@@ -29,6 +35,183 @@ import DisplayTable from "../../components/DisplayTable.astro"
Beam Solutions
+ Torsion of thin-walled hallow shafts builds on this content.
+
+
+ In general, the maximum shear stress is given by
+
+
+
+ For thin-walled shafts
+
+
+
+ where
+
+
+
+ Note that is NOT the cross sectional area of the hollow shaft!
+
+
+
+
+Ductile materials generally fail in shear
+
+
+
+
+
+ Axial – maximum shear stress at \(45^o\) angle
+ Torsion – maximum shear stress at \(0^o\) angle
+
+
+
+
+ Brittle materials are weaker in tension than shear
+
+
+
+
+
+ Axial – maximum normal stress at \(0^o\) angle
+ Torsion – maximum normal stress at \(45^o\) angle
+
+
+
+
+
+
+
+
+Failure of a material depends on (1) nature of loading and (2) type of material. There are theories that can predict material failure for complex states.
+
+
+
+
+
+ If a material is ductile, failure is defined by yield stress () and occurs at max shear stress ().
+
+
+
+
+
+
+
+ Consider a material subjected to uniaxial tension and is loaded to the yield point, then the following is true:
+
+
+
+
+Shear stress is responsible for causing the ductile material to yield and is the tensile yield strength.
+
+
+
+
+
+
+
+
+
+
+
+ Failure still occurs at max shear.
+
+
+
+
+
+ The equations for and can be plotted to show the failure surface. Loading conditions that occur outside of the surface are when the material fails.
+
+
+
+ If and have the same signs:
+
+
+
+
+
+ If and have opposite signs:
+
+
+
+
+
+
+
+
+
+Ductile materials likely do not fail due to stresses that only result in a volume change. It is hypothesized that failure is driven by distortion strain energy.
+
+
+
+ All elastic deformations can be broken down into volumetric and distortional deformations.
+
+
+
+ The total strain energy in a material is broken into these same parts.
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ The equating equation be plotted to show the failure surface. The elliptical surface is the Von Mises surface overlaid with the Teresca surface. Loading conditions that occur outside of the surface are when the material fails.
+
+
+
+
+
+
+ The following figure shows the failure surfaces of Tresca and von Mises criteria in term of the principal stresses in 3 dimensions.
+ The 2D figure above can be found cenered around the origin in this plot.
+ The shape of the 2D failure surface remains the same, but moves location depending on .
+ Just like in 2D, regions outside the surfaces are considered to have failed.
+
+
+
+
+Note: Click and drag to rotate, scroll to zoom, and click legend entries to toggle.
+
+
+
+
+For brittle materials, failure is caused by the maximum tensile stress and NOT compressive stress.
+
+
+
+
+
+
+
+
+ The and equations can be plotted to show the failure surface. Loading conditions that occur outside of the surface are when the material fails.
+
+ Brittle fracture can be difficult to predict, so use this theory with caution
+
+
+
+BEAM DEFLECTION CONTENT BELOW --> where does this go?
+
Goal: Determine the deflection and slope at specified points of beams and shafts.
diff --git a/src/pages/sol/failure_theories.astro b/src/pages/sol/failure_theories.astro
deleted file mode 100644
index a10f2101c..000000000
--- a/src/pages/sol/failure_theories.astro
+++ /dev/null
@@ -1,155 +0,0 @@
----
-import Layout from "../../layouts/Layout.astro"
-import Image from "../../components/Image.astro"
-import Section from "../../components/Section.astro"
-import SubSection from "../../components/SubSection.astro"
-import SubSubSection from "../../components/SubSubSection.astro"
-import InlineEquation from "../../components/InlineEquation.astro"
-import DisplayEquation from "../../components/DisplayEquation.astro"
-import BlueText from "../../components/BlueText.astro"
-import RedText from "../../components/RedText.astro"
-import Plotly from "../../components/Plotly.astro"
----
-
-
-
-
-
-
-
-Failure of a material depends on (1) nature of loading and (2) type of material. There are theories that can predict material failure for complex states.
-
-
-
-
-
- If a material is ductile, failure is defined by yield stress () and occurs at max shear stress ().
-
-
-
-
-
-
-
- Consider a material subjected to uniaxial tension and is loaded to the yield point, then the following is true:
-
-
-
-
-Shear stress is responsible for causing the ductile material to yield and is the tensile yield strength.
-
-
-
-
-
-
-
-
-
-
-
- Failure still occurs at max shear.
