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First Class Wednesday, January 20, 2010 07:55 Statics: a subject in mechanics Mechanics: the study of forces on bodies and their resulting motion (Isaac Newton a great contributor) Emphasis on bodies which are not moving (or moving at constant speed) TAM 212: dynamics - moving bodies Biomechanics: muskoskeletal mechanics TAM 251: Introductory Solid Mechanics - strength of materials - study of how things deform - twist, compress, stretch… ME 340: dynamics of buildings ME 310: fluid dynamics ME 330: behavior of materials ME 360: controls - robotics ME 320: heat transfer ME 370: design ME 350: design for manufacturability Notes Page 1 Syllabus Wednesday, January 20, 2010 08:19 Forces and bending moments Multiple bodies Beams, structural analysis, trusses Friction Can switch to TAM 211 at this point Forces on submerged bodies Notes Page 2 No course website yet - on mechse website will be Grades on Compass HMWKS due Fridays at 08:00 Ms Thomas's know what language we speak Notes Page 3 Precision: 3 significant figures ALWAYS (like IB math) Exams are during lectures (not in this room) Notes Page 4 Vectors, Newton's Laws of Motion, Gravity, Calculations, and Analysis Friday, January 22, 2010 07:56 A1-A5 (vectors) in book Read Newton's Three Laws of Motion 1. A particle at rest or moving with constant velocity will remain in this state unless acted upon by an unbalanced force. 2. A [article acted upon by an unbalanced force F experiences an acceleration a that is proportional to the particle mass, m. 3. The mutlual forces of action and reaction between two particles are equal opposite and collinear. Newton's law of gravitational attraction The mutual force F of gravitation between two particles of mass m1 and m2 is given by Weight is the force exerted by the earth on a particle at the earth's surface: Units of Measurement System Length Time Mass Force SI m s kg N US Customary (FPS) ft s slug lb Newton and slug are derived units Unit conversion - 1 ft = 0.3048m (exact) - 1 slug = 14.59 kg - 1 lb = 4.448 N Numerical Calculations - Dimensional homogeneity Equations must be dimensionally homogenous Notes Page 5 Equations must be dimensionally homogenous Significant Figures - Use 3 significant figures for final answers, more (usually 4) for intermediate calculations This is engineering convention (assume all dimensions are to 3 sf if not specified, 1 ft = 1.00 ft) General Procedure for analysis 1. Read the problem carefully, write it down carefully. 2. Draw given diagrams neatly and construct additional figures as necessary. 3. Apply principles needed. 4. Solve problem symbolically, then substitute numbers. Proved proper units throughout. Check significant figures. Box the final answer(s).. 5. See if answer is reasonable. Starting Monday: class meets in 228 Natural History Building Notes Page 6 Forces Monday, January 25, 2010 07:57 Office hours are now online - all are in 429 Grainger Force: the action of one body on another - can be treated as a vector, since forces obey all the rules that vectors do. Forces have magnitude, direction, and sense. Problem: Finda unit vector in the xy plane that is perpendicular to the force 3i-4j+12k N. Answers in back of book: don't necessarily have all solutions - if there are two explain that Problem: Find two 80 lb forces whose sum is the force 40i lb. Notes Page 7 Find a vector that is simultaneously in the plane of F1 and F2 and in the plane of F3 and F4. Discussions this week HMWK due Friday Notes Page 8 Forces, components, and division thereof Wednesday, January 27, 2010 07:58 Forces: the action of one body on another - magnitude, sense, and direction - vectors Force components: Law of sines and cosines can be used in 2-d problems, but using force components is usually more straightforward. - Can also solve problems graphically Adding vectors - simple - Pick a good system - but keep it once you have it PROBLEML Determine the magnitude of force F so that the resultant force of the three is as small as possible. What is the minimum magnitude? Question: how do we know this is the minimum rather than the maximum? - Cannot be a maximum, since the force could be large as possible - Note: could solve it geometrically, but much more difficult Force along a line: often a force is described as acting along a line - often said to be a line connecting A and B. Example: the window is held open by cable AB. Determine the length of the cable and express the 30N force at A along cable. Notes Page 9 Cantilever bending - force vertical on lug nut Components of forces parallel and perpendicular to a line: - The scalar component A // of a vector A along a line with unit vector u is given by A // =Acos(theta)=A.u. Thus the vector component is A//=A//u=(A.u)u, Aperp=A-A//. Problem: Determine the angle theta betweeen the two cables. Determine the projected component of the Force F=12lb in direction of AC Notes Page 10 Equilibrium, Free body diagrams, and Idealizations Friday, January 29, 2010 07:59 Equilibrium: - According to Newton's first law, a particle will be in equilibrium if and only if, ΣF=0, that is the sum of all forces on a particle is zero. - Generally, in 3D: Equilibrium requires: ΣF = ΣFxi +ΣFy j+ΣFzk =0. - can be separated into vector equations. - Note: this is valid in inertial reference frames - basically one in which we can regard as fixed - Many problems are also in only 1 or 2 dimensions - the remaining can be ignored Free Body diagram - A free body diagram is a drawing of a body, or part of a body which