Solid Mechanics

• Solid Mechanics
  • References
  • 3D Systems of Forces
    • Fundamental Vectors
    • How to find 3D moments
  • Trusses
    • Method of Joints
    • Method of sections
  • Basic Geometric Properties
    • Centroid
    • Moment of inertia
    • Plastic moment
    • Strain
    • Sign Convention
    • Relationship between V, BM, and w
    • Simple Areas
  • Shear
    • Complementary property of shear
    • The Shear Formula
    • Shear Flow
    • Shear Center
  • Axial
    • Statically indeterminate analysis
  • Bending
    • Bending moment diagram
    • 2D Axial Due to Bending
    • 3D Axial Due to Bending
    • Orientation of Neutral Axis
  • Torsion
    • Important formulas
    • Statically indeterminate
    • Thin-walled section
  • Combined Loadings
  • Deflection
    • Integration Approach
    • Moment-Area Method
  • Analysis of Stress
    • Uniaxial Stress
    • 2D Stress
    • 3D Stress
  • Analysis of Strain
    • Strain gauges
    • Stress-strain relationship
  • Strain Energy
  • Theories of Failure
    • Maximum Shear Stress Theory - Tresca Criterion
    • Energy Distortion Theory - Von Mises Criterion
    • Maximum Normal Stress Theory - Rankine's Theory
    • Additional Failure Theories
  • Failure of Columns
  • Energy Methods
    • Normal Stress Strain Energy
    • Shear Stress Strain Energy
  • Virtual Work
    • Uniaxial Virtual Work
    • Bending Virtual Work
    • Torsion Virtual Work
    • Other VW Applications
    • VW for Indeterminate Structure
    • VW using moment area approach
  • Frames
    • Frame Deflected Shape
  • Influence Line
  • 2D Design Examples

References

Course Notes

Textbook

Solutions 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

3D Systems of Forces

Fundamental Vectors

3D Force System = a situation in which there is triaxial loading. Forces are expressed as vectors, which are composed of smaller vectors.

$$\vec F = \vec{F_x} + \vec{F_y} + \vec{F_z} = F_x\cdot\hat{i} + F_y\cdot\hat{j} + F_z\cdot\hat{k}$$

$$\vec{F} = F\cdot\hat{u}$$

Coordinate Direction Angles = the angles between forces along specified axes and the overall force vector. $\alpha$ represents the angle between $\vec{F_x}$ and $\vec F$, $\beta$ represents the angle between $\vec{F_y}$ and $\vec F$, and $\gamma$ represents the angle between $\vec{F_z}$ and $\vec F$.

$$cos\alpha = \frac{A_x}{A},\ cos\beta = \frac{A_y}{A},\ cos\gamma = \frac{A_z}{A}$$

Position Vector = vector with two points in reference to each other rather than the origin. Algebraically, the position vector has $X = (X_b-X_a)$ etc.

$$\vec r_{AB} = (X_b - X_a)\hat{i} + (Y_b - Y_a)\hat{j} + (Z_b - Z_a)\hat{k}$$

Magnitude = numerical quantity of the force vector, found using the magnitudes of the forces along each respective axis.

$$|\vec{F}| = F = \sqrt{F_x^2 +F_y^2 + F_z^2}$$

Direction Unit Vector = vector indicating the direction of the force.

$$\hat{u} = \begin{bmatrix} cos{\alpha}\\ cos{\beta}\\ cos{\gamma}\\ \end{bmatrix} = \frac{\vec F}{F} = \frac{\vec r_{AB}}{r_{AB}}$$

How to find 3D moments

In order to find the moments, take the cross product between position vector and the force vector $\vec{r}\times\vec{F}$. This can be determined using the below.

$$\det{\begin{vmatrix} i & j & k \\ r_x & r_y & r_z \\ F_x & F_y & F_z \\ \end{vmatrix}}\tag{Moment Equation}$$

If you need the moment about a specified axis, you can use one of the below shortcut formulas.

$$r_yF_z - r_zF_y\tag{Moment about x axis}$$

$$r_zF_x - r_xF_z\tag{Moment about y axis}$$

$$r_xF_y - r_yF_x\tag{Moment about z axis}$$

Right Hand Rule = with the thumb curled in the direction of the specified axis, the fingers curl in the direction of positive moment along that axis.

Trusses

There are a couple big assumptions made in method of trusses:

1. All joints are pins (hinges)
2. All members are weightless

• If weight is accounted for, must put half of the member's weight as a vertical force on each end of the member

1. All loads are applied at joints only

• Accurate assumption for roof and bridge trusses

Method of Joints

The method of joints requires knowledge of a couple important concepts.

The first is, reactions at joints are opposite of that felt in the member itself. This is because a member needs to be in equilibrium ($\sum{F}=0$). In other words, if a member resists an external force, it will apply a force onto the joint in the direction that keeps the joint in equilibrium. However, then the joint applies an opposite force on the member to keep it in equilibrium.

Secondly, if there is a member that can apply a force at a joint that no other member can oppose, it is considered a zero-force member. This means that no force will be applied from it.

When evaluating each joint, they are typically assumed to be in tension to speed up the process. However, this is not necessary. With this assumption

Procedure of analysis (Example):

1. Find all external reactions
2. Start at a joint that has one unknown. Solve for forces along the x direction and y direction.
3. After solving for force along member, mark with arrows on the member whether it is in compression or tension.
4. Continue along truss until all forces are known.

In order to solve for full forces in one step, multiply force by unit vector in direction of analysis and solve for actual force.

Method of sections

• Used to save time
• One can only cut a maximum of 3 members at a time

Procedure for analysis (Example):

1. Find all external reactions
2. Take a cut such that it is possible to solve for an unknown, drawing forces assumed in tension
3. Solve for unknown. Repeat cuts untin necessary information found

Basic Geometric Properties

Centroid

$$\bar x = \frac{\sum(A\prime \bar x\prime)}{\sum(A\prime)} \tag{X axis centroid}$$

$$\bar y = \frac{\sum(A\prime \bar y\prime)}{\sum(A\prime)} \tag{Y axis centroid}$$

$$\bar x = \frac{\int_A \tilde x dA}{\int_A dA} \tag{Centroid of Area}$$

$$\bar x = \frac{\int_V \tilde x dV}{\int_V dV} \tag{Centroid of Volume}$$

The geometric centroid identifies the plane in which the elastic neutral axis is found. This is different than the plastic neutral axis which is determined 100% by area. For completely symmetric cross-sections, these axes are the same.

Moment of inertia

Moment of inertia = provides resistance against changing the rotational speed of rotating body. In a sense, it is analagous of mass in rotational situations. Units in form eg. $lb\cdot in^2$. dM represents small quantity of mass, and r represents distance of small mass from the axis.

There are two forms of moment of inertia: mass and area. These are often confused, but they are relatively simple to distinguish.

If someone says "moment of inertia", they are referring to mass, or first moment of inertia. This is referred to by I in calculations.

$$I = \int r^2 dM \tag{Moment of inertia}$$

$$I = I_{cm} + md^2 \tag{Parallel Axis Theorum Mass}$$

While the MOI equation is cool, MOIs are found on wikipedia for common shapes.

