PYRAMID IAS ACADEMY KARAIKUDI Max. Mark 75 ‘Max Duration 45 min (a) Both & and Rare true and ie the corect explanation of & (@) Both A and Rare true but Ris nota corect explanation of A (6) Alstrue but Ris false (4) Ais fale but Ris tue 1. If we use link support in a structural system, then how many unknowns would we have? al b)2 oo aa 2. A rigid prop is one which a) permits 50% of free deflection, that would have: ‘occurred if the prop were not there b) does not permit a lo the plane of prop ©) does not offer any reaction d) supports the entire load and relieves all other supports completely displacement perpendicular 3. Which of the following methods of structural analysis is a force method? 4) Slope deflection method b) Moment distribution method Column analogy method ——_d) Alll of these by 4) Using equations of static alone +b) Equation of compatibility alone «) Ignoring all deformations and assuming, th®™\ structure to be rigid < d) Using the equations of staties and the nces¥éry numbers of compatibility equations 5. Match List -1 with List - 11 ‘>> Tist-I |<) ( tat-l ‘Ay Space trusses _[1) Continuous structure B) Plate element | 3) Discrete structure ©) Arches 3)Statically Y| determinate structure 'D) Simply 4) One-dimensional supported beams _| dependent structure a) A-4, Bol, C-2,D-3 OA3,B-4, C1, D-2 b) A-2, Bal, C-4, D-3 d) A-1, B-2, C-3, D-4 6, What is the degree of kinematic indeterminacy of the frame shown in figure? Neglect axial deformation. | eI FLOOR, Baek COMPLEX Sil RAM NAGAR. KOTTANYUR, KARATKUDI-60 106 CONTACT 8431610, 85249 7970, $9521 60010, 4. Statically indeterminate structures can be solved \ i R aa HR 910 ) 08 7.Which of the following is carried by truss members? a) flexural load axial load b) bendlinys load d) shear load 8, Consider the following’statements. In the beam shown iftFig.foF all positions of load W (except x= 0) L.bending moment is paximum at B 2bending moment js maximum under load 3.deflection ig zet0 at A 4.deflection is Zero at B \ w @B) DA Hw as Of these statements: a)Tand 3 are correct b) 2and4 are correct Vand 4 are correct d) 1,3 and4 are correct 9. A cantilever beam of span'’ carries a point load of ‘w' al free end and it is propped at a distance of /4 from the free end then the prop reaction is a) 4w/2 by aw/2 ow a)w/2 10. The fixed end moment of end A of the fixed beam shown below Ww Ww : A B U3 U3 oe of ot 11. Statement (I): If degree of fixity at supports is lessened, the maximum hogging moment at the ends will decrease, Statement (II): If degree of fixity at supports is lessened, the maximum sagging moment at mid-span decreases.| 0.25 and 0.75 [22 ie single-span beam fixed at both ends sinks by A at one end, the fixed-end moment developed is | 90 1) 6ELa/P | orev 4) 12B14/# 13. Simply supported beam AB with udl is propped at centre. Then the prop reaction is equal to a) 5wl/8. b)3wi/8 —Qwi/8 dw | | | 14: The disibution factors for members AE and AC | ofthe box section are a) 05 and 05 b) Oo and 04 4d) and zero 15, For the structure shown in figure, the ratio of relative stiffness for AB and AC is se 2 we \ YY 16. Relative stiffness of a member in wjoint, whose other end is pinned, for calculating the distribution factor in moment distribution method is given by (where Tand L are specie ‘moment of Sneia nd the length ofthe member) a) i/L <) b)OSI/L ©) 0.25 1/L \ )0.751/L 17. The rotational stiffness of a prismatic beam clement is, 4) directly proportional toils area of cross-section ») inversely proportional to its area of cross-section ©) directly proportional to its length 4) inversely proportional to its length ae of 4 18. The fixed support condition in a real becomes in the conjugate beam a condition as a) fixed b) roller hinged d) free 19. Statement (I): Slope-deflection