all the forces acting on the body are shown. Relevant dimensions may also be required - No judgment regarding equilibrium or lack of it is made when constructing the diagram. Forces not directly acting on the body are not shown. Idealization: - Pulleys are usually regarded as frictionless, then the tension in a rope or cord around the pulley is the same on either side - Springs are usually regarded as linearly elastic then the tension is proportional to the change in length, s. Notes Page 11 Example Problems Friday, January 29, 2010 08:03 In reality either rope could break, as there is a distribution of ropes break at slightly different values - they follow a normal curve Notes Page 12 Notes Page 13 Systems of Particles Monday, February 01, 2010 07:59 Some practical engineering problems involve the use of statics of interacting or interconnected particles. To solve them, we use Newton's first law, ΣF=0, on multiple free body diagrams of particles or groups of particles. Notes Page 14 If you want to find the value of a force you must draw a free body diagram that includes that force. Example Problems Monday, February 01, 2010 07:59 The five ropes can each take 1500 N without reaking. How heavy can W be without breaking any. A man having a weight of 175 lb attempts to lift himself using one of the two methods shown, determine the total force he must exert on bar AB in each case and the normal reaction he exerts on the platform at C. The platform has a weight of 30 lb. Notes Page 15 Never forget to draw the overall free body diagram - it simplifies the problems greatly The collar can slide along the rod without friction. The spring, which is attached to the collar and to the ceiling exerts a force kΔ proportional to its stretch Δ, where k is the modlus of the spring. If k=50lb/in, W=50 lb and the collar weiths 20 lb find how much the string is stretched in the given position. Notes Page 16 Moment of a force Wednesday, February 03, 2010 07:57 Hour exam next Friday - cumulative so far - Vectors, Application of vectors… - It will be in 2 different rooms - here and 100 MSEB - there will be assigned seating The moment, MP, of a force, F about a point P is equal to the product, MP=Fd, where d is the perpendicular distance from P to the line of action of the force. The force F can be acting at any point along this line of action. Principle of transmittivity: F can be acting anywhere along the line of action Notes Page 17 Line of action of F Example Problems Wednesday, February 03, 2010 08:01 Three forces act on the plate. Determine the sum of the moments of the thre forces about point P. Note: a torque is a special kind of moment Notes Page 18 Notes Page 19 Moments of Forces and Couples Friday, February 05, 2010 08:00 This is used in post hole augers - things to dig holes, you put it in and then twist it - apply equal and opposite forces to turn it Moment of a Force about a Line: - For a point: MP=rxF - Ml=MP.ul - Ml==(MP.ul)ul - Note the perpendicular one is the total minus the parallel Couples: - A couple is a pair of equal, opposite and noncollinear forces - The moment of a couple is given by M=Fd (scalar) - M=rxF, it is independent of the origin you choose - If several couples act, MR=Σ(rxF). - This is a special case: note that the total force is equal to 0 Notes Page 20 Example problems Friday, February 05, 2010 08:00 Example: determine moment about Oa axis Determine the couple moment - express as a cartesian vector Notes Page 21 Notes Page 22 Equipollent force systems Monday, February 08, 2010 07:58 A force system is a collection of forces and couples applied to a body. Two force systems are said to be equipollent if they have the same resultant force, ΣF and the same resultant moment, ΣMP with respect to any point P. Here, for each system. ΣMP denotes the moment of all applied couples plus the sum of moments of applied forces with respect to point P Note: equivalent is often used instead of equipollent. Pronunciation: ee-quee-pah-lent Notes Page 23 Hour exam: - Assigned seating - On Friday - Homework-type problems, plus understanding questions - No calculators or electronics - Bring self, pencil, eraser - Through Friday's lecture - Do not need to use given-find-solution, only solution - 3 problems - Either pen or pencil is OK - pencil preferable - 50 minutes long, starts at precisely 8:00 Example problems Post tensioning ram on a forklift Monday, February 08, 2010 07:58 Can rephrase as: Replace the force and couple system by its resultant at point Q. Replace the force and couple system by an equipollent force and couple moment at point Q. Note: it does not matter where this couple is applied for 8k Notes Page 24 Resultants of Force Systems Check your seat before exam Course notes posted outside 429 Grainger Wednesday, February 10, 2010 07:57 Resultants: - A resultant is an equipollent force system consisting of only one force Fr and one couple MrP: Must specify P Fr=(ΣF) 1 MrQ=(ΣM)1 F can be anywhere along its line of action, couple can be applied anywhere Notes Page 25 Example problems Wednesday, February 10, 2010 07:58 The belt passing over the f=pulley is subjected to two forces each having a magnitude of 40 N, force F1 acts in the -k direction. Teplace thse by a resultant force system at point A. theta is 45 degrees. Notes Page 26 Hour Exam Monday, February 15, 2010 08:06 Make sure to explain work - use words Also simplifying important Exams by Friday hopefully Notes Page 27 Reductions Autonomous Materials Group Monday, February 15, 2010 07:59 Reduction classes: - Single resultant force (often possible) Concurrent forces with no couples Coplanar forces with couples perpendicular to force plane Parallel forces with couples perpendicular to forces - Screwdriver (always possible) - consists of a single force and a couple in the same direction A resultant is an equipollent force system consisting of only one force, Fr and one couple MrP - must specify point Single resultant forces - If FR is perpendicular to MrP with the force not zero, then an equipollent system consisting of only a single force can always be found. 