Geometric Stiffness Effects in 2D and 3D Frames


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1 Geometric Stiffness Effects in D and 3D Frames CEE 41. Matrix Structural Analsis Department of Civil and Environmental Engineering Duke Universit Henri Gavin Fall, 1 In situations in which deformations are not infinitesimall 1 small, linear elastic analses ma not capture the true structural response. For largestrain problems, results of finite deformation analsis is significantl more accurate than linear, infinitesimal deformation analsis. Incorporating the details of finite deformation, the analsis ma also be applied to a buckling analsis of the structural sstem. In the derivation of the linear elastic stiffness matrix for frame elements, the potential energ function includes strain energ due to bending, axial and shear deformation effects. Axial effects are decoupled from shear and bending effects in the resulting linear elastic stiffness matrices. 3 In finite deformation analsis, on the other hand, the potential energ function includes additional terms, which accounts for the interaction between the axial load effects on the frame element and the lateral deformation of the frame element. These effects are often called P effects. We will separate the potential energ function U into an elastic part U E (which contains the infinitesimal strain energ) and a geometric part, U G (which includes the interaction of lateral deformations and axial loads). The linear elastic strain energ results in the same frame element stiffness matrices k E that we have found previousl. So, this document focuses onl on the geometric component of the potential energ function. From this geometric part of the potential energ, we will derive the geometric stiffness matrix k G. As in the finite deformation analsis of trusses, we need to know the deformation of the structure in order to find the internal axial loads, but we need to know the internal axial loads to determine the geometric stiffness matrix and the deformations. This chickenandegg problem can be solved with the same tpe of NewtonRaphson iteration approach which we used previousl for finite deformation analsis of trusses. 1 infinitesimal means arbitraril close to zero, as in infinitesimal calculus. finite means neither infinite nor infinitesimal, as in a finite distance. 3 The rows and columns corresponding to axial effects (1st and 4th) have nonzero elements onl in the 1st and 4th columns and rows. Also, the rows and columns corresponding to bending and shear effects have nonzero elements onl in the nd, 3rd, 5th, and 6th columns and rows.
2 CEE 41. Matrix Structural Analsis Duke Universit Fall 1 H.P. Gavin 1 Deformed shape of a frame element in bending To start with, we need to introduce the deformed shape of a frame element. The deformed shape of a frame element, h(x), subjected to endforces, q, is a cubic polnomial. A cubic polnomial ma be written in a powerpolnomial form as follows: h(x) = a + a 1 x + a x + a 3 x 3. (1) ikewise, the slope of the beam, h (x) ma be expressed h (x) = a 1 + a x + 3a 3 x. () The polnomial coefficients, a,, a 3, satisf the end displacements and rotations of the Figure 1. The deformed shape of a beam, h(x) is assumed to be a cubic function of x. beam. Assuming small rotations, tan θ θ, and neglecting shear deformation effects, h() = u a = u h () = u 3 a 1 = u 3 h() = u 5 a + a 1 + a + a 3 3 = u 5 h () = u 6 a 1 + a + 3a 3 = u 6 These four equations with four unknowns (a,, a 3 ) have the solution a = u (3) a 1 = u 3 (4) a = 3(u 5 u )/ (u 3 + u 6 )/ (5) a 3 = (u 5 u )/ 3 + (u 3 + u 6 )/. (6)
