Utilizamos tu perfil de LinkedIn y tus datos de actividad para personalizar los anuncios y mostrarte publicidad más relevante. Puedes cambiar tus preferencias de publicidad en cualquier momento.
Próxima SlideShare
Cargando en…5
×

# Engineering Mechanics

21.317 visualizaciones

In Engineering Mechanics the static problems are classified as two types: Concurrent and Non-Concurrent force systems. The presentation discloses a methodology to solve the problems of Concurrent and Non-Concurrent force systems.

• Full Name
Comment goes here.

Are you sure you want to Yes No
• Inicia sesión para ver los comentarios

### Engineering Mechanics

1. 1. ENGINEERING MECHANICS K Chaitanya Mummareddy M.Tech, IITG Associate Professor and HOD of Mechanical Engineering Malineni Perumallu Educational Society’s group of Institutions Guntur Copyright 2014 M K Chaitanya 1
2. 2. System of Forces: Several forces acting simultaneously upon a body System of Forces Noncoplanar Coplanar 2D Concurrent Parallel 3D Nonconcurrent Concurrent General Copyright 2014 M K Chaitanya Parallel Nonconcurrent General 2
3. 3. Coplanar System of Forces 2D Copyright 2014 M K Chaitanya 3
4. 4. Non-Coplanar System of Forces 3D Copyright 2014 M K Chaitanya 4
5. 5. Method of approach to solve CONCURRENT and NON CONCURRENT FORCE Systems Problem 2-D 3-D Non-concurrent Concurrent Equilibrium Not Equilibrium (Resultant) (Unknowns) (Resultant) 1. Parallelogram Law Fx=0 2. Triangle law Fy=0 1. Choose a reference Point Not Equilibrium 3. Polygon law 4. Method of projections 2. Shift all the forces to a point Equilibrium (Unknowns) ΣFx=0 Concurrent Non Concurrent Not equilibrium Equilibrium (Resultant) (Unknowns) --- --- --- -R = F1+F2+F3 ΣFy=0 ΣMz=0 3. Find the resultant force and couple at that point --- --R= 0 --- --- -- -F1+F2+F3,,= 0 ΣFx=0 ΣFy=0 ΣFz=0, 4. Reduce the forcecouple system to a single force Not Equilibrium (Resultant) --- -- -- -R= F1+F2+F3…. --- --- --- --M = M1+M2+M3.... --- --rOPͯ R = M Equilibrium (Unknowns) --- --R= 0 --- --M= 0 ΣFx=0 , ΣMx=0 ΣFy=0, ΣMy=0 ΣFz=0, ΣMz=0 Copyright 2014 M K Chaitanya 5
6. 6. Method of approach to solve Coplanar (2D) problems Problem 2-D Non-concurrent Concurrent Not Equilibrium Not Equilibrium Equilibrium (Resultant) Equilibrium (Resultant) (Unknowns) (Unknowns) 1. Choose a reference Point 1. Parallelogram Law 2. Triangle law 3. Polygon law 4. Method of projections Fx=0 Fy=0 ΣFx=0 2. Shift all the forces to a point ΣFy=0 3. Find the resultant force and couple at that point ΣMz=0 4. Reduce the force-couple system to a single force Copyright 2014 M K Chaitanya 6
7. 7. Problem Determine the resultant of the following figure Problem – 2D- Concurrent - Resultant Copyright 2014 M K Chaitanya 7
8. 8. Problem solution Force Mag x  comp  F1 150 150 cos 30  F2 80  80 sin 20  F3 110 0  F4 100  100 cos 15 F Resultant is R  x  199.1 F 2 X y  comp  150 Sin30  80Cos 20  110  100 sin 15 F y   FY  14.3 2 R  199.12  14.32 Direction is 14.3 N tan   199.1 N Copyright 2014 M K Chaitanya R  199.6N   4.1 8
9. 9. Problem The resultant of the four concurrent forces as shown in Fig acts along Y-axis and is equal to 300N. Determine the forces P and Q. F F y x 0  R  300 N Problem – 2D- Concurrent - Resultant Copyright 2014 M K Chaitanya 9
10. 10. Problem solution Force Mag  F1 800  F2 380  F3 Q  F4 P F F y x 0  R  300 N F x F x  comp 800  380  QSin45  PSin 50 y  comp 0 0  QCos45  PCos 50  800  380  QSin45  P sin 50  0 y  QCos45  PCos 50  R  300 P = 511 N Q =- 40.3N Copyright 2014 M K Chaitanya 10
