BEAMS: STATICALLY INDETERMINATE
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1 LECTURE EMS: STTICLLY INDETERMINTE Third Edition. J. Clark School of Engineering Department of Civil and Environmental Engineering 18 Chapter 9.5 b Dr. Ibrahim. ssakkaf SPRING 003 ENES 0 Mechanics of Materials Department of Civil and Environmental Engineering Universit of Marland, College Park LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 1 eams Introduction Our previous analsis was limited to staticall determinate beams. beam, subjected onl to transverse loads, with more than two reaction components, is staticall indeterminate because the equations of equilibrium are not sufficient to determine all the reactions.
2 LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. eams Introduction In all of the problems discussed so far, it was possible to determine the forces and stresses in beams b utilizing the equations of equilibrium, that is F M F (9) LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 3 eams Introduction Coplanar Force Sstem F M F
3 LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 4 eams Staticall Determinate eam When the equations of equilibrium are sufficient to determine the forces and stresses in a structural beam, we sa that this beam is staticall determinate LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 5 eams eam When the equilibrium equations alone are not sufficient to determine the loads or stresses in a beam, then such beam is referred to as staticall indeterminate beam.
4 LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 6 eams Determinac of eams For a coplanar (two-dimensional) beam, there are at most three equilibrium equations for each part, so that if there is a total of n parts and r reactions, we have r = 3n, r > 3n, staticall determinate staticall indeterminate (30) LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 7 eams Eample 11 Classif each of the beams shown as staticall determinate of staticall indeterminate. If staticall indeterminate, report the degrees of of determinac. The beams are subjected to eternal loadings that are assumed to be known and can act anwhere on the beams.
5 LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 8 eams Eample 11 (cont d) I III II LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 9 eams Eample 11 (cont d) For part I: r 1 r r3 ppling Eq. 30, r = 3, n = 1, therefore, r = 3n, 3 = [3(1) = 3] staticall determinate
6 LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 10 eams Eample 11 (cont d) For part II: ppling Eq. 30, r 1 r = 5, n = 1, therefore, r r 3 r 4 r 5 r > 3n, 5 > [3(1) > 3] staticall indeterminate to second degree LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 11 eams Eample 11 (cont d) For part III: Note: r 3 = r 6 and r 4 = r 5 r 1 r r 3 r4 r 5 r 5 ppling Eq. 30, r = 6, n =, therefore, r = 3n, 6 = [3() = 6] staticall determinate r 6 r 7
7 LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 1 How to determine forces and stresses of transversel loaded beam that is staticall indeterminate? In order to solve for the forces, and stresses in such beam, it becomes necessar to supplement the equilibrium equations with additional relationships based on an conditions of restraint that ma eist. LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 13 In such cases the geometr of the deformation of the loaded beam is used to obtain the additional relations needed for an evaluation of the reactions (or other unknown forces). For problems involving elastic action, each additional constraint on a beam provides additional information concerning slopes or deflections.
8 LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 14 Such information, when used with appropriate slope or deflection equations, ields epressions that supplement the independent equations of equilibrium. Finding the deflection curve for staticall indeterminate beams requires no new theories or techniques. The unknown eternal reactions ma be treated simpl as ordinar eternal loads. LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 15 The deflection caused b these eternal loads can be found b an of the methods previousl discussed: direct integration, singularit functions, superposition, or b use of beam deflection tables. Since the presence of the eternal reactions places geometrical restrictions on the deflection curve, there will alwas be a sufficient number of boundar conditions to find the unknown reactions.
