Lecture 8: Flexibility Method. Example

Transcription

1 ecture 8: lexibility Method Example The plane frame shown at the left has fixed supports at A and C. The frame is acted upon by the vertical load P as shown. In the analysis account for both flexural and axial deformations. The flexural rigidity EI is constant. The axial rigidity is also constant. Joint B is a rigid connection. The structure is statically indeterminate to the third degree. A released structure is obtained by cutting the frame at joint B and the released actions Q 1, Q and Q are redundants. ind the magnitude and direction of these redundants.

2 ecture 8: lexibility Method The displacements in the released structure caused by P and corresponding to Q 1, Q and Q are in the previous figures. The displacement D Q1 consists of the sum of two translations which are found by analyzing the released structure as a set of two cantilever beams and. irst analyze the cantilever beam. The load P will cause a downward translation at B and a clockwise rotation at B. There is no axial displacement thus 0 D Q 1 The displacements D Q and D Q also consist of two vertical displacements and two rotations. D Q 5P 48EI DQ P 8EI

3 ecture 8: lexibility Method Since there is no load on member, there will be no displacement at end B and D D D 0 Q1 Q Q The D Q matrix becomes D Q 0 5P 48EI P 8EI P 48EI We now want to determine the flexibility matrix. Consider the released structure with Q 1 1

4 ecture 8: lexibility Method The displacements corresponding to Q 1, Q and Q are shown as flexibility coefficients 11, 1 and 1. If both axial and flexural deformations are considered, the displacements at end B of member are AE The displacements at end B of member BC are EI EI The flexibility coefficients become AE 11 EI EI

5 ecture 8: lexibility Method The same analysis must be made with Q 1 or frame sections and : EI 0 1 EI AE 0 0 : 1 Which leads to flexibility coefficients 1 0 EI AE EI

6 ecture 8: lexibility Method Once again from frame sections and (Q = 1) : EI 0 1 EI : EI 1 0 EI leading to flexibility coefficients 1 EI EI EI EI

7 ecture 8: lexibility Method Assembly of the flexibility matrix leads to 0 EI EI EI 0 EI EI EI EI EI Now let P 10 K 1 ft 144 inches E I 0, in 4 ksi A 10 in

8 ecture 8: lexibility Method With these numerical values and D Q EI 144 0, , in / kip in / kip The axial compliance (flexibility) of each component is quite small relative to the flexural compliance (flexibility). We will ignore the axial compliance of the beam and the column when assembling the flexibility matrix. The inverse of compliance is stiffness. This is equivalent to stating that the axial stiffness of the beam and column is so large relative to flexural stiffness of the beam and column that axial displacements are negligible.

9 ecture 8: lexibility Method Omitting axial deformations leads to the following flexibility matrix When the flexibility matrix is inverted we obtain

10 ecture 8: lexibility Method With -1 and D Q we can compute the unknown redundants utilizing the matrix equation i.e., 1 Q D Q D Q Q kips 4.06 kips 90.0 kip inches Note that the displacements associated with the redundants in the original structural, represented by the matrix {D Q } are zero. Because joint B is a rigid connection one can rationalize the rotation D Q is zero from this assumption. The other two displacements are zero because the axial stiffness of the beam and column is so large.

11 ecture 8: lexibility Method Example The plane truss shown to the left is statically indeterminate to the second degree. The horizontal reaction at support B (positive to the right) and axial force in bar AD (positive in tension) are selected as redundants. ind these redundants. The cut bar remains part of the released structure since its deformations must be included in the calculations of displacements in the released structure.

12 ecture 8: lexibility Method It should be noted that Q consists of a pair of forces acting on the released structure. A displacement corresponding to Q consists of the relative translation of the ends of the bar AD. When the ends of bar AD displace toward one another the displacements are in the direction of Q and thus are positive. When the ends move away the displacements are negative. The first step in the analysis is determining the displacements that correspond to Q 1 and Q in the released structure due to external loads. These displacements are denoted D Q1 and D Q and are depicted in the previous slide. Assuming that all the members have the same axial stiffness, then from application of Castigliano s theorem D Q 1 P.88 1 P D Q P Please verify these quantities for homework. Note the minus signs.

13 ecture 8: lexibility Method The next step will be the determination of the displacements associated with Q 1 and Q in the released structure due to unit loads at Q 1 and Q, i.e., determine the flexibility coefficients. The flexibility coefficient 11 is the displacement corresponding to Q 1 and caused by a unit value of Q 1. Thus The flexibility coefficient 1 is the displacement corresponding to Q and caused by a unit value of Q 1. Thus

14 ecture 8: lexibility Method The flexibility coefficient is the displacement corresponding to Q and caused by a unit value of Q. Thus The flexibility coefficient 1 is the displacement corresponding to Q 1 and caused by a unit value of Q. Thus

15 ecture 8: lexibility Method The flexibility matrix is The inverse of this matrix is There are no support displacements in the truss. Thus the displacement in the structure corresponding to Q 1 is D Q1 In addition, the displacement in the structure corresponding to Q consists of a relative displacement of the cut ends of bar AD. In the original, or primary structure, the cut ends of bar AD occupy the same location in space before loads are applied. After loads are applied the cut ends move to a different location in space, but relative to each other, no translation takes place between the cut ends, i.e, the bar does not separate. Thus D Q 0 0

16 ecture 8: lexibility Method Substituting the inverse of the flexibility matrix into leads to 1 Q D Q D Q Q P P.88 AE P AE.88 The minus sign for Q indicates that member AD is in compression