1. 1 MPA104 Mechanical Design Engineering Statics Duncan Price IPTME, Loughborough University © Copyright: D.M.Price@lboro.ac.uk (2006)

2. 2 Mechanics of Materials Mechanics of Materials can be divided into three categories:  Mechanics of Rigid Bodies  Statics – bodies at rest  Dynamics – bodies in motion  Mechanics of Deformable Bodies  Mechanics of Fluids

3. 3 Statics Statics is thoroughly used in the analysis of structures, for instance in architectural and structural engineering. Strength of materials is a related field of mechanics that relies heavily on the application of static equilibrium.

4. 4 Course Content 12 Lectures: 1-5Forces, springs Free body diagrams Resolution of forces (tutorial sheet) Equilibrants and resultants of forces (tutorial sheet) 7Class test (calculators needed) 8-11Levers, moments, reactions and centre of gravity (tutorial sheet) 12 Online test

5. 5 Newton’s Laws  Sir Isaac Newton (1642-1729)  Principia 1687  Formulated three laws on which all conventional motion is based

6. 6 Newton I A particle remains at rest or continues to move at a constant speed in a straight line unless there is a constant force acting on it.  The most important law  The one that most people don’t understand  The only one that doesn’t have an equation

7. 7 Newton II The resultant force on a particle is equal to the rate of change of momentum of the particle. The form F=ma is only valid if the mass is constant.

8. 8 Newton III The forces of action and reaction between interacting bodies are equal in magnitude and opposite in direction.  The force of the Earth on the Sun has the same magnitude as the force of the sun on the earth  The force of a tennis ball on a racket has the same magnitude as the force of the racket on the ball

9. 9 Newton’s Law of Gravitation (1) F = G M m / r 2 Where: M & m are particle masses G is the universal constant of gravitation (6.673 x 10-11 m 3 /kg s 2 ) r is the distance between the particles. F in Newtons (kg m/s 2 )

10. 10 Newton’s Law of Gravitation (2) On Earth: F = m g Where: m is the mass of the body in question (in kg) g = 9.81 m/s 2

11. 11 Spring Forces (1) If a spring is stretched from L 0 to L it exerts a force on the object to which it is attached: F = K (L – L 0 ) K = spring constant = force required to stretch spring by unit length (N/m). K depends on spring material and design. LOLO L FF

12. 12 Spring Forces (2) F = K (L – L 0 ) is also known as Hooke’s Law: The force exerted by a spring is proportional to its extension. Slope = F/ (L – L 0 ) = K F (L-L o )

13. 13 Free body diagrams (1)  FBD is an essential step in the solution of all problems involving forces on bodies  it is a diagram of the external surface of the body - not interested in internal forces  all other bodies in contact with the one we are interested in are replaced by vectors

14. 14 Free body diagrams (2) Sketch of person standing mg R1R1 R2R2 F=ma R 1 +R 2 -mg=ma, but no acceleration so, R 1 +R 2 =mg

15. 15 Free body diagrams (3) sketch mg T free body diagram

16. 16 Free body diagrams (4) Rules:  clear sketches  draw in the correct orientation  show all forces acting on the body  don’t show any internal forces between different parts of the body  show the forces not the components

17. 17 Representation of a force by a vector A force has the following characteristics:  Magnitude  Direction  Point of application A quantity which has magnitude and direction is a vector. A quantity which has magnitude only is a scalar.

18. 18 Vector addition F1F1 F2F2 F3F3 F1F1 F2F2 F3F3 FTFT These three forces act together on the particle. Their effect is equivalent to a single force which is the vector sum of the forces. F T is the resultant of the forces F 1, F 2 and F 3 FT=F1+F2+F3FT=F1+F2+F3

19. 19 Trigonometry (1) A B C  Pythagoras Theorem: A 2 = B 2 + C 2 Internal angles of a triangle add up to 180° 90 -  90°

20. 20 Trigonometry (2) A (hypotenuse) B (opposite) C (adjacent)  sin e (  ) = B/A cos ine (  ) = C/A tan gent (  ) = B/C

21. 21 Trigonometry (3) sin(45°) = 1/√2 = 0.707 cos(45°) = 1/√ 2 = 0.707 tan(45°) = 1/1 = 1 sin-1(0.707) = 45° cos-1(0.707) = 45° tan -1 (1) = 45° √2 m 1 m 45°

22. 22 Trigonometry (4) sin(30°) = 1/2 = 0.5 cos(30°) = √3/2 = 0.866 tan(30°) = 1√3 =0.577 sin(60°) = √3/2 = cos(30°) cos(60°) = 1/2 = sin(30°) tan(60°) = √3/1 = 1.73 Generally: cos(  ) = sin(90°-  ) tan(  ) = sin(  )/cos(  ) sin 2 (  ) + cos 2 (  ) = 1 1 m 2 m √3 m 30° 60°

