STRUCTURAL ANALYSIS - I UNIT-I DEFLECTION OF DETERMINATE STRUCTURES

STRUCTURL NLYSIS - I UNIT-I DEFLECTION OF DETERMINTE STRUCTURES 1. Why is it necessary to compute defections in structures? Computation of defection of structures is necessary for the foowing reasons: a. If the defection of a structure is more than the permissibe, the structure wi not ook aesthetic and wi cause psychoogica upsetting of the occupants. b. Exessive defection may cause cracking in the materias attached to the structure. For exampe, if the defection of a foor beam is excessive, the foor finishes and partition was supported on the beam may get cracked and unserviceabe. 2. What is meant by cambering technique in structures? Cambering is a technique appied on site, in which a sight upward curve is made in the structure/beam during construction, so that it wi straighten out and attain the straight shape during oading. This wi consideraby reduce the downward defection that may occur at ater stages. 3. Name any four methods used for computation of defections in structures. 1. Doube integration method 2. Macauay s method 3. Conjugate beam method 4. Moment area method 5. Method of eastic weights 6. Virtua work method- Dummy unit oad method 7. Strain energy method 8. Wiiot Mohr diagram method 4. State the difference between strain energy method and unit oad method in the determination of defection of structures. In strain energy method, an imaginary oad P is appied at the point where the defection is desired to be determined. P is equated to zero in the fina step and the defection is obtained. In unit oad method, an unit oad (instead of P) is appied at the point where the defection is desired. 5. What are the assumptions made in the unit oad method? ` 1. The externa & interna forces are in equiibrium. 2. Supports are rigid and no movement is possibe. 3. The materias is strained we with in the eastic imit. 6. Give the equation that is used for the determination of defection at a given point in beams and frames. Defection at a point is given by, δ I = M x m x dx EI 0 Where M x = moment at a section X due to the appied oads m x = moment at a section X due to a unit oad appied at that point I and in the direction of the Desired dispacement EI = fexura rigidity

7. Write down the equations for moments due to the externa oad for beam shown in Fig. X 2 X 3 50KN X 1 B x 5m R X 1 R B x X 2 x X 3 10m Portion Mx Limits C R x 0 to 4 CD R x - 50(x-4) 4 to 5 DB R x - 50(x-4) 5 to 10 8. Distinguish between pin jointed and rigidy jointed structure. S.no Pin jointed structure Rigidy jointed structure 1. The joints permit change of ange The members connected at a rigid joint wi Between connected member. maintain the ange between them even 2. The joints are incapabe of transferring any moment to the connected members and vice-versa. under deformation due to oads. Members can transmit both forces and moments between themseves through the joint. 3. The pins transmit forces between Connected member by deveoping shear. Provision of rigid joints normay increases the redundancy of the structures. 9. What is meant by therma stresses? Therma stresses are stresses deveoped in a structure/member due to change in temperature. Normay, determine structures do not deveop therma stresses. They can absorb changes in engths and consequent dispacements without deveoping stresses. 10. What is meant by ack of fit in a truss? One or more members in a pin jointed staticay indeterminate frame may be a itte shorter or onger than what is required. Such members wi have to be forced in pace during the assembing. These are caed members having Lack of fit. Interna forces can deveop in a redundant frame (without externa oads) due to ack of fit. 11. Write down the two methods of determining dispacements in pin jointed pane frames by the unit oad concept. The methods of using unit oads to compute dispacements are, i) dummy unit oad method. ii) Using the principe of virtua work.

12. What is the effect of temperature on the members of a staticay determinate pane truss? In determinate structures temperature changes do not create any interna stresses. The changes in engths of members may resut in dispacement of joints. But these woud not resut in interna stresses or changes in externa reactions. 13. Distinguish between deck type and through type trusses. deck type is truss is one in which the road is at the top chord eve of the trusses. We woud not see the trusses when we ride on the road way. through type truss is one in which the road is at the bottom chord eve of the trusses. When we trave on the road way, we woud see the web members of the trusses on our eft and right. That gives us the impression that we are going` through the bridge. 14. Define static indeterminacy of a structure. If the conditions of statics i.e., ΣH=0, ΣV=0 and ΣM=0 aone are not sufficient to find either externa reactions or interna forces in a structure, the structure is caed a staticay indeterminate structure. 15. Briefy outine the steps for determining the rotation at the free end of the cantiever oaded as shown in Fig. W X ns: x W B B B 1 a. Mx = -Wx X b. m x = -1 c. d. θ B = M x m x dx = (-Wx) (-1) dx 0 EI EI 0 16. The horizonta dispacement of the end D of the porta frame is required. Determine the reevant equations due to the unit oad at appropriate point. 4m 30 KN B 3m 3m C E D

