# Module 4. Analysis of Statically Indeterminate Structures by the Direct Stiffness Method. Version 2 CE IIT, Kharagpur

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1 Module Analysis of Statically Indeterminate Structures by the Direct Stiffness Method Version CE IIT, haragpur

2 esson 7 The Direct Stiffness Method: Beams Version CE IIT, haragpur

3 Instructional Objectives After reading this chapter the student will be able to. Derive member stiffness matrix of a beam element.. Assemble member stiffness matrices to obtain the global stiffness matrix for a beam.. Write down global load vector for the beam problem.. Write the global load-displacement relation for the beam. 7. Introduction. In chapter, a few problems were solved using stiffness method from fundamentals. The procedure adopted therein is not suitable for computer implementation. In fact the load displacement relation for the entire structure was derived from fundamentals. This procedure runs into trouble when the structure is large and complex. However this can be much simplified provided we follow the procedure adopted for trusses. In the case of truss, the stiffness matrix of the entire truss was obtained by assembling the member stiffness matrices of individual members. In a similar way, one could obtain the global stiffness matrix of a continuous beam from assembling member stiffness matrix of individual beam elements. Towards this end, we brea the given beam into a number of beam elements. The stiffness matrix of each individual beam element can be written very easily. For example, consider a continuous beam ABCD as shown in Fig. 7.a. The given continuous beam is divided into three beam elements as shown in Fig. 7.b. It is noticed that, in this case, nodes are located at the supports. Thus each span is treated as an individual beam. However sometimes it is required to consider a node between support points. This is done whenever the cross sectional area changes suddenly or if it is required to calculate vertical or rotational displacements at an intermediate point. Such a division is shown in Fig. 7.c. If the axial deformations are neglected then each node of the beam will have two degrees of freedom: a vertical displacement (corresponding to shear) and a rotation (corresponding to bending moment). In Fig. 7.b, numbers enclosed in a circle represents beam numbers. The beam ABCD is divided into three beam members. Hence, there are four nodes and eight degrees of freedom. The possible displacement degrees of freedom of the beam are also shown in the figure. et us use lower numbers to denote unnown degrees of freedom (unconstrained degrees of freedom) and higher numbers to denote nown (constrained) degrees of freedom. Such a method of identification is adopted in this course for the ease of imposing boundary conditions directly on the structure stiffness matrix. However, one could number sequentially as shown in Fig. 7.d. This is preferred while solving the problem on a computer. Version CE IIT, haragpur

4 Version CE IIT, haragpur

5 In the above figures, single headed arrows are used to indicate translational and double headed arrows are used to indicate rotational degrees of freedom. 7. Beam Stiffness Matrix. Fig. 7. shows a prismatic beam of a constant cross section that is fully restrained at ends in local orthogonal co-ordinate system x' y' '. The beam ends are denoted by nodes j and. The x' axis coincides with the centroidal axis of the member with the positive sense being defined from j to. et be the length of the member, A area of cross section of the member and inertia about 'axis. I is the moment of Two degrees of freedom (one translation and one rotation) are considered at each end of the member. Hence, there are four possible degrees of freedom for this member and hence the resulting stiffness matrix is of the order. In this method counterclocwise moments and counterclocwise rotations are taen as positive. The positive sense of the translation and rotation are also shown in the figure. Displacements are considered as positive in the direction of the coordinate axis. The elements of the stiffness matrix indicate the forces exerted on Version CE IIT, haragpur

6 the member by the restraints at the ends of the member when unit displacements are imposed at each end of the member. et us calculate the forces developed in the above beam member when unit displacement is imposed along each degree of freedom holding all other displacements to ero. Now impose a unit displacement along y' axis at j end of the member while holding all other displacements to ero as shown in Fig. 7.a. This displacement causes both shear and moment in the beam. The restraint actions are also shown in the figure. By definition they are elements of the member stiffness matrix. In particular they form the first column of element stiffness matrix. In Fig. 7.b, the unit rotation in the positive sense is imposed at j end of the beam while holding all other displacements to ero. The restraint actions are shown in the figure. The restraint actions at ends are calculated referring to tables given in lesson Version CE IIT, haragpur

7 In Fig. 7.c, unit displacement along y' axis at end is imposed and corresponding restraint actions are calculated. Similarly in Fig. 7.d, unit rotation about ' axis at end is imposed and corresponding stiffness coefficients are calculated. Hence the member stiffness matrix for the beam member is [ ] EI = EI EI EI EI EI EI EI (7.) The stiffness matrix is symmetrical. The stiffness matrix is partitioned to separate the actions associated with two ends of the member. For continuous beam problem, if the supports are unyielding, then only rotational degree of freedom Version CE IIT, haragpur

8 shown in Fig. 7. is possible. In such a case the first and the third rows and columns will be deleted. The reduced stiffness matrix will be, [ ] EI = EI EI EI (7.) Instead of imposing unit displacement along y' at j end of the member in Fig. 7.a, apply a displacement u ' along y' at j end of the member as shown in Fig. 7.5a, holding all other displacements to ero. et the restraining forces developed be denoted by, q q and q. q, Version CE IIT, haragpur

