Statically Indeterminate Beams 2

8/9/2019 Statically Indeterminate Beams 2

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STATICALLY INDETERMINATE BEAMS

Elevated concrete roadway crossing a river. Roadways of this type can be modeled as

continuous beams resting on several supports. Continuous beams are statically indeterminate; they

are analyzed by considering deflections in addition to equilibrium.

A beam is statically indeterminate if the number of support reactions exceeds the number

of independent equilibrium equations. In general, two equilibrium equations are available for a

beam supporting lateral loads ( 0 0, A being an arbitrary point). Hence, astatically determinate beam has two support reactions, which is the minimum number needed to

keep the beam in equilibrium. Additional reactions, being nonessential for equilibrium, are known


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as redundant reactions. The number of redundant reactions is called the degree of indeterminacy

of the beam.

In our study of axial and torsional loading, we found that the solution of statically

indeterminate problems requires the analysis of compatibility of deformation as well as

equilibrium. For beams, the compatibility equations are derived from the constraints imposed on

the elastic curve by the supports.

We note that each support reaction corresponds to a constraint imposed by the support. For

example, a simple support provides a force that imposes the deflection constraint. A built-in

support provides two reactions: a force imposing the constraint on deflection, and a couple

imposing the rotational constraint. Thus, the number of support constraints and the number of

reactions are always equal.

DOUBLE INTEGRATION METHOD

Recall that in the method of double integration, we derived the equation for the elastic

curve of the beam by integrating the differential equation two times, resulting in

+ +

If the beam is statically determinate, it has two support reactions and thus two constraints

on its elastic curve. Because the reactions can be computed from the equilibrium equations, the

conditions of constraint are available to compute C1and C2, the two constants of integration. In a

statically indeterminate beam, each redundant reaction represents an additional unknown.

However, there is also an additional constraint associated with each redundancy, which, when

substituted into deflection equation, provides an extra equation.


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For example, the simply supported beam in Fig. 1 (a) is statically determinate. It has two

deflection constraints ( 0 0) and two support reactions (RAand RB), as shown inthe figure. The reactions can be determined from the equilibrium equations, so that the constraints

can be used to compute the constants C1and C2in the deflection equation.

Figure 1: Examples of determinate and indeterminate beams

By building in the support at A, as shown in Fig. 1 (b), we introduce the additional

constraint 0 and the reactive couple MA (a redundant reaction). Therefore, the beam isstatically indeterminate of degree one. The number of unknowns is now five: three support


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reactions (RA, RB, and MA) and two constants of integration (C1 and C2). The number of available

equations is also five: two equilibrium equations and three constraints shown in the figure.

In Fig. 1(c), we have added another support at C that has a small initial gap. Assumingthat the beam makes contact with the support at C when the loading is applied, we see that the

support introduces another redundant reaction RC and the corresponding constraint .Since there are now two redundant support reactions, the degree of static indeterminacy of the

beam is two. The number of available equations for determining the six unknowns (RA, RB, RC,

MA, C1, and C2) is also six: two equations of equilibrium and the four constraint conditions shown

in the figure.

The above discussion assumes that M is the global expression for the bending moment

(applicable to the entire beam). If the beam is divided into two or more segments with different

expressions for M, double integration will result in additional constants of integration. However,

there will be an equal number of new constraints in the form of continuity conditions (deflections

and slopes must be continuous across the junctions between the segments). Clearly, the method of

double integration can become tedious for statically indeterminate beams with multiple segments,

unless M is expressed in terms of bracket functions.


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SAMPLE PROBLEM

Determine all the support reactions for the propped cantilever beam in the figure.

SOLUTION

Equilibrium: The free-body diagram of the beam, shown below, yields the equilibrium

equations:

0 ; + 72000 ()

0 ; + (12) 7200(6) 0 ()

Fig (a)

Because there are three support reactions (RA, RB, and MA) but only two independent

equilibrium equations, the degree of static indeterminacy is one.


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Compatibility: A third equation containing the support reactions is obtained by analyzing

the deformation of the beam. We start with the expression for the bending moment, obtainable

from the free-body diagram below:

+ 6 0 0 2 .

Fig (b)

Substituting M into the differential equation for the elastic curve and integrating twice, we

get:

+ 300.

+

2 100 + .

2 +

6 25 + + .

