(1) Class Test Solution (STRUCTURE) Answer key. 31. (d) 32. (b) 33. (b) IES MASTER. 34. (c) 35. (b) 36. (c) 37. (b) 38. (c) 39.

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1 () ass Test Soution (STRUTUR) nswer key. (b). (b). (c). (a) 5. (b) 6. (a) 7. (c) 8. (c) 9. (b) 0. (d). (c). (d). (d). (c) 5. (b, d) 6. ( ) 7. (c) 8. (d) 9. (b) 0. (c). (a). (a). (b) (b) 5. (b) 6. (b) 7. (d) 8. (c) 9. (d) 0. (a). (d). (b). (b). (c) 5. (b) 6. (c) 7. (b) 8. (c) 9. (d) 0. (b). (b). (a). (c). (c) 5. (a) 6. (b) 7. (c) 8. (d) 9 (a) 50. (c) 5. (b) 5. (c) 5 (b) 5. (d) 55. (c) 56. (b) 57. (c) 58. (c) 59. (c) 60. (c) IS STR 6. (b) 6. (d) 6. (b) 6. (d) 65. (a) 66. (a) 67. (b) 68. (a) 69. (a) 70. (c) 7. (b) 7. (a) 7. (d) 7. (c) 75. (a) Regd. office : -6, (Upper asement), Katwaria Sarai, New ehi-006 Phone : ob. : , mai: ies_master@yahoo.co.in, info@iesmaster.org

2 () IVI NGINRING SS TST (STRUTUR) SS TST [STRUTUR] SOUTIONS. (b). (b). (c). (a) No. of cuts in (i) required No. of restraint required S 9 No. of cuts in (ii) required No. of restraint required S Unknown dispacement are,,,, i.e., K 5 y The unknown dispacements are 5. (b) 6. (a),,,,,, y Since and y y ence K 7 No. of cuts required No. of restraint required S G IS STR,,,,,, c G G Gy y 7. (c) 8. (c) here m 9 no. of joints j 6 here m j the pin jointed truss is internay determinate and stabe hence it is a perfect frame if m > j, the truss is internay indeterminate and stabe if m < j, truss is internay unstabe for simpe truss W W W W W 5 R R Nos. of reactions (unknown) (ie. R, R, R, & R ) but nos. of equiibrium equations avaiabe (ie. one 0) R y R 0 & another degree of indeterminacy If a body is sufficienty constraint by eterna reaction such that rigid body movement of structure does not occur, then the structure is said to be stabe eternay. Necessary condition for this is that : There shoud be three reactions that are neither concurrent nor parae (in pane structure). Reactions shoud be non-parae, nonconcurrent and non-copanar for space structure (concurrent means meeting at a singe point). Parae reaction : Incined oading wi ead to rigid body movement. ence unstabe eternay. Regd. office : -6, (Upper asement), Katwaria Sarai, New ehi-006 Phone : ob. : , mai: ies_master@yahoo.co.in, info@iesmaster.org

3 SS TST (STRUTUR) IVI NGINRING () 9. (b) oncurrent Reaction : ence unstabe O (R ) & / In 0. (d). (c) O O + O (R ) / R O R / 0 R 8 8 R R is very sma ied end moment at due to U.. W (cockwise) ied end moment due to sinking of support 6. 6 W 96 IS STR W 8 ence, fiing end moment at W W W 8 8 efection at in eam ompression in coumn. (d). (d) (50 R) I (50 R) R. R R 50 R R (50 R) 6 7R 800 R kn V kn. (c) w 8 w w w 8 0 w 6 w 8 R R W 8 ampes of orce ethod w w astigiano s Theorm (ethod of east Work) Strain energy method. aperon s three moment equations (used 0 Regd. office : -6, (Upper asement), Katwaria Sarai, New ehi-006 Phone : ob. : , mai: ies_master@yahoo.co.in, info@iesmaster.org

4 () IVI NGINRING SS TST (STRUTUR) in continous beam anaysis) oumn anaogy method (used in rigid frames with fied supports) eibiity matri method ampes of dispacement ethod 5. (b, d) Sope defection method oment distribution method Stiffners matri method Kani s ethod 0kN/ m 0kN/ + 0 m 0 0 or, gain, 0 rom (i) c IS STR...(i)...(ii) 6. ( ) or, + 80 ig. (i) in fig. (i) W in fig. (ii) W ig. (ii) (Note the symmetricity of fig. (ii).) ig. (ii) is nothing but ig. (i) + its mirror image In fig. (ii) (c) W...() W 8 W W 8 sin5 5 o (as ) sin Regd. office : -6, (Upper asement), Katwaria Sarai, New ehi-006 Phone : or sma 5 o ob. : , mai: ies_master@yahoo.co.in, info@iesmaster.org

