solving statically indeterminate stucture by stiffnes method
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solving statically indeterminate stucture by stiffnes method

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solving statically indeterminate stucture by stiffnes method solving statically indeterminate stucture by stiffnes method Presentation Transcript

  • CE-416 Pre-stressed Concrete Lab Sessional Presented By: ABU SYED MD. TARIN Id No:10.01.03.020 Course teachers: Munshi Galib Muktadir Sabreena Nasrin Department Of Civil Engineering Ahsanullah University of Science & technology 12/17/2013 1
  •  Stiffness method is an efficient way to solve complex determinant or indeterminant structures . It will introduced that is a modern method for structural analysis. which is a powerful engineering method and has been applied in numerous engineering fields such as solid mechanics and fluid mechanics. it is also called the displacement method. 12/17/2013 2
  •  A statically indeterminate structure that the reaction and internal forces cannot be analyzed by the application of the equation of static alone . The indeterminacy of the structure may be either external ,internal or both . The space structute is externally indeterminate if the number of the reaction components is more than six. 12/17/2013 3
  •  Identify degree of kinematic indeterminacy (doki)  Apply restraints and make it kinematically determinate  Apply loads on the fully restraint structure and calculate forces.  Apply unknown displacements to the structure one at a time keeping all other displacements zero and calculate forces corresponding to each dof.  Write the equilibrium equations. solve the equation in matrix form and obtain the value of unknown displacements.  Calculate other reactions.(by super position) 12/17/2013 4
  •       The number of possible directions that displacements or forces at a node can exist in is termed a degree of freedom (dof ). Some examples are: Plane truss: has 2 degrees of freedom at each node: translation/forces in the x and y directions. Beams: have 2 degrees of freedom per node: vertical displacement/forces and rotation/moment. Plane Frame: has 3 degrees of freedom at each node: the translations/forces similar to a plane truss and in addition, the rotation or moment at the joint. Space Truss: a truss in three dimensions has 3 degrees of freedom: translation or forces along each axis in space. Space Frame: has 6 degrees of freedom at each node: translation/forces along each axis, and rotation/moments about each axis. 12/17/2013 5
  •  For frame ,doki=8*3-(3+2)=19  For beam, doki=4*2-(2+1)=5  For truss, doki=6*2-(2+2+1)=7 12/17/2013 6
  •  For the case of symmetry & antysymmetry, use of modified stiffness makes the problem easier. Previously, it was derived that stiffness factor =3EI/L when the far end is hinged. This is also modified stiffness.  Modified Stiffness K’=2EI/L=K/2 (For symmetry). K’=6EI/l=3k/2 (For Antisymmetry). K’= 3EI/L (When Far End Hinged) Determination Of Stiffness Factor  Case(i)When the far end is hinged: K=4EI/L.  Case(ii)When the far end is fixed : MAB=3EI/L. 12/17/2013 7
  • Direct Stiffness Method for Truss Analysis  The members are straight, slender, and prismatic. The crosssectional dimensions are small in comparison to the member lengths.  The joints are assumed to be frictionless pins (or internal hinges).  The loads are applied only at the joints in the form of concentrated forces. Direct Stiffness Method for Frame Analysis  The members are slender and prismatic. They can be straight or curved, vertical, horizontal, or inclined.  The joints can be assumed to be rigid connection, frictionless pins (or internal hinges), or typical connections.  The loads can be concentrated forces or moments that act at joints or on the frame members, or distributed forces acting on the members. 12/17/2013 8
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  • Solution  In the first step identify the degrees of freedom of the frame .The given frame has three degrees of freedom (i) Two rotations as indicated by U1 and U2  (ii) One horizontal displacement of joint B and C as indicated by U3  In the next step make all the displacements equal to zero by fixing joints B and C as shown in Fig.23.5c. On this kinematically determinate structure apply all the external loads and calculate reactions corresponding to unknown joint displacements . 12/17/2013 10
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