Preliminaries: Beam eflections Virtual Work There are several methods available to calculate deformations (displacements and rotations) in beams. They include: Formulating moment equations and then integrating to find rotations and displacements Moment area theorems for either rotations and/or displacements Virtual work methods Since structural analysis based on finite element methods is usually based on a potential energy method, we will tend to use virtual work methods to compute beam deflections. The theory that supports calculating deflections using virtual work will be reviewed and several examples are presented. 1
Consider the following arbitrarily loaded beam Identify M m ~ M(x) Moment at any section in the beam due to external loads m(x) Moment at any section in the beam due to a unit action m y I Stress acting on d due to a unit action
The force acting on the differential area d due to a unit action is ~ f ~ d The stress due to external loads is The displacement of a differential segment d by dx along the length of the beam is m y I M y I d dx dx E M y E I dx 3
The work done by the force acting on the differential area d due to a unit action as the differential segment of the beam (d by dx) displaces along the length of the beam by an amount is ~ dw f m y I M m y E I d dx The work done within a differential segment (now by dx) due to a unit action applied to the beam is the integration of the expression above with respect to d, i.e., d M y E I dx dw c c T B M m y E I d dx W differnetialsegment Mm EI c c T B y d dx Mm I EI dx Mm EI dx 4
The internal work done along the entire length of the beam due to a unit action applied to the beam is the integration of the last expression with respect to x, i.e., The external work done along the entire length of the beam due to a unit action applied to the beam is x M m x WInternal dx EI 0 With W External 1 W External 1 or the deformation () of the a beam at the point of application of a unit action (force or moment) is given by the integral on the right. W 0 0 Internal M M x m EI x m EI x x dx dx 5
Example 6.1 6
Example 6. Flexibility Coefficients by virtual work 7
Perspectives on the Flexibility Method In 1864 James Clerk Maxwell published the first consistent treatment of the flexibility method for indeterminate structures. His method was based on considering deflections, but the presentation was rather brief and attracted little attention. Ten years later Otto Mohr independently extended Maxwell s theory to the present day treatment. The flexibility method will sometimes be referred to in the literature as Maxwell-Mohr method. With the flexibility method equations of compatibility involving displacements at each of the redundant forces in the structure are introduced to provide the additional equations needed for solution. This method is somewhat useful in analyzing beams, frames and trusses that are statically indeterminate to the first or second degree. For structures with a high degree of static indeterminacy such as multi-story buildings and large complex trusses stiffness methods are more appropriate. Nevertheless flexibility methods provide an understanding of the behavior of statically indeterminate structures. 8
The fundamental concepts that underpin the flexibility method will be illustrated by the study of a two span beam. The procedure is as follows 1. Pick a sufficient number of redundants corresponding to the degree of indeterminacy. Remove the redundants 3. etermine displacements at the redundants on released structure due to external or imposed actions 4. etermine displacements due to unit loads at the redundants on the released structure 5. Employ equation of compatibility, e.g., if a pin reaction is removed as a redundant the compatibility equation could be the summation of vertical displacements in the released structure must add to zero. 9
Example 6.3 The beam to the left is statically indeterminate to the first degree. The reaction at the middle support R B is chosen as the redundant. The released beam is also shown. Under the external loads the released beam deflects an amount B. second beam is considered where the released redundant is treated as an external load and the corresponding deflection at the redundant is set equal to B. R B 5 8 w 10
more general approach consists in finding the displacement at B caused by a unit load in the direction of R B. Then this displacement can be multiplied by R B to determine the total displacement lso in a more general approach a consistent sign convention for actions and displacements must be adopted. The displacements in the released structure at B are positive when they are in the direction of the action released, i.e., upwards is positive here. The displacement at B caused by the unit action is 3 B 48EI The displacement at B caused by R B is δ B R B. The displacement caused by the uniform load w acting on the released structure is Thus by the compatibility equation B B R B 0 B 4 5 w 384 EI R B B B 5 8 w 11
