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- 1. P a g e | 172 Prepared by R.Vijayakumar, B.Tech (CIVIL), CCET, Puducherry STRUCTURAL ANALYSIS – 1 UNIT – 1 1. Name any two force methods to analyze the statically indeterminate structures. Column analogy method Flexibility matrix method Method of consistent deformation Theorem of least work 2. Write the general steps of the consistent deformation method. By removing the restraint in the direction of redundant forces, released structure (which is a determinate structure) is obtained. In this released structure, displacements are obtained in the direction of the redundant forces. Then the displacement due to each redundant force is obtained and the conditions of displacement compatibility are imposed to get additional equations. Solution for these equations gives the values of redundant forces. Then the released structure subjected to these known forces gives the forces and moments in the structure. 3. Give example of beams of one degree static indeterminacy.
- 2. P a g e | 173 Prepared by R.Vijayakumar, B.Tech (CIVIL), CCET, Puducherry In general, 𝐸 = 𝑟 − 𝑒 For this case, 𝑟 = 4 𝑎𝑛𝑑 𝑒 = 3 ∴ 𝐸 = 4 – 3 = 1 4. Define degree of kinematic indeterminacy (or) Degree Of Freedom. It is defined as the least no of independent displacements required to define the deformed shape of a structure. There are two types of DOF Joint type DOF Nodal type DOF 5. Differentiate external redundancy and internal redundancy. In pin jointed frames, redundancy caused by too many members is called internal redundancy. Then there is external redundancy caused by too many supports. When we introduce additional supports/members, we generally ensure more safety and more work (in analysis). 6. Why to provide redundant members? To maintain alignment of two members during construction To increase stability during construction To maintain stability during loading (Ex: to prevent buckling of compression members) To provide support if the applied loading is changed To act as backup members in case some members fail or require strengthening Analysis is difficult but possible
- 3. P a g e | 174 Prepared by R.Vijayakumar, B.Tech (CIVIL), CCET, Puducherry 7. What are the different methods used to analyze indeterminate structures? Finite element method Flexibility matrix method Stiffness matrix method 8. What are statically indeterminate structures? Give example. If the conditions of statics i.e., ΣH=0, ΣV=0 and ΣM=0 alone are not sufficient to find either external reactions or internal forces in a structure, the structure is called a statically indeterminate structure. 9. Define consistent deformation method. This method is used for the analysis of indeterminate structure. This method is suitable when the number of unknown is one or two. When the number of unknown becomes more, it is a lengthy method. 10.Define primary structure. A structure formed by the removing the excess or redundant restraints from an Indeterminate structure making it statically determinate is called primary structure. This is required for solving indeterminate structures by flexibility matrix method.
- 4. P a g e | 175 Prepared by R.Vijayakumar, B.Tech (CIVIL), CCET, Puducherry 11.Write the formulae for degree of indeterminancy. Two dimensional in jointed truss (2D truss) 𝑖 = (𝑚 + 𝑟) − 2𝑗 Two dimensional rigid frames/plane rigid frames (2D frame) 𝑖 = (3𝑚 + 𝑟) − 3𝑗 Three dimensional space truss (3D truss) 𝑖 = (𝑚 + 𝑟) − 3𝑗 Three dimensional space frame (3D frame) 𝑖 = (6𝑚 + 𝑟) − 6𝑗 Where, m = number of members r = number of reactions j = number of joints 12.What is the effect of temperature on the members of a statically determinate plane truss? In determinate structures temperature changes do not create any internal stresses. The changes in lengths of members may result in displacement of joints. But these would not result in internal stresses or changes in external reactions. 13.Define internal and external indeterminancy. Internal indeterminacy (I.I) is the excess no of internal forces present in a member that make a structure indeterminate.
