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4.4 - Internal determinacy. A structure is internally determinate if the forces and moments

in all of the members can be determined based only on the equations of static equilibrium.Indeterminate structures require equations from strength of materials to introduce the

stiffness of the internal members in order to solve for these forces.

If a structure is composed of a single rigid body and if it is externally determinate, it ispossible to determine whether or not it is internally determinate. If a single rigid body

structure is externally indeterminate, it is also internally indeterminate. Fig. 4.4.1 shows

examples of single rigid body structures that are externally determinate and indeterminate.

(a) (b)

Fig. 4.4.1 - Externally determinate (a) and externally indeterminate (b) structures.

If a structure is composed of multiple rigid bodies (i.e. it is a compound structure) and if

it is externally determinate, the structure can be broken into individual rigid bodies. It isthen possible to determine whether or not each of the rigid bodies is internally

determinate. If any rigid body is externally indeterminate, it is also internally

indeterminate. Fig. 4.4.2 shows compound structures that are externally determinate and

indeterminate.

(a) (b)

Fig. 4.4.2 - Externally determinate (a) and externally indeterminate (b)

compound structures.

Once a rigid body, be it from a single rigid body structure or from a compound structure,

has been determined to be externally determinate, it is possible to assess its internal

determinacy based on simple rules. These rules, however, differ for different basic types

of structures. In the next section, we will look at analysis of trusses and the rulesassociated with determinacy, and in the following section we will look at beams andframes and the rules associated with determinacy.

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5 - Analysis of trusses.

A truss meets the following criteria:

(1) All components are connected at joints by pins, and

(2) loads are applied only at the pins.

If this is the case, M is imposed to be zero at every joint, resulting in the force in the barbeing directed along the axis of the bar. The two conditions that remain to be satisfied are

Fh= 0 and Fv= 0. This means that there will be 2j equations where j = number ofjoints. The number of unknowns will be the forces in the n truss components plus the

three reactions. Thus the number of unknowns is n + 3.

For a truss to be statically determinate, the number of equations must equal the number of

unknowns, or:

2j = n + 3 statically determinate

If there are more unknowns than equations, the truss is statically indeterminate:

2j < n + 3 statically indeterminate

The degree of indeterminacy is given by:

n + 3 2j degree of indeterminacy

5.2 - Approaches to analysis. Two principal approaches are available to analyze theforces in trusses: the method of joints and the method of sections. When deciding on how

to approach the analysis of a truss, several questions must be asked:

(1) Do you have a computer available or are you doing the analysis with a calculator? Ifyou are doing a hand calculation, the method of sections might be the best. However, ifthe member for which you want to calculate the force is near the truss support, the

method of joints might work well.

(2) Do you only need the forces in one or two members or do you want to solve for forcesin many members? The method of sections often works well when you are calculating the

forces in only one or two members. For forces in many members, the method of joints

probably works best.

(3) Do you plan on changing the loadings on the truss or do you only want to look at one

loading condition? If you plan on changing the loadings on the truss, you should use acomputer and apply the method of joints.

(4) Do you want to calculate the reactions first and then calculate the forces in the

components? If you plan on changing the loading on the truss, you should use a computerto perform a method of joints analysis, and leave the supports as unknowns when writing

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the equations. This is because changing the loading will change the forces in the supports

as well as the forces in the components.

5.3 - Method of joints for calculator solutions. Any time you are doing an analysis using a

calculator, you have to make your computations as efficient as possible. This means

avoiding simultaneous equations whenever possible. To do this, you usually need to startyour analysis at a joint that has as few components coming in as possible. Remember, the

force in each component is an unknown.

Also, there are often components in the trusses with zero force in them. If the force is

known to be zero, then that component does not add an additional unknown even if it

connects to the joint that you are analyzing.

5.3.1 - Determining zero force components. Sometimes inspection can be used to

determine components that have zero force. Fig. 5.3.1 shows a truss having zero force

members that can be determined by inspection.

Fig. 5.3.1 - Truss having zero force components.

Because there is no load at D, Fvshows component CD to have zero force. Then, by Fh

we can show that component ED is also zero force. Finally, Fvat joint E shows the force

is CE to be zero.

This leads to two rules for determining zero force components:

Rule 1: - If no external force is applied to a joint where only two components meet, both

components must be zero force.

Rule 2: - If no external load is applied to a joint where three components meet, two of

which are collinear, the component that is not collinear is a zero force component.

Further inspection of the truss in Fig. 5.3.1 shows that the zero force members previously

found result in propagation of zero force to other members. For example Fhat joint Eshows EF to also be a zero force component.

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Rule 2 applied to joint H shows BH to be a zero force component.

Finally, Fvat the roller at G shows component AG to be zero force.

Thus, there are five zero force components in the truss shown in Fig. 5.3.1.

5.3.2 - Exercises.

Determine the zero forces components for the trusses and loadings shown below. Give

reasons based on Rules 1 and 2. For each zero force component, discuss whether or notthe component serves a purpose. Under what conditions could the component be

removed?

(a)

(b)

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(c)

Assume support A is a pin, and support B is a roller.Assume all component connections are pins.

(d)

Fig. 5.3.2 Trusses with zero force members.

5.3.3 Determining forces by inspection. A somewhat special case of the method of

joints is determining forces in components by inspection. This is really just a matter of

picking out joints where relatively few (hopefully only two) non-zero force componentscome together, and where it is easy to calculate the force in one of the components

without resorting to solving simultaneous equations. Fig. 5.3.3 shows such a case.

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Fig. 5.3.3 Truss component forces by inspection.

In Fig. 5.3.3 it can be seen that member AB is the only member at joint B that can take a

vertical force. Thus the 30kip load must be borne by YAB, and thus FAB= 5/3(30) =50kips. This leads directly to FBC= (4/5)FAB= 4/5(50) = 40kips. The reaction at C is thendetermined to be 40kips directed horizontally to the left. Also note that AC is a zero force

member, therefore the vertical reaction at A is 30kips. Finally, summing horizontals give

the reaction at A to be 40kips directed horizontally to the right.

5.3.3 - Exercises. Take full advantage of zero force members and calculation of forces by

inspection to calculate the forces in the trusses shown below. In particular, find all forces

that you can find without solving simultaneous equations.

30kN

(a)

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(b)

(c)

(d)

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(e)

(f)

Fig. 5.3.4 Analysis by zero force and inspection.

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