-
-
-
-
-
- The equations for and can be plotted to show the failure surface. Loading conditions that occur outside of the surface are when the material fails.
-
-
-
- If and have the same signs:
-
-
-
-
-
- If and have opposite signs:
-
-
-
-
-
-
-
-
-
-Ductile materials likely do not fail due to stresses that only result in a volume change. It is hypothesized that failure is driven by distortion strain energy.
-
-
-
- All elastic deformations can be broken down into volumetric and distortional deformations.
-
-
-
- The total strain energy in a material is broken into these same parts.
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
- The equating equation be plotted to show the failure surface. The elliptical surface is the Von Mises surface overlaid with the Teresca surface. Loading conditions that occur outside of the surface are when the material fails.
-
-
-
-
-
-
- The following figure shows the failure surfaces of Tresca and von Mises criteria in term of the principal stresses in 3 dimensions.
- The 2D figure above can be found cenered around the origin in this plot.
- The shape of the 2D failure surface remains the same, but moves location depending on .
- Just like in 2D, regions outside the surfaces are considered to have failed.
-
-
-
-
-Note: Click and drag to rotate, scroll to zoom, and click legend entries to toggle.
-
-
-
-
-For brittle materials, failure is caused by the maximum tensile stress and NOT compressive stress.
-
-
-
-
-
-
-
-
- The and equations can be plotted to show the failure surface. Loading conditions that occur outside of the surface are when the material fails.
-
- Brittle fracture can be difficult to predict, so use this theory with caution
-
-
-
-
\ No newline at end of file
diff --git a/src/pages/sol/intro.astro b/src/pages/sol/intro.astro
new file mode 100644
index 000000000..077662b49
--- /dev/null
+++ b/src/pages/sol/intro.astro
@@ -0,0 +1,48 @@
+---
+import Layout from "../../layouts/Layout.astro"
+import Image from "../../components/Image.astro"
+import Section from "../../components/Section.astro"
+import SubSection from "../../components/SubSection.astro"
+import SubSubSection from "../../components/SubSubSection.astro"
+import Itemize from "../../components/Itemize.astro"
+import Item from "../../components/Item.astro"
+import InlineEquation from "../../components/InlineEquation.astro"
+import DisplayEquation from "../../components/DisplayEquation.astro"
+import Row from "../../components/Row.astro"
+import Col from "../../components/Col.astro"
+import CalloutCard from "../../components/CalloutCard.astro"
+import CalloutContainer from "../../components/CalloutContainer.astro"
+import Warning from "../../components/Warning.astro"
+---
+
+
+
+
+
+
+
+
+
+
diff --git a/src/pages/sol/material_properties.astro b/src/pages/sol/material_properties.astro
index 612c6ce6e..2b45a47ae 100644
--- a/src/pages/sol/material_properties.astro
+++ b/src/pages/sol/material_properties.astro
@@ -28,12 +28,11 @@ import Warning from "../../components/Warning.astro"
-
-
-
-
-
-
-
- If you want more details on this topic, consider taking a course in continuum mechanics.
-
-
-
-
-
- Bold symbols, such as and ,
- represent tensors rather than scalar quantities.
-
-
-
- Additionally, some quantities might be represented with indicial notation. When indices on the right hand side of the equation are repeated,
- this represents a summation where the index values range from 1 to 3.
- For example,
- represents
-
- This is a shorthand to make writing long summations easier.
-
-
-
-
-
-
-Generally, the state of stress a 3D material can be represented as a second-rank tensor (i.e. a matrix) with components.
-
- Infinitesimal Stress element
-
-
-
-Using indicial notation, component of this tensor is denoted as . The same can be said for a material strain tensor.
-
-
- Infinitesimal Strain Element
-
-
-
-
-Using indicial notation, component of this tensor is denoted as . Note
-that both the stress and strain tensors are symmetric, that is and
-.
-
-
-
-
-
-For our purposes, lets bound the discussion of material stress-strain behavior to only the elastic range, that is, deformation is linear proportional applied load.
-This is Hooke’s law. The relationship between stress and strain is referred to as the constitutive relationship of a material. In the most general case, this relationship is defined as a fourth-rank tensor with components.
-This elasticity tensor, C, relating the stress and strain tensors is shown in indicial notation (rather than a matrix operation).
+
+ADD CONTENT ABOUT ASSUMPTIONS
-
- This is short form representation of a summation, known as indicial notation. Whenever there are repeated indices on the right hand side of an equation,
- as with k and l above (repeated in and ), these indices represent
- a summation over three dimensions. So the example can be rewritten as shown below, where and .