23. 23 Sine and Cosine functions  sin(  ) cos(  )

24. 24 Tangent function  tan(  )

25. 25 Resolving forces (1) Forces can be broken down into any number of component forces It is often convenient to choose two perpendicular directions for resolution F FyFy F = F x +F y F = (F x 2 +F y 2 ) 1/2 Scalar magnitudes FXFX

26. 26 Resolving forces (2) F FxFx FyFy  F x =F cos  F y =F sin   =tan -1 (F y /F x ) If the components are perpendicular, they may be added independently F1F1 F2F2 F3F3 FT=F1+F2+F3FT=F1+F2+F3 F Tx =F 1x +F 2x +F 3x F Ty =F 1y +F 2y +F 3y

27. 27 Resolving forces (3) 500 N 30° F x =500 cos 30° = 433 N F y =500 sin 30° = 250 N

28. 28 Resolving forces (4) F Tx = F 1x +F 2x +F 3x = 200 – 50 cos 30° + 80 cos 80° = 171 N F Ty = F 1y +F 2y +F 3y = 0 – 50 sin 30° – 80 sin 80° = - 104 N (i.e. downwards) F 1 = 200 N F 2 = 50 N F 3 = 80 N 30° 80°

29. 29 Resolving forces (5) F x = 171 N F y = 104 N F T = √(F x 2 +F y 2 ) = √(171 2 + 104 2 ) = 200 N  = tan -1 (F y /F x ) = tan -1 (-104/171) = -31°  FTFT

30. 30 Resolving forces (6) 30° 75° 100 N A BC horizontal components: F AC cos75° = F AB cos30° F AC = F AB (cos30°/cos75°) vertical components: F AC sin75° + F AB sin30° = 100 N F AB (cos30°/cos75°) sin75° + F AB sin30° = 100 F AB = 26.8 N F AC = 89.7 N

31. 31 Resultant If two or more forces at a point they can be replaced by a single force known as a resultant.

32. 32 Equilibrium When two or more forces act upon a body and are so arranged that the body remains at rest or moves at a constant speed in a straight line*, the forces are said to be in equilibrium. * i.e. Newton’s 1 st law

33. 33 Equilibrant If two or more forces act at a point and are not in equilibrium a force equal in magnitude and opposite in direction to their resultant must be applied to restore equilibrium. Such a force is called the equilibrant.

34. 34 Equilbrant and resultant a e 30 N 50 N 10 N 20 N ae = resultant ea = equilibrant

35. 35 “Y-hangs” What happens if we change the angle? 90° 750 N A B C Horizontally ACsin45° = BCsin45° AC = BC Vertically ACcos45° + BCcos45°=750 N AC + BC =750 N/cos45° 2 AC = 750 N/cos45° AC = 530 N = BC

36. 36 Trusses (1) One of the basic methods to determine loads in individual truss members is called the Method of Joints. Each joint is treated as a separate object and a free-body diagram is constructed for the joint.

37. 37 Trusses (2) 30°45° A B C 500 N Horizontal forces ABcos30° = ACcos45° AC = ABcos30°/cos45° Vertical forces ABsin30° + ACsin45° = 500 N ABsin30° + AB(cos30°/cos45°)sin45° = 500 N AB = 366 N AC = 448 N

38. 38 Trusses (3) 30°45° A B C 366 N Horizontal forces BC = ABcos30° = ACcos45° = 316 N Vertical forces at B R B = ABsin30° = 183 N Vertical forces at C R C = ACsin45° = 317 N Check R B + R C = 183 N + 317 N = 500 N 448 N

39. 39 Moment of a force The product of the force and the perpendicular distance from the line of action of the force to the axis. Moment about a = F d (units N m) d F a

40. 40 Principle of Moments (1) If a body is at rest under the action of several forces, the total clockwise moment of the forces about any axis is equal to the anticlockwise moment of the forces about the same axis.

41. 41 Principle of Moments (2) 2 m 1.5 m x N60 N a b y N Moments about a: 60  2 = x  1.5 x = 120/1.5 = 80 N Moments about b: (2 + 1.5)  80 = 2  y y = 280/2 = 140 N (ignoring mass of beam) Or more simply: y = 60 + 80 = 140 N

42. 42 Principle of Moments (3) 2 m 1.5 m x N60 N a b y N Now let the beam weigh 5 kg Additional force of 49 N acting centrally. Moments about a: (60  2) + (49  0.25) = x  1.5 x = 144.5/1.5 = 88.2 N Reaction of support y = 60 + 49 + 88.2 = 197.2 N 49 N1.75 m