If someone says "second moment of inertia", they are referring to area. This is used in beam design, with units of $mm^4$.

$$I_{xx} = \int y^2 dA \tag{Second Moment of inertia X Axis}$$

$$I_z = I_x + Ar^2 \tag{Parallel Axis Theorum Area}$$

Circular sections have a polar moment of inertia, J, derived from the second moment of inertia (above). By integrating this with polar coordinates between 0-2$\pi$, and knowing that $\rho$ is the radius, the below is true

$$J = \frac{\pi c^4}{2}\tag{Polar MOI Solid Section}$$

$$J = \frac{\pi}{2}\cdot (C_o^4-C_i^4)\tag{Polar MOI Hollow Section}$$

The polar moment of inertia is also known as $I_z$, and is the sum of $I_x$ and $I_y$. Thus, if you want to find the $I_y$ of a hollow, circular section, you can either divide $J$ by two or integrate the second MOI equation between 0-$\pi$.

Plastic moment

The plastic moment, Mp is the moment at which the entire cross-section has reached it's yield stress. It is either calculated about the plastic neutral axis $c$ or about the top.

A couple of assumptions are made before Mp is calculated.

1. The top of the cross-section is assumed in compression, the bottom in tension
2. Since yielding occurs @ $\sigma_y$, the stress is assumed to be uniformly so (100% yielded)

Procedure for analysis (Example):

1. Assume a location for the plastic neutral axis
2. Plot strain and stress diagram next to cross-section
3. Determine expressions in terms of $c$ for each respective segment of cross-section
4. Use equilibrium equations to isolate for $c$
5. Check to verify that $c$ was assumed in the correct location
6. Once $c$ is known, use it to find resultant forces
7. Solve for Mp

Strain

$$\varepsilon = \frac{\Delta L}{L_1}\tag{Longitudinal Strain}$$

$$\gamma = \theta\cdot\rho\tag{Shear Strain}$$

• maximum strain when $\rho$ is equal to c, or the point at the surface of rod

$$E = \frac{\sigma}{\varepsilon}\tag{Young's Modulus}$$

$$G = \frac{\tau}{\gamma}\tag{Shear Modulus}$$

• both of the moduli are a ratio between stress and strain

Sign Convention

Relationship between V, BM, and w

Given the above, the area load can be assumed to be uniform even if it is not. Using $\sum{F-y}=0$, it is found that $V-(V+dV)\omega dx = 0$. This is rearranged to:

$$\frac{dV}{dx}=-\omega$$

In other words. the slope of the shear force diagram is the negative distributed load intensity at that point. Now, when solving moments, the equation $Vdx - dM - \frac{\omega dx^2}{2}=0$ is found. Since dx is so small, it's square is assumed zero. Rearranging yields

$$\frac{dM}{dx}=V$$

In other words, the BMD slope is equal to the shear force at that point

Simple Areas

Shear

Complementary property of shear

$$\tau_{xy} = \tau\prime_{xy} = \tau_{yx} \tag{Complementary Property of Shear}$$

The Complementary Property of Shear means that every face of the cube element must be in equilibrium. Hence, the opposite side of the cube experiences the opposing direction of shear. $\tau_{xy}$ refers to shear originating at the x-axis and going in the direction of the y-axis.

While this can be useful, how does one find the average maximum shear stress? This is found along the neutral axis using the shear formula.

The Shear Formula

$$\tau = \frac{VQ}{It} \tag{Shear Formula}$$

• Q is geometric property taken about the neutral axis. It represents the sum of the areas above a specified point multiplied by the distance between their centroids and the neutral axis.
  • Since Q contains a factor of $\bar{y}$, it is parabolic, as it maximizes in the middle and minimizes at extrema.
• I is MOI of entire cross-section about neutral axis
• t is thickness just above point
• V = cross-sectional shear

Note: there are a few assumptions that make this formula's use limited. The formula assumes that shear stress is uniform a cross section's thickness. This is not the case. $\tau$ increases parabolically towards its extremities. This is gets worse with b/h ratios greater than 0.5. Fortunately, for webs of wide-flange sections, this is very accurate. However, it is inaccurate for flanges of wide-flange sections.

Shear Flow

Shear flow is the force per unit length that nails or glue are designed to resist. Shear flow plots are most valid for thin sections, as the shear can then be assumed to be uniform across the cross-sectional thickness.

$$q = \frac{VQ}{I} \tag{Shear Flow}$$

• V = shear force
• I = MOI about entire cross-section
• Q = $\sum{\bar{y}\cdot A}$ of either the hatched region or the white region. See below:

In order to find the force applied along a part of a cross section, use the below formula

$$F = \int_0^s{q ds}$$

There are few important properties of shear flow:

• Shear flow is zero at endpoints. This is simply because in the calculation, there is no area.
• Shear flow increases/decreases linearly across a horizontal. This is because the $\bar{y}$ is constant across a horizontal, but the area changes linearly across.
• Shear flow is equal on both sides where direction shifts. In other words $q_1 = q_2$ where a horizontal transitions to a vertical component of a member.
  • If a junction is present in an I bar or something similar where there is shear flowing from 2 sides, the vertical component adds them both.
• Shear flow is parabolic on verticals. This is because as $\bar{y}$ decreases, the area increases, leading to a quadratic.
• Shear flow maximizes at the neutral axis
• Shear flow cannot be calculated across the N.A. The cut must end at the N.A.
• Shear flow enters the cross section above the N.A. and exits it below the N.A.
• Remember: $\sum{F}=0$ and $\sum{F_y}=V$

How to design according to shear flow:

$$q = \frac{F_{bolt}}{nail spacing (s)}\tag{Design for nails}$$

$$q = \tau_{glue}\cdot{b_{glue}(thickness)}\tag{Design for glue}$$

Shear Center

The shear center, $e$, is the location along the neutral axis where $\sum{M} = 0$. It can be solved either by taking a moment about the point, or by taking a moment another point. If the latter is used, the force applied at e is V.

Axial

$$\sigma = E\cdot\varepsilon\tag{Axial stress}$$

Statically indeterminate analysis

$$\delta^{F_x} = \sum{\frac{F_{x}L}{AE}} \tag{Force Deformation Formula}$$

$$\delta^{T} = \alpha\Delta{T}L \tag{Thermal Deformation Formula}$$

• $\alpha$ is the linear coefficient of thermal expansion, and is unique to the material

Procedure for indeterminate analysis:

1. Allow for uncontrolled thermal expansion
2. Statics. For these scenerios, DO NOT ASSUME TENSION
3. Combatibility equations. If there is a gap, include it here
4. Constitutive law
5. Solve equations

Bending

Bending moment diagram

Method 1: Shear method

1. Find reactions
2. Draw shear force diagram
3. Find area of shapes.. point load translates to a rectangle, uniform area load translates to triangle, changing area load translates to parabola. BMD indicates moment as follows: if the shear is positive and increasingly positive, the moment will be positive with increasing slope. If shear is positive but increasingly negative, the monment will be positive with decreasing slope. ETC.

