L_-_Equivalent_System

L_-_Equivalent_System - FORCE SYSTEM RESULTANTS Equivalent...

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Unformatted text preview: FORCE SYSTEM RESULTANTS Equivalent System LEARNING OBJECTIVES Be able to find an equivalent force-couple system for a system of forces and couples PRE-REQUISITE KNOWLEDGE Units of measurement Trigonometry concepts Vector concepts Rectangular component concepts Couple moment concept Cross product and dot product EQUIVALENT SYSTEM An equivalent force and moment system replaces a complex system of forces and moments acting on a rigid body with a simplified one having the same external effects. Several topics are addressed including: 1. Replacement of force systems with no moment 2. Replacement of force systems with moments 3. Replacement of force systems with or without moments with a single resultant force 4. Replacement of force systems with or without moments with a wrench EXAMPLE 1 Replace the given forces with an equivalent system SOLUTION 1 Y RY F X RX F 11.6 kN-m Equivalent Force System SOLUTION 1 Y RY F 34.8 kN-m RX X F Equivalent Force System EXAMPLE 1 A FORCE SYSTEM PASSING THROUGH A REFERENCE POINT Sliding a force along a given action line does not affect its external effects on a rigid body (Principle of Transmissibility) = F SYSTEM WITH FORCES NOT PASSING THROUGH A REFERENCE POINT Force acting at point A can be represented with the same force acting at point O and a couple moment = = = = = EXAMPLE 1 Replace the force system acting on the beam by an equivalent force and couple moment acting at point A and then at point B SOLUTION 1 Y Vectors of forces : The 2.5 kN force F2.5 = {-2.0i - 1.5 j} kN X The 3.0 kN force F3.0 ={-3.0 j} kN The 1.5 kN force F1.5 = {1.5 sin 30i - 1.5 cos 30 j} kN The resultant force is the same at points A or B F = F2.5 +F1.5 +F3.0 SOLUTION 1- Contd. Y 5.8j kN 34.8k kN-m Recall F2.5 = {-2.0i - 1.5 j} kN F3.0 ={-3.0 j} kN F1.5 = {1.5 sin 30 i - 1.5 cos 30 j} kN 1.25i kN X MA = r F F = {( - 2.0 + 1.5 sin 30) i + ( - 1.5 - 1.5 cos 30 - 3.0 ) j} kN F = {-1.25i + -5.80 j} kN = ( 2i ) ( -2.0i-1.5 j) + ( 6i ) (1.5 sin 30i - 1.5 cos 30 j) + ( 8i ) ( - 3.0 j) = -3k - 9 cos 30k - 24k = -34.8 k kN - m or 34.8 k kN - m (clockwise) SOLUTION 1- Contd. 5.8j kN 34.8k kN-m 5.93 1.25i kN X kN Y F = {-1.25i + -5.80 j} kN The magnitude and inclination angle of the resultant force F are F= ( - 1.25) -1 2 + ( - 5.80 ) = 5.93 kN 2 - 5.80 and = tan = 77.8 ( in third quadrant ) - 1.25 SOLUTION 1- Contd. y 5.93 kN x 11.6k kN-m Couple moment at point B MB = r F = ( - 6i ) ( - 2.0i - 1.5 j) + ( - 2i ) (1.5 sin 30i - 1.5 cos 30 j) + ( 0i ) ( - 3.0 j) = 9k + 3 cos 30k - 0 = 11.6 k kN - m or 11.6 k kN - m (counter clockwise) SYSTEM WITH FORCES AND COUPLE MOMENTS = = Forces and couple moments can be represented with a resultant force and a resultant moment, The resultant force F = F R The resultant moment M RO = M C + M O Where MC is the summation of all couple moments and MO is the summation of all moments due to forces about the reference point EXAMPLE 2 Determine the resultant force and the resultant moment at point C SOLUTION 2 The resultant force F R = -300 j - 200 j - 400 j - 200i = {-200i - 900 j} lb Y X Summation of moments due to all forces about point C M O = ( - 7i + 9 j) ( - 300 j) + ( - 4i + 9 j) ( - 200 j) + ( 9 j) ( - 400 j) + ( 7 j) ( - 200i ) M O = 2100k + 800k + 0 + 1400k = (4300k ) lb - ft Summation of the couple moments The resultant moment about point C M C = (600k) lb - ft M RO = MO + MC = 4300k + 600k = (4900k ) lb - ft Or M R O 4900 lb - ft (counterclockwise) SIMPLIFICATION TO A SINGLE RESULTANT FORCE Systems that can be simplified to a single force : 1. Concurrent force systems, All forces passing through the same point 