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UNIT 2 EQUILIBRIUM :FREE BODYDIAGRAMStructure

2.1 IntroductionObjectives

2.2 Types of Supports and Constraints2.2.1 Plane Structures2.2.2 Space Structures

2.3 Free Body Diagrams2.4 Equilibrium of Coplanar Forces2.4.1 Concument Forces2.4.2 Non-concurrent Forces2.4.3 Beam Reactions for Various Types of Loads

2.5 Summary2.6 Key Words2.7 Answers to SAQs

2.1 INTRODUCTIONIn Unit 1, you have learnt the technique of reducing any system of forces to ibforce-couple system at some arbitrary point 0. hen this resultant f o r e and ths! re~ultantcouple both are equal to zero, the system is said to be in equilibrium. In this unit, you willstudy the necessary and sufficient conditions for the equilibrium of a rigid body. This willenable you to determine the unknown forces applied to the rigid body in equilibrium orunknown reactions exerted on it by its supports. This will help you to analyse anystructure which is the first step in the design of structures.ObjectivesAfter studying this unit, you should be able to

classify various types of supports and constraints,draw free body diagrams,understand conditions of equilibrium,determine the unknown forces acting on a body, andfind beam reactions for various types of loads.

2.2 TYPES OF SUPPORTS AND CONSTRAINTSIn engineering practice, you might have observed that the bodies are ~ 0 ~ e c t e dith thesurroundings in a number of standard types. These connections and supports may be inthe form of rollers, rockers, ball and socket joints, frictional surfaces, short links, cables,pins, hinges etc. They restrict the movement of the bodies in particular direction byoffering reactions. In some cases, it is easier to ascertain the direction of reactions byconsidering the direction of possible movement in absence of the support. Consider, forexample, a rigid body kept on a table. If the table would not have been there, the rigidbody would have falled down due to gravity. The table is restricting the downwardmovement of the body. This is possible if it is able to offer a force in upward direction.Thus, you can ascertain that the direction of reaction offered by the table is upward.2.2.1 Plane S t r ~ z 2 ~ r e sA plane structure lies in one plane e.g. a thin flat plate, an assembly of straight bars lyingin one plane, an arch, a simply supported beam, a cantilever. Some plane structures areconstrained by supports that permit no rigid body movement upon application of loads.The reactions offered by various supports connecting plane structures can be groupedunder three categories :

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1. Reactloas equllvalenttoa force with known h e f actionFigure2.1 (a) illustrates some of the supports in this category. They include rollers,rockers, fricticmlcss surfaces, short links, cables, collarson frictionless rods andfrictionless pins in slots. IBese supports prevent motion inone direction only. Thedirection of reaction can thereforebe ascertained easily.

Rocker

Sbori link C da r onMctionlesa rod

FrictionlessSuliPee

~ctioolessinin dot

2. Reactions equllvalentto a force of unknown directionA frictionless hinge cx pin, knifeedge support, rough surfaces are some of theexamples of this category shown in Figure 2.l(b).These supports prevent translationof the freebody in-alldirections. However, rotation of the body about the connectionispossible. The reactions offered by these supports involve two unknowns :magnitude and direction. They are usually represented by the components along twomutually perpendicular directions say horizontal and vertical.

3. Reactions equivalent to a force and a coupleA beam built in a wall as shown in Figure 2.1(c) is an example of this category. Sucha support offers full restraint against rotationand translational movement. Reactionsof this group involve three unknowns. Generally these are represented by twocomponents of the force along x and y directions and the moment of the couple.