Method 2: Point method

• Remember: moment at the free end of a cantilever is zero because it is static

At Waterloo, we learn the Canadian Method. In this sense, you are finding the moment being resisted by it internally, not applied to it. In this sense, use sign convention for if it is a left cut or a right cut to determine if the resisting moment is positive or not. This must be used in order to fulfill the statement $M = \int{V}$ with positive sign convention. If confused, see 11-4

At other schools, they may learn the British Method, which is the moment being applied to the beam. This, in many ways, makes more sense.

1. Choose an end and graph the moment at that point
2. Take cuts at points, find moment about the points.

While one could easily find moment about the points using the Canadian Method, Ben's method is quicker. Here is Ben's method: if the cut is a right cut, internal M = + clockwise moment being applied to it. If cut is a left cut, internal M = + counterclockwise moment being applied to it.

For design, it is important to know that maximum bending moment occurs where the shear is zero. This is because shear is the derivative of bending moment, and when the moment is at a maximum (slope = 0), the shear is equal to zero. For a simply supported beam, the maximum bending moment can be calculated at mid-span, since shear is zero there. The below formula simplifies this calculation:

$$M_{mid} = \frac{wL^2}{8}$$

• w is uniform load along beam. If multidirectional load, break w down into $w_y$ and $w_z$ to solve for $M_z$ and $M_y$ respectively

2D Axial Due to Bending

$$\sigma_{x} = -\frac{My}{I} \tag{Simple Flexture Formula}$$

• M = internal moment at specified point
• y = distance from centroidal axis
• I = MOI of cross-sectional area about the neutral axis

The flexture formula simply tells the normal stresses due to moment at any point y from the neutral axis. It is known that normal stress due to bending changes linearly along the cross-section, and it is treated as such.

The negative sign is used because normal sign convention suggests the top in compression and the bottom in tension. Thus, with a negative y on the bottom, the negatives would cancel out to yield a positive normal force. The maximimum force, $\sigma_{max}$, is found at the point furthest from the centroidal axis

3D Axial Due to Bending

$$\sigma_{x} = -\frac{M_{z}\cdot y}{I_{z}}+\frac {M_{y}\cdot z}{I_{y}}\tag{Complete Flexture Formula}$$

• The complete flexture formula assumes a positive sign convention when the top left corner is in tension, according to the above axis.
• The Right Hand Rule can help decipher tension or compression. Point your thumb in the direction of the axis above. Your fingers will curl in the direction that the moment curls. Where the top of the moment arrow exists, compression exists.

One can use the complete flexture formula to plot the stress-distribution along the cross-section.

Procedure to plot stress distribution:

1. Find unknowns in complete flexture formula ($M_z$,$M_y$,$I_z$,$I_y$
2. Find coordinates of corners of cross-section
3. Set-up flexture formula with knowns from step 1. Plug-in each coordinate to find stress at that point.
4. If stress due to axial is present, add it to all the calculated values.
5. Draw distribution. Start at the first coordinate and work your way around. If the stress value is negative, this means it is in compression. Thus, the arrows under the triangle will point towards the component of the cross section

• At a single point, stress is the same magnitude and form vertically and horizontally, similar to the complementary property of shear.

Orientation of Neutral Axis

In order to find the orientation of the neutral axis, you must first be able to plot the $\sigma_{x}$ distribution along the cross-section. It is this force, which is inherintly caused by multi-axial moments, that causes the change in N.A. because the N.A. axis exists along the plane between where the $\sigma_{x}$ is equal to zero. Hence, a sketch can really help when figuring out the orientation of the axis. If this is understood, here is the derivation of the neutral axis change.

Important variables: $\theta$ is the angle measured from the positive z-axis clockwise to the $M_{R}$, $M_{z} = M_{R}\cos\theta$, $M_{y} = M_{R}\sin\theta$, $M_{R} = \sqrt{M_z^2+M_y^2}$, and $\alpha$ is the angle of the N.A.

We know that at the neutral axis, axial stress due to bending is equal to zero. Thus, by setting the Complete Flexture Formula equal to zero, you will isolate for the below

$$\frac{y}{z} = \frac{M_y\cdot I_z}{M_z\cdot I_y}$$

Knowing the above variables, and since we know that the angle between y and z is $\alpha$, we can say

$$\tan\alpha = \frac{I_z}{I_y}\tan\theta$$

Torsion

Important formulas

$$\tau_{T} = G\cdot\gamma\tag{Shear Stress}$$

$$\tau_{T} = \frac{T\rho}{J}\tag{Shear Stress}$$

• very similar to flexture formula

$$\phi_{\frac{C}{A}} = \frac{1}{G\cdot J}\sum{TL}\tag{Angle of Twist}$$

$$\theta = \frac{T}{GJ}\tag{Angle of Twist per length}$$

Statically indeterminate

Procedure to solve statically indeterminate problem (Example):

1. Create a combatibility equation
2. Set up the calculation
3. Take cuts to find torque
4. Solve for polar moment of inertia
5. Solve equation for unknowns
6. Sketch T vs x graph
7. Solve for maximum shear stress
8. Draw stress circle to reveal findings

Thin-walled section

The method of thin-walled sections is special. In its case, $\tau$ is the average shear stress at the mid-thickness of the wall. Since larger thickness reduces the accuracy of average shear, only thin-walls are allowed. There are 2 requirements for thin-wall analysis:

1. Non-circular section
2. Closed section

Below are some specialized formulas:

$$\tau = \frac{T}{2\cdot t\cdot Am}\tag{Shear Stress}$$

$$J = \frac{4\cdot Am^2}{\oint_0^s{\frac{1}{1}ds}}\tag{Polar MOI}$$

$$\phi = \sum{\frac{T\cdot L}{4\cdot Am^2\cdot G}\oint_0^s{\frac{1}{t}ds}}\tag{Angle of Twist}$$

$$q = \frac{T}{2Am}\tag{Shear Flow}$$

• Am is simply the area of the midsection of the cross section, or halfway across the thickness
• The line integral is the sum of lengths along the outline of the Am divided by their respective thicknesses

Combined Loadings

Combined loading analysis is simply 3d analysis. It analyzes $\sigma_x$ from axial and bending, as well as $\tau$ due to shear and torsion.

Deflection

Integration Approach

The integration method often involves one or two cuts, about which one will find an equation for M(x) at that point. However, if one is given the area load $\omega$, no cut is necessary.

$$EI\frac{d^4 y}{d x^4}= -\omega\tag{Area Load Equation}$$

$$EI\frac{d^3 y}{d x^3}=V\tag{Shear Equation}$$

$$EI\frac{d^2 y}{d x^2}=M_O\tag{Moment Equation}$$

$$EI\theta(x) = \int{M_O dx} + C\tag{Slope Equation}$$

$$EIy(x)=\int{\int{M_O+C_1dx}} + C_2\tag{Deflection Equation}$$

Procedure to solve using integration method (Example):

1. Check for points where slope or deflection is known.
   • Deflection is typically zero wherever there is a support
   • Slope has a negative/positive relationship where cuts meet.
2. Solve for reactions at supports
   • Includes both moments and loads at supports.
3. Take a cut any distance x between an end of the beam and a reaction and find Mo equation.
   • Every change in loading requires another cut
4. Integrate equations according to the following equation.
5. Plug in knowns and solve.