ns: X X B C E x x 4m X X X X X x x 1 D ppy unit force in the horizonta direction at D. m x vaues are tabuated as beow: Portion m x Limits DC 1x 0 to 4m CE 1 4 0 to 3m EB 1 4 3 to 6m B 1x 0 to 4m 17. Differentiate the staticay determinate structures and staticay indeterminate structures? S.No staticay determinate structures staticay indeterminate structures 1. Conditions of equiibrium are sufficient to anayze the structure Conditions of equiibrium are insufficient to anayze the structure 2. Bending moment and shear force is independent of materia and cross sectiona area. Bending moment and shear force is dependent of materia and independent of cross sectiona area. 3. No stresses are caused due to temperature change and ack of fit. Stresses are caused due to temperature change and ack of fit. 18. Define : Trussed Beam. beam strengthened by providing ties and struts is known as Trussed Beams. 19. Define: Unit oad method. The externa oad is removed and the unit oad is appied at the point, where the defection or rotation is to found. 20. Give the procedure for unit oad method. 1. Find the forces P1, P2,. in a the members due to externa oads. 2. Remove the externa oads and appy the unit vertica point oad at the joint if the vertica defection is required and find the stress. 3. ppy the equation for vertica and horizonta defection.

UNIT-II INFLUENCE LINES 1. Where do you get roing oads in practice? Shifting of oad positions is common enough in buidings. But they are more pronounced in bridges and in gantry girders over which vehices keep roing. 2. Name the type of roing oads for which the absoute maximum bending moment occurs at the midspan of a beam. (i) Singe concentrated oad (ii) ud onger than the span (iii) ud shorter than the span (iv) so when the resutant of severa concentrated oads crossing a span, coincides with a concentrated oad then aso the maximum bending moment occurs at the centre of the span. 3. What is meant by absoute maximum bending moment in a beam? When a given oad system moves from one end to the other end of a girder, depending upon the position of the oad, there wi be a maximum bending moment for every section. The maximum of these bending moments wi usuay occur near or at the midspan. The maximum of maximum bending moments is caed the absoute maximum bending moment. 4. Where do you have the absoute maximum bending moment in a simpy supported beam when a series of whee oads cross it? When a series of whee oads crosses a simpy supported beam, the absoute maximum bending moment wi occur near midspan under the oad W cr, nearest to midspan (or the heaviest oad). If W cr is paced to one side of midspan C, the resutant of the oad system R sha be on the other side of C; and W cr and R sha be equidistant from C. Now the absoute maximum bending moment wi occur under W cr. If W cr and R coincide, the absoute maximum bending moment wi occur at midspan. 5. What is the absoute maximum bending moment due to a moving ud onger than the span of a simpy supported beam? When a simpy supported beam is subjected to a moving ud onger than the span, the absoute maximum bending moment occurs when the whoe span is oaded. M max max = w 2 8 6. State the ocation of maximum shear force in a simpe beam with any kind of oading. In a simpe beam with any kind of oad, the maximum positive shear force occurs at the eft hand support and maximum negative shear force occurs at right hand support. 7. What is meant by maximum shear force diagram? Due to a given system of roing oads the maximum shear force for every section of the girder can be worked out by pacing the oads in appropriate positions. When these are potted for a the sections of the girder, the diagram that we obtain is the maximum shear force diagram. This diagram yieds the design shear for each cross section. 8. What is meant by infuence ines? n infuence ine is a graph showing, for any given frame or truss, the variation of any force or dispacement quantity (such as shear force, bending moment, tension, defection) for a positions of a moving unit oad as it crosses the structure from one end to the other. 9. What are the uses of infuence ine diagrams? (i) Infuence ines are very usefu in the quick determination of reactions, shear force, bending moment or simiar functions at a given section under any given system of moving oads and (ii) Infuence ines are usefu in determining the oad position to cause maximum vaue of a given function in a structure on which oad positions can vary.