1. Concurrent forces (no couples) i. Act at the same point, just sum the forces 2. Coplanar, with couples perpendicular to plane i. Add the forces, that will be the force ii. Find MrP and then divide by the force and move the force over that much - that will be the single force resultant 3. Parallel forces with couples perpendicular i. Add the forces for the resultant force ii. Find moment about P, then move the force over by MrP divided by the force iii. Common with a plate with wires and weights Screwdriver - Most general case - always possible - Consists of a force and a couple in the same direction - push and twist Move the force over in order to cancel the perpendicular force - then it will just be the parallel moment and the force Notes Page 28 Example problem Monday, February 15, 2010 08:42 Replace the 3 forces by a screwdriver Notes Page 29 Distributed forces Screwdriver is referred to as a wrench by some texts. Wednesday, February 17, 2010 08:16 Distributed forces - In structural analysis we are often presented with a distributed load (force/unit length) w(x) and we need to find the equipollent loading F. Notes Page 30 Example Problems Wednesday, February 17, 2010 08:16 Replace the distributed load by an equipollent resultant force and specify its location on the beam measured from the pin at C An engineer measures the forces exerted by the soil on a 10m section of a building foundation and finds they are described by Determine a. The magnitude of the total force exerted on the foundation b. The magnitude of the moment about A due to the distributed load. Notes Page 31 Simple Distributed Loading Friday, February 19, 2010 07:59 Total force given by area under loading diagram It is located at the centroid of the area Notes Page 32 Equilibrium of a rigid body Friday, February 19, 2010 08:10 We regard a rigid body as a collection of particles: Force equilibrium: Let Fi=the resultant external force on particle i Let fij=the internal force on particle i by particle j Let fji=the internal force on particle j by particle i In general: 2D problems have 3 unknowns, 3 equations (2 forces, 1 moment) 3D problems have 6 unknowns, 6 equations (3 forces, 3 moments) Notes Page 33 Example problems Friday, February 19, 2010 08:28 Draw the free-body diagram of the automobile, which is being towed at constant velocity up the incline using the cable at C. The automobile has a mass of 5 Mg and a center of mass at G. The wheels are free to roll. Explain the signifficance of each force on the diagram Notes Page 34 Types of support Friday, February 19, 2010 08:34 Type Description Reactions Unknowns Roller/slider 1 Pin 2 Fixed 3 Notes Page 35 Free body diagram Friday, February 19, 2010 08:37 Notes Page 36 Applications of Equilibrium Monday, February 22, 2010 07:59 Sum of forces and sum of moments are zero. These are external forces on the extended body - the internal ones don't count for overall equilibrium Can prove that if the sum of the moments about a point is zero and the sum of the forces is zero => moment equilibrium is satisfied for any point. Notes Page 37 Example Problems Monday, February 22, 2010 07:59 The cantilevered jib crane is used to support the load of 780 lb. If the trolly T can be placed anywhere between 1.5 ft and 7.5 ft, determine the maximum magnitude of the reactions at the supports A and B. The supports are collars that allow the crane to rotate freely about the vertical axis. The collar at B supports a force in the vertical direction, but the one at A does not Easier to assume all forces are going in the same way, will then get a negative answer. Notes Page 38 Two and Three force members Monday, February 22, 2010 08:18 In many situations bodies are subject to only two or three forces and no couples. Two force members: - The forces must be equal, opposite, and collinear - Shape of body doesn't matter Three force members: - The three forces must be either concurrent or parallel and must be coplanar Notes Page 39 Problem Monday, February 22, 2010 08:26 It is possible for the 20N half cylinder to be in equilibrium on the smooth plane for only one value of the angle phi. For that angle find the tention in the cord as a function of theta and check your answer in the limiting cases theta is 0 and pi/2. Half cylinder weight W is suspended from a smooth vertical wall by a cord. Find the tension T in the cord as a function of theta and check the limiting cases Notes Page 40 Problems with extended bodies - 3D Wednesday, February 24, 2010 08:00 Plate, has weight 5500 lb. Determine the tension of each cable when when held as shown Often can get a quick answer by looking at moments about a line Member is supported by a pin at A and cable BC. If load is 300 lb, determine the components of reactions at supporsts Notes Page 41 Two marbles, radius R and wewight W are placed inside a hollow thin -walled tube diameter D. D<4R. Find minimum weight of tube so it will not turn over. Notes Page 42 Notes Page 43 More problems Friday, February 26, 2010 07:57 