method is a displacement method of analysis. | Il FLOOR, Baek COMPLEX, SRI RAM NAGAR, KOTTAIYUR KARAIKUDLA30 106, CONTACT set 31610, 5249 70570, 58591 C010 pees Hoppots S N CODE: 2024/ REVISION TEST 1 / SENIOR BATCH Statement (Il: Forces are the basic unknowns in the slope-deflection method 20. The Final moment at the end A, in a beam AB due to rotations 0A, 0B and downward settlement of support at Bis given by 4) Man= Mian [264 + 85 - =] by Mas= Mran- [204 + 88 -<] 6) Man= Mras £2![204 + 85 - =] €) Man= Mrao +728, + 260 21. What is the vertical deflection of joint C of the frame shown below? 2) PL{AE by 2PL/AE ¢ , @ E NRL Bab 1) 3PL/Al oy © 2, The bending moment diagram of the frame shown in figure will be wo? an 23. A two hinged semicircular arch of radius K carries a point load eccentrically on the span at a section whose radius vector is at ‘«’ with the horizontal. The horizontal thrust at each support is w w a= 7 x sina w d)—sin’ a 7 b), 3 t cos7a 24. Calculate the horizontal thrust for the two hinged semicircular arch loaded with point load at its crownH SEER a) W/2n b) ARW/3a. QW /a )W/n 25. A three hinged arch(parabolic) of span ‘L’ and ‘crown rise ‘h’ carries a uniformly distributed super imposed load of intensity ‘W’ per unit length. The hinges are located on two abutments at the same level. The third hinge at a quarter span from the left abutment. The horizontal thrust on the abutment is ye at om 26.A thee hinged parabolic arch of span ‘Ly’ and cena rise We caves point load W a0 Xm honzontlly fom te It hinge The normal thst “T’ at the section under the load if the horizontal thet is Hand vertical eation Vi a) Hcos® - V sin ® b)Hsin® +Vcos® Seas +Vsine WS ‘Okment 27. Rib shortening in arches reduce a) bending moment b) horizon ©) normal thrust &) radial ake: S Pport of the 28. ILD for the shear force at the su cantilever beam is ~N a ke = L—~ » ° 4) | eT FLOOR, Blk COMPLEX Sl RAM NAGAR, KOTTAIYUR, KARAIKUDIA30 108, CONTACT S464 31610, 5249 7070, 9521 «010 HSRC Vecee we) ‘DS. camyover Moment at end B due to moment M BT CODE. 2024/ REVISION TEST 1 / SENIOR BATCH. 29. Influence lines are drawn for structur a) Of any type b) Stutically determine ©) Pin-jointed truss d) None of the 30. The positive and negative shear force (KN) of 4 given influence line diagram shown below ok a : J bm 073 = d) 25,75 13,28 475,50) 75,75 31. The influence lines fof BM, SF or stress are of — for statically indeterminate structures Heurvilinear djeurves 2) Sight tines Parabolic 32, The miimuit bending moment due to tain of ‘wheel Toads on a simply supported girder a) alvays ofc ur atthe centre of span Batyeaysorcur atthe end span 1 aeajs occur under a wheel load Cg) alvays occur inthe middle third of span dpoled tend Afar th given beans " ¢—___j x . a)+M b)-M ) #M/2 d) -M/2 34. Which law states that "If two forces acting simultaneously at a point be represented in ‘magnitude and direction by the two adjacent sides of a parallogram, then the resultant of these two forces is represented in magnitude and direction by the diagonal of that parallogram originating from that point 1) lamie’s theorem b) triangle law of force ©) parallelogram law of force d) polygon law of force 35. The rate of change of momentum of a moving body is directly proportional to the impressed force and takes place in the direction of the force applied a) Newtons first law of motion b) Newtons second law of motion ©) Newtons third law of motion 4) Newtons viscous law of motionSS [ 36. What is the relationship between each force, if three concurrent forces acting on a body according to Lami’s theorem? a) Directly. proportional to the sine of the angle between the other two forves b) Inversely: proportional to the cosine of the angle between the other two forces ©) Directly