Chapter one DETERMINACY & INDETERMINACY OF STRUCTURES Introduction A structure refers to a system of connected parts used to support a load. Important examples related to civil engineering include buildings, bridges, and towers; and in other branches of engineering, ship and aircraft frames, tanks, pressure vessels, mechanical systems, and electrical supporting structures are important. hen designing a structure to serve a specified function for public use, the engineer must account for its safety, esthetics, and serviceability, while taking into consideration economic and environmental constraints. Classification of structures Structures in general can be classified as 1. Skeletal structures Idealized to a series of straight or curved lines e.g. Roof trusses, lattice girders, building frames etc. 2. Surface structures Idealized to a plane or curved surface e.g., slabs, shells etc. 3. Solid structures massive structures having all the three dimensions considerably large e.g. Dam, solid retaining walls etc. Skeletal structures can further be classified as 1. Pin jointed structures Considered to be of frictionless joints connected by pins, members are subjected only to axial tension and/or axial compression due to the externally applied loads being applied only at the joints 2. Rigid jointed structures Angles between the joined members remain unchanged even after loads being applied, members may be subjected to bending moments, twisting moments, shear forces and axial forces Skeletal frames may also be classified as 1. Plane frame All members and the applied loads are assumed to be in one plane 2. Space frame (three dimensional frame) All members do not lie in one plane and there by forces/loads also do not fall in one plane
Idealization of the Structure The techniques used to represent various structural systems by line drawing to analyse the structures. Idealized models used in structural analysis that represent pinned and fixed supports and pin-connected and fixed-connected joints. Actual structure
Supports and connection- Conventional Representations A body sustaining loads can be supported in several ways. The common types of supports are; (i) Roller Support: - sustains force in only one direction. Fig. Typical roller supports (one reactive force) Translation is prevented in the vertical direction but not in the horizontal direction. These kinds of supports allow rotation of the body. (ii) Hinge or Pin Support: - restrains the member from translating (moving) in any direction of the plane, but it does not prevent rotation. Fig. Hinged supports (two reactive forces) (iii) Fixed or Built in Support At the fixed support, the body (member) can neither translate nor rotate. Therefore, a fixed support resists both a moment and a force in any direction. Fig. Fixed support (three reactive forces) (iv) Cable Support: - prevent motion only along its axes and only if the motion tends to put the cable in tension. A cable can exert a pull but not a push.
(v) Ball and Socket (Three dimensional hinge supports) Rotation is allowed in any direction but not translation. (vi) Rigid Support in space (Three dimensional):- neither permits rotation nor translation in any direction. Equations of static equilibrium 6 reactions- 3 force and 3 moment reactions in the three axes (x-y-z). According to cartesian coordinate system, equations of static equilibrium are written as ΣF x = ΣF y = ΣF z = 0 Algebraic sums of components of all external forces along x, y and z axes = 0 ΣM x = ΣM y = ΣM z = 0 Algebraic sums of components of all moments along x, y and z axes = 0 External forces can be divided in to two groups: 1. Applied loads, 2. Reactive forces For the static equilibrium of plane structures three equations of equilibrium are enough to be satisfied. They are, ΣF x = 0, ΣF y = 0 and ΣM z = 0. For the static equilibrium of space structures all the above mentioned six equations of equilibrium are to be satisfied. 1.1. Free body diagram A free body diagram is a diagrammatical representation of the isolated body or combination of bodies treated as a single body, showing all forces applied to it by mechanical contact with other bodies that are imagined to be removed. It is a concept to visualize the internal forces of a body. The free body diagram is the most important single step in the solution of problems in mechanics.
Examples FBD (i) F3 F2 F1 M F3 F2 F1 H mass m R mg (ii) w w P B C P B C 5m 5m A D H A A D 4m R A R D (iii) FBDs for internal forces 4m a/ b/ x x = x x = P P P P P P P Geometric Stability & Static Determinacy of Trusses, beams and frames Stability depends on: Stability of structure Number of support reaction On the arrangements of the support reaction, but not depend on the strength of individual members If the structure is said to be stable: Can support any possible system of applied load Have sufficient number of reaction to prevent motion or translation The arrangements of the support reaction are non-concurrent and non- parallel.
