This blog post will contain my notes from the final sections of chapter 2.
2.10: Stress Concentrations
Stress concentrations: High stresses in very small regions of a bar.
Stress Concentration Factors:
The intensity of a stress concentration is usually determined by the ratio between maximum stress and normal stress, this is depicted by the stress-concentration factor K: \[K = \frac{{{\sigma _{\max }}}}{{{\sigma _{nom}}}}\] \({\sigma _{nom}} = \frac{P}{{ct}}\) = nominal stress (where ct is the net area at the cross section of the hole.)
2.11: Nonlinear Behavior
Perfect Plasticity: perfectly plastic regions on a stress-strain curve continue until the strains are 10 or 20 times larger than the yield strain.
A material having a stress-strain diagram with theses types of characteristics is called an elastoplastic material.
2.12: Elastoplastic Analysis
Yield displacement is the downright displacement of the bar at the yield load and is equal to the elongation of the inner bar when first releasing yield stress \({\sigma _y}\) :\[{\delta _Y} = \frac{{{\sigma _y}{L_2}}}{E}\] The plastic displacement \({\delta _Y}\) at the instant the load just reaches the plastic load \({P_p}\) and is equal to the elongation of the outer bars at the instant they reach yield stress.
\[{\delta _P} = \frac{{{\sigma _y}{L_1}}}{E}\] Now Compare \({\delta _P}\) with \({\delta _y}\) and get the ratio: \[\frac{{{\delta _P}}}{{{\delta _y}}} = \frac{{{L_1}}}{{{L_2}}}\] My next blog posts will be about my first few days at my internship and some of the interesting and new experiences I will have had working in a lab.
Thanks for reading,
-Nick Thompson
Sunday, February 26, 2017
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:
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:
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.
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
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
Thursday, February 23, 2017
Notes on Chapter 2 Sections 2.3, 2.4, 2.5
Towards the end of last week I finished reading sections 2.3 through 2.5 in Chapter 2. This post will give a summary of the important topics through those specific sections.
2.3 :
Section 2.3 dealt with changing material lengths under nonuniform conditions.
When a linearly elastic material is loaded only at the ends, the change in length of the material can be obtained through the equation:
2.3 :
Section 2.3 dealt with changing material lengths under nonuniform conditions.
When a linearly elastic material is loaded only at the ends, the change in length of the material can be obtained through the equation:
\(\partial = \frac{{PL}}{{EA}}\)
Where P is the load on the bar, L is the length of the bar, E is the modulus of elasticity, and A is the cross sectional are.
However, 2.3 deals with this equation in a more general sense and not just solely when it is used for linearly elastic materials.
Bars consisting of Prismatic Segments:
The change in length of the bar when loaded can be obtained from the equation:
\(\partial = \sum\limits_{i = 1}^n {\frac{{{N_i}{L_i}}}{{{E_i}{A_i}}}} \)
Where i is the numbering index for various segments, n is the total number of segments, and \({{N_i}}\) is the internal axial force in segment i.
Bars with continuously Varying Loads or Dimensions:
Given that: \[d\partial = \frac{{N(x)dx}}{{EA(x)}}\]
The elongation of the entire bar is obtained by integrating over the entire length with respect to x.
\[\partial = \int\limits_0^L {d\partial = \int\limits_0^L {\frac{{N(x)dx}}{{EA(x)}}} } \]
2.4 :
Section 2.4 deals with statically indeterminate structures.
A structure is classified as statically determinate if its reactions and internal forces can be determined solely from free-body diagrams and equations of equilibrium (i.e \(\sum F = 0\)). It is important to note that the forces of statically determinate structures can be found without knowing the properties of the materials.
However, most structures are more complex that a bar and their reactions and internal forces cannot be found by statics alone. These types of structures are classified as statically indeterminate. In order to analyze these structures, supplemental equations must be used in addition to displacement equations.
2.5 :
Sections 2.5 deals with Thermal effects, Misfits, and Prestrains.
External loads are not the only sources of stresses and strains in a structure. Other sources include thermal effects, which arise from temperature changes, misfits, which result from imperfections in construction, and prestrains, produced by initial deformations.
Thermal effects:
Changes in temperature produce expansion or contraction of structural materials, resulting in thermal strains and thermal stresses.
The equation for thermal strain \({\varepsilon _T}\) is proportional to the temperature change \(\Delta T\):
\[{\varepsilon _T} = \alpha (\Delta T)\]
Where \(\alpha\) is the coefficient of thermal expansion.
Temperature-displacement relation calculates the increase in any dimension by the equation:
\[\partial ={ \varepsilon_T}L= \alpha(\Delta T)L\]
Misfits:
Misfit members are members with slightly different measurements than those which were intended in their design and consequently do not fit properly in their structure.
Prestrains:
Prestrains occur in a system when misfit members are used in construction of a given structure. These misfit members introduce strains and stresses into the structure before any loads are applied to it and are therefore called "Prestrains" and "Prestresses."
If a structure is statically determinate, small misfits in one or more members will not produce strains or stresses in the overall structure. However, in statically indeterminate structures, small misfits do introduce strains and stresses as the overall structure is not free to adjust to those misfit members.
Next post will pertain to my notes on sections 2.6, 2.7, 2.8, and 2.9.
Thanks for reading,
-Nick Thompson
Notes on the Final Sections of Chapter 1 and Beginning of Chapter 2
Last week I finished taking notes on Chapter 1 with a re-introduction to linear elasticity and Hooke's law and an introduction to Poisson's ration. This blog post will contain my general notes on the topics above as well as the contents of Chapter 2.
