ChapterSixStiffness：六章刚度.doc

Chapter Six: stiffness6.1 Member end releasesJoint stiffness is expressed in the master matrix for a structure, but two situations exist that may cause them to vary. First, a joint that is being utilized as a support may be released. For example, the support becomes slotted or pinned. Second, a member can be physically released from a joint for one or more of the six possible end displacements. When looking at a joint release, the joint is fully designed as a support before the release. This creates the reaction components. When a member release occurs, the member is released in some direction and the stiffness contribution that member was making to the joint changes or goes to zero. Take note that the member is released in the local system. Example 6.1-1Example 6.1-26.2 Non-prismatic members6.2.1 Rotation and TranslationFrom Castiglianos:Since I varies, let:Therefore:In matrix form, the flexibility matrix:Solving for PA & MA by the cofactor method:In Stiffness form:Transmit to the B end:By equilibrium:Carry-over factor:With rotation only:The distribution factor is a ratio of the rotational stiffness of a member to the sum of the rotational stiffness of all members at the joint. Rotational stiffness, the stiffness due to rotation at point A:Distribution Factor:Rotation and deflection at point B can be derived using the same method. The stiffness matrix is symmetric:In Stiffness form:Transmit to the B end:By equilibrium:Carry-over factor:With rotation only:The distribution factor is a ratio of the rotational stiffness of a member to the sum of the rotational stiffness of all members at the joint. Rotational stiffness, the stiffness due to rotation at point A:Distribution Factor:6.2.2 Fixed-end MomentsUniformly distributed load:let, therefore, By equilibrium: And, Uniformly distributed load:Concentrated Load:For either case, let, In matrix form:By equilibrium:Example 6.2.2-1Example 6.2.2-2Example 6.2.2-3Example 6.2.2-4Total Two-Dimensional Non-Prismatic Stiffness Matrix X-Z6.3 Shear stiffness6.3.1 Shear stiffnessFrom statics, Example 6.3.-1Example 6.3.-2Two-Dimensional Shear Stiffness MatrixThree-Dimensional Shear Stiffness Matrix6.3.2 Shear areasor, Example 6.3.2-16.4 Geometric stiffnessAn ordinary stiffness analysis, whether it includes shear deformations or not, makes no adjustments for the changing geometry of a loaded structure. Forces and moments are calculated from the original positions of the joints, not from their deformed positions. Elastic buckling, which is a function of joint deformations, is therefore impossible to predict using ordinary stiffness analysis. A procedure to include member and joint deformations in force and moment calculations has been developed by assuming a deformed shape and calculating the additional moment such a deformation would cause. The first is a beam used to derive the ordinary elastic stiffness matrix: the moment in the beam is a function of only the end shears and moments, as given by the following equation:The second is a beam showing bending deformations: the internal moment is a function of not only the end shears and moments, but of also the axial force multiplied by the beams lateral deflection, y: This additional moment, the product of axial force (Pix) and lateral deflection (y), is usually called the “P-delta effect”. To derive a stiffness matrix that includes the P-delta effect, one must consider the equilibrium of the deformed beam. Using the principle of superposition, one can first consider a beam with a known deflection while the rotation is held to zero. The internal bending moment in the beam is found from equilibrium:If y is assumed to be a cubic function, the four known boundary conditions give the following solution:And the internal moment is:As in the basic elastic stiffness derivation, it is assumed that axial shortening caused only by the axial force, Pix. This geometric stiffness derivation, therefore, only considers . From Castiglianos theorem, the rotation and deflection of the free end are:The derivatives of the expressions with respect to shear force and moment are:It should be noted that rotation and deflection are functions of moment only, since shear deformations are ignored in this derivation. Setting to zero and solving for Piy and Miz in terms of the deflection gives two terms in the stiffness matrix:Now consider a beam with a known rotation while the deflection is held to zero. Again, assuming y to be a cubic, the internal bending moment is found from the boundary conditions:And the rotation and deflection from Castiglianos theorem are:Setting to zero provides the other two terms in the stiffness matrix:Each term in the stiffness matrix contains two values. One value is a function of deformation only , and is the same as for the basic stiffness matrix. The second value is a function of both deformation and axial force (Pix); it is an additional “geometric” stiffness term. These equations can be written in matrix form as follows:The matrix on the left- the basic elastic stiffness matrix- will be called , and the matrix on the right- the geometric stiffness matrix- will be called , so that this equation can be rewritten:Where, is a vector of forces is a vector of corresponding vector of deformationsThrough appropriate transformations, keeping in mind that axial deformation , and torsional rotation are unaffected by the geometric stiffness component, this equation can be expanded to include the 12x12 elastic and geometric stiffness matrices and the force and deformation vectors. Example 6.4-1Example 6.4-2Geometric Stiffness Matrix6.5 TorsionSt Venant and Warping:According to St. Venants principle, the stress distribution on sections depends primarily on the magnitude of torque and not on the stress distrib