Example Lecture 12: The Stiffness Method Prismatic Beams. Consider again the two span beam previously discussed and determine

Size: px

Start display at page:

Download "Example Lecture 12: The Stiffness Method Prismatic Beams. Consider again the two span beam previously discussed and determine"

Octavia French
5 years ago
Views:

1 Example 1.1 Consider again the two span beam previously discussed and determine The shearing force M1 at end B of member B. The bending moment M at end B of member B. The shearing force M3 at end B of member BC. The bending moment M4 at end B of member BC. The force R1 at support. The couple R at support. Force R3 at support B. Force R4 at support C. P1 P M P P P P P 3 1

2 Member end actions in the restrained structure will be denoted by a vector { M }. Keep in mind the beams below are really one restrained beam with the cantilever support in the middle of the beam (as indicated). The member end-actions are treated as if they were support reactions for each beam segment. M1 P P M3 P M Pa b P 4 M4 P 8

3 We can use the same approach when we analyze the restrained structure after unit displacements are applied. The corresponding member end actions, denoted by the matrix { MD } are given below when a unit rotation is applied at B. MD11 6EI MD31 6EI MD 4EI 1 MD41 4EI 3

4 The corresponding member end actions associated with a unit rotation is applied at C are MD1 MD 0 MD3 0 MD4 6EI EI thus MD EI

5 P EI P EI P M D MD M M EI P D The superposition principle leads to the following matrix equation from a previous solution which leads to 5

6 Turning our attention to beam reactions in the restrained structure, denoted by a vector [ R ] once again the beams below are really one restrained beam with the cantilever supports replaced with forces and moments. The two beams are treated as cantilevercantilever beams. R1 P R3 P R 4 P P P R P 8 R3 P R 3 R3 R3 3 P 6

7 We also analyze the restrained structure after unit displacements are applied. The corresponding reactions, denoted by the matrix [ RD ] are given below when a unit rotation is applied at B. RD11 6EI RD31 6EI RD 41 6EI RD 1 EI RD31 6EI 6EI 6EI RD

8 The corresponding reactions associated with a unit rotation is applied at C are RD1 RD RD3 0 RD3 RD3 6EI RD 4 6EI thus RD EI

9 P EI P EI P R D RD R R EI P D The superposition principle leads to the following matrix equation From a previous solution which leads to 9

10 Example 1. 10

11 Example 1.3 For the two span beam previously discussed determine the unknown displacement at joints B and C. In addition find the member end-actions as well as the reactions. The unknowns are identified as 11

12 Using the following restrained structure The actions in the restrained structure due to applied loads corresponding to the previously identified displacements are D 1 P 1 8 P 8 P 8 D P P thus D P 8 4 1

13 pplying a unit rotation at B in the restrained structure, i.e., leads to the following stiffness coefficients 4EI 4EI S11 S 11 S 11 8EI S 1 6EI 13

14 pplying a unit translation at C in the restrained structure, i.e., leads to the following stiffness coefficients S 1EI 3 S 1 6EI 14

15 Thus S EI which leads to S EI 3 4 With 0 D and 1 D S D D then D 30EI P 8 4 P 40EI

16 Once again, member end actions in the restrained structure will be denoted by a vector [ M ]. Keep in mind the restrained beam is treated as two cantilever beams side by side. thus P1 P 1 M 1 P M 8 4 P 4 M 4 P 16

17 We analyze the restrained structure after unit displacements are applied. The corresponding member end actions, denoted by the matrix [ MD ] are given below when a unit rotation is applied at B. MD 11 6EI 4EI MD 1 When a unit translation is applied at C then MD 0 MD1 0 17

18 Thus 3 EI MD 0 0 Using superposition and previous results leads to D M M MD P

19 Next we turn our attention to beam reactions in the restrained structure subject to the applied loads, denoted by a vector [ R ]. Keep in mind the restrained beam is treated as two cantilever beams side by side. R 1 P1 P R P 1 8 P 4 R 3 P1 P 3P R 4 P 8 P 8 19

20 Thus 8 P R 8 1 We analyze the restrained structure after unit displacements are applied. The corresponding member end actions, denoted by the matrix [ MD ] are given below when a unit rotation is applied at B. RD 11 6EI EI RD 1 EI RD 41 6EI 6EI RD31 RD31 RD31 0 0