- 5. P a g e | 176 Prepared by R.Vijayakumar, B.Tech (CIVIL), CCET, Puducherry External indeterminacy (E.I) is the excess no of external reactions in the member that make a structure indeterminate. Indeterminacy (i) = I.I + E.I E.I = r – e; I.I = i – EI Where, r = no of support reactions and e = equilibrium conditions e = 3 (plane frames) and e = 6 (space frames) 14.What are the requirements to be satisfied while analyzing a structure? Equilibrium condition Compatibility condition Force displacement condition 15.Define degree of indeterminacy. The excess number of reactions take make a structure indeterminate is called degree of indeterminancy. Indeterminancy is also called degree of redundancy. Indeterminancy consists of internal and external indeterminancies. It is denoted by the symbol ‘i’. Degree of redundancy (i) = I.I + E.I Where, I.I = Internal indeterminancy E.I =External indeterminancy
- 6. P a g e | 177 Prepared by R.Vijayakumar, B.Tech (CIVIL), CCET, Puducherry 16.Differentiate the statically determinate structures and statically indeterminate structures. S. NO STATICALLY DETERMINATE STRUCTURES STATICALLY INDETERMINATE STRUCTURES 1. Conditions of equilibrium are sufficient to analyze the structure Conditions of equilibrium are insufficient to analyze the structure 2. Bending moment and shear force is independent of material and cross sectional area Bending moment and shear force is dependent of material and independent of cross sectional area 3. No stresses are caused due to temperature change and lack of fit Stresses are caused due to temperature change and lack of fit 4. Extra conditions like compatibility of displacements are not required to analyze the structure. Extra conditions like compatibility of displacements are required to analyze the structure along with the equilibrium equations. UNIT – 2 1. Distinguish between plane truss and plane frame. Plane frames are two-dimensional structures constructed with straight elements connected together by rigid and/or hinged connections. Frames are subjected to loads and reactions that lie in the plane of the structure.
- 7. P a g e | 178 Prepared by R.Vijayakumar, B.Tech (CIVIL), CCET, Puducherry If all the members of a truss and the applied loads lie in a single plane, the truss is called a plane truss. 2. What is meant by cambering technique in structures? Cambering is a technique applied on site, in which a slight upward curve is made in the structure/beam during construction, so that it will straighten out and attain the straight shape during loading. This will considerably reduce the downward deflection that may occur at later stages. 3. Give the procedure for unit load method. Find the forces P1, P2, ……. in all the members due to external loads Remove the external loads and apply the unit vertical point load at the joint if the vertical deflection is required and find the stress Apply the equation for vertical and horizontal deflection 4. What are the assumptions made in unit load method? The external & internal forces are in equilibrium Supports are rigid and no movement is possible The materials are strained well within the elastic limit 5. Why is it necessary to compute deflections in structures? Computation of deflection of structures is necessary for the following reasons: If the deflection of a structure is more than the permissible, the structure will not look aesthetic and will cause psychological upsetting of the occupants.
- 8. P a g e | 179 Prepared by R.Vijayakumar, B.Tech (CIVIL), CCET, Puducherry Excessive deflection may cause cracking in the materials attached to the structure. For example, if the deflection of a floor beam is excessive, the floor finishes and partition walls supported on the beam may get cracked and unserviceable. 6. Define unit load method. The external load is removed and the unit load is applied at the point, where the deflection or rotation is to found. 7. Distinguish between pin jointed and rigidly jointed structure. S. NO PIN JOINTED STRUCTURE RIGIDLY JOINTED STRUCTURE 1. The joints permit change of angle Between connected members. The members connected at a rigid joint will maintain the angle between them even under deformation due to loads. 2. The joints are incapable of transferring Any moment to the connected members and vice- versa. Members can transmit both forces and Moments between themselves through the joint. 3. The pins transmit forces between Connected members by developing shear. Provision of rigid joints normally increases the redundancy of the structures. 8. What are the conditions of equilibrium? The three conditions of equilibrium are the sum of horizontal forces, vertical forces and moments at any joint should be equal to zero.