There are 4 courses in biomechanics [email protected] A vertical force of 80 lb acts on the crankshaft. Determine the horizontal equilibrium force, P, that must be applied to the handle and the x,y,z components of force at the smooth journal bearing A and the thrust bearing B. The bearings are properly aligned and exert only force reactions on the shaft. Journal bearing - allows it to turn around Thrust bearing - prevents it from sliding either - put a thing on the other end Brake pads used to be made of asbestos, now made of a polymer like that in bulletproof vests (N, H, and O) The wall crane supports a load of 700 lb. Determine the horizontal and vertical components of reactions at the pins A and D. Also, what is the force in the cable at the winch W? Can pick FD in direction of BD as there are only 2 forces on BD so forces are equal, opposite, and collinear. Notes Page 44 Notes Page 45 Introduction to structures Friday, February 26, 2010 08:43 Structural equilibrium - Talbot laboratory, Titan formwork builds formwork for concrete - build a multistory building Must be lightweight, easy to assemble Notes Page 46 Trusses Monday, March 01, 2010 07:59 Mode of failure is not necessarily unique Buckling: slender member under compression bends - very important One test is worth a thousand expert opinions Trusses are structures formed by straight, pin connected (2-force-members) bars that are loaded only at the joints. Both 2D and 3D trusses are used widely Two primary methods to find loads in members - Methods of joints - Method of sections Always first set the overall structure in equilibrium Method of joints 1. Set the entire structure into external equilibrium 2. Set each joint in the structure into force equilibrium only, moving from joint to joint. Assume all internal loads are tensile. (if negative then compressive) Notes Page 47 Example Problems Monday, March 01, 2010 07:59 Truss subject to loading shown Distance between rails on railroad is standard at 4 ft 8.5 inches Bob Miller Determine force in member FG of truss - state tension or compression Notes Page 48 Notes Page 49 Trusses: method of sections Failure point is often the joint, especially dissimilar or composite material Long slender members are very easy to break in bending Wednesday, March 03, 2010 08:01 Method. 1. Set the body into external equilibrium 2. Cut the structure at a section of interest into two separate pieces and set either part into force and moment equilibrium. Assume all internal loads are tensile. Notes Page 50 Example problems Wednesday, March 03, 2010 08:09 The Howe bridge truss is subjected to the loading shown. Determine the force in members DE, EH, and HG and state if the members are in compression/tension. Determine the force in OE, LE and LK of the Baltimore truss, state if members in tension or compression Notes Page 51 Notes Page 52 Frames and Machines 17 March: Hour Exam 2 Can stay and learn TAM 211 material Prob. A Wednesday or Thursday night for TAM 210 exam Friday, March 05, 2010 07:59 Here, the members can be truss elements, beams, pulleys, cables, and other components. Generally, if the structure has moving parts, it is called a machine; otherwise it is called a frame. The general solution method is the same: 1. Set the entire structure into external equilibrium. This step will generally produce more unknowns than there are relevant equations of equilibrium 2. Isolate various part(s) of the structure, setting each part into equilibrium. The sought forces or couples must appear in one or more free-body diagrams. 3. Solve for the requested unknowns. Notes Page 53 Example Problems Friday, March 05, 2010 08:01 Determine threactions at supports A and B. The tower truss has weight 575 lb, centre of gravity at G. Rope is used to hoist it vertical. Compute the compressive force along the shear leg, the tension in BC. Notes Page 54 Note: tensions in the ropes will not be the same Note: can draw more FBD's and calculate the rest of forces (which were not requested here) Notes Page 55 Internal Forces 1st two are forces 2nd two are moments Monday, March 08, 2010 07:58 Cutting members at internal points reveals internal forces and moments: Force/Moment Diagram _____________________________________________________________ Normal (axial) Transverse shear Torsional (torque) Or twisting moment Bending NOTE: forces and moments are equal and opposite on opposite sides of cut. Positive directions are denoted above - convention is similar to that of tension. Notes Page 56 Example problem Monday, March 08, 2010 08:10 Three torques act on the shaft. Determine the internal torque at A, B, C, and D Determine the normal force, shear force, and bending moment at a section passing through point D of the two member frame. Notes Page 57 Note: these re valid only at point D. In general, N,V, and M vary throughout the structure. (here N does not) Needed to set body into external equilibrium. In general, V and M will vary Problem: The shaft is supported by a thrust bearing at A and a journal bearing at B. The shaft will fail when the maximum moment is 5 kip.ft. If L=10ft determine the largest uniform distributed load the shaft will support. Notes Page 58 Shear force and bending moment distributions Examination next week - notes on wall outside 429 Grainger. Wednesday, March 10, 2010 07:59 For beams, the shear force V and bending moment M are generally functions of position x along the beam. The relations for V(x) and M(x) can be found from force and moment equilibrium, respectively. Diagrams: Notes: - The assumed