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mxco2uðz; tÞot2¼ 0 ð9ÞIn an identical fashion, the dynamic equilibrium relationship for motion in the x–z plane yields the secondgoverning differential equation asX nxj¼1GAxjo2uðz; tÞoz2þX nxj¼1GAxjyjo2uðz; tÞoz2 X nxj¼1mxjo2uðz; tÞot2 X nxj¼1mxjyjo2uðz; tÞot2¼ 0 ð10Þwhere nx is the number of frames running in the x direction and yj is the distance of frame j from the shearcentre, S. GAxj and mxj are the effective shear rigidity in the x direction and uniformly distributed mass per Eqs. (17) are now solved and posed in dynamic stiffness form. Although each equation was developed indi-vidually from a consideration of the planar shear beam of Fig. 5, they now describe the motion of a three-dimensional, shear–torsion coupled beam whose co-ordinate system and sign convention are shown inFig. 6. This beam (exact finite element) will replace a typical segment of the original, asymmetric, three-dimen-sional frame structure. The whole of the original structure can then be reconstituted by assembling the exactfinite elements corresponding to each segment in the usual way.Eq. (17) is solved on the assumption of harmonic motion, so that the instantaneous displacements can bewritten asuðz; tÞ¼ UðzÞ sinxt ð18aÞvðz; tÞ¼ V ðzÞ sinxt ð18bÞuðz; tÞ¼ UðzÞ sinxt ð18cÞwhere U(z), V(z)and U(z) are the amplitudes of the sinusoidally varying displacements and x is the circularfrequency. Substituting Eq. (18) into Eq. (17) and re-writing in non-dimensional form givesU00ðnÞþ x2k2xUðnÞ ycx2k2xUðnÞ¼ 0 ð19aÞV 00ðnÞþ x2k2yV ðnÞþ xcx2k2yUðnÞ¼ 0 ð19bÞU00ðnÞ ð1=r2mÞycx2k2uUðnÞþð1=r2mÞxcx2k2uV ðnÞþ x2k2uUðnÞ¼ 0 ð19cÞwherek2x ¼ mL2=GAx ð20aÞk2y¼ mL2=GAy ð20bÞk2u ¼ r2mðmL2=GJÞ ð20cÞandn ¼ z=L ð20dÞEqs. (19a)–(19c) can be re-written in the following matrix formD2þ x2k2x 0 ycx2k2x0 D2þ x2k2yxcx2k2y ð1=r2mÞycx2k2u ð1=r2mÞxcx2k2u D2þ x2k2u26 6 437 7 5UðnÞV ðnÞUðnÞ26 437 5 ¼ 0 ð21Þin which D = d/dn.Eq. (21) can be combined into one equation by eliminating either U, V or U to give the sixth-order differ-ential equationD2þ x2k2x 0 ycx2k2x0 D2þ x2k2yxcx2k2y ð1=r2mÞycx2k2u ð1=r2mÞxcx2k2u D2þ x2k2u W ðnÞ¼ 0 ð22Þwhere W = U, V or U.The solution of Eq. (22) is found by substituting the trial solution W(n)=esnto yield the characteristicequationb2þ k2x 0 yck2x0 b2þ k2yxck2y yck2u xck2u r2mðb2þ k2uÞ ¼ 0 ð23Þwhere b2=(s/x)2.Eq. (23) is a cubic equation in the frequency parameter b2and it can be proven that it always has threenegative real roots. Let these three roots be b21, b22 and b23, where b2j(j = 1,2,3) are all real and positive.Thereforesx 2¼ b2jgiving s ¼ ixbj ðj ¼ 1; 2; 3Þ where i ¼ffiffiffiffiffiffiffi 1pð24ÞIt follows that the solution of Eq. (22) can be written in the formW ðnÞ¼ C1 cos b1xn þ C2 sin b1xn þ C3 cos b2xn þ C4 sin b2xn þ C5 cos b3xn þ C6 sin b3xn ð25ÞEq. (25) represents the solution for U(n), V(n) and U(n), since they are all related via Eq. (21). They can bewritten inpidually as The Wittrick–Williams algorithm (Williams and Wittrick, 1970; Wittrick and Williams, 1971) has beenavailable for over thirty years and has received considerable attention. The algorithm states thatJ ¼ J0 þ sfKg ð37Þwhere J is the number of natural frequencies of the structure exceeded by some trial frequency, x*, J0 is thenumber of natural frequencies that would still be exceeded if all members were clamped at their ends so as tomake the nodal displacement vector D = 0 and s{K} is the sign count of the structure stiffness matrix K. s{K}is defined in reference (Wittrick and Williams, 1971) and is equal to the number of negative elements on theleading diagonal of the upper triangular matrix obtained from K, when x = x*, by the standard form of Gausselimination without row interchanges.From the definition of J0, it can be seen thatJ0 ¼XJm ð38Þwhere Jm is the number of natural frequencies of a member, with its end clamped, which have been exceededby x*, and the summation extends over all members of the structure. In the present case it is possible to deter-mine the value of Jm symbolically, using a direct approach, as follows.The end conditions for a clamped–clamped member ared1 ¼ d2 ¼ 0 ð39ÞIf Eq. (39) is substituted into Eq. (30) it is clear that the condition for non-trivial solutions isE00E I0CS ¼ 0 ð40ÞHowever, it is easy to show that the left-hand determinant can never be zero. Thus, noting that the right-handdeterminant is that of a lower triangular matrix, Eq. (40) is only satisfied when the product of its significantleading diagonal terms is zero, i.e.Y 3 6. Special case: uniform structuresWhen all storeys of a frame can be considered to be identical, the whole frame may be modelled as a single,substitute shear–torsion beam, which is clamped at one end and free at the other. The end conditions for sucha beam ared1 ¼ 0 ð44aÞp2 ¼ 0 ð44bÞEq. (44b) can be written in the following form using Eqs. (28)U0ðn ¼ 1ÞV 0ðn ¼ 1ÞU0ðn ¼ 1Þ26 437 5 ¼ 0 or d02 ¼ 0 ð45Þwhere d02 is the derivative of the vector of displacement functions when n =1.If Eqs. (44a) and (45) are substituted into Eq. (30), suitably differentiated, it is clear that the condition fornon-trivial solutions isb1b2b3x3E00E I0SC ¼ 0 ð46ÞHowever, it is easy to show that only the right-hand determinant can pass through zero for non-trivial solu-tions. Thus, noting that it has the form of a lower triangular matrix, Eq. (46) is only satisfied when the productof its significant leading diagonal terms is zero, i.e.Y 3j¼1Cbjx ¼ 0