Concurrent Force System SPP.pptx

Charactteristics of forces;
Vector to represent forces;
Classification of forces;
What is force system;
Principles of forces;
Resultant of forces;
Components of forces;
Solved numericals;
examples;
Solved problems;
excercise;

1. Concurrent Force Systems Lecture-03 CL-101 ENGINEERING MECHANICS B. Tech Semester-I Prof. Samirsinh P Parmar Mail: [email protected] Asst. Professor, Department of Civil Engineering, Faculty of Technology, Dharmsinh Desai University, Nadiad-387001 Gujarat, INDIA

2. Objectives To understand the basic characteristics of forces To understand the classification of force systems To understand some force principles To know how to obtain the resultant of forces in 2D and 3D systems To know how to obtain the components of forces in 2D and 3D systems

3. Characteristics of forces Force: Vector with magnitude and direction Magnitude – a positive numerical value representing the size or amount of the force Directions – the slope and the sense of a line segment used to represent the force Described by angles or dimensions A negative sign usually represents opposite direction Point of application A point where the force is applied A line of action = a straight line extending through the point of application in the direction of the force The force is a physical quantity that needs to be represented using a mathematical quantity

4. Example  i 1000 N j Point of application Line of action magnitude direction

5. Vector to represent Force A vector is the mathematical representation that best describes a force A vector is characterized by its magnitude and direction/sense Math operations and manipulations of vectors can be used in the force analysis

6. Free, sliding, and fixed vectors Vectors have magnitudes, slopes, and senses, and lines of applications A free vector The application line does not pass a certain point in space A sliding vector The application line passes a certain point in space A fixed vector The application line passes a certain point in space The application point of the vector is fixed

7. Vector/force notation The symbol representing the force  bold face or underlined letters The magnitude of the force  lightface (in the text book, + italic) A  A or A  A

8. Classification of forces Based on the characteristic of the interacting bodies: Contacting vs. Non-contacting forces Surface force (contacting force) Examples: » Pushing/pulling force » Frictions Body force (non-contacting force) Examples: » Gravitational force » Electromagnetic force

9. Classification of forces Based on the area (or volume) over which the force is acting Distributed vs. Concentrated forces Distributed force The application area is relatively large compare to the whole loaded body Uniform vs. Non-uniform Concentrated force The application area is relatively small compare to the whole loaded body

10. What is a force system? A number of forces (in 2D or 3D system) that is treated as a group: A concurrent force system All of the action lines intersect at a common point A coplanar force system All of the forces lie in the same plane A parallel force system All of the action lines are parallel A collinear force system All of the forces share a common line of action

11. The external and internal effects A force exerted on the body has two effects: External effects » Change of motion » Resisting forces (reactions) Internal effects » The tendency of the body to deform  develop strain, stresses If the force system does not produce change of motion » The forces are said to be in balance » The body is said to be in (mechanical) equilibrium

12. External and internal effects Example 1: The body changes in motion Example 2: The body deforms and produces (support) reactions  The forces must be in balance F Not fixed, no (horizontal) support a F Fixed support Support Reactions

13. Principle for force systems Two or more force systems are equivalent when their applications to a body produce the same external effect Transmissibility Reduction = A process to create a simpler equivalent system to reduce the number of forces by obtaining the “resultant” of the forces Resolution = The opposite of reduction to find “the components” of a force vector  “breaking up” the resultant forces

14. Principle of Transmissibility Many times, the rigid body assumption is taken  only the external effects are the interest The external effect of a force on a rigid body is the same for all points of application of the force along its line of action  

15. Resultant of Forces – Review on vector addition Vector addition Triangle method (head-to-tail method) Note: the tail of the first vector and the head of the last vector become the tail and head of the resultant  principle of the force polygon/triangle Parallelogram method Note: the resultant is the diagonal of the parallelogram formed by the vectors being summed R  A  B  B  A R A B B A R

16. Resultant of Forces – Review on geometric laws Law of Sines Laws of Cosines c 2  a 2  b 2  2 ab co s  b 2  a 2  c 2  2 ac co s  a 2  b 2  c 2  2 ac co s  A B C c a b   

17. Resultant of two concurrent forces The magnitude of the resultant (R) is given by 1 2 1 2 R 2  F 2  F 2  2 F F cos  1 2 1 2 The direction (relative to the direction of F 1 ) can be given by the law of sines R 2  F 2  F 2  2 F F co s  R s in   F 2 s in  Pay attention to the angle and the sign of the last term !!!

18. Resultant of three concurrent forces and more Basically it is a repetition of finding resultant of two forces The sequence of the addition process is arbitrary The “force polygons” may be different The final resultant has to be the same

19. Resultant of more than two forces The polygon method becomes tedious when dealing with three and more forces It’s getting worse when we deal with 3D cases It is preferable to use “rectangular-component” method

20. Example Problem 2-1 Determine: The resultant force ( R) The angle  between the R and the x-axis Answer: The magnitude of R is given by R 2  90 2  60 2  2 ( 900) ( 600) co s 4 R  1413.3  141 3 lb The angle  between the R and the 900-lb force is given by s in   s i n ( 18  4 ) 600 1413.3   15.83 6 o The angle  therefore is   15.83 6  3 5  50. 8

21. Example Problem 2-2 Determine The resultant R The angle between the R and the x-axis

22. Another example If the resultant of the force system is zero, determine The force F B The angle between the F B and the x-axis

23. Force components

24. Resolution of a force into components The components of a resultant force are not unique !! The direction of the components must be fixed (given) R  A  B  ( G  I )  H  C  D  E  F

25. Steps: Draw lines parallel to u and v crossing the tip of the R Together with the original u and v lines, these two lines produce the parallelogram The sides of the parallelogram represent the components of R Use law of sines to determine the magnitudes of the components Parallel to v How to obtain the components of a force (arbitrary component directions)? Parallel to u 900 sin110 o F v s i n 2 5 o F u s i n 4 5 o   900 s i n 4 5 o sin110 900 s i n 2 5 sin110 o  405 N F v   677 N F u 

26. Example Problem 2-5 Determine the components of F = 100 kN along the bars AB and AC Hints: Construct the force triangle/parallelogram Determine the angles    Utilize the law of sines

27. Another example Determine the magnitude of the components of R in the directions along u and v, when R = 1500 N

28. Rectangular components of a force What and Why rectangular components? Rectangular components  all of the components are perpendicular to each other (mutually perpendicular) Why? One of the angle is 90 o ==> simple Utilization of unit vectors Rectangular components in 2D and 3D Utilization of the Cartesian c.s. A r b it r a r y rectangular

29. The Cartesian coordinate system The Cartesian coordinate axes are arranged following the right-hand system (shown on the right) The setting of the system is arbitrary, but the results of the analysis must be independent of the chosen system x Department of Mechanical Engineering y z

30. Unit vectors A dimensionless vector of unit magnitude The very basic coordinate system used to specify coordinates in the space is the Cartesian c.s. The unit vectors along the Cartesian coordinate axis x, y and z are i , j , k , respectively The symbol e n will be used to indicate a unit vector in some n- direction (not x, y, nor z) Any vector can be represented as a multiplication of a magnitude and a unit vector A  A e n  A e n B   B e n   B e n A is in the positive direction along n B is in the negative direction along n e n  A  A A A Department of Mechanical Engineering

31. The rectangular components of a force in 2D system While the components must be perpendicular to each other, the directions do not have to be parallel or perpendicular to the horizontal or vertical directions x y F y = F y j F x = F x i i j F  F  F x  F y  F x i  F y j F  F 2  F 2 x y   tan  1 F y F x F x  F cos  F y  F s i n 

32. F F F F F F z z y y x x z y x z z y y F 2  F 2  F 2 F  F  F cos  F  F cos  F x  F cos  x  1  1  1   cos   cos   cos The rectangular components in 3D systems F n F F F x i  F y j  F z k  e   F x i  F y j  F z k F  F e n F  F x  F y  F z x y z F y = F y j F x = F x i F z = F z k F i k j e n  z  x  y e n  co s  x i  co s  y j  co s  z k

33. Dot Products of two vectors A  B  B  A  A B co s   AB co s   A B It’s a scalar !!! Special cosines: Cos o = 1 Cos 30 o = ½ √3 Cos 45 o = ½ √2 Cos 60 o = 0.5 Cos 90 o =

34. The dot product can be used to obtain the rectangular components of a force (a vector in general) A t  A  A n The component along e t Remember, e n and e t are perpendicular A n  ( A  e n ) e n A n  A n e n A n  A  e n  A co s  n ( m a gnitud e ) (the vectorial component in the n direction) The component along e n Dot products and rectangular components

35. Cartesian rectangular components The dot product is particularly useful when the unit vectors are of the Cartesian system (the i, j, k ) x y F y = F y j F x = F x i i j F  F x  F  i  F co s  F y  F  j  F co s ( 90   )  F sin  Also, in 3D, F z  F  k 90-  F  F x  F y  F x i  F y j  ( F  i ) i  ( F  j ) j Department of Mechanical Engineering

36. More usage of dot products … Dot products of two vectors written in Cartesian system The magnitude of a vector (could be a force vector), here A is the vector magnitude A  A  A 2 cos  A 2  A A  A A  A A x x y y z z The angle between two vectors (say between vectors A and B ) A  B  A x B x  A y B y  A z B z     AB  1  A x B x  A y B y  A z B z    cos

37. The rectangular components of arbitrary direction  F x co s  x n  F y co s  y n  F z co s  zn n  ( F x i  F y j  F z k )  e F n  F  e n F  F x  F y  F z  F x i  F y j  F z k F  F n e n  F t e t e n  co s  x n i  co s  y n j  co s  z n k z  F x i  e n  F y j  e n  F z k  e n x y F y = F y j F x = F x i F z = F z k F i k j e n  zn  xn  yn F t F n Can you show the following?