10. Draw the infuence ine diagram for shear force at a point X in a simpy supported beam B of span m. 1 X B x (-x) (-x) + x/ 11. Draw the ILD for bending moment at any section X of a simpy supported beam and mark the ordinates. 1 X B x (-x) (-x) 12. What do you understand by the term reversa of stresses? In certain ong trusses the web members can deveop either tension or compression depending upon the position of ive oads. This tendancy to change the nature of stresses is caed reversa of stresses. 13. State Muer-Bresau principe. Muer-Bresau principe states that, if we want to sketch the infuence ine for any force quantity (ike thrust, shear, reaction, support moment or bending moment) in a structure, (i) We remove from the structure the resistant to that force quantity and (ii) We appy on the remaining structure a unit dispacement corresponding to that force quantity. The resuting dispacements in the structure are the infuence ine ordinates sought. 14. State Maxwe-Betti s theorem. In a ineary eastic structure in static equiibrium acted upon by either of two systems of externa forces, the virtua work done by the first system of forces in undergoing the dispacements caused by the second system of forces is equa to the virtua work done by the second system of forces in undergoing the dispacements caused by the first system of forces. 15. What is the necessity of mode anaysis? (i) When the mathematica anaysis of probem is virtuay impossibe. (ii) Mathematica anaysis though possibe is so compicatedand time consuming that the mode anaysis offers a short cut. (iii) The importance of the probem is such that verification of mathematica anaysis by an actua test is essentia. 16. Define simiitude. Simiitude means simiarity between two objects namey the mode and the prototype with regard to their physica characteristics: Geometric simiitude is simiarity of form Kinematic simiitude is simiarity of motion Dynamic and/or mechanica simiitude is simiarity of masses and/or forces.

17. State the principe on which indirect mode anaysis is based. The indirect mode anaysis is based on the Muer Bresau principe. Muer Bresau principe has ead to a simpe method of using modes of structures to get the infuence ines for force quantities ike bending moments, support moments, reactions, interna shears, thrusts, etc. To get the infuence ine for any force quantity, (i) remove the resistant due to the force, (ii) appy a unit dispacement in the direction of the (iii) pot the resuting dispacement diagram. This diagram is the infuence ine for the force. 18. What is the principe of dimensiona simiarity? Dimensiona simiarity means geometric simiarity of form. This means that a homoogous dimensions of prototype and mode must be in some constant ratio. 19. What is Begg s deformeter? Begg s deformeter is a device to carry out indirect mode anaysis on structures. It has the faciity to appy dispacement corresponding to moment, shear or thrust at any desired point in the mode. In addition, it provides faciity to measure accuratey the consequent dispacements a over the mode. 20. Name any four mode making materias. Perspex, pexigass, acryic, pywood, sheet aradite and bakeite are some of the mode making materias. Micro-concrete, mortar and paster of paris can aso be used for modes. 21. What is dummy ength in modes tested with Begg s deformeter. Dummy ength is the additiona ength (of about 10 to 12mm) eft at the extremities of the mode to enabe any desired connection to be made with the gauges. 22. What are the three types of connections possibe with the mode used with Begg s deformeter. (i) Hinged connection (ii) Fixed connection (iii) Foating connection 23. What is the use of a micrometer microscope in mode anaysis with Begg s deformeter. Micrometer microscope is an instrument used to measure the dispacements of any point in the x and y directions of a mode during tests with Begg s deformeter.