- 9. P a g e | 180 Prepared by R.Vijayakumar, B.Tech (CIVIL), CCET, Puducherry (i.e.) ∑H = 0; ∑V = 0; ∑M = 0 9. Define trussed beam. A beam strengthened by providing ties and struts is known as Trussed Beams. 10.Define ‘deck’ and ‘through’ type truss. A deck type is truss is one in which the road is at the top chord level of the trusses. We would not see the trusses when we ride on the road way. A through type truss is one in which the road is at the bottom chord level of the trusses. When we travel on the road way, we would see the web members of the trusses on our left and right. That gives us the impression that we are going` through’ the bridge. 11.What is meant by lack of fit in a truss? One or more members in a pin jointed statically indeterminate frame may be a little shorter or longer than what is required. Such members will have to be forced in place during the assembling. These are called members having Lack of fit. Internal forces can develop in a redundant frame (without external loads) due to lack of fit. 12.Give any two situations where sway will occur in portal frames. Eccentric or Unsymmetrical loading Non-uniform section of the members
- 10. P a g e | 181 Prepared by R.Vijayakumar, B.Tech (CIVIL), CCET, Puducherry 13.What are the different types of forces acts on a frame? Dynamic load Static load 14.What is meant by settlement of supports? Support sinks mostly due to soil settlement. Rotation of ‘fixed’ ends can happen either because of soil settlement or upheaval of horizontal or inclined fixed ends. Fixed end moments induced in beam ends because of settlement or rotation of supports. 15.What is a rigid joint? The members connected at a rigid joint will maintain the angle between them even under deformation due to loads. Members can transmit both forces and moments between themselves through the joint. Provision of rigid joints normally increases the redundancy of the structures. 16.Write down the assumptions made in portal method. The point of contra-flexure in all the members lies at their middle points Horizontal shear taken by each interior column is double the horizontal shear taken by each of exterior column 17.Write down the assumptions made in cantilever method. The point of contra-flexure in all the members lies at their middle points
- 11. P a g e | 182 Prepared by R.Vijayakumar, B.Tech (CIVIL), CCET, Puducherry The direct stress or axial stress in the columns due to horizontal forces, are directly proportional to their distance from the centroidal vertical axis of the frame 18.What are the methods used to analyze the beam when it settle at supports? Kani’s method Moment distribution method Slope deflection method 19.Differentiate symmetry and anti-symmetry frames. SYMMETRY FRAME ANTI-SYMMETRY FRAME For symmetric loading, Symmetric quantities like bending moment, displacements are symmetrical about the centroidal vertical axis. For anti-symmetric loading, Symmetric quantities like bending moment, displacements are zero at the point of centroidal vertical axis. Anti-symmetric quantities like slope and shear force are zero at the point of centroidal vertical axis. Anti-symmetric quantities like slope and shear force are distributed about the centroidal vertical axis. 20.What is meant by thermal stress? Thermal stresses are stresses developed in a structure/member due to change in temperature. Normally, determinate structures do not develop thermal stresses. They can absorb changes in lengths and consequent displacements without developing stresses.
- 12. P a g e | 183 Prepared by R.Vijayakumar, B.Tech (CIVIL), CCET, Puducherry 21.Write any two important assumptions made in the analysis of trusses? The frame is a perfect frame The frame carries load at the joints All the members are pin-joined 22.Differentiate perfect and imperfect trusses? The frame which is composed of such members, which are just sufficient to keep the frame in equilibrium, when the frame is supporting an external load, is known as perfect frame. Hence for a perfect frame, the number of joints and number of members are given by, 𝑛 = 2𝑗 − 3 A frame in which number of members and number of joints are not given by 𝑛 = 2𝑗 − 3 is known as imperfect frame. This means that number of members in an imperfect frame will be either more or less than 2𝑗 − 3 23.Write the difference between deficient and redundant frames? If the number of members in a frame are less than (2𝑗 − 3), then the frame is known as deficient frame. If the number of members in a frame are more than (2𝑗 − 3), then the frame is known as redundant frame. UNIT – 3 1. What are the assumptions made in slope deflection method? This method is based on the following simplified assumptions. All the joints of the frame are rigid, (i.e.) the angle between the members at the joints does not change, when the members of frame are loaded.
