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# ME 270 Exam 1 Study Guide ME 270

Purdue

GPA 3.1

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## Popular in Basic Mechanics 1

## Popular in Mechanical Engineering

This 5 page Study Guide was uploaded by Abby Frazier on Saturday September 24, 2016. The Study Guide belongs to ME 270 at Purdue University taught by Professor Daniel Hoyniak in Fall 2016. Since its upload, it has received 62 views. For similar materials see Basic Mechanics 1 in Mechanical Engineering at Purdue University.

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Date Created: 09/24/16

ME 270 Exam 1 Study Guide 1 STUDY GUIDE ME 270 EXAM 1: Learning objectives: LECTURE 1: • Newton’s Laws -1 Law: every particle remains at rest, or continues to move in a straight line with ! ! ! ! ∑ F =0 ∑ M =0 uniform motion, if there are no unbalanced forces!! !! -2 Law: the rime rate of change of the linear momentum of a particle is proportional to the unbalanced forces acting on it and occurs in the direction in which the force acts F = d mv! !∑ dt( ) -3 Law: To every action there is an equal and opposite reaction—two bodies acting on each other are equal in magnitude and opposite in direction • Basic units/unit conversions/numerical accuracy -SI system of units: mass (kg), length (m), time (s), 2 1lbs)s 1slug!= ft -British Gravitational system of units: force (lb), length (ft), time (s) and -If x and y are defined to three significant digits, and t to only two, the answer for z will be accurate only to two significant digits • Basic trig operations a b c c =a +b −2abcos γ ( ) sin( ) =sin( )= sin( ) Law of cosines: !! Law of sines: !! • Basic vector addition: ???? = ???? + ???? ; ???? = ???? ????▯+ ???? ????▯+ ???? ???? ▯ ▯ ▯ ▯ ???? = ???? = ???? ▯ + ????▯ + ???? ▯ ???? = ???? + ???? = (???? + ▯ )???? ▯ ???? + ???? ????▯+ (????▯+ ???? )???? ▯ ▯ LECTURE 2: • Position vectors – used to define the position in space a point has with respect to a refrence point rAB = <(B x A )Y (B –yA ),y(B -ZA )> z • Magnitude of a position vector – length of a position vector ▯ ▯ ▯ |rAB = (???? − ▯ ) +▯(???? −???? ) ▯ (????▯− ???? ) ▯ ▯ • Unit vectors – dimensionless vector that is used to describe the direction of a vector u = r /|r | AB AB AB • Force vectors – represents the magnitude and direction of a force F AB = |FABu AB = FAB AB • Direction angles for a vector ME 270 Exam 1 Study Guide 2 Theta x arccos(r AB,x|rAB) --> angle between i and r vectors Theta y arccos(r AB,y|rAB) --> angle between j and r vectors Theta z arccos(r AB,z|rAB) --> angle between k and r vectors • Make engineering estimates of these quantities LECTURE 3: • Dot product – results in a scalar quantity A · B = B · A = |A||B|cos(theta) • Projection of a vector in another direction A proj (A ·u)u -if A · B >0 then the projection of B is in the direction of A -if A · B <0 then the projection of B is in the opposite direction of A -if A · B = 0 then theta = 90 and A is perpendicular to B - if A · B = |A||B| then theta = 0 and A is parallel to B • The angle between two vectors Theta = arccos(A ·u/|A||B|) • Make engineering estimates of these quantities LECTURE 4/5: • Drawing free body diagrams representing forces acting on a particle -provides a coordinate system to establish a solution methodology -provides a graphical display of all forces/moments acting on the rigid body -provides a record of geometric dimensions needed for establishing moments of the forces -external: applied forces which are typically known or prescribed such as forces due to gravity, springs, or cables -reaction: those forces that result from the external forces—required to keep the body from moving • Determine the forces required for particle equilibrium—can solve for up to two unknowns ∑F = ∑F x +∑F y 0 • Making engineering estimates of the forces required for particle equilibrium LECTURE 6: • Use vector cross product to calculate the angle between two vectors A x B = (A B – B A )i + (A B – B A )j + (A B – B A )k y z y z x z x z x y x y |A x B| = |B x A| = |A||B|sin(theta) àthe components of B perpendicular to A multiplied by the magnitude of A or vise versa • Use vector cross product to define a vector normal to the plane defined by two vectors Unit normal = (A x B)/|A x B| • Make engineering estimates of these quantities LECTURE 7: • Use vector product to calculate moment of a force about a point M o r oA x F = (rxFy– r y xk àpositive moment = counterclockwise, negative moment = clockwise |M O = |roA|F|sin(theta) -can determine the direction of moment using right hand rule -the point A can be any point along the line of action of the foce vector ME 270 Exam 1 Study Guide 3 -use trig relationships to determine moments for 2D problems -use vector cross product to determine moments about a point for 3D problems • Determine the equivalent force-couple for a system of moments o Force couples are a pair of forces that are equal in magnitude, parallel, and opposite in direction |M| = |F|d or M = rab F -d is the perpendicular distance between forces -rabs any position vector between the lines of action of the two forces -force couples cause no net force -the moment due to a force couple is the same regardless of the point the moment was summed about • Make engineering estimates of the moment LECTURE 8: • Inspect the supports of a rigid body in order to determine the nature of the reactions o Cable—one unknown; reaction is a tension force which acts AWAY from the member in the direction of the cable o Weightless link—one