Engineering Mechanics: Statics in SI Units, 12eeportfolio.lib.ksu.edu.tw/user/T/H/T093000099... · 2011-05-19 · Force Summation. i j k N m i j k j k k X k M M M M r XF r XF i j

Force System Resultants4

Engineering Mechanics: Statics in SI Units, 12e

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Chapter Objectives

• Concept of moment of a force in two and three dimensions• Method for finding the moment of a force about a specified

axis.• Define the moment of a couple.• Determine the resultants of non-concurrent force systems• 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• Cross Product• Moment of Force – Vector Formulation• Principle of Moments• Moment of a Force about a Specified Axis• Moment of a Couple• Simplification of a Force and Couple System• Further Simplification of a Force and Couple System• Reduction of a Simple Distributed Loading

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4.7 Simplification of a Force and Couple System

• An equivalent system is when the external effects are the same as those caused by the original force and couple moment system

• External effects of a system is the translating and rotating motion of the body

• Or refers to the reactive forces at the supports if the body is held fixed

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4.7 Simplification of a Force and Couple System

• Equivalent resultant force acting at point O and a resultant couple moment is expressed as

• If force system lies in the x–y plane and couple moments are perpendicular to this plane,

MMM

FF

OOR

R

MMM

FF

FF

OOR

yyR

xxR

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4.7 Simplification of a Force and Couple System

Procedure for Analysis• Establish the coordinate axes with the origin located at

point O and the axes having a selected orientation• Force Summation• Moment Summation

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Example 4.16

A structural member is subjected to a couple moment M and forces F1 and F2. Replace this system with an equivalent resultant force and couple moment acting at its base, point O.

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Solution

Express the forces and couple moments as Cartesian vectors.

mNkjkjM

Njiji

rrNuNF

NkF

CB

CBCB

.}300400{53500

54500

}4.1666.249{)1.0()15.0(

1.015.0300

)300()300(

}800{

22

2

1

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Solution

Force Summation.

mNkji

kjikXkkj

FXrFXrMMMM

Nkji

jikFFF

FF

BCOCRo

R

R

.}300650166{

04.1666.24911.015.0)800()1()300400(

}8004.1666.249{

4.1666.249800

;

21

21

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4.8 Further Simplification of a Force and Couple System

Concurrent Force System• A concurrent force system is where lines of action of all the

forces intersect at a common point O

FFR

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4.8 Further Simplification of a Force and Couple System

Coplanar Force System• Lines of action of all the forces lie in the same plane • Resultant force of this system also lies in this plane

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4.8 Further Simplification of a Force and Couple System

Parallel Force System• Consists of forces that are all parallel to the z axis• Resultant force at point O must also be parallel to this axis

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4.8 Further Simplification of a Force and Couple System

Reduction to a Wrench• 3-D force and couple moment system have an equivalent

resultant force acting at point O • Resultant couple moment not perpendicular to one another

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Example 4.18

The jib crane is subjected to three coplanar forces. Replace this loading by an equivalent resultant force and specify where the resultant’s line of action intersects the column AB and boom BC.

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Solution

Force Summation

NkN

kNNF

FFkNkN

kNkNF

FF

Ry

yRy

Rx

xRx

60.260.2

6.0545.2

;25.325.3

75.1535.2

;

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Solution

For magnitude of resultant force,

For direction of resultant force,

16.4

)60.2()25.3()()( 2222

RyRxR

kN

FFF

7.38

25.360.2tantan 11

÷

÷÷

Rx

RyFF

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Solution

Moment Summation Summation of moments about point A,

my

mkNmkN

mkNmknkNykN

MM ARA

458.0

)6.1(5450.2)2.2(

5350.2

)6.0(6.0)1(75.1)0(60.2)(25.3

;

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Solution

Moment Summation Principle of Transmissibility

mx

mkNmkN

mkNmknxkNmkN

MM ARA

177.2

)6.1(5450.2)2.2(

5350.2

)6.0(6.0)1(75.1)(60.2)2.2(25.3

;

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