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~~SLIDESHOW~~ | ~~SLIDESHOW~~ | ||

- | ====== Chapter 12 - Rotation I: Kinematics and Dynamics ====== | + | ====== Chapter 12 - Static equilibrium, Elasticity and Fracture ====== |

- | ===== Polar coordinates for rotational motion ===== | + | Equilibrium describes a state where both the linear acceleration and angular acceleration of an object or system are zero. This means the object is either at rest or moving with constant velocity. |

- | {{angulardisplacement.png}} | + | Here we will focus on situations where the system is a rest, so called static equilibrium, as opposed to dynamic equilibrium, where the system has constant velocity. |

- | If we consider two points on a turning disc at distance $r_{1}$ and $r_{2}$ from the center of the disc we can make a number of observations. | + | ===== Conditions for Equilibrium ===== |

- | First the distance traveled by the points is quite different, points that are far from the center go through a greater distance. | + | For an object to have no translational acceleration |

- | However, the angular displacement is the same for both points, suggesting that angular displacement is a good variable to describe the motion of the whole rotating object. | + | $\Sigma \vec{F}=0$ |

- | We would like to be able to have a straight forward relationship between the actual distance traveled and the angular displacement, which we can have if we express our angular displacement in [[wp>Radian|radians]]. | + | and for this to be true the sum of **each component** of the forces must be zero. |

- | As the relationship between the arc length on a circle and the angle which it subtends in radians is | + | An object may have no net force on it, but be able to turn, which means for an object to be in equilibrium |

- | $l=r\theta$ | + | $\Sigma \tau = 0 $ |

- | a change in the angular displacement of $\Delta\theta$ results in a change in the displacement | + | Where the torques can be calculated around any axis. Some axes are more sensible than others, a good idea is to choose an axis at place where a force is acting (as the torque due to that force around that axis will be zero.) |

- | $\Delta l=r\Delta\theta$ | + | ===== 2 Balls on a rod ===== |

- | ===== Angular velocity===== | + | {{2ballsonarod.png}} |

- | For change in angular displacement $\Delta \theta$ in a time interval $\Delta t$ we can define an average angular velocity | + | In the above example we calculate the torques around the pivot point. This is a sensible choice, but we could have also chosen, for example, an axis that goes through the second mass $m_{2}$. |

- | $\bar{\omega}=\frac{\Delta \theta}{\Delta t}$ | + | In this case |

- | As we are now accustomed, we can also define an instantaneous velocity | + | $\Sigma \tau=m_{1}g4L-({m_{1}+m_{2}})g3L=0$ |

- | $\omega=\lim_{\Delta t\to 0}\frac{\Delta \theta}{\Delta t}=\frac{d\theta}{dt}$ | + | $m_{1}gL=m_{2}g3L$ |

- | The units of angular velocity can be expressed as either $\mathrm{rad\,s^{-1}}$ or $\mathrm{s^{-1}}$ | + | $\large m_{2}=\frac{m_{1}}{3}$ |

- | ===== Angular acceleration ===== | ||

- | Previously we only considered circular motion in which the angular velocity $\omega$ remained constant with time, but of course we can also define an angular acceleration | + | ===== Crane boom with arm at 90 degrees ===== |

- | $\bar{\alpha}=\frac{\Delta \omega}{\Delta t}$ | ||

- | $\alpha=\lim_{\Delta t\to 0}\frac{\Delta \omega}{\Delta t}=\frac{d\omega}{dt}$ | + | There are four forces which act on the arm. Forces do not need to be directed along the arm as it is a rigid object. On the other hand, the force due to tension **must** be directed along the rope (otherwise it would change it's shape). To take the weight of the arm in to account we can treat the weight of the rod as if it acts through the center of mass of the object, which we will approximate as being halfway along the arm. |

- | The units of angular acceleration can be expressed as either $\mathrm{rad\,s^{-2}}$ or $\mathrm{s^{-2}}$ | + | {{boomarmwitharmweigt.png}} |

- | ===== 12.P.008 ===== | ||

- | ===== Equations of motion for rotational motion ===== | + | Sum of forces on arm |

- | At the beginning of the course we derived, using calculus, a set of equations for motion under constant acceleration. | + | $F_{hx}=F_{Tx}$ |

