Power Engineering 4A Unit 1 Chapter 2 Forces & Moments 2019 PDF

Power Engineering 4A Unit #1 Chapter 2 Forces & Moments Learning Outcome When you complete this chapter you should be able to: Perform calculations involving forces and moments and determine when a system of forces are in equilibrium. Learning Objectives Here is what you should...

Power Engineering 4A Unit #1 Chapter 2 Forces & Moments Learning Outcome When you complete this chapter you should be able to: Perform calculations involving forces and moments and determine when a system of forces are in equilibrium. Learning Objectives Here is what you should be able to do when you complete each objective: Define the moment of a force and its units Determine the direction and calculate the magnitude of the moment of a force. Introduction When a force is applied to an object that is either stationary or in motion then that force will change the objects motion. Forces may also change an objects direction of travel. The term “moment” means the product of a force times the distance from a centre point. This chapter deals with forces that produce moments on beams that are fixed at some point or points. If fixed at only 1 point a beam would be able to rotate around that fixed point based on the force applied. A practical example of understanding moments is suspending a steam pipe with hangers at the proper locations along this pipe. Understanding scaler and vector quantities will assist you in these moments. Objective #1 Define the moment of a force and its units Force Is the push or pull exerted on a body which may make the body move, bring it to rest or change it’s direction. – pulling force of a transport truck causes it to move whereas the pushing force on brake pads on the rotating wheels will cause it to stop Force When several “balanced” forces are acting on a body and no movement take place the body and system of forces are said to be in equilibrium 10N 10N Moment of Force https://ca.video.search.yahoo.com/yhs/search;_ylt=AwrEzeICBW1d1UIAKAY87olQ;_ylu=X3oDM TBncGdyMzQ0BHNlYwNzZWFyY2gEdnRpZAM-;_ylc=X1MDMTM1MTIxNjcwMARfcgMyBGFjdG4D Y2xrBGNzcmNwdmlkA1o5LmNYakV3TGpGSGlSYTRXdXpHd2dBV09UY3VNUUFBQUFEQnIwaFA EZnIDeWhzLXJvdHotMDAxBGZyMgNzYS1ncARncHJpZAM1ZlpvR05aQVMwbVFNelNCOHB5VW 5BBG5fcnNsdAM2MARuX3N1Z2cDMARvcmlnaW4DY2EudmlkZW8uc2VhcmNoLnlhaG9vLmNvb QRwb3MDMARwcXN0cgMEcHFzdHJsAwRxc3RybAMxOQRxdWVyeQNtb21lbnQlMjBvZiUyMGZv cmNlBHRfc3RtcAMxNTY3NDI1OTk5?p=moment+of+force&ei=UTF-8&fr2=p%3As%2Cv%3Av%2 Cm%3Asa&fr=yhs-rotz-001&hsimp=yhs-001&hspart=rotz#id=3&vid=898c5d3af6e2afb02bdaf56f8 17872ac&action=view Moment of Force A force acting any distance from that point will tend to cause a rotation around that point. The moment or turning moment of a force around a point is equal to a force multiplied by the perpendicular distance from the line of force to the point or Moment = Force x perpendicular distance Moment of Force A force of 300 N acts at a perpendicular distance of 2 m from the right of a point. What is the moment ? Moment = F x pd = 300 N x 2m = 600 Nm (Newton meter) clockwise  Moment of Force A force of 600N is acting as shown in the diagram below, What would be the turning moment of the force around Point “A”? 600 N 5.3 m A 5.0 m  Moment of Force 5.3 m 600 N A 5.0 m Moment = F x pd =600 N x 5.0 m = 3000 Nm Since the force is acting in a downward or Clockwise direction the Moment is said to be 3000 Nm Clockwise Moment of Force Find the turning moment in the following diagram? 100 N 5.5 m 700 N 6m Forces and Moments 5.5 100 N m 700 N 6.0 m Since 100 N is acting in the opposite direction to the 700 N the total force is 600 N counter clockwise Therefore Moment = F x pd 600 N x 6 m = 3600 Nm CCW Objective #2 Determine the direction and calculate the magnitude of the moment of force Direction of a Moment of Force A force, acting at a distance from a point, will produce or tend to produce rotation around a point with the point as centre. This rotation is either clockwise or counter clockwise as show above Direction of a Moment of Force If forces A and B are equal it is in equilibrium. If A is greater than B rotation is down If B is greater than A rotation is upward Direction of a Moment of Force Fulcrum A single support about which a bar is free to rotate. Equilibrium – Upward