Force System Resultants 4 Engineering Mechanics: Statics in SI Units, 12e 2011/3/23 1
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Force System Resultants4
Engineering Mechanics: Statics in SI Units, 12e
2011/3/23 1
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
2011/3/23 2
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
2011/3/23 3
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
2011/3/23 52
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
2011/3/23 53
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
2011/3/23 54
2011/3/23 55
2011/3/23 56
2011/3/23 57
2011/3/23 58
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.
2011/3/23 59
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
2011/3/23 60
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
2011/3/23 61
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
2011/3/23 62
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
2011/3/23 63
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
2011/3/23 64
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
2011/3/23 65
2011/3/23 66
2011/3/23 67
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.
2011/3/23 68
Solution
Force Summation
NkN
kNNF
FFkNkN
kNkNF
FF
Ry
yRy
Rx
xRx
60.260.2
6.0545.2
;25.325.3
75.1535.2
;
2011/3/23 69
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
2011/3/23 70
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
;
2011/3/23 71
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
;
2011/3/23 72
2011/3/23 73
2011/3/23 74
2011/3/23 75
2011/3/23 76
Related Documents
