ME 201 Engineering Mechanics: Statics. Final Exam Review

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1 ME 201 Engineering Mechanics: Statics inal Exam Review

2 inal Exam Testing Center (Proctored, 1 attempt) Opens: Monday, April 9 th Closes : riday, April 13 th Test ormat 15 Problems 10 Multiple Choice (75%) 5 ill-in (25%) May be multiple parts/answers to each Problem No Time Limit Allowable Resources One side of one 8.5 x 11 sheets of notes Calculators

3 inal Exam This inal Exam is CLOSED BOOK/NOTES with the exception of the following allowable resource materials: * One side of one 8.5" x 11" sheet of paper * Calculators * Excel or Mathematica may be used as calculator (no external files) You may NOT consult with anyone who has not taken the exam concerning the Exam. You may NOT use other resources such as the Internet, computers, homework, or textbooks. I have read and commit to following the instructions above.

4 Evaluation Weighting Item Weight Practice Problems 10% Homework Exercises 15% Mastery Quizzes 34% Teach One Another/Grad Plan 7% inal Exam 34%

5 ME 201 Engineering Mechanics: Statics Chapter 1: Introduction Newton s Laws of Motion Units

6 Units of Measurement SI US, English Length m - meters ft feet Time s seconds s seconds Mass kg kilograms slug orce N - newton lb pounds

7 ME 201 Engineering Mechanics: Statics Chapter 2 Vectors Resultants

8 Resolution of a orce into Rectangular Components y Vector notation: = x i + y j Components: x = cos θ y = sin θ O θ x y 2 2 x y x

9 Equilibrium of a Particle 0 In vector notation: x i + y j = 0 Or, for this equation to be satisfied x = 0 y = 0

10 Direction Cosines cos x cos y cos z z x cos y z cos z z cos x α y β γ y y z y x x x

11 Direction Cosines Relationships U x i y j z k U = cosα i + cosβ j + cosγ k 2 1 cos cos 2 cos 2

12 ME 201 Engineering Mechanics: Statics Chapters 3&4 More Vectors Moments System of orces

13 Moment of a orce Scalar ormulation M O = d Where d = moment arm or perpendicular distance from axis O to the line of action of the force Line of action d O

14 Position Vectors General Case any 2 points A & B: r AB = (x B -x A ) i + (y B -y A ) j + (z B -z A ) k Z r AB B A Y X

15 Cross Product C = A x B = A B sin θ U C U C C=AB sin θ AxB i Ax j Ay k Az A θ B Bx By Bz Taking minors (with j negative) =(A y B z -A z B y )i (A x B z -A z B x )j +(A x B y -A y B x )k

16 Moment of a orce Vector ormulation M O = r x where r = position vector from O to any point lying on the line of action of Magnitude of M Direction M = r sin θ = (r sin θ) = d Perpendicular to the plane containing r and such that the direction of M is specified by right-hand rule Vector Notation M O =(r y z -r z y )i (r x z -r z x )j +(r x y -r y x )k

17 Dot Product Dot Product is a scalar operation used in projecting vectors onto a given axis A B = A B cos θ A B = A x B x + A y B y + A z B z Where 0º θ 180º cos A x B x A A y B y B A B z z θ A A B B A B

18 Projection of a Vector onto a Given Axis The scalar projection of A along a line is determined from the dot product of A and the unit vector U which defines the direction of the line A U = A U cos θ = A cos θ θ U = A U A A U

19 Moment of a orce about a Specified Axis Remember And M O = r x r is vector from O to line of action of defining a unit vector U A and perform a dot product operation M O = U a (r x ) rx M a x u r r i x x ax x r j y u r y y ay y k r z u r z z az z

20 Moment of a Couple: Scalar M = d Where is magnitude of one force d is the perpendicular distance (or moment arm) between the forces direction by right-hand rule + - d -

21 Moment of a Couple: Vector M = r x Where r is position vector between the lines of action of the forces is one forces Think of taking the moment of one of the forces about a point lying on the line of action of the other force Result is referred to as a free vector it can act at any point x z r - y

22 ME 201 Engineering Mechanics: Statics Chapter 5 Equilibrium

23 Equilibrium or equilibrium Balance of forces Prevents translation Balance of moments Prevents rotation =0 M =0 We will use both scalar (2D) and vector (3D) analysis methods