- 13. P a g e | 184 Prepared by R.Vijayakumar, B.Tech (CIVIL), CCET, Puducherry Between each pair of the supports the beam section is constant. 2. Define fixed end moment. The moments at the fixed joints of loaded member are called fixed end moment. 3. Write down the slope deflection equation for a fixed end support. 𝑀𝐴𝐵 = 𝑀 𝐹𝐴𝐵 + 2𝐸𝐼 𝑙 [ 2𝜃 𝐴 + 𝜃 𝐵 + 3𝛿 𝑙 ] 4. What are the moments induced in a beam member, when one end is given a unit rotation, the other end being fixed. What is the moment at the near end called? When 𝜃 = 1, 𝑀𝐴𝐵 = 4 𝐸𝐼 𝑙 , 𝑀 𝐵𝐴 = 2 𝐸𝐼 𝑙 𝑀𝐴𝐵 Is the stiffness of AB at B 5. Define the term sway. Sway is the lateral movement of joints in a portal frame due to the unsymmetrical in dimensions, loads, moments of inertia, end conditions, etc. Sway can be prevented by unyielding supports provided at the beam level as well as geometric or load symmetry about vertical axis.
- 14. P a g e | 185 Prepared by R.Vijayakumar, B.Tech (CIVIL), CCET, Puducherry 6. What are the situations where in sway will occur in portal frames? Eccentric or unsymmetrical loading Unsymmetrical geometry Different end conditions of the column Non-uniform section of the members Unsymmetrical settlement of supports A combination of the above 7. What is the ratio of sway moments at column heads when one end is fixed and the other end hinged? Assume that the length and M.I of both legs are equal. Assuming the frame to sway to the right by δ Ratio of sway moments = 𝑀 𝐵𝐴 𝑀 𝐶𝐷 = − ( 6 𝐸𝐼 𝛿 𝑙2 ) − ( 3 𝐸𝐼 𝛿 𝑙2 ) = 2 8. A beam is fixed at its left end and simply supported at right. The right end sinks to a lower level by a distance ‘∆’ with respect to the left end. Find the magnitude and direction of the reaction at the right end if ‘l’ is the beam length and EI, the flexural rigidity.
- 15. P a g e | 186 Prepared by R.Vijayakumar, B.Tech (CIVIL), CCET, Puducherry 𝑀𝐴 (𝑑𝑢𝑒 𝑡𝑜 𝑠𝑖𝑛𝑘𝑖𝑛𝑔 𝑜𝑓 𝐵) = 3 𝐸𝐼 𝛿 𝑙2 9. What are the symmetric and anti-symmetric quantities in structural behavior? When a symmetrical structure is loaded with symmetrical loading, the bending moment and deflected shape will be symmetrical about the same axis. Bending moment and deflection are symmetrical quantities. 10.How many slope deflection equations are available for a two span continuous beam? There will be 4 nos. of slope-deflection equations are available for a two span continuous beam. 11.What are the quantities in terms of which the unknown moments are expressed in slope-deflection method? In slope-deflection method, unknown moments are expressed in terms of Slope (θ) Deflection (∆) 12.The beam shown in figure is to be analyzed by slope-deflection method. What are the unknowns and to determine them. What are the conditions used?
- 16. P a g e | 187 Prepared by R.Vijayakumar, B.Tech (CIVIL), CCET, Puducherry Unknowns: 𝜃 𝐴, 𝜃 𝐵, 𝜃 𝐶 Equilibrium equations used: 𝑀𝐴𝐵 = 0 𝑀 𝐵𝐴 + 𝑀 𝐵𝐶 = 0 𝑀 𝐶𝐵 = 0 13.How do your account for sway in slope deflection method for portal frames? Because of sway, there will be rotations in the vertical members of a frame. This causes moments in the vertical members. To account for this, besides the equilibrium, one more equation namely shear equation connecting the joint-moments is used. 14.Write down the equation for sway correction for the portal frame shown in figure. 𝑆ℎ𝑒𝑎𝑟 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 = 𝑀𝐴𝐵 + 𝑀 𝐵𝐴 𝑙1 + 𝑀 𝐶𝐷 + 𝑀 𝐷𝐶 𝑙2 = 0 15.Who introduced slope-deflection method of analysis? Slope-deflection method was introduced by Prof. George A. Maney in 1915.