-
-
-
+
-Expanding over these two summations yields:
-
-
+
-What this means is there can be as many as 9 components of strain and 9 material coefficients that contribute to each stress component.
-(You can see the benefit of indicial notation here.)
+
-
+Loading in this region results in elastic behavior, meaning the material returns to its original shape when unloaded. This region of the diagram is mostly a straight line, limited by the proportional limit. This region ends at (yielding). For , the diagram is linear, and the behavior is elastic. For , the diagram is nonlinear, but the behavior is still elastic.
-
-However, not all 81 components of this tensor are independent, because of symmetry of stress and strain tensors there are only 21 independent components. If
-the material is assumed to be isotropic (having the same behavior in all directions) this is further reduced to just 2 independent components (for instance,
- Young’s modulus and , Poisson’s ratio). With this simplification, the 3D isotropic elastic material constitutive relation can be simplified.
+Hooke's law is used for small deformations in the elastic region. The Young's modulus (also known as the elastic modulus) is the slope .
-
-
-
-
-Here the engineering shear strains are introduced.
-This, along with identification of shear stresses
-as
-allows for the expression for pure axial stress & strain in one direction to be represented as .
-Rearranging the terms, relation can be expressed in a more general form.
-
-
-
-Similarly, the expression for pure shear stress & strain can be expressed as .
-
-Again, this can be expressed in a more general form.
-
-
-
-
-
-
-
-
-
-A stress-strain diagram is the relationship of normal stress as a function of normal strain. One way to collect these measurement is a uniaxial tension test in which a specimen at a very slow, constant rate (quasi-static). A load and distance are measured at frequent intervals.
-
-
-
-
-
-
-
- Things that effect the material properties include imperfections, different composition, rate or loading, or temperature.
-
-
-
- Stress-strain diagram showing key properties
-
-
-
-Loading in this region results elastic behavior, meaning the material returns to its original shape when unloaded. This region of the diagram is mostly a straight line, limited by the proportional limit). This region ends at (yielding). For , the diagram is linear, and the behavior is elastic. For , the diagram is nonlinear, but the behavior is still elastic.
-
-
- Young's Modulus
-
-
-
- Hooke's law is used for small deformations in the elastic region. The Young's modulus is the slope .
@@ -226,8 +90,9 @@ Loading in this region results elastic behavior, meaning the ma
+
-Shear Modulus
+
Shear stress strain diagram
@@ -235,18 +100,78 @@ Loading in this region results elastic behavior, meaning the ma
-In the above equation, \(\nu\) refers to Poisson's ratio, further explained in a subsequent section. Only two of the three material constants (ie; , , ) are independent in isotropic materials.
+In the above equation, \(\nu\) refers to Poisson's ratio. Only two of the three material constants (ie; , , ) are independent in isotropic materials.
-
+
-
+
-Stresses above the plastic limit () cause the material to permanently deform.
+Poisson's ratio
- Yield Strength
+ As illustrated in the figure above, when an object is being compressed it expands outward, and when stretched it will become thinner. Depending on the material, it may experience more or less lateral deformation for a given amount of axial deformation. The ratio that governs the amount of lateral strain per unit of axial strain is called Poisson's ratio.
+
+
+
+
+
+ Consider a cylinder with volume . We know that . Since the diameter is dependant on the change in length \(dx\) due to Poisson's ratio, we get
+
+ We use the chain rule to find the change in volume
+
+
+
+
+ By definition,
+
+
+
+
+
+ Returning to our first equation,
+
+
+
+
+
+ Since \(\partial V \geq 0\) and \(\nu > 0\) in nicely behaved materials, \( 0 < \nu \leq 0.5\). In practice \(\nu = 0.5\) only if the material is incompressible, thus in most materials \(\nu < 0.5\).
+
+
+
+
+
+
+
+
+ A negative Poisson's ratio is possible when the material structure is non-trival.
+
+
+
+
+
+
+
+
+
+ADD CONTENT ABOUT DUCTILITY
+
+
+
+
+
+Stresses above the plastic limit () cause the material to permanently deform.
+
Perfect plastic or ideal plastic: well-defined , stress plateau up to failure. Some materials (e.g. mild steel) have two yield points (stress plateau at ). Most ductile metals do not have a stress plateau; yield strength is then defined by the 0.002 (0.2%) offset method.
