Presentation on theme: "ENGR-1100 Introduction to Engineering Analysis"— Presentation transcript:
1 ENGR-1100 Introduction to Engineering Analysis Lecture 10
2 Previous Lecture Outline Vector representation of a moment:- Moment of a force about a point.-Moment of a force about a line.
3 Lecture outlineCouples.Resolution of a force into a force and a couple.
4 A CoupleA system of forces whose resultant force is zero but the resultant moment about a point is not zero.FF
5 A CoupleA system of forces whose resultant force is zero but the resultant moment about a point is not zero.MA=|F2|dMB=|F1|dxzF1AOF2BdIf |F1 | = | F2| :MA= MB=Fdy
6 The sum of the moment of two forces about any point O is: M0= r1 X F1+ r2 X F2xzF1AOF2BdrABr2r1Since F2 equals -F1:M0= r1 X F1+ r2 X (-F1)=(r1–r2)X F1= rABX F1Therefore:M0= rABX F1 = F1d en
7 The Characteristic of a Couple 1) The magnitude of the moment of the couple.2) The sense (direction of rotation) of the couple.3) The orientation of the moment of the couple.zdF1ABF2Oyx
8 Couple Transformation Translation to a parallel positionRotation of a couple
9 Changing the magnitude and distance provided the product F Changing the magnitude and distance provided the product F.d remains constant.
10 The Resultant Couple C= SCx + SCy+ SCz =SCx i + SCy j + SCz k The magnitude of the couple:|C|= SCx2 +SCy2 +SCz2The couple can also be written as:C=C eWhere:e=cos (qx) i+ cos (qy) j +cos (qz) kThe direction:qx=cos-1(Cx/C); qy=cos-1(Cy/C); qz=cos-1(Cz/C);
11 Example P4-82Determine the moment of the couple shown in Fig. P4-82 and the perpendicular distance between the two forces760 NA760 N200 mmB350100 mm
12 Solution760 N350BA200 mm100 mmFA = -760 cos(350) sin(350) =-622 i – j NrBA = -0.1 i j mMB=i j k= 168 k Nm|MB|= Mx2 + My2 + Mz2 = 168 Nmd=M/F=168/760=0.22 m
13 Class Assignment: Exercise set 4-81 please submit to TA at the end of the lectureDetermine the moment of the couple shown in Fig.P4-81and the perpendicular distance between the two forces.Solution:MA= 3030 k in lbd=8.66 in
14 Class Assignment: Exercise set 4-83 please submit to TA at the end of the lectureTwo parallel forces of opposite sense, F1 = (-70i - 120j - 80k) lband F2 = (70i + 120j + 80k) lb, act at points B and A of a bodyas shown in Fig. P Determine the moment couple and the perpendicular distance between the two forces.Solution:MA= 320 i -920 j k ft lbd=9.17 ft
15 Resolution of a force into force and a couple dOFOF-FM0F
16 Example P4-104Replace the 350 N force shown in Fig. P4-104 by a force at point B and a couple. Express your answer in Cartesian form.C
17 Solution rBC = 0.1 i + 0.25 j m rBC FC -FC FC = 350 cos(400) sin(400)= i – 225 j NC=i j k= k NmCCBFC
18 Class Assignment: Exercise set 4-102 please submit to TA at the end of the lectureReplace the 600-N force shown in Fig. P-102 by a force at point A and a couple. Express your answer in Cartesian vector form.Solution:CA= k Nm
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