3 Geometric Stiffness Effects in D and 3D Frames 3 You should be able to confirm this solution for the polnomial coefficients. Note that the cubic deformation function h(x) ma also be written as a weighted sum of cubic polnomials. h(x) = u b (x) + u 3 b 3 (x) + u 5 b 5 (x) + u 6 b 6 (x), (7) The weights u i are simpl the set of local element displacements and the functions b i (x) are each cubic functions in x. These cubic shape functions represent beam deformations due to a unit applied displacement in the corresponding coordinate (onl). Neglecting shear deformation effects, the frame element shape functions are the Hermite polnomials, b (x) = 1 3 (x/) + (x/) 3 b () = 1 b () = b () = b () = b 3 (x) = x (1 x/) b 3 () = b 3() = 1 b 3 () = b 3() = b 5 (x) = 3 (x/) (x/) 3 b 5 () = b 5() = b 5 () = 1 b 5() = b 6 (x) = (x/) (x/ 1) b 6 () = b 6() = b 6 () = b 6() = 1 Equations (1) and (7) are two different was of expressing exactl the same equation, h(x). The finite element method makes use of the form of equation (7). To complete the picture, for axial deformations, (which contribute to transverse deformations onl through the axial load in the geometric stiffness matrix), b 1 (x) = (1 x/) b 4 (x) = (x/). You should confirm that with the given definitions of b i (x), and the coefficients a i, that equations (1) and (7) are equivalent. Now, to introduce how this assumed deformation shape function can be used to find a potential energ function, let s recall the internal elastic strain energ of a beam due to bending effects, U B = 1 M (x) dx. (8) EI Now, since the curvature of the beam is M(x)/(EI) and assuming infinitesimal deformation, h (x) is practicall the same as the curvature, and U B = 1 M(x)h (x) dx (9) = 1 EI h (x) h (x) dx. (1) Equation (8) is an expression of the strain energ in terms of the internal bending moment; equation (9) is an expression of the strain energ in terms of the internal bending moment and the assumed cubic deformation function; and equation (1) is an expression of the strain energ in terms of the assumed cubic deformation function onl. In the finite element method it is common to express the potential energ using forms like equation (1). If the internal bending moment M(x) does indeed generate the assumed deformation function h(x) then all three forms of the elastic strain energ are exactl equivalent and completel interchangeable.
4 4 CEE 41. Matrix Structural Analsis Duke Universit Fall 1 H.P. Gavin Power polnomial basis Hermite polnomial basis 1 x/ 1 x/ 1 x/ 1 x/ 1 x/ 1 x/ 1 x/ 1 x/ 1 x/ 1 x/
5 Geometric Stiffness Effects in D and 3D Frames 5 Axial load effects in transverselloaded frame elements Turning now to our problem of determining the potential energ associated with axial loads and transverse displacements, recall the elastic strain energ due to axial loads: U A = 1 N (x) EA dx. (11) Since the incremental displacement du within a segment of length dx is du = N(x) EA dx, the internal strain energ due to axial effects ma be written U A = 1 N(x) du. (1) In this example, du is the elastic extension of the beam due to axial loads. If the beam element also has transverse displacements, h(x), the end of the beam will also displace in the direction of the axial load. With the usual smallangle approximation, sin θ θ and tan θ θ, Figure shows that du = dh dh dh = dx dx dh dx (13) dx Figure. Axial end displacement effects due to transverse displacement, ignoring deformation.
6 6 CEE 41. Matrix Structural Analsis Duke Universit Fall 1 H.P. Gavin 3 Geometric stiffness of frame elements The previous section shows that the potential energ due to axial loads, N(x) and transverse displacements, h(x), is U G = 1 N(x) dh dh dx. (14) dx dx If the axial load N(x) is constant over the length of the beam then the tensile force is T = N(x) = const., and U G = 1 T h (x) h (x) dx. (15) Substituting, and carring out the integral leads to the potential energ function in terms of transverse end displacements, u and u 5, and end rotations, u 3 and u 6, U G = T 3 ( u 3u 6 3u 5 u 3 3u 5 u 6 + 3u u 3 + 3u u 6 +18u 5 36u 5 u + 18u + u 3 + u 6). (16) Invoking Castigliano s theorem, the partial derivative of the potential energ function with respect to a displacement coordinate is the force in the direction of that displacement coordinate. The end forces due to geometric stiffness effects can then be found as follows: q = U G u = T 3 (36u + 3u 3 36u 5 + 3u 6 ) q 3 = U G u 3 = T 3 (3u + 4 u 3 3u 5 u 6 ) q 5 = U G u 5 = T 3 ( 36u 3u u 5 3u 6 ) q 6 = U G u 6 = T 3 (3u u 3 3u u 6 ). Writing these expressions in matrix form, we arrive at the geometric stiffness matrix for a frame element: q 1 u 1 6 q 6 q 3 = T u u 3, (17) q 4 u 4 q u 5 q 6 u where the tension in the beam is given b T = EA(u 4 u 1 )/. The geometric stiffness matrix for a D (planar) frame element in local coordinates is: 6 6 k G = T (18)