2. Coplanar force systems, All two-dimensional problems 3. Parallel force systems, All parallel forces but passing through different point CONCURRENT FORCE SYSTEMS All forces passing through the same point FR = F F2 F3 F1 P = P COPLANAR FORCE SYSTEMS All two-dimensional problems = = EXAMPLE 3 Replace the loading on the frame by a single resultant force and specify where its line of action intersects member AB measured from point A SOLUTION 3 The resultant force F Y X R = -300 j - 200 j - 400 j - 200i = {-200i - 900 j} lb Summation of moments due to all forces and couple moments about point A M + ( 7i - 2 j) ( - 200i ) + 600k = 0 - 600k - 2800k - 400k + 600k = -3200k lb - ft A = ( 0 ) ( - 300 j) + ( 3i ) ( - 200 j) + ( 7i ) ( 400 j) SOLUTION 3 Distance, d from point A to the line of action of the resultant force can be determined using cross product as follows X Y - 3200k = ( di ) X( - 200i - 900 j) - 3200k = -900dk 3200 d= = 3.56 ft 900 M A = ( di ) X ( FR ) All parallel forces but passing through different point PARALLEL FORCE SYSTEMS = = EXAMPLE 4 If F1=20 kN and F2=50 kN, replace the system of forces by a single resultant force and specify its location (x,y) SOLUTION 4 The resultant force =20 kN F R = -20k - 50k - 20k - 50k = {-140k} kN =50 kN Summation of moments due to all forces about the origin M + (11j) ( - 20k ) + (10i + 13 j) ( - 50k ) = 200 j + 200 j - 150i - 220i + 500 j - 650i = {-1020i + 900 j} kN - m O = (10i ) ( - 20k ) + ( 4i + 3 j) ( - 50k ) SOLUTION 4 Contd. =20 kN =50 kN Location (x,y) from origin to the line of action of the resultant force - 1020i + 900 j = -140 yi + 140 xj 900 - 1020 x= = 6.43m and y = = 7.29m 140 - 140 M O = -1020i + 900 j = ( xi + yj) ( FR ) = ( xi + yj) ( - 140k ) QUIZ 5 If F1=30 kN and F2=60 kN, replace the system of forces by a single resultant force and specify its location (x,y) QUIZ 5 A) X = 5.25 m and Y = 7.88 m B) X = 7.88 m and Y = 6.25 m C) X = 7.88 m and Y = 5.25 m D) X = 6.25 m and Y = 7.88 m QUIZ 5 The resultant force =30 kN F R = -20k - 50k - 30k - 60k = {-160k} kN =60 kN Summation of moments due to all forces about the origin M + (11j) ( - 30k ) + (10i + 13 j) ( - 60k ) = 200 j + 200 j - 150i - 330i + 600 j - 780i = {-1260i + 1000 j} kN - m O = (10i ) ( - 20k ) + ( 4i + 3 j) ( - 50k ) QUIZ 5 Contd. =30 kN =60 kN Location (x,y) from origin to the line of action of the resultant force - 1260i + 1000 j = -160 yi + 160 xj 1000 - 1260 x= = 6.25m and y = = 7.88m 160 - 160 M O = -1260i + 1000 j = ( xi + yj) ( FR ) = ( xi + yj) ( - 160k ) SIMPLIFICATION TO A WRENCH Most 3-D systems cannot be simplified to a single force, but they can be simplified to a wrench. A wrench or a screw is a couple moment collinear with the resultant force. EXAMPLE 5 Replace the three forces by a wrench and also specify the location (x,y) where its line of action intersects the plate SOLUTION 5 Step 1 Determine the resultant force F 500i + 300 j + 800k 500 2 + 300 2 + 800 2 R = {500i + 300 j + 800k} N Step 2 - Determine the directional cosine of FR u FR = = 0.5051i + 0.3030 j + 0.8081k Since FR and the resultant moment are parallel, they have the same directional cosines SOLUTION 5 - continued Step 3 Calculate the moments at point P in the x, y and z directions by multiplying each force by its moment arm 1. Around the x axis through point P, the 800k N force has a moment arm of (4 - y), the other two forces create no moment in that direction. M Px = 800(4 - y ) SOLUTION 5 - continued Step 3 Calculate the moments at point P in the x, y and z directions by multiplying each force by its moment arm 2. Around the y axis through point P, the 800k N force has a moment arm of x, the other two forces create no moment in that direction. M Py = 800 x SOLUTION 5 - continued Step 3 Calculate the moments