If the structure is unstable: Insufficient number support reaction (which means have more possibility of equation than unknown support reaction). The arrangements of the reaction component are concurrent and parallel Static determinacy of structures Statically indeterminate structures are those structures which cannot be analyzed with the help of equations of static equilibrium alone. A structure is statically indeterminate when it possesses more members or is supported by more reactive restraints than are strictly necessary for stability and equilibrium. In analysis of statically indeterminate structures, the number of unknowns is greater than the number of independent equations derived from the conditions of static equilibrium. Additional equations based on the compatibility of deformation must be written in order to obtain sufficient number of equations for the determination of all the unknowns. The number of such additional equations required for the determination of all the unknowns is known as the degree of static indeterminacy or degree of redundancy of the structure. Difference between determinate and indeterminate structures S. No. Determinate Structures Indeterminate Structures 1 Conditions of equilibrium are not Equilibrium conditions are fully adequate adequate to fully analyze the to analyze the structure. structure. 2 Bending moment or shear force at any section is independent of the material property of the structure. Bending moment or shear force at any section depends upon the material property. 3 The bending moment or shear force at any section is independent of the crosssection or moment of inertia. The bending moment or shear force at any section depends upon the crosssection or moment of inertia. 4 Temperature variations do not cause stresses. Temperature variations cause stresses. 5 No stresses are caused due to lack of fit. Stresses are caused due to lack of fit. 6 Extra conditions like compatibility of Extra conditions like compatibility of displacements are required to analyze displacements are not required to the structure along with the analyze the structure. equilibrium equations.
Determinacy in Truss The use of beams/plate-girders does not always provide the most economic or suitable structural solution when spanning large openings. In buildings which have lightly loaded, long span roofs, when large voids are required within the depth of roof structures for services, when plated structures are impractical, or for aesthetic/architectural reasons, the use of roof trusses, lattice girders or space-frames may be more appropriate. Such trusses/girders/frames, generally, transfer their loads by inducing axial tension or compressive forces in the individual members. Degree of External indeterminacy in truss External indeterminacy is related to how stably the support system exerts reaction components. It can be calculated as the number of external reaction components in excess of the number of equilibrium equations required for the static determinacy or minimum number of reaction components required for stability. A structure is usually externally indeterminate or redundant if the reactions at the supports cannot be determined by using the available equations of equilibrium. The external indeterminacy (E) can thus be said as Total number of reaction components (R) minimum number of reaction components required for stability (r) I.e. E = R r The degree of external indeterminacy is E = (R 6) for space structures and E = (R 3) for plane structures Degree of internal indeterminacy in truss Internal indeterminacy is related to the members that are more in number than that is required for stability. hen there is more number of members in the structure, there could be more numbers of stress resultants that are unknowns to be found for the complete analysis. It can be said that the internal indeterminacy is based on the number of additional members present in the frame than that is required for a determinate structure, in the case of pin jointed structures and is based on the excess number of internal stress resultants in the case of rigid jointed structures. In a pin jointed structure the number of internal stress resultants is only one, ie either tension or compression. Hence instead of saying as number of excess stress resultants it can be said as number of excess members. For pin jointed plane structures to be internally determinate, the equation to be satisfied is m = 2j 3; where m = number of members and j = number of joints.
The basic pin jointed plane internally determinate structure can be a triangular structure with 3 members and 3 joints. See fig. 1.1. Fig 1.1 Fig 1.2 Further the pin jointed plane internally determinate structure can be created by adding 2 members and 1 joint. See fig 1.2. Hence the equation to be satisfied for the structure to be determinate can be coined as number of members (m) = 2 times number of joints (j) first three joints. (Of course, for the basic triangular structure, No. of members = number of joints) The basic pin jointed space truss to be determinate should contain at least 6 members and 4 joints. See fig. 1.3. Further expansion of such determinate space truss can be done by adding 3 members and one joint. See fig. 1.4. Fig 1.3 Fig 1.4 For the basic structure as in fig 1.3, the governing equation can be framed as m = 3j 6. The same equation is applicable for further addition of members and joints to form determinate space structure. The degree of internal indeterminacy in case of pin jointed structures can then be written as I = m (2j 3) for plane trusses & I = m (3j 6) for space trusses Generally the degree of internal indeterminacy in case of rigid frames is determined as the number of unknown internal stress resultants minus the number of equilibrium equations. Total degree of Indeterminacy Total degree of indeterminacy or redundancy is equal to the number by which the unknowns (ie. Reaction components as well as stress resultants) exceed the condition equations of equilibrium. The excess restraints are called as redundant.
