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ENGR-1100 Introduction to Engineering Analysis

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ENGR-1100 Introduction to Engineering Analysis

Lecture 9

Previous Lecture Outline

- Moment of a force.
- Principle of moments and Varignon’s law.

Lecture Outline

- Vector representation of a moment:
- - Moment of a force about a point.
- -Moment of a force about a line.

Right-Hand System

The Cartesian coordinates axes are arranged as a right-hand system.

Multiplication of Cartesian Vectors

- The scalar (or dot) product.
- The vector (or cross) product.

A

q

B

0< q <1800

The Scalar (or dot) Product

The scalar product of two intersecting vectors

A•B=B•A=AB cos(q)

A•B = Ax Bx + Ay By + Az Bz

The angle between the vectors can be found by doing the dot product and dividing the result by the scalar component of AB

0< q <1800

The vector (or cross) product

The vector product of two intersecting vectors A and B, by definition, yields a vector C that has the magnitude that is the product of the magnitude of vectors A and B and the sine of the angle between them with a direction that is perpendicular to the plane containing vectors A and B.

AxB=-BxA=(AB sin(q))ec

C

B

q

A

It is often more convenient to use the vector approach to simplify moment calculation.

O

r

d

F

a

A

M0=r X F = | r || F |sin a e

Where: r is the position vector from point O to a point A on the line if action of the forceF.

e – unit vector perpendicular to plane containing r and F.

O

r2

d

r1

F

a2

A1

a1

A2

Does the moment depends on the location of point A along the line of action of F?

d = r1sin a1

= r2sin a2

M0=r X F = | r || F |sin a e

=F (rsin a) e

=F d e

z

F

The location of point A

A

rAB

rA= xAi+ yAj+ zA k

B

rB

rA

zA

zB

The location of point B

y

xB

yB

rB= xBi+ yBj+ zB k

xA

yA

Since:

x

rA= rB+ rAB

Finding the Position Vector

r= rAB = rA-rB= (xAi+ yAj+ zA k ) - (xBi+ yBj+ zB k ) =

(xA- xB) i+ (yA- yB) j+ (zA- zB) k

y

F

A

r

j

O

x

i

i j k

rx ry 0

Fx Fy 0

M0= = (rx Fy - ry Fx)k= Mzk

(rx Fy - ry Fx)k= Mzk

Two Dimensional Case

F= Fxi+ Fyj

r= rxi+ ryj

M0=r X F = (rxi+ ryj) X (Fxi+ Fyj) =

rx Fx(ix i ) + rx Fy(ix j ) + ry Fx(jx i ) + ry Fy(jx j ) =

z

k

r

A

O

y

i

j

x

Three Dimensional Case

F

F= Fxi+ Fyj+ Fzk

r= rxi+ ryj + rzk

M0=r X F = (ryFz-rzFy)i+(rzFx-rxFz)j +(rxFy-ryFx)k

= Mxi +Myj +Mzk

We can also use the determinant form:

The scalar components of the moment:

i j k

rx ry rz

Fx Fy Fz

Mx= (ryFz-rzFy)

My=(rzFx-rxFz)

Mz= (rxFy-ryFx)

M0= = (ryFz-rzFy)i+(rzFx-rxFz)j +(rxFy-ryFx)k

= Mxi +Myj +Mzk

M0=r X F = (ryFz-rzFy)i+(rzFx-rxFz)j +(rxFy-ryFx)k

= Mxi +Myj +Mzk

|Mo|= Mx2 + My2 + Mz2

The magnitude of the moment:

The moment can also be written as:

Mo=M0 e

Where:

e=cos (qx) i+ cos (qy) j +cos (qz) k

The direction:

qx=cos-1(Mx/M); qy=cos-1(My/M); qz=cos-1(Mz/M);

F

z

rB=rA+rBA

B

rBA

A

M0=(rA+rBA) X F =

rA

rB

k

O

y

i

j

M0= (rAX F)

x

The position vector can be drawn to any point on

the line of action of the force

Does the moment depend on the location of the point along the line of action of F in the three dimensional case?

M0=rB X F

M0= (rAX F)+(rBAX F )

But rBA is collinear with F

The principle of a moment; proof

M0=r X F

But:

R =F1+F2 …. Fn

Therefore:

M0= r X R= r X ( F1+F2 …. Fn )=

(r XF1)+(r X F2 )+..…+(r XFn )

Thus:

M0=MR=M1+M2+…..+Mn

z

B

F

13 in

O

y

12 in

15 in

A

x

P4-47

Example P4-47

A force with magnitude of 928 lb acts at a point in a body as shown in Fig P4-47. Determine the moment of the force about point O.

z

B

F

13 in

Fz= F cos(qz) =928 *{13/23.2}=520.1 lb

O

y

12 in

15 in

rOB= 13k[in]

A

x

Solution

dAB= 122 +152 +132 =23.3 in

Fx= F cos(qx) =928 *{(–12)/23.2}=-480 lb

Fy= F cos(qy) = 928 *{(–15)/23.2}=-600 lb

F= -480i - 600j+ 520.1k[lb]

z

B

i j k

0013

-480-600520

F

13 in

M0=

O

y

12 in

15 in

A

x

F= -480i - 600j+ 520.1k lb

rOB= 13kin

M0= [0-13*(-600)] i - [0-13*(-480)] j +0 k

M0= 7800i - 6240jlb•in

Class Assignment: Exercise set 4-48

please submit to TA at the end of the lecture

A force with a magnitude of 860 N acts at a point in a body as shown in Fig. P4-48. Determine

a) The moment of the force about point C.

b)The perpendicular distance from the line of action of the force to point C.

- Solution:
- Mc=283i -155.7j +119.7k Nm
- d=0.401m

B

en

MOB=(MOen) en =[(rXF) en] en= MOBen

A moment of a force about a line

MOB= MOen = (rXF) en =

en

i j k

rx ry rz

Fx Fy Fz

enxeny enz

rx ry rz

Fx Fy Fz

MOB= MOen = (rXF) en =

Where: enx, eny and enz are the Cartesian components of the unit vector

MOB=(MOen) en =[(rXF) en] en= MOBen

B

C

200 mm

F

O

y

220 mm

240 mm

A

x

Example P4-62

The magnitude of the force Fin the figure is 595 N. Determine the scalar component of the moment at point O about line OC.

dAB= 2202 +2002 =297.3 mm

Fz= F cos(qz) =595 *{–200/297.3} = 400.2 N

Solution

Fx= F cos(qx) =595 *{(–220)/297.3} = -440.3 N

Fy= F cos(qy) = 595 *{0/297.3}=0 N

F= -440.3i +0 j + 400.2kN

rOA = 220 i+ 240 jmm

i j k

220 240 0

-440.3 0 400.2

M0=

F= -440.3i +0 j + 400.2kN

rOA = 220 i+ 240 jmm

M0= [240*400.2-0] i - [220*400.2-0] j + [0-240*(-400.2)] k Nmm

M0= 96i - 88j + 105.7kNm

dOC= 2202 +2002 =297.3 mm

z

M0= 96i - 88j + 105.7kNm

Mo

B

C

200 mm

F

O

y

220 mm

eOC= 220/297.3i +0 j + 200/297.3k

240 mm

A

x

eOC= 0.74i +0 j + 0.673k

MOC= MOeOC = 96*0.74-88*0

+105.7*0.673

MOC= 142.2 Nm

Class Assignment: Exercise set 4-61

please submit to TA at the end of the lecture

The magnitude of the force Fin Fig. P4-61 is 450 lb. Determine the scalar component of the moment at point B about line BC.

Solution:

MBC=2.78 in.kip