Example 6.4 If a structure is statically indeterminate to more than one degree, the approach used in the preceeding example must be further organized and more generalized notation is introduced. Consider the beam to the left. The beam is statically indeterminate to the second degree. statically determinate structure can be obtained by releasing two redundant reactions. Four possible released structures are shown. 1
The redundants chosen are at B and C. The redundant reactions are designated Q 1 and Q. The released structure is shown at the left with all external and internal redundants shown. Q1 is the displacement corresponding to Q 1 and caused by only external actions on the released structure Q is the displacement corresponding to Q caused by only external actions on the released structure. Both displacements are shown in their assumed positive direction. 13
We can now write the compatibility equations for this structure. The displacements corresponding to Q 1 and Q will be zero. These are labeled Q1 and Q respectively Q1 Q 1 F11Q1 F1Q 0 Q Q F1Q1 FQ 0 In some cases Q1 and Q would be nonzero then we would write Q1 Q 1 F11Q1 F1Q Q Q F1Q1 FQ 14
where: The equations from the previous page can be written in matrix format as FQ Q Q { Q } - vector of actual displacements corresponding to the redundant { Q } - vector of displacements in the released structure corresponding to the redundant action [Q] and due to the loads [F] - flexibility matrix for the released structure corresponding to the redundant actions [Q] {Q} - vector of redundants Q Q1 Q Q Q1 Q Q Q1 Q F F F 11 1 F F 1 15
The vector {Q} of redundants can be found by solving for them from the matrix equation on the previous overhead. F Q Q Q 1 Q F Q Q To see how this works consider the previous beam with a constant flexural rigidity EI. If we identify actions on the beam as P P M P P P P3 1 P Since there are no displacements imposed on the structure corresponding to Q 1 and Q, then Q 0 0 16
The vector [ Q ] represents the displacements in the released structure corresponding to the redundant loads. These displacements are Q 3 13P 1 Q 4EI 97P 48EI 3 The positive signs indicate that both displacements are upward. In a matrix format Q 3 P 48EI 6 97 17
The flexibility matrix [F ] is obtained by subjecting the beam to unit load corresponding to Q 1 and computing the following displacements 3 F11 F1 3EI 3 5 6EI Similarly subjecting the beam to unit load corresponding to Q and computing the following displacements 3 5 F1 F 6EI 3 8 3EI 18
The flexibility matrix is The inverse of the flexibility matrix is s a final step the redundants [Q] can be found as follows 16 5 5 6 3 EI F 5 5 16 7 6 3 1 EI F 64 69 56 97 6 48 0 0 5 5 16 7 6 3 3 1 1 P EI P EI F Q Q Q Q Q 19
The redundants have been obtained. The other unknown reactions can be found from the released structure. isplacements can be computed from the known reactions on the released structure and imposing the compatibility equations. iscuss the following sign conventions and how they relate to one another: 1. Shear and bending moment diagrams. Global coordinate axes 3. Sign conventions for actions - Translations are positive if the follow the direction of the applied force - Rotations are positive if they follow the direction of the applied moment 0
Example 6.5 three span beam shown at the left is acted upon by a uniform load w and concentrated loads P as shown. The beam has a constant flexural rigidity EI. Treat the supports at B and C as redundants and compute these redundants. In this problem the bending moments at B and C are chosen as redundants to indicate how unit rotations are applied to released structures. Each redundant consists of two moments, one acting in each adjoining span. 1
The displacements corresponding to the two redundants consist of two rotations one for each adjoining span. The displacement Q1 and Q corresponding to Q 1 and Q. These displacements will be caused by the loads acting on the released structure. The displacement Q1 is composed of two parts, the rotation of end B of member B and the rotation of end B of member BC Similarly, 3 w P Q 1 4EI 16 EI such that P P P Q 16EI 16EI 8 EI w 3P Q 48EI 6P
The flexibility coefficients are determined next. The flexibility coefficient F 11 is the sum of two rotations at joint B. One in span B and the other in span BC (not shown below) F 3EI 3EI 11 3EI Similarly the coefficient F 1 is equal to the sum of rotations at joint C. However, the rotation in span C is zero from a unit rotation at joint B. Thus F 1 6EI 3
Similarly F 3EI 3EI 3EI F 1 6EI The flexibility matrix is F 6EI 4 1 1 4 The inverse of the flexibility matrix is F 1 EI 5 4 1 1 4 4
0 15 1 P w Q P w P w P P w EI EI F Q Q Q Q Q 1 6 8 10 6 3 48 0 0 4 1 1 4 5 1 1 40 7 60 P w Q s a final step the redundants [Q] can be found as follows and 5
Example 6.6 6