UNIT-III RCHES 1.What is an arch? n arch is defined as a curved girder, having convexity upwards and supported at its ends. The supports must effectivey arrest dispacements in the vertica and horizonta directions. Ony then there wi be arch action. 2.What is a inear arch? If an arch is to take oads, say W 1, W 2, and W 3 (fig) and a Vector diagram and funicuar poygon are potted as shown, the funicuar poygon is known as the inear arch or theoretica arch. q Q R W 1 D W 3 W 1 W 2 W 3 P Q R S O t P C O E S p W 2 H Vector Diagram s r T B Space Diagram The poar distance ot represents the horizonta thrust. The inks C, CD, DE, and EB wi be under compression and there wi be no bending moment. If an arch of this shape CDEB is provided, there wi be no bending moment. For a given set of vertica oads W 1, W 2..etc., we can have any number of inear arches depending on where we choose O or how much horizonta thrust (ot) we choose to introduce. 3.State Eddy s theorem. Eddy s theorem states that The bending moment at any section of an arch is proportiona to the vertica intercept between the inear arch (or theoretica arch) and the centre ine of the actua arch. BM x = Ordinate O 2 O 3 x scae factor X W 2 W 1 W 3 o 2 o 3 ctua arch Theoretica arch x o 1 X 4.What is the degree of static indeterminacy of a three hinged paraboic arch? For a three hinged paraboic arch, the degree of static indeterminacy is zero. It is staticay determinate.

5.Expain with the aid of a sketch, the norma thrust and radia shear in an arch rib. H B H N R Let us take a section X of an arch. (fig (a) ). Let θ be the incination of the tangent at X. If H is the horizonta thrust and V the vertica shear at X, from the free body of the RHS of the arch, it is cear that V and H wi have norma and radia components given by, N= H cosθ + V sinθ R= V cosθ - H sinθ 6.Which of the two arches, viz. circuar and paraboic is preferabe to carry a uniformy distributed oad? Why? Paraboic arches are preferaby to carry distributed oads. Because, both, the shape of the arch and the shape of the bending moment diagram are paraboic. Hence the intercept between the theoretica arch and actua arch is zero everywhere. Hence, the bending moment at every section of the arch wi be zero. The arch wi be under pure compression which wi be economica. 7.What is the difference between the basic action of an arch and a suspension cabe? n arch is essentiay a compression member which can aso take bending moments and shears. Bending moments and shears wi be absent if the arch is paraboic and the oading uniformy distributed. cabe can take ony tension. suspension bridge wi therefore have a cabe and a stiffening girder. The girder wi take the bending moment and shears in the bridge and the cabe, ony tension. Because of the thrusts in the cabes and arches, the bending moments are consideraby reduced. If the oad on the girder is uniform, the bridge wi have ony cabe tension and no bending moment on the girder. 8.Under what conditions wi the bending moment in an arch be zero throughout. The bending moment in an arch throughout the span wi be zero, if (i) The arch is paraboic and (ii) The arch carries uniformy distributed oad throughout the span. 9.Draw the ILD for bending moment at a section X at a distance x from the eft end of a three hinged paraboic arch of span and rise h. M x = µ x Hy µ x Hy (+) (-) x(-x)/ x(-x)/ 10. Indicate the positions of a moving point oad for maximum negative and positive bending moments in a three hinged arch.

Considering a three hinged paraboic arch of span and subjected to a moving point oad W, the position of the point oad for a. Maximum negative bending moment is 0.25 from end supports. b. Maximum positive bending moment is 0.211 from end supports. 11. Draw the infuence ine for radia shear at a section of a three hinged arch. Radia shear is given by F x = H sinθ - V cosθ, where θ is the incination of tangent at X. x cosθ sinθ 4r x cosθ 12. Sketch the ILD for the norma thrust at a section X of a symmetric three hinged paraboic arch. Norma thrust at X is given by P = H cosθ + V sinθ, where θ is the incination of tangent at X. x sinθ (-x)sinθ 4y c cosθ 13. Distinguish between two hinged and three hinged arches. S.No. Two hinged arches Three hinged arches 1. Staticay indeterminate to first degree Staticay determinate 2. Might deveop temperature stresses Increase in temperature causes increase in centra rise. No stresses. 3. Structuray more efficient Easy to anayse. But in costruction, the centra hinge may invove additiona expenditure. 4. Wi deveop stresses due to sinking of supports Since this is determinate, no stresses due to support sinking. 14. Expain rib-shortening in the case of arches. In a two hinged arch, the norma thrust which is a compressive force aong the axis of the arch wi shorten the rib of the arch. This in turn wi reease part of the horizonta thrust. Normay, this effect is not considered in the anaysis (in the case of two hinged arches). Depending upon the importance of the work we can either take into account or omit the effect of rib shortening. This wi be done by considering (or omitting) strain energy due to axia compression aong with the strain energy due to bending in evauating H.