Examples 1 Plane Trusses Determine the external, internal and total degree of indeterminacy and conditions of stability of the trusses shown in fig 1.12 through 1.15 4 8 y 2 3 5 7 9 11 1 6 10 Fig. 1.12 x External Indeterminacy: Total number of external reaction components for general loading (R) = 8 Number of equilibrium equations (r) = 3 Degree of External indeterminacy E = R r = 8 3 = 5 Internal indeterminacy: Number of members = m = 11 Number of joints = j = 7 Degree of internal indeterminacy I = m (2j r) = 11 2(7) + 3 = 14 14 = 0 Hence internally determinate Total degree of indeterminacy = E + I = 5 + 0 = 5 Example 2 4 6 12 15 1 3 5 7 9 10 14 16 18 11 2 8 13 17 Fig. 1.13 00 External Indeterminacy: Total number of external reaction components for general loading (R) = 3 Number of equilibrium equations (r) = 3
Degree of External indeterminacy E = R r = 3 3 = 0 Hence externally determinate Internal indeterminacy: Number of members = m = 18 Number of joints = j = 10 Degree of internal indeterminacy I = m (2j r) = 18 2(10) + 3 = 21 20 = 1 Total degree of indeterminacy = E + I = 0 + 1 = 1 1 M=0 4 7 10 12 15 6 3 5 9 11 A 8 2 14 16 13 17 18 19 20 External Indeterminacy: Fig 1.16 Total number of external reaction components for general loading (R) = 4 Number of equilibrium equations + one condition equation (r) = 3 + 1 = 4 Degree of External indeterminacy E = R r = 4 4 = 0 Internal indeterminacy: Number of members = m = 20 Number of joints = j = 12 The number of equilibrium equations + one condition equation ie (r) = 3+1 = 4 The condition equation at joint A : ΣM at the joint A = 0 Degree of internal indeterminacy I = m (2j r) = 20 2(12) + 4 = 24 24 = 0 Hence the structure is internally stable Total degree of indeterminacy = E + I = 0 + 0 = 0 ie the structure is determinate Condition equations In indeterminate structural analysis, the number of equilibrium equations may be used to construct some limited number of equations to solve for limited number of unknowns only. here special internal conditions of construction exist, it is possible to write additional
statically equations for use in the determination of unknowns. Such equations are called condition equations. Some of such condition equations can be (a) bending moment at an internal hinge is zero, (b) vertical deflection at firm supports is zero, (c) algebraic sum of bending moments at the ends of members joining at a joint is zero etc. Determinacy of Beams A stable structure should have at least three reactive components, but not always sufficient, for external stability of a 2D structure, which are non-concurrent and non-parallel. Cantilever beam No. of external reactions = 3 No. of equilibrium equns = 3 Stable & determinate Hence stable and determinate Simply supported beam Stable & determinate. No. of external reactions = 3 No. of equilibrium equns = 3 Hence stable and determinate Simply supported with two side over hang or Propped Cantilever beam Stable & determinate. No. of external reactions = 3 No. of equilibrium equns = 3 Hence stable and determinate External indeterminacy in beams 3 + 2 = 5
Stable and indeterminate to two degrees No. of reactions possible = 5 No. of Equations of equilibrium available = 3 Degree of External indeterminacy = 5-3 = 2 Fixed beam 3 + 3 = 6 Stable and indeterminate to three degrees No. of reactions possible = 6 No. of Equations of equilibrium available = 3 Degree of External indeterminacy = 6-3 = 3 Simply supported beam with hinged supports 2 + 2 = 4 Stable and indeterminate to one degree No. of reactions possible = 4 No. of Equations of equilibrium available = 3 Degree of External indeterminacy = 4-3 = Continuous beam o o Determinacy of Frame 3 + 1 + 2 + 2 = 8 Stable & externally indeterminate to 5th degree
Frames are often used in building and are composed of beams and columns that are either pin or fixed connected like truss, frame s extend in two or three dimensional. The loading on a frame causes bending of its members, and if it has rigid joint connections, this structure is generally indeterminate from the moment interaction between the beams and columns at the rigid joints. Degree of indeterminacy for multi-storeyed frame Method 1 Consider the two bay two storeyed frame shown in the fig External indeterminacy: Total number of external reactions = 9 Hence the degree of external indeterminacy = R r = 9 3 = 6 G H I D E F A B C Internal indeterminacy: Consider the upper storey. Let the number of vertical members in the storey be x. Number of internal reactions corresponding to x vertical members is 3x. Since three statical equations of equilibrium are available the degree of internal redundancy for this storey is 3x 3 = 3(x 1) If there are n number of storeys and each storey has x vertical members, then the degree of internal redundancy would be 3(x 1)(n 1). If the number of vertical members is different for different storeys, say x 1 for the first storey and x 2 for the second story etc., Total degree of internal indeterminacy = 3(x 2 1)+3(x 3 1)+.. +3(x n 1) = 3(x 2 + x 3 +..+ x n ) 3(n 1) (x 2 + x 3 +..+ x n ) = Total number of columns in the upper storeys excluding the first storey. = m (say) Therefore degree of internal indeterminacy = 3m 3(n 1) = 3(m n + 1)
