1 6 three member sozen

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1.6 Truss with three members

Figure 1-6-1 shows a truss, with a pin support at B and a roller support at C. Note that the two

supports are connected with a horizontal member. Select this type of truss in GOYA-T, and apply

an upward force, FAy, to equilibrium athe topnodeA. You will find that the horizontal member

shortens and the right support moves to the left.

Fig. 1-6-1 Truss of three members

The reason of the shortening is may be explained as follows. First, look at Fig. 1-6-2 showing the

free-body diagram of equilibrium athe topnodeA. ItEquilibrium conditions helps determine the

axial forces in the inclined members:

P1= P

2=

FAy/ 2

sin60=

FAy

3(1.4.1)

Fig. 1-6-2 Equilibrium at node A

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Second, look at Fig. 1-6-3 showing equilibrium at the lower right nodethe free-body diagram at

support C. Note that the reactionfrom the floor is in the vertical direction. AThe horizontal

component cannot exist because of the roller. As a result, the axial force in the bottom member

is:

P3

= P2

c o s6 0 = FAy

2 3(1.6.1)

The negative sign represents that the axial force is compressive. The bottom member shortens.

Figure 1-6-3b also leads to the reaction of the floor. It is downward and the direction is

represented by the negative sign:

RCy = P2 sin60 = FAy

2(1.6.2)

Fig. 1-6-3 Equilibrium at support C

Similarly, the equilibrium at support B is shown in Fig. 1-6-4. The reaction from the floorat B is

same as that at Cthe right support:

RBy = P1 sin60 = FAy

2(1.6.3)

Fig. 1-6-4 Equilibrium at support B

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Apply a force FAxpulling to the right at the uppernode Aas shown in Fig. 1-6-5. You will find

that the horizontal member elongates and support C moves to the right.

Fig. 1-6-5 Horizontal force to the truss

Look at Fig. 1-6-6 showing equilibrium at node A. It indicates that the axial force in the left

memberAB isin tensileon:

P1 =FAx (1.4.5)

The axial force in member AC is incompressiveon:

P2

= FAx (1.4.6)

Fig. 1-6-6 Equilibrium at the top node

Next, look at Fig. 1-6-7 showing equilibrium the free-body diagram at the lower rightnode C.

Again, the reaction at C from the flooris vertical because of the presence of the roller. The axial

force in the bottom member is tensile:

P3

= P2co s60 =

FAx

2(1.6.4)

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This tensile force elongates the bottom member, moving node C to the right. The free-body

diagram in Fig.ure 1-6-7b can also be used to determine the reaction at Calso leads to the

reaction from the floor, which is upward indicated by its representedby thepositive sign:

RCy

= P2

sin 60 = 32FAx (1.6.5)

Fig. 1-6-7 Equilibrium at support C

Similarly, equilibrium at theThe free-body diagram for leftsupport B is shown in Fig. 1-6-8.

From that we can determine, which leads to the x- and y- components of the reaction. from the

floor:

RBx = P1 c o s6 0 P3 = FAx

2

FAx

2= FAx (1.6.6)

RBy = P1 sin60 = 3

2FAx (1.6.7)

The minus signs indicate representthat the reactions are to the left and downward, respectively.

Fig. 1-6-8 Equilibrium at support B

Thus,Aall reaction forces can be and wereare determined considering equilibrium at the nodes.

However, we can calculate the reactions directly if we consider the equilibrium of the truss as a

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whole. Note that the number of unknown variables in Fig. 1-6-5 is three (RBx, RBy, andRCy). Thus,

we need three equations. The following two conditions would be obvious.

1. The sum of the x-components of the forces acting on the structure is should bezero ( X= 0):

this leads toRBx +Fx = 0.

2. The sum of the y-components of the forces acting on the structureshould be is zero ( Y= 0):

this leads toRBy +RCy = 0.

The third condition may be called the principle of the lever, discovered by Archimedes, an

ancient Greek philosopher. In the case of the lever scale of Fig. 1-6-9, the principle requires

F1a1=F2 . To generalize the principle, we need to introduce a moment, an action that turns

an object around a point as shown in Fig. 1-6-10. The moment is defined as follows.

(Moment) = (Force) x (Distance) (1.6.8)Equation 1.6.8 is very important. You should understand and remember it. In this book, a

clockwise moment is defined as positive and an anti-clockwise moment is defined as negative.

For an object in equilibrium, the lever principle requires the following.

3. The sum of the moments acting on the structureshould beis zero ( M= 0).

In the case of the lever shown in Fig. 1-6-9, the principle leads to

M=F1a1+F2a2 (1.6.9),

where F1a1 andF2a2 represent, respectively, the anti-clockwise and clockwise moments around

the point B. In fact, the reference point can be anywhere else other than at point B , because the

lever would not rotate around any point. If the reference point is taken as shown in Fig. 1-6-11,

the principle leads to:

M=F1xF1+F2( )x+a1( )+F2x+a1+a2( ) (1.6.10)

You may find this equation reduces to Eq. 1.6.9.

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A

B

C

a1 a2

F1F2

F1+F2

Force

Distance

Reference

point

Fig. 1-6-9 Lever scale Fig. 1-6-10 Definition of moment

a1 a2

F12

F1+F2

x

Referencepoint

Fig. 1-6-11 Moment around an arbitrary point

Back to the problem of Fig. 1-6-5: the equilibrium of moments around the left support leads to:

sin 60 0 Ax CyM F L R L= = (1.6.11).