Joint isplacements, Member End ctions nd Reactions Previously we focused on finding redundants using flexibility (force) methods. Typically redundants (Q 1, Q,, Q n ) specified by the structural engineer are unknown reactions. Redundants are determined by imposing displacement continuity at the point in the structure where redundants are applied, i.e., we imposed FQ Q Q If the redundants specified are unknown reactions then after these redundants are found other actions in the released structure could be found using equations of equilibrium. When all actions in a structure have been determined it is possible to compute displacements by isolating the individual subcomponents of a structure. isplacements in these subcomponents can be calculated using concepts learned in Strength of Materials. These concepts allow us to determine displacements anywhere in the structure but usually the unknown displacements at the joints are of primary interest if they are non-zero.. 7
Instead of following the procedure just outlined we will now introduce a systematic procedure for calculating non-zero joint displacements, reaction, and member end actions directly using flexibility methods. Consider the two span beam below where the redundants Q 1 and Q have been computed previously in Example 6.4. The non-zero joint displacements J1 and J, both rotations, as well as reactions R1 and R. can be computed. We will focus on the joint displacements J1 and J first. Keep in mind that when using flexibility methods translations are associated with forces, and rotations are associated with moments. Reactions other than redundants will be denoted { R } and these quantities can be determined as well. The objective here is the extension of the flexibility (force) method so that it is more generally applied. 8
The principle of superposition is used to obtain the joint displacement vector { J }, which is a vector of displacements that occur in the actual structure. For the structure depicted on the previous page the rotations in the actual structure at joints B ( = J1 ) and C ( = J ) are required. When the redundants Q 1 and Q were found superposition was imposed on the released structure requiring the displacement associated with the unknown redundants to be equal to zero. In finding joint displacements in the actual structure superposition is used again and displacements in the released structure are equated to the displacement in the actual structure. Focusing on joint B, superposition requires Here J1 J 1 JQ 11Q1 JQ 1Q J1 = non-zero displacement (a rotation) at joint B in the actual structure, at the joint associated with Q 1 J1 = the displacement (a rotation) at joint B associated with J1 caused by the external loads in the released structure. JQ11 = the rotation at joint B associated with J1 caused by a unit force at joint B corresponding to the redundant Q 1 in the released structure JQ1 = the rotation at joint B associated with J1 caused by a unit force at joint C corresponding to the redundant Q in the released structure 9
Thus displacements in the released structure must be further evaluated for information beyond that required to find the redundants Q 1 and Q. In the released structure the displacements associated with the applied loads are designated { J } and are depicted below. The displacements associated with the redundants are designated [ JQ ] and are similarly depicted. In the figure to the right unit loads are shown applied at the redundants. These unit loads were used earlier to find flexibility coefficients [F ij ]. These coefficients were then used to determine Q 1 and Q. Now the unit loads are used to find the components of [ JQ ]. released structure 30
similar expression can be derived for the rotation at C ( = J ), i.e., Here J J JQ1 Q1 JQ Q J = non-zero displacement (a rotation) at joint C in the actual structure, at the joint associated with Q J = the displacement (a rotation) at joint C associated with J caused by the external loads in the released structure. JQ1 = the rotation at joint C associated with J caused by a unit force at joint B corresponding to the redundant Q 1 in the released structure JQ = the rotation at joint C associated with J caused by a unit force at joint C corresponding to the redundant Q in the released structure 31
The expressions J1 and J can be expressed in a matrix format as follows Q J J JQ where J J1 J J J1 J JQ11 JQ1 JQ JQ1 JQ and Q Q1 Q which were determined previously 3
In a similar manner we can find reactions via superposition For the first expression R1 R 1 RQ 11Q1 RQ 1Q R R RQ 1Q1 RQ Q R1 = the reaction in the actual beam at R = the reaction in the actual beam at R1 = the reaction in the released structure due to the external loads R = the reaction in the released structure due to the external loads RQ11 = the reaction at in the released structure due to the unit action corresponding to the redundant Q 1 RQ = the reaction at in the released structure due to the unit action corresponding to the redundant Q RQ1 = the reaction at in the released structure due to the unit action corresponding to the redundant Q 33
The expressions on the previous slide can be expressed in a matrix format as where R R1 R Q R R R R1 R RQ RQ11 RQ1 RQ RQ 1 RQ Q Q1 Q 34
