Unit-III

ELEMENTS OF CIVIL ENGINEERING AND

ENGINEERING MECHANICS

by

Prof. Karisiddappa, MCE, Hassan

COMPOSITION OF FORCES: The reduction of a given system of forces

to the simplest system that will be its equivalent is called the problem of

composition of forces.

• RESULTANT FORCE: It is possible to find a single force which will have the same effect as that of a number of forces acting on a

body. Such a single force is called resultant force.

• The process of finding out the resultant force is called composition of forces.

COMPOSITION OF CO-PLANAR CONCURRENT FORCE SYSTEM

COMPOSITION OF TWO FORCES: It is possible to reduce a given

system of forces i.e., two forces to the simplest system as its equivalent

(resultant force) with the help of parallelogram law of forces.

• LAW OF PARALLELOGRAM OF FORCES: If two forces, which act at a point be represented in magnitude and

direction by the two adjacent sides of a parallelogram drawn from one of its

angular points, their resultant is represented by the diagonal of the

parallelogram passing through that angular point, in magnitude and

direction.

RR

B C

A O

α

R

θ

αcos2 212

2

2

1 FFFFR ++=

RR

B C

A O

α

α

PROOF:

R

θ

D

Consider two forces F1 and F2 acting at point O as shown in

figure. Let α be the angle between the two forces.

Complete the parallelogram ACBO .Drop perpendicular CD to

OA produced. Let R be the resultant force of forces and

.Let θ be the inclination of the resultant force with the line of action of the force.

From triangle OCD,

+=

+=

=

++=

++=

+++=

++=

====

++=

+=

−

αα

θ

αα

θ

θ

α

α

ααα

αα

αα

cos

sintan

cos

sintan

tan

cos2

cos2

sincoscos2

)sin()cos(

,sin,cos,

)(

21

21

21

2

21

2

2

2

1

2

221

2

1

2

22

2

22

221

2

1

2

2

2

2

21

2

221

222

222

FF

F

FF

F

OD

CD

FFFFR

FFFFR

FFFFFR

FFFR

ROCFCDFADFOA

CDADOAOC

CDODOC

21

0

21

0

21

0

,180

,0

,90

FFR

FFR

FFR

−==

+==

+==

α

α

αIF IF IF

1F

1F

1F

2F

2F

2FR

• TRIANGLE LAW OF FORCES: If two forces acting simultaneously on a body are

represented by the sides of a triangle taken in order,

their resultant is represented by the closing side of

the triangle taken in the opposite order.

O

1F

2F

1F

2F

O A

B

R

• POLYGON LAW OF FORCES: If a number of concurrent forces acting simultaneously on a body ,are

represented in magnitude and direction by the sides of a polygon,

taken in order , then the resultant is represented in magnitude and

direction by the closing side of the polygon, taken in opposite order.

O

1F

2F

3F

4F

O

1F 2F

3F

4F

R

D

A

B

C

1R

2R

COMPOSITON OF FORCES BY

RESOLUTION(Principle of resolved parts)

• The components of each force in the system in two mutually perpendicular directions are found.

• Then, the components in each direction are algebraically added to obtain the two components.

• These two component forces which are mutually perpendicular are combined to obtain the resultant force.

1θ

4θ

2θ

3θ

X

Y

1F2F

3F

4F

Algebraic sum of the components of forces in X

direction

44332211 coscoscoscos θθθθ FFFFFx +−−=∑

Algebraic sum of the components of forces in Y

f direction

Now the system of forces is equal to two

mutually perpendicular forces ,

44332211 sinsinsinsin θθθθ FFFFFy −−+=∑

∑∑ YX FF &

=

+=

∑∑

∑ ∑

−

X

Y

YX

F

F

FFR

1

22

tanθ

NUMERICAL PROBLEMS

1. Determine the magnitude and direction of the resultant of the two forces of magnitude 12 N and 9 N acting at a point ,if the

angle between the two forces is

GIVEN:

NF 121 = NF 92 =030=α

0

0

01

21

21

022

21

2

2

2

1

81.12

30cos912

30sin9tan

cos

sintan

3.20

30cos9122912

cos2

=

+=

+=

=

×××++=

++=

−

−

θ

θ

αα

θ

α

FF

F

NR

R

FFFFR

2.Find the magnitude of two equal forces acting at a point with an angle of 600 between them, if the resultant is equal to N330

GIVEN:

3.The resultant of two forces when they act at right angles is 10 N .Whereas, when they act at a angle of 600 , the resultant is N. Determine the magnitude of the two forces.