proportional to the cosine of the angle between the other tivo forces 4) Inversely proportional to the sine of the angle hetiween the other two forces 37. Match the list 1(section) with list 2 (center of gravity (from base) CODE 2024/ REVISION TEST 1 / SENIOR ©) yield stress «d) none of the aly 43, When a square or rectangular block subjected to ‘a shear load is in equilibrium, the shear stress in one plane is always associated with a comple- mentary shear stress (of equal value) in the other plane at degrees to it b) 60 ) 180, a5 9% 44, Relation between E, K and Cis given by a) E= 9KO)/(3K +0) b) E= GK + 0)/(6KC) WE= 6KO)/(K +20) d) E= GKQ)/GK +0) 45, Maximum stress (mu) induced in a bar of length List 1 tin? 1, rotating at an angular velocity w, is given by [x Cylinder a 2) 05 purl 1) 0.125 : Ope! pak Beone me Y a a 46. A body is subjected to normal stresses P; and Psin sphere two mutually perpendicular directions along with Disemisphere | fh ———]_ simple shear q,magfequm shear stress is given by eee eee re ESR AIBICIDD WAIICI Da Zest —7 AaB 2CADA DAABICIDS » Se ) CES 38. The radius of gyration for rectangle is at hs 9! dL. str¥sse¥, the normal stress on a plane will be greater co : Ms than’ the average principal stress, if the inclination of the law of gravitation, 1 The gravitational force is directly proportional to sum of the two objects ey 2.The gravitational forces is directly prdpoRtional to thence beeen there” RN Which of the above statements is’ tp a) only 1 byonly RS ©) both Tand 2 4) none af ie above ey 40. This theorem states thatthe moment of inertia of a planar body about ah axis perpendicular to its plane is equal to'tke sum of its moments of inertia about two perpendieslar axes concurrent with the perpendicular axis and lying in the plane of the body 2) parallel axis theorem ) perpendicular force theorem © perpendicular axis theorem 4) hase area theorem 39. Consider the following statement “meee a. represents the resistance developed by a unit area of cross-section. 4) Unit stress © Either of the above b) Total stress 4) All of the above 42. There is always a limiting value of load up to which the strain totally disappears on the removal of load; the stress corresponding to this load is called 2) elastic limit 'b) unit stress {I FLOOR, Bak COMPLEX, Skt RAM NAGAR KOTTANYUR, KARAIKUDI-30 106.CONTACT 864610 8524997, 95276010 wet patamilio am Nogapolscm, prramuhasuademy cya ameasSbation (A) :In the analysin of two dimensional ‘that plano with the plane of maximum principal stress is less than 45° Reason (R)} The horizontal coordinate of the point on the Mohr's circle represents the stress on the given plane which is greater than the coordinate off the centre of Mohs circle 48. At a point in a steel member, the major principal stress is 2000 kg/cm? and the minor principal stress is compressive. If the uni-axial tensile yield stress is 2500kg/em2, then the magnitude of the minor principal stress at which yielding will commence, according to the maximum shearing stress theory, is a) 1000 kg/em2 2000 kg/em2 ©) 2500 kg/em2 4) 500 kg/cm? 49. A simply supported beam of 16 m efective span carries the concentrated loads of kN, 5 kN and 3 kN. at distances 3, 7 and 11 m respectively from the left support. Calculate maximum shearing force and bending moment. £4) Smus=-15KN, ,Moux=-49kN-m 1) Syux-18KN,,,Maya=-71KN-m, ©) Sans -7KN»Mo=33.3KN-m. ©) St 7KN/Myys™33.3KN-m, 450. The establishes a relationship between the radius of curvature to which the beam bends, the bending,(torsion equa (©) either (a) or (6) deen (U) none of the above SHA rectangular beam 20 ca deep by 10m wid subjected to maximum bending moment of 500 KNm. Determine the maximum stress. in the beam(MN/m?). If the value of E for the material is 200 GN/m?, ay350 h)750 850 )950 52. For no tension in the section, the eccentrici must not exceed mes a) (k)/d b) @ky/a o) @k)/d a) k/va (where, d= depth of the section, k= radiusof gyration) 53. The Radius of kernel of a circular section is ad/2b)d/4 ous days 54, In case of a rectangular section 2) Kanes = 0.5 Cea ) tins ©) tun 3/2 mean 12 mean 55. A square section with side x of a beam is sub- jected to a shear force S, the magnitude of shear CODE: 2024/ REVISION TEST 1 / SENIOR BATCH, Which of these statements are correct? a)land3—b)2and4 «land? d) Sanda A.The seer o hollow shaft forthe same keg matexialand ‘Weight is_—___asolidshaft. 4) best bjmonethin opullo enecfhabowe 60, In case of a hollow shaft the average torsional energy/anit volume is given by bisa De Ye one dy 61, A solid shaft is (o transmit a torque of 80 kNm. If not to exceed 45 MN/m?, find the maximum diameter of the shaft. a)380mm = b)280mm_———« 208mm «308mm 62. Strain energy (U) caused by bending is given by the relation 7 a) U= J M/ (28 ds) B)U=f (NA), (2B1 ax USS (Me/(ED ax 4) VAY 0M9)/ GED dx 63, 8a¥axial pull of 50 KN is suddenly applied to a stress at the top edge of the square is steel rod 2m long and 10 em? in cross-section. ays ws oe Zee. Calculate the strain energy(Nm) that can be A 56. Shear span is defined as the zone where, a) bending moment is zero y os iSeonskant Er cone 4) bending moment is constant \ ‘ 57. A igh bar is supported by el 2 shoven in the given figure. ~y son or The deflection of the point B will be b) 20 mm downward ) 40 mm downward a) 10mm upward ©) 5mm upward 58. Consider the following statements regarding a beam of uniform cross-section simply supported at its ends and carrying a concentrated load at one of its middle third points: be 1 Its deflection under the load wil maximum. 2. The bending moment under the load will be maximum, 3. The deflection at span will be maximum. 4. The slope at the nearest support will be |e FLOOR, ReK COMPLEX, Skt RAM NAGAR, KOTTAIYUR, KARAIKUDI-G3) 106 CONTACT S864 161, 8524970970, 9521 010 Mem nop, prams the midpoint of the \ Py absorbed, if E= 200GN / (m’) 20 50 0150 4)200 64. The ratio of equivalent length of the column to the minimum radius of gyration is called (2) Poisson's ratio _(b) slenderness ratio (6) factor of safety (d) none of the above 65. A hollow cast iron column rigidly fixed at one end and pin-pointed at the other end, has 150 mm outer diameter and 120 mm inner diameter. Its Iength is 6 m and E may be taken as 90 GN/m? Calculate the critical load of this column by Euler formula a)720Kn ——BY820KN-)724KNd)S24KN 66. theory is suitable for brittle materials. 1) Maximum strain energy b) Maximum shear stress theory ‘© Maximum principal stress theory d) Distortion energy theory. 67. Aa per IS: 456-1978, the ratio of stress in concrete to design purposes is taken as @)0.67 b) 0.567 oar 4)0.138 68. Consider the following cases in the design of reinforced concrete members in flexureee” i) Over-reinforced section ‘Tension failure ) Compression failure iv) Under-reinforced section Which of the above cases are considered for safe design of R.C members in flesure? a) ()and (x) only ——-b) (x) andl (iv) only «) (ai) and (Gv) only d) (i) andl (it) only 69, Flexural collapse in over-reinforced beams is due to 4) Primary compression failure by Secondary compression failure «) Primary tension failure ) Bond failure 70. Match List I with List 11 [Reinforcement Type) ] List I (Anchorage Requirement) A. Footing slab, tensile |1. La into the reinforcement support B. Cantilever beam, tensile reinforcement 2.6 @ for 135° bend ©. Seiplysuppaned Bess, [Lalo The tensile port Reinforcement D. Bean, shear sirup [Tarot TRE Column face ECD a13 42 ' hi2as wo jaan \ ai2a3 ene etter tiet ee a Pb compression the values of bond stresd hal