15. Define the effect of yieding of support in the case of an arch. Yieding of supports has no effect in the case of a 3 hinged arch which is determinate. These dispacements must be taken into account when we anayse 2 hinged or fixed arches under U = H instead of zero H U = V instead of zero V Here U is the strain energy of the arch and H and V are the dispacements due to yieding of supports. 16. Write the formua to cacuate the change in rise in three hinged arch. Change in rise 2 + 4r 2 α T 4r where = span ength of the arch r = centra rise of the arch α = coefficient of therma expansion T = change in temperature 17. In a paraboic arch with two hinges how wi you cacuate the sope of the arch at any point? Sope of paraboic arch = θ = tan -1 4r ( 2x) 2 where θ = Sope at any point x (or) incination of tangent at x. = span ength of the arch r = centra rise of the arch 18. How wi you cacuate the horizonta thrust in a two hinged paraboic arch if there is a rise in temperature? Horizonta thrust = α TEI y 2 dx 0 where = span ength of the arch y = rise of the arch at any point x α = coefficient of therma expansion T = change in temperature E = Young s Moduus of the materia of the arch I = Moment of inertia 19. What are the types of arches according to the support conditions? i. Three hinged arch ii. Two hinged arch iii. Singe hinged arch iv. Fixed arch (or) hingeess arch 20. What are the types of arches according to their shapes? i. Curved arch ii. Paraboic arch iii. Eiptica arch iv. Poygona arch

UNIT-1V SLOPE-DEFLECTION METHOD 1. What are the assumptions made in sope-defection method? (i) Between each pair of the supports the beam section is constant. (ii) The joint in structure may rotate or defect as a whoe, but the anges between the members meeting at that joint remain the same. 2. How many sope defection equations are avaiabe for a two span continuous beam? There wi be 4 nos. of sope-defection equations, two for each span. 3. What is the moment at a hinged end of a simpe beam? Moment at the hinged ends of a simpe beam is zero. 4. What are the quantities in terms of which the unknown moments are expressed in sopedefection method? In sope-defection method, unknown moments are expressed in terms of (i) sopes (θ) and (ii) defections ( ) 5. The beam shown in Fig. is to be anaysed by sope-defection method. What are the unknowns and, to determine them, what are the conditions used? B C Unknowns: θ, θ B, θ C Equiibrium equations used: (i) M B = 0 (ii) M B + M BC = 0 (iii) M CB = 0 6. How do you account for sway in sope defection method for porta frames? Because of sway, there wi br rotations in the vertica members of a frame. This causes moments in the vertica members. To account for this, besides the equiibrium, one more equation namey shear equation connecting the joint-moments is used. 7. Write down the equation for sway correction for the porta frame shown in Fig. The shear equation (sway correction) is M B + M B + M CD + M DC = 0 D 8. Write down the sope defection equation for a fixed end support. B C D The sope defection equation for end is M B = M B + 2EI 2θ + θ B + 3 Here θ = 0. Since there is no support settement, = 0. M B = M B + 2EI θ B + 3

9. Write down the equiibrium equations for the frame shown in Fig. B C Unknowns : θ B, θ C Equiibrium equations : t B, M B + M BC = 0 h t C, M CB + M CD = 0 P Shear equation : M B + M B Ph + M CD + M DC + P = 0 D 10. Who introduced sope-defection method of anaysis? Sope-defection method was introduced by Prof. George.Maney in 1915. 11. Write down the genera sope-defection equations and state what each term represents? B Genera sope-defection equations: M B = M B + 2EI 2θ + θ B + 3 M B = M B + 2EI 2θ B + θ + 3 where, M B, M B = Fixed end moment at and B respectivey due to the given oading. θ, θ B = Sopes at and B respectivey = Sinking of support with respect to B 12. Mention any three reasons due to which sway may occur in porta frames. Sway in porta frames may occur due to (i) unsymmetry in geometry of the frame (ii) unsymmetry in oading or (iii) Settement of one end of a frame. 13. How many sope-defection equations are avaiabe for each span? Two numbers of sope-defection equations are avaiabe for each span, describing the moment at each end of the span. 14. Write the fixed end moments for a beam carrying a centra cockwise moment. M B /2 /2 Fixed end moments : M B = M B = M 4 15. State the imitations of sope defection method. (i) It is not easy to account for varying member sections (ii) It becomes very cumbersome when the unknown dispacements are arge in number.