@@ -259,9 +184,9 @@ Stresses above the plastic limit (Atoms rearrange in plastic region of ductile materials when a higher stress is sustained. Plastic strain remains after unloading as permanent set, resulting in permanent deformation. Reloading is linear elastic up to the new, higher yield stress (at A') and a reduced ductility.
- Ultimate Strength
-
+
+
The ultimate strength () is the maximum stress the material can withstand.
@@ -273,11 +198,9 @@ Stresses above the plastic limit (), the middle of the material elongates before failure.
-
-
-
-
+
+
Also called fracture or rupture stress () is the stress at the point of failure for the material. Brittle and Ductile materials fail differently.
@@ -312,103 +235,195 @@ Stresses above the plastic limit (For this reason, concrete is almost always reinforced with steel bars or rods whenever it is designed to support tensile loads.
+
- Strain Energy builds on this content in Engineering Materials.
+ Fatigue builds on this content in Engineering Materials and Mechanical Design.
- Deformation does work on the material: equal to internal strain energy (by energy conservation).
+ SN curve examples
+
+ If stress does not exceed the elastic limit, the specimen returns to its original configuration. However, this is not the case if the loading is repeated thousands or millions of times. In such cases, rupture will happen at stress lower than the fracture stress - this phenomenon is known as fatigue.
+
+
+
+
+
+ Strain Energy builds on this content in Engineering Materials.
-
+ Deformation does work on the material: equal to internal strain energy (by energy conservation).
- Energy density.
+
- Can be generalized to any deformation: areas under stress-strain curves
+ Energy density.
-
+ Can be generalized to any deformation: areas under stress-strain curves
+
+
+
+
+
+
+
+
+
+ - Isotropic: material properties are independent of the direction
+ - Anisotropic: material properties depend on the direction (ie; composites, wood, and tissues)
+
+
+
+
+
+
+ADD CONTENT HERE
+
+
+
+
+
+ADD CONTENT HERE
+
+
+
+
+
+
+
+
+ If you want more details on this topic, consider taking a course in continuum mechanics.
-
+
- Fatigue builds on this content in Engineering Materials and Mechanical Design.
+ Bold symbols, such as and ,
+ represent tensors rather than scalar quantities.
+
- SN curve examples
+ Additionally, some quantities might be represented with indicial notation. When indices on the right hand side of the equation are repeated,
+ this represents a summation where the index values range from 1 to 3.
+ For example,
+ represents
- If stress does not exceed the elastic limit, the specimen returns to its original configuration. However, this is not the case if the loading is repeated thousands or millions of times. In such cases, rupture will happen at stress lower than the fracture stress - this phenomenon is known as fatigue.
+ This is a shorthand to make writing long summations easier.
+
-
+
+Generally, the state of stress a 3D material can be represented as a second-rank tensor (i.e. a matrix) with components.
+
+ Infinitesimal Stress element
+
-
-
-
+
+Using indicial notation, component of this tensor is denoted as . The same can be said for a material strain tensor.
+
-
- - Isotropic: material properties are independent of the direction
- - Anisotropic: material properties depend on the direction (ie; composites, wood, and tissues)
-
+ Infinitesimal Strain Element
+
-
+
-
+Using indicial notation, component of this tensor is denoted as . Note
+that both the stress and strain tensors are symmetric, that is and
+.
-Poisson's ratio
+
- As illustrated in the figure above, when an object is being compressed it expands outward, and when stretched it will become thinner. Depending on the material, it may experience more or less lateral deformation for a given amount of axial deformation. The ratio that governs the amount of lateral strain per unit of axial strain is called Poisson's ratio.
+
+For our purposes, lets bound the discussion of material stress-strain behavior to only the elastic range, that is, deformation is linear proportional applied load.
+This is Hooke’s law. The relationship between stress and strain is referred to as the constitutive relationship of a material. In the most general case, this relationship is defined as a fourth-rank tensor with components.
+This elasticity tensor, C, relating the stress and strain tensors is shown in indicial notation (rather than a matrix operation).
+
-
+
+ This is short form representation of a summation, known as indicial notation. Whenever there are repeated indices on the right hand side of an equation,
+ as with k and l above (repeated in and ), these indices represent
+ a summation over three dimensions. So the example can be rewritten as shown below, where and .
+
+
+
-
+Expanding over these two summations yields:
+
-
- Consider a cylinder with volume . We know that . Since the diameter is dependant on the change in length \(dx\) due to Poisson's ratio, we get
-
- We use the chain rule to find the change in volume
+
-
+What this means is there can be as many as 9 components of strain and 9 material coefficients that contribute to each stress component.