sign conventions for F and M0 must be honored - Only a concentrated force will cause a jump in V(x) - Only a concentrated moment will cause a jump in M(x) - A concentrated force and a concentrated moment can be applied at the same point though this condition is not a requirement. Notes Page 59 Example Problem II Friday, March 12, 2010 08:30 The beam consists of three segments pin connected at B and E. Draw the shear force and bending moment diagrams for the beam. Notes Page 60 Diagrams Friday, March 12, 2010 08:42 Notes Page 61 Problem Wednesday, March 10, 2010 08:15 Express the x,y,z components of the internal loading along the rod as a function of y where 0<=y<=4 Notes Page 62 Diagram Friday, March 12, 2010 08:02 From problem on Monday Diagrams: Notes Page 63 Shear-force and bending moment diagrams Monday, March 15, 2010 07:59 The shear force V and bending moment M are generally functions of position x along a beam. The distributions V(x) and M(x) for statically determinate problems can always be found by - Setting the beam in external force and moment equilibrium, then - Cutting the beam at arbitrary points and setting portions into force and moment equilibrium The distributions can then be plotted. Differential and integral relations for q, V, and M - q: distributed load (positive up) - V: shear force (positive up) - M: bending moment (positive z) - dV/dx=-q - dM/dx=-V Be very careful of sign conventions Notes Page 64 Example problem Monday, March 15, 2010 08:01 The beam will fail when the maximum moment Mmax is 30 kip.ft or the maximum shear, Vmax is 8 kips. Determine the largest distributed load, w the beam will support Notes Page 65 Diagrams Monday, March 15, 2010 08:27 Notes Page 66 HE 2 Friday, March 19, 2010 07:59 Euler derived the moment equilibrium condition Torsion is principal deformation mode in axially loaded helical spring. Notes Page 67 Friction Friday, March 19, 2010 07:59 Friction forces are the tangential components of force between bodies in contact. Here we study only dry friction, also called Coulomb friction: The coefficient of static friction is somewhat larger than the coefficient of kinetic friction. Both depend on the materials in contact. The friction force f always acts to oppose relative motion. Notes Page 68 Example problem Friday, March 19, 2010 08:07 A roll of paper has a uniform weight of 0.75 lb and is suspended from the wire hanger so that it rests against the wall. If the hanger has a negligible weight and the bearing at O can be considered frictionless, determine the minimum force P needed to start turning the roll. The coefficient of static friction between the wall and the paper is 0.25. Notes Page 69 Notes Page 70 Wedges Monday, March 29, 2010 07:58 These are tapered pieces used to locate and hold objects in place. Usually, friction is involved. Inserting and removing are not the same: Inserting: Friction ALWAYS acts AGAINST you. Smaller force to remove than to insert, as N A opposes you going in, but helps you go out. Removing: Force equilibrium is always required. Moment equilibrium is ordinarily not considered. Note: the sign of P for removing a wedge is not necessarily positive as shown, since N A has a component in the direction of +P. Notes Page 71 Belts Monday, March 29, 2010 08:04 Belts are flexible members that can transmit tangential force to pulleys, pins, brake drums, trees, posts, capstans, and other rough surfaces: An important feature of belts is that the tension T is generally not constant in the region of contact and therefore, in general, T2 != T1. Assume belt is slipping to the left. Consider a differential element: Notes Page 72 Example Problem Monday, March 29, 2010 08:24 The truck, which has a mass of 3.4 Mg is to be lowered down the slope by a rope that is wrapped around a tree. If the wheels are free to roll and the man at A can resist a pull T of 300 N, determine the minimum number of turns the rope should be wrapped around the tree to lower the truck at a constant speed. The coefficient of kinetic friction between the treee and rope is 0.3 Note: if the man is smart he will wrap the tree more than twice around. Otherwise he may lose control. Notes Page 73 Example Problem 2 Monday, March 29, 2010 08:33 The uniform 50 lb beam is supported by the rope, which is attached to the end of the beam, wraps over the rough peg and is then connected to the 100 lb block. If the coefficient of static friction between the beam and block and between the rope and peg is 0.4, determine the maximum distance d that the block can be placed from A and still remain in equilibrium. Assume the block will not tip. Notes Page 74 Screws Wednesday, March 31, 2010 08:00 Use same concept as wedges Think of screw as wedge wrapped around a rod The equipollent force pushing the entire screw thread along the contact mean radius r is M/r. Consider square (Acme) threads only. Different conditions apply for tightening and loosen Author does this Note: there are self loosening screws (and other cases) - that is if the coefficient of friction is less than tangent of alpha, a negative moment is required to keep the screw from unraveling. Notes Page 75 Example Problem Wednesday, March 31, 2010 08:20 The two blocks under the double wedge are brought together using a left and right square threaded screw. If the mean diameter is 20 mm, the lead is 5 mm and the coefficient of friction is 0.4, determine the torque needed to draw the blocks together. The coefficient of friction between each block and its surfaces of contact is 0.4. Notes Page 76 Note: spreading