11. 11. Problem Determine the resultant of the following figure Problem – 2D- Non Concurrent - Resultant Copyright 2014 M K Chaitanya 11
12. 12. Resultant of General forces in a plane – Coplanar non-concurrent Step 1: Choose a reference point Step 3: Find the resultant force and moment of forces about Copyright 2014 O Step 2: Shift all the forces to a point Step 4: Reduce resultant force and moment to a single force M K Chaitanya 12
13. 13. Problem solution: Resultant – Non-concurrent general forces in a plane Step:2: Shift all forces to point A Step:1: Choose A as reference Point Step 3: Find resultant force and couple Step:4: Reduce it to a single force x = 1880/600 x = 3.13m Copyright 2014 M K Chaitanya 13
14. 14. Problem Determine the resultant of the following figure Problem – 2D- Non Concurrent - Resultant Copyright 2014 M K Chaitanya 14
15. 15. Problem solution Example: Step:1 Step:3 Resultant – Non-concurrent general forces in a plane Determine the resultant force of the non-concurrent forces as shown in plate and distance of the resultant force from point ‘O’. Step:2 Step:4 Copyright 2014 M K Chaitanya 15
16. 16. Problem Find tension in the string and reaction at B Problem – 2D - Concurrent - Equilibrium Copyright 2014 M K Chaitanya 16
17. 17. EQUATIONS OF EQUILIBRIUM Copyright 2014 M K Chaitanya 17
18. 18. Types of supports and reaction forces (2D) Copyright 2014 M K Chaitanya 18
19. 19. Problem solution Find T, Rb FBD of C W=30N Since the body is in equilibrium and the forces are concurrent …… ∑FX=0; Rb - T Cos600 = 0 ……………1 T= 34.64N ∑FY=0; T Sin600 – W= 0 ……………….2 Rb = 17.32N Copyright 2014 M K Chaitanya 19
20. 20. Problem Find the reactions at A,B,C,D AND at F, Given W=100N Problem – 2D - Concurrent - Equilibrium Copyright 2014 M K Chaitanya 20
21. 21. W=100N Find Ra, Rb and Rc Since the body is in equilibrium and the forces are concurrent …… ∑FX=0; Rf Cos600 - Rc Sin600 = 0 ………1 ∑FY=0; Rf Sin600 + Rc Cos600 - W(100) = 0…...2 Rf=86.6N Rc=50N Since the body is in equilibrium and the forces are concurrent …… ∑FX=0; -Rb sin600 - Rf Cos600 + Ra = 0 ………..…1 ∑FY=0; Rb Cos600 - Rf Sin600 – W = 0…………..2 Copyright 2014 M K Chaitanya Rb = 350N Ra = 346.4N 21
22. 22. Problem Find T1, T2 , T3 and θ Problem – 2D - Concurrent - Equilibrium Copyright 2014 M K Chaitanya 22
23. 23. Problem Find T1, T2 , T3 and θ Problem solution: FBD of A FBD of B Copyright 2014 M K Chaitanya 23
24. 24. Problem solution Since the body is in equilibrium and the forces are concurrent …… T1=48.8N ∑FY=0; T1 Cos350 - W(40N) = 0…...2 FBD of A ∑FX=0; T2 – T1Sin 350 = 0 ……….…1 T2=28.0N Since the body is in equilibrium and the forces are concurrent …… ∑FX=0; -T2 + T3Sin θ0 = 0 ……….…1 FBD of B ∑FY=0; T3 Cos θ0 - W(50N) = 0…...2 Copyright 2014 M K Chaitanya T3=57.3N θ=29.3N 24
25. 25. Types of supports and reaction forces (2D) Copyright 2014 M K Chaitanya 25
26. 26. Problem Determine the reactions at A and B Problem – 2D - Non Concurrent - Equilibrium Copyright 2014 M K Chaitanya 26
27. 27. Problem solution Since the body is in equilibrium and the forces are general forces then…… ∑FX=0; Rax = 0 ……………………….…1 Rby=30N ∑FY=0; Ray + Rby - 40 = 0…...........2 Ray=10N ∑MA=0; (Rby*L) – 40*(3L/4) = 0……3 Rax=0N Copyright 2014 M K Chaitanya 27