- | $v= v_{0}+at$ | + | $mg+m_{a}g=F_{Ty}+F_{hy}$ |

- | $x= x_{0}+v_{0}t+\frac{1}{2}at^2$ | + | Sum of torques on arm around hinge |

- | $v^{2}=v_{0}^2+2a(x-x_{0})$ | + | $F_{Ty}l=F_{T}l\sin\theta=mgl+m_{a}g\frac{l}{2}$ → $F_{T}\sin\theta=mg+\frac{1}{2}m_{a}g$ |

- | We can equally derive similar equations for our rotational quantities. Indeed as we can see that the relationships between the new rotational quantities we have now are exactly the same as those between the translational quantities we can simply rewrite the translational motion equations in terms of rotational variables. | + | $F_{T}=\frac{mg+\frac{1}{2}m_{a}g}{\sin\theta}$ |

- | $\omega= \omega_{0}+\alpha t$ | + | Compressive force on arm is $F_{hx}=F_{Tx}=(mg+\frac{1}{2}m_{a}g)\frac{{\cos\theta}}{\sin\theta}=\frac{mg+\frac{1}{2}m_{a}g}{\tan\theta}$ |

- | $\theta= \theta_{0}+\omega_{0}t+\frac{1}{2}\alpha t^2$ | + | ===== Crane boom with rope at 90 degrees ===== |

- | $\omega^{2}=\omega_{0}^2+2\alpha(\theta-\theta_{0})$ | + | {{boomarmrope90.png}} |

- | ===== 12.P.005 ===== | + | Forces on arm |

- | ===== 12.P.026 ===== | + | |

- | ===== From angular to tangential quantities ===== | + | $F_{hy}=mg+m_{a}g$ |

+ | $F_{hx}=F_{Tx}$ | ||

- | A point at distance $r$ from the center of rotation will have at any time a tangential velocity of magnitude | + | Torques on arm |

- | $v=\omega r$ | + | $F_{T}l\sin\phi=(mgl+m_{a}g\frac{l}{2})\sin(90^{o}-\phi)=(mg+\frac{1}{2}m_{a}{g})l\cos\phi$ |

- | and a tangential acceleration (not to be confused with the centripetal acceleration) of | + | $F_{T}=F_{Tx}=F_{hx}=\frac{mg+\frac{1}{2}m_{a}{g}}{\tan\phi}$ |

- | $a=\alpha r$ | + | Compressive force on spring |

- | ===== 12.P.038 ===== | + | $\frac{F_{hx}}{\cos\phi}=\frac{mg+\frac{1}{2}m_{a}{g}}{\sin\phi}$ |

- | ===== Useful relationships concerning the angular velocity ===== | + | ===== Crane boom with arm and rope at angle ===== |

- | Rotational motion, when not accelerated, can be considered to be a form of periodic motion, and so relationships between the angular velocity, frequency and period are useful. | + | {{boomarmgeneral.png}} |

- | $T=\frac{2\pi}{\omega}$ | + | Forces on arm |

- | $f=\frac{\omega}{2\pi}$ | + | $F_{hy}+F_{Ty}=F_{hy}+F_{T}\sin\theta=mg+m_{a}g$ |

- | $\omega=2\pi f$ | + | $F_{hx}=F_{Tx}=F_{T}\cos\theta$ |

- | We'd also like to be able to express the centripetal acceleration in terms of $\omega$ | + | Torques on arm |

- | $a_{R}=\frac{v^{2}}{r}=\frac{(\omega r)^{2}}{r}=\omega^{2}r$ | + | $F_{T}l\sin(\phi+\theta)=(mgl+m_{a}g\frac{l}{2})\sin(90^{o}-\phi)=(mg+\frac{1}{2}m_{a}{g})l\cos\phi$ |

- | ===== About that acceleration.. ===== | + | $F_{T}=\frac{(mg+\frac{1}{2}m_{a}{g})\cos\phi}{\sin(\phi+\theta)}$ |

- | {{totalrotacc.png}} | + | $F_{Tx}=\frac{(mg+\frac{1}{2}m_{a}{g})\cos\phi\cos\theta}{\sin(\phi+\theta)}=F_{hx}$ |

