Chapter -4- Force System Resultant … · To present methods for determining the resultants of non-concurrent force systems. To ... In 2D problems, moment of the force is found about

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Ishik University / Sulaimani

Civil Engineering Department

Chapter -4-

Force System Resultant 1

Ishik

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Ishik

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CHAPTER OBJECTIVES

To discuss the concept of the moment of a force and show

how to calculate it in two and three dimensions.

To provide a method for finding the moment of a force about

a specified axis.

To define the moment of a couple.

To present methods for determining the resultants of non-

concurrent force systems.

To indicate how to reduce a simple distributed loading to a

resultant force having a specified location.

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CHAPTER OUTLINE

Moment of a Force – Scalar Formation

Principle of Moments

Moment of a Force about a Specified Axis

Moment of a Couple

Equivalent System

Resultants of a Force and Couple System

Reduction of a Simple Distributed Loading

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4.1 MOMENT OF A FORCE – SCALAR FORMATION

Moment of a force about a point or axis – a measure of the

tendency of the force to cause a body to rotate about the

point or axis.

Case 1

Consider horizontal force Fx,

which acts perpendicular to

the handle of the wrench and is

located dy from the point O.

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Fx tends to turn the pipe about the z axis.

The larger the force or the distance dy, the greater the

turning effect.

Torque – tendency of

rotation caused by Fx

or simple moment (Mo) z

Moment axis (z) is

perpendicular to shaded

plane (x-y)

Fx and dy lies on the

shaded plane (x-y)

Moment axis (z) intersects

the plane at point O

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Case 2

Apply force Fz to the wrench.

Pipe does not rotate about z axis.

Tendency to rotate about x axis.

The pipe may not actually rotate

Fz creates tendency for rotation

so moment (Mo) x is produced.

Moment axis (x) is perpendicular to shaded plane (y-z).

Fz and dy lies on the shaded plane (y-z).

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Case 3

Apply force Fy to the wrench.

No moment is produced about point O.

Lack of tendency to rotate as line of action passes through O.

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In General

Consider the force F and the point O which lies in the shaded

plane.

The moment MO about point O, or about an axis passing

through O and perpendicular to the plane, is a vector quantity.

Moment MO has its specified magnitude and direction.

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Magnitude

For magnitude of MO, MO = Fd where

d = moment arm or perpendicular distance from the axis at

point O to its line of action of the force.

Units for moment is N.m.

Direction

Direction of MO is specified by using

“right hand rule”.

- fingers of the right hand are curled

to follow the sense of rotation when

force rotates about point O.

- Thumb points along the moment axis

to give the direction and sense of the moment

vector.

- Moment vector is upwards and perpendicular

to the shaded plane.

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Direction

MO is shown by a vector arrow with a curl to

distinguish it from force vector.

Example (Fig b)

MO is represented by the

counterclockwise curl, which

indicates the action of F.

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Direction

Arrowhead shows the sense of

rotation caused by F Using the

right hand rule, the direction and

sense of the moment vector points

out of the page.

In 2D problems, moment of the force

is found about a point O.

Moment acts about an axis perpendicular to the plane

containing F and d .

Moment axis intersects the plane at point O.

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Resultant Moment of a System of Coplanar Forces

Resultant moment, MRo = addition of the moments of all the

forces algebraically since all moment forces are collinear.

MRo = ∑Fd

taking clockwise to be negative.

o A clockwise curl is written along the equation to

indicate that a positive moment if directed along the +

z axis and negative along the – z axis

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Moment of a force does not always cause rotation.

Force F tends to rotate the beam clockwise about A

with moment.

MA = FdA

Force F tends to rotate the beam counterclockwise

about B with moment.

MB = FdB

Hence support at A prevents the rotation.

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Example 4.1

For each case, determine the moment of the force about point O.

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Solution;

Line of action is extended as a dashed line to establish moment

arm d.

Tendency to rotate is indicated and the orbit is shown as a

colored curl.

)(.5.37)75.0)(50()(

)(.200)2)(100()(

CWmNmNMb

CWmNmNMa

o

o

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Solution;

)(.0.21)14)(7()(

)(.4.42)45sin1)(60()(

)(.229)30cos24)(40()(

CCWmkNmmkNMe

CCWmNmNMd

CWmNmmNMc

o

o

o

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Example 4.2

Determine the moments of the 800N force acting on the frame

about points A, B, C and D.

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Solution;

Scalar Analysis

Line of action of F passes through C

)CCW(m.N400)m5.0)(N800(M

m.kN0)m0)(N800(M

)CW(m.N1200)m5.1)(N800(M

)CW(m.N2000)m5.2)(N800(M

D

C

B

A

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Example 4.3

Determine the resultant moment of the four forces

acting on the rod shown in figure about point O. Ish

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4.2 PRINCIPLES OF MOMENTS

The guy cable exerts a force F on the

pole and creates a moment about the

base at A,

MA = Fd

If the force is replaced by Fx and Fy

at point B where the cable acts on

the pole, the sum of moment about

point A yields the same resultant

moment. 25

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Fy create zero moment about A

MA = Fxh

Apply principle of

transmissibility and slide the

force where line of action

intersects the ground at C, Fx

create zero moment about A

MA = Fyb

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Example 4.3

The force F acts at the end of the angle bracket. Determine the

moment of the force about point O.

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Solution

Method 1:

MO = 400sin30°N(0.2m)-400cos30°N(0.4m)

= -98.6N.m

= 98.6N.m (CCW)

As a Cartesian vector,

MO = {-98.6k}N.m

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Solution

Method 2:

Express as Cartesian vector

r = {0.4i – 0.2j}N

F = {400sin30°i – 400cos30°j}N

= {200.0i – 346.4j}N

For moment,

mNk

kji

FXrMO

.6.98

04.3460.200

02.04.0

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Example 4.4

The member is subjected to a force of F=6 kN. If θ= 45º,

determine the moment produced by F about point A.

Ans;

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Example 4.5

The two boys push on the gate with forces of FA=30 Ib and FB=50

Ib as shown. Determine the moment of each force about C. Which

way will the gate rotate, clockwise or counterclockwise? Neglect the

thickness of the gate.

Ans;

The gate will

rotate C.C.W.