statics chapter four moment by laith batarseh home next previous end

Chapter Four

Laith Batarseh

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MOMENT

By

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MOMENT SCALAR

Definition

Moment is defined as the tendency of a body lies under force to rotate about a point not on the line of the action of that force (i.e. there is a distance between the force and the rotation point )

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The acting forceMoment arm

Description

Moment depends on two variables:

Moment is a vector quantity

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MOMENT SCALAR

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Description

ForceArmTendency to rotate

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MOMENT SCALAR

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Tendency for rotation

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MOMENT SCALAR

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Magnitude

FD

Moment magnitude (M) = F.D

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MOMENT SCALAR

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Direction

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MOMENT SCALAR

Solving procedures

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1. Define the magnitudes of force (F) and arm (D)

2. Assume the positive direction (eg. Counter clock wise)

3. Find the magnitude of moment (M) as F.D

4. Give the moment the correct sign according to the tendency for rotation

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MOMENT SCALAR

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Example [1]

Find the moment caused by the following forces about point O

100 N

0.5m

2m

(b)

O

100 N

0.5m

2m

(a)

O

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MOMENT SCALAR

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Example [1]

Assume the CCW direction is the positive direction.

100 N

0.5m

2m

(b)

O

100 N

0.5m

2m

(a)

O

Branch (a) Mo = F.d = -(100N)(0.5m) Mo=-50 N.m=50N.m CWBranch (b) Mo=F.d = (100N)(2m) Mo=200 N.m CCW

+

+

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Principle Of Moments

Principle of Moments

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some times called Vrigonon’s theorem (Vrigonon is French

mathematician 1654-1722).

State that the moment of a force about a point equals the summation

of the moments created by the force components

In two dimensional problems: the magnitude is found as M = F.d and

the direction is found by the right hand rule

In three dimensional problems: the moment vector is found by M =rxf

and the direction is determined by the vector notation (ie. i,j and k

directions)

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Example [1]


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Example [1]

Mo,1 = 100 sin(30) (10) = 500 N.m

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Mo,2 =- 100 cos(30) (5) =- 433N.m+

M = Mo,1+Mo,2=500-433=67N CCW

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Example [2]

1. Force analysis

100 cos(40)1.2 m

0.3 m

100 sin(40)

120 cos(60)

120 sin(60)

2. Moment calculations

∑ M = (100 cos(40))(1.5) –( 120 cos(60))(1.2) =43 N.m CCW +

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MOMENT SCALAR

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Moment resultant

F1

F2F3

d1

d2 d3

M1O

M2

M3

Mo = ∑Mo = M1 + M2 – M3 = F1d1+F2d2 – F3d3 +

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MOMENT SCALAR

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Example [2]


2m

3m

5m

1m30o

100 N

50 N

60 N

75 NO

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MOMENT SCALAR

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Example [2]

2m

3m

5m

1m30o

100 N

50 N

60 N

75 NO

CWmNmNM

M

o

o

.45.272.45.272

1)30cos(755)30sin(750503602100

+

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MOMENT SCALAR

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Exercise


100 N

300 N

5m

2m 45o

30oO 0.3m

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MOMENT SCALAR

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Exercise

CWmNmNM

M

o

o

.897.897

5)45sin(3003.0)45cos(3002)30sin(100

+

100 N

300 N

5m

2mO 0.3m

300 sin (45)N

300 cos (45)N

100sin (30)N

100cos (30)N

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MOMENT VECTOR

Cross product

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Cross product is a mathematical operation can be done on vectors

Cross product for one time is done for two vectors

The cross product of two vectors is a vector perpendicular to the plane

of A and B

The notation of vector A cross vector B is: C = AxB where C is the

resultant vector from the cross product

the vector C can be represented as : C =CUc where Uc is a unit vector in a

direction perpendicular to the plane that contains both A and B.