Since there are three support reactions, we also have three support constraints. Applying

these constraints to the elastic curve, shown by the dashed line in Fig. (a), we get

1. |= 0 ( ) 02.|= 0 ( ) 03.|= 0 ( )

(12)

2 + (12)

3 25(12) 0 ()


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The solution of Eqs. (a)(c) is

Because the results are positive, the reactions are directed as shown in Fig. (a).

MOMENT-AREA METHOD

The moment-area method is well suited for deriving the compatibility equations for

statically indeterminate beams. If the total number of support reactions is n, the degree of static

indeterminacy of the beam is 2. A total of n equations are available for computing the supportreactions: two equilibrium equations and 2compatibility equations to be obtained from themoment-area theorems.

SAMPLE PROBLEM

The beam AB in the figure is built in at both ends and carries a uniformly distributed load

over part of its length. Compute all the support reactions acting on the beam.


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SOLUTION

The free-body diagram of the entire beam in Fig. (a) contains four unknown end reactions: the

forces RAand RBand the couples MAand MB. Because there are only two independent equilibrium

equations, the beam is statically indeterminate of degree two. It follows that two compatibility

equations are required for the solution.

Fig (a)

Equilibrium:From the free-body diagram in Fig. (a), the two independent equilibrium equations

are:

0 + 5400 ()

0 (12) +540(4.5) 0 ()

Compatibility:The elastic curve of the beam is shown in Fig. (b). Because the slope at each end

is horizontal due to the built-in supports, we conclude that the change in the slope between A and

B is zero. From the bending moment diagram by parts in Fig. (c), the first moment-area theorem

gives us:


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Fig (b)

Fig (c)

/ ] 12 (12) 12 13 (9)(2430) 0 ()

A second compatibility equation is obtained by noting that the tangential deviation /ofB with respect to A is zero (we could also have used the condition / 0). Applying the secondmoment-area theorem using the bending moment diagram in Fig. (c), we obtain

/ ] /

12 (12)(12)(4) 12(6) 13 (9)(2430)(2.25) 0 ()

Solving Eqs. (a)(d) gives

. . .


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BUCKLING OF LONG STRAIGHT COLUMNS

A building frame made of beams and

columns. A slender column may fail by

buckling well before the stress reaches

the yield point.

The term column is applied to a member

that carries a compressive axial load.

Columns are generally subdivided into

the following three types according to

how they fail:

Short columns fail by crushing (e.g.,

yielding). Even if loaded eccentrically, a

short column undergoes negligible

lateral deflection, so that it can be

analyzed as a member subjected to combined axial loading and bending.

Long columns fail by buckling. If the axial load is increased to a critical value, the initially

straight shape of a slender column becomes unstable, causing the column to deflect

laterally and eventually collapse. This phenomenon, which is known as buckling, can occur

at stresses that are smaller (often much smaller) than the yield stress or the proportional

limit of the material.


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Intermediate columns fail by a combination of crushing and buckling. Because this

mechanism of failure is difficult to analyze, intermediate columns are designed using

empirical formulas derived from experiments.

In analysis, these columns are treated as beams subject to axial load and bending, but with

one major difference: The effect of lateral deflections on equilibrium is no longer ignored. In other

words, the free-body diagrams are drawn using the deformed rather than the undeformed column.


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COMBINED LOADINGS

Concrete columns with protruding steel reinforcing rods. The reinforcement increases the

strength of concrete beams and columns. In particular, reinforcing steel is essential in resisting

tension, since the tensile strength of concrete is negligible.

In many structures the members are required to resist more than one kind of loading (combined

loading). These can often be analyzed by superimposing the stresses and strains cause by each load

acting separately.

Superposition of stresses and strains is permissible only under the following conditions:

a. The stresses and the strains must be a linear function of the applied loads (Hookeslaw

must be obeyed and the displacements must be small).

b. There must be no interaction between the various loads.


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Figure 2: (a) Rectangular bar carrying axial and lateral loads; (b)-(d) stress

distribution obtained by superimposing stresses due to axial load and bending;

(e) stress distribution f P/A > |M|c / I

Figure 2 (a) shows a bar of rectangular cross section that carries lateral loading and an axial

force P acting at the centroid of the cross section. If P were acting alone, it would cause the

uniformly distributed axial stress as

/on the typical cross section m-n of the bar, as shown

in Fig. 2 (b). The bending stress that results from the lateral loading acting by itself would be /, where M is the bending moment acting at section m-n. This stress is shown in Fig. 2 (c).When the axial and lateral loads act simultaneously, the stress s at any point on section m n is

obtained by superimposing the two separate effects:

+

which results in the stress distribution shown in Fig. 2 (d).


