1206 - Concepts in Physics Wednesday, November 04th
Request from Mark Brown Those of you who will miss the LAB tomorrow, please let Mark B. know - either stopping by or a quick email will be fine. State you name and the reason
Just wondering... The solutions for assignment #3 and #4 have been taken off the wall across my office during reading week This left some students without the possibility to look at them This is not a good way to do things, please think about others as well
Notes You will get the midterm exams back on Friday (after class) Any other notes about that will be on Friday as well After today s class you have everything you need to complete assignment #5 Remember that only 1 point for #5 (8) is part of the 25 points that count as 100%
We will talk about deformation today deformations magnified 100
Elastic deformation and Young s Modulus A spring returns to its original shape when the force compressing or stretching it is removed. In fact, all materials become distorted in some way when they are squeezed or stretched and many of them return to their original shape (rubber). We call these material elastic. To understand this a little deeper, we can make a picture (model) of how a certain material looks like at the atomic level. For a solid, we can assume, that the forces between atoms act like springs. Elastic behavior has its origin in the forces that atoms exert on each other. These atomic-level springs help the material to return to its original shape, once the forces that caused the deformation are removed. The force that holds atoms of a solid together is particularly strong, therefore considerable force must be applied to stretch it. Experimentally it has been shown, that the magnitude of the force can be expressed by the following relation, provided that the amount of stretching is small compared to the original length: F = Y (ΔL/L0)A where A is the cross-sectional area of (for example) a rod, ΔL the increase in length, L0 the original length and F the magnitude of the stretching force applied perpendicularly to the surface at the end. The term Y is a proportionality constant called Young s modulus. Thomas Young lived (1773-1829). Young s moduli are given in units of N/m 2 Typical values for solids are 10 8-10 11
Example In a balancing act, a performer supports the combined weight (1640 N) of a number of colleagues. Each thighbone of this performer has a length of 0.55m and an effective crosssectional area of 7.7 x 10-4 m 2. The Young s Modulus for bone compression is 9.4 x 10 9 N/m 2. Determine the amount by which each thighbone compresses under the extra weight. Each thighbone supports half the weight, therefore F = 0.5 * 1640N = 820 N. The Young s modulus for bone compression is given as well as the length and the crosssectional area. So, we can find the amount by which the additional weight compresses the thighbone ΔL. F = Y (ΔL/L0)A ---> ΔL = FL0/YA ΔL = (820N)(0.55m)/(9.4 x 10 9 N/m 2 * 7.7 x 10-4 m2) = 6.2 x 10-5 m This is a very small change.
Shear deformation and the shear modulus It is possible to deform a solid object in a way other than stretching or compressing. For example, imagine a thick book with a hardcover laying on a table. When you use your hand to push it along the top cover will be shifted relative to the pages below and the stationary bottom cover. The resulting deformation is called shear deformation and occurs because of the combined effect of the force F applied (by the hand) to the top of the book and the force -F applied (by the table) to the bottom of the book. The direction of the forces are parallel to the covers of the book, each of which has an area A. These two forces have equal magnitudes, but opposite direction, so the book remains in equilibrium. The magnitude F of the force needed to produce an amount of shear ΔX for an object with thickness L0 ΔX F = S (ΔX/L0)A L where A is the cross-sectional area and S is the constant of proportionality called the shear modulus with units N/m 2 (like the Young s modulus). Typical shear modulus are of the order of 10 9-10 11.
Example: A block of Jell-O is resting on a plate. The block is 0.030 m high, 0.070 m wide and 0.070 m long. A person is pushing tangentially across the top surface with a force of F = 0.45 N The top surface moves a distance ΔX = 6.0 x 10-3 m relative to the bottom surface. What is the shear modulus of Jell-O? The person applies a force that is parallel to the top surface of the Jell-O block. The shape of the block changes. We can use the formula that give the magnitude of the force to determine the shear modulus, since everything else is given. F = S (ΔX/L0)A --> S = FL0/ΔXA S = [(0.45N)(0.030m)]/[(6.0 x 10-3 m)(0.070m)(0.070m)] = 460 N/m 2 This is a fairly small number, reflecting the fact that Jell-O can be deformed easily.