- 17. P a g e | 188 Prepared by R.Vijayakumar, B.Tech (CIVIL), CCET, Puducherry 16.Write down the equilibrium equations for the frame shown in figure. Unknowns: 𝜃 𝐵, 𝜃 𝐶 Equilibrium equations used: 𝑀 𝐵𝐴 + 𝑀 𝐵𝐶 = 0 𝑀 𝐶𝐵 + 𝑀 𝐶𝐷 = 0 𝑆ℎ𝑒𝑎𝑟 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 = 𝑀𝐴𝐵 + 𝑀 𝐵𝐴 − 𝑃ℎ 𝑙 + 𝑀 𝐶𝐷 + 𝑀 𝐷𝐶 𝑙 + 𝑃 = 0 17.Write down the general slope-deflection equations and state what each term represents. General slope deflection equations: 𝑀𝐴𝐵 = 𝑀 𝐹𝐴𝐵 + 2𝐸𝐼 𝑙 [ 2𝜃 𝐴 + 𝜃 𝐵 + 3𝛿 𝑙 ] 𝑀 𝐵𝐴 = 𝑀 𝐹𝐵𝐴 + 2𝐸𝐼 𝑙 [ 2𝜃 𝐵 + 𝜃 𝐴 + 3𝛿 𝑙 ] Where, MFAB, MFBA = Fixed end moment at A and B respectively due to given loading 𝜃 𝐴, 𝜃 𝐵 = Slopes at A and B respectively 𝛿 = Sinking of support A with respect to B 18.How many slope-deflection equations are available for each span? Two numbers of slope-deflection equations are available for each span, describing the moment at each end of the span.
- 18. P a g e | 189 Prepared by R.Vijayakumar, B.Tech (CIVIL), CCET, Puducherry 19.In a continuous beam, one of the support sinks. What will happen to the span and support moments associated with the sinking of support. Let support D sinks by 𝛿. This will not affect span moments. Fixed end moments (support moments) will get developed as under 𝑀 𝐹𝐶𝐷 = 𝑀 𝐹𝐷𝐶 = − 6 𝐸𝐼 𝛿 𝑙1 2 𝑀 𝐹𝐷𝐸 = 𝑀 𝐹𝐸𝐷 = − 6 𝐸𝐼 𝛿 𝑙2 2 20.What is the basis on which the sway equation is formed for a structure? Sway is dealt with in slope-deflection method by considering the horizontal equilibrium of the whole frame taking into account the shears at the base level of columns and external horizontal forces. 𝑇ℎ𝑒 𝑠ℎ𝑒𝑎𝑟 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 𝑖𝑠 𝑀 𝐴𝐵 + 𝑀 𝐵𝐴 – 𝑃ℎ 𝑙 + 𝑀 𝐶𝐷 + 𝑀 𝐷𝐶 𝑙 + 𝑝 = 0 21.State the limitations of slope-deflection method. It is not easy to account for varying member sections It becomes very inconvenience when the unknown displacements are large in number This method is advantageous only for the structures with small Kinematic indeterminacy
- 19. P a g e | 190 Prepared by R.Vijayakumar, B.Tech (CIVIL), CCET, Puducherry The solution of simultaneous equation makes the method tedious for annual computations 22.Why slope-deflection method is called a ‘displacement method’? In slope-deflection method, displacements (like slopes and displacements) are treated as unknowns and hence the method is a ‘displacement method’. 23.Define Flexural rigidity of beams. The product of young’s modulus (E) and moment of inertia (I) is called Flexural Rigidity (EI) of Beams. The unit is Nmm2 . 24.Define constant strength beam. If the flexural Rigidity (EI) is constant over the uniform section, it is called Constant strength beam. 25.Define continuous beam. A Continuous beam is one, which is supported on more than two supports. For usual loading on the beam hogging (- ive) moments causing convexity upwards at the supports and sagging (+ ive) moments causing concavity upwards occur at mid span. 26.What are the advantages of continuous beam over simply supported beam? The maximum bending moment in case of continuous beam is much less than in case of simply supported beam of same span carrying same loads.
