SIMPLIFICATION OF FORCE AND COUPLE SYSTEMS & THEIR
SIMPLIFICATION OF FORCE AND COUPLE SYSTEMS& THEIR FURTHER SIMPLIFICATIONToday’s Objectives:Students will be able to:a) Determine the effect of moving aforce.b) Find an equivalent force-couplesystem for a system of forces andcouples.In-Class Activities: Check Homework Reading QuizApplicationsEquivalent SystemsSystem ReductionConcept QuizGroup ProblemSolving Attention Quiz
APPLICATIONSWhat are the resultant effects on the person’s handwhen the force is applied in these four different ways?Why is understanding these differences important whendesigning various load-bearing structures?
APPLICATIONS (continued)Several forces and a couple momentare acting on this vertical section ofan I-beam. ?For the process of designing the Ibeam, it would be very helpful ifyou could replace the various forcesand moment just one force and onecouple moment at point O with thesame external effect? How willyou do that?
SIMPLIFICATION OF FORCE AND COUPLE SYSTEM(Section 4.7)When a number of forces and couplemoments are acting on a body, it iseasier to understand their overall effecton the body if they are combined into asingle force and couple moment havingthe same external effect.The two force and couple systems arecalled equivalent systems since theyhave the same external effect on thebody.
MOVING A FORCE ON ITS LINE OF ACTIONMoving a force from A to B, when both points are on thevector’s line of action, does not change the external effect.Hence, a force vector is called a sliding vector. (But theinternal effect of the force on the body does depend on wherethe force is applied).
MOVING A FORCE OFF OF ITS LINE OF ACTIONBWhen a force is moved, but not along its line of action, there isa change in its external effect!Essentially, moving a force from point A to B (as shown above)requires creating an additional couple moment. So moving aforce means you have to “add” a new couple.Since this new couple moment is a “free” vector, it can beapplied at any point on the body.
SIMPLIFICATION OF A FORCE AND COUPLE SYSTEMWhen several forces and couple momentsact on a body, you can move each forceand its associated couple moment to acommon point O.Now you can add all the forces andcouple moments together and find oneresultant force-couple moment pair.
SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM(continued)WR W1 W 2(MR)o W1 d1 W2 d2If the force system lies in the x-y plane (a 2-D case), then thereduced equivalent system can be obtained using the followingthree scalar equations.
FURTHER SIMPLIFICATION OF A FORCE ANDCOUPLE SYSTEM (Section 4.8) If FR and MRO are perpendicular to each other, then the systemcan be further reduced to a single force, FR , by simply movingFR from O to P.In three special cases, concurrent, coplanar, and parallel systemsof forces, the system can always be reduced to a single force.
EXAMPLE IGiven: A 2-D force systemwith geometry as shown.Find: The equivalent resultantforce and couplemoment acting at A andthen the equivalentsingle force locationmeasured from A.Plan:1) Sum all the x and y components of the forces to find FRA.2) Find and sum all the moments resulting from moving eachforce component to A.3) Shift FRA to a distance d such that d MRA/FRy
EXAMPLE I (continued) FRx 50(sin 30) 100(3/5) 85 lb FRy 200 50(cos 30) – 100(4/5) 163.3 lb MRA 200 (3) 50 (cos 30) (9)– 100 (4/5) 6 509.7 lb·ftFR ( 852 163.32 )1/2 184 lbFR tan-1 ( 163.3/85) 62.5 The equivalent single force FR can be located at a distance dmeasured from A.d MRA/FRy 509.7 / 163.3 3.12 ft
EXAMPLE IIGiven: The slab is subjected tothree parallel forces.Find:Plan:1) Find FRO Fi FRzo kThe equivalent resultantforce and couplemoment at the origin O.Also find the location(x, y) of the singleequivalent resultantforce.2) Find MRO (ri Fi) MRxO i MRyO j3) The location of the single equivalent resultant force is givenas x – MRyO/FRzO and y MRxO/FRzO
EXAMPLE II (continued)FRO {100 k – 500 k – 400 k} – 800 k NMRO (3 i) (100 k) (4 i 4 j) (-500 k) (4 j) (-400 k) {–300 j 2000 j – 2000 i – 1600 i} { – 3600 i 1700 j }N·mThe location of the single equivalent resultant force is given as,x – MRyo / FRzo (–1700) / (–800) 2.13 my MRxo / FRzo (–3600) / (–800) 4.5 m
READING QUIZ1. A general system of forces and couple moments acting on arigid body can be reduced to a .A) single forceB) single momentC) single force and two momentsD) single force and a single moment2. The original force and couple system and an equivalentforce-couple system have the same effect on a body.A) internalC) internal and externalB) externalD) microscopic
ZCONCEPT QUIZ S1. The forces on the pole can be reduced toa single force and a single moment atpoint .A) PB) QD) SE) Any of these points. R QC) R PYX2. Consider two couples acting on a body. The simplest possibleequivalent system at any arbitrary point on the body will haveA) One force and one couple moment.B) One force.C) One couple moment.D) Two couple moments.