- | The total acceleration of an object in accelerated rotational motion will be the vector sum of two perpendicular vectors, the tangential acceleration $\vec{a}_{tang}$, and the radial acceleration $\vec{a}_{R}$. | + | Compressive force on spring |

- | The magnitude of the total acceleration is | + | $\frac{F_{hx}}{\cos\phi}=\frac{(mg+\frac{1}{2}m_{a}{g})\cos\theta}{\sin(\phi+\theta)}$ |

- | $\vec{a}=\sqrt{a_{tang}^2+a_{R}^2}=\sqrt{\alpha^2r^2+\omega^{4}r^2}=r\sqrt{\alpha^2+\omega^4}$ | ||

- | and it is directed an angle | ||

- | $\arctan\frac{\alpha}{\omega^{2}}$ away from the radial direction. | + | ===== Hanging sign problem ===== |

- | ===== 12.P.031 ===== | + | {{f11finalq3fig.png}} |

+ | After numerous unfortunate incidents involving polar bears and tourists who have seen too many Coke<sup>TM</sup> ads, Santa has decided to install a new sign at the North Pole. To do this he uses a uniform metal rod of length 0.8 m and a cable which he attaches 20 cm from the end of the metal rod. The new sign has mass 3 kg and hangs from the end of the rod, while the rod itself has mass 1 kg. The cable is attached the top of the pole and the rod is held by a hinge 30 cm down from the top of the pole. | ||

- | ===== Pseudovector representation of angular velocity and acceleration ===== | + | What angle θ does the cable make with the rod? |

- | As we know, translational velocity and acceleration are vector quantities. While we have defined angular velocity and acceleration we can see that they can represent points that are moving in directions that change with time, which would make them difficult to represent with vectors defined in cartesian coordinates. | + | What is the tension in the cable? |

- | We can however consider an axial vector, ie. a vector that points in the direction about which rotation occurs as a good way of representing the direction of these rotational quantities. | + | What are the horizontal and vertical components of the force exerted by the hinge on the rod? Take the positive directions to be up and to the right. |

- | We give the direction of rotation around the axes according to a [[wp>Right-hand_rule|right-hand rule]]. | + | ===== Hanging sign solution ===== |

- | {{rhgriprule.png}} | + | {{f11finalq3fig.png}} |

+ | $\theta=\tan^{-1}\frac{30}{60}=26.57^{o}$ | ||

+ | Vertical forces, up is positive $T\sin\theta+Fh_{y}=m_{sign}g+m_{rod}g$ | ||

+ | Horizontal forces, right is positive $-T\cos\theta+Fh_{x}=0$ | ||

+ | Torques $0.6T\sin\theta=0.8m_{sign}g+0.4m_{rod}g$ | ||

+ | $T=\frac{0.8\times3\times9.8+0.4\times1\times9.8}{0.6\times\sin26.57^{o}}=102.24\mathrm{N}$ | ||

+ | $Fh_{y}=4.981-102.24\sin26.57^{o}=-6.53\mathrm{N}$ | ||

+ | So force points down. | ||

- | ===== Combining translation motion with rotational motion - rolling ===== | + | $Fh_{x}=102.24\cos26.57^{o}=91.4\mathrm{N}$ |

- | /*<html> | + | |

- | <object width="640" height="384" type="application/x-shockwave-flash" data="RollingDiscCycloids.swf"> | + | |

- | | + | |

- | | + | |

- | </object> | + | |

- | </html>*/ | + | |

- | Let's consider a wheel that [[http://www.upscale.utoronto.ca/GeneralInterest/Harrison/Flash/ClassMechanics/RollingDisc/RollingDisc.html|rolls without slipping]]. For an object to roll without slipping there is a frictional requirement; it is actually not possible for an object to roll on a frictionless surface, it would instead slide. | + | Force points to the right |

+ | ===== Will the ladder slip? ===== | ||

- | The point around which the wheel turns, which is also it's center of mass, is seen to execute translational motion with velocity $\vec{v}$ | + | {{ladderslip.png}} |