- 20. P a g e | 191 Prepared by R.Vijayakumar, B.Tech (CIVIL), CCET, Puducherry In case of continuous beam, the average bending moment is lesser and hence lighter materials of construction can be used to resist the bending moment. UNIT – 4 1. Explain moment distribution method (Hardy cross method). This method is first introduced by Professor Hardy Cross in 1932. It is widely used for the analysis of indeterminate structures. It uses an iterative technique. The method employs a few basic concepts and a few specialized terms such as fixed end moments, relative stiffness, carry over, distribution factor. In this method, all the members of the structure are first assumed to be fixed in position and fixed end moments due to external loads are obtained. 2. Define distribution factor. When several members meet at a joint and a moment is applied at the joint to produce rotation without translation of the members, the moment is distributed among all the members meeting at that joint proportionate to their stiffness. 𝐷𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 = 𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑠𝑡𝑖𝑓𝑓𝑛𝑒𝑠𝑠 𝑆𝑢𝑚 𝑜𝑓 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑠𝑡𝑖𝑓𝑓𝑛𝑒𝑠𝑠 𝑎𝑡 𝑡ℎ𝑒 𝑗𝑜𝑖𝑛𝑡 If there are three members, 𝐷𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟𝑠 = 𝑘1 𝑘1+ 𝑘2+ 𝑘3 , 𝑘2 𝑘1+ 𝑘2+ 𝑘3 , 𝑘3 𝑘1+ 𝑘2+ 𝑘3
- 21. P a g e | 192 Prepared by R.Vijayakumar, B.Tech (CIVIL), CCET, Puducherry 3. Define carry over factor. A moment applied at the hinged end B “carries over” to the fixed end A, a moment equal to half the amount of applied moment and of the same rotational sense. C.O =0.5 4. What is the difference between absolute and relative stiffness? Absolute stiffness is represented in terms of E, I and l, such as 4EI / l. Relative stiffness is represented in terms of ‘I’ and ‘l’, omitting the constant E. Relative stiffness is the ratio of stiffness to two or more members at a joint. 5. In a member AB, if a moment of -10kN.m is applied at A, what is the moment carried over to B? Carry over moment = Half of the applied moment ∴ Carry over moment to B = -10/2 = -5 kN.m 6. Define Stiffness factor. It is the moment required to rotate the end while acting on it through a unit rotation, without translation of the far end being 𝑆𝑖𝑚𝑝𝑙𝑦 𝑠𝑢𝑝𝑝𝑜𝑟𝑡𝑒𝑑 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦 (𝑘) = 3 𝐸𝐼 𝑙 𝐹𝑖𝑥𝑒𝑑 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦 (𝑘) = 4 𝐸𝐼 𝑙 Where, E = Young’s modulus of the beam material I = Moment of inertia of the beam L = Beam’s span length
- 22. P a g e | 193 Prepared by R.Vijayakumar, B.Tech (CIVIL), CCET, Puducherry 7. Define carry over moment. It is defined as the moment induced at the fixed end of the beam by the action of a moment applied at the other end, which is hinged. Carry over moment is the same nature of the applied moment. 8. What is the sum of distribution factors at a joint? Sum of distribution factors at a joint = 1. 9. What is the moment at a hinged end of a simple beam? Moment at the hinged end of a simple beam is zero. 10.A rigid frame is having totally 10 joints including support joints. Out of slope-deflection and moment distribution methods, which method would you prefer for analysis? Why? Moment distribution method is preferable. If we use slope-deflection method, there would be 10 (or more) unknown displacements and an equal number of equilibrium equations. In addition, there would be 2 unknown support moments per span and the same number of slope-deflection equations. Solving them is difficult. 11.What are the limitations of moment distribution method? This method is eminently suited to analyze continuous beams including non-prismatic members but it presents some difficulties when applied to rigid frames, especially when frames are subjected to side sway Unsymmetrical frames have to be analyzed more than once to obtain FM (fixed moments) in the structures This method cannot be applied to structures with intermediate hinges