Fixed mpporiFigun 2 1 c)

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2.2.2 Space StructuresA space structure is a structure in a three-dimensional space. ?bere are six fundamentalmotions possible viz. translation in x, y atld z directim s and rotations about tbe x , y and zaxes. &pending on tbe con8traint.8 provided by the supports the number of unknownreactions vary &om one to six.Frictionless surfaces, cables and ball suppo rts prevent translation in one direction onlyand thus exert a single force of known line af action. The magnitude of the reaction is theonly unknown in such cases. Wheels on rails or rollers on rough surfaces involve twounknown reaction components as motion in two directions is restricted. 111case of balland socket joints or rough surfaces in direc t conta ct there a re three urnknowns as theyprevent translation in three directions. A universal joint which does not allow rotationabout one axis will involve 4 urlknovlns viz. three componen ts along x , y and zdirections and a couple. A fixed suppcrrt does not allow my motion, neither translationnor rotation. This involves maximum number of reaction components : hree forcecomponents and three couples.

Ball and socket(Three unknowns) Universal joint(foonr unlmowns)Fixed 1upp0tl(six ~ W M )

2.3 FREE BODY DIAGRAMSTo study any structure, you should be in a position to identify all the forces acting on it.A structure can be divided in parts for simplicity. If you separate a body from Itssurroundings and draw a diagram to represent that body and ndicate all the forces fromthe surroundings that act on it; such a diagram is called a free-body d iagram.Example 2.1

Consider the propped cantilever shown in Figure 2.3 (a). To draw the free-bodydiagram for the beam, isolate the beam from the surroundings. l[he fixed and thehinge supports will be replaced by constraining forces offered by them. All externalloads will also be replaced by forces exerted on it and then the resulting diagram iscalled free-body diagram of the beam.

Space DiagramVA 1 1 R~

he-Body Diagram

Free-Body Diagram for the beam in Figure 2.3 b) shows(i) beam AB (ii) constraining forcesHA VA nd MA ffered by the fixed support atA, (iii ) constraining forces Rg ffered by the roller support B , nd (iv) the extkrnalload acting at C.

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Example 2.2A 3m long boom is held by a ball and socket joint at A and by two cables CD andBE It carries a load of 20 kN at B. Draw Free-Body Diagram for the boomAB.

Y

@)Agure 2.4

SolutionFree-Body Diagram for.theboom AB in Figure 2.4(b) shows(i)boomAB (ii) constraining forces A,, Ay andAz due to ball and socket support at A,(iii) tension TcD in cable CD, and (iv) tension TBEn cableBE.

SAQ 1(1) Draw free-body diagram for the member AB n following cases :

(a ) 201dV. (b) 4 0 kN

Smoothwall1For ladder AB , ength 1= 2.5 m.Assume self weight 200 Nconcentrated at C.

B Rough FloorhAgure25 c)

B

0.8 m

Joint A i s a11 A hinged joint

y axis(e l

z axisRk r r 2.5(4 Figure25 e)

--Cx axis

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In Figure 2.5(e), memberAB is supported by a ball and socket atB and leans against Eqdibriua rh odya smooth wall atA. Cord CD is attached to the mid-point of AB . m.ypm' ty axis

z xis

In Figure 2.5(f),DF and CE are cables. JointA is ball and socket joint.(2) Identify the reaction components possible in case of following supports andconnections:

(A) Frictionless pin in slot(B) Hinge joint(C) Wheel on rail(D) Hinge and bearing supporting axial thrust and radial load (Figure 2.6)(E) Ballandsocketjoint(F) Fixed support.

2.4 EQUILIBRIUM OF COPLANAR FORCESConsider a rigid body acted upon by a number of coplanar forces. These forces will causethe rigid body either

(A) move in a particular direction without rotating, or(B ) rotate at its own place without moving, or(C) rotate about itself and at the same time move in any particular direction, or(D) remain completely at rest.