This results inRCy = FAx sin 60 , which is equivalent to the solution what weobtained previously,

Eq. (1.6.5). But this process makes itis much easier to calculate the reactions directly rather than

considering the equilibrium of each node. than considering the equilibrium at each node.

Exercise using GOYA-T: Assume that each member of the truss shown in Fig. 1-6-12 fails

at the axial force of 100 N both in tension and compression. Apply forces in seven directions one

at a time as shown in the figure and develop a table of the loads, axial forces and reaction

corresponding to the failure of the truss.

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FAx FAy P1 P2 P3

P1 P2

P3

RCy

RCy

+100 0 +99 99 +50 +87

(1)

(2)

(3)(4)(5)

(6)

(7)

Direction

(1)

(2)

(3)(4)

(5)

(6)

(7)

Fig. 1-6-12 Apply force in various directions and check the limit

Exercise 1-6-1: Plot the safe -domain of the truss shown in Fig. 1-6-12 assuming that the

members fail at 100 N both in tension and compression.

Solution: The axial forces of the left and right members,P1 andP2, were determined in Section 1-

4.

P1

= FAx +FAy

3(1.4.3) again

P2= FAx +

FAy

3(1.4.4) again

The axial force of the bottom member,P3, is determined adding Eqs. (1.6.1) and (1.6.4).

P3=FAx

2

FAy

2 3(1.6.12)

Thus, we obtain the shaded safe domain in Fig. 1-6-13.

FAx(N)

FAy(N)

P1=100

0 100

200

P2=100

P3=100P3=100

P2=100P1=100

100 3

Fig. 1-6-13 Safe domain

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Note that the conditions ofP3 = +100 (the broken lines in the figure) do not affect the shaded

region. In fact, Eqs. 1.6.12 and 1.4.4 leads toP3 = P2/2, which indicates that the axial force ofthe

bottom memberAC is alwaysa half of that ofthrightmemberBC. Therefore, the truss members

need not have an equal cross-sectional areas: it is more economical if we make the cross-

sectional area of thebottom memberAC a half the cross-sectional areaof those of each ofthe

uppermembers AB and BC as shown illustrated in Fig. 1.6.14.

60

FAx

FAy

Fig. 1.6.14 Economical design

Exercise 1-6-2: Calculate the reactions when if external forces FAx and FAy are

simultaneously applied simultaneously to the truss at node A as shown in Fig. 1-6-15. Assume

that the reactions in the figure have positive signs.

Fig. 1.6.15 Right-angled truss

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Hi, Master. I do not understand the reason why we should calculate the reactions.

There are two reasons. First, we need to check whether theground can sustain the forces from the truss. It would be

dangerous if the support goes downfails.

Oh, yes, obviously. Thanks. Bye.

Wait. There is another important reason. Do Yyou know the reason why we need

tofor determiningcalculate the axial forces?

Well, to check the safety of the truss. If an axial force exceeds a limit, the truss will fail.

Good. Later on, you will learn sophisticated trusses

consisting of socomprising many members. In such

trusses, you will have great difficultiesa big troubleif you

try to calculate the axial forces using the equilibrium conditions at

each node only.. If you calculate the reactionsbeforehandfirst, it is

much easier. As you know, all the three reactions are obtained from X= Y= M= 0, three

simultaneous equations with three unknowns.

You should be rigorous, Master. For a truss supported by a

pin and a roller, the number of the reactions is three, and

therefore, X= Y= M= 0 is enough. For the other

cases, however, it maynot bes not enough. For example, a trusssupported by two pins has four reactions, and we need another

equation.

You are right, Joan. But lets ignore such trusses for a while. We may talk about it next

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year. Anyhow, important thing is that external forces and reactions always satisfy X= Y=

M= 0.

Oh, you mean that I can treat the reactions and the external forces equally.

Exactly. Both the reactions and the external forces work from outside the truss. In

Section 1-2, you learned that an external force and a reaction work from outside making

an axial force in a rod. Its a similar story.

+++++++ Foradvanced interested readers: the moment produced by a force is a vector. ++++++

The moment produced by a force can be defined more elegantly. Here, vectors are represented

using bold letters such as a orF. You know that a sum of two vectors is a vector. How about a

product of two vectors? Mathematicians have invented two kinds of products: an inner product,

which makes a scalar, and an outer product, which makes a vector. The latter applies to the

definition of the moment of a force. Assume a heavy disc on the x-y plane as shown in Fig. 1-6-

16, which may rotates around the z-axis. An external force F is applied to the disc at the point A,

which location is represented by a vectora. Both a and F are on the x-y plane. According to Fig.

1-6-10, the moment of the force F around the point O is defined as the product of

and the

distance from the point O to the force,

asin :

M=aFsin ,

which equals the area of the parallelogram made by a and F. Note that if the force is in the radial

direction (i.e. = 0), the force will not turn the disc (M= 0). As the angle increases up to 90

degrees, the disc tends to turn easily. Also note that the moment has an axis of rotation (the z-

axis). Like a screw in Fig. 1-6-17, the moment of a force is a vectorhaving both a magnitude and

a direction. Mathematicians call it an outer product and express it as follows.

M=a

A Ssimilar notion appears in electromagnetism. The other definition of the product of vectors (an

inner or scalar product) also plays an important role in mechanics, representing energy or work,

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which is a scalar. In Fig. 1-6-18, the force F and the displacement u makes the following work

W.

W=Fucos

O

F

M

M

uay

x

z

axF

A

Fig. 1-6-16 Force on a disc Fig. 1-6-17 Screw Fig. 1-6-18 Work

or friction energy

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1 6 three member sozen

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