- 23. P a g e | 194 Prepared by R.Vijayakumar, B.Tech (CIVIL), CCET, Puducherry UNIT – 5 1. What is the value of rotation moment at a fixed end considered in Kani’s method? 𝑀𝐴𝐵 = 2𝐸 𝐾𝐴𝐵 𝜃 𝐴 𝑀 𝐵𝐴 = 2𝐸 𝐾 𝐵𝐴 𝜃 𝐵 2. What are the fundamental equations of Kani’s method? ∑𝑀𝑖𝑗 = ∑𝑀 𝐹𝑖𝑗 + 2 ∑𝑀𝑖𝑗 ′ + ∑𝑀𝑗𝑖 = 0 ∑𝑀𝑖𝑗 ′ = − 1 2 ( ∑𝑀 𝐹𝑖𝑗 + ∑𝑀𝑗𝑖 ′ ) 3. What are the limitations of Kani’s method? Gasper Kani of Germany gave another distribution procedure in which instead of distributing entire moment in successive steps, only the rotation contributions are distributed Basic unknown like displacements which are not found directly 4. What are the advantages of Kani’s method? Hardy Cross method distributed only the unbalanced moments at joints, whereas Kani’s method distributes the total joint moment at any stage of iteration The more significant feature of Kani’s method is that the process is self- corrective. Any error at any stage of iteration is corrected in subsequent steps
- 24. P a g e | 195 Prepared by R.Vijayakumar, B.Tech (CIVIL), CCET, Puducherry Framed structures are rarely symmetric and subjected to side sway, hence Kani’s method is best and much simpler than other methods like moment distribution method and slope displacement method 5. State Miller-Breslau principle. Miller-Breslau principle states that, if we want to sketch the influence line for any force quantity like thrust, shear, reaction, support moment or bending moment in a structure, We remove from the structure the resistant to that force quantity We apply on the remaining structure a unit displacement corresponding to that force quantity. The resulting displacements in the structure are the influence line ordinates sought. 6. Define rotation factor. Rotation factor in Kani’s method is akin to distribution factor in moment distribution method. Actually, 𝑢 = − 0.5 × 𝐷𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 7. Define displacement factor. ∆𝑖𝑗 Is the “displacement factor” for each column, similar to 𝑢𝑖𝑗 we adopted earlier for rotation factor. Actually, ∆𝑖𝑗 = −1.5 𝐷𝐹 and is applicable to column only.
- 25. P a g e | 196 Prepared by R.Vijayakumar, B.Tech (CIVIL), CCET, Puducherry 8. Brief about Kani’s method of analysis. Kani’s method of analyzing indeterminate structures, particularly, building frames was developed in Germany in the year 1947 by Dr. Gasper Kani. Like moment distribution it is a method of solving slope deflection equations by an iterative method. Hence, this will fall under the category of stiffness method wherein the level of difficulty would be decided by the degrees of freedom (and not the degree of redundancy). 9. What are the basic principles of compatibility? Compatibility is defined as the continuity condition on the displacements of the structure after external loads are applied to the structure. 10.Define Kani’s method and how it is better than MDM. Kani’s method is similar to the MDM in that both these methods use Gauss Seidel iteration procedure to solve the slope deflection equations, without explicitly writing them down. However the difference between Kani’s method and the MDM is that Kani’s method iterates the member end moments themselves rather than iterating their increment Kani’s method essentially consists of a single simple numerical operation performed repeatedly at the joints of a structure, in a chosen sequence. 11.Write the procedure for Kani’s method. While solving structures by this method the following steps may be kept in mind. Compute all fixed end moments
- 26. P a g e | 197 Prepared by R.Vijayakumar, B.Tech (CIVIL), CCET, Puducherry Compute and tabulate all rotation factors for all joints that would have rotation. Fixed ends will not have rotation factors. Nor rotation contributions either to the same (fixed end) or to the opposite end. Extreme simply supported ends will initially get a fixed end moment. Iterative process can be formed. (Or) Fixed end moment Rotation factor Resultant restraint moment Iteration cycle Final moment 12.What are the methods of analyzing building frame? Cantilever method Factor method Portal method