5 SS TST (STRUTUR) IVI NGINRING (5) sin5 sin5 sin 5 8. (d) 9. (b) ( ) P P 5 0 kn m m R m 0 kn m 6m IS STR R (i) y 0 R R (ii) R R R 0... (iii) 0 0. (c) 6 R R... (iv) rom equation (iv) and equation (iii) kn Initiay ( )...() 6 / i the end R 0 ( ) 7 / R 0...() Regd. office : -6, (Upper asement), Katwaria Sarai, New ehi-006 Phone : ob. : , mai: ies_master@yahoo.co.in, info@iesmaster.org

6 (6) IVI NGINRING SS TST (STRUTUR). (a). (a). (b) P R 7 > 00 radian 0m 7 7 m y inspection, the sway woud be towards right. P. (b) 00 0 ember Stiffness O O O O IS STR 0 knm istribution factor for O oment in member O at O 00.8 arry over factor Therefore, bending moment at 5. (b) 6. (b) kn-m.8 Tota Re ative Jo int ember Re ative Stiffness Stiffness istribution factor (I) 8 / 8 (I) / Since end is free end, member has zero reative stiffness / / / ppying sope defection for beam θ θ iagram due to fied end moment P P P P 6 Regd. office : -6, (Upper asement), Katwaria Sarai, New ehi-006 Phone : ob. : , mai: ies_master@yahoo.co.in, info@iesmaster.org

7 SS TST (STRUTUR) IVI NGINRING (7) P P 8 6 P 6 7. (d) 8. (c) 9. (d) 0. (a) P P P 6 5P 8 The horizonta member of the frame wi have both the ends at reative eve difference of. It is simiar to ied end moment of the horizonta 6Δ member. IS STR istribution factor for I I I 7 7 istribution factor for 7 7. (d) istribution factor for istribution factor for. (b) R 6 0 R. 6 R 6 X 0 6 PQ PR 7 : 7 R 6 (for point contrafeure) Tension in member PQ Tension in member PR y ami s theorem at point P. PR 05 o 0 o P 5 o PQ PR sin5 PQ sin0 sin05 PQ sin0. sin ree body diagram at point Q, PQ 5 o Q QR Regd. office : -6, (Upper asement), Katwaria Sarai, New ehi-006 Phone : ob. : , mai: ies_master@yahoo.co.in, info@iesmaster.org

8 (8) IVI NGINRING SS TST (STRUTUR) 0 QR QP cos 5 o. (b) 5 kn 50 kn b 50 kn 50 kn. (c) I 5 50 QR QR m 6 m N I J G K m b 8 m 8 m 8 m N I IJ (a) (b) Section bb JK rom Section bb using fig. (b) I y K K IS STR K () 0 K 6.67 kn K 6.67 kn () 0 (.5) (.5) R 6 R 7.07 kn 5. (b) Taking moment of a forces about sin + R 0 sin R kn 0 v R + R I 0 0 R kn R I R I 8 kn R 8 kn o a cut between & G T 8 T 5m take moment about I G I 8 T G m T T T 6. (T) take moment about 0 kn T G 5 0 T G (om.) Regd. office : -6, (Upper asement), Katwaria Sarai, New ehi-006 Phone : ob. : , mai: ies_master@yahoo.co.in, info@iesmaster.org

9 SS TST (STRUTUR) IVI NGINRING (9) 6. (c) cos 60º+P G P Z or, P P + 50 U Using castigiano s theorem (nd theorem) 7. (b) The nature of the forces in members have been depicted in the above figure. To cacuate, horizonta difection of the joint, appying a force p at. m 500 kn 60º m 60º 60º P R R 0 R 50 kn R 50 kn t joint c, cos 0º + cos 0º 500 IS STR and Sin 0º sin 0º kn 500 kn t joint, Sin 60º 50 and 8. (c) 5 kn kn U P P. (P 50) mm P 0 5 kn 5 kn 0 kn 7.5 kn 7.5 kn 7.5 kn G 0 kn 7.5 kn 5.5 kn or getting horizonta defection at, and oad, in the horizonta direction is appied R R 5.5 kn R kn Using castigiano s method, n pi i p i i i i... (i) Where p i in member forces in the i th member due to combined actrion of eterna forces and. fterward, setting 0 in equation Regd. office : -6, (Upper asement), Katwaria Sarai, New ehi-006 Phone : ob. : , mai: ies_master@yahoo.co.in, info@iesmaster.org