forces = Downward forces – Forces acting to the right = forces acting to the left – Clockwise moments = counter clockwise moments  Direction of a Moment of Force https://ca.video.search.yahoo.com/yhs/search;_ylt=AwrEzeLOBW1dITQAsSA87olQ;_ylu=X3oDMTBncGdyMzQ0BHNlY wNzZWFyY2gEdnRpZAM-;_ylc=X1MDMTM1MTIxNjcwMARfcgMyBGFjdG4DY2xrBGNzcmNwdmlkA3Q1Lnp4REV3TGpG SGlSYTRXdXpHd2dCZU9UY3VNUUFBQUFETjRHM1IEZnIDeWhzLXJvdHotMDAxBGZyMgNzYS1ncARncHJpZANRUlBS MDFrdlJnaWo0QjhBb3RhNGtBBG5fcnNsdAM2MARuX3N1Z2cDMARvcmlnaW4DY2EudmlkZW8uc2VhcmNoLnlhaG9vL mNvbQRwb3MDMARwcXN0cgMEcHFzdHJsAwRxc3RybAMyNwRxdWVyeQNmdWxjcnVtJTIwYW5kJTIwZXF1aWxpYnJ pdW0EdF9zdG1wAzE1Njc0MjYxODI-?p=fulcrum+and+equilibrium&ei=UTF-8&fr2=p%3As%2Cv%3Av%2Cm%3Asa&fr= yhs-rotz-001&hsimp=yhs-001&hspart=rotz#id=3&vid=8645fc16153abbe60853ea7268357247&action=view Direction of a Moment of Force If d1 is ½ of d2 and force B ½ of force A then this situation is equilibrium Direction of a Moment of Force Where d1 = 10m d2 = 20 m A = 60 N B = 30 N Prove this is in Equilibrium Direction of a Moment of Force Clockwise moments = Counter clockwise moment F x pd = F x pd 60N x 10m = 30N x 20m 600Nm = 600 Nm Therefore system is in equilibrium Direction of a Moment of Force Find the direction of rotation in the following D1 = 11.75m and A is 22 N D2 = 26m and B is 14 N Direction of a Moment of Force CW moment = F x pd = 22N x 11.75m = 258.5 Nm CCW moment = F x pd = 14N x 26m = 364 Nm Therefore movement will be in the CCW (upward) direction Direction of a Moment of Force A bar of negligible mass is supported at A and loaded as shown below. What would be the clockwise and counter clockwise moments and which direction will the bar rotate 100 N 40 N A 1m 2m Direction of a Moment of Force 100 N 40N A 1m 2m CCW moment = F x pd = 100N x 1M = 100Nm CW moment =F x pd = 40N x 2 m = 80NM Therefore the bar will move in the CCW direction Direction of a Moment of Force A bar of negligible mass is supported on a fulcrum as shown. What force is required at the right to maintain equilibrium? ? 50 N A 1m 2m Direction of a Moment of Force 50 N ? A 1m 2m CW moments = CCW moments F x 2m = 50 N x 1m F = 50N /2 F = 25 N Direction of a Moment of Force A bar of negligible mass is supported on a fulcrum as shown. How many kg are required at the left to maintain equilibrium? 27 kg ? kg A 21 m 14 m  Direction of a Moment of Force 27 kg ? kg A 21 m 14 m 27 kg to a force F =ma = 27 x 9.81 = 264.87N CW = CCW 264.87N x 14m = F x 21m F = 3708.18Nm 21m F = 176.58 N F=mxa m = F / 9.81 m = 176.58 / 9.81 m = 18 kg Direction of a Moment of Force A bar of negligible mass is supported on a fulcrum as shown. What distance is required on the right to maintain equilibrium? 71 N 33N A 19 m ?m Direction of a Moment of Force 71 N 33N A 19 m ?m CW =CCW F x pd = F x pd 71 N x pd = 33 N x 19 m pd = 33 N x 19m 71 N pd = 8.83 m Beams A beam may be defined as a rigid member or bar, supported in some way so that it is capable of carrying a load or system of loads. A simply supported beam where it is free to bend without restriction Beams Point Loads – If loads on a beam can be considered to be concentrated at various points, they are called Point Loads Point Load Point Load Beam Support Support Beams ( Reaction at Supports ) For equilibrium of a beam , the external forces on the beam must be resisted by forces at the supports Point Load Point Load Beam Support or R 1 Support or R2 Point Loads Support Support A 6m beam is supported at each end and carries point loads as shown above. Calculate the reaction at each support to achieve equilibrium Point Loads Moments may be taken from either support (R1) or support (R2) Moments from R2 = F x pd = 50 N x 2 m = 100 Nm = F x pd = 100 N x 5 m = 500 Nm Total moments at R2 =600 Nm Moments for R1 = R1 x 6m CW = CCW R1 x 6m = 600 Nm R1 = 600 Nm 6m R1 =100 N Point Loads Upward force = Downward force R1 + R2 = 100N + 50N 100 + R2 = 150N R2 = 150N – 100N R 2 = 50 N Therefore the force acting at R1 =100N and the force acting at R2 =50N Point Loads To Prove take moments from opposite side or R1 CW =CWW (100 N x 1 m) + (50 N x 4 m) = 6 m R2 N 100 Nm + 200 Nm = 6 m R2 N R2 N = 300 Nm 6m R2 = 50 N Knowledge Questions Chapter #2 Knowledge Questions Chapter #2 Knowledge Questions Chapter #2 Knowledge Questions Chapter #2