blocks apart would require a different (smaller) value of M. Notes Page 77 Centroids Friday, April 02, 2010 07:58 The centroid of a shape (volume, area, or line) is the point C at which all the shape may be regarded as concentrated, such that the first moment of the concentrated shape with respect to any point in space is the same as that of the given shape: Notes Page 78 Example Problems Friday, April 02, 2010 08:05 Locate the centroid of the shaded area bounded by the parabola and the line y=a Locate the centroid z of the volume segment. Notes Page 79 HE 2: y=0.9x+20 - adjusted Avg: 75.7, median 76.3 For every dV with a positive value of y (or x) there is another dV with a negative value of y (or x). Notes Page 80 Another problem Monday, April 05, 2010 08:00 The anatomical center of gravity G of a person can be determined by using a scale and a rigid board having a uniform weight W1 and length l. With the person's weight W known, the person lies down on the board and the scale reading P is recorded. From this show how to calculate the location x bar. Discuss the best place l 1 for the smooth support at B in order to improve the accuracy of this experiment. Best place for l 1 is probably l/2, since then (l/2-l 1)W1 vanishes. Then errors in W1 become irrelevant. Versabar: weighing structures and lifting them Notes Page 81 Composite bodies Monday, April 05, 2010 08:12 If shapes are joined together the volumes (or areas or lines) and first moments add: Can also consider cutouts or cavities by subtracting that volume and moment from above equation (common: plate with a hole drilled in it) Notes Page 82 Example Monday, April 05, 2010 08:16 Determine the location x bar of the centroid of the solid made from a cone cylinder and hemisphere. Notes Page 83 Another problem Monday, April 05, 2010 08:32 When material is subtracted the corresponding V i effectively changes sign. Notes Page 84 Example: bridfeeder 5,7,9 rule for birdfeeders 5 feet off the ground, 7 feet from something sideways, 9 feet above Wednesday, April 07, 2010 08:08 The bird-feeder pole is made from a continuous length of uniform rod. Determine the location (x bar,ybar) of the centroid C. If have two bodies the centroid of the composite must lie along the line connecting the two individual centroids Notes Page 85 Moment of Inertia of an Area Friday, April 09, 2010 07:56 There are also moments of inertia of mass (dynamics) Motivation: beam bending: Consider a beam in bending under the action of a bending moment M: When doing testing, must approximate the actual field conditions Will bend (if pure moment is applied) into the arc of a circle Tension at bottom, compression at top, nothing in the centre (centroid) this is called the neutral surface In introduction to Solid Mechanics (TAM 251), it will be shown that the neutral surface passes through the centroid of the cross sectional area and that the stress σ varies linearly with distance y from this surface. Shear is not particularly important here Notes Page 86 Example Friday, April 09, 2010 08:19 Determine the moment of inertia, I x for a rectangular area of width b and height h about the x axis through its centroid. Calculate moment of inertia about the bottom Note that this is much LARGER than before (4x larger to be precise) Determine the moment of inertia for the shaded area about the x axis Notes Page 87 Note: can also be solved from taking horizontal strips, but more difficult Notes Page 88 Moments of inertia of an area continued Monday, April 12, 2010 07:58 Types: There are several types of second moment or moment of inertia of an area: Ordinary Cross Polar The commonly encountered ones are the ordinary and polar types. Note that J P=Ix+Iy for any cross section - this is useful for finding J P when Ix and Iy are known or easily calculated separately. Note: calculate the moments means the two ordinary moments Notes Page 89 Example problems Monday, April 12, 2010 08:05 Determine the moments of inertia Ix and Iy and polar moment of inertia JO, for a thin strip. Assume that t/l << 1. Application Determine the distribution of shear stress, tau, in a linear elastic circular rod of radius R subjected to a torque T: Notes Page 90 Numeric integration, Simpson's rule Monday, April 12, 2010 08:30 Notes Page 91 Example Monday, April 12, 2010 08:41 Determine moment of inertia for the shaded area about the x axis. Use simpson's rule to evaluate the integral. Notes Page 92 The parallel-axis theorem Wednesday, April 14, 2010 07:58 Powerful relations are easily derived for moments of inertia of an area about axes parallel to the centroidal axes, x 1, y1: Note that x 1, y1 must pass through the centroid. Also, the moments of inertia are seen to be a minimum about the centroidal axes. Notes Page 93 Examples Wednesday, April 14, 2010 08:18 Determine the moment of inertia Ix of a rectangle about its edge given hat Ixc=bh3/12 aboutits centroid. Determine the moment of inertia xc of a parabolic sector aboutits centroidal axis. Notes Page 94 Question: What shape of cross section of a given area A maximizes I xc for a given maximum width b and height h. Must be symmetric about the axis and as far away as possible An I-beam to take shear and keep it from falling apart Note: fall apart and have shear if not having I part Notes Page 95 Composite areas Friday, April 16, 2010 08:00 For composite areas add areas, first moments, and second moments algebraically: Where + is used if an area is added and - I sused if area is subtraced. Usually the location is calculated with respect to the