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ENGINEERING MECHANICS
K Chaitanya Mummareddy
M.Tech, IITG
Associate Professor and
HOD of Mechanical Engineering
Malineni P...
System of Forces: Several forces acting simultaneously upon a body
System of
Forces
Noncoplanar

Coplanar
2D

Concurrent

...
Coplanar System of Forces 2D

Copyright 2014 M K Chaitanya

3
Non-Coplanar System of Forces 3D

Copyright 2014 M K Chaitanya

4
Method of approach to solve CONCURRENT and NON CONCURRENT FORCE Systems

Problem

2-D

3-D
Non-concurrent

Concurrent
Equi...
Method of approach to solve Coplanar (2D) problems

Problem

2-D

Non-concurrent

Concurrent

Not Equilibrium

Not Equilib...
Problem

Determine the resultant of the following figure

Problem – 2D- Concurrent - Resultant
Copyright 2014 M K Chaitany...
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

Resulta...
Problem

The resultant of the four concurrent forces as shown in Fig acts
along Y-axis and is equal to 300N. Determine the...
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
...
Problem

Determine the resultant of the following figure

Problem – 2D- Non Concurrent - Resultant
Copyright 2014 M K Chai...
Resultant of General forces in a plane –
Coplanar non-concurrent
Step 1: Choose a reference point

Step 3: Find the result...
Problem solution:

Resultant – Non-concurrent general forces in a plane

Step:2: Shift all forces to point A

Step:1: Choo...
Problem

Determine the resultant of the following figure

Problem – 2D- Non Concurrent - Resultant
Copyright 2014 M K Chai...
Problem solution
Example:

Step:1

Step:3

Resultant – Non-concurrent general forces in a plane

Determine the resultant f...
Problem

Find tension in the string and reaction at B

Problem – 2D - Concurrent - Equilibrium
Copyright 2014 M K Chaitany...
EQUATIONS OF EQUILIBRIUM

Copyright 2014 M K Chaitanya

17
Types of supports and reaction forces (2D)

Copyright 2014 M K Chaitanya

18
Problem solution

Find T, Rb

FBD of C

W=30N

Since the body is in equilibrium and the forces are concurrent ……
∑FX=0; Rb...
Problem

Find the reactions at A,B,C,D AND at F, Given W=100N

Problem – 2D - Concurrent - Equilibrium
Copyright 2014 M K ...
W=100N
Find Ra, Rb and Rc
Since the body is in equilibrium and the forces are
concurrent ……
∑FX=0; Rf Cos600 - Rc Sin600 =...
Problem

Find T1, T2 , T3 and θ

Problem – 2D - Concurrent - Equilibrium
Copyright 2014 M K Chaitanya

22
Problem

Find T1, T2 , T3 and θ

Problem solution:

FBD of A

FBD of B
Copyright 2014 M K Chaitanya

23
Problem solution
Since the body is in equilibrium and the forces are
concurrent ……
T1=48.8N

∑FY=0; T1 Cos350 - W(40N) = 0...
Types of supports and reaction forces (2D)

Copyright 2014 M K Chaitanya

25
Problem

Determine the reactions at A and B

Problem – 2D - Non Concurrent - Equilibrium

Copyright 2014 M K Chaitanya

26
Problem solution

Since the body is in equilibrium and the forces are general
forces then……
∑FX=0; Rax = 0 ……………………….…1

R...
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...
Problem

Determine the reaction at C and the reaction at E, Given P=200N

From the freebody diagram of AB .Since the body ...
From the freebody diagram of DE .Since the body is in equilibrium then…… ……
∑FX=0; Rx - RE Sin600= 0 …………..…1
∑FY=0; Ry – ...
Problem

A roller weighting 2000N rests on a inclined bar weighting 800N as
shown in Fig. Assuming the weight of bar negli...
Problem
FBD of Figure

Concurrent

Non Concurrent

Copyright 2014 M K Chaitanya

32
Problem

Equilibrium of Concurrent force system

From the free body diagram of ball .Since the body is in equilibrium then...
Problem

Equilibrium of Non Concurrent force system

From the freed body diagram of ROD .Since the body is in equilibrium ...
Method of approach to solve NON COPLANAR (3D) STATIC problems
Problem

3-D

Concurrent

Non Concurrent

Not equilibrium

E...
Resultant of Concurrent Forces In 3D

R  FAB  FAC  FAD




R  FAB .



R  FAB .

AB
AC
AD
 FAC .
 FAD .
AB
AC
AD...
Problem

Determine the Resultant acting at A

Problem – 3D- Concurrent - Resultant
Copyright 2014 M K Chaitanya

37
Example:1

Determine the Resultant acting at A

R  Fab  Fac
Fab  FAB .





{-2i - 6 j  3 k}

Fab  840 *

Fac  FAC...
R  - 60i - 1080j  480k
Magnitude of Resultant is

 (-60)2  (10802 )  4802  1183.3N

 60
Cos x 
; x  930
1183.3...
Resultant of Concurrent Forces In 3D

r

OP

RMO

Copyright 2014 M K Chaitanya

40
Problem

Determine the Resultant non concurrent of the
forces (3D)

Problem – 3D- Non Concurrent - Resultant
Copyright 201...
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 ...
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...
Problem
Given: A 1500N plate, as shown, is supported
by three cables and is in equilibrium.
Find: Tension in each of the c...
Equilibrium of 3D concurrent forces
--- --R= 0
--- --- -- -F1+F2+F3….= 0
ΣFx=0
ΣFy=0
ΣFz=0,

Copyright 2014 M K Chaitanya
...
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...
F AB  FAB .





F AB  FAB *

FAC  FAC .

(-6)  (12 )  4
2

.

AC

{-4i - 12 j  6 k}
(-4)2  (12 2 )  (6) 2

...
FAB  FAC  FAD  W  0

Solving the three simultaneous equations gives
FAB = 858 N

FAC = 0 N
FAD = 858 N

Copyright 2014...
Problem

Example – 3D Equilibrium

Three cables are used to tether a balloon as shown. Determine the
vertical force P exer...
Types of supports and reaction forces (3D)

Copyright 2014 M K Chaitanya

50
Types of supports and reaction forces (3D)

Copyright 2014 M K Chaitanya

51
Problem
Given: Determine the reaction in the string CD and reactions at B. A ball and socket
joint at A

Problem – 3D- Non...
Equilibrium of 3D Non concurrent forces
--- --R= 0
--- --- -- -F1+F2+F3….= 0
--------R=0 =
M

--0
--- -- ----- ------ -F1+...
Problem Solution

Copyright 2014 M K Chaitanya

54
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  4.5...
References
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•
•
•
•
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•
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•

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•

Engg. Mechanics ,Timoshenko & Young.
2.Engg. Mechanics, R.K. Bansal , Laxmi publicati...
THANK YOU
Copyright 2014 M K Chaitanya

57
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Engineering Mechanics

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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.

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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

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