7 Geometric Stiffness Effects in D and 3D Frames 7 The coordinate transformation process is identical to the process carried out before for the elastic element stiffness matrix. The coordinate transformation matrix, T, is T = c s s c 1 c s s c 1, (19) where s and c are the sine and cosine of the counterclockwise angle from global element coordinate number 1 to the frame element. Here we are making the approximation that the deformed inclination of the frame element is approximatel the same as the original inclination of the frame element. The element stiffness matrix in global coordinates is found b appling the coordinate transformation matrix. 6 6 s sc s s 6sc s K G = T T k G T = T sc c c 6 sc c c 1 s c s c s 6sc s 6 6 s sc s sc 6 c c 6 sc 6 c c s c s c It is not hard to confirm this expression for K G, and ou should feel encouraged to do so. The assembl of the structural stiffness matrix K s with elastic and geometric effects proceeds exactl as with the elastic stiffness matrix. ()
8 8 CEE 41. Matrix Structural Analsis Duke Universit Fall 1 H.P. Gavin 4 Derivation of stiffness coefficients directl from the strain energ function It is common to derive the coefficients of a stiffness matrix directl from the strain energ function. Note that the i, j component of the stiffness matrix is k ij = u j q i = k ji = and that the i th component of the end force, q i, is q i = u i U. Therefore, the stiffness coefficients ma be written k ij = U u i u j. u i q j, If the stiffness matrix to be determined is for bending effects onl, then, as seen before, U = U B = 1 EI h (x) h (x) dx. Now, since integration and differentiation are both linear operations, it does not matter which is done first, integration or differentiation. Therefore, the stiffness coefficient ma be written, k Eij = 1 EI h (x) u i h (x) u j dx. The elastic stiffness matrix incorporating bending effects onl ma be determined directl from this expression. ikewise, the geometric stiffness matrix ma be determined directl from k Gij = 1 T h (x) u i h (x) u j dx. The coefficients for the elastic stiffness matrix and the geometric stiffness matrix for frame elements in three dimensions are derived using this method in the remaining sections of this document.
9 Geometric Stiffness Effects in D and 3D Frames 9 5 Cubic shape functions for beams including shear deformations Consider the twelve local coordinates of a three dimensional frame element. The transverse Figure 3. The twelve local coordinates of a threedimensional frame element. displacements in the local x plane, h (x), of an elastic beam ma be separated into a shearrelated component, h s (x) and a bendingrelated component h b (x), h (x) = h s (x) + h b (x). (1) The shear force at the end of the beam in the local direction, q ma be found in terms of the beam enddisplacements in the local direction and the endrotations about the local z axis, q = k u + k 6 u 6 + k 8 u 8 + k 1 u 1. The effective shear strain is simpl h s(x) = q GA s = 1 GA s (k u + k 6 u 6 + k 8 u 8 + k 1 u 1 ). () The internal bending moment, M z (x), due to the effects of the end displacements and end rotations is simpl M z (x) = q x q 6, and the curvature is approximatel h b(x) = 1 EI z [ (k u + k 6 u 6 + k 8 u 8 + k 1 u 1 ) x (k 6 u + k 66 u 6 + k 68 u 8 + k 6 1 u 1 ) ]. (3)
