engineering mechanics

ENGINEERING MECHANICS

Copyright 2014 M K Chaitanya

K Chaitanya MummareddyM.Tech, IITG

Associate Professor and HOD of Mechanical Engineering

Malineni Perumallu Educational Society’s group of Institutions

Guntur

1

System of Forces: Several forces acting simultaneously upon a body

System of Forces

Coplanar

2D

Concurrent Non-concurrent

Parallel General

Non-coplanar

3D

ConcurrentNon-

concurrent

Parallel General

Copyright 2014 M K Chaitanya 2

Coplanar System of Forces 2D


Non-Coplanar System of Forces 3D


Method of approach to solve CONCURRENT and NON CONCURRENT FORCE Systems

Problem

2-D

Concurrent

Not Equilibrium

(Resultant)

1. Parallelogram Law

2. Triangle law

3. Polygon law

4. Method of projections

Equilibrium

(Unknowns)

Fx=0

Fy=0

Non-concurrent

Not Equilibrium

(Resultant)

1. Choose a reference Point

2. Shift all the forces to a point

3. Find the resultant force and couple at that point

4. Reduce the force-couple system to a single force

Equilibrium

(Unknowns)

ΣFx=0

ΣFy=0

ΣMz=0

3-D

Concurrent

Not equilibrium

(Resultant)

--- --- --- --R = F1+F2+F3

Equilibrium

(Unknowns)

--- ---R = 0

--- --- -- --F1+F2+F3,,= 0

ΣFx=0

ΣFy=0

ΣFz=0,

Non Concurrent

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


Method of approach to solve Coplanar (2D) problems

Problem

2-D

Concurrent

Not Equilibrium

(Resultant)

1. Parallelogram Law

2. Triangle law

3. Polygon law

4. Method of projections

Equilibrium

(Unknowns)

Fx=0

Fy=0

Non-concurrent

Not Equilibrium

(Resultant)

1. Choose a reference Point

2. Shift all the forces to a point

3. Find the resultant force and couple at that point

4. Reduce the force-couple system to a single force

Equilibrium

(Unknowns)

ΣFx=0

ΣFy=0

ΣMz=0


Determine the resultant of the following figure

Problem – 2D- Concurrent - ResultantCopyright 2014 M K Chaitanya

Problem

7

22 3.141.199 R

N6.199R

15sin10015cos100100

1100110

208020sin8080

3015030cos150150

4

3

2

1

F

F

CosF

SinF

compycompxMagForce

1.199 xF 3.14 y

F

22

FF YXRResultant is

N1.199

N3.14tan 1.4

Direction is


Problem solution

8

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.

Problem – 2D- Concurrent - ResultantCopyright 2014 M K Chaitanya

Problem

0 xF

NRFy

300

9

5050

4545

0380380

0800800

4

3

2

1

PCosPSinPF

QCosQSinQF

F

F

compycompxMagForce

050sin45380800 PQSinFx

3005045 RPCosQCosFy

P = 511 NQ =- 40.3N


Problem solution

0 xF

NRFy

300

10


Problem – 2D- Non Concurrent - Resultant


Problem

11

Resultant of General forces in a plane –Coplanar non-concurrent

Step 2: Shift all the forces to a point

Step 3: Find the resultant force and moment of forces about O

Step 4: Reduce resultant force and moment to a single force

Step 1: Choose a reference point


Resultant – Non-concurrent general forces in a plane

Step:2: Shift all forces to point A

Step:4: Reduce it to a single forceStep 3: Find resultant force and couple

Step:1: Choose A as reference Point


x = 1880/600x = 3.13m

Problem solution:

13




Problem

14

Example:

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:4Step:3

Step:1


Problem solution

15

Problem – 2D - Concurrent - Equilibrium

Find tension in the string and reaction at B


Problem

16

EQUATIONS OF EQUILIBRIUM


Types of supports and reaction forces (2D)



FBD of C

Since the body is in equilibrium and the forces are concurrent ……

∑FX=0; Rb - T Cos600 = 0 ……………1

∑FY=0; T Sin600 – W= 0 ……………….2

W=30N

Find T, RbProblem solution

T= 34.64N

Rb = 17.32N

19


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


Problem

20


W=100NFind Ra, Rb and Rc


∑FX=0; Rf Cos600 - Rc Sin600 = 0 ………1

∑FY=0; Rf Sin600 + Rc Cos600 - W(100) = 0…...2

Rf=86.6NRc=50N


∑FX=0; -Rb sin600 - Rf Cos600 + Ra = 0 ………..…1

∑FY=0; Rb Cos600 - Rf Sin600 – W = 0…………..2

Rb = 350NRa = 346.4N 21



Problem Find T1, T2 , T3 and θ

22


FBD of A FBD of B

Find T1, T2 , T3 and θProblem

Problem solution:

23


FBD of A


∑FX=0; T2 – T1Sin 350 = 0 ……….…1

∑FY=0; T1 Cos350 - W(40N) = 0…...2


∑FX=0; -T2 + T3Sin θ0 = 0 ……….…1

∑FY=0; T3 Cos θ0 - W(50N) = 0…...2FBD of B

Problem solution

T1=48.8N

T2=28.0N

T3=57.3N

θ=29.3N

24


Problem – 2D - Non Concurrent - Equilibrium

Determine the reactions at A and BProblem

26


Since the body is in equilibrium and the forces are general forces then……

∑FX=0; Rax = 0 ……………………….…1

∑FY=0; Ray + Rby - 40 = 0…...........2

∑MA=0; (Rby*L) – 40*(3L/4) = 0……3

Problem solution

Rby=30N

Ray=10N

Rax=0N

27


Since the body is in equilibrium then…….

