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Unformatted text preview: FORCE SYSTEM RESULTANTS
Equivalent System LEARNING OBJECTIVES Be able to find an equivalent forcecouple system for a system of forces and couples PREREQUISITE 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 FRY X FRX 11.6 kNm Equivalent Force System SOLUTION 1
Y FRY 34.8 kNm FRX
X Equivalent Force System EXAMPLE 1
Y X Y X 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.
Recall Y 5.8j kN 34.8k kNm 1.25i kN X 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 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 ) ( 2.0i1.5j) + ( 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.
Y 5.8j kN 34.8k kNm 5.931.25i kN X kN 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.
11.6k kNm
Calculate the moment at point B 5.93 kN
y x 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
M2=r2xF2 =
M1=r1xF1 = 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) ALTERNATIVE SOLUTION 2
F F
RY RX = 300 j  200 j  400 j = 900lb = 200 lb FR = (900 2 + 200 2 ) 0.5 = 921.9544lb inclined at = 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
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 twodimensional 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 twodimensional 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
d Distance, d from point A to the line of action of the resultant force can be determined using cross product as follows
Y X  3200k = ( di ) X(  200i  900 j)  3200k = 900dk 3200 d= = 3.56 ft 900 M A = ( di ) X ( FR ) PARIS 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 Assume the location of the resultant force is at (x,y) from the origin  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 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
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
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
=30 kN Assume the resultant force is located at (x,y) from the 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
Most 3D 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 Nm 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 } Nm 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 lbft 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 ...
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 Summer '08
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