Moment-Area Method

1st Moment Area Theorum: the change in slope from A to B on the elastic curve equals the area under the $\frac{M}{EI}$ diagram between A and B

$$\theta DA = \theta_B-\theta_A = \int^D_A{\frac{M}{EI}}dx\tag{1st MA Theorum}$$

• $\theta DA$ is the area under the curvature diagram

$$\theta(A) = \frac{\Delta DA}{L_{beam}}\tag{General Curvature Slope}$$

2nd Moment Area Theorum: The tangential deviation of a point B from the tangent drawn at another point A equals the moment area of the $\frac{M}{EI}$ diagram bound by these points A and B about an axis through the first point B

$$\Delta DA = \int^D_A{\frac{M\bar{x}}{EI}}dx\tag{2nd MA Theorum}$$

• It is the moment-area under the curvature diagram
• $\Delta DA$ is the deflection from point tangential to D to point tangential to A. Otherwise, $\Delta AD$ is the deflection from a point tangential to A to a point tangential to D. The point of origination comes first.

Procedure to find maximum deflection using moment-area method:

1. Create bending moment diagram
   • Do not do shear force diagram first, takes too much time
   • Canadian method shows moment applied to the point externally. British method shows moment resisted at internal point. For courses sake, use Canadian method.
2. Draw deflection diagram below moment diagram
   • Maximum deflection will be on the side with the longest section of beam. This point will be named D for instance
3. Find curvature of endpoint with respect to starting position (eg. A)
4. Find slope at A by dividing curvature of length over the length
5. Knowing that the $\theta A$ is equivalent to $\theta DA$, it is okay to set $\theta A$ equal to the area of moments between the point of maximum deflection and an endpoint in an effort to discover where the maximum deflection occurs.

Analysis of Stress

Uniaxial Stress

Based on the above, several formulas can be derived:

$$A_{cut}=\frac{A}{cos\theta}$$

$$N=Pcos\theta\tag{Normal Force}$$

$$F=Psin\theta\tag{Shear Force}$$

Given these, the below can also be derived:

$$\sigma=\sigma_x\cos^2{\theta}\tag{Normal Stress}$$

$$\tau=\frac{1}{2}\sigma_x\sin{2\theta}\tag{Shear Stress}$$

Now, since $\sin{(90)}$ is 1, $\tau$ is maximized when $\theta = 45°$. This plane is called Lüder's Band, and proves the following:

$$\tau_{max}=\frac{1}{2}\sigma_x\tag{Max Shear Stress}$$

2D Stress

Plane stress: one in which the stresses at a point in the body act in one plane

Based on a force cube, it can be proven using moments that $\tau_{xy}$=$\tau_{yx}$, as long as it is in the linear elastic isotrophic region. However, inclined planes present a slightly larger challenge.

Using equilibrium equations, the below can be proven.

$${\sigma_x\prime}=\frac{\sigma_x +\sigma_y}{2}+\frac{\sigma_x -\sigma_y}{2}\cos{(2\theta)} + \tau_{xy}\sin{(2\theta)}\tag{X Normal Stress}$$

$${\sigma_y\prime}=\frac{\sigma_x +\sigma_y}{2}-\frac{\sigma_x -\sigma_y}{2}\cos{(2\theta)} - \tau_{xy}\sin{(2\theta)}\tag{Y Normal Stress}$$

$$\tau_{x\prime y\prime}=-\frac{\sigma_x -\sigma_y}{2}\sin{(2\theta)} + \tau_{xy}\cos{(2\theta)}\tag{Shear Stress}$$

$$\sigma_{ave}=\frac{\sigma_x + \sigma_y}{2}\tag{Average Normal Stress}$$

There are 2 planes, upon which the shear stress is 0. They are known as the Principal Planes, and are separated by $90°$. These planes can be calculated by $\theta = \frac{1}{2}\arctan{\frac{2\tau_{xy}}{\sigma_x - \sigma_y}}$ and the other is + $90°$. Interestingly enough, the normal stresses are also maximized/minimized along these planes. Thus, they are known as the principle stresses, and are denoted by $\sigma_1 = \sigma_{max}$ and $\sigma_2 = \sigma_{min}$.

Mohr's circle shows this relationship graphically. By setting the center of the circle as the average normal stress, and the below equation for the radius, Mohr's Circle can be plotted (Stress/Strain Example)

$$R = \sqrt{(\frac{\sigma_x - \sigma_y}{2})^2 + \tau_{xy}^2}$$

• $\tau$ is positive down. This can be determined based upon upon whether or not the shear causes a CCW moment on the force cube
  • $\tau_{max}$ is found at $45°$ to the principle planes. It is also the radius of the circle, and can be found by calculating $\frac{\sigma_1 - \sigma_2}{2}$. It is important to understand that this is the maximum in the x-y plane only
• $\theta$ is positive CCW. This is illustrated in the below

3D Stress

• assumes that $\sigma_1$ parallel to x-axis, $\sigma_2$ parallel to y-axis, and $\sigma_3$ parallel to z-axis.
• all locations of $\tau_{max}$ occur at $45°$

To determine the absolute maximum shear, one needs to be observant of the large radius of the 3D Mohr's Circle. This radius is equal to the absolute shear.

• $\sigma_1 > \sigma_2 > \sigma_3$, $\tau_{abmax} = \frac{\sigma_1 - \sigma_3}{2}$
  • $45°$ to xy plane
• $\sigma_1 > \sigma_2$, $\sigma_3 = 0$, $\tau_{abmax} = \frac{\sigma_1}{2}$ on xz plane
  • $45°$ to xy plane
• $\sigma_1$ is tensile, $\sigma_2$ is compressive $\sigma_3 = 0$, $\tau_{abmax} = \frac{\sigma_1 - \sigma_2}{2}$
  • $45°$ to xz plane

Analysis of Strain

Plane strain only: $\varepsilon_x$, $\varepsilon_y$, $\gamma_{xy}$

• $\varepsilon_z$, $\gamma_{xz}$, and $\gamma_{yz}$ = 0
• strains marked by rotation look like $\varepsilon_x\prime$

Strain can be determined strikingly similar to the way stress can be determined. The below Stress Transformation Equations outline this:

$$\varepsilon_x\prime = \frac{\varepsilon_x + \varepsilon_y}{2} + \frac{\varepsilon_x - \varepsilon_y}{2}\cos{2\theta} + \frac{\gamma_{xy}}{2}\sin{2\theta}\tag{Normal Strain X}$$

$$\varepsilon_y\prime = \frac{\varepsilon_x + \varepsilon_y}{2} - \frac{\varepsilon_x - \varepsilon_y}{2}\cos{2\theta} - \frac{\gamma_{xy}}{2}\sin{2\theta}\tag{Normal Strain Y}$$

$$\frac{\gamma_{x\prime y\prime}}{2} = - \frac{\varepsilon_x-\varepsilon_y}{2}\sin{2\theta}+\frac{\gamma_{xy}}{2}\cos{2\theta}\tag{Shear Strain}$$

Analysis can be performed much easier using the Mohr's Circle for strain. In a much similar way to the stress equivalent, it is plotted with average normal strain as the center and a radius calculated below:

$$\varepsilon_{avg} = \frac{\varepsilon_x + \varepsilon_y}{2}$$

$$R = \sqrt{(\frac{\varepsilon_x-\varepsilon_y}{2})^2 + (\frac{\gamma_{xy}}{2})^2}$$

• the sign of the principal angle for strain is dependent on it's relation to the x-axis. If it is CW, it is negative.