The value of the scalar quantity C is given as : C=A.B.sin(ϕ)

where ϕ is the angle between A and B.The cross product is controlled by the right-hand rule

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MOMENT VECTOR

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Graphical representation

ϕ

Uc

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MOMENT VECTOR

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Cartesian vector formulation

i=jxk

j=kxi

K=ixj

y

x

z

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MOMENT VECTOR

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Cartesian vector formulation

A = Ax i + Ay j + Az k

B = Bx i + By j + Bz k

AxB=(Ay Bz -Az By )i-(Ax Bz - Az Bx )j + (Ax By - Ay Bx )k

zyx

zyx

BBBAAAkji

x BA

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MOMENT VECTOR

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Moment – vector formulation

F

rθ

d

M

Ox

yMagnitude: Mo = rFsin(θ) = Fd

Direction: perpendicular to x-y plane (z-direction)

Matrix notation:

Resultant moment:MRo = ∑(rxF)

zyx

zyx

FFFrrrkji

x FrMo

Best for three dimensional

problems

Mo = rxF

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MOMENT VECTOR

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Example [1]


F = [5i + 10j + 6k]N

z

x

y3m

2m1m

O

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MOMENT VECTOR

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Example [1]

1. Formulate the position vector (r) : r = 3i+2j+1k

2. Find the moment vector (Mo) by matrix notation

kjikji

x 201326105123 FrMo

F = [5i + 10j + 6k]Nr = 3i+2j+1k

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MOMENT VECTOR

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Example [2]


xy

z

O

F = [-5i + 5j -5k]N

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MOMENT VECTOR

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Example [2]

1. Formulate the position vector (r) : r= 15i + 10j +6k

2. Find the moment vector (Mo) by matrix notation

kjikji

x 1254580555

61015

FrMo

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MOMENT VECTOR

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Summary

Moment is a vector can be found by cross product and

matrix notation

Matrix notation:

zyx

zyx

FFFrrrkji

x FrMo

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Principle of Moments

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some times called Vrigonon’s theorem (Vrigonon is French

mathematician 1654-1722).

State that the moment of a force about a point equals the summation

of the moments created by the force components

In two dimensional problems: the magnitude is found as M = F.d and

the direction is found by the right hand rule

In three dimensional problems: the moment vector is found by M =rxf

and the direction is determined by the vector notation (ie. i,j and k

directions)

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Example [1]


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Example [1]

Mo,1 = 100 sin(30) (10) = 500 N.m

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Mo,2 =- 100 cos(30) (5) =- 433N.m+

M = Mo,1+Mo,2=500-433=67N CCW

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Example [2]

The following is a gate. In which direction this gate will rotate?

100N

120N

1.2 m

0.3 m60o

40o

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Example [2]

1. Force analysis

100 cos(40)1.2 m

0.3 m

100 sin(40)

120 cos(60)

120 sin(60)

2. Moment calculations

∑ M = (100 cos(40))(1.5) –( 120 cos(60))(1.2) =43 N.m CCW +

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Example [3]


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Example [2]

r1= 15i + 10j +6k F1 = [5j]N kikji

x 753005061015 FrMo,1

r2= 15i + 10j -6k F2 = [-5j]N kikji

x 753005061015

FrMo,2

i60oM

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Moment of force about axis

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In many real cases, the force tendency to rotate is about a specified

axis.Example :

Moment components:

Mo,1 = (100)(10) (about y-axis)

Mo,2 = (100)(15) (about x-axis)

Mo,3 =0 (about z-axis)

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Magnitude

Scalar analysis: M = F.d

Vector analysis:

zyx

zyx

z,ay,ax,a

a

FFFrrr

uuux.M Frua

Where: ua is a unit vector defining the direction of a-axis and given as

ua = ua,x i + ua,y j + ua,z k

To find the moment vector (Ma): Ma. ua

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Example [1]

Mo,1 = 100 sin(30) (10) = 500 N.m

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Mo,2 =- 100 cos(30) (5) =- 433N.m+

M = Mo,1+Mo,2=500-433=67N CCW

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Example [2]

Determine the magnitude of the moments of the force F about the x, y, and z axes. Solve the problem

using a Cartesian vector approach

using a scalar approach.

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Example [2]

using a Cartesian vector approach

rAB = {5i + 4j -3k} m

For the axes: x, y and z the unit vectors are i, j and k respectively.

