Notes for Week of January 11
Notes for Week of January 11 CHEM1041
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This 7 page Class Notes was uploaded by Macen Notetaker on Friday January 15, 2016. The Class Notes belongs to CHEM1041 at University of Cincinnati taught by Dr. Waddell in Spring 2016. Since its upload, it has received 53 views. For similar materials see General Chemistry 2 in Chemistry at University of Cincinnati.
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Date Created: 01/15/16
BME 1030 Statics amp Basic Strengths of Materials Notes for Week of January 11 Chapter 1 Mechanics The branch of science that deals with the response of physical bodies to the action of forces Mechanics is comprised of the following three areas 1 Mechanics of Rigid Bodies a Statics b Dynamics i Kinematics ii Kinetics 2 Mechanics of Deformable Bodies a Strengths of Materials 3 Mechanics of Fluids a Fluid Statics b Fluid Dynamics This course will focus on Statics and Strengths of Materials Mechanics of Rigid Bodies Statics Deals with physical bodies that are acted on by balanced forces and as such are equilibrium This equilibrium is shown as the object being at rest or not accelerating Statics provides ways to determine support reactions and relationships between internal force distributions and external loads for stationary structures Dynamics Mechanics in relation to physical bodies not in equilibrium Kinematics Geometry of motion Mechanics of Deformable Bodies Shape or volume of physical body will change due to external forces Fluid Mechanics Mechanics of liquids andor gases at rest or in motion Fundamental Concepts and Principles Fundamental Quantities Space Geometric region where mechanics occurs Time Interval between two events Mass Quantitative measure of Inertia Inertia Property of a body that causes it to resist a change in motion Force Action of one body on another E BWJajn Me Forces are identified by their three main parts Point of Application Magnitude Direction Forces are vector quantities and as such are represented by an arrow Six Fundamental Principles 1 Parallelogram Law a If two vectors are placed with their tails together this point will represent the point of application of the Result vector The intersection of the two vectors lines of action will represent one end of the Resultant vector 2 Principle of Transmissibility a A vector can be translated along its own line of action and remain equivalent to the original vector 3 Newton s First Law Foundation of Statics a For a body at rest the sum of the forces is zero 4 Newton s Second Law Foundation for Dynamics a Force is the product of mass and acceleration Fma 5 Newton s Third Law a Force of Action Force of Reaction 6 Newton s Law of Gravitation Weight is given by the equation wmg where m is mass and g is gravitational acceleration This value is 981 ms2 in SI units and 322 fts2 in US Customary units Standard Units Quantity SI Unit US Unit Mass Kilogram kg Slug Length Meter m Foot ft Time Second sec Second sec Force Newton N Pound lbs 1Nkgn1Sluglbssec2 sec ft Stress load area Nm2 Pascal Pa Method of Problem Solving Most effective way to learn Statics as a student Solve a variety of problems The most effective engineer is one who has the ability to reduce a complicated problem to its simpler parts Chapter 2 Force Mechanical action exerted by one body on another Magnitude Amount or Size Direction Orientation and Sense Point of Application Origin Graphic Examples of Multiplication Division of a vector by a scalar If the lines of action of two vectors intersect then these vectors are concurrent Scalars amp Vectors Scalars Positive or Negative physical quantity completely specified by its magnitude For example length 3 m or mass 10 kg Vector Physical quantity expressed with magnitude and direction Graphically the parts of a vector can be identified as follows Magnitude is given by the length of the arrow Direction of the line of action is given by the arrows angle from the Xaxis Sense of direction is given the arrow s head or tip Multiplying a vector by a scalar I w 4 f 1139 x 2 7 be 39 7 y I l X A M Aquot I F x A W 9quot 745 2 Review of Plane Trigonometry Pythagorean Theorem 612 sz Sin 6 Opposite Hypotenuse Cos G Adjacent Hypotenuse Tan 6 Opposite Adjacent Sin 62 Cos 621 Law of Cosines A2 BZCZ ZBCcosa BZA2CZ 2ACcos b CZA2 BZ 2ABcosc Law of Sines A B C nanbnc Resolution of a Force into Components One Approach Force Triangle Another Approach Lines of Action Both of these examples show methods of determining both components of a vector marked Q and P Rectangular Components of a Force FFXFy FXFcosG FyFsinG F2 F02 Fm Tan 6 Fy FX Cartesian Vector Notation Using unit vectors to denote direction Finij If two vectors are added the components of the Resultant are equal to the sum of the respective components of the vectors added F1 F2 R x F1x F2x Ry F1y ng
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