- | In the reference frame of the ground a point on the edge of the wheel comes to rest when it is contact with the ground. However from the perspective of the center of the wheel, the motion of the point of the edge of the wheel is purely rotational with $\omega=\frac{v}{r}$ or $v=\omega r$. | + | The forces that act on the ladder of length $l$ are the weight of the ladder that acts down at it's center of mass, $m\vec{g}$, a normal force exerted by the wall, $\vec{f}_{NW}$, an normal force exerted by the ground $\vec{f}_{NG}$, and a force due to the friction of the ground, $\vec{F}_{Fr}$. The friction between the ladder and the wall is typically neglected. |

- | We can describe the displacement of a point on the edge of the wheel in the reference frame of the ground by adding the rotational motion of the point to the translational motion of the center of the wheel. This gives the equation | + | For equilibrium the sum of the forces in both the horizontal and vertical directions must be zero and also the sum of the torques around an appropriate axis must be equal to zero. |

- | $\vec{r}=vt\hat{i}+r\cos(\theta(t))\hat{i}+(r\sin(\theta(t))+r)\hat{j}$ | + | Balance of the forces in the horizontal direction implies that |

- | $\to \vec{r}=(r\omega t+r\cos(\omega t))\hat{i}+(r\sin(\omega t)+r)\hat{j}$ | + | $F_{NW}=F_{Fr}$ |

- | A path of this kind is called a [[wp>Cycloid|cycloid]] | + | and in the vertical direction |

- | Differentiating with respect to time give us the velocity | + | $F_{NG}=m_{l}g$ |

- | $\vec{v}=(r\omega-r\omega(\sin(\omega t)))\hat{i}+r\omega\cos(\omega t)\hat{j}$ | + | If we calculate the torques around the point at which the ladder touches the ground then they will be |

- | More on wheels and rolling at [[http://www.animations.physics.unsw.edu.au/jw/rolling.htm|Physclips]] | + | $F_{Nw}l\sin\theta-m_{l}g\frac{l}{2} \sin(90^{o}-\theta)=0$ |

+ | $F_{Nw}l\sin\theta-m_{l}g\frac{l}{2}\cos\theta=0$ | ||

- | ===== Torque ===== | + | This can be written in terms of the frictional force by using the equation we got from the horizontal forces. |

- | In the same way as a force causes linear acceleration, $\vec{F}=m\vec{a}$, there must be an analogous quantity and equation related to angular acceleration. From experience, for example when we [[http://www.usna.edu/MathDept/website/courses/calc_labs/wrench/TorqueWrench.html|use a wrench]] (or a spanner in civilized parts of the world) we can expect that the angular equivalent of Force, which is called [[wp>Torque|Torque]], will depend on the amount of Force applied, the distance at which it is applied and the direction of its application. | + | |

- | {{torque.png}} | + | $F_{fr}l\sin\theta-m_{l}g\frac{l}{2}\cos\theta=0$ |

- | We define the torque due to a force $F$ applied at a distance $R$ from a pivot point as either | + | The maximum force that can be provided by friction is $\mu F_{NG}$ and so the maximum safe angle is when this frictional force is provided |

- | $\tau\equiv RF_{\perp}$ or $\tau\equiv R_{\perp}F$ | + | $\mu F_{NG}l\sin\theta_{max}-m_{l}g\frac{l}{2}\cos\theta_{max}=0$ |

- | and we can calculate the magnitude of the torque using | + | $\mu mgl\sin\theta_{max}-m_{l}g\frac{l}{2}\cos\theta_{max}=0$ |

- | $\tau=RF\sin\theta$ | + | $\tan\theta_{max}=\frac{1}{2\mu}$ |

- | The units of torque are $\mathrm{Nm}$ | + | Of course if a person is standing on the ladder, their weight would need to be taken in to account both in the balance of the vertical forces and the torques. |

- | ===== 12.P.041 ===== | ||

- | ===== Production of Torque in an engine ===== | ||

- | You have probably heard of the word torque mostly in the context of car engines. | + | ===== Stable and Unstable equilibrium ===== |