unknown; reaction is a force which acts alone the axis of the link o Roller—one unknown; reaction is a force which acts perpendicular to the surface at the point of contact o Roller/pin in confined smooth slot—one unknown; the reaction is a force which acts perpendicular to the surface at the point of contact o Smooth contacting surface—one unknown; the reaction is a force which acts perpendicular to the surface at the point of contact o Member pin connected to collar on smooth rod—one unknown; the reaction is a force which acts perpendicular to the rod o Smooth pin or hinge—two unknowns; the reactions are two components of force, or the magnitude and direction of the resultant force o Member fixed connected to collar on smooth rod—two unknowns; the reactions are the couple moment and the force which acts perpendicular to the rod o Fixed support—three unknowns; the reactions are the couple moment and the two force components, or the couple moment and the magnitude and direction of the resultant force • Use that information to draw a free body diagram LECTURE 9: • Evaluate the unknown reactions holding a rigid body in equilibrium by solving the equations of static equilibrium 2D Equations: 1. Two force equatios in the x &y directions and a moment equation in the z direction ∑F =x0 ∑F x 0 ∑M =o0 2. One force equation, in the x or y direction, and two moment equaitons about different points ∑F =x0 ∑M =B0 ∑M = c 3. Three moment equations about different points that cannot be collinear ME 270 Exam 1 Study Guide 4 ∑M =A0 ∑M =B0 ∑M =c0 • Recognize situations of static indeterminacy and partial and improper constraint on the basis of the solvability of the equations of static equilibrium o Static indeterminacy—when a system has more constraints than is necessary to hold the system in equilibrium o Static determinacy—when a system has a sufficient number of constraints to prevent motion without any redundancy o Partial constraint—when there is an insufficient number of reaction forces to prevent motion of the system o Improper constraint—when a system has a sufficient number of reaction forces but one or more are improperly applied so as not to prevent motion of the system LECTURE 10/11: • Evaluate the unknown reactions holding a rigid rigid body in equilibrium by solving the equations of static equilibrium 3D Equations for Static Equilibrium: F R ∑F = 0 M RO = ∑M o 0 -6 component equations for 3D equilibrium ∑F x 0 ∑F =y0 ∑F z 0 ∑M =x0 ∑M =y0 ∑M =z0 1. Cable—one unknown; reaction is a force which acts from the member in the known direction of the cable 2. Smooth surface support—one unknown; the reaction is a force which acts perpendicular to the surface at the point of contact 3. Roller—one unknown; the reaction is a force which acts perpendicular to the surface at the point of contact 4. Ball and socket—three unknowns; the reaction are three rectangular force components 5. Single journal bearing—four unknowns; the reactions are two force and two couple- moment components which act perpendicular to the shaft 6. Single journal bearing with square shaft-five unknowns; reactions are two force and three couple-moment components* 7. Single thrust bearing—five unknowns; reactions are three force and two couple- moment components* 8. Single smooth pin—five unknowns; the reactions are three force and two couple- moment components* 9. Single hinge—five unknowns; the reactions are three force and two couple-moment components* 10. Fixed support—six unknowns; the reactions are three force and three couple-moment components *Note: The couple moments are generally not applied if the body is supported elsewhere LECTURE 12: • Determine the equivalent loading and its location of a given distributed load ME 270 Exam 1 Study Guide 5 o The distributed load may be replaced by a single concentrated load -the magnitude of the concentrated load is equal to the area under the load curve W = Area = integral W(x)dx from 0àL -the line of action passes through the centroid of that area x̄= (integral xw(x)dx from 0àL)/(integral w(x)dx from 0àL) • Evaluate the support reactions resulting from the equivalent load • Make engineering estimates of the equivalent loads and its location LECTURE 13: • Determine the volume, mass, centroid, and center or mass of a composite structure o Centroid = center of a line, area, or volume o Center of mass = gravitational center of a line, area, or volume o The centroid and center of mass coincide when the density is uniform throughout the part Centroid by composite parts: ▯ ▯ ▯ Line: L = ▯▯▯ ???? ▯ ; Lx̄= ▯▯▯ ???? ▯???? ▯ ▯ = ▯▯▯ ???? ▯???? ▯ ▯ Area: A = ▯ ???? ; Ax ̄ ▯ ???? (???? ) ; A???? = ▯ ???? (???? ) ▯▯▯ ▯ ▯ ▯▯▯▯ ▯ ▯ ▯ ▯▯▯▯ ▯ ▯ ▯ Volume: V = ▯▯▯ ????▯ ; Vx̄= ▯▯▯???? ▯???? )▯ ▯ ; Vx ̄ ▯▯▯???? ▯???? ▯ ▯ Center of mass by composite parts: ▯ ▯ m = ▯▯▯▯???? ▯ = ▯▯▯???? ▯ ▯ mx = (????▯)???? ▯ = ▯▯▯(???? ▯???? ▯ ▯ ▯▯▯ m???? = ▯ (???? )???? = ▯ (???? )???? ???? ▯▯▯ ▯ ▯ ▯▯▯ ▯ ▯ ▯ m???? = ▯ (???? )???? = ▯ (???? )???? ???? ▯▯▯ ▯ ▯ ▯▯▯ ▯ ▯ ▯ • Estimate the volume, mass, centroid and center of mass of a body LECTURE 14: • Determine the volume, mass, centroid, and center of mass using integral calculus Line: L = integral dl L???? = integral (???? ▯dl L???? = integral (???? ▯dl Area: A = integral dA A???? = integral (???? ▯dA A???? = integral (???? )▯A Volume: V = integral dV V???? = integral (???? ▯dV V???? = integral (???? )▯V • Make an engineering estimate of the volume, mass, centroid and center of mass of a body

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