+(You can see the benefit of indicial notation here.)
-
- By definition,
-
+
-
+
+However, not all 81 components of this tensor are independent, because of symmetry of stress and strain tensors there are only 21 independent components. If
+the material is assumed to be isotropic (having the same behavior in all directions) this is further reduced to just 2 independent components (for instance,
+ Young’s modulus and , Poisson’s ratio). With this simplification, the 3D isotropic elastic material constitutive relation can be simplified.
-
- Returning to our first equation,
-
+
-
+
-
- Since \(\partial V \geq 0\) and \(\nu > 0\) in nicely behaved materials, \( 0 < \nu \leq 0.5\). In practice \(\nu = 0.5\) only if the material is incompressible, thus in most materials \(\nu < 0.5\).
-
-
-
+Here the engineering shear strains are introduced.
+This, along with identification of shear stresses
+as
+allows for the expression for pure axial stress & strain in one direction to be represented as .
+Rearranging the terms, relation can be expressed in a more general form.
-
+
-
+Similarly, the expression for pure shear stress & strain can be expressed as .
+
+Again, this can be expressed in a more general form.
+
+
+
+
+
+
+
+
+
+A stress-strain diagram is the relationship of normal stress as a function of normal strain. One way to collect these measurement is a uniaxial tension test in which a specimen at a very slow, constant rate (quasi-static). A load and distance are measured at frequent intervals.
+
+
+
+
+
+
- A negative Poisson's ratio is possible when the material structure is non-trival.
+ Things that effect the material properties include imperfections, different composition, rate or loading, or temperature.
-
+ Stress-strain diagram showing key properties
+
+
diff --git a/src/pages/sol/pressure_vessels.astro b/src/pages/sol/pressure_vessels.astro
index 17ff35acc..09235d6ec 100644
--- a/src/pages/sol/pressure_vessels.astro
+++ b/src/pages/sol/pressure_vessels.astro
@@ -27,67 +27,5 @@ import BlueText from "../../components/BlueText.astro"
-
+
+
+
+
+
+
+
+
+
+REWORD CONTENT
+ Saint-Venant's Principle: slender beam case
+
+ Notation
+
+Stress analysis very near to the point of application of load . Saint-Venant's principle: the stress and strain produced at points in a body sufficiently removed* from the region of external load application will be the same as the stress and strain produced by any other applied external loading that has the same statically equivalent resultant and is applied to the body within the same region.
+
+ *farther than the widest dimension of the cross section
+
+
+
+ADD CONTENT HERE
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Axially Varying Properties
+For non-uniform load, material property and cross-section area:
+
+
+
+
+
+
+
+
+
+
+ Assume variations with are "mild" (on length scale longer than cross-sectional length scales)
+
+
+
+
+
+ - Draw a FBD
+
+ - Equilibrium equations: force balance and moment balance
+
+ - Constitutive equations: stress-strain or force-displacement relations
+
+ - Compatibility equations: geometric constraints
+
+
+
+ Misfit Problems
+A misfit problem is one in which there is difference between a design distance and the manufactured length of a material. Some misfits are created intentionally to pre-strain a member. (e.g. spokes in a bicycle wheel or strings in a tennis racket). This type of problem neither modifies the equilibrium equations (1) nor the force-extension relations, (2) but the compatibility equations, (3) need to be modified.
+
+
+
+
+
+ Stress concentration factors build on this content in engineering materials and machine failure.
+
+
+ Stress concentraions
+
+
+ The stress concentration factor is the highest at lowest cross-sectional area.
+
+
+ - Found experimentally
+
+ - Solely based on geometry
+
+
+
+
+
+
+
+
+Torsion refers to the twisting of a specimen when it is loaded by couples (or moments) that produce rotation about the longitudinal axis. Applications include aircraft engines, car transmissions, and bicycles, etc.
+
+
+
+
+ Torsion lines
+
+
+
+ - For circular shafts (hollow and solid): cross-sections remain plane and undistorted due to axisymmetric geometry
+ - For non-circular shafts: cross-sections are distorted when subject to torsion
+ - Linear and elastic deformation
+
+
+Gears:
+
+ - Gears are perfectly rigid.
+ - The rotation axis is perfectly fixed in space.
+ - Gear teeth are evenly spaced and perfectly shaped so there is no gap to create lost motion.
+ - Gear tooth faces are perfectly smooth so there is no slip.
+ - Mated gears twist through the same arc length.