In a similar manner we can find member end actions via superposition For the first expression M1 M 1 MQ 11Q1 MQ 1Q M M MQ1Q1 MQQ M 3 M3 MQ31Q1 MQ3Q M 4 M4 MQ41Q1 MQ4Q M1 = is the shear force at B on member B M1 = is the shear force at B on member B caused by the external loads on the released structure MQ11 = is the shear force at B on member B caused by a unit load corresponding to the redundant Q 1 MQ1 = is the shear force at B on member B caused by a unit load corresponding to the redundant Q The other expressions follow in a similar manner. 35
The expressions on the previous slide can be expressed in a matrix format as follows where M 1 M M M 3 M 4 Q M M MQ M 1 M M MQ M3 M4 MQ11 M Q1 MQ31 MQ41 MQ1 MQ MQ3 MQ4 Q Q1 Q The sign convention for member end actions is as follows: + when up for translations and forces + when counterclockwise for rotation and couples 36
Example 6.7 Consider the two span beam to the left where it is assumed that the objective is to calculate the various joint displacements J, member end actions M, and end reactions R. The beam has a constant flexural rigidity EI and is acted upon by the following loads P1 P M P P P 3 P P 37
Consider the released structure and the attending moment area diagrams. The (M/EI) diagram was drawn by parts. Each action and its attending diagram is presented one at a time in the figure starting with actions on the far right. 38
From first moment area theorem 1 P 1 P J 1 1 1.5 0.5 EI EI P 1 P EI EI 5P 4EI J 1 P 1 3P 3 EI EI P 1 P EI EI 13P 8EI 10 13 P J 39 8EI
Consider the released beam with a unit load at point B JQ 11 1 EI EI JQ 1 1 EI EI 40
Consider the released beam with a unit load at point C JQ 1 leading to 1 EI 3 EI 1 JQ 1 EI EI 1 3 JQ EI 1 4 41
Previously in Example 6.4 with Q P 69 56 64 Q J J JQ then the displacements J1 and J are J P 10 1 3 P 69 8EI 13 EI 1 4 56 64 P 17 11EI 5 4
P P P P F R R Y 0 1 1 3 0 P P P P P M R R Using the following free body diagram of the released structure Then from the equations of equilibrium 43
Using a free body diagram from segment B of the entire beam, i.e., then once again from the equations of equilibrium F Y 0 M1 P P M1 0 M B 0 M P P P M 3P 44
Using a free body diagram from segment BC of the entire beam, i.e., then once again from the equations of equilibrium F Y M3 0 0 M3 P P M B 0 M4 P M 4 P P 45
0 3 0 P P M R P P Thus the vectors M and R are as follows: Member end actions in the released structure. Reactions in the released structure. 46
Finally with Q R R RQ then knowing [ R ], [ RQ ] and [Q] leads to R P P 1 1 69 P 56 64 P 107 56 31 47
In a similar fashion, applying a unit load associated with Q 1 and Q in the previous cantilever beam, we obtain the following matrices MQ 1 0 0 0 1 1 1 1 RQ 48
Similarly, with Q M M MQ and knowing [ M ], [ MQ ] and [Q] leads to M 0 3P 1 1 0 P 69 0 56 0 164 P 0 5 P 0 56 64 36 49
Summary Of Flexibility Method The analysis of a structure by the flexibility method may be described by the following steps: 1. Problem statement. Selection of released structure 3. nalysis of released structure under loads 4. nalysis of released structure for other causes 5. nalysis of released structure for unit values of redundant 6. etermination of redundants through the superposition equations, i.e., F Q Q QS QS Q QT QP QR 1 Q F Q QS 50
7. etermine the other displacements and actions. The following are the four flexibility matrix equations for calculating redundants member end actions, reactions and joint displacements where for the released structure Q J JS ll matrices used in the flexibility method are summarized in the following tables JQ Q M M MQ Q R R JS J JT RQ JP JR 51
MTRIX ORER EFINITION QT, Q Q Q JQ, QP QR q x 1 q x 1 q x 1 q x q q x 1 Unknown redundant actions (q = Number of redundant) isplacements in the actual structure Corresponding to the redundant isplacements in the released structure corresponding to the redundants and due to loads isplacements in the released structure corresponding to the redundants and due to unit values of the redundants isplacements in the released structure corresponding to the redundants and due to temperature, prestrain, and restraint displacements (other than those in Q) QS q x 1 QS Q QT QP QR 5
MTRIX ORER EFINITION J J Q, JT, JP JR j x 1 j x 1 j x 1 j x 1 Joint displacement in the actual structure (j = number of joint displacement) Joint displacements in the released structure due to loads Joint displacements in the released structure due to unit values of the redundants Joint displacements in the released structure due to temperature, prestrain, and restraint displacements (other than those in Q) JS j x 1 JS J JT JP JR F q x q Matrix of flexibility coefficients 53
MTRIX ORER EFINITION M M m x 1 m x 1 Member end actions in the actual structure (m = Number of end-actions) Member end actions in the released structure due to loads MQ R R RQ m x q r x 1 r x 1 r x q Member end actions in the released structure due to unit values of the redundants Reactions in the actual structure (r = number of reactions) Reactions in the released structure due to loads Reactions in the released structure due to unit values of the redundants 54