Let F1 and F2 be the two forces,

Given – when α =900

R = 10N

When α =600 R = N

We have,

When α =900

Squaring both sides 100= F12 + F2

2 (1)

When α =600

0212

2

2

1 60cos2148 FFFF ++=

5.02 212

2

2

1 ×++ FFFF

sayFFF ,21 ==

060,330 == αNR

NF

FR

FFFR

FFFFR

FFFFR

30

3

60cos2

cos2

222

022

21

2

2

2

1

=

=

++=

××++=

++= α

148

148

αcos2 212

2

2

1 FFFFR ++=

2

2

2

110 FF +=

squaring both sides

148 = F1

2 + F2

2 (2)

substituting (1) in (2)

148 = 100+F1F2

F1F2 = 48 (3)

squaring equation (3),we get

F12 + F2

2 = 48

2 (4)

From (1) F22

= 100 – F1

2 (5)

Subtracting (5) in (4)

( )

( )( )

NFNF

F

F

F

F

FF

FF

6&8

64

1450

19650

504850

48100

48100

21

2

1

2

1

22

1

2222

1

22

1

4

1

22

1

2

1

==

=

=−

=−

+−=−

−=−

=−

5.The 26 KN force is the resultant of two forces. One of the force is as

shown in figure .Determine the other force.

4.Find the magnitude and direction of the resultant

force for the system of concurrent forces shown

below.

N20

N25

N30

N35

030045

040

X

Y

∑∑

−=

−−=

NF

F

X

X

70.30

40cos3545cos3030cos20 000

∑∑

=

−++=

NF

F

Y

Y

72.33

40sin3545sin302530sin20 000

( ) ( )NR

R

FFRYX

60.45

72.3370.3022

22

=

+−=

+= ∑ ∑

0

1

1

68.47

70.30

72.33tan

tan

=

=

=

−

−

∑∑

θ

θ

θX

Y

F

F

∑ XF

∑ YFR

θ

0

y 26kN

12 5 10kN 3 4

X

0

Let F be magnitude of unscnorm force with Fx and Fy as its

components in x and y directions.

Component of R in x directions 13 θ1 12

Rx = 26 x cos θ1 = 26 x 5/13 = 10kN 5

Component of R in y direction

Ry = 26 x sin θ1 = 26 x 12/13 = 24kN

Component F and 10kN in X direction

= Fx +10 cos θ2 5 = Fx + 10x 4/5 = Fx +8 θ2 3

4

Component of F and 10kN in y direction

= Fx + 10 x Sin θ2 = Fy + 10 x 3/5 = Fy + 6

Using R/x = /Fx

10 = Fx +8

24 = Fy + 6

Fx = 2kN, Fy = 18kN

But F = √Fx2+Fy2 = √22 + 182 F = 18.11kN

θ2 = tan -1

(Fy /Fx) = tan -1

(18/2) = 83.660

θ2 = 83.660 ( inclination of F w.r.t x – axis)

6.Three forces act at a point in a plate as shown in figure. If the

resultant of these forces is vertical, find the resultant force and

angle α..

100N 160N α. 120 N α 0

Since the resultant force is vertical, algebraic sum of horizontal components of these must

be equal to zero.