+
+
+
+
+ Stress distribution
+
+
+
+
+
+ Geometry of Deformation
+
+ Deformation
+
+The angle of twist increases as x increases.
+
+
+
+where is the radius and is a point between 0 and .
+
+
+
+
+
+ Hence, the shear strain ():
+
+ - is proportional to the angle of twist
+ - varies linearly with the distance from the axis of the shaft
+ - is maximum at the surface
+
+
+ Torsional Stiffness
+
+
+
+
+
+
+
+
+
+The shear stress in the elastic range varies linearly with the radial position in the section.
+
+
+Note: shaft under torque rotating at angular speed transmits power .
+
+
+
+ Angle of Twist
+
+From observation:
+
+ - The angle of twist of the shaft is proportional to the applied torque
+ - The angle of twist of the shaft is proportional to the length
+ - The angle of twist of the shaft decreases when the diameter of the shaft increases
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Power transmition can be used to find shear stress in gear shafts.
+
+
+
+
+ Note: 1 hp = 550 ft.lb/s
+
+
+
+ Gear system
+
+Gear system problems can be solved using the torsion equations.
+
+ Common assumptions:
+
+
+
+ - Gears are perfectly rigid.
+ - The rotation axis is perfectly fixed in space.
+ - Gear teeth are evenly spaced and perfectly shaped so there is no gap to create lost motion.
+ - Gear tooth faces are perfectly smooth so there is no slip.
+ - Mated gears twist through the same arc length.
+
+
+To solve:
+
+
+ - Draw a FBD for each shaft.
+ - Use the torsion equilibrium equations of each shaft to find the recation forces and torques.
+ - Use the shear stress equation to find the stresses in the shaft.
+
+
+The angle of twist between two mating gears is related through the gear ratio.
+
+
+
+
+
+
+
+
+Transverse loads applied to a horizontal straight beam will cause it to deflect primarily up or down. This type of deformation is referred to as bending.
+
+A beam in bending will develop normal (tensile and compressive) bending stresses throughout the beam. The magnitude of these stresses depends on the internal bending moment in the beam as well as the beam's cross-section geometry.
+
+ Pure Bending
+
+
+
+Take a flexible strip, such as a thin ruler, and apply equal forces with your fingers as shown. Each hand applies a couple or moment (equal and opposite forces a distance apart). The couples of the two hands must be equal and opposite. Between the thumbs, the strip has deformed into a circular arc. For the loading shown here, just as the deformation is uniform, so the internal bending moment is uniform, equal to the moment applied by each hand.
+
+
+
+
+
+
+Bending diagram
+
+ -
+ Plane sections remain plane no shear stress/strains.
+
+ Therefore:
+ Also, traction free boundary conditions yields...
+
+ -
+ There is a Neutral axis between the top and the bottom where the length does not change.
+
+
+ -
+ The beam deforms into a circular arc where the top surface () is in compression , and the bottom surface () is in tension .
+
+ - Any point in the beam is in a state of uniaxial normal stress.
+ - Finding stresses is a statically indeterminate problem.
+
+
+
+
+ADD CONTENT HERE
+
+
+
+
+
+EDIT CONTENT PLACEMENT
+ Boundary Conditions
+
+Below are common support conditions and loading conditions that are used to model beams.
+
+Pin supports allow rotation but not translation; fixed supports restrain both translation and rotation of the beam at that location.
+
+Statically Determinate Beams
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+ Statically Indeterminate Beams
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+
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+ Loadings
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+Maximum normal stress
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+ Composite beams builds on this content.
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+
+ Recall that does not depend on the material properties of the beam, and is based only on the assumptions of geometry done so far.
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+ In non-homogeneous beams, we can no longer assume that the neutral axis passes through the centroid of the composite section. We should now determine that location…
+
+
+ After obtaining the TRANSFORMED CROSS SECTION, we get
+
+
+ Therefore, the neutral axis passes through the centroid of the transformed cross section.
+
+ Note: the widening () or narrowing () must be done in a direction parallel to the neutral axis of the section, since we want y-distances to be the same in the original and transformed section, so that the distance y in the flexural formula is unaltered.
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+From equilibrium: the centroid is located at , i.e., the neutral axis passes through the centroid of the section.
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+ To evaluate the maximum absolute normal stress, denoting “c” the largest distance from the neutral surface, we use:
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+ Note: the ratio I/c depends only upon the geometry of the cross section. This ratio is called the elastic section modulus and is denoted by S.