9 LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 16 General Rules Each staticall indeterminate beam problem has its own peculiarities as to its method of solution. ut there are some general rules and ideas that are common to the solution of most tpes of beam problems. LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 17 General Rules (cont d) These general rules and guidelines are summarized as follows: 1. Write the appropriate equations of equilibrium and eamine them carefull to make sure whether or not the beam problem is staticall determinate or indeterminate. Eq. 30 can help in the case of coplanar problems.. If the problem is staticall indeterminate, eamine the kinematic restraints to determine
10 LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 18 General Rules (cont d) the necessar conditions that must be satisfied b the deformation of the beam. 3. Epress the required deformations in terms of the loads or forces. When enough of these additional relationships have been obtained, the can be adjoined to the equilibrium equations and the beam problem can then be solved. LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 19 The Integration Method Consider the simpl supported cantilever beam that is subjected to a uniforml distributed load w w L Figure 35
11 LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 0 The Integration Method Drawing the free-bod diagram of the beam (see Fig. 36), we note that the reactions involve four unknowns, while onl three equilibrium equations are available, namel F M F (31) LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 1 The Integration Method w R L L/ wl Figure 36 M w R L R
12 LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. eams Consider beam with fied support at and roller support at. From free-bod diagram, note that there are four unknown reaction components. Conditions for static equilibrium ield F F M The beam is staticall indeterminate. lso have the beam deflection equation, EI = d M 0 0 ( ) d + C1 + C which introduces two unknowns but provides three additional equations from the boundar conditions: t, θ t = L, LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 3 The Integration Method The problem is obviousl indeterminate to the first degree because we have three unknown reactions and onl three equations of equilibrium. We know that in staticall indeterminate problem, the reactions ma be obtained b considering the deformation of the structure involved.
13 LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 4 The Integration Method We should, therefore, proceed with the computation of the slope and deformation along the beam. First, the bending moment M () at an given point of beam is epressed in terms of the distance from, the given load, and the unknown reactions. LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 5 The Integration Method Integrating in, epressions for the slope θ and the deflection, which contain two additional unknowns, namel, the constants of integration C 1 and C. ut altogether si equations are available to determine the reactions and the constants C 1 and C.
14 LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 6 The Integration Method These si equations are: The three equilibrium equations (Eq. 31) The three equations epressing that the boundar conditions are satisfied, i.,e., that slope and deflection at are zero, and that the deflection at is zero (Fig. 37). Thus the reactions at the supports ma be determined, and the equations for the elastic curve ma be obtained. LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 7 The Integration Method w [, θ ] [, ] L [ = L, ] Figure 37. Deflected Shape of the eam and the oundar Conditions
15 LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 8 Illustrative Eample using the Integration Method Determine the reactions at the supports for the simpl supported cantilever beam of Figure 35 in terms of w and L. w L LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 9 Illustrative Eample using the Integration Method Equilibrium Equations: From the free bod diagram of Fig. 38, we write F F M ; ; ; R R M + R R wl L + 1 wl (3a) (3b) (3c)
16 LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 30 Illustrative Eample using the Integration Method (cont d) R M L/ wl w R L R Figure 38. Free-bod Diagram for the eam LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 31 Illustrative Eample using the Integration Method (cont d) Equation of Elastic Curve: or Drawing the free-bod diagram of a portion of the beam (C) as shown in Fig. 39, we write + M C ; 1 M ( ) = w 1 M ( ) w + R M M + R (33)
17 LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 3 Illustrative Eample using the Integration Method (cont d) w R M R w w / C L V M() Figure 38. Free-bod Diagram for the portion C of the eam LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 33 Illustrative Eample using the Integration Method (cont d) Equating the epression for M() of Eq.33, to the curvature times EI, and integrating twice, gives 1 EI = w + R M EIθ = EI = w + R EI = w + R M M + C C + C (34a) (34b) (34c)
18 LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 34 Illustrative Eample using the Integration Method (cont d) Referring to boundar conditions shown in Fig. 37, we make, θ in Eq. 34b,, = 0 in Eq. 34c, and conclude that C 1 = C. Thus, Eq. 34c can be rewritten as follows to the elastic curve epression: 4 3 w R M EI = + (35) 4 6 LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 35 Illustrative Eample using the Integration Method (cont d) ut the third boundar condition requires that for = L. Therefore, substituting these values into Eq. 35, gives 4 wl R L EI(0) = or 3M R wl M L (36)
19 LECTURE 18. EMS: STTICLLY INDETERMINTE (9.5) Slide No. 36 Illustrative Eample using the Integration Method (cont d) Solving this equation simultaneousl with the three equilibrium equations (Eq. 3), the reactions at the supports are determined as follows: R M = 1 8 wl R R 5 = wl 8 3 = wl 8
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