Linear Elasticity:
Many structural materials such as wood, plastics, and most metals behave both elastically and linearly when an initially loaded. The stress-strain curves of these materials, consequently, begin with a straight line that passes through the origin.
A material is therefore said to be linearly elastic when it exhibits elastic behavior and possesses a linear relationship between stress and strain.
Linearly elastic materials are very important in engineering because structures and machines which are designed in this linearly elastic region do not permanently deform when yielding occurs.
Hooke's Law:
English scientist Robert Hooke was the first person to scientifically investigate the elastic properties of materials. He measured the stretching of long wires supporting weights and was able to establish the linear relationship between applied loads and the resulting elongation.
Hooke's Law describes the linear relationship between stress and strain for a bar in simple tension or compression through the equation:
Linear Elasticity:
Many structural materials such as wood, plastics, and most metals behave both elastically and linearly when an initially loaded. The stress-strain curves of these materials, consequently, begin with a straight line that passes through the origin.
A material is therefore said to be linearly elastic when it exhibits elastic behavior and possesses a linear relationship between stress and strain.
Linearly elastic materials are very important in engineering because structures and machines which are designed in this linearly elastic region do not permanently deform when yielding occurs.
Hooke's Law:
English scientist Robert Hooke was the first person to scientifically investigate the elastic properties of materials. He measured the stretching of long wires supporting weights and was able to establish the linear relationship between applied loads and the resulting elongation.
Hooke's Law describes the linear relationship between stress and strain for a bar in simple tension or compression through the equation:
\(\sigma = E\varepsilon \)
Where \(\sigma\) is the axial stress, \(\varepsilon\) is axial strain and \(E\) is the modulus of elasticity for the given material.
Poisson's Ratio:
The ratio between the lateral strain \(\varepsilon '\) and the axial strain \(\varepsilon\) acting on an object is known as Poisson's ratio.
This ration can be denoted as:
Poisson's Ratio:
The ratio between the lateral strain \(\varepsilon '\) and the axial strain \(\varepsilon\) acting on an object is known as Poisson's ratio.
This ration can be denoted as:
\(v = - \frac{{Lateral strain}}{{Axial strain}} = - \frac{{\varepsilon '}}{\varepsilon }\)
The minus sign is inserted into the equation to counteract the opposite signs of the lateral and axial strains when acting on a material.
Starting Chapter 2:
Chapter 2 of the Mechanics of Materials textbook pertains to axially loaded members. an axially loaded member is any structural component subjected only to tensile or compressive forces. After an intro in section 2.1, section 2.2 deals with the changing lengths of axially loaded members. This sections gives information on springs, prismatic bars, and cables.
My next blog post will deal with my notes last week of sections 2.3 through 2.5
Thanks for reading,
-Nick
Sunday, February 12, 2017
Progress on my textbook readings
For my second blog post I will talk about how I am progressing through my independent reading of the textbooks I was provided as well as things I am finding interesting along the way.
While I was not introduced to many new topics in the Engineering Mechanics: Statics textbook, as the first two chapters mainly consisted of a basic re-introduction to Newton's laws, Newtonian Gravitation, and vector properties and calculations. The Mechanics of Materials textbook as already provided me with new information about several key aspects regarding the behavior of materials.
Through my readings of the Mechanics of Materials textbook, so far, I have learned several fundamental concepts about different types of stresses and strains and also the key differences between static and dynamic testing of a material.
Aside from basic knowledge, I have also started solving basic stress-strain problems in the textbook using the several stress and strain equations.
As I continue to progress through the textbook, I provide additional information that I find interesting or just learned about.
Thanks for reading,
-Nick Thompson
Thursday, February 9, 2017
Learning about several introductory concepts regarding the mechanics of materials, stress-strain responses, dynamic and static testing, and failure analysis
Over the past few days I have been studying several introductory concepts in preparation for my internship at ASU with Dr. Mobasher.
Dr. Mobasher provided me with two textbooks and links which will give insight on some of the fundamental characteristics and physical properties of materials as well as different means to analyse them.
I have been watching Youtube videos by Dr. Yiheng Wang, a professor at Lone Star College in Texas, about Engineering Mechanic, specifically statics. The link the channel can be found below.
The two textbooks provided by Dr. Mobasher are tilted Mechanics of Materials Sixth Edition by James M. Gere, and Engineering Mechanics, Statics, Fourth Edition by Anthony Bedford and Wallace Fowler.
My next post will be about things I find interesting during my readings, as well as thoughts and more details about my internship.
Thanks for reading,
-Nick Thompson
Channel link: https://www.youtube.com/playlist?list=PLLbvVfERDon1pceRKOjAxiqFTEvghmZKh
Dr. Mobasher provided me with two textbooks and links which will give insight on some of the fundamental characteristics and physical properties of materials as well as different means to analyse them.
I have been watching Youtube videos by Dr. Yiheng Wang, a professor at Lone Star College in Texas, about Engineering Mechanic, specifically statics. The link the channel can be found below.
The two textbooks provided by Dr. Mobasher are tilted Mechanics of Materials Sixth Edition by James M. Gere, and Engineering Mechanics, Statics, Fourth Edition by Anthony Bedford and Wallace Fowler.
My next post will be about things I find interesting during my readings, as well as thoughts and more details about my internship.
Thanks for reading,
-Nick Thompson
Channel link: https://www.youtube.com/playlist?list=PLLbvVfERDon1pceRKOjAxiqFTEvghmZKh
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