Volume deformation and the bulk modulus So far we have been talking about compressive forces along one dimension. It is also possible to apply compressive forces so that the size of every dimension (length, width, and depth) decreases, leading to a decrease in volume. This over compression occurs, for example, when an object is submerged in a liquid, and the liquid presses inward everywhere on the object. The forces acting in such situation are applied perpendicular to every surface, and we speak of the perpendicular force per unit area. The magnitude of the perpendicular force per unit area is called the pressure P. This should be familiar, since we have discussed this for fluids before. P = F/A (unit for pressure is N/m 2 = Pa) Suppose we change the pressure on an object by an amount ΔP (final - initial). This change in pressure causes a change in volume of the object by an amount ΔV. Experimentally one finds that the change in pressure is directly proportional to the fractional change in volume ΔV/V0: ΔP = -B (ΔV/V0) The proportionality constant B is known as the bulk modulus. The minus sign is necessary because and increase in pressure always causes a decrease in volume.
Stress, Strain and Hooke s law Over the last couple of slides we have defined amounts of forces that are needed for a given amount of deformation. They have common features, even so they describe different types of deformation. We will re-write them here: F/A F/A ΔP = Y = S = -B ΔL/L0 ΔX/L0 ΔV/V0 The left side of each equation is the magnitude of the force per unit area required to cause an elastic deformation and is in general called stress. The right side of each equation involves the change in a quantity divided by a quantity (unitless ratio) and is referred to as the strain that results from the applied stress. Stress is proportional to Strain Stress is directly proportional to strain, this relationship was first discovered by Robert Hooke (1635-1703) and is knows as Hooke s law.
There are limits to the proportionality We had this before for the specific case of springs Stress Strain Springs are also a good example for the proportionality limit. Once you have stressed (applied force) to a spring beyond its elastic limit, it will not return to its original shape and can t be used as a spring anymore.
Thermal Stress Do you remember the example of the buckling sidewalk? If the concrete slabs had not buckled upward, they would have been subjected to immense forces from the buildings. The forces needed to prevent a solid object from expanding must be strong enough to counteract any change in length that would occur due to a change in temperature. Although the change in temperature maybe small, the forces - and hence the stresses - can be enormous.
Example A steel beam is used in the roadbed of a bridge. The beam is mounted between two concrete supports when the temperature is 23 C, with no room provided for thermal expansion. What compressional stress must the concrete supports apply to each end of the beam, if they are to keep the beam from expanding when the temperature rises to 42 C. Concrete support Beam Concrete support Stress = F/A = Y ΔL/L0 where Y is the Young s modulus. If the steel beam were free to expand because of the change in temperature, the length would change by ΔL = αl0δt. Because the concrete supports do not permit any expansion, they must supply a stress to compress the beam by an amount ΔL. Stress = Y ΔL/L0 = Y (αl0δt)/l0 = YαΔT For steel, the values of Young s modulus and the coefficient of linear expansion are Y = 2.0 x 10 11 N/m 2 and α = 12 x 10-6 ( C) -1 respectively. The change in temperature is ΔT = 42C - 23C = 19C. The thermal stress is Stress = YαΔT = (2.0 x 10 11 N/m 2 )(12 x 10-6 ( C) -1 )(19 C) = 4.6 x 10 7 N/m 2
Bimetallic strip How does your electric water boiler know when to turn itself off when the water has reached a certain temperature? Bimetallic strips are used for that purpose (and many others). A bimetallic strip is made from two think strips of metal that have different coefficients of linear expansion - therefore they behave differently when heat (or cooling) is applied. Often brass (α = 19 x 10-6 (C ) -1 ) and steel (α = 12 x 10-6 (C ) -1 ) are selected. The two pieces are welded together. When the bimetallic strip is heated, the brass (green), having the larger value of α, expands more than the steel (red). Since the two metals are bonded together, the bimetallic strip bends into an arc. When the strip is cooled, the bimetallic strip bends into the opposite direction