Strain gauges

The stress transformation equations are written differently for strain gauge analysis:

$$\varepsilon_x\prime = \varepsilon_x\cos^2{\theta} + \varepsilon_y\sin^2{\theta} + \gamma_{xy}\sin{\theta}\cos{\theta}$$

$$\varepsilon_y\prime = \varepsilon_x\sin^2{\theta} + \varepsilon_y\cos^2{\theta} - \gamma_{xy}\sin{\theta}\cos{\theta}$$

$$\gamma_{x\prime y\prime} = -2(\varepsilon_x - \varepsilon_y)\sin{\theta}\cos{\theta} + \gamma_{xy}(\cos^2{\theta} - \sin^2{\theta})$$

One can solve for the normal strains by creating three of equation 1 above; 1 for each gauge respectively.

• using the CAST rule, one can determine if an angle is negative or positive
• if one of the gauges is along an axis, it is known to be the strain along that axis
• afterwards, plug determined values into Mohr's Circle for strain to solve for principal strain

Stress-strain relationship

When stress is applied in the x-direction, $\varepsilon_x = \frac{\sigma_x}{E}$ and $\varepsilon_y = \frac{-\nu\sigma_x}{E}$. Due to this relationship, which is variable-swapped with a stress in the y-direction, the below relationship is true for situations where there is both $\sigma_x$ and $\sigma_y$.

$$\varepsilon_x = \frac{\sigma_x}{E} - \frac{\nu\sigma_y}{E}$$

$$\varepsilon_y = \frac{\sigma_y}{E} - \frac{\nu\sigma_x}{E}$$

Thus, Hooke's law can be examined in the 2D case by multiplying both sides by E, such that $E\varepsilon_x = \sigma_x - \nu\sigma_y$ and $E\varepsilon_y = \sigma_y - \nu\sigma_x$

Similarly, when shear stress is present, and it is assumed that it doesn't affect the normal x and y strains and that normal stresses have no affect on shear strain, the below is true, which is another 2D Hooke's law.

$$\gamma_{xy} = \frac{\tau_{xy}}{G}$$

Now in the 3D case, stresses in the z direction must also be examined. This is seen below, and can be altered with variable swapping according to the plane:

$$E\varepsilon_x = \sigma_x - \nu{(\sigma_y + \sigma_z)}$$

$$G\gamma_{yz} = \tau{yz}$$

• at $\sigma_1$ and $\sigma_2$, the shear stress is equal to zero, and there is thus no shear strain

Principal strains occur in the direction parallel to principal stresses in 3D.

Below are some fundamental formulas to relate material properties.

$$\frac{E}{G} = 2(1+\nu)$$

$$\nu = \frac{Fr^2\phi}{4T\delta} - 1$$

Strain Energy

When an elastic body is deformed through the action of external forces, recoverable energy is stored within the body in the form of strain energy. Since energy is conserved, external work = internal work.

$$W=\int_0^{\delta_1} Pd\delta = u\tag{Work}$$

In the case of a spring, we let $\frac{K}{1} = {P}{\delta}$. Through integration, the below is true:

$$u=\frac{1}{2}P\delta\tag{Linear Spring}$$

In the uniaxial case, by letting $\varepsilon = \frac{\delta}{L}$, $\sigma=\frac{P}{A}$, and $Ld\varepsilon = d\delta$, work is found to be $W=\int_0^{\varepsilon} \sigma d\varepsilon$. By dividing by the volume, we can solve for strain energy density. This is also known as the strain energy per unit volume.

$$u = \int_0^{\varepsilon} \sigma d\varepsilon\tag{Strain Energy Density}$$

When integrated, this yields the below

$$u = \frac{1}{2}\sigma\varepsilon$$

In the event of pure shear stress, the below is true

$$u=\frac{1}{2}\tau\gamma$$

Now, in the event of 3D stress states, there should be xyz components of $\sigma$, $\tau$, $\varepsilon$, and $\gamma$ respectively. Assuming Hooke's law is obeyed, the strain energy density can be considered as a result of all of these effects, as seen below:

$$u=\frac{1}{2}[\sigma_x\varepsilon_x + \sigma_y\varepsilon_y + \sigma_z\varepsilon_z + \tau_{xy}\gamma_{xy} + \tau_{yz}\gamma_{yz} + \tau_{xz}\gamma_{xz}]\tag{3D Energy Density}$$

It is important to recognize that strain energy density is a scalar quantity, and is independent of xyz axes. It is then much easier to analyze using the principle directions, as there would be no shear stress or strain. Assuming shear is zero, the 3D Hooke's laws can be substituted into the 3D Energy Density Equation, making it only a function of $\sigma$ and not of $\varepsilon$. See below:

$$u = \frac{1}{2E}[(\sigma_1^2 + \sigma_2^2 + \sigma_3^2)-2\nu (\sigma_1\sigma_2 + \sigma_2\sigma_3 + \sigma_1\sigma_3)]$$

• the first part of this equation is associated with $\Delta$ in volume
• the second part of this equation is associated with distortion, or change in shape

In the figure, stress state (ii) causes volumetric change, but no change in orientation of axes. By letting $\sigma_{avg}=\frac{\sigma_1 + \sigma_2 + \sigma_3}{3}$, the following can be derived:

$$u_v = \frac{1-2\nu}{6E}[\sigma_1 + \sigma_2 + \sigma_3]^2\tag{Volumetric Strain Energy}$$

State (iii) causes no volumetric change, but it does cause distortion. There are two formulas for this: 3D and plane stress

$$u_d = \frac{1+\nu}{6E}[(\sigma_1-\sigma_2)^2 + (\sigma_2-\sigma_3)^2 + (\sigma_3-\sigma_1)^2]\tag{3D Distortion Strain E}$$

$$u_d = \frac{1+\nu}{6E}[\sigma_1^2 - \sigma_1\sigma_2 + \sigma_2^2]\tag{Plane Stress Distortion Strain E}$$

• in the event that the shear modulus $G$ is known, $\frac{1+\nu}{6E}$ can be replaced with $\frac{1}{12G}$

Theories of Failure

There are many causes of failure. The most common are:

• yielding of ductile material (steel) $\sigma_f=\sigma_u$
• failure of brittle material (concrete) $\sigma_f=\sigma_y$
• excessive elastic or plastic deformation
• buckling in columns. This is often rapid and without warning

Most information related to material strength is obtained from tension tests (metals) and compression tests (brittle materials). Because it is time-consuming to analyze combined stresses using multiple tests, they are calculated using stress theories

Maximum Shear Stress Theory - Tresca Criterion

This theory agrees with ductile tests, is simple, and is based on the observation that yielding of ductile materials are initiated by slipping along Lüder lines. The tresca hexagon is an important reference, as no failure occurs within the hexagon.