28. 28. Problem A man raises a 10 kg joist, of length 4 m, by pulling on a rope. Find the tension in the rope and the reaction at A. Since the body is in equilibrium then……. ∑FX=0; Rx – Tc*Cos200= 0 ……….…1 ∑FY=0; Ry - Tc* Sin200 - W(98.1) = 0….........2 Tc=82N Rx=77.1N Ry=126.14N ∑MA=0; -(W*L/2) + (Tc*Cos200*4Sin450) –(Tc*Sin200*4*Cos450) = 0……3 Copyright 2014 M K Chaitanya 28
29. 29. Problem Determine the reaction at C and the reaction at E, Given P=200N From the freebody diagram of AB .Since the body is in equilibrium then…… …… ∑FX=0; Rx + P*Cos300= 0 …………..…1 RC=150N ∑FY=0; Ry + Rc- PSin300 = 0…….........2 ∑MA=0; (Rc* 2.44) - (PSin300* 3.66) = 0……….3 CONTINUES…….. Copyright 2014 M K Chaitanya 29
30. 30. From the freebody diagram of DE .Since the body is in equilibrium then…… …… ∑FX=0; Rx - RE Sin600= 0 …………..…1 ∑FY=0; Ry – RC+ RECos600 = 0…….........2 RE=100N ∑MD=0; (RC*1.22) - (RECos600 * 3.66) = 0……….3 Copyright 2014 M K Chaitanya 30
31. 31. Problem A roller weighting 2000N rests on a inclined bar weighting 800N as shown in Fig. Assuming the weight of bar negligible, determine the reactions at D and C and reaction in bar AB. Problem – 2D – (Both Concurrent & Non Concurrent - Equilibrium) Copyright 2014 M K Chaitanya 31
32. 32. Problem FBD of Figure Concurrent Non Concurrent Copyright 2014 M K Chaitanya 32
33. 33. Problem Equilibrium of Concurrent force system From the free body diagram of ball .Since the body is in equilibrium then…… …… ∑FX=0; R1 – R2*Sin300= 0 …………..…1 R1=1154.7N ∑FY=0; R2 *Cos300 -2000 = 0…….........2 R2=2309.4N CONTINUES…….. Copyright 2014 M K Chaitanya 33
34. 34. Problem Equilibrium of Non Concurrent force system From the freed body diagram of ROD .Since the body is in equilibrium then…… …… ∑FX=0; -Hc + R2*Sin300= 0 …………..…1 ∑FY=0; RD + Vc -R2*Cos300 - 800 = 0…….........2 ∑MC=0; (RD* 5 Cos300) - (800* 2.5 Cos 300) - R2*2 = 0……….3 Copyright 2014 M K Chaitanya Hc=1154.7N Vc=1333.3N RD=1466.7N 34
35. 35. Method of approach to solve NON COPLANAR (3D) STATIC problems Problem 3-D Concurrent Non Concurrent Not equilibrium Equilibrium (Resultant) (Unknowns) --- --- --- -R = F1+F2+F3…… --- --R= 0 --- --- -- -F1+F2+F3….= 0 ΣFx=0 ΣFy=0 Not Equilibrium Equilibrium (Resultant) (Unknowns) --- -- -- -R= F1+F2+F3…. --- --- --- --M = M1+M2+M3.... --- --R= 0 --- --rOPͯ R = M --- --M= 0 ΣFx=0 , ΣMx=0 ΣFz=0, ΣFy=0, ΣMy=0 ΣFz=0, ΣMz=0 Copyright 2014 M K Chaitanya 35
36. 36. Resultant of Concurrent Forces In 3D R  FAB  FAC  FAD   R  FAB .  R  FAB . AB AC AD  FAC .  FAD . AB AC AD AB  FAC .   AC  FAD .  R   Fx i +  Fy j +  Fz k cos  x  cos  y  Copyright 2014 M K Chaitanya cos  z   Fx R Fy R  Fz R 36 AB
37. 37. Problem Determine the Resultant acting at A Problem – 3D- Concurrent - Resultant Copyright 2014 M K Chaitanya 37
38. 38. Example:1 Determine the Resultant acting at A R  Fab  Fac Fab  FAB .   {-2i - 6 j  3 k} Fab  840 * Fac  FAC . (-2)  ( 6 )  3 2 Magnitude of Resultant is 2 2   Fac  420 * R  Fab  Fac AB AC {3i - 6 j  2 k} (3) 2  ( 6 2 )  2 2 R  - 60i - 1080j  480k  (-60)2  (10802 )  4802  1183.3N Copyright 2014 M K Chaitanya 38 .
39. 39. R  - 60i - 1080j  480k Magnitude of Resultant is  (-60)2  (10802 )  4802  1183.3N  60 Cos x  ; x  930 1183.3  1080 Cos y  ; y  155.80 1183.3 480 Cos z  ; z  660 1183.3 Copyright 2014 M K Chaitanya 39
40. 40. Resultant of Concurrent Forces In 3D r OP RMO Copyright 2014 M K Chaitanya 40