16.Why is sope-defection method caed a dispacement method? In sope-defection method, dispacements (ike sopes and dispacements) are treated as unknowns and hence the method is a dispacement method. 17. Define degrees of freedom. In a structure, the numbers of independent joint dispacements that the structures can undrgo are known as degrees of freedom. 18. In a continuous beam, one of the support sinks. What wi happen to the span and support moments associated with the sinking of support. C D E 1 2 Let support D sinks by. This wi not affect span moments. Fixed end moments (support moments) wi get deveoped as under M CD = M DC = -6EI 1 2 M DE = M ED = -6EI 1 2 19. rigid frame is having totay 10 joints incuding support joints. Out of sope-defection and moment distribution methods, which method woud you prefer for anaysis? Why? Moment distribution method is preferabe. If we use sope-defection method, there woud be 10 (or more) unknown dispacements and an equa number of equiibrium equations. In addition, there woud be 2 unknown support momentsper span and the same number of sope-defection equations. Soving them is difficut. 20. What is the basis on which the sway equation is formed for a structure? Sway is deat with in sope-defection method by considering the horizonta equiibrium of the whoe frame taking into account the shears at the base eve of coumns and externa horizonta forces. The shear condition is M B + M B Ph + M CD + M DC + P = 0

UNIT-V MOMENT DISTRIBUTION METHOD 1. What is the difference between absoute and reative stiffness? bsoute stiffness is represented in terms of E, I and, such as 4EI /. Reative stiffness is represented in terms of I and, omitting the constant E. Reative stiffness is the ratio of stiffness to two or more members at a joint. 2. Define: Continuous beam. Continuous beam is one, which is supported on more than two supports. For usua oading on the beam hogging ( - ive ) moments causing convexity upwards at the supports and sagging ( + ve ) moments causing concavity upwards occur at mid span. 3. What are the advantages of Continuous beam over simpy supported beam? 1. The maximum bending moment in case of continuous beam is much ess than in case of simpy supported beam of same span carrying same oads. 2. In case of continuous beam, the average bending moment is esser and hence ighter materias of construction can be used to resist the bending moment. 4. In a member B, if a moment of 10 KNm is appied at, what is the moment carried over to B? Carry over moment = Haf of the appied moment Carry over moment to B = -10/5 = -5 KNm 5. 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 caed? When θ = 1, B M B = 4EI M B = 2EI θ = 1 M B is the stiffness of B at B. 6. beam is fixed at and simpy supported at B and C. B = BC =. Fexura rigidities of B and BC are 2EI and EI respectivey. Find the distribution factors at joint B if no moment is to be transferred to support C B C Joint B: Reative stiffness: I 1 = 2I for B. K B = 2 3 x I 1 = 3I for BC K BC = ¾ = 0.75 4 4 Distribution factors: DF for B: K B = 2 = 8/11 = 0.727 K B + K BC 2 + 0.75 DF for BC: K BC = 0.75 = 3/11 = 0.273 K BC + K B 2 + 0.75