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+The magnitudes of stress and strain vary along the cross section. The magnitudes increase as the point of intrest moves away from the neutral axis. The maximum is at the point farthest away from the neutral axis.
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+This variation also changes as more modes are added.
+
+Elastic range: bending moment is such that the normal stresses remain below the yield strength.
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+Constitutive Relationship:
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+Force Equilibrium:
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+Moment Equilibrium:
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+ Stress due to eccentric loading found by superposing the uniform stress due to a centric load and linear stress distribution due to a pure bending moment.
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+ Validity requires stresses below proportional limit (elastic region), deformations have negligible effect on geometry, and stresses not evaluated near points of load application.
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+Transverse loading creates corresponding longitudinal shear stresses which will act along longitudinal planes of the beam
+
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+ADD CONTENT HERE
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+ When a shear force is applied, it tends to cause warping of the cross section. Therefore, when a beam is subject to moments and shear forces, the cross section will not remain plane as assumed in the derivation of the flexural formula. However, we can assume that the warping due to shear is small enough to be neglected, which is particularly true for slender beams ().
+
+
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+ Transverse loading applied to a beam results in normal and shearing stresses in transverse sections, as well as a bending moment.
+
+
+ The distribution of normal and shearing stresses satisfies the following where is the cross sectional area.
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+ When shearing stresses are exerted on the vertical faces of an element, equal stresses must be exerted on the horizontal faces.
+
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+ Longitudinal shearing stresses must exist in any member subjected to transverse loading.
+
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+ Symmetry of stress: transverse xy stress implies longitudinal yx stress.
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+Shear stress varies throughout the cross-section of the beam where the maximum shear stress is at the neutral axis and is zero at the edges. Shear stress can be represented as an average across the beam.
+
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+Shear stress can also be evaluated at any given point () from the z-axis.
+
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+
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+ (A) An element of width in a bending beam has a FBD of the bending moment stress distribution.
+
+ (B) An element distance away from the neutral axis has a FBD of the bending moment stress distribution For this element to be in equilibrium, must be present.
+
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+ (C) Simplified FBD. The width of this element is
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+ - : shear force
+ - : First area moment of inertia of cut section ( )
+ - : Second area moment of inertia of whole cross-sectional area
+ - : beam thickness
+
+
+Rectangular Beam
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+Maximum shear stress occurs at the neutral axis.
+
+
+
I-Beam
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+The neutral axis of an I-beam is the center. To find the I, use the parallel axis theorem. When determining the stress distribution, break the beam up into sub-sections of constant thickness for easier calculations. The stress distribution is a discontinuous parabola.
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+ shear stresses in thin-walled members builds on this content.
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+ The variation of shear flow across the section depends only on the variation of the first moment.
+
+
+
+ For a box beam, q grows symmetrically. From the edge to where the beam becomes hollow, the flow increases linearly and from where it becomes hollow to the neutral axis, the flow increases parabolically.
+
+
+ For a u-channel beam, the shear flow increases linearly on the sides from the edge to the base, and on the base it increases from the sides to the neutral axis.
+
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+
+In some cases, it is more useful to look at shear flow () through a structure than internal shear stress.
+
+L: Shear flow direction. R: Shear flow magnitude
+
+For an I-beam, the shear flow increases symmetrically on top from the end to the neutral axis in the y-direction, and on the side increases from the flanges to the neutral axes.
+
+Shear flow is the a measure of force per length.
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+Shear flow is most applicable in design situations such as built-up beams.
+Built-up beams are beams that have been put together with either an adhesive (glue, epoxy, etc) or fasteners (nails, bolts, etc). Calculating shear flow allows for determining how much adhesive or how often a fastener is needed.
+
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+Adhesives supply resistance along the length of the contacting parts. Determine the minimum shear strength at these contacting/weak points using the shear stress equation.
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+ The vertical shear is is .
+ Find the minumum required shear strength for the glue ().
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+ : shear force
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+ : First area moment of inertia of cut section ( )
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+ : Second area moment of inertia of whole cross-sectional area
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+
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+ : length of connected region
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+ Solve.
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+Fasteners supply resistance at fixed intervals. Use the shear flow formula to analyze the beam.
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+ The nails are at an interval of of . The shear force is .
+ Find the shear flow () and the load at each fastener ().
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+ Shear flow.
+
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+ Force in the nail.