at point P in the x, y and z directions by multiplying each force by its moment arm 3. Around the z axis, the 500k N force has a moment arm of y and the 300j N force has a moment arm of (6 x), the 800k N force creates no moment in that direction. M Pz = 500 y + 300(6 - x) SOLUTION 5 - continued Step 4 - Let the magnitude of the resultant moment be MRP, then its MPX, MPY, and MPZ components can be calculated using the directional cosines as follows: MPX = 0.5051 MRP= 800(4 - y) MPY = 0.3030 MRP= 800x MPZ = 0.8081 MRP= 500y + 300(6 - x) SOLUTION 5 - continued Step 5 Solve the 3 simultaneous equations MRP = 3070.9 N-m X = 1.16 m; and Y = 2.06 m ALTERNATIVE SOLUTION 5 As before, the resultant force FR M O = ( 0) ( 500i ) + ( 4 j) ( 800k ) + ( 6i + 4J ) ( 300 j) = { 3200i + 1800k } N-m FR = {500i + 300 j + 800k} N The resultant moment MO due to all forces about the origin M O = ( 0) ( 500i ) + ( 4j) ( 800k ) + ( 6i + 4j) ( 300j) M O = { 3200i + 1800k } N - m ALTERNATIVE SOLUTION 5 - continued Recall that F1F2=(F1)(F2)(cos ), where is the angle between F1 and F2. If F2 is a unit vector, then F2 is equal to 1.0 and F1F2=(F1)(cos ), which is the component of F1 along F2. Based on the above, the component of MO along FR (labeled M||) can be given as: FR M = M O F R FR and M = M O F R FR F R ALTERNATIVE SOLUTION 5 - continued Substituting yields: 500i + 300 j + 800k M = ( 3200i +1800k ) = 3071 N - m 2 2 2 500 + 300 + 800 500i + 300 j + 800k M = 3071 2 2 2 500 + 300 + 800 = {1551i + 931j + 2481k } N - m M = M O - M = ( 3200i +1800k ) - (1551i + 931j + 2481k ) = {1649i - 931j - 681k} N - m, M = 2012 N - m ALTERNATIVE SOLUTION 5 - continued In order to eliminate M, FR must be moved to (x , y) location such that it creates M about the origin. The (x , y) can be calculated as follows: M = ( xi + yj) ( FR ) Substituting yields 1649i - 931j - 681k = ( xi + yj) ( 500i + 300 j + 800k ) 1649i - 931j - 681k = 800 yi - 800 xj + 300 xk - 500 yk Solving for x and y yields 1649 = 800 y or y = 1649 = 2.061 m 800 - 931 - 931 = -800 x or x = = 1.164 m - 800 CLASS PROBLEM Replace the forces by a wrench, Specify the magnitude of the force & couple moment for the wrench and the point P(y, z) where its line of action intersects the plate. CLASS PROBLEM - SOLUTION Step 1 Determine the resultant force F R = {-40i - 60 j - 80k} lb Step 2 - Determine the directional cosine of FR u FR = - 40i - 60 j - 80k 40 + 60 + 80 2 2 2 = -0.3714i - 0.5571j - 0.7428k Since FR and the resultant moment are parallel, they have the same directional cosines CLASS PROBLEM - SOLUTION Step 3 Calculate the moments at point P in the x, y and z directions by multiplying each force by its moment arm 1. Around the x axis through point P, the -60j lb force has a moment arm of (12 - z) ft, and the -80k lb has a moment arm of (y) ft. M Px = 720 - 60 z + 80 y SOLUTION 5 - continued 2. Around the y axis through point P, the -40i lb force has a moment arm of z ft, the other two forces create no moment in that direction. M Py = 40 z CLASS PROBLEM - SOLUTION 3. Around the z axis, the 40i lb force has a moment arm of (12 - y) ft, the other two forces create no moment in that direction. M Pz = 480 - 40 y CLASS PROBLEM - SOLUTION Step 4 - Let the magnitude of the resultant moment be MRP, then its MPX, MPY, and MPZ components can be calculated using the directional cosines as follows: MPX = - 0.3714 MRP= 720 60z + 80y MPY = - 0.5571 MRP= 40z MPZ = - 0.7428 MRP= 480 40y CLASS PROBLEM - SOLUTION Step 5 Solve the 3 simultaneous equations MRP = - 624 lb-ft Z = 8.69 ft; and Y = 0.41 ft The negative sign of MRP indicates that its direction is opposite to that of FR CLASS PROBLEM - Alternative Solution As before, the resultant force FR F R = {-40i - 60 j - 80k} lb The resultant moment MO due to