CE371 Structural Analysis II Lecture 5:

Transcription

1 CE371 Structural Analysis II Lecture 5: ) Preliminary Remarks 15.2) Beam-Member Stiffness Matrix 15.3) Beam-Structure Stiffness Matrix 15.4) Application of the Stiffness Matrix.

2 15.1) Preliminary Remarks Applying the stiffness method on beams requires that we subdivide the beams into beam finite elements (subelements) that are a) Prismatic b) Free of Loads In order to satisfy these conditions, beam finite elements are selected so that their nodes are located at points where Member is supported (supports) Members are connected External loads are applied Cross-section suddenly changes Displacement is to be determined The global coordinate system x, y, z is selected such one node at the origin and all other nodes in the beam have positive coordinates. local coordinates x, y, z will be selected so that x is directed towards the far end node. 1

3 Kinematic Indeterminacy If we only consider the effects of bending and shear, then each node can have only two dof s: a) Vertical displacement (N y, F y ) b) Rotation (N θ, F θ ) When numbering these dof s, we ll give the lower code numbers to free f dof s and higher code numbers to supported s dof s: Free f dof s: known (prescribed) forces, unknown displacements Supported f dof s: unknown forces, known (prescribed) displacements The number of f dof s is the degree of kinematic indeterminacy. 4 th degree kinematic indeterminacy 5 th degree kinematic indeterminacy 2

4 Internal Hinges & Sliders When there is an internal hinge, beam ends at both sides of the hinge will have same deflection but different rotations. Thus, the node must be represented by 3 dof s instead of 2: Deflections: 1 Rotations: 2 When there is an internal slider, beam ends at both sides of the slider will have different deflection, but same rotation. Thus, the node must be represented by 3 dof s instead of 2: Deflections: 2 Rotations: 1 3

5 Intermediate Loadings Beams are frequently loaded with distributed or point loading between its joints. To satisfy condition (b), that is members must be free of load, any distributed load or concentrated load be nodes must be converted to a set of equivalent nodal loads. Equivalent nodal loads, q E, are assumed to be equal but opposite to the fixed-end reactions, q F, so that q E = -q F, where the fixed reactions are taken from the following figure. The beam end forces can now be calculated as q = kk + q F (1a) or q = kk q E (1b) or q q F = kk (1c) or q + q E = kk (1d) 4

6 15.2) Beam-Member Stiffness Matrix Consider the following beam: Where: q and d Ny N y : near end vertical force and displacement q and d Nθ N θ : near end rotational force and displacement q and d Fy F y : far end vertical force and displacement q and d Fθ F θ : far end rotational force and displacement The stiffness component of each dof can be evaluated by fixing all dof s except one at a time, which is displaced by 1 unit, and calculating the force resulting from this displacement. 5

7 Beam-Member Stiffness Components Consider the following modes for the beam-member: d = 1 Ny d = 1 Fy d Nθ = 1 d Fθ = 1 Since q q F = k, then q q F Ny N y q q F Nθ N θ q q F Fy F y q q F Fθ F θ = 2EE L 6 3 L 2 L 3 L 6 3 L 2 L 3 L 2 3 L 6 L 2 3 L 6 L L 1 3 L 2 d Ny d Nθ d Fy d Fθ (2) 6

8 15.3) Beam-Structure Stiffness Matrix Once every beam-member stiffness matrix and fixed-endforce is determined, the global stiffness equation can be assembled as then partitioned as following: F Q f Q f Q s Q = K ff K ff D f (3) F K s ff K ss D s The free dof s displacements can now be found from D f = K ff 1 Q f Q f F K ff D s (4) The supported dof s forces (reactions) can be found from Q s = K fs D s + K ss D s + Q s F (5) The end forces can now be calculated from Eq. (2). 7

9 Problem 15.1 Determine the moments at 1 and 3. Assume 2 is a roller and 1 and 3 are fixed. EI is constant. 8

10 Problem 15.2 Determine the moments at 1 and 3 if the support 2 moves upward 5 mm. Assume 2 is a roller and 1 and 3 are fixed. EI = 60(10 6 ) N.m 2. 9

11 Problem 15.8 Determine the reactions at the supports, and the shear on the right node of member 1 and on the left of member 2. EI is constant. 10

12 Problem 15.8 Determine the reactions at the supports, and one the right end of member BC and on the left end of member CD. Assume A is fixed. EI is constant. 11