Stiffness method of structural analysis
- 1. SUBJECT :- STRUCTURAL ANALYSIS - 2 MATRIX ANALYSIS (STIFFNESS METHOD)
- 2. STIFFNESS METHOD • This method is a powerful tool for analyzing indeterminate structures. One of its advantages over the flexibility method is that it is conducive to computer programming. • Once the analytical model of structure has been defined, no further decisions are required in the stiffness method in order to carry out the analysis. • Stiffness method the unknowns are the joint displacements in the structure, which are automatically specified. • In the stiffness method the number of unknowns to be calculated is the same as the degree of kinematic indeterminacy of the structure. • Stiffness = Action Displacement
- 3. The essential features of stiffness method :- • This method is also known as displacement method or equilibrium method. • This method is a matrix version of classical generalized slope-deflection method. • Kinematically indeterminate structures are solved using this method. • Joint displacements are treated as primary unknowns in this method. • Numbers of unknowns id equal to the degree of kinematic indeterminacy of the structure. • The unknown joint displacements for a particular structure are uniquely defined. • Conditions of joint equilibrium are used to form equations in unknown displacement.
- 4. ACTIONS AND DISPLACEMENT • The term action and displacement are the fundamental concepts in structural analysis. An action(force) is most commonly a single force or couple. However, an, action may be also a combination of force and couple, a distributed loading, or a combination of these actions. • In addition to actions that are external to a structure, it is necessary to deal also with internal actions. These actions are the bending moment, shear force, axial force and twisting moment.
- 5. • The cantilever beam is subjected at end B to loads in the form of action 𝑃1 and 𝑀1. At the fixed end A the reactive force and reactive couple are denoted 𝑅 𝐴 and 𝑀𝐴, respectively. In calculating the axial force N, bending moment M, and shear force V at any section of the beam such as midpoint, it is necessary to consider the static equilibrium of the beam. One possibility to construct a free body diagram of the right-hand half of the beam, as show in fig-b. in so doing, it is evident that each of the internal actions appears in the diagram as a single force or couple.
- 6. • There situations, however, in which the internal actions appear as two forces or couples. This case occurs most commonly in structure analysis when a “release” is made at some point in a structure as shown in a fig for a continuous beam. If the bending moment is released at joint B of the beam, the result is the same as if a hinge were placed in the beam at the joint. In the order to take account of B.M. in the beam, it must be considered as consisting of two equal and opposite couples 𝑀 𝐵 that act on the left and right hand positions of the beam with the hinge at B.
- 7. • A displacement, which is most commonly a deflection or a rotation at some point in a structure. A deflection refer to the distance moved by a point in the structure, and a rotation means the angle of rotation of the tangent to the elastic curve at a point.
- 8. • Action is noted by A and displacement is noted by D. • Portrays a cantilever beam subjected to action 𝐴1, 𝐴2 and 𝐴3. The displacement corresponding to 𝐴1 and due to all loads acting simultaneously is denoted by 𝐷1 in fig-a, similarly, the displacements corresponding to 𝐴2and 𝐴3 are denoted by 𝐷2 and 𝐷3.
- 9. • Now consider the cantilever beam subjected to action 𝐴1 only the displacement corresponding to 𝐴1 in this beam is denoted by 𝐷11. The significance of the two subscripts is as follows. • the first subscript indicates that the displacement correspond to action 𝐴1 and the second indicates that the cause of the displacement is action 𝐴1. In a similar manner, the displacement corresponding to 𝐴2 in this beam is demoted by 𝐷21, where the first subscript shows that the displacement correspond to 𝐴2 and the second shows that it is caused by 𝐴1.also show in fig-b is the displacement 𝐷31 corresponding to the couple 𝐴3 and caused by 𝐴1. • 𝐷11 = 𝐴1 𝐿3 24𝐸𝐼 𝐷21 = 5𝐴1 𝐿3 48𝐸𝐼 𝐷31 = 𝐴1 𝐿2 8𝐸𝐼
- 10. SUPERPOSITION • In using the principle of superposition it is assumed that certain action and displacements cause other action and displacements to be developed in the structure. • In general terms principle states that the effect produced by several causes can be obtained by combining the effects due to the individual causes.
- 11. • The beam is subjected to load 𝐴1 and 𝐴2 which produce various action and displacement through out the structure. • for reaction 𝑅 𝐴, 𝑅 𝐵 and 𝑀 𝐵 are developed at the support, and displacement D is produced at the midpoint of the beam. The effect of the action 𝐴1 and 𝐴2 acting separately are shows in fig-b and fig-c.
- 12. • The beam has constant flexural rigidity EI and is subjected to the loads 𝑃1 , M, 𝑃2, and 𝑃3 . since rotation can occur at joints B and C ,the structure is kinematically indeterminate to the second degree when axial deformation are neglected. Let the unknown rotation at these joints be 𝐷1 𝑎𝑛𝑑 𝐷2, respectively, and assume that counterclockwise rotations are positive . These unknown displacement may be determined by solving equations of superposition for the action at joint B and C, described in the following discussion. • The restrained structure which is obtained by this means is shown in fig-b and consist of two fixed end beams. The restrained structure is assumed to be acted upon by all of the loads except those that correspond to the unknown displacement , thus, only the loads 𝑃1 , 𝑃2, and 𝑃3 are shows in fig-b. all loads that correspond to the unknown joints displacement, such as the couple Min this example, are taken into account later. The moments 𝐴 𝐷𝐿1 and 𝐴 𝐷𝐿2 are the action of the restrained corresponding to 𝐷1 and 𝐷2, respectively, and caused by loads acting on the structure .
- 13. • For example, the restrained action 𝐴 𝐷𝐿1 Is the sum of reactive moments at B due to the load 𝑝1 acting on member AB and the reactive moment at B due to the 𝑃2 Acting on member BC.
- 15. EXAMPLE • K.I. = 2 Let, θ 𝐵 = 𝐷1 θ 𝑐 = 𝐷2 AD = actions in actual structure corresponding to redundant AD1 = 0 AD2 = 0 ADL = actions in restrained structure due to loads corresponding to redundant. ADL1 = 𝑤𝑙 8 - 𝑤𝑙 8 = 24 ∗10 8 - 12 ∗10 8 = 15KN.m ADL1 = 𝑤𝑙 8 = 12 ∗10 8 = 15KN.m
- 16. • 𝑠11 = 4 𝐸𝐼 10 + 4 𝐸𝐼 10 = 0.8EI 𝑠21 = 2 𝐸𝐼 10 = 0.2EI 𝑠12 = 2 𝐸𝐼 10 = 0.2EI 𝑠22 = 4 𝐸𝐼 10 = 0.4EI S = EI 0.8 0.2 0.2 0.4 S = 0.8*0.4 – 0.2*0.2 = 0.28EI • 𝑠−1 = 1 S adjS = 1 0.28EI 0.4 −0.2 −0.2 0.8 D = -𝑠−1 * ADL = 1 0.28EI 0.4 −0.2 −0.2 0.8 * 15 15 θ 𝐵 = -10.71/EI θ 𝑐 = -32.14/EI