Notes on Sections 2.6, 2.7, and 2.8 in Chapter 2

This blog post is a continuation of my notes of Chapter 2 and will contain sections 2.6, 2.7, 2.8.

2.6: Stress on Inclined Sections

Stress Elements:
A stress element is an isolated element of a material which depicts the stresses acting on all faces of that elements.
The dimensions of a stress element are assumed to be infinitesimally small and are therefore drawn to a large scale.

Stresses on Inclined Sections:
Observing the stresses acting on inclined section of an object provides a more complete picture. Since the stresses are uniform throughout the entire bar, the stresses acting over the inclined section must be equally distributed.

When observing the inclined section of an object, you must first specify the orientation of the inclined section. Orientation is usually established by the angle $\theta$ between the x-axis and the normal to the section.

To find the stresses acting on a section, the forces must be broken up into components. These components are the normal force N and shear force V, which is tangential to the plant of the object. The force components can be expressed as:

$N = P\cos \theta$     $V = P\sin\theta$

Since ${\sigma _\theta } = \frac{N}{{{A_1}}} = \frac{P}{A}{\cos ^2}\theta$ and ${\tau _\theta } =  - \frac{V}{{{A_1}}} =  - \frac{P}{A}\sin \theta \cos \theta$
The normal and shear stresses can be defined as:
${\sigma _\theta } = {\sigma _x}{\cos ^2}\theta  = {\sigma _x}(1 + \cos 2\theta )$

${\tau _\theta } =  - {\sigma _x}\sin \theta \cos \theta  =  - \frac{{{\sigma _x}}}{2}(\sin 2\theta )$


Maximum Normal and Shear Stresses:
${\sigma_\theta } = {\sigma_x}$ when ${\theta} = 0$

As ${\theta }$ increases or decreases, the normal stress diminishes until ${\theta }$ = $\pm$ ${90^ \circ }$ where it becomes zero, because there are no normal stresses on sections parallel to the longitudinal axis. The maximum normal stress occurs at ${\theta }$ = 0 and is: \[{\sigma_{max}} = {\sigma_x}\]
When ${\theta }$ = $\pm$ ${45^ \circ }$, the normal stress is one half the maximum value.

The maximum shear stresses have the same magnitude:

${\tau _{\max }} = \frac{{{\sigma _x}}}{2}$

2.7: Strain Energy

2.7 looks at strain energy from its simplest form , through axially loaded members subjected to static loads. A static load is one that has no dynamic or inertial effects due to motion.

Work is therefore defined as:  $$W = \int\limits_0^\delta  {{P_1}d{\delta _1}}$$

Strain energy is equal to work so it is therefore equal to the work equation stated above. 

Elastic and Inelastic Strain Energy:

Elastic strain energy is strain energy recovered during unloading. 

Inelastic strain energy is strain energy that is permanently lost during the unloading process. 

Linearly Elastic Behavior:

$$U = W = \frac{{P\delta }}{2}$$  

This equation describes the strain energy, U, stored in a bar

Since $\delta  = \frac{{{P^2}L}}{{2EA}}$, in a linearly elastic bar strain energy can take either of the following forms: 

$U = \frac{{{P^2}L}}{{2EA}}$ or $U = \frac{{EA{\delta ^2}}}{{2L}}$

Displacements caused by a single load:

Since U = W = $\frac{{P\delta}}{2}$, this questions can be easily rearranged to solve for displacement:

$$\delta  = \frac{{2U}}{P}$$  

2.8: Impact Loading

Loads can be classified as static or dynamic depending on whether they remain constant or vary with time.
-A static load is applied slowly so it causes no vibrational or dynamic effects in the structure.
-Dynamic loads take many forms - some are apllied and removed suddenly (impact loads), others persist for long periods of time and continuously vary in intensity (fluctuating loads)

Maximum elongation of the Bar:

Starting by equating the potential energy lost to maximum strain energy, we may get to maximum elongation of a bar with several derivations.
Our starting equation:
$$w(h + {\delta _{\max }}) = \frac{{EA{\delta _{\max }}}}{{2L}}$$
And max elongation is found to be :
$${\delta _{\max }} = \sqrt {2h{\delta _{st}}}  = \sqrt {\frac{{m{v^2}L}}{{EA}}}$$
Maximum Stress in a Bar:
$${\sigma _{\max }} = \frac{{E{\delta _{\max }}}}{L}$$
Through several derivations and substitutions, we arrive at the final equation which describes the maximum stress which a bar can receive:
$${\sigma _{\max }} = \sqrt {\frac{{m{v^2}E}}{{AL}}}$$

Impact Factor:
The Impact factor is known as the ratio between the dynamic response of a structure and the static response (for the same load) :
Impact Factor = $\frac{{{\delta _{\max }}}}{{{\delta _{st}}}}$

My next blog post will be on my notes on the final sections of Chapter 2. Sections 2.10, 2.11, and 2.12.

Thanks for reading,

-Nick Thompson