Mx = i . (rAB x F) My = j . (rAB x F) Mz = k . (rAB x F)

m.N144105345

001M x

m.N5

4105345

010M y

m.N30

4105345

100M z

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Example [2]

using a scalar approach

Mx = ∑Mx = 10(3) – 4(4) = 14 N.m

My = ∑My = -5(3) + 4(5) = 5 N.m

Mz = ∑Mz = -5(4) + 10(5) = 30N.m

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Moment of couple

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Couple is the moment generated by two forces has the same

magnitude and opposite direction.

Definition

F

-F

d

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Moment of couple

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This analysis is considered for 2-D

problems

The magnitude of moment is found

by: M=F.d where F is the force

magnitude.

the direction of couple moment is

perpendicular to the plain that

contain the F and d and it is found by

the right hand rule.

Scalar analysis

-FF

d

M

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Moment of couple

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Example on scalar analysis

F2d1

d2

-F2-F1

F1

M1 = F1 . d1 M2 = F2 . d2

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Moment of couple

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Vector analysis

This analysis is considered for 3-D

problems

The moment vector is found by:

M=F x r where r is the position vector

directed between the forces F and –F

Note that the moment vector is

dependent on the position vector

directed between the forces F and –F

(r).

Derivation: M = rB x F + rA x –F = (rB - rA) x F But: (rB - rA) = r Then: M = rx F

F-F

O

r

rArB

A

B

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Moment of couple

Example [1]

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Find the resultant moment couple produced by the following forces. All dimensions in m

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Moment of couple

Example [1]

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Solution:1. By traditional moment analysis

∑Mo = -(150)(7)-(150)(7) – (200)(9)-(200)(9) = -5700 N.m +

2.By cpouple moment analysis

∑Mc = -(150)(14) – (200)(18) = -5700 N.m +

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Moment of couple

Example [2]

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Find the moment couple produced by the following force. All dimensions in m

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Moment of couple

Example [2]

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1. F = 150 j 2. r = 3k 3. M = r x F = 150 j x 3k = {450N.m} i

M

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Moment of couple

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in many of life applications, an equivalent couple is required to solve some technical problems such as space and size. Equivalent couples are the couples that have the same magnitude and same direction As you can see, the relation between the forces and the arm distances in equivalent coupels is reverse (for example, as we reduce the moment arm, the required force for equivalent couple increases)

Equivalent couples

F1

-F1

d1 d2

F2

-F2

=F1.d1 = F2.d2

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Moment of couple

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Resultant couple is the vectorial summation of the couples act on the body as you can see. In simple situation as shown in the figure, the parallelogram is used

to sum the moments and in more complicated cases or three dimensional problems, the Cartesian notation is used.

Resultant couple

M1 M2

M1

M2

MR

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Moment of couple

Example [1]

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The member shown in the figure is subjected to three coupling forces : 150, 200 and 100 N. the moment arms are shown.

If All dimensions in m, find the resultant couple moment produced by the following forces.

150 N

150 N

100N

100N

200N

200N

Solution:

First, the Moment direction:

Second, the calculate the coupled moments

M1 = -(200)(13) = -2600 N.m

M2 = -(150)(7) = -1050 N.m

M3 = - (100)(8) = -800 N.m

Finally, calculate the moment sum (resultant)

MT = -2600-1050 -800=-4450 N.m

MT = 4450 N.m (clockwise)

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Moment of couple

Example [1]

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Moment of couple

Example [2]

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The member shown in the figure is subjected to two coupling forces : 200 and 100 N. the moment arms are shown.

Find the resultant couple moment produced by the following forces.

3m

3m

6m

0.6m

45o 100 N

100 N200 N

45o

200 N

Solution:


Second, the calculate the coupled moments M1 = -(100)(0.6) = -60 N.m

M2,x = (200 cos(45))(0.6) = 85 N.m

M2,y = -(200 sin(45))(3) = -424 N.m

Finally, calculate the moment sum (resultant) MT = --60+85-242=-399 N.m

MT =399 N.m (clockwise)

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Moment of couple

Example [2]

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Moment of couple

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For three dimensional problems, it is better to use the Cartesian

notation (I, j and k) to represent the monuments.

The moment in a direction not one of the principle axes (x, y or z)

can be represented as: M = Mu. Where u is a unit vector in the

direction of the moment M.