In our case, m = 3, n = 2, The degree of internal indeterminacy = 3(3 2 + 1) = 6 Total degree of Indeterminacy = 6 + 6 = 12 Example 1 Determine the degree of indeterminacy for the following multistoreyed frames given in fig. 1.18 and 1.19 G H D E F A B Fig 1.18 C External indeterminacy: Total number of external reactions = 3 + 2 + 1 = 6 Hence the degree of external indeterminacy = R r = 6 3 = 3 Internal indeterminacy: Total number of columns in the higher storeys = m = 2 Number of storeys = n = 2 Degree of internal indeterminacy = 3(m n + 1) = 3(2 2 + 1) = 3 Total degree of indeterminacy = 3 + 3 = 6 F G H External indeterminacy: Total number of external reactions = 3 + 3 + 2 = 8 Hence the degree of external indeterminacy = R r = 8 3 = 5 D E Internal indeterminacy: A Fig B 1.19 C Total number of columns in the higher storeys = m = 2 Number of storeys = n = 2 Degree of internal indeterminacy = 3(m n + 1) = 3(2 2 + 1) = 3 Total degree of indeterminacy = 5+3 = 8
Reading assignments:- method 2, method 3,method 4 and method 5 Degree of kinematic indeterminacy A skeletal structure is said to be kinematically indeterminate if the displacement components of its joints cannot be determined by compatibility equations alone. In the case of kinematic indeterminate structure, the number of unknown displacement components is greater than the number of compatibility equations. For these structures additional equations based on equilibrium must be written in order to obtain sufficient number of equations for the determination of all the unknown displacement components. The number of these additional equations necessary for the determination of all the independent displacement components is known as the degree of kinematic indeterminacy or the degree of freedom. Pin jointed frame: Each joint of a pin jointed plane frame has two independent displacement components (translation in x and y directions). The number of compatibility equations can be said as equal to the number of constraints imposed by the support conditions (R). Hence, Kinematic indeterminacy of a pin jointed plane frame = 2j R Similarly for pin jointed space frame, the independent displacement components per joint are 3 (translation in x, y and z directions) Hence, Kinematic indeterminacy of a pin jointed space frame = 3j R Rigid jointed frame: Each joint of a rigid jointed plane frame has three independent displacement components (translation in x and y directions & rotation in z direction). The number of compatibility equations (c) can be said as equal to the number of constraints imposed by the support conditions (R) plus other factors such as the inextensibility of members. If there are m number of members which are inextensible, Kinematic indeterminacy of a rigid jointed plane frame = 3j c = 3j (R + m) Similarly for rigid jointed space frame, the independent displacement components per joint is 6 (translation and as well as rotation in x, y and z directions) Hence, Kinematic indeterminacy of a rigid jointed space frame = 6j c = 6j (R + m) Determine the kinematic indeterminacy of the pin jointed plane frame shown in the fig. 1.22 Kinematic indeterminacy of a pin jointed plane frame = 2j R In our problem, number of joints j = 6,
Number of restrictions imposed by supports = 3 Hence Kinematic indeterminacy = (2 * 6) 3 = 9 Fig.1.22 Determine the kinematic indeterminacy of the pin jointed space frame shown in the fig. 1.23 Kinematic indeterminacy of a pin jointed space frame = 3j R In our problem, number of joints j = 4, Number of restrictions imposed by supports = 6 Hence Kinematic indeterminacy = (3 * 4) 6 = 6 Fig.1.23 Determine the kinematic indeterminacy of the fixed beam with internal hinges shown in the fig. 1.24, without considering inextensibility of members (Assume members are extensible) Kinematic indeterminacy of plane rigid frame = 3j R In our problem, number of joints j = 4, Number of restrictions imposed by supports R = 6 Fig.1.24 Hence Degree of freedom = (3 * 4) 6 = 6 (The 6 degree of freedom are the two translations in x and y directions and one rotation in z direction of the two internal hinges) Determine the kinematic indeterminacy of the plane rigid frame shown in the fig. 1.25, considering inextensibility of members (Assume members are inextensible) G H D E F A B C Fig.1.25 Kinematic indeterminacy of a rigid jointed plane frame = 3j c = 3j (R + m) In our problem, number of joints j = 8,
Number of restrictions imposed by supports R = 6 Number of inextensible members = 8 (5 column members & 3 beam members) Hence Degree of Kinematic indeterminacy = (3 * 8) (6 + 8) = 10 Analysis of statically determinant trusses and frames (a) Analysis of statically determinant trusses The magnitude and sense of these forces can be determined using standard methods of analysis such as the method of sections, the method of joint-resolution, the method of tension coefficients or the use of computer software. The first two methods indicated are summarized and illustrated here. 1. Method of Sections The method of sections involves the application of the three equations of static equilibrium to two-dimensional plane frames. The sign convention adopted to indicate ties (i.e. tension members) and struts (i.e. compression members) in frames is as shown in Figure The method involves considering an imaginary section line which cuts the frame under consideration into two parts A and B as shown in Figure 3.4. Since only three independent equations of equilibrium are available any section taken through a frame must not include more than three members for which the internal force is unknown. Consideration of the equilibrium of the resulting force system enables the magnitude and sense (i.e. compression or tension) of the forces in the cut members to be determined. Example: Pin-Jointed Truss A pin-jointed truss supported by a pinned support at A and a roller support at G carries three loads at joints C, D and E as shown in Figure 3.2. Determine the magnitude and sense of the forces induced in members X, Y and Z as indicated.
