Chapter 3

LECTURE 4Statically Indeterminate StructuresAdvantages & DisadvantagesFor a given loading, the max stress and deflection of an indeterminate structure are generally smaller than those of its statically determinate counterpart.Statically indeterminate structure has a tendency to redistribute its load to its redundant supports in cases of faulty designs or overloading.1Statically Indeterminate StructuresAdvantages & Disadvantages (contd)Although statically indeterminate structure can support loading with thinner members & with increased stability compared to their statically determinate counterpart, the cost savings in material must be compared with the added cost to fabricate the structure since often it becomes more costly to construct the supports & joints of an indeterminate structure.One has to careful of differential displacement of the supports as well.2Statically Indeterminate StructuresMethod of AnalysisTo satisfy equilibrium, compatibility and force-displacement requirements for the structure.Force MethodDisplacement Method3Force Method of Analysis: General ProcedureConsider the beam shown in Fig. 10.3(a)From free-body diagram, there would be 4 unknown support reactions3 equilibrium equationsBeam is indeterminate to first degreeTo obtain the additional equation, use principle of superposition & consider the compatibility of displacement at one of the supportsThis is done by choosing one of the support reactions as redundant & temporarily removing its effect on the beam4Force Method of Analysis: General ProcedureThis will allow the beam to be statically determinate and stable.Here, we will remove the rocker at B.As a result, the load P will cause B to be displaced downward as shown in Fig. 10.3(b).By superposition, the unknown reaction at B causes the beam at B to be displaced upward, Fig. 10.3(c).5Force Method of Analysis: General ProcedureFig 10.3

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5-2 Force Method of Analysis: General ProcedureAssuming +ve displacement act upward, we write the necessary compatibility equation at the rocker as:

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5-2 Force Method of Analysis: General ProcedureUsing methods in Chapter 8 or 9 to solve for B and fBB, By can be found.Reactions at wall A can then be determined from equation of equilibrium.The choice of redundant is arbitrary.8Force Method of Analysis: General ProcedureThe moment at A, Fig. 10.4(a) can be determined directly by removing the capacity of the beam to support moment at A, replacing fixed support by pin support.As shown in Fig .10.4(b), the rotation at A caused by P is A.The rotation at A caused by the redundant MA at A is AA, Fig. 10.4(c).9Force Method of Analysis: General ProcedureFig 10.4

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Force Method of Analysis: General Procedure11

Maxwells Theorem of Reciprocal Displacement: Bettis LawThe displacement of a point B on a structure due to a unit load acting at point A is equal to the displacement of point A when the load is acting at point B.

Proof of this theorem is easily demonstrated using the principle of virtual work.12

Maxwells Theorem of Reciprocal Displacement: Bettis LawThe theorem also applies for reciprocal rotations.The rotation at point B on a structure due to a unit couple moment acting at point A is equal to the rotation at point A when the unit couple is acting at point B.13Example 10.1Determine the reaction at the roller support B of the beam shown in Fig. 10.8(a)EI is constant.Fig 10.814

Example 10.115

Example 10.1 - solutionPrinciple of superpositionBy inspection, the beam is statically indeterminate to the first degree.The redundant will be taken as By.Fig. 10.8(b) shows application of the principle superposition.We assume By acts upward on the beam.16Example 10.1 - solutionCompatibility equation

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Example 10.4Draw the shear & moment diagrams for the beam shown in Fig. 10.11(a).EI is constant.Neglect the effects of axial load.

18Example 10.4Fig 10.11

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Example 10.4Fig 10.11

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Example 10.4 - solutionPrinciple of SuperpositionSince axial load is neglected, the beam is indeterminate to the second degree.The 2 end moments at A & B will be considered as the redundants.The beams capacity to resist these moments is removed by placing a pin at A and a rocker at B.The principle of superposition is applied to the beam as shown in Fig. 10.11(b).21Example 10.4 - solutionCompatibility equationsReference to points A & B, Fig 10.11(b) requires22

Example 10.4 - solutionCompatibility equations (contd)The required slopes and angular flexibility coefficients can be determined using standard tables.23

Example 10.4 - solutionCompatibility equations (contd)

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Example 10.4 - solutionUsing these results, the end shear are calculated, Fig. 10.11(c).The shear and moment diagrams plotted.

25Example 10.5Determine the support reactions on the frame shown in Fig. 10.13(a).EI is constant.26Example 10.5Fig 10.13

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AExample 10.5Fig 10.13

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Example 10.6 - solutionPrinciple of SuperpositionThe frame is statically indeterminate to the first degree.By choosing the reaction at B to be redundant, the pin at B is replaced by a roller.The principle applied is as shown in Fig. 10.13(b).29Example 10.6 - solutionCompatibility EquationReference to point B in Fig 10.13(b) requires

The terms B and fBB will be computed using the method of virtual work.The frames x coordinates & internal moments are shown in Figs. 10.13(c) and 10.13(d).30

Example 10.6 - solutionCompatibility Equation (contd)31

Example 10.6 - solutionCompatibility Equation (contd)32