The resultant moment can be found finally by vector addition for

the moments vectors

3-D problems

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Moment of couple

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3-D problems

M1 M2

x

z

y

θ

Example:

The figure show an object subjected to

two moments: M1 and M2. as you can

see M2 has an angle θ from the y-axis.

The moment M1 can be represented as:

M1= M1i.

While the moment

M2 can be represented as:

M2 = M2 (0i + cos(θ) j + sin(θ) k)

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Simplification of a force and couple system

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In many of situation where there a group of forces and moments acting on an object, it is seem more convenient to reduce the large number of forces and moments to one force and one moment.

Physical meaning: replacing a system of forces and moments by a system of one force and one moment.

Condition: the external effects produced by the forces and moments on the body for the original system are the same of the single force and moment in the new simplified system

Equivalent couples

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Simplification conditions

B A BF F-F F

Note: the acting force can be transport from one position to another on its line of action (i.e. force vector)

A B A B

F

M = F.d

F

Note: the force acts on a member can be transport from one position to another on a line perpendicular force vector by adding the moment generated by the original force (i.e. M=F.d)

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F1

O

F2

r1r2

M

Assume an object as shown in Fig.a is subjected to two forces ( F1 and F2) and one moment M. The forces F1 and F2 has a position vectors r1 and r2 respectively from the rotation point O to the line of action for each force.

Fig.a

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To convert the previous system into one force-moment system we must:•first move each force to the point of rotation O. this step include adding the moments produced by both forces (M1 and M2 respectively )at the rotation point as shown in the Fig.b.

Fig.b

O

F2 M

M1 = r1 x F1M2 = r2 x F2

F1

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•Then all forces and moments are summed using the following formulas:

FR = ∑F MR,O = ∑MO + ∑M

Fig.c

O

FRMR,o

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Example [1]

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Figure below shows a plate a group of forces (100, 150, 200 and 300 N)

If All dimensions in m, Simplify the following force system to single force and moment system about point O

Solution:

First, calculate the resultant force FR

∑Fx = 300 – 100 = 200 N

∑Fy = 200 + 150 = 350 N

Second, calculate the resultant moment MR

MR =-(200)(13) – (300)(4) –(150)(5) – (100)(4) = 4950 N.m

Finally, represent the new force and moment on the original system

as shown.

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Example [1]

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NFR 403350200 22 o60

200350tan 1

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Example [2]: couple resultant

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The system shown in the figure is subjected to two coupling forces : 200 and 100 N. the moment arms are as shown.

Simplify the following force system to single force and moment system about point O

3m

3m

6m

0.6m

45o 100 N

100 N200 N

45o

200 N399 N.m

Solution:


Second, calculate the coupled moments M1 = -(100)(0.6) = -60 N.m

M2,x = (200 cos(45))(0.6) = 85 N.m

M2,y = -(200 sin(45))(3) = -424 N.m

Finally, calculate the moment sum (resultant) MT = -60+85-242=-399 N.m

MT =399 N.m (clockwise)

Note that the resultant force (FR ) = 0 which is true for all the coupled forces systems

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Example [2]

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Example [3]

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The three forces act on the pipe

assembly. If F1 = 50 N and F2 = 80 N,

replace this force system by an

equivalent resultant force and couple

moment acting at O. Express the

results in Cartesian vector form..

Solution:

First, calculate the resultant force FR

Finally, calculate the resultant moment MR using cross product

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Example [3]

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kk ]210[8050180 NFR

mN

kjikjikjix

.}22515{

500005.02

800005.025.1

180000025.1

ji

FrMo

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Concurrent forces: forces that’s lines of action intersect at a common point

Concurrent forces are simply summed to find FR and as seen the moment is zero due to the passing of forces lines of action through the rotation point

Special cases:

F1 F2

F4F3

OFR

O=

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Coplanar forces: forces share the same plane

Coplanar forces produce moments about the point of rotation and are summed to find FR . All the moments produced by the acting forces are summed to find the equivalent moment M.