centroid Notes Page 96 Examples Friday, April 16, 2010 08:02 Determine the location of the centroid ybar of the beam constructed from the two channels and the cover plate. If each channel has a cross sectional area A c of 11.8 in2 and a moment of inertia IxC about a horizontal axis passing through its own centroid Cc of 349 in^4. Determine the moment of inertia I x1 for the beam's cross sectional area about the x 1 axis. Notes Page 97 Example Friday, April 16, 2010 08:15 Determine the moment of inertia I x for the composite area about the x axis. 153.66-1.865*1.865*31.53=43.9916 Notes Page 98 Practical example Friday, April 16, 2010 08:30 The moments of inertia, Ix for the box beam and the wide flange I-beam are the same: Can calculate by cutting it apart and then putting it back as in the box beam. Railroad tracks are manufactured in quarter mile long sections - the factory is typically 2 miles long. The rounded fillet is important and does affect the result by approximately 3-4%. Notes Page 99 Virtual Work Eyjafjallajokull Monday, April 19, 2010 07:58 Rearth=4000 miles Atmosphere: Highest mountain:~4 miles Irregularities are very small Ronald Adrian - turbulence: Particle Image Velocimetry (sp?) The principle of virtual work allows some complex structural problems to be solved easily, and leads to powerful techniques such as the finite-element method in structural mechanics Basic Concepts Dot product Consider two vectors A and B If A = 0, then A.B=0 for any B More interestingly, if A.B = 0 for any (and all) B, then A=0 Work Work given by dot product Virtual displacements: a conceptually possible displacement or rotation of all or part of a system of particles. Kinematic constraints need not be honored Virtual work The virtual work of the resultant external force, ΣF acting on ta system of particles undergoing a uniform virtual displacement, δu δW=ΣF.δu Good description in textbook Observe that if the system is in external force equilibrium, ΣF=0 and therefore δW=0 for any uniform virtual displacement δu. The more interesting result however is that: If δW for any uniform δu, then the sum of forces equals 0 and threfore the system is in force equilibrium Also, for moment equilibrium. Here the resultant Σrx. The virtual work of the resultant momenta acting through a virtual rotation Then if δW=0 for any rotation then the sum of moments equals zero and therefore it is in moment equilibrium Notes Page 100 Problems Monday, April 19, 2010 08:23 The thin rod of weight W rests agains the smooth wall and floor. Determine the magnitude of force P needed to hold it in equilibtium Note: the effort required to solve this problem by the principle of virtual work is about the same as that by the equations of equilibrium for a free-body diagram of the rod. If a force P of 30 lb is applied perpendicular to the handle of the toggle press determine the compressive force developed at C. Let thetat equal 30 degrees. Notes Page 101 Notes Page 102 Virtual Work continued Wednesday, April 21, 2010 07:59 Principle of Virtual work Let δW be the virtual work done by external forces and couples on a system during a virtual displacement δu. Then: If δW=0 for any virtual translation δu, then ΣF=0, it is in force equilibrium If δW=0 for any virtual rotation δθ then ΣM=0, it is in moment equilibrium In practice a mix of virtual translations and rotations is used to solve equilibrium problems using the principle of virtual work Notes Page 103 Think of virtual work as a double exposure on an old camera Example 1 Wednesday, April 21, 2010 08:02 Each member of the pin connected mechanism has a mass of 8 kg. If the spring is unstretched when theta is 0, determine a. The angle for equilibrium if K=2500N/m, and M=50N.m b. The required stiffness k so that the mechanism is in equilibrium when theta is 30 degrees and M=0 Notes Page 104 Notes Page 105 Example 2 Wednesday, April 21, 2010 08:31 Violating constraints This is permissible - constraints can be violated, there is NO reason why not Notes Page 106 Example 3 Wednesday, April 21, 2010 08:44 A horizontal force acts on the endo f the link. Dtermine the angles theta 1 and theta 2 for equilibrium. Each link is uniform and has a mass m. Notes Page 107 Another problem Friday, April 23, 2010 07:51 The crankshaft is subjected to a torque M of 50 N.m. Determine the horizontal compressive force F applied to the piston a. For 0°<θ<90° b. When θ=60° Notes Page 108 This is approximately sinusoidal - connecting arm complicates this Notes Page 109 Problem - lamp Friday, April 23, 2010 08:28 Determine the mass of A and B required to hold the 400g desk lamp in balance for any angles, theta and phi. Neglect te weight of the mechanism and thes size of the lamp Notes Page 110 Problem: rods in well Monday, April 26, 2010 07:54 Rods AB and BC have centers of mass located at their midpoints. If all contacting surfaces are smooth and BC has a mass of 100 kg, determine the appropriate mass of AB required for equilibrium. Notes Page 111 Spring Monday, April 26, 2010 08:21 The mechanism consists of the four pin connected bars and three springs, each having a stiffness k and an unstretched length l0. Determine the horizontal forces P that must be applied to the pins in order to hold the mechanism in the horizontal position for equilibrium. Also, it is possible that the springs can only operate in tension - in that case 2lsin(theta) must be greater than or equal to l naught. Notes Page 112 Hydrostatic Forces Wednesday, April 28, 2010 07:57 Definition: fluid: a substance that cannot support a shear