• if plane stress or 3D with that $\sigma_3 = 0$ and the normal streses have the same sign, $\tau_{max} = \frac{\sigma_y}{2}=\frac{\sigma_1}{2}$
• if $\sigma_3 = 0$ and the stresses have opposite signs, $\tau_{max} = \frac{\sigma_1 - \sigma_2}{2}$

Energy Distortion Theory - Von Mises Criterion

This theory is in favour of ductile materials, and is based on failure occuring when the strain E of distortion at any point in a stressed body reaches the same value of failure from the tensile test

• in the case where $\sigma_3 = 0$, the equations form an ellipse equation. Failure cannot occur within the ellipse. There is also another 3D equation for the event that $\sigma_3\neq{0}$

$$\sigma_1^2 + \sigma_2^2 - \sigma_1\sigma_2 = \sigma_y^2\tag{Plane Stress Von Mises}$$

$$(\sigma_1 - \sigma_2)^2 + (\sigma_2 - \sigma_3)^2 + (\sigma_1 - \sigma_3)^2 = 2\sigma_y^2\tag{3D Von Mises}$$

Maximum Normal Stress Theory - Rankine's Theory

This theory predicts failure when the maximum principal normal stress reaches the value of the axial failure strss in tension or compression test. It is in favour of brittle materials, and allows any combination of $\sigma_1$ and $\sigma_2$ within Rankine's box to be safe. Perhaps at a fault, it assumes failure in tension and compression occur at the same stress

It is important to recognize that in uniaxial load, all 3 theories yield the same failure prediction. Otherwise, Von Mises and Tresca are similar, but Rankine is much different

Additional Failure Theories

Mohr's Failure Criterion predicts brittle failure, and requires a tension, compression, and a shear test to be performed. It can be expensive

Saint-Venant Theory predicts failure when the largest principal normal strain reaches $\varepsilon_{lim}$ in the axial tensile test. However, since Poisson's ratio causes tension to reduce strain in the perpendicular direction, it overestimates the strength of the material

Maximum Total Energy Theory considers failure when the total strain energy exceeds that of its predicted yield energy. However, it doesn't agree with hydrostatic experiments

Failure of Columns

Short stocky columns fail by crushing from compression, when stress $\frac{P}{A}$ exceeds limit value

Long, slender columns fail suddenly from a change in configuration shape once $P = P_{crit}$

What is instability?

Stable equilibrium: a small horizontal disturbance acting on the ball will not cause excessive lateral displacement

Neutral equilibrium: an infinite number of adjacent equilibrium states exist

Unstable equilibrium: a slight lateral disturbance will cause large movement of the ball from its original equilibrium state

• this is similar to axial loaded columns

When $P<P_{crit}$, it is straight and stable. If $P=P_{crit}$, moment will develop so that it is straight but not stable. If $P>P_{crit}$, large lateral deflection will occur so that it is unstable. This is necessary so that all configurations remain in equilibrium.

Given that $\frac{M}{EI}=\frac{d^2v}{dx^2}$ and that the moment caused at a cut is force applied times the deflection distance ($P\cdot{v}$), the general solution is solved below:

$$v=A\cos{\sqrt{\frac{P}{EI}}}x + B\sin{\sqrt{\frac{P}{EI}}}x\tag{Deflection}$$

• A and B integration constants
• B and A can be zero when 0=x or L=x, due to boundary conditions. In this case, there is no deflection and the column remains straight
• if one is zero but not the other, it is indeterminate and B (or A) can take on any value

$v = B\sin{\frac{P}{EI}}x$ is only possible when the A term is equal to zero. Thus, by setting it equal to zero and solving for P, there are many critical values for P. The lowest of these is below:

$$P_{crit} = \frac{n^2\pi^2EI}{L^2}\tag{Euler Formula}$$

• The case of one node (n) is when then there are only the end supports. However, if there are any supports in the middle, each of these is considered an additional node. Therefore, each additional support increases the $P_{crit}$ significantly. It is clear that n=1 is most common
• Check both $I_x$ and $I_y$. Buckling will occur about the weak axis

By dividing the critical P by the area, the critical buckling stress can be derived, as below:

$$\sigma_{crit} = \frac{\pi^2EI}{\frac{A}{I}L^2}\tag{Critical Buckling Stress}$$

$\sqrt{\frac{I}{A}}=r$ is known as the radius of gyration. When subbed in, it yields

$$\sigma_{crit}=\frac{\pi^2E}{(\frac{L}{r})^2}$$

• $\frac{L}{r}$, E, and $\sigma_y$ are normally given in the steel handbook

The radius of gyration is the distance such that if the object's mass were concentrated at r, it would yield the equivalent MOI as the original object. $I=Ar^2$

• The effective length is the length of a pinned-pinned column of the same EI value that would yield the same critical load. It is dependent on the boundary conditions. If the body is not supported by a pin-pin system, the critical load will be affected by the effective length, and calculated by the below:

$$P_{crit} = \frac{\pi^2EI}{L_e^2}$$

• $\sigma_{crit}$ is altered by the $L_e$ as well. $\frac{L_e}{r}$ is known as the slenderness ratio

While it can sometimes be said that the P is loaded in the centre of the column, it is more often shifted left or right. This causes a moment due to eccentricity e.

