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Lecture 10 ENGR-1100 Introduction to Engineering Analysis.

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Presentation on theme: "Lecture 10 ENGR-1100 Introduction to Engineering Analysis."— Presentation transcript:

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2 Lecture 10 ENGR-1100 Introduction to Engineering Analysis

3 Previous Lecture Outline Vector representation of a moment: - Moment of a force about a point. -Moment of a force about a line.

4 Lecture outline Couples. Resolution of a force into a force and a couple.

5 A Couple A system of forces whose resultant force is zero but the resultant moment about a point is not zero. F F

6 A Couple A system of forces whose resultant force is zero but the resultant moment about a point is not zero. y x z F1F1  O F2F2  d M A =|F 2 |dM B =|F 1 |d If |F 1 | = | F 2 | : M A = M B =Fd

7 The sum of the moment of two forces about any point O is: M 0 = r 1 X F 1 + r 2 X F 2 Since F 2 equals -F 1 : M 0 = r 1 X F 1 + r 2 X (-F 1 ) x z F1F1  O F2F2  d r AB r2r2 r1r1 Therefore: M 0 = r AB X F 1 = F 1 d e n =(r 1 –r 2 )X F 1 = r AB X F 1

8 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. x z F1F1  O F2F2  d y

9 Couple Transformation Translation to a parallel position Rotation of a couple

10 Changing the magnitude and distance provided the product F. d remains constant.

11 C=  C x +  C y +  C z  C x i +  C y j +  C z k The Resultant Couple The magnitude of the couple: |C|=  C x 2 +  C y 2 +  C z 2 The direction:  x  cos -1 (C x /C);  y  cos -1 (C y /C);  z  cos -1 (C z /C); The couple can also be written as: C=C eC=C e e=cos (  x ) i+ cos (  y ) j +cos (  z ) k Where:

12 Example P4-82 Determine the moment of the couple shown in Fig. P4-82 and the perpendicular distance between the two forces 760 N      mm  mm

13 Solution 760 N      mm  mm F A = -760 cos(   ) sin(   ) =-622 i – j N r BA = -0.1 i j m MB=MB= i j k = 168 k Nm |M B |= M x 2 + M y 2 + M z 2 = 168 Nm d=M/F=168/760=0.22 m

14 Class Assignment: Exercise set 4-81 please submit to TA at the end of the lecture Determine the moment of the couple shown in Fig.P4-81 and the perpendicular distance between the two forces. Solution: M A = 3030 k in lb d=8.66 in

15 Two parallel forces of opposite sense, F 1 = (-70i - 120j - 80k) lb and F 2 = (70i + 120j + 80k) lb, act at points B and A of a body as shown in Fig. P4-83. Determine the moment couple and the perpendicular distance between the two forces. Class Assignment: Exercise set 4-83 please submit to TA at the end of the lecture Solution: M A = 320 i -920 j k ft lb d=9.17 ft

16 Resolution of a force into force and a couple  d  F F -F   M0M0 F

17 Example P4-104 Replace 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

18 Solution F C = 350 cos(   ) sin(   ) = i – 225 j N C=C= i j k = k Nm C C CBCB FCFC FCFC -F C r BC = 0.1 i j m r BC

19 Class Assignment: Exercise set please submit to TA at the end of the lecture Replace 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: C A = k Nm


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