Special cases:

F1 F2

F4F3

OFR

OM

=

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Parallel forces system:

F2

F1

F3

F4 O

zFR = ∑F

O

zMR,O

b

aFR = ∑F O

z

b

a

d

A reverse process can be done to transform the single force – moment system into a single force with moment arm from the rotation point

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Establish the coordinate system (x, y and z axes). It is preferred to put the origin of this system at the rotation point. Force summation

find the resultant force by summing the acting forces. You may resolve the forces to their rectangular components.

Moment summation The resultant moment is the summation of the moments acting on the body and the moments produced by the acting forces.

Analysis procedures

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In three dimensional systems, we can find an equivalent force

and moment. However, in general cases the moments and

force are not perpendicular to each other. Because of that, it

become impossible to reduce the system to single force with

moment arm from the rotation point.

Special cases:

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Example [1]

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Replace the following forces-moment system to a single force system.

5m

10 kN 7 kN

30o

2m

8m

Solution:

First, calculate the resultant force F

∑Fx = 10 + 7cos(30) = 16.06kN

∑Fy = - 7 sin(30) = 3.5 kN

Second, calculate the resultant moment MR

MR =-(7sin30)(5) -(7cos30)(8)-(10)(8-2) = 126 kN.m

Finally, you can represent the new force and moment on the original

system.

Xo= MR/FR = 126/16.43 = 7.67 m from the base point

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Example [1]

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kNFR 43.165.306.16 22 o3.12

06.165.3tan 1

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Example [2]

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Replace the following forces-moment system to a single force system.

3m3m3m

x

z

y

200N

300 N600 N

4m

Solution:

First, calculate the resultant force F

FR = 300 + 600 – 200 = (700k) N

Second, calculate the resultant moment Mx and My

Mx =(300)(0) – (200)(3) + (600)(6) = 3000 N.m = FRy → y = 4.29m

My =-(300)(3) + (200)(0) - (600)(4) = 3300N.m = FRx→ x = 4.71m

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Example [2]

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Solution:

Finally, you can represent the new force on the original system.

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Example [2]

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z

y700 N

4.29m4.72 m

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Distributed Loads

Uniform loading along a single axis

x

w

L

w = w(x)

x

w

L

dF = dA

x

dxw = w(x)

Distributed force is a force acting on a line or surface of the rigid body. The value of this force (w) is represented by a function in terms of dimensions. For example: w(x).

The force function could be linear or none linear

We can represent the distributed force by a single. To do that we first take an infinitesimal segment of the distributed force (dF) which equal the infinitesimal segment of the area under the force function as you can see on the screen

Statics

Distributed Loads

Magnitude of resultant force

x

FRw

w = w(x)

'x

Let us first assume a distributed force w(x) acting on the member as shown in the fig. Now, assume there is an equivalent force called FR for the distrusted force w(x) and it is located at a distance equal x’The magnitude of FR can obtained by integrating the function w(x) over the distance x:

AdAdxxwF

AR .

As you can see from the above equation that the magnitude of FR equal the area under the curve w(x)

Statics

Distributed Loads

Location of resultant force

The location of the resultant force (d) can be found using the principle of centroid (will be discussed later) as:

x

FRw

w = w(x)

'x

A

A

L

L

dA

dAx

dxxw

dxxxwx

.

.

.'

For this stage, the location of the centroid for the given shape will be given

Statics

Distributed Loads

Analysis procedures

To analyze the distributed forces you have to follow the following procedures:

distributed load is defined as function w = w(x) with

unit of N/m or lbf/ft.

The effect of distributed load is simplified by single

concentrated force acts at certain point in the body

The resultant force equals the area under the loading

diagram and acts on the centroid of this area

Statics

Distributed Loads

Example 1:

x

w200 N/m

2m(a)

x

w 200 N/m

3m(b)

Determine the magnitude and the location of the equivalent resultant force acting on the beam shown in the figures.

