Lecture 8

ENGR-1100 Introduction to Engineering Analysis

Lecture outline

• Moment of a force

• Principle of moments and Varignon’s law

Non Concurrent Forces• For concurrent forces R= 0 was sufficient for equilibrium.• For non concurrent forces the particle idealization is no

longer valid.• For this case the condition R= 0, while still necessary for

equilibrium, is not sufficient.• A second restriction is related to tendency of a force to

produce rotation (angular acceleration) of a body.• This gives rise to the concept of moment of a force:

F

F

Moment M0 of Force F about AA

• Magnitude of Moment:

M0= |F|d

• Direction of Moment is along axis A-A

• That is, moment is a vector along the direction of A-A:

M0= |F|d eAA

where eAA is unit vector along axis A-A

Fd

M0

•In 2D, we work in the plane perpendicular to the axis that contains F. Point O is intersection of axis A-A with that plane.

•The magnitude of the moment is:

M0 = |F|d

where: F = force; and d = perpendicular distance from the line of action of the force to the axis (point O).

•Units of Moment M0 : lb•ft or lb•in in U.S. Customary system; and N•m or kN•m in SI units.

Fd

M0

•In 2D, we don’t need the complication of treating M0 as a vector. We treat M0 as a scalar (it is understood that it is a vector perpendicular to plane).

•The magnitude of moment M0 in the plane is:

M0=|F|d

• The sign of M0 is:

(+) when counterclockwise (figure)

(-) when clockwise (opposite to figure)

Example P4-2

MA=|F|dA=225*0.6 = 135 Nm

MB=|F| dB=225*0.4 = 90 Nm

MC=|F| dC=225*0.8 = 180 Nm

MA= 135 Nm

MB= 90 Nm

MC= 180 Nm

C

B

Class Assignment: Exercise set 4-3please submit to TA at the end of the lecture

MA1=|F1|dA=500*30 = 15,000 lb•in

MB1=|F1| dB=500*20 = 10,000 lb•in

MA1= 15 kip•in

MB1= -10 kip•in

MB2=|F2| dA=300*30 = 9,000 lb•in

MC2=|F2| dC=300*25 = 7,500 lb•in

MB2= 9 kip•in

MB= -7.5 kip•in

Example P4-10

A C

B

L

200 mm

160mm

dac

Cos (600) = 200/L

Cos (600) = dac/(L-160)

L = 200/ Cos (600)

L=160+dac/Cos (600)

dac=200-160 Cos (600)=120 mm

A C

B

L

200 mm

160mm

dac

dac=120 mm

MA=|F|dAC=500*0.12 = 60 Nm

MB=|F| dB=500*0.2 = 100 Nm

F

MA= -60 Nm

MB= -100 Nm

Principle of moments:

• The moment M0 of the resultant R of a system of forces with respect to any axis or point is equal to the vector sum of the moment of the individual forces of the system with respect to the same axis or point.

R= F1+ F2+…..+ Fn

M0= Rdr=F1d1+ F2d2+…..+ Fndn

Application of the principle: Varignon’s Theorem

A

y

B

x

R

•From Geometry:

4) R cos =

R Cos

A Cos

A cos

B Cos

+ B cos

MR=MA+ MB

From equations 1, 2, 3, and 4:

h

1) MR= R.d=R(h cos )

d

2) MA= A.a=A(h cos a

3) MB= B.b=B(h cos b

Example P4-18

FFx

Fy

Fx=F cos(320) = 250 cos(320) = 212 N

Fy=F sin(320) = 250 sin(320) = 132.5 N

Solution

Using Varignon’s Theorem

MA= Fx *0.25 + Fy *0.21 = 80.8 Nm

Class Assignment: Exercise set 4-7please submit to TA at the end of the lecture

Two forces are applied to a beam as shown in the figure. Determine the moments of forces F1 and F2 about point A.

F2 = 175 lb

F1 = 250 lb

60o

x

y

3 ft

AB

Answer:MA1 = 650 ft.lb

MA2 = -525 ft.lb

Class Assignment: Exercise set 4-11please submit to TA at the end of the lecture

Determine the moment of the 350 lb force shown in the figure about points A and B.

350 lb

40o

x

y

10 in

A

C Answer:MA=5.47 in.kip

MB=2.25 in.kip

B

12 in

Class Assignment: Exercise set 4-19please submit to TA at the end of the lecture

Determine the moment of the 750-lb force shown in Fig. P4-19 about point O.

Vector Representation of a Moment

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

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

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

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

Fd

r

Finding the Position Vector

r= rAB = rA-rB = (xA i + yA j + zA k ) - (xB i + yB j + zB k )

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

rA= rB+ rAB

Since: x

y

z

F

xA

xB

yA

yB

zAzB

rAB

rB= xB i + yB j + zB k

The location of point BrB

rA= xA i + yA j + zA k

The location of point A

rA

Two Dimensional Case

F= Fx i + Fy j

r= rx i + ry j

x

y

r

F

O

A

j

iM0=r X F = (rx i + ry j) X (Fx i + Fy j) = rx Fx ( i x i ) + rx Fy ( i x j ) + ry Fx ( j x i ) + ry Fy ( j x j ) =

(rx Fy - ry Fx ) k= Mz k

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

i j k rx ry 0Fx Fy 0

Fd

r1

Sin 1= d / r1

r2

Sin 2= d / r2

d = r1Sin 1 = r2Sin 2