10 (0) IVI NGINRING SS TST (STRUTUR) 9. (d) (i) Sign convention p p i i i i i at 0 Tensie force is taken as +ve and compressive force is taken as ve. If defection cacuated in equation (i) is the, it is in the direction of appied oad at the point of desired defection. Otherwise opposite to the direction of appied bad. onsidering joint at, G cos G 7.5 kn G sin kn at joint, G cos G.5 Thus, members having term invoved in their interna forces are member G and member G. Thus these two members ony wi be invoved in finding horizonta defection at point. Other remaining members have no. term invoved in their interna forces and on differention w.r.t force, wi give zero term. cos 5 sin 0.6 m IS STR 6.87º m abrication error case (ack of fit case) ember engths are generay made onger or shorter than actua required ength in order to introduce member forces which wi compensate for defection due to the dead oad. ue to this ack of fit, joint defection from origina eve wi take pace. This joint defection in this case is cacuated using virtua work principe.. uidi Where d i fabrication error in i th member Sign connection d i +ve if member is onger than normay epected d i ve if member is shorter than normay epected Tension the ompression ve +ve if defection in the direction of appied unit oad ve if defection opposite to the direction of appied unit oad. ere in question, vertica defection is found outr at point c, a unit oad at in vertica direction is appied and member forces one found out. t joint, sin and cos. t joint, t Joint, m 5m ember u (mm) U(mm) mm (down) Regd. office : -6, (Upper asement), Katwaria Sarai, New ehi-006 Phone : ob. : , mai: ies_master@yahoo.co.in, info@iesmaster.org

11 SS TST (STRUTUR) IVI NGINRING () 0. (b) et unit oad be at a distance from free end ( )/ / ( ) diagram Infuence ine ordinate for defection wi be (moment about the free end). (b) y ( ) ( ) ( ) ( ) ( ). 6 When oad is on at a distance from Now, R /5 c R 6 R /5 IS STR R 5 R 5 t /. Now, when oad is on at a distance y from R 6 y R y/6 y 6 R y/ t y / y /. When oad is in at y distance from. (a) 0 R (y ) y (y ) 6 t y 6 0 y 8 8 (). / y / ut section as shown : U U U U U 5 5 / 6 m m When unit oad is at R, R U (Tension) 6 5 m Regd. office : -6, (Upper asement), Katwaria Sarai, New ehi-006 Phone : ob. : , mai: ies_master@yahoo.co.in, info@iesmaster.org

12 () IVI NGINRING SS TST (STRUTUR) When unit oad is at R R. (c) U (ompression). Shear in Pane o o R R 0 S in Pane o 0 o R S in Pane o o R S in Pane o 0 R S in Pane o IS STR o R 0 S in Pane o 0. (c) 5. (a) oment at R o cos. o os o I for force in member o U is obtained bymutipying the ordinate of I for shear in pane o by sec aso by dividing the ordinate of I for moment at by cos o uer resau theorem is vaid for both staticay determinate and indeterminate structures. 6. (b) n infuence ine represents the variation of either the reaction, shear, moment or defection at a specified point in a member as a concentrated unit force moves over the member. Infuence ines represent the effect of a moving oad ony at a specified point on a member whereas shear and moment diagram represents the effect of fied oads at a points aong the member. Thus infuence ine heps in deciding at a gance where shoud the moving oads be paced on the structure so that it creates greatest infuence at the specified point. for K 8 & K R cos Rsec Regd. office : -6, (Upper asement), Katwaria Sarai, New ehi-006 Phone : ob. : , mai: ies_master@yahoo.co.in, info@iesmaster.org

13 SS TST (STRUTUR) IVI NGINRING () 7. (c) 8. (d) ence option (b) is correct. K K I.5I 6 (I) 55. (c) 56. (b) 57. (c) 6 K 9. (a) 50. (c) 5. (b) 5. (c) 5. (b) 5. (d) K K or equiibrium of point, the forces must be baanced. rawing reactions at point or baanced forces, member shoud be under tension whie member shoud be under compression as shown in above figure. ence, neither bar nor is subjected to bending. () () P () IS STR 6 R 0 m right 0 0R 5 h 5 m R R... () 0 eft 0R 5 + P (0 )...() rom () and () 0R 0 R + P (0 )...() R + R P... () rom () and () 0R 0 (P R ) + P (0 ) or 0R 0P + 0P P 0P P P (0 ) R 0 P 0 gain, 0R 5 + P(0 ) 0 P P P P 0 or 5 K or, 5 P K + 8 P 0 Regd. office : -6, (Upper asement), Katwaria Sarai, New ehi-006 Phone : ob. : , mai: ies_master@yahoo.co.in, info@iesmaster.org

