa]$ where $F\,:\,Force,\,A\,=\,Area$.

Strain means the difference between the initial length and final length for compared to the initial length, such as $\epsilon\,=\,\frac{L_0\,-\,L_f}{L_0}$. where $L_0$ is initial length of a specimen of a certain material, and $L_f$ is final length of the specimen.

The mechanics of materials are the set of study for the relations between them.

-2. Preliminaries - statics

• Equations of Equilibrium

Firstly, mechanics of materials deal with the static state, so we have two main formulas such that

$$\begin{align}& \sum \mathbb{F}\,=\,0 \\ & \sum \mathbb{M}_O \,=\,0\end{align}$$

In the most simple coordinates, in other words, cartesian coordinates, generally we deal with the xyz, so the above formula are expressed 6 formulas like below:

$$\begin{align} & \sum F_x\,=\,0, \sum F_y\,=\,0, \sum F_z\,=\,0 \\ & \sum M_x\,=\,0, \sum M_y\,=\,0, \sum M_z\,=\,0\end{align}$$

If you’re in the two-dimensional situation, above 6 formulas are to be reduced to 3 formulas such as:

$$\begin{align}&\sum F_x\,=\,0 \\& \sum F_y\,=\,0 \\& \sum M_O\,=\,0 \end{align}$$

• Support Reactions

In statics, support options help us to solve the problem properly, sometimes are critical to define whether the problem could be solved.

Fig.1 - Support options -3. Forces used to analyze the stress and strain

• Normal force: Force acts perpendicular to the area.
• Shear force: Force makes the body to slide over one another.
• Torsional moment (or torque): The moment makes the body twist according to the perpendicular axis to the surface area.
• Bending moment: The moment makes the body bended

Abstract descriptions are below:

Fig.2 - Force and moments acting on the body

In many cases, we’re going to deal with this picture:

Fig.3 - Coplanar force and moment -4. Stresses - Normal and shear stress

Fig.4 - Z-direction stresses

• Normal Stress: stress for the normal force to a surface. For [Fig.4], $\sigma_z\,=\displaystyle{\lim_{\Delta A \to 0}}\frac{\Delta F_z}{\Delta A}$

• Shear Stress: stress for the shear force to a surface. For [Fig. 4], $\tau_{zx}\,=\displaystyle{\lim_{\Delta A \to 0}}\frac{\Delta F_x}{\Delta A}$, $\tau_{zy}\,=\displaystyle{\lim_{\Delta A \to 0}}\frac{\Delta F_y}{\Delta A}$

So, stresses are generally expressed like below:

Fig.5 - General expressions of stresses -5. Premise: the condition of material which we're getting to analyze

• Homogeneous: the property of material which has the same physical and mechanical properties throughout its volume

• Isotropic: the property of material which has the same properties in all directions, then when the load P is applied to the bar through the centroid of its cross-sectional area, the bar will deform uniformly throughout the central region of its length.

  Note: Anisotropic material: have different properties in different directions / however it also deform uniformly when subjected to the axial load P.

  

Fig. 6 - Uniform deformation of homogeneous and isotropic material -6. Average Normal Stress

• if a bar is taken and divided into the large number of infinitesimally small specimen, it has been undergone uniform deformation, and it means directly that the cross section be subjected to a constant normal stress distribution just like [Fig. 7.]

Fig. 7 - Infinitesimally small specimen -

• Each small area $\Delta\,A$ on the cross section is subjected to a force $\Delta N\,=\,\sigma\,\Delta\,A$, [Fig. 8.], and the sum of these forces acting over the entire cross-sectional area must be equivalent to the internal resultant force $\mathbb{P}$ at the section. If we let $\Delta N\,\rightarrow\sigma\,dA$ and therefore $\Delta N\,\rightarrow dN $, then, recognizing $\sigma$ is constant, we have

Fig. 8 - Average Normal Stress $$\begin{align}+\uparrow F_{R_z}\,=\,\sum F_z ; &\,\,\,\,\, \int\,dN\,=\,\int_A \sigma\,dA \\ &N\,=\,\sigma\,A \\ & \sigma\,=\frac{N}{A}\end{align}$$

$\sigma$: average normal stress at any point on the cross-sectional area

$N$: internal resultant normal force

$A$: cross-sectional area of the bar where $\sigma$ is determined

• The stress distribution in Fig. 1–12 indicates that only a normal stress exists on any small volume element of material located at each point on the cross section. so we get the formula: $\begin{align}\sum F_z\,=\,0; & \,\,\,\,\,\,\,\sigma(\Delta A)\,-\,\sigma'(\Delta A)\,=0 \\ & \sigma\,=\,\sigma' \end{align}$

Fig. 9 - Internal Equilibrium

• It says that the normal stress components on the element must be equal in magnitude but opposite in direction. // this is based on the assumption that the material is subjected to ==uniaxial stress== and by this analysis the component is determined whether is subjected to the tension or compression.

7. Average Shear Stress

$\tau_{avg}\,=\,\frac{V}{A}$ where

$\tau_{avg}\,=\,$average shear stress at the section, which is assumed to be the same at each point on the section

$V\,=\,$internal resultant shear force on the section determined from the equations of equilibrium

$A\,=\,$area of the section

• The analysis starts from this figure:

Fig. 10 - Analysis of shear stress

Fig. 11 - Shear stress flow in the block $$\sum F_y\,=\,0;\,\,\,\,\,\tau_{zy}(\Delta x \Delta y)\,-\,\tau_{zy}'\Delta x \Delta y\,=\,0$$ $$\sum M_x\,=\,0;\,\,\,\,\,-\tau_{zy}(\Delta x \Delta y_)\Delta z)\,+\,\tau_{yz}(\Delta x \Delta z)\Delta y\,=\,0$$

and so, all four shear stresses must have equal magnitude and be directed either toward or away from each other at opposite edges of the element,

• References

R.C. Hibbeler, “Mechanics of materials”, Pearson, 10th ed., ch.1

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