Inetedges a) Decreased by 25% by Ihetedges by 20% Jeeta pier 72. Statement()}: In ops \RE—Beam, bond stress ftevelope ts dive tpg wba cnd Eaanal and mechanical resis . Statement (If: Inadggquacy of bond strength can be compensated by prdviding end anchorage in the reinforcing, bars 73.In the design of two-way slab restrained at all edges, torsional reinforcement required is 4) 0.75 times the area of steel provided at midspan in the same direction. b) 0.375 times the area of steel provided at midspan in the same direction. ©) 0.75 times the area of steel provided in the shorter span Nil 74. Match List 1 List II in connection with ‘moment capacities for different condition and types [Ill FLOOR, BK COMPLEX, SRI RAM NAGAR, KOTTAIYUR, KARAIKUDI-0 106, CONTACT 8186451610, 85249 7970, 59521 010 ‘eve pyramiliosraemy Blogspot com, pramiiiasalemeymalltnn CODE. 2024/ REVISION TEST 1/ SENIOR 3A.» of slabs as per yield line theory and select the answe using the codes given below ar) List (Type of slab) ] List capacity) x Fquilateral | y,¥ triangular s | [Bo Simply supported [> 2" square slab 7% — C. Square slab fixed al all edges ___j Codes: ; aA2BICS by A3BAC- J A1B3C2 AB B2C4 75. Critical section for shear in case of flat slabs is at a distance of a) effective depth of Stak from periphery column/drop panel bb) /2 from peripheryrot column capita ©) al the drop panel of slab d) at the peripheryoreptumn of 1/ drop panel 76. The raal moment atthe boundary of a simply supported citctlar slab of radius R, subjected to a unifdxmly distributed load of w will be c see By wR2/12 «) 3WR2/16 dl) zer0 y 77. inan isolated reinforced concrete column footing of “effective depth d, the stress in punching shear is checked 4a) al the centre of the column 1) atthe face of the column ala distance d/2 away from the face of the column, d) ata distance d/2 away from the centre of the column 178. How is the depth of footing for an isolated column governed? 1. By maximum bending moment 2. By shear force 3. By punching shear Select the correct answer using the code given below: 8. 2and 3 only b Land 2 only ©. Tand 3 only 4.1,2and3 79. A reduction factor C, to load carrying capacity of along column is given by a) C,=1.25~ 35 b) Gat OG125-3 Gas where Le is effective length of column and b is width of column 80, The diameter of ties in a column should be a) more than or equal to one fourth of diameter of main bar ») more than or equal to 5 mm ©) more than 5 mm but less than one-fourth of diameter of main bar 1[a more than 5 mm and also more than one-fourth of diameter of main bar 81, While designing combined footing, the resultant of 'e column loads passes through the centre of gravity of the footing slab such that the net soil pressure obtained is 2. Paabal b Trapezoid ¢. Uniform, : dd. Non-aniform 82.1f the modular ratio is m, steel ratio is r and overall depth of a beam isd, the depth of the critical neutral axis of the beam, is a) (m/m-r}d © (mte/ mjd b) (m/mtr)d a) (-m/m)d. 83. Assertion A: In the working stress method, the stress in compression steel is laken as 15. x modular ratio x slress in surrounding concrete Reason R: The modulus of elasticity of concrete in compression is only two-thirds of thal in tension 84.1n a reinforced concrete, cantilever retaining wall, the maximum base pressure occurs at a) Toe of the wall 'b) Heel of the wall «) Mid-point of bases slab 4) Below the stem in base slab c 85. Hoop tension in a RCC water tank will bes. resisted by- a) Steel alone ) Concrete ©) Partially by concrete and partially by steel dd) None of the above KAY \S 86. In limit State Method of desigo& RCC flexural members, the centre ty of the compressive forces acting'on the section from the top most fiber of concer distance of a) 0.57 ofthe depth dfngutral axis 1b) 