21 When a unit translation is applied at C we obtain RD 4 6EI RD 1 RD 0 RD3 RD3 RD3 0 1EI 3 1EI 3 thus RD EI

22 Using superposition and previous results leads to D R R RD P

23 Example 1.4 3

24 Joint Stiffness Matrix: Multi-Span Beams stiffness coefficient at any joint of a structure is composed of the sum of the stiffnesses of members which frame into the joint. It is convenient to sum the member stiffnesses in some systematic fashion. Consider the following prismatic member: This sub-component is taken from a three-dimensional frame so that all possible forces and moments can be depicted at each joint. The member is fully restrained and it is convenient to adopt an orthogonal axes oriented to the member. 4

25 The x m -axis is a centroidal axis, the x m - y m plane and the x m - z m plane are principle planes of bending. We assume that the shear center and the centroid of the member coincide so that twisting and bending are not coupled. The properties of the member are defined in a systematic fashion. The length of the member is, the cross sectional area is x. The principle moments of inertial relative to the y m and z m axes are I y and I z, respectively. The torsional constant J (= I y + I z only for a circular cross section, aka, the polar moment of inertia only for this special case) will be designated I x in the figures that follow. The member stiffnesses for the twelve possible types of end displacement are summarized on the next two pages. 5

26 6

27 7

28 The stiffness matrix for the structure depicted in the last two overheads is a 1x1 matrix denoted S m, and each column represents the actions caused by a separate unit displacement. Each row corresponds to a figure in the previous two overheads. 8

The member stiffness matrix needed for other structural subcomponents, e.g., continuous beams and or members from a plane frame (-d), are of lesser order than the matrix on the previous overhead.

29 The member stiffness matrix needed for other structural subcomponents, e.g., continuous beams and or members from a plane frame (-d), are of lesser order than the matrix on the previous overhead. This is a result of only certain end displacements considered in the analysis of the structure. Consider one member of a continuous beam between supports denoted as j and k. The x m, y m, and z m axes are taken in the directions shown in the figure so that the x m -y m plane is the plane of bending for this beam. In a continuous beam there are four significant types of displacements at a joint the corresponds to four types of actions in the beam. These displacements are shown at the left and are identified as vectors 1 through 4 in the figure. These displacements correspond to a shear force and bending moment at each joint. 9

30 The corresponding member stiffness matrix is of order 4x4 as shown below. The elements of this matrix are obtained from cases (), (6), (8) and (1) in a previous overhead. No torsion No axial deformation No biaxial bending If these are included then frames can be analyzed using this approach. For a continuous beam that have pin supports, i.e., no translational displacements at the joints, only the rotations shown in figure (c) are possible. In this case the first and third rows and columns of S m above would be eliminated and the reduced stiffness matrix appears as follows: fully restrained support 30

31 Example 1.5 The concept of an overall joint stiffness matrix will be explained in conjunction with the two span beam shown below. We have analyzed this beam before but here no loads are applied on the structure. The restrained structure and the six possible joint displacements are labeled. Keep in mind that the axial stiffness is assumed to be large relative to flexural stiffness. 31

n over-all stiffness matrix, [S j ], can

This matrix involves application of unit

32 n over-all stiffness matrix, [S j ], can be generated which contains terms for all possible joint displacements, including those restrained at the supports. This matrix involves application of unit displacements as shown in the figures on this overhead. 3

33 Each column in the stiffness matrix below corresponds to the restraint reactions produced for each unit displacements application. Note that the matrix is square and symmetric. Expansion of its determinant would demonstrate that the matrix is singular due to the fact that certain rows and columns are linear combinations of one another. 33

34 The matrix on the previous page was partitioned according to whether the displacements identified in the restrained structure, i.e., were free to displace in the unrestrained structure With partitioning the matrix can be considered to have four elements as follows 34

35 The upper left partition, S, is a square, symmetric matrix that corresponds to the unknown displacements. The inverse of this matrix is used in the expression {D} = [S] -1 ({ D } { D }) The lower left partition, {S RD }, is rectangular matrix contains actions that correspond to the support restraints. This partition gives the reactions for the structure due to unit values of the unknown displacements. Thus and {S RD } = { RD } { R } = { R } + {S RD }{D} The upper right partition, {S DR }, is simply the transpose of {S RD }. The lower right hand partition, {S RR }, is square and symmetric. It contains actions corresponding to support restraints due to unit displacements in the restrained structure. The matrices {S DR } and {S RR }will be used to analyze structures with special support displacements. The stiffness matrix S is only a small portion of S J. This is a consequence of the fact that this particular structure is highly restrained to begin with. In large structures having many joints and few supports, the matrix S constitutes a large portion of S J. 35

Generic Numbering Systems In the previous section of notes the joint displacements were numbered in a convenient order, i.e., translations proceeded rotations at each joint.