41. 41. Problem Determine the Resultant non concurrent of the forces (3D) Problem – 3D- Non Concurrent - Resultant Copyright 2014 M K Chaitanya 41
42. 42. Transfer all forces about O to get a force and couple system R   F  500k - 300k  200k - 50k R  350k M M M O  (r OA  50k )  (r OB  200k )  (r OC  300k ) O  ((0.35j  0.5i)  50k )  (0.5i  200k )  ( 0.35j  300k ) O  87.5i - 125j Copyright 2014 M K Chaitanya 42
43. 43. Reduce the force couple system to a single force r OP  R  MO (x i  y j  z k)  (350k)  87.5i - 125j x  0.375m, y  0.25m Copyright 2014 M K Chaitanya 43
44. 44. Problem Given: A 1500N plate, as shown, is supported by three cables and is in equilibrium. Find: Tension in each of the cables. Problem – 3D- Concurrent - Equilibrium Copyright 2014 M K Chaitanya 44
45. 45. Equilibrium of 3D concurrent forces --- --R= 0 --- --- -- -F1+F2+F3….= 0 ΣFx=0 ΣFy=0 ΣFz=0, Copyright 2014 M K Chaitanya 45
46. 46. y W FBD of Point A: x z FAB FAC FAD The particle A is in equilibrium, hence FAB  FAC  FAD  W  0 Copyright 2014 M K Chaitanya 46
47. 47. F AB  FAB .   F AB  FAB * FAC  FAC . (-6)  (12 )  4 2 . AC {-4i - 12 j  6 k} (-4)2  (12 2 )  (6) 2   AD {6i - 12 j  4 k} (6) 2  (12 2 )  (4) 2 W  1500j Copyright 2014 M K Chaitanya 2   FAD  FAD * {-6i - 12 j  4 k} 2 FAC  FAC * FAD  FAD . AB 47
48. 48. FAB  FAC  FAD  W  0 Solving the three simultaneous equations gives FAB = 858 N FAC = 0 N FAD = 858 N Copyright 2014 M K Chaitanya 48
49. 49. Problem Example – 3D Equilibrium Three cables are used to tether a balloon as shown. Determine the vertical force P exerted by the balloon at A knowing that the tension in cable AB is 60 N. y P x z FAB FAC FAD FAB  FAC  FAD  P  0 Copyright 2014 M K Chaitanya 49
50. 50. Types of supports and reaction forces (3D) Copyright 2014 M K Chaitanya 50
51. 51. Types of supports and reaction forces (3D) Copyright 2014 M K Chaitanya 51
52. 52. Problem Given: Determine the reaction in the string CD and reactions at B. A ball and socket joint at A Problem – 3D- Non Concurrent - Equilibrium Copyright 2014 M K Chaitanya 52
53. 53. Equilibrium of 3D Non concurrent forces --- --R= 0 --- --- -- -F1+F2+F3….= 0 --------R=0 = M --0 --- -- ----- ------ -F1+F2+F3….= 0 M1+M2+M3….= 0 ΣFx=0 ΣFy=0 ΣFx=0 , ΣFz=0, ΣMx=0 ΣFy=0, ΣMy=0 ΣFz=0, ΣMz=0 Copyright 2014 M K Chaitanya 53
54. 54. Problem Solution Copyright 2014 M K Chaitanya 54
55. 55. M 0 (r  T A 1 CD )  (r 2  2i )  (r 3  ( B y j  Bz k )  0 (6 j - 3k)  TCD CD  (6 j  2.5k)  2i  (6 j  4.5i)  ( B y j  Bz k )  0 CD TCD  2.83kN BY  4.06kN BZ  0.417 kN AX  3.05kN AY  1.556kN AZ  1.250kN Copyright 2014 M K Chaitanya 55
56. 56. References • • • • • • • • • • • • Engg. Mechanics ,Timoshenko & Young. 2.Engg. Mechanics, R.K. Bansal , Laxmi publications 3.Engineering Mechanics,Fedinand.L.Singer , Harper – Collins. 4. Engineering Mechanics statics and dynamics, A Nelson, Mc Gra Hill publications 5. Engg. Mechanics Umesh Regl, Tayal. 6. Engineering Mechanics by N H Dubey 7. Engineering Mechanics , statics – J.L.Meriam, 6th Edn – Wiley India Pvt Ltd. 8. Engineering Mechanics , dynamics – J.L.Meriam, 6th Edn – Wiley India Pvt Ltd. 9. Mechanics For Engineers , statics - F.P.Beer & E.R.Johnston 5th Edn Mc Graw Hill Publ. 10.Mechanics For Engineers, dynamics - F.P.Beer & E.R.Johnston – 5th Edn Mc Graw Hill Publ. 11. www.google.com 12. http://nptel.iitm.ac.in/ Copyright 2014 M K Chaitanya 56
57. 57. THANK YOU Copyright 2014 M K Chaitanya 57