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The maximum tensile and compressive stresses on a cross section depend, of course, upon

the relative magnitudes of the two terms in the equation. When drawing the stress distribution in

Fig. 2 (d), we assumed that / ||/, theentire cross section would be in tension, as shown in Fig. 2 (e) (if P were a compressive force, then

the entire cross section would be in compression). If the cross section were not symmetric about

the neutral axis, the distances to the top and bottom fibers would, of course, have to be considered

when sketching stress distributions such as those in Figs. 2 (c) and (d).

Figure 3: When beam deflections are large, the contribution of the axial load to

the bending moment cannot be neglected.

The superposition implied in equation is valid only when the deformation of the bar is sufficiently

small so that displacements can be neglected in the computation of M. Referring to Fig. 3, we see

that if displacement is not neglected, P contributes to the bending moment at section m-n by the

amount , where is the lateral displacement of the bar at that section.

Consequently, we have

()


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where, as before, M is to be interpreted as the bending moment due to lateral loading acting alone.

We now see that the equation is valid only if is small compared to M. Note that if P is tensile(positive), its moment reduces the bending stress. The opposite effect occurs when P is

compressive (negative), when its moment increases the bending stress. These effects are negligible

for most structural members, which are usually so stiff that the additional bending stresses caused

by P can be ignored. However, in slender compression members (columns), the effects can be very

significant, requiring more exact methods of analysis.

Before the equations can be applied, equilibrium analysis must be used to determine the

axial load P and the bending moment M at the cross section of interest. When the normal stresses

at a particular cross section are needed, a free-body diagram exposing the force system at that

section will suffice. However, if the maximum normal stress is to be found, axial force and bending

moment diagrams will usually be required to locate the critical section.


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SAMPLE PROBLEMS

The wood beam ABCD in the fig. (a) carries two vertical loads. The beam is supported by

a pin at A and the horizontal cable CE. Determine the magnitude of the largest stress (tensile or

compressive) in the beam and its location. Neglect the weight of the beam.

Fig (a)

SOLUTION

Preliminary CalculationsThe cross-sectional dimensions of the beam are b = 6 in. and h

= 10 in., which yield the following cross-sectional properties:

6(10) 60

12 6(10)

12 500


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Equilibrium Analysis The free-body diagram of the beam is shown in Fig. (b). From the

equilibrium equation

Fig (b)

0 66000(4) 3000(12) 0

we obtain T =10 000 lb for the tension in the cable. The reactions at A can now be computed

from

0, which yields Ah=10 000 lb and Av= 9000 lb.


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To determine the axial force and bending moment at any location in the beam, it is

convenient resolve the forces in Fig. (b) in directions that are parallel and perpendicular to the

beam. The results are shown in Fig. (c). The equilibrium analysis of the beam is now completed

by constructing the axial force and bending moment diagrams in Figs. (d) and (e), respectively.

Computation of the Largest StressBecause the axial force is negative (compressive)

everywhere in the beam, the maximum compressive stress in the beam has a larger magnitude than

the maximum tensile stress. Inspection of the axial force and bending moment diagrams leads us

to conclude that the largest compressive stress occurs either on the cross section immediately

below point B or on the cross section immediately below point C. Which stress is larger can be

determined by computing the stresses at both sections.

At the section immediately below B, we have P = -13 400 lb and M = 6000 lb- ft. The

maximum compressive stress occurs at the top of the section, where the compressive stresscaused by P adds to the maximum compressive stress caused by the positive bending moment. If

we use y = h / 2 = 5 in , this stress is

13400

60 (600012)(5)

500 943


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The axial force and the bending moment that act at the section immediately below C are

P = -9800 lb and M = -12 000 lb -ft. For this case, the maximum compressive stress occurs atthe bottom of the section, where the compressive stress caused by P adds to the maximum

compressive stress caused by the negative bending moment. Using y = -h / 2 = -5 in yields

9800

60 (1200012)(5)

500 1603

Comparing the two values, we see that the largest stress in the beam has the magnitude

and it acts at the bottom of the section just below C.