36 Generic Numbering Systems In the previous section of notes the joint displacements were numbered in a convenient order, i.e., translations proceeded rotations at each joint. lso, free displacements were numbered before constrained displacements. Consider the arbitrary numbering system below, the sort of numbering system an end user of RIS or STDS might impose on the analysis. If all matrices were generated conforming to the arbitrary numbering system we could lose some, if not all, of the partition definitions developed in the last section of notes. What is required of RIS and STDS is the ability to take an arbitrary numbering system like the one above and transform it back to the numbering system which segregates matrix elements associated with degrees of freedom from those associated with support constraints. 36

37 The S J matrix for the arbitrary numbering system is the 6 by 6 matrix shown below. 37

38 In order for this S J matrix to be useful the actual degrees of freedom and support constraints in the structure must be recognized. If the fourth and sixth rows are switched to the first and second rows, while all others move downward, we obtain the following matrix: 38

39 Next the fourth and sixth column are moved to the first and second column, while all other columns move to the right without changing order. This rearrangement produces the S J matrix we had previously, i.e., Software algorithms must have the capability to track degrees of freedom and perform the necessary matrix manipulation in order to identify pertinent information. 39

40 Matrix ssembly Continuous beams considered here are prismatic, rigidly connected to each beam segment and supported at various points along the beam. Joints are selected at points of support, at any free end, and changes in cross section (i.e., the beam is prismatic). continuous beam having m members and m+1 joints is depicted in figure (a) to the left. Support restraints of two types may exist at any joint in a continuous beam. These are restraints against rotation and/or restraints against translations. We will only consider flexural deformations. Torsion and axial displacements are not considered. Thus only two displacements can occur at each joint. 40

41 Given the numbering system in figure (b) the translation at a particular joint is numbered prior to a rotation and it follows that the number of translations is equal to the number of joints minus one, while the rotation is twice the joint number. Thus at joint j the translations and rotation are number j-1 and j respectively. It is evident that the total number of possible joint displacements is twice the joints (or n j ). If the total number of support restraints against translation and rotations is denoted n r, then the actual degrees of freedom are n n j n r m n r Here n is the number of degrees of freedom. 41

42 To relate the end displacements of a particular member to the displacements of a joint, consider a typical member in figure (c) below. The member end displacements are numbered j1, j, k1 and k3 and correspond to end displacements 1,, 3 and 4 in figure (b). The new notation helps facilitate computer programming. The four end displacements correspond to the four joint displacements as follows: j1 j 1 k1 k 1 j j k k Since j and k are equal numerically to i and (i+1), then: j1 i 1 k1 i 1 j i k i This indexing system is necessary to construct the joint stiffness matrix [S j ] 4

The analysis of continuous beams consists of establishing the stiffness matrix and the load matrix. The most important matrix generated is the overall joint stiffness matrix [S j ].

43 The analysis of continuous beams consists of establishing the stiffness matrix and the load matrix. The most important matrix generated is the overall joint stiffness matrix [S j ]. The joint stiffness matrix consists of contributions from the beam stiffness matrix [S m ]. It is convenient to assess the contributions for one typical member i and repeat the process for members 1 through m. So the next step involves expressing the stiffness coefficients shown in the figure to the left in terms of the various member stiffnesses that contribute to the joint stiffnesses. 43

44 This next step requires that the member stiffnesses be obtained from the matrix below: For example the contribution to the joint stiffness (S j ) j1, j1 from member i-1 is the stiffness S m33 for that member. Similarly, the contribution to (S j ) j1,j1 from member i is the stiffneess S m11 from member i 44

45 In general the contribution of one member to a particular joint stiffness will be denoted by appending the member subscript to the member stiffness itself. From this discussion one can see that the joint stiffness matrix coefficients are generated by the following expressions: S J S j j M 33 S 1, 1 i1 M11 i S S S J j, j1 S J S k1, j1 M 31 i S J S M 41 i k, j1 M 43 i1 M 1 i which represent the transfer of elements of the first column of the member stiffness matrix [S m ]to the appropriate location in the joint stiffness matrix [S j ] 45

46 Expressions analogous to the previous expressions are easily obtained for a unit rotation about the z axis at joint j: S J S j j M 34 S 1, i1 M1 i S S S J j, j S J S k1, j M 3 i S J S M 4 i k, j M 44 i1 M i Expressions analogous to a unit y displacement at joint k are: S J S 13 j1, k M i S S J j, k1 S J S 1, 1 M 33 S k k i M11 i1 S J S M 43 SM 1 i 1 k, k1 M 3 i i 46

47 Finally the expressions for a unit z rotation at joint k are: S J S 14 j1, k M i S S J j, k S J S 1, 1 M 34 S k k i M1 i1 S J S M 44 SM i i 1 k, k M 4 i The last 4 sets of equations show that the sixteen elements of the 4x4 member stiffness matrix [S M ] i for member I contribute to the sixteen of the stiffness matrix [S J ] coefficients in a very regular pattern. This pattern can be observed in the figure on the next overhead. 47

For this structure the number of joints is seven, the number of possible joint displacements is fourteen, and the joint stiffness matrix [S J ] is dimensionally 14x14.