The moment caused is equal to $M=-P(v+e)$. Now, since the maximum v is found in the centre of the beam, the 2 different expressions of v can be derived:

$$v_{max} = e[\sec{(\sqrt{\frac{P}{EI}}\cdot{\frac{L}{2}})}-1]$$

$$v_{max} = e[\sec{(\sqrt{\frac{P}{P_{crit}}}\cdot{\frac{\pi}{2}})}-1]$$

Now, it is known that the maximum stress is a sum of axial and bending, such that $\sigma = \frac{P}{A} + \frac{Mc}{I}$. Knowing that $M = P(v_{max} + e)$ and that $\frac{I}{A} = r^2$, the below can be derived:

$$\sigma_{max} = \frac{P}{A}[1+\frac{ce}{r^2}\sec{(\frac{L}{2}\cdot{\sqrt{\frac{P}{EI}}})}]$$

$$\sigma_{max} = \frac{P}{A}[1+\frac{ce}{r^2}\sec{(\frac{\pi}{2}\cdot{\sqrt{\frac{P}{P_{crit}}}})}]\tag{Secant Formula}$$

• if $\sigma_{max} = \sigma_y$, the column begins to fail by yielding in the extreme concave fibre. Further loading will cause a gradual spread of the yielding zone through the section and very little load can be carried beyond initial yield
• superposition cannot be applied in max stress because relationship between stress and P is nonlinear
• FS must be applied to P, not to stress

Energy Methods

Normal Stress Strain Energy

The words Energy and Work are like synonyms. As external forces are applied, E associated w/ deformation increases by the amount of work done

• For an elastic system, the work done on the system is stored in a form of elastic energy. The elastic energy is fully recoverable. This is represented by strain energy density u and strain energy U

It is known in the uniaxial case, $u=\frac{1}{2}\sigma_1\varepsilon_1$. Taking this over the entire volume, the strain energy can be found (below):

$$U = \frac{1}{2}\int{\frac{\sigma_1^2}{E}}dV\tag{Uniaxial Strain Energy}$$

• for the case of a uniform bar in tension or compression, $\sigma = \frac{P}{A}$. By subbing this into the above, letting $dV=A\cdot{dx}$, with bounds on x between $0$ and $L_{bar}$, the below is found:

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

• for a truss (several members), the total strain E is found by summing the strain E of each member together

If there normal stress due to bending moment, it is known that $\sigma = \frac{My}{I}$. By letting $dV=dA\cdot{dx}$, with the bounds on x between $0$ and the $L_{beam}$, $y^2dA$ simplifies to I on the numerator. Thus, the below is true:

$$U = \frac{1}{2}\int_0^L{\frac{M^2}{EI}}dx\tag{Bending Strain Energy}$$

Shear Stress Strain Energy

For shear, it is known that $dU = \frac{1}{2}\tau\gamma dV$, with $\gamma = \frac{\tau}{G}$. By setting $dV = dA\cdot{dx}$, it can be found that $U = \frac{1}{2}\int\frac{\tau^2 A}{G}dx$. Since $\tau = \frac{V}{A}$, the below can be proven:

$$U = \frac{1}{2}\int_0^L{\frac{V^2}{GA}dx}\tag{Shear Strain Energy}$$

However, this assumes that shear stress is uniformly distributed, as $\tau_{avg}$ was used. This assumption is incorrect. In order to adjust this, a factor of K (often 1.2) is multiplied by the value of the shear strain energy.

Most often, the bending strain energy is >> shear strain energy since beams are typically slender. Thus, unless a short deep beam is being analyzed, strain energy due to shear is typically ignored.

• $\frac{U_{shear}}{U_{bending}}~0.02$. Thus for slender beams, if shear is ignored, there will be an error of 0.02.

The case of shear due to torsion is also different. Since $\tau = \frac{T\rho}{J}$, and $J=\int{\rho^2 dA}$, the below can be found:

$$U = \frac{1}{2}\int_0^L{\frac{T^2}{JG}dx}\tag{Torsion Strain Energy}$$

• if T is constant along the shaft, $U = \frac{T^2L}{2GJ}$

Complementary Energy is the sum of the work from all of the applied loads. This corresponds to the magnitude of the loads multiplied by the deformations that they respectively caused. Since internal work = external work, strain energy equals deformation. Thus, the below is true:

$$W = \sum_{i=1}^n{\int_0^{\delta}P_id\delta_i}$$

By differentiating both sides, the below can be said:

$$\frac{\partial{U}}{\partial\Delta_i} = P_i\tag{Castigliano's 1st Theorum}$$

Complementary Work $W_c$ is not real, but is the area between the strain energy curve and the vertical component. As such, $W = \sum_{i=1}^n{\int_0^{\delta_i}dP_i}$. It is true that $U_c = W_c$ since external must be equal to internal. This can be used to prove the Deflection Theorum below:

$$\frac{\partial{U_c}}{\partial{P_i}} = \delta_i\tag{Castigliano's 2nd Theorum}$$

• only useful for linear systems. This is because in the linear case, $U_c = U$

• the $i$ represents the point of interest where the deflections are found
• if there is load applied at point of interest, a dummy load can be applied at $i$ in the same direction. Similarly, a dummy moment can be applied in order to calculate slope
• in some cases, it is simpler to differentiate with respect to $P_i$ or $M_i$ before integration

Procedure for Castigliano Analysis:

1. If required, apply dummy load or moment
2. Use superposition to create an equation for the beam or situation
3. Integrate and solve

Virtual Work

• conservation of E cannot be applied to calculate deformation at any points where load is not applied. Thus, a load must be applied to calculate $\Delta$ or $\theta$
• to calculate $\Delta$ at point A, an imaginary virtual external load (Q) is applied at A in the direction of the displacement. This in turn causes a virtual internal load (f).
• in order for equilibrium to be satisfied, external virtual work must be equal to internal virtual work. (Virtual external F * real external displacement = virtual internal Fv * real internal displacement)

$$\Delta = \sum{f\cdot{dL}}$$

• 1 = external virtual load Q
• $\Delta$ = external displacement caused by actual load
• $f$ = internal virtual load $dL$ = internal real displacement caused by internal real load
• the forces must be a form of moment to calculate the deformation in a form of rotation $\theta$
• either set can be real or imaginary
• external imaginary load known as unit load because it is set to 1

$$Q\Delta = fx\tag{Combatibility}$$

For a linear system, F = Kx, where K is the spring stiffness. It is known that $K = \frac{EA}{L}$$. Thus, it can be said that $x = \frac{F}{K}$. Thus, by setting Q = 1,

$$\Delta = \frac{f\cdot{F}}{K}\tag{Real Deflection}$$

• more generally, $\Delta = fx$
• x is real displacement caused by external real forces, temperature changes, misfit of component parts, and support settlements
• valid for nonlinear system
• if a rotation at a point on a lienar structure is required, $\Delta$ is an angle and Q is a torque (of unit value) applied in the direction of $\Delta$

Uniaxial Virtual Work

If f is a virtual member force in equilibrium with external virtual applied load:

$$\Delta = \frac{F\cdot{f}\cdot{L}}{AE}$$

• for a truss, the sum of energy is taken along all members
• sum all member components, then divide resultant by EI

Procedure for analysis:

1. Calculate all internal member forces caused by the applied load P
2. To find $\Delta$, apply a unit vertical force at C. Find member forces in terms of P
3. Calculate the internal virtual member forces caused by the applied unit load at C by subbing in P=1

• if the sign of $\Delta$ is positive, its direction is similar to that of unit load

Bending Virtual Work

Given that $d\theta = \frac{Mm}{EI}dx$, letting m = unit virtual internal moment due to virtual external load, it is true that $d(IVW)=\frac{M\cdot{m}}{EI}dx$. Thus:

$$IVW_T = \int_0^L{\frac{Mm}{EI}dx}$$

Procedure for analysis:

1. Set a virtual unit moment at the point where rotation is to be solved for
2. Draw bending moment diagram for virtual load to identify the regions of no virtual moment, to simplify the virtual load calculations
3. Find actual moment along the beam
4. Integrate in the $IVW$ equation