160 cos α – 120 – 100 sin α = 0

120 + 100 sin α = 160 cos α

6 + 5 sin α = 8cos α

Squaring both the sides

(6+5 sine α )2 = (8 cos α )

2

36 + 60 sin α + 25 sin2α = 64 (1-sin

2 α)

25 sin2 α +64 sin

2 α + 60sin α = 64-36

89 sin2 α + 60 sin α = 28

Sin2 α + 0.674 sin α =0.315

(sin α + 0.337)2

= 0.315 + 0.3372

= 0.428

sin α + 0.337 = √0.428 = 0.654

sin α = 0.654 – 0.337 = 0.317

α = sin-1

(0.317) = 18.50

Resultant force R = Σ Fy

= 160 sin α + 100 cos α

= 160sin 18.50 + 100 cos 18.50

R = 145.60 N

7.ABCDE is a regular hexagon. Forces 90 N,P,Q,240 N and 180 N act along

AB,CA,AD,AE and FA respectively as shown in the figure. Find the forces

P and Q such that the resultant force is zero.

C D B P Q E 90N 300 300 240N 300 600 300 A 180N F X

Since the resultant force is equal to zero, Σ Fx = 0 and Σ Fy = 0

Σ Fx = -180 +240 cos 300 + Q cos 60

0 – p

cos 90

0 + 90 cos 120

0 = 0

-180 + 207.85 + 0.5 Q – 45 =0

0.5Q = 17.15

Q = 34.308N

Σ Fy = 180 sin00+240

sin30

0 + Q sin60

0 – P + 90 sin120

0 = 0

120 + 34.308 x sin 600 – P + 90 sin 120

0 = 0

P = 227.654 N

Moment of force F about O= F x a

= AB x OC

= twice the area of triangle OAB

Thus moment of F about O= 2 x Area of triangle OAB

COMPOSITION OF COPLANAR NON-

CONCURRENT FORCE SYSTEM

MOMENT OF A FORCE: Moment is defined as the product of the magnitude of the force and

perpendicular distance of the point from the line of

action of the force.

GEOMETRICAL REPRESENTATION OF MOMENT

Consider a force F represented ,in magnitude and direction

by the line AB. Let O be a point about which the moment

of the force F is required. Let OC be the perpendicular

drawn. Join OA and OB

c

FA

B

O

a

VARIGNON’S PRINCIPLE OF MOMENTS:

If a number of coplanar forces are acting simultaneously on a particle, the algebraic sum

of the moments of all the forces about any point is equal to the moment of their resultant

force about the same point.

PROOF:

For example, consider only two forces F1 and F2

represented in magnitude and direction by AB and AC as shown in figure below.

Let O be the point, about which the moments are taken. Construct the parallelogram

ABCD and complete the construction as shown in fig.

By the parallelogram law of forces, the diagonal AD represents, in magnitude and

direction, the resultant of two forces F1 and F2, let R be the resultant force.

By geometrical representation of moments

the moment of force about O=2 Area of triangle AOB

the moment of force about O=2 Area of triangle AOC

the moment of force about O=2 Area of triangle AOD

But,

Area of triangle AOD=Area of triangle AOC + Area of triangle ACD

Also, Area of triangle ACD=Area of triangle ADB=Area of triangle AOB

Area of triangle AOD=Area of triangle AOC + Area of triangle AOB

Multiplying throughout by 2, we obtain

2 Area of triangle AOD =2 Area of triangle AOC+2 Area of triangle AOB i.e., Moment of force R about O=Moment of force F1 about O + Moment of force F2

about O

Similarly, this principle can be extended for any number of forces.

CO

BA

D

1F

2FR

NUMERICAL PROBLEMS

1.For the non-concurrent coplanar system shown in fig below,

determine the magnitude , direction and position of the resultant

force with reference to A.

R

N35

N25

N50

N20

A

B

C

D

( )

∑∑

−=−−=

→=−=

NF

NF

Y

X

853550

52025

( )↓= N85

( ) NFFR YX 15.858552222 =−+=+= ∑∑

∑ XF

∑ YF

R

θmd

dR

525.215.85

75140

325435

=+

=

×+×=×

or

mx

x

53.285

75140

32543585

=+

=

×+×=×

∑ XF

∑ YF

R

θ

2.Determine the resultant of the force system acting on the plate

as

shown in figure given below wirh respect to AB and AD.

