ENGR-1100 Introduction to Engineering Analysis

ENGR-1100 Introduction to Engineering Analysis Lecture 10

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

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

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

A system of forces whose resultant force is zero but the resultant moment about a point is not zero. z d F1 A B F2 O x A Couple MA=|F2|d MB=|F1|d If |F1|= | F2| : MA= MB=Fd y

z Since F2 equals -F1: d F1 M0= r1 X F1+ r2 X (-F1) A rAB r1 F2 B r2 O Therefore: M0= rABX F1 = F1d en x The sum of the moment of two forces about any point O is: M0= r1 X F1+ r2 X F2 =(r1–r2)X F1= rABX F1

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. The Characteristic of a Couple z d F1 A F2 B O y x

Translation to a parallel position Rotation of a couple Couple Transformation

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

The Resultant Couple C= SCx+ SCy+ SCz=SCx i + SCy j + SCz k The magnitude of the couple: |C|= SCx2 +SCy2 +SCz2 The couple can also be written as: C=Ce Where: e=cos (qx) i+ cos (qy) j +cos (qz) k The direction: qx=cos-1(Cx/C); qy=cos-1(Cy/C); qz=cos-1(Cz/C);

Example P4-82 Determine the moment of the couple shown in Fig. P4-82 and the perpendicular distance between the two forces 760 N A 760 N 200 mm B 350 100 mm

|MB|= Mx2 + My2 + Mz2 = 168 Nm 760 N A 760 N 200 mm B i j k -0.10.2 0 -622 -435.90 350 100 mm MB= = 168 k Nm d=M/F=168/760=0.22 m Solution FA = -760 cos(350) - 760 sin(350) =-622 i – 435.9 jN rBA = -0.1 i + 0.2 jm

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: MA= 3030 k in lb d=8.66 in

Class Assignment: Exercise set 4-83 please submit to TA at the end of the lecture Two parallel forces of opposite sense, F1 = (-70i - 120j - 80k) lb and F2 = (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. Solution: MA= 320 i -920 j +1100 k ft lb d=9.17 ft

-F O M0  O  F d F F Resolution of a force into force and a couple

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

-FC rBC rBC = 0.1 i + 0.25 jm FC C i j k 0.10.25 0 268.1 -2250 CB C= = -89.5 k Nm FC Solution C FC = 350 cos(400) - 350 sin(400) = 268.1 i – 225 jN

Class Assignment: Exercise set 4-102 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: CA= -193.9 k Nm

ENGR-1100 Introduction to Engineering Analysis