The last case is that of the state of equilibrium. To move the body in a particular directionthere must exist a resultant force and to rotate there must exist a resultant couple. If theresultant force and the resultant couple both are absent the body can neither move norrotate. It will remain at rest. Let us study some typical cases.2.4.1 Concurrent ForcesIf a number of coplanar concurrent forces are acting on a body theq there may exist asingle resultant force passing through the point of intersection of all the concurrentforces. But there will not exist any couple. The resultant forcemay have two componentssayR, and R, wherex and y are any two mutually perpendicular co-ordiate axes. If thebody is to be at rest then you must get both these components equal to zero. Theconditions for equilibrium in case of coplanar concurrent forces are, therefore, as follows :

Rx=OR,,=0

But R, = z FX and R, = z F,,: z F , = O and z F Y = O

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equal t-17xm.(2) The algebraic sum of the resolved parts of all forces in a direction at rightangles to the fist direction must be equal to zero.

Example 2.3Four coplanar concurrent forces act at a point and keep it at rest. These are shown inFigure 2.7. Determine the forcesP and Q.

SolutionLet us assume that forcesP and Q act away from the point 0. As the point 0 enlainsat rest we can apply equations of equilibrium

~ F , = o

As P is positive, the assumed direction is correct. Therefore,P acts away from0.

: P sin 60 + 1260 sin 60 - Q = 0

As Q is also positive, the assumed direction of Q is correct. Therefore, Q acts awayfrom0.Example 2.4

A smooth sphere weighing 200 N is resting as shown in Figure 2.8. Determine thereactions at the supports.Solution

Let us fust ascertain the directions of reactions. As the wall is vertical and smooth thereaction atA will be horizontal i.e. norm1 to the wall. Similarly, the reaction at Bwill be normal to the line inclined at 60' to the horizontal. Both these reactionswillpass through 0 he centre of sphere because these are normals drawn to the tangentsof the sphere atA andB. Theweight of the cylinder can be assumed concentrated at0. hus, three concurrent forces keep the sphere at rest. Let us apply conditionsd

I equilibrium.I C F , = O: RA - RB cos8 --- 0 8 = 30" by the geometryof the figure

: = 0.866RB~ F ~ = o

RB sin 30 - W = 0

The same problem can be solved using Lami's theorem which states, "If three forcesacdng{XI body keep it at rest then each force is proportional to the sine of the anglebetween the other two forces'' {Figure2.9).

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RA - RB--- - wsin ( 90"+ 0 ) sin 90" sin ( 180"- 0')Putting the known values, we get,

.'. RA = 200 cos '0 = 346.4 Nsin 30'RB= ---00 - 400.0 Nsin 30'

SAQ 2Determine the minimum value of force P required just to start the wheel over thestep 300mm high. The diameter of the wheel is 1.2 m and the weight is 800N. Alsofind the direction of P.(Note that the reaction offered by the ground is zero when the wheel is just on hepoint of moving over the step.) 1

2.4.2 Non-concurrent ForcesIf non-concurrent coplanar forces are acting on a body, their resultant may be in the formof a couple in addition to a resultant force. If a body is to remain at rest then there shouldneither be a resultant force nor a resultant moment, i.e.

Therefore, in addition to the two conditions of equilibrium as in case of concurrent forcesthere is one more condition of equilibrium which can be stated in words as :"me algebraic sum of moments of all the fotces about any point in their plane mustbeequal to zero".

Example 2.5A boardABCD is held in position as shown in Figure 2.11 by a cable RE and hinge atA. If the weight of the bard is 5kN, etermine the reactions at hingeA am! !hetension Tin the cable.

Eqdibrlom :Free BodyD i m

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SolutionLet the components of the reaction atA be HA and VAas shown in the Figure 2.11.'Ihe board is at rest under the action of four forces HA ,VA T and W.Taking moments of all forces aboutA ,we get,

... HA =T CQS30(On putting, T= 5 IrN) HA = 5 x 0.866

= 4.33 kNx F Y = 0

: VA + T sin 30' - W = 0: VA = W - T sin 30

= 5 - 5 x 0 5= 2.5 kN

RA == d4.332 + 2 . 5 0 ~= 5 kN

Now

= 30.?,,action Ahas - ,aitude of 5 kN and is inclined at 30 to thehorizontal.