14 () IVI NGINRING SS TST (STRUTUR) y question, 0 - P 0 R P 0 0 0V 0 V 0 kn V kn (c) 59. (c) m R P right 5 m R R R h 5 m 0.5 R 5.5 R... (i) y question; tan θ R.5R.5 R tanθ 0. V 0m IS STR kn/m 0 y y 0m 0 m V 0 V 0 0 or kn 0 kn 0 ending moment at 0 m from y y c 0 y 0 (0 0) 0 7.5m at 0 from. 0 + V 0 y kn 0 If be the incination of any point on the centre ine of an arch, then the vertica shear V and horizonta thrust woud combine to produce a radia shear R and a norma thrust N at the section. Then (or) R andp (or) N V cos sin V sin cos Radia shear force at 0m R Radia shear force V cos sin where V Net vertica shear force at 0 m from orizonta thrust tan dy d yc 0 tan tan Regd. office : -6, (Upper asement), Katwaria Sarai, New ehi-006 Phone : ob. : , mai: ies_master@yahoo.co.in, info@iesmaster.org

15 SS TST (STRUTUR) IVI NGINRING (5) V 0 V / 0 0 kn Radia shear force, sin R V cos sin 0 cos Norma thrust at 0 m from Norma thrust, 60. (c) N V sin cos V 0 sin cos kn R X N X If be the incination of any point on the centre ine of an arch, then the vertica shear V and horizonta thrust woud combine to produce a radia shear R and a norma thrust N at the section. Then (or) R X V cos sin andp (or) N V sin cos The behaviour of a structure subjected to horizonta forces depends on its height to width ratio. The deformation in ow-rise structures where the height is smaer than its width, is characterised predominanty by shear deformation. In high rise buiding where height is severa times greater than its atera dimensions, is dominated by bending action. There are two methodsnamey porta method and canti ever method to anayse the structures subjected to horizonta oading. Porta ethod : IS STR The porta method is an approimate anaysis for anaysing buiding frames subjected to atera oads such as wind oads/ seismic forces. Since, shear deformations are dominant in ow rise structures, the method makes simpyfying assumptions regarding horizonta shear in coumns. ach bay of a structure is treated as a porta frame and horizonta force is distributed equay among them. ssumptions in Porta ethod :. The points of infection are ocated at the mid-height of each coumn above the first foor. If the base of the coumn is fied, the point of infection is assumed at mid heght of the ground foor coumns as we otherwise, it is assumed at the hinged coumn base.. Points of infection occur at mid span of beams.. Tota horizonta shear at any foor is distributed among the coumns of that foor such that the eterior coumns carry haf the force carried by the inner coumns. The basis of this third assumption in the frame is composed of individua portas having one bay ony. antiever ethod : This method is appicabe to high rise structures. This is based on the simpyfing assumptions regarding the aia force in coumns. ssumptions in antiever ethod. The points of infection are ocated at the mid-height of each coumn above the first foor. If the base of the coumn is fied, the point of infection is assumed at mid height of the groun d foor coumns as we. Otherwise it is assumed at the hinged coumn base.. Points of infection occur at mid span of beams.. The basic assumption of the method can be stated as the aia force in the coumn at any foor is ineary proportiona to its Regd. office : -6, (Upper asement), Katwaria Sarai, New ehi-006 Phone : ob. : , mai: ies_master@yahoo.co.in, info@iesmaster.org