all forces about the origin M O = ( 0) ( - 80k ) + (12 j) ( - 40i ) + (12 j + 12k ) ( - 60 j) M O = { 720i + 480k } lb - ft CLASS PROBLEM - Alternative Solution Recall that F1F2=(F1)(F2)(cos ), where is the angle between F1 and F2. If F2 is a unit vector, then F2 is equal to 1.0 and F1F2=(F1)(cos ), which is the component of F1 along F2. Based on the above, the component of MO along FR (labeled M||) can be given as: FR M = M O F R FR and M = M O F R FR F R CLASS PROBLEM - Alternative Solution Substituting yields: - 40i - 60 j - 80k M = ( 720i + 480k ) = -624 lb - ft 2 2 2 40 + 60 + 80 - 40i - 60 j - 80k M = -624 2 2 2 40 + 60 + 80 = { 231.7i + 347.6 j + 463.5k } lb - ft M = M O - M = ( 720i + 480k ) - ( 231.7i + 347.6 j + 463.5k ) = {488.3i - 347.6 j + 16.5k} lb - ft , M = 599.6 lb - ft CLASS PROBLEM - Alternative Solution In order to eliminate M, FR must be moved to (y , z) location such that it creates M about the origin. The (y , z) can be calculated as follows: M = ( yj + zk ) ( FR ) Substituting yields {488.3i - 347.6 j + 16.5 k} = ( yj + zk ) ( - 40i - 60 j - 80k ) 488.3i - 347.6 j + 16.5 k = -80 yi + 60 zi - 40 zj + 40 yk Solving for y and z yields - 347.6 = 8.69 ft - 40 16.5 16.5 = 40 y or y = = 0.41 ft 40 - 347.6 = -40 z or z = QUESTION A general system of forces and couple moments acting on a rigid body can be reduced to a ___ . A) two moments B) single moment C) single force and two moments D) single force and a single moment QUESTION The original force and couple system and an equivalent force-couple system have the same _____ effect on a body. A) internal C) internal and external B) external D) microscopic QUESTION The forces on the pole can be reduced to a single force and a single moment at point ____ . A) P C) R B) Q D) S Z S R Q P Y E) All of the above X QUESTION Consider two couples acting on a body. The simplest possible equivalent system at any arbitrary point on the body will have A) one force and one couple moment B) one force C) one couple moment D) two couple moments QUESTION y 30 lb 1' 1' 30 lb x P 40 lb For this force system, the equivalent system at P is ___________ . A) FRP = 40 lb (along +x-dir.) and MRP = +60 ft lb B) FRP = 0 lb and MRP = +30 ft lb C) FRP = 30 lb (along +y-dir.) and MRP = -30 ft lb QUESTION Consider three couples acting on a body. The equivalent systems will be _______ at different points on the body. A) Different B) The same C) zero D) None of the above. QUESTION If F1 and F2 are perpendicular to each other, then F1F2 = _______ A) F1F2 B) 1.0 C) 0.0 D) None of the above. QUESTION If F1 and F2 are parallel to each other, then F1F2 = _______ A) F1F2 B) 1.0 C) 0.0 D) None of the above. FEEDBACK What is the level of your understanding regarding moment about an axis? The magnitude of a moment about axis a-a is ua-a (r x F) The moment vector about axis a-a is (ua-a (r x F)) ua-a A) Have no idea B) Somewhat C) Great deal What is the level of your understanding about couple moment? The moment due to a couple is constant about any points The moment vector due to a couple = r x F; where r = position vector pointing from one of the couple force to the other and F = the couple force that r is pointing to The magnitude of a moment due to couple = F.d; where d = perpendicular distance from one force to the other A) Have no idea B) Somewhat C) Great deal FEEDBACK FEEDBACK What is the level of your understanding regarding equivalent systems? A system with forces passing through a reference point A system with forces not passing through a reference point A system with forces and couple moments A system's simplification to a single resultant force A system's simplification to a wrench A) Have no idea B) Somewhat C) Great deal ...
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