7. Define: Moment distribution method.( Hardy Cross method). It is widey used for the anaysis of indeterminate structures. In this method, a the members of the structure are first assumed to be fixed in position and fixed end moments due to externa oads are obtained. 8. Define: Stiffness factor. It is the moment required to rotate the end whie acting on it through a unit rotation, without transation of the far end being (i) Simpy supported is given by k = 3 EI / L (ii) Fixed is given by k = 4 EI / L Where, E = Young s moduus of the beam materia. I = Moment of inertia of the beam L = Beam s span ength. 9. Define: Distribution factor. When severa members meet at a joint and a moment is appied at the joint to produce rotation without transation of the members, the moment is distributed among a the members meeting at that joint proportionate to their stiffness. Distribution factor = Reative stiffness / Sum of reative stiffness at the joint If there is 3 members, Distribution factors = k 1, k 2, k3 k 1 + k 2 + k 3 k 1 + k 2 + k 3 k 1 + k 2 + k 3 10. Define: Carry over moment and Carry over factor. Carry over moment: It is defined as the moment induced at the fixed end of the beam by the action of a moment appied at the other end, which is hinged. Carry over moment is the same nature of the appied moment. Carry over factor (C.O): moment appied at the hinged end B carries over to the fixed end, a moment equa to haf the amount of appied moment and of the same rotationa sense. C.O =0.5 11. Define Fexura Rigidity of Beams. The product of young s moduus (E) and moment of inertia (I) is caed Fexura Rigidity (EI) of Beams. The unit is N mm 2. 12. Define: Constant strength beam. If the fexura Rigidity (EI) is constant over the uniform section, it is caed Constant strength beam. 13. What is the sum of distribution factors at a joint? Sum of distribution factors at a joint = 1. 14. Define the term sway. Sway is the atera movement of joints in a porta frame due to the unsymmetry in dimensions, oads, moments of inertia, end conditions, etc.

15. Find the distribution factor for the given beam. L B L C L D Joint Member Reative stiffness Sum of Reative stiffness Distribution factor B 4EI / L 4EI / L (4EI / L) / (4EI / L) = 1 B B BC 3EI /L 4EI / L 3EI /L + 4EI / L = 7EI / L (3EI / L) / (7EI / L )= 3/7 (4EI / L) / (7EI / L) = 4/7 C CB CD 4EI / L 4EI / L 4EI / L + 4EI / L =8EI / L (4EI / L) / (8EI / L) =4/8 (4EI / L) / (8EI / L)= 4/8 D DC 4EI / L 4EI / L (4EI / L)/ (4EI / L) = 1 16. Find the distribution factor for the given beam. L ( 3I) B L (I) C Join Member Reative stiffness Sum of Reative stiffness Distribution factor B 4E (3I ) / L 12EI / L (12EI / L) / (12EI / L) = 1 B B BC 4E( 3I) /L 4EI / L 12EI /L + 4EI / L = 16EI / L (12EI / L) / (16EI / L )= 3/4 (4EI / L) / (16EI / L) = 1/4 C CB 4EI / L 4EI / L (4EI / L) / (4EI / L) =1 17. Find the distribution factor for the given beam. B L C L D Join Member Reative stiffness Sum of Reative stiffness Distribution factor B B BC 0( no support) 3EI / L 3EI / L 0 (3EI / L ) / ( 3EI / L) =1 C CB CD 3EI / L 4EI / L 3EI /L + 4EI / L = 7EI / L (3EI / L) / (7EI / L )= 3 / 7 (4EI / L) / (7EI / L) = 4 / 7 D DC 4EI / L 4EI / L (4EI / L) / (4EI / L) =1

18. What are the situations where in sway wi occur in porta frames? a. Eccentric or unsymmetric oading b. Unsymmetrica geometry c. Different end conditions of the coumns d. Non-uniform section of the members e. Unsymmetrica settement of supports f. combination of the above 19. What is the ratio of sway moments at coumn heads when one end is fixed and the other end hinged? ssume that the ength and M.I of both egs are equa. ssuming the frame to sway to the right by δ δ δ Ratio of sway moments = B C - 6EIδ M B = 2 = 2 M CD - 3EIδ 2 D 20. beam is fixed at its eft end and simpy supported at right. The right end sinks to a ower eve by a distance with respect to the eft end. Find the magnitude and direction of the reaction at the right end if is the beam ength and EI, the fexura rigidity. M (due to sinking of B) = 3EI 2 21. What are symmetric and antisymmetric quantities in structura behaviour? When a symmetrica structure is oaded with symmetrica oading, the bending moment and defected shape wi be symmetrica about the same axis. Bending moment and defection are symmetrica quantities.