10 1 CEE 41. Matrix Structural Analsis Duke Universit Fall 1 H.P. Gavin in which small angles are assumed. B computing the potential energ function for shear and bending deformations, U = 1 EI z (h b(x)) dx + 1 GA s (h s(x)) dx, and taking the partial derivatives of the potential energ function with respect to the displacements coordinates u, u 6, u 8 and u 1, rows, 6, 8, and 1 of the elastic stiffness matrix ma be computed. The shape of the deformed beam ma therefore b found b integrating equations () and (3) to obtain h s (x) and h b (x) and b solving for the constants of integration using the end conditions. So doing, h (x) = h s(x) + h b(x) = 1 q + 1 ( ) 1 GA s EI z q x q 6 x + C 1. Inserting the end condition h () = u 6 the constant of integration, C 1, is EI z u 6. Integrating again, h (x) = h s (x) + h b (x) = 1 q x + 1 ( 1 GA s EI z 6 q x 3 1 ) q 6x + u 6 x + C. Inserting the boundar condition h() = u, and noting that 1/(GA s ) = Φ /(1EI z ), the deformed shape of the beam becomes h (x) = 1 ( 1 EI z 6 q x 3 1 q 6 x 1 1 Φ q x) + u 6 x + u. The transverse deformation shape function for a beam with bending and shear deformation in the x plane. h (x) = Φ { [x 3 3x Φ x + 3 (1 + Φ )] u + [x 3 ( + Φ /)x + 3 (1 + Φ /)x] u 6 + [ x 3 + 3x + Φ x] u 8 + [x 3 (1 Φ /)x Φ 3 x/] u 1 }. (4) For bending and shear deformations in the x z plane, the shape function ma be found using an analogous method, while respecting the righthand coordinate sstem. The transverse deflection, h z (x) will consist of shear and bending components, h z (x) = h s (x) + h b (x). The endshear force and the endbending moment arise for end displacements in the local z direction and end moments about the local axis, q 3 = k 33 u 3 + k 35 u 5 + k 39 u 9 + k 3 11 u 11,
11 Geometric Stiffness Effects in D and 3D Frames 11 and The effective shear strain is q 5 = k 53 u 3 + k 55 u 5 + k 59 u 9 + k 5 11 u 11. h s(x) = q 3 GA sz = 1 GA sz ( k 33 u 3 + k 35 u 5 + k 39 u 9 + k 3 11 u 11 ), and the bending curvature is approximatel h b(x) = 1 EI (q 3 x + q 5 ) = 1 EI [ (k 33 u 3 + k 35 u 5 + k 39 u 9 + k 3 11 u 11 ) x +(k 53 u 3 + k 55 u 5 + k 59 u 9 + k 5 11 u 11 ) ], where small angles are again assumed. Integrating h b(x) and combining with h s(x), h z(x) = 1 q ( ) 1 GA sz EI q 3 x + q 5 x + C 1. Inserting the end condition, h z() = u 5 gives C 1 = EI u 5. Integrating again, h z (x) = 1 q 3 x + 1 ( 1 GA sz EI 6 q 3 x ) q 5 x u 5 x + C. Now inserting the end condition h z () = u 3 gives C = u 3 and noting that 1/(GA sz ) = Φ z /(1EI ) the deformed shape ma be written h z (x) = 1 ( 1 EI 6 q 3 x q 5 x 1 1 Φ z q 3 x) u 5 x + u 3. Finall, the shape function for a frame element bending and shear in the local x z plane is h z (x) = Φ z { [x 3 3x Φ z x + 3 (1 + Φ z )] u 3 + [ x 3 + ( + Φ z /)x 3 (1 + Φ z /)x] u 5 + [ x 3 + 3x + Φ z x] u 9 + [ x 3 + (1 Φ z /)x + Φ z 3 x/] u 11 }. (5) Note that this expression is equivalent to equation (4) except for the fact that Φ z replaces Φ and that the signs of the u 5 and u 6 shape functions are reversed, as are the signs of the u 11 and u 1 shape functions. This is consistent with the righthand coordinate sstem. For axial displacements, the shape function is the same as for a truss, ( h x (x) = 1 x ) u 1 + x u 7. (6) ikewise, for torsional displacements, the shape function is analogous to the axial displacement shape function ( h θx (x) = 1 x ) u 4 + x u 1. (7)
12 1 CEE 41. Matrix Structural Analsis Duke Universit Fall 1 H.P. Gavin 6 The 3D elastic stiffness matrix for frame elements including shear and bending effects Differentiating and integrating the shape functions derived above, the threedimensional elastic stiffness matrix for frame elements in local coordinates including bending and shear deformation effects is: k E = EA 1EI z 6EI 3 (1+Φ ) z 1EI 6EI 3 (1+Φ z) (1+Φ z) GJ 6EI (4+Φ z)ei (1+Φ z) (1+Φ ) (1+Φ z) (4+Φ )EI z (1+Φ ) 6EI z (1+Φ ) EA 1EI z 6EI 3 (1+Φ ) z 1EI 6EI 3 (1+Φ z) (1+Φ z) GJ 6EI ( Φ z)ei (1+Φ z) 6EI z (1+Φ ) (1+Φ ) (1+Φ z) ( Φ )EI z (1+Φ ) EA 1EI z 6EI 3 (1+Φ ) z 1EI 6EI 3 (1+Φ z) (1+Φ z) GJ 6EI ( Φ z)ei (1+Φ z) (1+Φ ) (1+Φ z) ( Φ )EI z (1+Φ ) 6EI z (1+Φ ) EA 1EI z 6EI 3 (1+Φ ) z 1EI 6EI 3 (1+Φ z) (1+Φ z) GJ 6EI (4+Φ z)ei (1+Φ z) 6EI z (1+Φ ) (1+Φ ) (1+Φ z) (4+Φ )EI z (1+Φ ), where Φ = 1EI z GA s, and Φ z = 1EI GA sz.