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EqPllibrillm t Fra BodyDiagramNote :You may altempt the same problem considering theboard to be subjected to 3forces only viz. iq4 ,Tand W. Ae th e e fo rws keep the body at rest, these mustbe concurrent. FfnQ he point of concurrence&ndget the values of unknownsaA& T, US@F x = O a n d x F y =O

knowing the direction of RA,$% ma y also use Lami's theorem either.2.43 Beam Reactions for Various Typesof b a d a'Ihe conditions of equilibrium for rigid bodies can be used to f ind the rewlcdons of a beamDepending on support conditions, beams are classified as cantilever, simply supported,continuous, propped cantilever and fixed beam. If a beam is fixed at one end and free atthe other it is called a cantilever. As at fixed end neittiet &am lation nor rotation ispossible there will be three unlcnown reaction components HA ,VA and M A There arethree equatione of equilibrium. Hence, these unknow ns can be found out. Similfuly, thereactions in case of s iaply supported beam can also be found out. In simply supportedbeam or in a beam hinged at one etkl and roller at other end, the number of unknowns islimited to three. Hence these beams are called statically determinate beams. You candetermine the reactiohs using the sta tic conditions of equilibr ium. But if the number ofunknowns exceed three as in the case of contiriuttusbeams, propp3d cantilever and fixedbeams, you cannot determine the un kn ow q just by m aking use of BfatidsU equations ofequilibrium. Therefore, these beams are called statically indeterminate beams.Hefe inthis unit you will learn to find the reactions in case of statically determinate beams only.A beam is a structu ral member generally carry ing transverse loads. These loads cSVnbeconcentrated, uniformly distributed or triangular in nature. Let us study each type ofloading by solving some problems.Example 2.6

A cantilever AB, 1.8 m long is fixed at A and carriesWarmly distributed load of20 lcNlm over its enti re length and a point load of 30 kW at the fret?end. Determinethe reactions at A.

SolutionLet the reaction components atA be HA ,VA and MA as shown in Figure 2.12. Letus replace the uniformly distribu ted load of 20 kN/m by a single force of20 x 1.8 = 36 kN acting at the centre of AB i.e. at 0.9 m fromA .There is nohorizontal force acting on the be am

= 66kNTaking moments at A ,we get,

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Therefore, the reaction atA consists of a vertical force acting upwards of themagnitude 66 kN and an anticlockwise moment of 86.4 kN-m.Example 2.7. A beam AB is hinged at A and is supported at C. It is loaded as shown in .Figure 2.13. Find out the reactions at A and C.

Figure 213 I RCSolution

Let the reaction components at A be HAand VA as sh o w in Figure 2.13.The reactionat Cwill be acting vertically upwards. As the beam is atirest under the action of theforces , the conditions of equilibrium can be applied.Taking mom ents of a11 forces about A, we get,

Note :Moments due to HA,VAand 40 cos 60' about A are zero as they pass throughpoint A .Clockwise moments are taken as positive.

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The reaction at A has a magnitudeof 49.672kN and is inclined at 66' 15' 24" withrespect to horizontal whereas reaction at C is acting vertically upwards and has amagnitude of 67.172kN.

SAQ 3Determine the reactions of the beams loaded as shown in following figures.

figure2.14 (a)

Equilibrium t Free BodyDiagnm

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Figure2.14 (c)

/ BNote : nternal

/ A C D E hinge at @

2.5 SUMMARYIn this unit, yo u have learnt to classify various types of sup ports and constraints, rollers,rockers, ball and soc ket joints, frictional surfaces, short links, cab les, pins, hinges are theexam ples of support connections. These connections offer resistance to the m ovement ofthe bodies wh ich are called reactions. Following tables give the various types ofconnections and their reactions

Table 1 : Supports for Rigid Bodies (Plane Structures)SI. Typesof connection ReactionNo..-1. Cable A tension force (pull) acting away from the member alongthe direction of the cable.I 2. Link A force acting along the axis of the link. I3. Roller4. Rocker