Step 1: Evaluate the support reactions. It is not necessary to know any information regarding the frame members at this stage other than dimensions as shown in Figure 3.3, since only externally applied loads and reactions are involved. Apply the three equations of static equilibrium to the force system: Step 2: Select a section through which the frame can be considered to be cut and using the same three equations of equilibrium determine the magnitude and sense of the unknown forces (i.e. the internal forces in the cut members).
It is convenient to assume all unknown forces to be tensile and hence at the cut sectiontheir direction and lines of action are considered to be pointing away from the joints (refer to Figure 3.4). If the answer results in a negative force this means that the assumption of a tie was incorrect and the member is actually in compression, i.e. a strut. The application of the equations of equilibrium to either part of the cut frame will enable the forces X(FDE), Y(FEI) and Z(FHI) to be evaluated. Note: The section considered must not cut through more than three members with unknown internal forces since only three equations of equilibrium are applicable. Consider part A:
These answers can be confirmed by considering Part B of the structure and applying the equations as above. 2. Method of Joint Resolution Considering the same frame using joint resolution highlights the advantage of the method of sections when only a few member forces are required. In this technique (which can be considered as a special case of the method of sections), sections are taken which isolate each individual joint in turn in the frame, e.g. In Figure 3.6 four sections are shown, each of which isolates a joint in the structure as indicated in Figure 3.7.
Since in each case the forces are coincident, the moment equation is of no value, hence only two independent equations are available. It is necessary when considering the equilibrium of each joint to do so in a sequence which ensures that there are no more than two unknown member forces in the joint under consideration. This can be carried out until all member forces in the structure have been determined. Consider Joint G: Consider Joint F: substitute for calculated values, i.e. FFG (direction of force is into the Joint) Consider Joint H: substitute for calculated values, i.e. FGH and FFH
Consider Joint E: substitute for calculated values, i.e. FEF and FEH Analysis of statically determinant frames In the case of statically determinate frames, only the equations of equilibrium are required to determine the member forces. They are often used where there is a possibility of support settlement since statically determinate frames can accommodate small changes of geometry without inducing significant secondary stresses. Analysis of such frames is illustrated as follow: Example 1 Determine the reactions at the support for the frame shown in Fig. below. Solution Free-Body Diagram The free-body diagram of the frame is shown in Fig. 3.18(b). Note that the trapezoidal loading distribution has been divided into two simpler, uniform, and triangular, distributions whose areas and centroids are easier to compute
Static Determinacy The frame is internally stable with r ¼ 3. Therefore, it is statically determinate. Support Reactions By applying the three equations of equilibrium, we obtain. Example 2 Statically Determinate Frame A asymmetric portal frame is supported on a roller at A and pinned at support D as shown in Figure 5.3. For the loading indicated. Determine the support reactions.
Apply the three equations of static equilibrium to the force system Assuming positive bending moments induce tension inside the frame MB= (6.0 4.0ĩ(2.0)= 48.0 knm MC=+(46.5 3.0) (40.0 4.0)= 20.50 knm The values of the end-forces F1 to F8 can be determined by considering the equilibrium of each member and joint in turn. Consider member AB: Consider joint B:
Consider member BC: Consider member CD: Check joint C: The axial force and shear force in member CD can be found from: Axial load=+/ (Horizontal force Cosα)+/ (Vertical force Sinα) Shear force=+/ (Horizontal force Sinα)+/ (Vertical force Cosα) The signs are dependent on the directions of the respective forces. Member CD: