L_-_Equivalent_System - FORCE SYSTEM RESULTANTS RESULTANTS...

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Unformatted text preview: FORCE SYSTEM RESULTANTS RESULTANTS Equivalent System Equivalent LEARNING OBJECTIVES LEARNING Be able to find an equivalent force-couple Be system for a system of forces and couples system PRE-REQUISITE KNOWLEDGE PRE-REQUISITE Units of measurement Trigonometry concepts Vector concepts Rectangular component concepts Couple moment concept Cross product and dot product EQUIVALENT SYSTEM EQUIVALENT 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 Y X Y X SYSTEM WITH FORCES NOT PASSING THROUGH A REFERENCE POINT THROUGH Force acting at point A can be represented with the same force acting at point O and a couple moment = = = = = EXAMPLE 1 EXAMPLE Replace the force system acting on the beam by an Replace equivalent force and couple moment acting at point A and then at point B point SOLUTION 1 SOLUTION Y Vectors of forces : The 2.5 kN force The F2.5 = {-2.0i - 1.5 j} kN kN X The 3.0 kN force The F3.0 ={-3.0 j} kN kN The 1.5 kN force The F1.5 = {1.5 sin 30°i - 1.5 cos 30° j} kN kN The resultant force is the same at points A or B The F = F2.5 +F1.5 +F3.0 SOLUTION 1- Contd. SOLUTION Y .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-1.5j) + ( 6i ) × (1.5 sin 30°i - 1.5 cos 30° j) + ( 8i ) × ( - 3.0 j) = -3k - 9 cos 30°k - 24k = -34.8 k kN - m or 34.8 k kN - m (clockwise) SOLUTION 1- Contd. SOLUTION Y 5.8j kN 34.8k kN-m F = {-1.25i + -5.80 j} kN 5.931.25i kN X kN The magnitude and inclination angle θ of the resultant force F are F= ( − 1.25) -1 2 + ( − 5.80 ) = 5.93 kN 2 - 5.80 and θ = tan = 77.8° ( in third quadrant ) - 1.25 SOLUTION 1- Contd. SOLUTION 11.6k kN-m Calculate the moment at point B 5.93 kN y x MB = ∑ r × F = ( - 6i ) × ( - 2.0i - 1.5 j) + ( - 2i ) × (1.5 sin 30°i - 1.5 cos 30° j) + ( 0i ) × ( - 3.0 j) = 9k + 3 cos 30°k - 0 = 11.6 k kN - m or 11.6 k kN - m (counter clockwise) SYSTEM WITH FORCES AND COUPLE MOMENTS COUPLE M2=r2xF 2 = M1=r1xF 1 = Forces and couple moments can be represented with a resultant force and a resultant moment, The resultant force F = The F R The resultant moment M RO = ∑ M C + ∑ M O ∑ Where ΣMC is the summation of all couple moments and ΣMO is the summation of all moments due to forces about the reference point point EXAMPLE 2 EXAMPLE Determine the resultant force and the resultant moment Determine at point C at SOLUTION 2 SOLUTION The resultant force ∑F R = -300 j - 200 j - 400 j - 200i = {-200i - 900 j} lb Y X Summation of moments due to all Summation 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 Summation The resultant The moment about point C point ∑ M C = (600k) lb - ft ∑M RO = ∑ MO + ∑ MC = 4300k + 600k = (4900k ) lb - ft Or M R O 4900 lb - ft (counterclockwise) ALTERNATIVE SOLUTION 2 ALTERNATIVE ∑F ∑F RY RX = -300 j - 200 j - 400 j = −900lb = -200 lb FR = (900 2 + 200 2 ) 0.5 = 921.9544lb inclined at α = tan −1 (900 / 200) = 257.47 o from the x axis Y X ∑M O = 600 + (300)(7) + (200)(4) + (400)(0) + (200)(7) = (4900k) lb − ft counterclockwise SIMPLIFICATION TO A SINGLE RESULTANT FORCE RESULTANT Systems that can be simplified to a single force : 1. Concurrent force systems, All forces passing Concurrent through the same point through 2. Coplanar force systems, All two-dimensional Coplanar problems problems 3. Parallel force systems, All parallel forces but Parallel passing through different point passing CONCURRENT FORCE SYSTEMS SYSTEMS All forces passing through the same point FR = ∑ F F2 F3 ● F1 P = ●P COPLANAR FORCE SYSTEMS SYSTEMS All two-dimensional problems = = EXAMPLE 3 EXAMPLE Replace the loading on the frame by a single resultant force Replace and specify where its line of action intersects member AB measured from point A measured SOLUTION 3 SOLUTION The resultant force ∑F Y X R = -300 j - 200 j - 400 j - 200i = {-200i - 900 j} lb Summation of moments due to all Summation 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 SOLUTION d Distance, d from point A to the Distance, line of action of the resultant force can be determined using force cross product as follows cross Y X - 3200k = ( di ) X( - 200i - 900 j) - 3200k = -900dk 3200 d= = 3.56 ft 900 ∑M A = ( di ) X ( ∑FR ) PARIS PARIS All parallel forces but passing through different point PARALLEL FORCE SYSTEMS PARALLEL = = EXAMPLE 4 EXAMPLE If F1=20 kN and F2=50 kN, replace the system of forces by a single =50 resultant force and specify its location (x,y) resultant SOLUTION 4 SOLUTION The resultant force =20 kN ∑F R = -20k - 50k - 20k - 50k = {-140k} kN =50 kN Summation of moments due to Summation 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. SOLUTION =20 kN =50 kN Assume the location of Assume the resultant force is at (x,y) from the origin (x,y) - 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 QUIZ If F1=30 kN and F2=60 kN, replace the system of forces by a single =60 resultant force and specify its location (x,y) resultant QUIZ QUIZ 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 QUIZ The resultant force =30 kN ∑F R = -20k - 50k - 30k - 60k = {-160k} kN =60 kN Summation of moments due Summation 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 QUIZ =30 kN Assume the resultant force is Assume located at (x,y) from the origin origin =60 kN - 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 SIMPLIFICATION Most 3-D systems cannot be simplified to a Most single force, but they can be simplified to a wrench. A wrench or a screw is a couple moment collinear with the resultant force. moment EXAMPLE 5 EXAMPLE Replace the three forces by a wrench and also specify the Replace location (x,y) where its line of action intersects the plate location SOLUTION 5 SOLUTION Step 1 – Determine the resultant force resultant ∑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 SOLUTION Step 3 – Calculate the moments at point P in the x, y and z directions by multiplying each force by its moment arm its 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 SOLUTION Step 3 – Calculate the moments at point P in the x, y and z directions by multiplying each force by its moment arm by 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 SOLUTION Step 3 – Calculate the moments at point P in the x, y and z directions by multiplying each force by its moment arm by 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 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 = 0.5051 MRP= 800(4 - y) MPY = 0.3030 MRP= 800x MPZ = 0.8081 MRP= 500y + 300(6 - x) SOLUTION 5 - continued SOLUTION Step 5 – Solve the 3 simultaneous equations MRP = 3070.9 N-m X = 1.16 m; and 1.16 Y = 2.06 m 2.06 ALTERNATIVE SOLUTION 5 ALTERNATIVE As before, the resultant As force FR force 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 origin M O = ( 0) × ( 500i ) + ( 4j) × ( 800k ) + ( 6i + 4j) × ( 300j) M O = { 3200i + 1800k } N - m ALTERNATIVE SOLUTION 5 - continued continued ALTERNATIVE Recall that F1●F2=(F1)(F2)(cos θ), where θ is the angle )(cos ), between F1 and F2. If F2 is a unit vector, then F2 is equal to 1.0 and F1●F2=(F1)(cos θ), which is the component of F1 along )(cos F2. Based on the above, the component of MO along FR Based (labeled M||) can be given as: (labeled FR M = M O • F R FR and M = M O • F R FR F R ALTERNATIVE SOLUTION 5 - continued continued ALTERNATIVE 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 continued ALTERNATIVE In order to eliminate M┴, FR must be moved to (x , y) location such that it creates M┴ about the origin. The (x location , y) can be calculated as follows: y) 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 CLASS 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 CLASS Step 1 – Determine the resultant force resultant ∑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 CLASS Step 3 – Calculate the moments at point P in the x, y and z directions by multiplying each force by its moment arm by 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 SOLUTION 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 CLASS 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 CLASS 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 CLASS Step 5 – Solve the 3 simultaneous equations MRP = - 624 lb-ft Z = 8.69 ft; and 8.69 Y = 0.41 ft 0.41 The negative sign of MRP indicates that its RP direction is opposite to that of FR direction CLASS PROBLEM - Alternative Solution Alternative As before, the resultant As force FR force ∑F R = {−40i − 60 j − 80k} lb The resultant moment MO due to all forces about the origin to M O = ( 0) × ( − 80k ) + (12 j) × ( − 40i ) + (12 j + 12k ) × ( − 60 j) M O = { 720i + 480k } lb − ft CLASS PROBLEM - Alternative Solution Alternative CLASS Recall that F1●F2=(F1)(F2)(cos θ), where θ is the angle Recall )(cos ), between F1 and F2. If F2 is a unit vector, then F2 is equal to 1.0 and F1●F2=(F1)(cos θ), which is the component of )(cos ), F1 along F2. Based on the above, the component of MO along FR (labeled M||) can be given as: (labeled FR M = M O • F R FR and M = M O • F R FR F R CLASS PROBLEM - Alternative Solution Alternative CLASS Substituting yields: Substituting − 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 Alternative CLASS In order to eliminate M┴, FR must be moved to (y , z) location such that it creates M┴ about the origin. The (y location , z) can be calculated as follows: z) 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 y QUESTION 30 lb 1' 30 lb 1' 40 lb x P• For this force system, the equivalent system at P is ___________ . A) FRP = 40i lb and MRP = +60 ft-lb B) FRP = 0i lb and MRP = +30 ft-lb C) FRP = 30j lb and MRP = -30 ft-lb D) FRP = 40i lb 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 F1●F2 = _______ A) F1F2 B) 1.0 C) 0.0 D) None of the above. QUESTION If F1 and F2 are parallel to each other, then F1●F2 = _______ 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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