Statics

Distributed Loads

Example 1:

x

w200 N/m

2m

Solution: part a

FR = area under the loading diagram

FR = (200 N/m) (2m) = 400 N

(x’) at the center of load rectangle

x‘ = 1m

x

w

1m

400 NYou can note that the centroid of a rectangular area is its geometric center

Statics

Distributed Loads

Example 1:

Solution: part b

Similar to part aFR = area under the loading diagram

FR = (1/2)(200 N/m) (3m) = 300 N

(x’) at the centroid of triangle load

x‘ = (2/3)(3) = 2m

You can note that the centroid of a triangular area is located at a distance equal 1/3 of its height fro its base.

x

w 200 N/m

3m

x

w

2m

300 N

Statics

Distributed Loads

Example 2:

Determine the magnitude and the location of the equivalent resultant force acting on the beam shown in the figure.

x

w 250 N/m

w = (30)x2 N/m

2.5m

Statics

Distributed Loads

Example 2:

Solution: magnitude

x

w156.25 N

NF

xdxxdAF

R

AR

25.15630

35.230

330.30

33

5.2

0

35.2

0

2

x

w 250 N/m

w = (30)x2 N/m

2.5m

Try to solve it by your self and verify the solution

Statics

Distributed Loads

Example 2:

Solution: location

x

w156.25 N

1.875 m

x

w 250 N/m

w = (30)x2 N/m

2.5m

m

x

x

dxxx

dA

dAxx

A

A

875.125.156

430

25.156

.30.

5.2

0

4

5.2

0

2

Try to solve it by your self and verify the solution

Statics

Distributed Loads

Combined distributed loads

In this lecture we will learn how find the resultant force from a combined distributed forces

If you have a several disturbed loads , you can find the resultant force for the whole combination by finding the resultant force from each distributed force and then sum the resultant forces to obtain one equivalent force

Statics

Distributed Loads

Example 1:


x

w400N/m

300N/m

2m 4m 2m

Statics

Distributed Loads

Example 1:

Solution: first load


FR = (0.5)(400 N/m) (2m) = 400 N

(x’) at the centroid of triangle load

x‘ = (1/3)(2) = 2/3m

x

w

2/3 m

400N

Statics

Distributed Loads

Example 1:

Solution: second load


FR = (400 N/m) (4m) = 1600 N

(x’) at the centroid of load area

x‘ = 2+(0.5)(4)= 4m

x

w

4m

1600N

Statics

Distributed Loads

Example 1:

Solution: third load


FR = (300 N/m) (2m) = 600 N

(x’) at the centroid of load area

x‘ = 2+4+(0.5)(2)= 7m

x

w

7m

600N

Statics

Distributed Loads

Example 1:

Solution: representing all forces

w

2/3m 4m

7m

600N1600N400N

Statics

Distributed Loads

Example 1:

Solution: representing all forces

w

x’=4.2m

2600NFR =∑F = 400 + 1600 + 600 = 2600 N

To find the location of this force we must use the technique of simplifying of a force and couple moment you learn previously

MR = FR * x’ → x’ = MR/FR

MR =(400)(2/3) + (1600)(4) + (600)(7) = 10867 N.m

x‘ = 10867/2600 = 4.2 m

Try to solve the same problem if the second load (400N/m) acts on the lower surface of the body. Be aware to the sum of forces and the direction of resultant moment

Statics

Distributed Loads

Example 2:


250 N/m

200 N/m 400 N/m

60o 45o

5m

5m 10m

Statics

Distributed Loads

Example 2:

1250 N

1000 N

4000 N

60o 45o

5m

5m 10m

∑Fx = -(4000)(cos45) + (1000)(cos30) = -2198 N∑Fy = -1250 – (4000)(sin45) – (1000)(sin30) = -4578 NFR = {(-2198)2 + (-4578)2}1/2 = 5078 N Ө = tan-1(-4578/-2198) = 64o

Statics

Distributed Loads

Example 2:

to find the location of the resultant force, we can take the moment about a certain point in the body. Let us take point O.

∑Mo = -(1000)(cos30)(0.5)(5sin60) – (1000)(sin30)(0.5)(5cos60) – (1250)(2.5+5cos60) – (4000)(sin45)(5cos60+5+(0.5)10cos45) + (4000)(cos45)(0.5)(10sin45) = -29963 N.m

x‘ = Mo/FR = 29963 /5078 = 5.9 m from point O

O

1250 N

1000 N

4000 N

60o 45o

5m

5m 10m(4000)(cos45)

(4000)(sin45)

(1000)(cos30)

(1000)(sin30)

statics chapter four moment by laith batarseh home next previous end

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