Torsion Virtual Work

$$IVW_T = \frac{T\cdot{tL}}{GJ}$$

Other VW Applications

Deflection due to temperature

• requires $\alpha$, T and L

Since small angle and small deformation $\theta$, $d\theta = \frac{\alpha\Delta{T}dx}{d}$, as the tan component is equal to the angle itself. However, from before we know that $d\theta = \frac{M}{EI}dx$ and $\Delta = \int\frac{M\cdot{m}}{EI}dx$ or $1\cdot{\Delta} = \int{md\theta}$

1. To find $\Delta_A$ due to thermal effects, apply unit load at point of interest.
2. Solve for m term in terms of this unit load, and replace $d\theta$ with $\frac{\alpha\Delta{T}dx}{d}$. There will be an additional negative sign in this equation due to sign convention

Discrepancies in member lengths for a truss

• assumes +ve is longer and -ve is shorter
• let $\frac{F_iL_i}{E_iA_i} = e_i$

1. Apply unit load at point of interest
2. Calculate internal forces due to virtual work from unit load $f_i = 1$
3. Sub into equation $\Delta_A = \sum{f_i\cdot{e_i}}$, with $e_i$ representing the discrepancy in length. $e$ is the real discrepancy, while $f$ is the virtual internal force

Arch Deformation

• from previous, $M(\theta) = PR\sin\theta$. For virtual work, this is $m(\theta) = 1\cdot{R}\sin\theta$
• As it is known that $1\cdot{\Delta} = \int_0^L{\frac{Mm}{EI}ds}$, sub the values for M and m in to get the below

$$\Delta = \frac{PR^3}{EI}\int_0^{\pi}\sin^2\theta{d\theta}$$

• this boils down to $\frac{\pi PR^3}{2EI}$, which is comparable to Castigliano's Method

VW for Indeterminate Structure

• there are 2 categories of indeterminate structures. Redundancy can be either a support reaction or an internal force. A determinate structure is formed by removing this reduncancy by either taking away the extra support or by cutting the redundant member. Thus, the primary structure (determinate) is analysed bearing the combatibility situation
  • to determine if a beam is indeterminate, find the total number of reactions and the total number of available equations. The difference is the degree of indeterminancy
  • to determine if a truss is determinate, $r+b=2n$, with r being reactions, b being bars, and n being joints. If $r+b>2n$, statically indeterminate

In the case of a beam

1. Find M(x)
2. Apply a unit load at point of interest, and solve for m(x)
3. It is known that $1\cdot{\Delta_A} = \int_0^L{\frac{Mm}{EI}}dx$. Due to indeterminancy, use combatibility equation $\Delta_A = 0$.

In the case of a truss

1. Primary Structure: cut one of the members (the redundent member)
2. Find actual forces carried by all of the members
3. Secondary Structure: Apply unit force at point of interest (along cut) to solve for displacement of cut member
4. Solve for virtual force carried by each member, and solve for deflection using the IVW formula for trusses
5. Knowing that the deflection of the cut member is actually zero (since there is a support), name reaction force R. Since $\frac{Rf_iL_i}{A_iE_i}$ is the elongation of respective members due to R (multiply $f_i$ by R), we can let deformation equal to the sum of deformation due to other real forces plus $R\sum\frac{f_i^2L_i}{A_iE_i}$. Since deformation is zero, isolate for R

VW using moment area approach

Since it is known that $1\cdot{\Delta} = \int{\frac{mM}{EI}dx}$, we can prove that deflection is equal to the real moment area times the x value on secondary structure corresponding to both the slope of the secondary structure BMD and the centroid of the area of primary structure from either side, all divided by EI.

Procedure for analysis

1. Draw BMD for primary and secondary structure
2. Set up equality to solve for x terms
3. Apply and solve

Frames

• constructed with rigid joints instead of hinged joints. Joints are moment-resisting
• frames are subjected to BM, shear and axial forces
• the degree of stability and determinacy of frames are determined by comparing the number of unknowns with the number of available independent equations

If b is number of members, r is number of reactions, n is number of joints, s is special equations, there will be $3b+r$ available unknowns. The amount of available equations are as follows:

$$3n+s>3b+r\tag{Unstable Frame}$$

$$3n+s=3b+r\tag{Statically Determinate}$$

$$3n+s<3b+r\tag{Statically Indeterminate}$$

Procedure for analysis:

1. Solve for external reactions
2. Take cuts in between joints, members, and supports and solve for BM,V,and N along each point
3. Plot axial, shear, and BM diagrams along members

Frame Deflected Shape

• original angle of a rigid joint must be preserved
• original length of each member is unchanged (little axial deformation in compare to flexural deformation)
• deflected shape must satisfy boundary conditions
• curvature must be consistent with the moment
• where moment = zero, inflection point is seen on deflected shape

Influence Line

The influence line is a line along a beam that indicates the shear, moment, or reactions at the specified point when the load (unit of 1) moves across the beam. The line is always linear.

To find reaction influence line, simply take cuts along the length of the beam.

To find shear influence line (Example):

1. Draw shear force diagram with the unit load applied at the point of interest. From this, there are two options.
2. a) Connect corners of shear-force diagram with ends of beam, or b) Take mid-way cuts to determine which side is negative and positive
3. If complicated, start at point of interest and then do endpoints, followed by midpoints
4. To find actual shear at point of interest, multiply the applied forces by the amount of the influence lines at each position and sum them together.

To find moment influence line:

1. Apply load at point of interest, then endpoints and midpoints to create moment influence line
2. To find actual moment at point, multiply the applied forces by the amount of the moment lines at each position and sum them together.

2D Design Examples

To find internal 2D shear, moment, and axial reactions (Example):

1. Draw FBD
2. Solve for reactions
3. Take cuts at desired points and solve

To design plate with a bolt (Example):

• In order to design this plate with a bolt properly, one must test for bearing of plate (punching failure), tension of plate (tensile failure), shear of plate (also punching), and shear of bolt. All of this is assuming that allowable bearing, shear, and tensile stresses are given

1. Check failure of the plate in tension. Use $\sigma_{allow} = N/A$ at the smallest cross-section.
2. Check failure of the plate by bearing.
   1. Project curved area of the bolt to a rectangle.
   2. Use $\sigma_{allow} = \frac{N}{Projected A}$ at the bolt hole
3. Check failure of plate by shear (remember bolt is in double shear)
   1. Calculate area that would experience shear if bolt pushed through
   2. Use $\tau_{allow} = V/A$ to solve
4. Check failure of bolt by shear. Remember that in equilibrium, the shear force felt by the bolt is equal to the force applied to the bar
   1. Use $\tau_{allow} = V/A$ to solve

How to find required diameter of pin given required average maximum shear stress (Example):

1. Find all external forces being applied at the pins
2. Find resultant forces being applied
3. Check if pin is in double shear. If so, divide resultant force by 2
4. Use formula $\tau_{avg} = V/A$ and isolate for pin diameter.