- | This [[http://auto.howstuffworks.com/auto-parts/towing/towing-capacity/information/fpte4.htm|animation]] from How Stuff Works illustrates how the torque on the crankshaft is produced in a [[wp>Four-stroke_engine|4 stroke engine]]. | + | An object in static equilibrium will not move from it's equilibrium position unless it is disturbed by an additional force. The question of stability concerns what happens when equilibrium is **slightly** disturbed. If the equilibrium is stable the system will return to it's equilibrium position, but if it instead moves away from the equilibrium then we say the equilibrium is unstable. If the small perturbation moves the system to a new equilibrium we can say the equilibrium is neutral. |

+ | We can demonstrate the different kinds of equilibrium with the two balls on a stick we looked at earlier. | ||

+ | ===== Circus Trick ===== | ||

- | ===== Torque as a vector ===== | ||

- | When we talked about torque before we looked at how to calculate it's magnitude. More correctly it is a vector $\vec{\tau}$ that is defined from a cross product of force $\vec{F}$ and the displacement from the center $\vec{r}$ | + | This [[http://www.youtube.com/watch?v=K5QGoP6bXaY|circus trick]] looks pretty difficult. |

- | $\vec{\tau}=\vec{r}\times\vec{F}$ | ||

- | {{torquevector.png}} | + | ===== Conditions for stable equilibrium===== |

- | ===== Vector cross product ===== | + | But actually it isn't! |

- | The [[wp>Cross_product|cross product]] of two vectors $\vec{A}$ and $\vec{B}$ is another vector | + | An object whose center of gravity is below it's point of support will be in stable equilibrium so [[http://www.youtube.com/watch?v=FKMTRsOSKQU&feature=player_embedded|riding a unicycle on a high wire]] is easy if there is weight suspended from it like the one at [[http://www.cosi.org/|COSI]]. |

- | $\vec{C}=\vec{A}\times\vec{B}$ | + | This is also the reason that [[http://en.wikipedia.org/wiki/File:Maria_Spelterini_at_Suspension_Bridge.jpg|tightrope artists]] will sometimes hold a heavy pole that bends down at the ends, to lower their center of gravity and increase their stability. |

- | The magnitude of the vector $\vec{C}$ is $AB\sin\theta$ | + | ===== Corbel arch ===== |

- | The direction of $\vec{C}$ is perpendicular to the plane which contains both $\vec{A}$ and $\vec{B}$. Use of the [[http://upload.wikimedia.org/wikipedia/commons/d/d2/Right_hand_rule_cross_product.svg|right hand rule]] can help you determine this direction. | ||

- | If we consider the vectors in unit vector notation, $\vec{A}=A_{x}\hat{i}+A_{y}\hat{j}+A_{z}\hat{k}$ and $\vec{B}=B_{x}\hat{i}+B_{y}\hat{j}+B_{z}\hat{k}$ the cross product can be expressed as the [[wp>Determinant|determinant]] of a matrix | + | An object can be in stable equilibrium if a vertical line from the center of gravity lies within it's center of support. |

- | $\vec{A}\times\vec{B}=\left| \begin{array}{ccc} | + | We will demonstrate this concept by building a [[wp>Corbel_arch|corbel arch]]. |

- | \hat{i} & \hat{j} & \hat{k} \newline | + | |

- | A_{x} & A_{y} & A_{z} \newline | + | |

- | B_{x} & B_{y} & B_{z} \end{array} \right| = \hat{i}\left| \begin{array}{cc} | + | |

- | A_{y} & A_{z}\newline | + | |

- | B_{y} & B_{z} \end{array} \right| | + | |

- | - | + | |

- | \hat{j}\left| \begin{array}{cc} | + | |

- | A_{x} & A_{z}\newline | + | |

- | B_{x} & B_{z} \end{array} \right| | + | |

- | + | + | |

- | \hat{k}\left| \begin{array}{cc} | + | |

- | A_{x} & A_{y}\newline | + | |

- | B_{x} & B_{y} \end{array} \right|$ | + | |

- | $=(A_{y}B_{z}-A_{z}B_{y})\hat{i}+(A_{z}B_{x}-A_{x}B_{z})\hat{j}+(A_{x}B_{y}-A_{y}B_{x})\hat{k} | + | {{corbelarch.png}} |