Part B 1. pin-jointed frame shown in Fig. is carrying a oad of 6 tonnes at C. Find the vertica as we as horizonta defection at C. Take area of member B as 10cm 2 and those of members C and BC as 15cm 2. E = 2 x 10 3 t/cm 2. 4 m 4 m B 3 m 6 t 2. Using the method of virtua work, determine the horizonta dispacement of a point C of the frame shown in Fig. Take E = 2 x 10 5 N/mm 2, I = 4 x 10 6 mm 4. 10KN/m B C 2.5m 4m 30KN 2.5m 3. Using Infuence ine diagram, find (i) maximum Bending moment (ii) maximum positive and negative shears at 4m from eft support of a simpy supported irder of span 10m, when a train of 4 whee oads 10KN, 15KN, 30KN and 30KN spaced at 2m, 3m and 3m respectivey cross the span with the 10KN oad eading. 4. Draw the Infuence ine diagram for shear force and bending moment for a section at 5m from eft support of a simpy supported beam, 20m ong. Hence cacuate the maximum B.M and S,F at the section. Due to uniformy distributed roing oad of ength 8m and intensity 10KN/m run. 5. For the span shown in Fig., obtain the bending moment at a section P, 20m from, due to given oads in the position indicated. so determine the position of the oads for maximum bending moment of section P and the vaue of maximum bending moment. 8t 8t 16t 18t 17t 6m 5m 6m 7m 20m P 30m 6. uniformy distributed oad of 5t/m, onger than span, ros over a beam of 25m span. Using infuence ine, determine the maximum shear force and bending moment at a section 10m from the eft end support. 7. system of concentrated oads shown in Fig. Ros from eft to right across a beam simpy supported over a span of 10m, the 10KN oad eading. For a section 4m from the eft support, determine maximum shear force and bending moment. 8. ud of 40t/m covers eft hand haf of the span of a two hinged paraboic arch, span 36m and centra rise 8m. Determine the position and magnitude of maximum bending moment. so find shear force and norma thrust at the section. ssume that the moment of inertia at a section varies as secant of sope at the section.

9. three hinged paraboic arch of span 40m and rise 8m carries a ud of 30KN/m over the eft haf span. (i) nayse the arch and draw the bending moment diagram. (ii) so evauate the thrust and shear force at a section 10m from eft hinge. 10. three hinged circuar arch of span 16m and rise 4m is subjected to two point oads of 100KN and 80KN at eft and right quarter span points respectivey. Find the reaction at the supports. Find aso BM, radia shear and norma thrust at 6m from the eft support. 11. three hinged paraboic arch of span 30m has its supports at depths of 4m and 16m beow crown C. The arch carries a oad of 80KN at a distance of 5m to the eft of C and a second oad of 100KN at 10m to the right of C. Determine te reactions at supports amd BM under the oads. 12. three hinged paraboic arch has a horizonta span of 30m with a centra rise of 5m. a point oad of 10KN moves across the span from eft to right. Cacuate the maximum positive and negative moments at a section 8m from yhe eft hinge. so, cacuate the position and magnitude of the absoute maximum BM that may occur in the arch. 13. three hinged paraboic arch has a horizonta span of 36m with a centra rise of 6m. point oad of 8KN moves across from eft to right. Cacuate the maximum sagging and hogging BM at the section 9m from the eft hinge. Cacuate aso the position and amount of absoute maximum BM that may occur on the arch. 14. two hinged paraboic arch of span 25m, rise 6m is subjected to a ud of 15KN/m over the eft haf span and a point oad of 25KN T 9.5m from the right support. Find the support reactions, BM, radia shear and norma thrust at 4m from the eft support. 15. three hinged paraboic arch is subjected to a ud of 10KN/m for the eft haf portion. Using ILD, find the BM, radia shear and norma thrust at a section 4m from the eft support. 16. beam BC supported on a coumn BD is oaded as shown in Fig. nayse the frame by sope defection method and draw bending moment diagram. 3t/m 3t B C 4m 2m 3I I 4m I D 17. continuous beam of constant moment of inertiais oaded as shown in Fig. Find the support moments and draw bending moment diagram. 4000N 1m B C D 8m 3m 3m 3m 18. nayse the continuous beam shown in Fig. by moment distribution method and draw the bending moment diagram. 2KN/m 5KN 8KN B C D 6m 3m 2m 2.5m 2.5m I 2I I