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.

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
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2. System of Forces: Several forces acting simultaneously upon a body
System of
Forces
Noncoplanar
Coplanar
2D
Concurrent
Parallel
3D
Nonconcurrent
Concurrent
General
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Parallel
Nonconcurrent
General
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3. Coplanar System of Forces 2D
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4. Non-Coplanar System of Forces 3D
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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
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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
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7. Problem
Determine the resultant of the following figure
Problem – 2D- Concurrent - Resultant
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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
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R  199.6N
  4.1
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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
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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
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11. Problem
Determine the resultant of the following figure
Problem – 2D- Non Concurrent - Resultant
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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
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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
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14. Problem
Determine the resultant of the following figure
Problem – 2D- Non Concurrent - Resultant
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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
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16. Problem
Find tension in the string and reaction at B
Problem – 2D - Concurrent - Equilibrium
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17. EQUATIONS OF EQUILIBRIUM
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18. Types of supports and reaction forces (2D)
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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
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20. Problem
Find the reactions at A,B,C,D AND at F, Given W=100N
Problem – 2D - Concurrent - Equilibrium
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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
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Rb = 350N
Ra = 346.4N
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22. Problem
Find T1, T2 , T3 and θ
Problem – 2D - Concurrent - Equilibrium
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23. Problem
Find T1, T2 , T3 and θ
Problem solution:
FBD of A
FBD of B
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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
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T3=57.3N
θ=29.3N
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25. Types of supports and reaction forces (2D)
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26. Problem
Determine the reactions at A and B
Problem – 2D - Non Concurrent - Equilibrium
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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
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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
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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……..
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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
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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)
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32. Problem
FBD of Figure
Concurrent
Non Concurrent
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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
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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
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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
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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 
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cos  z 
 Fx
R
Fy
R
 Fz
R
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AB

37. Problem
Determine the Resultant acting at A
Problem – 3D- Concurrent - Resultant
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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
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.

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
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40. Resultant of Concurrent Forces In 3D
r
OP
RMO
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41. Problem
Determine the Resultant non concurrent of the
forces (3D)
Problem – 3D- Non Concurrent - Resultant
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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
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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
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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
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45. Equilibrium of 3D concurrent forces
--- --R= 0
--- --- -- -F1+F2+F3….= 0
ΣFx=0
ΣFy=0
ΣFz=0,
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46. y
W
FBD of Point A:
x
z
FAB FAC
FAD
The particle A is in equilibrium, hence
FAB  FAC  FAD  W  0
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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
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2


FAD  FAD *
{-6i - 12 j  4 k}
2
FAC  FAC *
FAD  FAD .
AB
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48. FAB  FAC  FAD  W  0
Solving the three simultaneous equations gives
FAB = 858 N
FAC = 0 N
FAD = 858 N
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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
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50. Types of supports and reaction forces (3D)
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51. Types of supports and reaction forces (3D)
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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
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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
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54. Problem Solution
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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
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56. References
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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/
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57. THANK YOU
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