13 Geometric Stiffness Effects in D and 3D Frames 13 7 Formulation of the geometric stiffness matrix from the cubic shape functions For the elements of the geometric stiffness matrix in rows,6,8, and 1, the potential energ function is U G = 1 T h (x) h (x) dx, where h (x) is given b equation (4). ikewise, for rows 3,5,9, and 11, the potential energ function is U Gz = 1 T h z(x) h z(x) dx, where h z (x) is given b equation (5). For rows 1 and 7, the potential energ function is where h x (x) is given b equation (6). U Gx = 1 T h x(x) h x(x) dx, 8 Torsion In the torsion of noncircular sections, torsional displacements result in axial deformation (warping) of the crosssection. In such cases, the work of the axial tension, T, moving through the warping displacements provides the potential energ function for rows 4 and 1, U Gθ = 1 T J x h A θx(x) h θx(x) dx, x where h θx (x) is given b equation (7) and J x is the torsional moment of inertia. The geometric stiffness coefficients ma then be found b forming the Hessian of the appropriate potential energ function, k Gij = U G u i u j.
14 14 CEE 41. Matrix Structural Analsis Duke Universit Fall 1 H.P. Gavin 9 The 3D geometric stiffness matrix for frame elements including shear and bending effects Differentiating and integrating the shape functions as described above, the threedimensional geometric stiffness matrix for frame elements in local coordinates including axial, bending, shear and torsional warping effects is: k G = T 6/5+Φ +Φ (1+Φ ) 6/5+Φ z+φ z (1+Φ z) /1 (1+Φ ) /1 (1+Φ z) J x A x /15+ Φ z/6+ Φ z /1 /1 /1 (1+Φ z) (1+Φ z) /15+ Φ /6+ Φ /1 (1+Φ ) (1+Φ ) 6/5 Φ Φ (1+Φ ) 6/5 Φ z Φ z (1+Φ z) /1 (1+Φ ) /1 (1+Φ z) Jx A x /3 Φ z/6 Φ z /1 /1 (1+Φ z) /1 (1+Φ ) (1+Φ z) /3 Φ /6 Φ /1 (1+Φ ) 6/5 Φ Φ (1+Φ ) 6/5 Φ z Φ z (1+Φ z) /1 (1+Φ ) /1 (1+Φ z) Jx A x /3 Φ z/6 Φ z /1 /1 /1 (1+Φ z) (1+Φ z) /3 Φ /6 Φ /1 (1+Φ ) (1+Φ ) 6/5+Φ +Φ (1+Φ ) 6/5+Φ z+φ z (1+Φ z) /1 (1+Φ ) /1 (1+Φ z) J x A x /15+ Φ z/6+ Φ z /1 /1 (1+Φ z) /1 (1+Φ ) (1+Φ z) /15+ Φ /6+ Φ /1 (1+Φ ), where T = EA(u 7 u 1 )/, Φ = 1EI z GA s and Φ z = 1EI GA sz.
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