+
+ Since the nails are uniformally distributed, each nail is assumed to supply a shearing force
+ that resists the shear flow in the beam. Along a single row of nails (refer to the free-body diagram above), the total force is:
+
+
+
+
+
+ Because there are two rows of nails, the total resistance load supplied by the nails is:
+
+
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+
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+ To find the shearing force in each nail, we equate the total force supplied by the nails with
+ the total shear flow acting across their total spacing, , to obtain:
+
+
+
+
+
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+
+
+
+ Internal forces of a member can be determined by creating an imaginary cut in a member, and then solving for the internal shear force, normal force, and bending moment (which, after the cut, become "external" forces that can be solved for).
+
+
+
+ To determine how the internal shear force and bending moment change throughout the member, shear force and bending moment diagrams are created. Shear force diagrams provide a graphical representation of the internal shear force within a member, and bending moment diagrams provide a graphical representation of the internal bending moment within a member.
+
+
+ Internal shear force (V), internal normal force (N), and internal moment (M).
+
+
+
+
+
+ Sign conventions for internal forces and moments
+ Sign conventions for internal forces and moments in 3D
+
+
+
+
+ Creating a shear force and bending moment diagram allows us to create a graphical representation of and as a function of the position along the beam, . Therefore when creating internal loading diagrams we are trying to write equations for and .
+
+
+
+ The general procedures for creating shear / bending moment diagrams are:
+
+ - Find support reactions
+ - Specify the x-coordinates
+ - Divide the beam into regions: one region for every change in loading
+ - Analyze each region by cutting inside each region and analyzing the free body diagram of the left side of the cut
+ - Apply equations of equilibrium to solve for \(V\) and \(M\) as functions of x
+
+
+
+
+
+
+
+
+
+ - When there is an external concentrated force or moment, there will be a "jump" in the shear force or bending moment diagram
+ - \(w(x)\), \(V\), and \(M\) are related via the following relationship:
+
+
+ Relationships between \(w(x)\), \(V\), and \(M\).
+
+
+
+
diff --git a/src/pages/sol/statically_indeterminate_problems.astro b/src/pages/sol/statically_indeterminate_problems.astro
new file mode 100644
index 000000000..9bc097017
--- /dev/null
+++ b/src/pages/sol/statically_indeterminate_problems.astro
@@ -0,0 +1,68 @@
+---
+import Layout from "../../layouts/Layout.astro"
+import Image from "../../components/Image.astro"
+import Section from "../../components/Section.astro"
+import SubSection from "../../components/SubSection.astro"
+import SubSubSection from "../../components/SubSubSection.astro"
+import InlineEquation from "../../components/InlineEquation.astro"
+import DisplayEquation from "../../components/DisplayEquation.astro"
+import BlueText from "../../components/BlueText.astro"
+import RedText from "../../components/RedText.astro"
+import Plotly from "../../components/Plotly.astro"
+import Itemize from "../../components/Itemize.astro"
+import Item from "../../components/Item.astro"
+---
+
+
-Dimensionless [mm/mm, in/in, or %]
+
+
+
+ Initial and final lengths of some section of the specimen are measured, perhaps by some handheld device such as a ruler. Axial strain computed directly by following formula:
+
+
+
+
+ Accurate measurements of strain in this way may require a fairly large initial length.
+
+
+
+
+
+
+ A clip-on device that can measure very small deformations. Two clips attach to a specimen before testing. The clips are attached to a transducer body. The transducer outputs a voltage. Changes in voltage output are converted to strain.
+
+
+ A tensile test in the Materials Testing Instructional Laboratory, Talbot Lab, UIUC
+
+
+
+
+
+Small electrical resistors whose resistance charges with strain. Change in resistance can be converted to strain measurement. Often sold as "rosettes", which can measure normal strain in two or more directions. Can be bonded to test specimen.
+
+ Rosette strain gauge arrangement and example
+
+
+
+
+
+A calibrated wire is set into vibration and its frequency is measured. Small changes in the length of the wire as a result of strain produce a measurable change in frequency, allowing for accurate strain measurements over relatively long gauge lengths.
+
+ Vibrating wire strain gauge attached to the side of a bridge
+
+
+
+
+
+
+
+Image placed on surface of test specimen. Image may consist of speckles or some regular pattern. Deformation of image tracked by digital camera. Image analysis used to determine multiple strain component.
+
+ Experiment set up. The diffuse light source consists of two fluorescent tube lights that produce white light, behind a translucent plastic sheet.
+
+ strain calculated through DIC of straight-curved specimen with an applied load of 114 N from TAM 456, UIUC.
+
+
+
-