Section 4.6 Moment of a Couple

Engineering Mechanics: Statics Fifteenth Edition Chapter 4 Force System Resultants Section 4.6 Moment of a Couple Copyright © 2022 Pearson Education, Inc. All Rights Reserved Moment of a Couple Today’s Objectives: a. define a couple, and, b. determine the moment of a couple. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Applications (1 of 1) When you grip a vehicle’s steering wheel with both hands and turn, a couple moment is applied to the wheel. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Moment of a Couple (1 of 2) A couple is defined as two parallel forces with the same magnitude but opposite in direction separated by a perpendicular distance “d.” The moment of a couple is defined as MO = F d (using a scalar analysis) or as MO = r  F (using a vector analysis). Here r is any position vector from the line of action of F to the line of action of F. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Moment of a Couple (2 of 2) The net external effect of a couple is that the net force equals zero and the magnitude of the net moment equals F * d. Since the moment of a couple depends only on the distance between the forces, the moment of a couple is a free vector. It can be moved anywhere on the body and have the same external effect on the body. Moments due to couples can be added together using the same rules as adding any vectors. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Example Ⅰ: Scalar Approach (1 of 2) Given: Two couples act on the beam with the geometry shown. Find: The magnitude of F so that the resultant couple moment is 1.5 kN  m clockwise. Mz = u • (r  F ). Plan: 1. Add the two couples to find the resultant couple. 2. Equate the net moment to 1.5 kN  m clockwise to find F. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Example Ⅰ: Scalar Approach (2 of 2) Solution: The net moment is equal to: Solving for the unknown force F, we get F = 2.33 kN Copyright © 2022 Pearson Education, Inc. All Rights Reserved Example Ⅱ: Vector Approach (1 of 2) Given: A 450 N force couple acting on the pipe assembly. Find: The couple moment in Cartesian vector notation. Plan: 1. Use M = r × F to find the couple moment. 2. Set r = rAB and F = FB. 3. Calculate the cross product to find M. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Example Ⅱ: Vector Approach (2 of 2) Solution: rAB = { 0.4 i } m 4 3 FB = {0 i + 450   j − 450   k } N 5 5 = {0 i + 360 j − 270 k } N M = rAB  FB = [{0 ( −270 ) – 0 ( 360 )}i – 4 ( −270 ) – 0 ( 0 ) j + 0.4 ( 360 ) – 0 ( 0 ) k ] N·m = {0 i +108 j + 144 k } N·m Copyright © 2022 Pearson Education, Inc. All Rights Reserved Group Problem Solving I (1 of 3) Given: Two couples act on the beam with the geometry shown and d = 4 f t. ee Find: The resultant couple Plan: 1. Resolve the forces in x and y-directions so they can be treated as couples. 2. Add these two couples to find the resultant couple. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Group Problem Solving I (2 of 3) The x and y components of the upper-left 50 lb force are: 50 lb (cos 30°) = 43.30 lb vertically up 50 lb (sin 30°) = 25 lb to the right Do both of these components form couples with their matching components of the other 50 force? No! Only the 43.30 lb components create a couple. Why? Copyright © 2022 Pearson Education, Inc. All Rights Reserved Group Problem Solving (3 of 3) Now resolve the lower 80 lb force: (80 lb)  3  , acting up   5 (80 lb)  4  , acting to the right 5   Do both of these components create a couple with components of the other 80 lb force? The net moment is equal to: Copyright © 2022 Pearson Education, Inc. All Rights Reserved Group Problem Solving Ⅱ (1 of 2) Given: F = {80 k} N and – F = {– 80 k} N Find: The couple moment acting on the pipe assembly using Cartesian vector notation. Plan: 1. Use M = r × F to find the couple moment. 2. Set r = rAB and F = {80 k} N. 3. Calculate the cross product to find M. Copyright © 2022 Pearson Education, Inc. All Rights Reserved Group Problem Solving Ⅱ (2 of 2) rAB = { (0.3 − 0.2) i + (0.8 − 0.3) j + (0 − 0) k } m = { 0.1 i + 0.5 j } m F = {80 k } Ν = {(40 − 0) i − (8 − 0) j + (0) k } N  m = { 40 i − 8 k } Ν  m Copyright © 2022 Pearson Education, Inc. All Rights Reserved Reading Quiz (2 of 2) 2. The moment of a couple is called a __________ vector. A. Free B. Spinning C. Fixed D. Sliding Copyright © 2022 Pearson Education, Inc. All Rights Reserved