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ECCENTRICALLY RIVETED CONNECTIONS

Bolted connection in a steel frame. The bolts must withstand the shear forces imposed on

them by the members of the frame. In eccentrically loaded joints, the connection is subjected to

applied loads that result in a line of action passing outside the center of rotation of the fastener

group. Some common examples are bracket-type connections, web splices in beams and girders,

and the standard beam connections. Because of the eccentricity of the applied load, the fastener

group is subjected to a shear force and a twisting moment. Both the moment and the shear force

result in shear stresses in the fasteners, and both of these effects have to be considered in

determining the capacity of the connection.


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For many years the analysis and design of eccentrically loaded fastener groups was based

on the assumption that rotation of the connection takes place about the center of gravity of the

fastener group and that the load versus deformation response of an individual fastener is linear. In

this method, the eccentric load is resolved into a shear load P acting through the centroid of the

fastener group and a torsional moment Pe, where e is the eccentricity of P with respect to the

centroid of the fastener group. The shear force acting through the centroid is assumed to be

distributed uniformly among the fasteners, as in other shear splices. The moment is assumed to

cause stresses in the fasteners that vary linearly with the distance from the fastener to the center of

gravity. The stress in any fastener is evaluated by vectorially adding the stress resulting from each

load component, that is, the con-centric shear force and the moment. The method further assumes

the connected plates to be rigid enough to remain essentially undeformed during twist. The

influence of the frictional resistance between the component parts of the connection is neglected.

Tests on eccentrically loaded riveted connections indicated that the elastic analysis yielded

a conservative design. On the basis of test results, the method was modified by introducing an

effective eccentricity, which is less than the actual eccentricity. Empirical formulas to determine

the effective eccentricity as a function of specific fastener patterns were developed. Reduction in

eccentricity yielded a factor of safety more compatible with the value used for shear alone. The

method is essentially based on the elastic behavior of the fastener group. Reducing the eccentricity

decreases the magnitude of the bending component and recognizes the actual strength of the joint

observed in tests.

Although use of the method just outlined produced safe designs using either the actual or

the so-called effective eccentricity, physical testing showed that the factor of safety with respect


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to the ultimate load was both variable and larger than that used for other types of bolted

connections.

Consider the group of bolts shown in Fig. 4 (a) subjected to a force F having an eccentricity

of e from the centroid C of the bolt group. To find the magnitude of shear force in each bolt, it is

convenient to decompose the loading in Fig 4 (a) to the statically equivalent loading in Fig. 4(b),

which in turn can be considered as the sum of the force F passing through the centroid C and a

couple of magnitude Fe as shown in Fig. 4 (c). The shear force in each bolt due to force F passing

through the centroid is equal to F divided by the number of bolts N. Now we calculate the shear

forces in the bolts due to the couple of magnitude Fe, using the following assumption:

1. The gusset plate is perfectly rigid.

2. The bolts are elastic.

Figure 4

The coupleMc = Fe causes the plate to rotate about the centroid of the bolt group. This rotation

produces strains in the bolts which are proportional to their distances from the centroid. Since


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bolts are assumed to behave elastically, shear forcesR1,R2, ,R5 developed in the bolts are

also proportional to their corresponding distance r1, r2, , r5 from the centroid (Fig 5). In other

words, if we denote the small angle of rotation of the plate by (in radian) and displacement of

boltj in the direction of shear forceRjby dj, we can write

(1) (2)

(3)

Figure 5

In Eqs. (2) and (3),K is a proportionality factor. By writing the moment equilibrium we obtain

=

= (4)

Deleting between Eqs. (2) and (4), we will have.

=


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This shear forceRj is normal to the line drawn from the centroid C to the boltj. It is often easier

to use the vertical and horizontal components of the shearRj. By selecting the origin of

coordinates at the centroid C

=

=

Finally, taking into account the contribution of the concentric force, we have

=

=


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SAMPLE PROBLEM

The plate shown in figure is fastened to the fixed member by five 10-mm-diameter rivets.

Compute the value of the loads P so that the average shearing stress in any rivet does not exceed

70 MPa

Solving for location of centroid of rivets:

Where


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The critical rivets are at distance r1from centroid:


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Mapua Institute of Technology

School of Civil, Environmental, and Geological Engineering

Research Paper

On

STRENGTH OF MATERIALS

(MEC32-1)

Statically Indeterminate Beams

Buckling of Long Straight Columns

Combined Loadings

Eccentrically Riveted Connections

Submitted by:

Cuevas, Dustin Glenn C.

2012104002

Submitted to:

Prof. Gerardo Usita


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Statically Indeterminate Beams 2

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