10N 5N

60

0 D 10Nm 30

0 C

3m A 4m B

14.14N 1 1 20 N

Σ Fx = 5cos300 + 10cos600 + 14.14cos450 = 19.33N Σ Fy = 5sin300 - 10sin600 + 14.14sin450 = -16.16N R = √( Σ Fx2 + Σ Fy2) = 25.2N θ= Tan-1(Σ Fx/ Σ Fy) θ= Tan-1(16.16/19.33) = 39.890 D C y A x θ B R 16.16N Tracing moments of forces about A and applying varignon’s principle of moments we get +16.16X = 20x4 + 5cos300x3-5sin300x4 + 10 + 10cos600x3

Θ 19.33N 19.33N

X = 107.99/16.16 = 6.683m Also tan39.89 = y/6.83 y = 5.586m.

3.The system of forces acting on a crank is shown in figure

below. Determine the magnitude , direction and the point of application of the resultant force.

500 N 150 700N 150 600 600 150mm 150 mm 150 Cos600=75mm Σ Fx = 500cos600 – 700 = 450N Σ Fy = 500sin600 = -26.33N R = √( Σ Fx2 + Σ Fy2) = √(-450)2 + (-2633)2 R = 267.19N (Magnitude) Σ Fx

θ θ= Tan-1(Σ Fx/ Σ Fy) = Tan-1(2633/450) R Σ Fy θ= 80.300 (Direction) ΣFx Θ x R ΣFy

Tracing moments of forces about O and applying varignon’s principle of moments we get -2633x x= -500x sin600x300-1000x150+1200x150cos600 -700x300sin600 X = -371769.15/-2633 X = 141.20mm from O towards left (position).

4.For the system of parallel forces shown below, determine the magnitude of the resultant

and also its position from A .

100N 200N 50N 400N

R

A B C D 1m 1.5m 1m

X

Σ Fy = +100 -200 -50 +400 = +250N ie. R = Σ Fy =250N ( ) Since Σ Fx = 0 Taking moments of forces about A and applying varignon’s principle of moments -250 x = -400 x 3.5 + 50 x 2.5 + 200 x 1 – 100 x 0 X = -1075/ -250 = 4.3m

5.The three like parallel forces 100 N,F and 300 N are acting as shown in figure below. If

the resultant R=600 N and is acting at a distance of 4.5 m from A ,find the magnitude of

force F and position of F with respect to A.

100N F 600 N 300N

A B C D 4.5m 2.5m

X

Let x be the distance from A to the point of application of force F

Here R = Σ Fy 600=100+F+300 F = 200 N

Taking moments of forces about A and applying varignon’s principle of moments, we get 600 x 4.5 = 300 x 7 + F x 200 x = 600 x 4.5 -300 x 7

X = 600/200 = 3m from A

6.A beam is subjected to forces as shown in the figure given below. Find the magnitude ,

direction and the position of the resultant force.

17kN 10kN 20kN 10kN 5kN

α

θ 4kN A B C D E 2m 3m 2m 1m

Given tan θ = 15/8 sin θ = 15/ 17 cos θ = 8/17 tan α = 3/4 sin α = 3/5 cos α = 4/5

Σ Fx = 4 +5 cos α – 17 cos θ = 4+5 x 4/5 – 17 x 8/17 Σ Fx = 0 Σ Fy = 5 sin α -10 +20 – 10 + 17 sin θ = 5 x 3/5 -10+20 – 10 + 17 x 15/17

Σ Fy = 18 kN ( )

Resultant force R = √ 2Fx 2 + Σ Fy2 = √ 0+182 R = 18 kN ( ) Let x = distance from A to the point of application R Taking moments of forces about A and applying Varignon’s theorem of

moments -18 x = -5 x sin α x 8 +10 x 7 -20 x 5 + 10 x 2 = -3 x 8 +10 x7 – 20 x 5 + 10 x 2

X = -34/-18 = 1.89m from A (towards left)