43. 43 Principle of Moments (4) 1.5 m ab y N Now support the unloaded beam at a and b: Moments about b: 49  1.25 = y  1.5 y = 61.25/1.5 = 40.83 N Therefore x = 49 – 40.83 = 8.17 N 49 N1.75 m x N 0.5 m

44. 44 Principle of Moments (5) 1.5 m ab y N Now support the unloaded beam at a, b and c: Moments about b: 49  1.25 = y  1.5 + z  3 Moments about a: 49  0.25 = x  1.5 - z  1.5 Moments about c: 49  1.75 = x  3 + y  1.5 49 N1.75 m x N 0.5 m c z N

45. 45 Principle of Moments (6) 1.75 m Now support the unloaded beam at c only… 49 N1.75 m c z N

46. 46 Centre of Gravity (1) The centre of gravity of an object is a point at or near the object through which the resultant weight of the object acts.

47. 47 Centre of Gravity (2) If a vertical line through the centre of gravity falls outside the base upon which the body relies for stability, overturning will result, unless precautions, such as tying the base down, are taken.

48. 48 Centre of Gravity (3) For a rectangle or square, the centre of gravity occurs at the centre of the section.

49. 49 Centre of Gravity (4) For a triangular section the centre of gravity occurs at a point 1/3 of the height from the base.

50. 50 Centre of Gravity (5) For a circular section the centre of gravity occurs at the centre of the circle. r

51. 51 Centre of Gravity (6) To determine the centre of gravity for a compound section: 1.Divide the section into several parts (i.e. rectangles and triangles, circles) so that the centre of gravity of each individual part is known. 2.Determine the area and position of centre of gravity of each part. 3.Take moments about a convenient axis to determine the centre of gravity of the whole body. This is based on the principle that, along any one axis (or in any one direction): when the distance is measured from the same point in each case.

52. 52 Centre of Gravity (7) Determine the position of the centre of gravity of the L-section shown below: 0.5 m 1 m 0.75 m

53. 53 Centre of Gravity (8) Divide up section into two rectangles, identify c.o.g of each relative to O and calculate area. 0.5 m 0.25 m Area 1 = 0.5 m 2 0.25 m 0.625 m Area 2 = 0.125 m 2 O Total area = 0.625 m 2

54. 54 Centre of Gravity (9) Let X = location in X axis of c.o.g 0.5 m 0.25 m 0.5 m 2 0.125 m 2 O Total area = 0.625 m 2 0.625 X = 0.5  0.5 + 0.125  0.25 X = 0.45 m

55. 55 Centre of Gravity (10) 0.5 m 2 0.25 m 0.625 m 0.125 m 2 O Total area = 0.625 m 2 Let Y = location in Y axis of c.o.g 0.625 Y = 0.125  0.625 + 0.5  0.25 Y = 0.325 m

56. 56 Centre of Gravity (11) Now let us drill a hole in the object, where is the new centre of gravity? 0.4 m 0.25 m Area 3 = 0.12564 m 2 Total area = 0.49936 m 2

57. 57 Centre of Gravity (12) Find new X and Y: 0.25 m0.12564 m 2 Total area = 0.49936 m 2 0.49936 X = 0.5  0.5 + 0.125  0.25 – 0.12564  0.25 X = 0.5 m 0.49936 Y = 0.5  0.25 + 0.125  0.625 – 0.12564  0.25 Y = 0.344 m

58. 58 Centre of Gravity (13) Another example:

59. 59 Centre of Gravity (14) Divide the section up into rectangles and triangles:

60. 60 Centre of Gravity (15) Select the first solution (but either method would give the same answer). Area 13x1.5/2=2.25m 2 Area 24x1.5=6.00m 2 Area 33x1.0=3.00m 2 Total =11.25m 2

61. 61 Centre of Gravity (16) Select two axes A-A and B-B, at the extreme edge of the figure. The distance to the centre of gravity of each section can then be calculated from these axes.

62. 62 Centre of Gravity (17) Let X be the horizontal distance to the centre of gravity of the whole figure measured from A-A and Y be the vertical distance to the centre of gravity of the whole figure measured from B-B. Taking moments about A-A: 11.25X = (2.25x2.25)+6.0x2.25)+(3.0x1.5) = 5.6+13.5+4.5 = 23.6 X = 23.6/11.25 = 2.1 m

63. 63 Centre of Gravity (18) Taking moments about B-B: 11.25Y = (2.25x6.0) + (6.0x3.0) + (3.0x0.5) = 13.5 + 18.0 + 1.5 = 33.0 Y = 33.0/11.25 = 2.9m

64. 64 Centre of Gravity (19) Now let us support the section at the middle and far right… What is the ratio of forces on each support? 3 m X Y 1.5 m

65. 65 Centre of Gravity (20) Moments about Y. 1.5 R X = 0.9 R X = 0.6 therefore R Y = 0.4 3 m X Y 1.5 m