48 For this structure the number of joints is seven, the number of possible joint displacements is fourteen, and the joint stiffness matrix [S J ] is dimensionally 14x14. The indexing scheme is shown down the left hand edge and across the top. The contributions of individual members are indicated in the hatched block., each of which is dimensionally 4x4. The blocks are numbered in the upper right corner to indicated the member associated with the block. The overlapping blocks are dimensionally x and denote elements that receive contributions from adjacent members. 48

To obtain this rearranged matrix, rows and columns of the original matrix have been switched in proper sequence in order t place the stiffnesses pertaining to the

49 Suppose that the actual beam has simple supports at all the joints as indicated in the figure below. The rearranged and partitioned joint stiffness matrix is shown at the lower right. To obtain this rearranged matrix, rows and columns of the original matrix have been switched in proper sequence in order t place the stiffnesses pertaining to the actual degrees of freedom in the first seven rows and columns. s an aid in the rearranging process, the new row and column designations are listed in the previous figure for the matrix along the right hand side and across the bottom. The rearranging process is consistent with the numbering system in the figure above. 49

50 In summary, the procedure followed in generating the joint stiffness matrix [S J ] consists of taking the members in sequence and evaluating their contributions one at a time. Then the stiffness matrix [S M ] i is generated, and the elements of this matrix are transferred to the [S J ] as indicated in the previous overheads. fter all members have been processed in this manner, the [S J ] matrix is complete. This matrix can be rearranged and partitioned in order to isolate the [S] matrix. The inverse of this matrix is then determined and the unknown displacements are computed. 50

51 Example 1.6 The continuous beam shown below is restrained against translation at support C and has a fixed end support at D. t joint B the flexural rigidity changes from EI to EI. Determine the stiffness matrix [S] for the structure.

52 The original member and joint numbering scheme is depicted in the previous picture. From the figure one can see that the number of members, m, is 3, the number of joints, n j, is four and the number of support restraints, n r is five. Thus the number of degrees of freedom is n n j n 3 4 The member properties and the indices are given in the table below. The moment of inertia and length of each member are input for the analysis. The remaining quantities are computed from the joint and member numbering scheme. r 5

53 Joint restraints for the beam are depicted in the table below. The index numbers for all possible displacements are listed for each joint. This is followed by a restraint list, in which the indicator 1 implies a restraint, and the indicator 0 implies no restraint or a degree of freedom. The last column depicts the index of the displacement that corresponds to a degree of freedom when the joint stiffness matrix [S J ] is rearranged

54 The member stiffness matrix for each member of the structure is given in the matrices below.

Elements from [S M ] 1, [S M ] and [S M ] 3 are transferred to the S J matrix as shown to the left. The various contributions from each member are delineated by dashed lines.

55 Elements from [S M ] 1, [S M ] and [S M ] 3 are transferred to the S J matrix as shown to the left. The various contributions from each member are delineated by dashed lines. The regions that overlap are where two members contribute to the matrix simultaneously. The joint numbering system from the original figure are shown running across the top and along the left of the matrix. The matrix will be rearranged based on the numbering system given in a previous overhead. That numbering system is running along the bottom and down the right hand side of the matrix.

The third, fourth, and sixth rows and columns are shifted to the first three rows and the first three columns. ll other rows and columns are shifted downward.

56 The third, fourth, and sixth rows and columns are shifted to the first three rows and the first three columns. ll other rows and columns are shifted downward. The rearrangement is shown in the matrix to the left. This matrix is partitioned, as indicated by the dashed lines in a manners similar to the matrix below:

57 From the previous overhead one can extract the 3x3 stiffness matrix [S], which is Inverting this matrix yields

Loads. Lecture 12: PRISMATIC BEAMS

Loads. Lecture 12: PRISMATIC BEAMS Loads After composing the joint stiffness matrix the next step is composing load vectors. reviously it was convenient to treat joint loads and member loads separately since they are manipulated in different