stress at rest (a solid can). Fluids can be gases or liquids Pascal's law - The pressure p at a point in a fluid is the same in all directions For that to be a valid FBD then it must be a solid, as it is sustaining shear stress Incompressible fluids - An incompressible fluid is one for which the mass density, ρ, is independent of the pressure, p. Liquids are considered incompressible. Gases are compressible but may be approximated as incompressible if the pressure variations are relatively small. - This is an idealization - there technically exists no such substance Pressure variation in a fluid at rest - For an incompressible fluid at rest, the pressure varies linearly with depth h: Notes Page 113 The Palimpsest Project The Archimedes Palimpsest - can look at his original writings - it is online (and at a museum) Problem Wednesday, April 28, 2010 08:27 The factor of safety for tipping of the concrete dam is defined as the ratio of the stabilizing moment about O due to the dam's wieight divided by the overturning moment about O due to the water pressure. Determine this factor if the concrete has a specific weight of 150 lb/ft^3 and for water 62.4 lb/ft^3 Notes Page 114 Hydrostatic Forces There are practice problems online Friday, April 30, 2010 08:00 Pressure varies linearly from surface This is gage pressure Pressure at a point is independent of direction in fluids Notes Page 115 Problem: gate Friday, April 30, 2010 08:03 The 2m wide rectangular gate is pinned at its center A and is prevented from rotating by the block at B. Determine the reactions at these supports due to hydrostatic pressure - density of water = 1.0 Mg/m^3 Note that this is independent of H Note that force at pin will be much larger than at block Notes Page 116 Problem: curved surface Friday, April 30, 2010 08:23 The arched surface AB is shaped in the form of a quarter circle. If it is 8m long, determine the horizonal and vertical components of the resultant force caused by the water acting on the surface, use 1Mg/m^3 for water density Notes Page 117 Grain bin Friday, April 30, 2010 08:42 Notes Page 118 Another hydrostatic problem Next lecture is a review of the course Monday, May 03, 2010 07:58 The gasoline tank is constructed with elliptical ends. Determine the resultant force F and its location (the center of pressure) yp on these ends if the tank is half full. Take γ=41lb/ft3. Notes Page 119 Notes Page 120 Final Exam: Information Wednesday, May 05, 2010 07:58 12 May 2010, 8 to 11 228 Natural History Building Comprehensive - no conflicts Should be able to work and check completely in 1.5 hours, but will have the full 3 hours. Work any 5 of the 6 problems given. Write OMIT on the problem to be omitted. No calculators. Only a pen or pencil is needed. (Can bring a ruler and eraser.) Explain you work, provide clear diagrams, work logically and neatly. Notes Page 121 FOLLOW INSTRUCTIONS Final Exam Review: Wednesday, May 05, 2010 08:03 Force vectors - Vector operations, law of gravitation, force vectors along a line, components parallel and perpendicular to a line Free body diagrams - Free body diagrams for particles; equilibrium - A drawing of a body or a part of a body on which ALL forces acting on the body are shown. PERIOD - CUT the body Idealizations - Frictionless pulley - T same on each side - Linearly elastic spring Moment of a force - About a point, and about a line; couples - MP=rxF (note: order matters) - even in case of couples - r from -F to F - Moment parallel and perpendicular to line Equipollence (equivalence) - The sum of moments and sum of forces are the same in two systems - Resultant: an equipollent system consisting of one force and one couple at a point specified - Special equipollent systems: Concurrent forces - no couples, resultant is force only Coplanar forces - couples normal to plane, resultant is force only Parallel forces - can create a force only resultant - Screwdriver (wrench) - always possible as long as force resultant does not vanish Resultant force and moment acting along line of that force Pushing and turning at one point Distributed forces - Often along lines or planes - must calculate equivalent force - Use Simpson's rule if only have numerical data Types of support - Roller/slider - 1 unknown (1 force) - Pin - 2 unknowns (2 force) - Fixed - 3 unknowns (2 force, 1 moment) - Others (hinges…) General equilibrium - extended bodies - Written by Bernoulli's and Euler - For equilibrium sum of forces and sum of moments equals ZERO - Two-force members - equal force in opposite direction - even if body is not straight - Three-force members - either concurrent or parallel (but parallel is technically concurrent at infinity) Trusses - Method of joints - Method of sections - really can only cut 3 members if going to find the forces - BUT can have zero force members - Could apply same idea to 3-D structures Frames, machines - Members which may be built in at ends - Frame - doesn't move; machine - can move Internal forces - Normal (axial) - Transverse shear - Torsional (torque) - Bending - Shear-force and bending-moment diagrams - graph them and shade under the curve - Differential and integral relations + jumps Friction - Coulomb friction; belts - Wedges; screws Centroids - Composite bodies Moments of inertia - Ordinary and polar - Parallel axis theorem, composite areas Virtual work - Imaginary displacements - Method of choice with many interconnected joints with no work done at joints Hydrostatic forces - Varies linearly from surface Notes Page 122 Grades will be posted app. May 19