A force acting perpendicular to the surface at the point ofcontact.A force acting perpendicular to the surface at the point ofcontact.( 5. ~modthsurface Same as roller. I

6. Pin or Hinge A force having two unknow ns : magnitude and direction orI two forces with unknown magnitudes in two mutuallyperpendicular known directions i.e., HA and VA. ,

7. Member fued connected A couple moment and the fo rce acting perpendicular to theto collar on smooth rod rod.8. Fixed support A couple moment and two unknown force components.

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Table 2 : uppor t s forRigid Bodies Space Structures) E q d b r i m : ree BodyDlsgnun1 Sl. TypesofwnnectionNo.

Reaction

1 1. Cable A tension force (pull) acting away from themember along the direction of the cable.2. Smooth surface or roller 0n.a smooth A force which acts perpendicular to thesurface surface at the point of contact.3. Ball and socket joint4. Single ournal bearing

/ 5. Single thrust bearing6. Single smooth pin

I 7. Single hinge1 8. Fired support

A force having three unknown reactioncomponents.Two force components and two couplemoments perpendicular to the shaft.W e e force components and two couplemoments.Three force components and two couplemoments.Three force components and two couplemoments.W e e force components and three couple .moments.

The fu st step in solving equilibrium problems is to draw free-body diagram. Isolate thebody from its surrounding. Replace the constraints by reactions. Show all forces actingon the member and apply the equ ations of static equilibriumIn the general case of a sy stem of equilibrium, there are six equations of staticequilibrium. These ensure that the resultant force and resultant mom ent both are zero.(1) CF,=O (2) , CF,=O (3 ) CF,=Owhere x, y and z are three mutually perpendicular axes.In particular, if the forc es are parallel and we take z axis parallel to them, then the fus t,second and last equations are identically satisfied (no force ex ist along x and y axes)Therefore, the equations of static equilibrium are reduced to three, viz.

If the forces are concurrent and if we cho ose the point of concurrence as the origin thenthe last three equations will always be satisfied and only three equations will be requiredto sd ve the problems. These. are

Similarly, in the case of concurre nt forces in a plane there will be only two equations ofstatic equilibrium :~ F , = o nd x F y = O

In case of parallel forces in a plane these equ ations will be

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where0 s the moment centre in tbe plane containing the parallel forces. 0may be anypoint in the plane. The z axis will be parallel to the direction of forces.2.6 KEY WORDSEquilibriumStatic Equilibrium

: A state of rest or of uniform motion with aconstant velocity.: A state of rest. The resultant of a l l forces actingon a body ia zero. Neither translation, norrotation i.e. Z F= 0 and Z =0.

Equations of Static Equilibrium : The conditions expressed mathematically tomaintain the static equilibrium.Free Brdy Diagram : A diagram showing a body isolated from thesurrounding, on which forces acting on the bodyare indicated by vectors.Support Reactions : The reactive forces and couple moments acting ata support.A force is developed by a support thatrestricts the translation of the body. A couplemoment is developed when the rotation of thebody is prevented.Beam : A smctural member designed to supportloadings applied perpendicular to its axis.Simply Supported Beam : A beam pinned at one end and supported onrollers at the other end.CantileverFixed Beam

: A beam fixed at one end and free at the other end.: A beam having fixed supports at both the ends.

2.7 ANSWERS TO SAQ sSAQ 1

(1)Free body diagrams for memberAB are as follows :(a) (b)

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(e)t'"

Bz

z xis F'igurefor Answer to SA Q 1(2) Answers may be verified from the preceding text.