ATTENTION QUIZ1. For this force system, the equivalent system at P is.A) FRP 40 lb (along x-dir.) and MRP 60 ft ·lbB) FRP 0 lb and MRP 30 ft · lbC) FRP 30 lb (along y-dir.) and MRP -30 ft ·lbD) FRP 40 lb (along x-dir.) and MRP 30 ft ·lby 1'P1'30 lb40 lb30 lbx
GROUP PROBLEM SOLVING IGiven: A 2-D force and couplesystem as shown.Find: The equivalent resultantforce and couplemoment acting at A.Plan:1) Sum all the x and y components of the two forces to find FRA.2) Find and sum all the moments resulting from moving eachforce to A and add them to the 1500 N m free moment to findthe resultant MRA .
GROUP PROBLEM SOLVING I (continued)Summing the force components: Fx 450 (cos 60) – 700 (sin 30) – 125 N Fy – 450 (sin 60) – 300 – 700 (cos 30) – 1296 NNow find the magnitude and direction of the resultant.FRA (1252 12962)1/2 1302 N and tan-1 (1296 /125) 84.5 MRA 450 (sin 60) (2) 300 (6) 700 (cos 30) (9) 1500 9535 N m
GROUP PROBLEM SOLVING IIGiven: Forces F1 and F2 areapplied to the pipe.Find: An equivalent resultantforce and couple momentat point O.Plan:a) Find FRO Fi F1 F2b) Find MRO MC ( ri Fi )where,MC are any free couple moments (none in this example).ri are the position vectors from the point O to any point on the lineof action of Fi .
GROUP PROBLEM SOLVING II (continued)F1 {– 20 i –10 j 25 k} lbF2 {–10 i 25 j 20 k} lbFRO {–30 i 15 j 45 k} lbr1 {1.5 i 2 j} ftr2 {1.5 i 4 j 2 k} ftThen, MRO ( ri Fi ) r1 F1 r2 F2MRO {ij k1.5 2 0-20 -10 25 ij k1.5 4 2-10 25 20} lb·ft {(50 i – 37.5 j 25 k ) (30 i – 50 j 77.5 k )} lb·ft {80 i – 87.5 j 102.5 k} lb·ft
a change in its external effect! Essentially, moving a force from point A to B (as shown above) requires creating an additional couple moment. So moving a force means you have to “add” a new couple. Since this new couple moment is a “fr
a change in its external effect! B Essentially, moving a force from point A to B (as shown above) requires creating an additional couple moment. So moving a force means you have to “add” a new couple. Since this new couple moment is a “
STATIC FORCE ANALYSIS WEEK 1 . COUPLE MOMENT OF COUPLE Couple: Two equal and opposite forces (F & F’) Moment of Couple: A vector normal to the plane of the couple (M R BA X F) Conditions of Equilibrium A system of bodies is
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