9 The forces are equal to, q = = (7.) u' ; q = u' ; q = u' ; q u' Now, give displacements u', u', u' and u' simultaneously along displacement degrees of freedom,, and respectively. et the restraining forces developed at member ends be q, q, q and q respectively as shown in Fig. 7.5b along respective degrees of freedom. Then by the principle of superposition, the force displacement relationship can be written as, q q q q = EI EI EI EI EI EI EI EI u' u' u' u' (7.) This may also be written in compact form as, {} [ ] { u' } q = (7.5) Version CE IIT, haragpur

10 7. Beam (global) Stiffness Matrix. The formation of structure (beam) stiffness matrix from its member stiffness matrices is explained with help of two span continuous beam shown in Fig. 7.6a. Note that no loading is shown on the beam. The orthogonal co-ordinate system xy denotes the global co-ordinate system. For the case of continuous beam, the x - and x' - axes are collinear and other axes ( y and y', and ' ) are parallel to each other. Hence it is not required to transform member stiffness matrix from local co-ordinate system to global co Version CE IIT, haragpur

11 ordinate system as done in the case of trusses. For obtaining the global stiffness matrix, first assume that all joints are restrained. The node and member numbering for the possible degrees of freedom are shown in Fig 7.6b. The continuous beam is divided into two beam members. For this member there are six possible degrees of freedom. Also in the figure, each beam member with its displacement degrees of freedom (in local co ordinate system) is also shown. Since the continuous beam has the same moment of inertia and span, the member stiffness matrix of element and are the same. They are, Global d. o. f ocal d. o. f [ '] ' ' ' ' ' ' ' ' = ' ' ' ' ' ' ' ' (7.6a) Global d. o. f ocal d. o. f 5 6 [ ] = 5 6 (7.6b) The local and the global degrees of freedom are also indicated on the top and side of the element stiffness matrix. This will help us to place the elements of the element stiffness matrix at the appropriate locations of the global stiffness matrix. The continuous beam has six degrees of freedom and hence the stiffness matrix is of the order 6 6. et [ ] denotes the continuous beam stiffness matrix of order 6. From Fig. 7.6b, may be written as, 6 [ ] Version CE IIT, haragpur

13 The loading on the beam shown in Fig. 7.8(a), is equal to the sum of Fig. 7.8(b) and Fig. 7.8(c). In Fig. 7.8(c), the joints are restrained against displacements and fixed end forces are calculated. In Fig. 7.8(c) these fixed end actions are shown in reverse direction on the actual beam without any load. Since the beam in Fig. 7.8(b) is restrained (fixed) against any displacement, the displacements produced by the joint loads in Fig. 7.8(c) must be equal to the displacement produced by the actual beam in Fig. 7.8(a). Thus the loads shown Version CE IIT, haragpur

14 in Fig. 7.8(c) are the equivalent joint loads.et, p, p, p, p, p5 and p6 be the equivalent joint loads acting on the continuous beam along displacement degrees of freedom,,,,5 and 6 respectively as shown in Fig. 7.8(b). Thus the global load vector is, p p p p p p 5 6 = Pab Pa w Pb w + Pba w + P w (7.8) 7.5 Solution of equilibrium equations After establishing the global stiffness matrix and load vector of the beam, the load displacement relationship for the beam can be written as, { } [ ]{ u} { P } { } P = (7.9) where is the global load vector, u is displacement vector and is the global stiffness matrix. This equation is solved exactly in the similar manner as discussed in the lesson. In the above equation some joint displacements are nown from support conditions. The above equation may be written as [ ] { p } { p } u = [ ] [ ] [ ] [ ] { u } u { u } (7.) { p } { } u where and denote respectively vector of nown forces and nown displacements. And { p u }, { u u } denote respectively vector of unnown forces and unnown displacements respectively. Now expanding equation (7.), Version CE IIT, haragpur

15 Since { } u [ ] { uu} [ ]{ u } [ ]{ u } [ ]{ u } { p } = + (7.a) p } = + (7.b) { u u is nown, from equation 7.(a), the unnown joint displacements can be evaluated. And support reactions are evaluated from equation (7.b), after evaluating unnown displacement vector. et, and 5 be the reactions along the constrained degrees of freedom as shown in Fig. 7.9a. Since equivalent joint loads are directly applied at the supports, they also need to be considered while calculating the actual reactions. Thus, p = p + 5 p5 [ ]{ } u u (7.) The reactions may be calculated as follows. The reactions of the beam shown in Fig. 7.9a are equal to the sum of reactions shown in Fig. 7.9b, Fig. 7.9c and Fig. 7.9d. Version CE IIT, haragpur

16 From the method of superposition, or Pb = + u + 6u6 (7.a) Pa = + u + 6u6 (7.b) w 5 = + P + 5u + 56u6 (7.c) Version CE IIT, haragpur

17 + + = u u P wl Pa Pb (7.a) Equation (7.a) may be written as, + + = u u P wl Pa Pb (7.b) Member end actions are calculated as follows. For example consider the first element.,,, q q q q [ ] + = u u b Pa Pa Pab Pb q q q q element (7.6) In the next lesson few problems are solved to illustrate the method so far discussed. Summary In this lesson the beam element stiffness matrix is derived from fundamentals. Assembling member stiffness matrices, the global stiffness matrix is generated. The global load vector is defined. The global load-displacemet relation is written for the complete beam structure. The procedure to impose boundary conditions on the load-displacement relation is discussed. With this bacground, one could analyse continuous beam by the direct stiffness method. Version CE IIT, haragpur

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