- | $ | + | |

- | ===== Properties of cross products ===== | ||

- | $\vec{A}\times\vec{A}=0$ | + | ===== True Arch ===== |

- | $\vec{A}\times\vec{B}=-\vec{B}\times\vec{A}$ | ||

- | $\vec{A}\times(\vec{B}+\vec{C})=(\vec{A}\times\vec{B})+(\vec{A}\times\vec{C})$ | + | The weakness of a corbel arch as compared to a true [[wp>Arch|arch]] which should contain a [[wp>Catenary|catenary]] curve, is that not all the tensile stresses are converted in to compressive stresses. What does this mean and why it is important? We need to know a little about the properties of materials. |

- | $\frac{d}{dt}(\vec{A}\times\vec{B})=\frac{d\vec{A}}{dt}\times\vec{B}+\vec{A}\times\frac{d\vec{B}}{dt}$ | + | {{truearch.png}} |

+ | ===== Stress ===== | ||

- | ===== 12.P.047 ===== | + | When an object is exposed to a external force, it can deform. The relative size of this deformation, is called a **strain** $\varepsilon=\frac{\Delta l}{l_{0}}$. Often, when dealing with material properties it can be more useful to consider the force as a force per unit area, or **stress** $\sigma=\frac{F}{A}$ |

- | ===== 12.P.048 ===== | ||

- | ===== Acceleration due to torque ===== | + | {{stresses.png}} |

- | We now want to look at the relationship between the torque which is applied and the angular acceleration it generates. Let's look at the case of a force $F$ applied at 90 degrees to an object of mass $m$ attached to a pivot point at distance $R$ | ||

- | {{leverarmI.png}} | + | [[http://www.doitpoms.ac.uk/tlplib/dislocations/dislocation_motion.php|These videos]] of [[http://www.doitpoms.ac.uk/tlplib/dislocations/bubbleraft.php|bubble rafts]] from [[http://www.doitpoms.ac.uk/index.html|DoITPoMs]] at Cambridge are useful for visualizing the way materials respond to the various kinds of stress. |

- | Newton's second law tells us that | + | ===== Elastic Deformation ===== |

- | $F=ma$ | + | Many materials exhibit elastic behavior, where they will recover from a strain to their original shape. |

- | If we multiply both sides by R we get | + | This can be divided in to two regimes, the linear elastic regime, below the proportional limit, where objects obey Hooke's Law |

- | $FR=maR$ | + | $F=k\Delta l$ |

- | and by remembering that $a=\alpha R$ we get | + | and their strain is linearly proportional to the stress. As the stress is increased this is followed by a non-linear elastic regime where the stress and strain are not linearly related. |

- | $\tau=mR^{2}\alpha$ | + | The upper limit of this non-linear elastic regime is the [[wp>Elastic_limit|elastic limit]]. Up to this limit the object will revert to it's original shape when the stress is removed, beyond it the object will deform in a non-reversible fashion, until it eventually breaks. |

- | ===== Moment of Inertia ===== | + | The material specific plot of stress vs strain is called a [[wp>Stress%E2%80%93strain_curve|stress strain curve]]. |

- | In an object with a common angular acceleration which consists of many masses at various distances from the center of rotation, each of which may be subject to a torque we can say that | + | {{metalyield.png}} |

- | $\sum\limits_{i}\tau_{i}=(\sum\limits_{i} m_{i}R_{i}^{2})\alpha$ | ||

- | As we discussed for forces, different masses can exert internal torques one each, but when summed over all the masses these cancel due to Newton's Third Law, meaning that our sum over all the torques on the masses $\sum\limits_{i}\tau_{i}$ becomes the sum of the external torques $\sum\tau$. We call the quantity $\sum\limits_{i} m_{i}R_{i}^{2}$ the moment of inertia $I$ and write the rotational equivalent of Newton's Second Law as | ||

- | $\sum \tau=I\alpha$ | + | ===== Elastic Moduli ===== |

- | This equation is valid for a rigid object around a fixed axis (ie. for this to be valid the masses can't move around with respect to the object during the motion as they would if the object was flexible, and the axis must stay in the same position with respect to the masses rotating around it). | + | There are a set of [[wp>Elastic_modulus|elastic moduli]] that link the deformation of an object to particular types of stress. |