16 (6) IVI NGINRING SS TST (STRUTUR) 6. (b) 6. (d) 6. (b) 6. (d) 65. (a) distance from the centroid of a the coumns at that eve. Sope defection method is a stiffness method in wh ich u nknow n joint dispacement are found out by appying the equiibrium condition at end. In sope defection equation, we use the principe of superposition by considering separatey the moments deveoped at each support due to each of the dispacement,, and then the oads, so dispacement at joint are independent. Kani s method Gaspar Kani s method of structura anaysis is simiar to cross moment distribution in that both these methods use Gauss-Seide iteration procedure to sove the sope defection equation without epicity writting them down. owever, whereas the moment distribution method obtains the unknowns (i.e., the end moments of the structura members) by iterating their increments, Kani s method iterates these unknown s themseves. This method essentiay consists of a singe, simpe numerica operation performed repeatedy by the joints of a structure in a chosen sequence. Resuts of any desired accuracy may be obtained by peforming this operation a sufficient number of times using the required number of significant digits. Kani s method is speciay usefu for the anaysis of mutistorey frames. It has the advantages of simpicity, speed, economy of time, abour and space and of accuracy. owever, perhaps, the two most attractive features of this method are as foows: IS STR It has a buit-in error eimination so that computationa errors automaticay disappear in subsequent operations. This aso makes possibe the introduction of any changes in oads or member engths 66. (a) 67. (b) 68. (a) 69. (a) 70. (c) 7. (b) 7. (a) 7. (d) 7. (c) that may become necessary during cacuations without necessitating a new anaysis. Such changes are inserted in the computationa scheme wherever required and the anaysis simpy continued. It requires ony one tabe of cacuations even for highy irreguar frames with mutipe side sway. ompared to most other methods, Kani s method invoves substantiay ess abour and time in the anaysis of such frames. The above advantages make Kani s method one of th e most pow erfu techniques appicabe to a types of continuous beams and frames. oment distribution method. This method consists of soving sope defection equations by successive approimation that may be carried out to any desired degree of accuracy. ssentiay, the method begins by assuming each joint of a structure is fied. Then by unocking and ocking each joint in succession, the interna moments at the joints are distributed and baanced unti the joints have rotated to their fina or neary fina positions. This method of anaysis is both repetitive and easy to appy. anua anaysis of gabe frames mosty uses the moment distribution or sope defection methods. These methods are usuay engthy and have no buit-in-error eimination capabiity. or a three-hinged arch subjected to vertica Regd. office : -6, (Upper asement), Katwaria Sarai, New ehi-006 Phone : ob. : , mai: ies_master@yahoo.co.in, info@iesmaster.org

17 SS TST (STRUTUR) IVI NGINRING (7) oads ony, the horizonta support reactions at the arch springings are equa and oppoiste and act inward. The vertica support reactions at the arch springings are equa to those of a simpy supported beam of identica ength with identica oads. V V a W V V or the above three hinged arch, the vertica reaction at support is obtained by considering moment eqiibrium about support I. ence 0 V Wa and V wa/ V W V w( a/). These vaues for V and V are identica to the reactions of a simpy supported beam of the same span as the arch with the same appied oad W. The horizonta thrust at the springings is determined from a free body diagram of the right haf of the arch. onsidering moment equiibrium about the crown hinge at ; 0 V / V This vaue for is identica to the bending moment at the centre of a simpy supported beam of the same ength with the same appied oad W mutpied by /. The bending moment in the arch at any point a distance from the eft support is given by V.y for IS STR a V ( ) y for a < where y is the height of the arch at a distance from the eft support. The epressions for bending moment may be considered as the superposition of the bending moment of a simpy supported beam of the same span with the same appied oad W pus the bending moment due to the horizonta thrust. Wa( a)/ a due to appied oad on a S.S. beam 75. (a) due to horizonta thrust / / of the three hinged arch ethod of joint In a panar-truss, at every point there are two conditon of equiibrium 0 and 0 Since a the members at a joint are assumed to pass through a singe point, moment about the joint wi aways be zero. ence, 0 wi not be of any consequence. Thus two unknowns can be found out from two equiibrium equations. ikewise, we can proceed to other joints and find out member forces by using equiibrium equation if no. of unknown forces at the joint are at most two in number. Regd. office : -6, (Upper asement), Katwaria Sarai, New ehi-006 Phone : ob. : , mai: ies_master@yahoo.co.in, info@iesmaster.org

(1) Class Test Solution (STRUCTURE) Answer key. 31. (d) 32. (b) 33. (b) IES MASTER. 34. (c) 35. (b) 36. (c) 37. (b) 38. (c) 39.

(1) Class Test Solution (STRUCTURE) Answer key. 31. (d) 32. (b) 33. (b) IES MASTER. 34. (c) 35. (b) 36. (c) 37. (b) 38. (c) 39. () ass Test Soution (STRUTUR) 7-09-07 nswer key. (b). (b). (c). (a) 5. (b) 6. (a) 7. (c) 8. (c) 9. (b) 0. (d). (c). (d). (d). (c) 5. (d) 6. (a) 7. (c) 8. (d) 9. (b) 0. (c). (a). (a). (b) (b) 5. (b) 6.