4.5 The framework element stiffness matrix

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1 45 The framework element stiffness matri Consider a 1 degree-of-freedom element that is straight prismatic and symmetric about both principal cross-sectional aes For such a section the shear center coincides with the centroid Displacements resulting from uniform aial deformation fleural deformation and twisting deformation are considered Displacements resulting from transverse shearing deformations and those resulting from out-of-plane (longitudinal) warping of a cross-section that torsional forces may cause are neglected Cases in which transverse shear and torsional warping may be necessary or desirable discussed briefly in Sec 46 ethods for including them in the analysis in Sec 76 and Sec 74 respectively The restrictions described earlier greatly simplifies the development of the element stiffness matri in that they yield an analytical problem in which a number of effects are uncoupled ie a situation in which a particular force vector causes a displacement only in the same vector direction To obtain the stiffness matri four separate cases are considered: n aial force member pure torsional member beam bent about one principal ais beam bent about the other principal ais November /6

2 451 ial force member (The stiffness matri for the aial force member in local coordinates was developed in Sec 41 (eqn (3)) and can also be written using eqn (5) or the results of E 4 Nevertheless it will be redone here as a further illustration of the fleibility-stiffness transformations of Sec 44) Consider the member in Fig 47 where a stable statically determinate support system has d been prescribed Determine [ Φ ] and [ ] { Ff} { F} { Fs} { F 1} { Δ f} { u} { Δs} { u1} { Fs } = [ Φ]{ Ff } ie { F 1} [ 1] { F} = = = = pplying F the only displacement is u =- [ ] [ 1] Φ =- F F u e d d d E ie u F E E E ù \ [ d] = ê ëe ú û -1 T [ k ] = [ Φ][ d] [ Φ] = [ + ] -1 E êë k ff ù= úû [ d] = [] 1 T -1 T E fs êë úû = êë sf úû = [ d] [ Φ] = [-1] E ss 1 ì F k f ff fs Δ f From eqn (411) ü ì ü = F êk s sf k ú î þ ë ss û îδ s þ E 1-1ù \ [ k] = (46a) ê ë-1 1 ú û and ì F ü E 1-1ùì u ü = F 1 ê 1 1 ú î þ ë- ûî u 1þ Reorder rows and columns ìf ü E 1-1ùìu ü 1 1 = F ê 1 1 ú ë- ûî u þ î þ (46b) November /6

3 45 Pure torsional member From elementary strength of materials for a shaft subjected to a pure torque the rate of twist ie the change in rotation of the cross-section about the longitudinal ais per unit length along the ais is where = torque at the section considered G modulus of rigidity (shear modulus) and J torsional constant - a geometric property of the cross-section with dimensions length units to the fourth power For the special case of the circular cylindrical shaft it is equal to the polar moment of inertia of the cross-section Consider the member in Fig 48 where a stable statically determinate support system has d been prescribed Determine [ Φ ] and [ ] { Ff} { } { Fs} { 1} { Δ f} { } { Δs} { 1} { Fs } = [ Φ]{ Ff } ie { } [ 1] { } The total displacement is = = = = =- [ ] [ 1] 1 Φ =- d d ie 0 0 ù \ [ d] = ê ë ú û -1 T [ k ] = [ Φ][ d] [ Φ] = [ + ] -1 êë k ff ù= úû [ d] = [] 1 T -1 T fs êë úû = êë sf úû = [ d] [ Φ] = [-1] ss 1 November /6

4 From eqn (411) ì F k f ff ü ì fs Δ ü f = F êk s sf k ú î þ ë ss û îδ s þ and \ [ k] = 1-1ù (47a) ê ë-1 1 ú û ì ü 1-1ùì ü î þ ë ûî þ = 1 ê-1 1 ú 1 Reorder rows and columns ì 1 ü 1-1ù ì 1 ü = ê 1 1 ú î þ ë- û î þ (47b) November /6

5 453 Beam bent about -ais (Since this section also introduces the basic force-displacement relationships for the fleural member it is appropriate to discuss certain fundamental aspects of beam fleure and to define conditions and terminology that will be used here and at a later time) The stresses and strains at any cross-section caused by bending about the - ais are directed along the - ais of the member The strain e is e y d v =- =- y (48) d where is the radius of curvature approimated by 1/( dvd / ) lso = Ee hence dv =- Ey (49) d (- sign in eqn (48) because y - ais is upward for beam with positive curvature assume tensile strain is positive) The direct stress has a ero stress resultant ( d = 0 ) oment about - ais ò =-ò yd (430) (ssume positive bending moments to be those that cause positive curvature) In GZ tet moments are frequently referred to as to the stresses of bending their relationship to the actual stresses being given by eqn (430) From eqns (49) and (430) dv E y d d = ò (431a) ie dv dv = E y d= EI d ò (431b) d where I = ò - moment of inertia of the section about the - ais yd (If in eqn (431b) is regarded as the stress of bending then dvd / is correspondingly the strain of bending The elastic coefficient connecting stress to strain is no longer just E ; it is now EI The factor I results from the integration of behavior on the cross-section just as is the integrated bending effect of the stresses ) November /6

6 The development of the stiffness matri of a beam element is somewhat more difficult than that for the aial and torsional elements The stress of bending is related to the strain of bending by a second-order rather than first-order differential equation (eqn (431b)) Integration produces two constants of integration that must satisfy the specified support conditions lso several possibilities eist for the stable statically determinate support that is needed for the generation of a fleibility matri The most obvious choices are: (1) to place simple supports at each end (the simple beam) and () to fi either end leaving the other free (the cantilever) Different degrees of freedom are suppressed in each case and consequently different fleibility matrices will be obtained But the resulting complete stiffness matri must be the same since it includes all degrees of freedom Consider the cantilever system shown in Fig 49 (In this case the choice of support conditions is immaterial; it s a matter of personal preference But there are cases in which certain choices are preferable from the standpoint of integration and satisfaction of support conditions) From eqn (414) ie { Δ f } [ d]{ Ff } = ìv ü ìf ü î þ î þ y = [ d] The terms in [ d ] are from Eample 43 [ d] ù 3 = EI 1 êë úû The remainder of the process of forming [ k ] is in Eample 44 ì F ü y ùì v ü 1 - ú ú 6 6 ú EI ú = (43) F y v î þ ê ë ú ûî þ November /6

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