∑FX=0; Rx – Tc*Cos200= 0 ……….…1

∑FY=0; Ry - Tc* Sin200 - W(98.1) = 0….........2

∑MA=0; -(W*L/2) + (Tc*Cos200*4Sin450) –(Tc*Sin200*4*Cos450) = 0……3

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.Problem

Tc=82N

Rx=77.1N

Ry=126.14N

28

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

∑FY=0; Ry + Rc- PSin300 = 0…….........2

∑MA=0; (Rc* 2.44) - (PSin300* 3.66) = 0……….3

CONTINUES……..


Problem

RC=150N

29

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

∑MD=0; (RC*1.22) - (RECos600 * 3.66) = 0……….3


RE=100N

30

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

Problem – 2D – (Both Concurrent & Non Concurrent - Equilibrium)

31


Problem

FBD of Figure

Concurrent

Non Concurrent

32

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

∑FX=0; R1 – R2*Sin300= 0 …………..…1

∑FY=0; R2 *Cos300 -2000 = 0…….........2

CONTINUES……..


Problem

Equilibrium of Concurrent force system

R1=1154.7N

R2=2309.4N

33

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


Problem

Equilibrium of Non Concurrent force system

Hc=1154.7N

Vc=1333.3N

RD=1466.7N

34

Method of approach to solve NON COPLANAR (3D) STATIC problems

Problem

3-D

Concurrent

Not equilibrium

(Resultant)

--- --- --- --R = F1+F2+F3……

Equilibrium

(Unknowns)

--- ---R = 0

--- --- -- --F1+F2+F3….= 0

ΣFx=0

ΣFy=0

ΣFz=0,

Non Concurrent

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



Resultant of Concurrent Forces In 3D

ADACAB F F F R

k zF+jy F+ix F

AD

AD.

AC

AC.

AB

AB.

...

FFF

FFF

ADACAB

ADACAB

R

R

RABACAB

R

zF

R

yF

R

xF

z

y

x

cos

cos

cos

36

Determine the Resultant acting at A

Problem – 3D- Concurrent - Resultant


Problem

37

Example:1 Determine the Resultant acting at A

. 3)6((-2)

k} 3 j 6 - {-2i * 840 F

. F

222ab

ABab F

AB

2)6((3)

k} 2 j 6 - {3i * 420 F

. F

222ac

ACac F

AC

480k 1080j -60i- R F F R acab

F F R acab

N3.1183480)1080((-60) 222 Magnitude of Resultant is


480k 1080j -60i- R

N3.1183480)1080((-60) 222

Magnitude of Resultant is

0

0

0

66;3.1183

480

8.155;3.1183

1080

93;3.1183

60

zz

yy

xx

Cos

Cos

Cos



rOP M OR

Resultant of Concurrent Forces In 3D

40


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


Problem

41


Transfer all forces about O to get aforce and couple system

k350R

50k-200k300k-500k F R

125j- 87.5i

)3000.35j()200.5i0()50 0.5i) .35j0((

)300()200()50(

M

M

M

O

O

OCOBOAO

kkk

kkk rrr

42


0.25m y 0.375m, x

125j-87.5i (350k) k) z jy i(x

Mr OOP

R

Reduce the force couple system to a singleforce

43


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

Problem

44

Equilibrium of 3D concurrent forces


--- ---R = 0

--- --- -- --F1+F2+F3….= 0

ΣFx=0

ΣFy=0

ΣFz=0,

45

z

W

x

y

FABFAC

FAD

FBD of Point A:

The particle A is in equilibrium, hence

0 W F F F ADACAB


. 4)12((-6)

k} 4 j 12 - {-6i * F

. F

222ABAB

AB

F

F

ABAB

)6()12((-4)

k} 6 j 12 - {-4i * F

. F

222ACAC

ACAC

F

F

AC

)4()12((6)

k} 4 j 12 - {6i * F

. F

222ADAD

ADAD

F

F

AD

1500j W


Solving the three simultaneous equations gives

FAB = 858 N

FAC = 0 N

FAD = 858 N

0 W F F F ADACAB


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.


z

P

x

y

FABFAC

FAD

0 P F F F ADACAB

Problem

49



50



51


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

Problem – 3D- Non Concurrent - Equilibrium

Problem

52

Equilibrium of 3D Non concurrent forces


--- ---R = 0

--- --- -- --F1+F2+F3….= 0

ΣFx=0

ΣFy=0

ΣFz=0,

--- ---R = 0

--- --- -- --F1+F2+F3….= 0

--- ---M = 0

--- --- -- --M1+M2+M3….= 0

ΣFx=0 , ΣMx=0

ΣFy=0, ΣMy=0

ΣFz=0, ΣMz=0

53


Problem Solution

54


kNA

kNA

kNA

kNB

kNB

kN

kBjBiCD

CDT

kBjBiT

Z

Y

X

Z

Y

zyCD

zyCD rrr

250.1

556.1

05.3

417.0

06.4

83.2T

0)(4.5i)j6(22.5k)j6(3k)-j6(

0)(()2()(

0

CD

321

AM

55

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/


THANK YOU


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