Expansion of holes - Example When we have a hole in a piece of solid that is heated, does is expand as well? Lets assume a pattern made out of tiles, 3 up and 3 to the side, where the central one has been left out. We can then think about this question on a single tile basis first and then put it back together after the heat has been applied. If every single tile expands when heated, which we know to be true, then the hole must have expanded as well. It will still have the size of one (now expanded tile) after heating. So holes expand in exactly the same way as the surrounding material. This conclusion is true for any solid and for any shape of hole. Thus, if follows that a hole in a piece of solid material expands when heated and contract when cooled, just as if it were filled with the material that surrounds it. If the hole is circular, we can use equation ΔL = αl0δt to find the change in any linear dimension. A gold ring (α=14 x 10-6 ( C) -1 ) has an inner diameter of 1.5 x 10-2 m and a temperature of 27 C. The ring falls into a sink of hot water with a temperature of 49 C. What is the change in the diameter of the hole in the ring? Change in temperature ΔT = 49 C - 27 C = 22 C α and L0 are given, therefore ΔL = αl0δt = (1.5 x 10-2 m)(14 x 10-6 ( C) -1 )(22 C) = 4.6 x 10-6 m
Volume thermal expansion So far we have been looking at linear expansion for one- or two-dimensional changes. The volume of a normal material increases as the temperature increases. Analog to linear thermal expansion, the change in volume ΔV is proportional to the change in temperature ΔT and to the initial volume V0, provided the change in temperature is not too large. The proportionality constant β, known as the coefficient of volume expansion can be used to write: ΔV = βv0δt Common unit for the coefficient of the volume expansion is (C ) -1 Thermal expansion usually will be measured at room temperature (20 C). The values for β for liquids are substantially larger than those for solids, because liquids typically expand more than solids for the same initial volumes and temperature changes. For most solids the coefficient of volume expansion is three times as much as the coefficient for linear expansion. If a cavity exists within a solid object, the volume of the cavity increases when the object expands, just as if the cavity were filled with the surrounding material. This is due to the same argument we have used to derive at the conclusion that a hole expands the same as the surrounding material.
Example - an automobile radiator A small plastic container, called the coolant reservoir, catches the radiator fluid that overflows when an automobile engine becomes hot. The radiator is made of copper (β = 51 x 10-6 (C ) -1 ), and the coolant has a coefficient of volume expansion of β = 4.10 x 10-4 (C ) -1. If the radiator is filled to its 15-quart capacity when the engine is cold (6.0 C), how much overflow from the radiator will spill into the reservoir when the coolant reaches its operating temperature of 92 C? When the temperature increases, both the coolant and the radiator expand. If they were to expand by the same amount, there would be no overflow. However, the liquid coolant expands more than the radiator, and the overflow volume is the amount of coolant expansion minus the amount of the radiator cavity expansion. When the temperature increases by 86 C, the coolant expands by: ΔV = βv0δt = (4.10 x 10-4 (C ) -1 )(15 quarts)(86 C ) = 0.53 quart When the temperature increases by 86 C, the radiator expands by: ΔV = βv0δt = (51 x 10-6 (C ) -1 )(15 quarts)(86 C ) = 0.066 quart Therefore the overflow volume is 0.53 quart - 0.07 quart = 0.45 quart
Special case - water Although most substances expand when heated, a few do not. For instance, if water is at its freezing point of 0 C, its volume decreases until the temperature reaches 4 C. Above 4 C water behaves normally (meaning as other solids and fluids) and its volume increases as the temperature increases. Because a given mass of water has a minimum volume at 4 C, see figure. The fact that water has its greatest density at 4 C, rather than at 0 C, has important consequences for the way in which a lake freezes. When the air temperature drops, the surface layer of water is chilled. As the temperature of the surface layer drops toward 4 C, this layer becomes more dense than the warmer water below. The denser water sinks and pushes up the deeper and warmer water, which in turn is chilled at the surface. This process continues until the temperature of the entire lake reaches 4 C. Further cooling of the surface water below 4 C makes it less dense than the deeper layers; consequently, the surface does not sink but stays on top. Once the cooling reaches 0 C, the formation of ice that floats on the water starts. Ice has a smaller density than water. Below the ice the temperature of the water however remains above 0 C. The sheet of ice acts as an insulator that reduces the loss of heat from the lake, especially if the ice is covered with a blanket of snow. As a result, lakes usually do not freeze solid and therefore fish and other aquatic life can survive.