- | We should note that the distance $R$ is the distance of each mass with respect to the axis of rotation, which need not be the center of mass, and is not an intrinsic property of an object! | + | [[wp>Young's_modulus|Young's modulus]], $E$, is the constant that relates stress along a direction to strain in that direction in the elastic regime. |

- | ===== 12.P.053 ===== | + | $\sigma=E\varepsilon$ |

+ | $\frac{F}{A}=E \frac{\Delta l}{l_{0}}$ | ||

- | ===== Two weights on a thin bar ===== | + | This formula can be applied to either tensile stress or compressive stress. |

+ | For shear stresses a different elastic modulus, the [[wp>Shear_modulus|shear modulus]], $G$ applies. $G$ is usually smaller than $E$. | ||

- | {{2wonab.png}} | + | $\sigma=G\varepsilon$ |

- | The moment of inertia of two point weights $m_{1}$ and $m_{2}$ on a bar at distance $R_{1}$ and $R_{2}$ from the axis of rotation is | + | $\frac{F}{A}=G \frac{\Delta l}{l_{0}}$ |

- | $I=m_{1}R_{1}^{2}+m_{2}R_{2}^2$ | + | Note: for this equation $\Delta l$ is perpendicular to $l_{0}$! |

- | The axis of rotation coincides with the center of mass only if $m_{1}R_{1}=m_{2}R_{2}$ | + | An object subject to uniform pressure change will have a volume change |

+ | $\frac{\Delta V}{V_{0}}=-\frac{1}{B}\Delta P$ | ||

+ | |||

+ | $B$ is the [[wp>Bulk_modulus|bulk modulus]]. | ||

+ | |||

+ | |||

+ | ===== Strength of materials ===== | ||

+ | |||

+ | Different materials, depending on their micro-structure have different [[wp>Strength_of_materials|strengths]] with relation to the different strains. By strength we refer to the maximum stress that can be withstood without the material breaking. | ||

+ | |||

+ | [[wp>Ultimate_tensile_strength|Ultimate tensile strength]] for ductile materials is usually determined by [[http://www.youtube.com/watch?v=-ejDNBojQd0&feature=related|necking]], brittle materials, like concrete typically have very low tensile strength, but have high compressive strength. Structures under compressive stress can [[http://www.youtube.com/watch?v=VzMn0t-1vDY&feature=related|collapse]] well before the material itself fails due to [[wp>Buckling|buckling]]. | ||

+ | |||

+ | Shear failure can be greatly assisted by imperfections that allow [[http://www.doitpoms.ac.uk/tlplib/dislocations/dislocation_glide.php|dislocation glide]]. | ||

+ | |||

+ | |||

+ | ===== Truss bridge ===== | ||

+ | |||

+ | A [[wp>Truss_bridge|truss bridge]] uses connected elements to spread the load out, some elements are under compression, others are under tension. In the diagram below we ignore the mass of the elements themseleves. | ||

+ | |||

+ | {{trussbridge.png}} | ||

+ | |||

+ | Frequently different materials are used for compression and tension elements. Long truss bridges get very heavy, a better design for these applications is a suspension bridge. | ||

+ | |||

+ | ===== Suspension bridge ===== | ||

+ | |||

+ | The simplest kind of [[wp>Suspension_bridge|suspension bridge]] is one that hangs under it's own weight in the shape of the [[wp>Catenary|catenary]], the equation for which is $y=a \cosh (\frac{x}{a})$. An inverted catenary is the ideal shape for an arch which supports only it's own weight. | ||

+ | |||

+ | A more normal suspension bridge, and also more typical arches, support weight besides their own and the ideal shape in these situations is closer to that of a parabola, in the case where the mass of the cables can be neglected the shape is entirely parabolic. | ||

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+ | ===== The best bridge ===== | ||

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+ | {{bridgenorth.png}}{{bridgesouth.png}} | ||

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+ | The [[wp>Sydney_Harbour_Bridge|Sydney Harbour Bridge]] is a steel [[wp>Through_arch_bridge|through arch bridge]]. As you can see it incorporates elements of both suspension and truss bridges. The deck itself is suspended from a trussed double arch. |