24 ree Body Diagrams 1. Sketch shape free from constraints & connections 2. Include external forces and couples Applied loadings Reactions at supports Weight of the body 3. Include dimensions

25 Common Support Reactions Simple Support Pin or Hinge Support ixed Support

26 Solving Problems Begin with BD (ree Body Diagram) Use Equations of Equilibrium to formulate equations containing unknown forces, reactions, & moments x =0 y =0 M o =0 Solve simultaneous equations (or look for equation with single unknown)

27 2-orce Members If the following conditions exist orce applied at only 2 points on member No couples Called a 2-force member Then The line of action passes through points A & B A is equal and opposite in magnitude and direction to B Rotational or moment equilibrium B B A A A B

28 Equilibrium in 3-Dimensions Support Reactions A force is developed by a support that restricts the translation of the attached member A couple is developed when rotation of the attached member is prevented = 0 x = 0 y = 0 z = 0 M = 0 Mx = 0 My = 0 Mz = 0

29 ME 201 Engineering Mechanics: Statics Chapter 6 Analysis of Structure Method of Joints Zero orce Members Method of Sections rames Machines

30 Simple Truss To Design members and connections of a truss, necessary to determine forces developed in each member Assumptions: All loadings are applied at the joints Members are joined together by smooth pins Assumptions satisfactory for bolted or welded joints

31 Method of Joints If truss is in equilibrium, each joint must also be in equilibrium Approach: Sketch ree Body Diagram of Entire Truss Solve for Reactions Sketch ree Body Diagrams of Each Joint 2 unknown limit per BD x =0 y =0 M o =0

32 Zero orce Members If only 2 members form a truss joint AND no external load or support reaction is applied to the joint, then the members must be Zero orce Members X X X X

33 Zero orce Members If 3 members form a truss joint for which 2 members are collinear, the 3 rd member is a Zero orce Member, provided no external force or support reaction is applied to the joint. X X

34 Method of Sections If a body is in equilibrium, then each part of the body must also be in equilibrium Approach Sketch ree Body Diagram of Entire Body Solve for Reactions Cut &Sketch ree Body Diagrams of Section(s) 3 unknown limit per BD x =0 y =0 M o =0 Look for Direct Solutions

35 rames & Machines Obtain solutions of rames and Machines by using Multiple ree Body Diagrams Note equal but opposite forces in opposing BDs Utilize properties of 2-force members where appropriate

36 ME 201 Engineering Mechanics: Statics Chapter 8 riction

37 riction Theory In general: μ N s = μ s N k = μ k N (for impending motion) (for sustaining motion, kinetic)

38 Guidelines for Solving riction Problems Use equations of equilibrium first μ N = μ N only if motion impends, or occurs Use care in determining direction of Always opposes motion/impending motion Opposite in 2 BD of same point

39 Guidelines for Solving riction Problems or sliding blocks N may not align with W Introduces Couple This distance may be found by summing M Can use to determine Slide vs. Tip With friction at multiple points need to examine motion at all points Simultaneous - only if physical link One slip vs. other not slip

40 ME 201 Engineering Mechanics: Statics Chapter 9 Centroids & Moments of Inertia

41 Centroids of Composite Areas Procedure Divide into numbered, simple areas ind centroid of each area Sum moment areas relative to origin X ~ xa A Y ~ ya A y 1 3 Shape Area x y xa ya x 3

42 Centroids And finally X ~ xa A A ~ xda A da Y ~ ya A A A ~ yda da

43 inding I of Composite Areas Procedure Divide into numbered, simple areas ind I c of each of each area ind composite centroid Use parallel axis theorem to find composite I I x I c Ad 2 y x

44 Moments of Inertia I x A y 2 da I y A x 2 da

45 Radius of Gyration Is defined as the distance from its moment of inertia axis at which the entire area could be considered as being concentrated without changing its moment of inertia A compressive member tends to buckle in the direction of its least radius of gyration k I A What are the units? in in 4 2 in

46 Preparing for the Exam

Unit 21 Couples and Resultants with Couples

Unit 21 Couples and Resultants with Couples Page 21-1 Couples A couple is defined as (21-5) Moment of Couple The coplanar forces F 1 and F 2 make up a couple and the coordinate axes are chosen so that