SAQ 2

Figure forAmwers to SAQ 2Let the reaction offered by the step on the wheel at the start of motion be R as shownin Figure for Answers to SAQ 2.The reaction offered by the groundhas been taken as zero as the wheel is just on thepoint of moving over the step.Since the wheel is in equilibrium, therefore,

CM * = OTaking moments about A,

PsinaAC + P c o s a O C = WACOC - 0.3 1P s i n a + P c o s a Since, sinLOAC =- - -OA 0.6 2

: LOAC = 30"1P sin a -t P cos a [ = 800 Therefore, -C = ~ n 3 ~ 0C =Jj

6 ~ s i n a Pcos a = 8 0 0 5

dPFor getting minimum value of force P, there should be- 0d a\

Then, dP- - - 8006 ~ C O S- sin a1d a = 0[ f i s i n a + cos a12

Equilibrium :E'ree BodyDiagram

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Now,F o r a = 6 0 " , P = 4 0 0 6 ~The student may verify the answer by applying Lami's theorem.

SAQ 3(a) _ 28kN

Let vertical and hor k~nta leactions atA beRAV andRAH respectively.MA sthe total moment coming on the fixed endA of cantileverbeam AB. Let usreplace the uniformly distributed load of16x 1.2= 19.2 kN acting at a distancebeam isat rest under the action of thebe applied to calculateRAV, AH and MA.

~ F , , = o: RAv - 28 sin 45"- 19.2 = 0

Taking moments aboutA, we get,

Let the reaction components atA beHAand VA s shown in figure. Thereactiona t P : - ,RBwill be acting vertically upwards. As thebeam is in

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equilibrium, the condition of equil ihbmcmbe amlied. The moment of the ~quilibrium ree B O ~ Ycouple acting at D can be calculated as 20 x 1= 20kN.m acting in clockwise Disglalndirection. The uniformly distributed load of 12 kN/m cm Ix replaced by asingle force of 12 x 2.5 = 30kN acting at 6.5 + = 7.75 m fromA.fiere is no hofizontal force acting on the beam. Then,

C F r = 0 il. H ~ s 0Taking mcmients of all forces about A, wegat,

~ M ~ = o32x 2 + 20 - kB 6.5+ (12 x 2.9) (7.75) = 064 + 20- 6.5 RB + 232.9 = b

RB = 64 + 96+332.5 - 4aa69 LN6.5Note :Moments due to HA and a W tA are zero as they pass tbrough point A .Clockwise moments have been taken as p08ititlb;

E F ~ = o;. VA- 32 t 48.69 -30 0

Therefore, the reaction at A is 13.31kNwhereas teaCti& fA Bid -8vertically upwards and has a magnitude of 48.69kN.

The vertical and horizontal reaction atA beRAV ndR A ~espectively.MA isthe total moment coming on the fixed end A of cantilever beamAB .Let USreplace the triangular loading by a single force of

1acting at a distance of5 1.2) = 0.4 m from AAs the beam is at rest, under the action of forces, the condition of equilibriumcan,be applied to calculateRAV RAHandMA .There isno horizontal force acting on the beam. Then,

Taking moments aboutA ,we getCM * = O

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Therefore, the reaction at A consists of a vertical fopcr acting upwards of themagnitude 14.4 kN and an anticlockwise moment of 5.76 kN.m.(4

MkN /m lOkN

There is an internal hinge at C from where the beam may be cut into twosections A C and CB as shown below :

Part - 1 Part - 2Now, the part I ana part L L;UI ue analyseu separately to get RAV RAH,MAand RE .Part 1This part now can be analysed as cantilever beam with Rc as a concentratedload at the free end. Here, Rc = 34 kN.Taking moment about A,

Now, taking m oments about C,~ M C = O

... R A v x 3 - 1 9 2 - 6 0 ~ 1 . 5= 0: RAv = 94 W

There fore, the rection at A consists of a vertical force acting upwards of themagnitude 94 kN and an anticlockwise moment of 192 kN-m whereas reactionat B is also acting vertically upwards and has a magnitude of 26 kN.Part 2This part can be analysed as simply supported beam.Taking moment about C,