38. Summarizing …. The components of a force resultant are not unique Graphical methods (triangular or parallelogram methods) combined with law of sinus and law of cosines can be used to obtain components in arbitrary direction Rectangular components are components of a force (vector) that perpendicular to each other The dot product can be used to obtain rectangular components of a force vector obtain the magnitude of a force vector (by performing self- dot-product) Obtain the angle between two (force) vectors

39. Example Problem 2-6 Find the x and y scalar components of the force Find the x’ and y’ scalar components of the force Express the force F in Cartesian vector form for the xy- and x’y’- axes

40. Example Problem 2-6   F x  F co s  F y  F cos ( 90   ) F x '  F cos  F y '  F cos ( 90   )   90  28  6 2 o   62  30  3 2 o F x  450 cos 62  21 1 N F y  450 s i n 62  39 7 N F x '  450 cos 32  38 2 N F y  450 s i n 32  23 8 N F  ( 21 1 i  39 7 j ) N  ( 38 2 e x '  23 8 e y ' ) N Writing the F in Cartesian vector form: Department of Mechanical Engineering

41. Example Problem 2-8 Department of Mechanical Engineering Find the angles  x ,  y , and  z  x is the angle between OB and x axis and so on ..) The x, y, and x scalar components of the force. The rectangular component F n of the force along line OA The rectangular component of the force perpendicular to line OA (say F t ) B

42. Example Problem 2-8 To find the angles: Find the length of the diagonal OB, say d – d = 5.831 m Use cosines to get the angles The scalar components in the x, y, and z directions: B z y o x 3 5.831 4 5.831 3 5.831  59. o  46.7 o  59.0  1   cos  1   cos  1   cos F x  F co s  x  12.86 2 kN F y  F co s  y  17.15 kN F z  F co s  z  12.86 2 kN F  (12.862 i  17.150 j  12.862 k ) kN Department of Mechanical Engineering

43. Example Problem 2-8 To find the rectangular component F n of the force along line OA: Needs the unit vector along OA Method 1 : Follow the method described in the book Method 2: utilize the vector position of A (basically vector OA) – Remember, that any vector can be represented as a multiplication of its magnitude and a unit vector along its line of application OA  r A  3 i  1 j  3 k r e  3 i  1 j  3 k  0.68 8 i  0.23 j  0.68 8 k 4.36   r A 3 i  1 j  3 k 3 2  1 2  3 2 A OA

44. Example Problem 8-2 F OA  F  e OA The scalar component of F along OA F OA  (12.862 i  17.150 j  12.862 k )  (0.688 i  0.230 j  0.688 k ) F OA  12.862  0.688  17.150  0.230  12.862  0.688  21.643 kN The vector component of F along OA F OA  ( F  e OA ) e OA  21.6(0.688 i  0.230 j  0.688 k )  14.8 6 i  4.9 7 j  14.8 6 k The vector component of F perpendicular to OA F t  F  F OA  (12.862 i  17.150 j  12.862 k )  (14.86 i  4.97 j  14.86 k )  (  2 i  12.1 8 j  2 k ) The scalar component of F perpendicular to OA F  | F |  | (  2 i  12.1 8 j  2 k ) |  (  2 ) 2  12.1 8 2  (  2 ) 2  12.5 kN t t OA t F 2  F 2  21.643 2  12.50 2  25 kN Check: F 

45. The Cartesian rectangular components of forces can be utilized to obtain the resultant of the forces x y F 1 F 2 F 1x F 2x F 2y F 1y Adding the x vector components, we obtain the x vector component of the resultant Adding the y vector components, we obtain the y vector component of the resultant The resultant can be obtained by performing the vector addition of these two vector components R x   F x  F 1 x  F 2 x R y   F y  F 1 y  F 2 y R  R x  R y  R x i  R y j Resultants by rectangular components

46. Resultants by rectangular components The magnitude of the resultant The angles formed by the resultant and the Cartesian axes All of the above results can be easily extended for 3D system The scalar components of the resultant R x  F 1 x  F 2 x  ( F 1 x  F 2 x ) i  R x i R y  F 1 y  F 2 y  ( F 1 y  F 2 y ) j  R y j R  R 2  R 2 x y R R R R y y x x  1  1   cos   cos

47. HW Problem Determine the non-rectangular components of R

48. HW Problem Determine the components of F 1 and F 2 in x-y and x’-y’ systems

49. HW Problem Express the cable tension in Cartesian form Determine the magnitude of the rectangular component of the cable force Determine the angle  between cables AD and BD Typo in the problem!!! B(4.9,-7.6,0) C(-7.6,-4.6,0) Don’t worry if you don’t get the solution in the back of the book

50. HW Problem e Determine the scalar components Express the force in Cartesian vector form Determine the angle  between th force and line AB

51. HW problems Given: F 1 = 500 lb, F 2 = 300 lb, F 3 = 200 lb Determine the resultant Express the resultant in the Cartesian format Find the angles formed by the resultant and the coordinate axes

52. HW Problem Given T 1 and T 2 are 650 lb, Determine P so that the resultant of T 1 , T 2 and P is zero