0.36 of the depth Of nettral axis ©) 042 of the Septet ret axis ) None of theses \S 87. As per IS 456: 2000, the final deflection due to all loads including the effect of temperature, creep and shrinkage should not normally exceed: a) span/250 b) span /350 ©) span/400 ) span/450 88. Ina rectangular beam of section bxd, subjected to ultimate torsional moment Tu, equivalent ultimate shear can be given as. a) Vu +1.6Tu/b b) Vu +16 Tu/ bd Vu 42Tu/b d) Yut Tu/ 2b | FLOOR, eK COMPLEX SRI RAM NAGAR, KOTTAIYUR, ARAIKUDI 430 10, CONTACT: 9486431610, 85219 70570, 99521 4010 [ornament eter CODE: 2024/ REVISION TEST 1 / SENIOR BATCH 89. As per IS 800: 2007, what should be the maximum slenderness ratio for the tension members in which reversal stress due to load other than wind or seismic forces occurs? a) 350 ») 400 ©) 580 d) 180 90. 1f the area of cross section of a single angle discontinuous strut is 30cm? and allowable working, stress corresponding to its slenderness ratio is 625 Kglem?, the safe load carrying capacity of the member is a) 10 tonnes ©) 15 tonnes b) 12 tonnes 4) 18 tonnes 91. Column footing is provided a) To spread the column load over a larger area b) To ensure that inlensitys of hearing pressure between the column footipg, and soil doos not exceed sermissible bearing, capacity of soil column footing: 4) Alloy jon are correct sa.tn nig” beings the horizontal beams Pein ttween the wall columns supporting & ral covering are called a) Siringers b) Girts ©) Trimmers 4d) Lintels y ‘493. The rolled steel-t-sections are most commonly used as beams because these provide: 1. Larger moment of inertia with less cross-sectional 2. Greater lateral stabil a) Only 1 and 2 b) Only 2 d) None of these 94, In a grillage footing the maximum shear force occurs at the a) Edge of grillage beam ©) Center of grillage beam b) Centre of base plate 4d) None of these 95. As per IS 800: 2007, what is the maximum pitch of bolts allowed in a lap joint between two steel plates (in compression) of equal of thickness, t? a) 2.5 times diameter of the hole }) 25 times diameter of the bolt ) 12 tor 200 mm whichever is less 4) 16 tor 200 mm whichever is less 96.Which one of the following factors does not affect the lateral buckling strength of a steel I-section undergoing bending about its major axis? a) Laterally unsupported length of the compression flange. b) Radius of gyration about the major axis of the section, ©) Boundary conditions at the ends _ —___$—@) Radius of gyration about the minor axis of th 97. In the case of a P.S.C. beam, to satisfy the Limit State of Serviceability in cracking, match List-1 with List-l and select the correct answer using the codes given below the lists: List-l A. Class 1 structure B, Class 2 structure . Class 3 structure List-l 1. No visible cracking 0; <3 N/mm? 2 Concrete section uncracked; crack width to be calculated to check 3. No cracking under service loads Codes: A c at b) 2 SENIOR BATCH 124/ REVISION TEST gs 2 1 98. In. pre-tensioning scheme, pre-stress load is transferred in a) a single stage proe )eithe e pr depending upon the magnitude of load 4d) the same manner as in post-tensioning % heme a b) multi stage process ir multi-sta 99. For a certain set of external loads, concordant profile in a prestressed beam represents to some scale the fluonce line diagram —_b) Shear forve dia ) Bending moment diagram 4) Wilhiot- Mohr diage 100, Assertion prestress in ensioned beams are (a): Lossed~in morthan the losses in tensioned beams, Reason (R): This i pa shortening jally due to the PYRAMID IAS ACADEMY The Best Coaching Centre for Civil Engineers. Karaikudi ‘Congratulations to our students for their selection as eer in various departments through SSC JE-CIVIL 2022 Exam FLOOR Bick COMPLEX SRI RAM NAGAR KOTTANUR, A dadrn 3 2 rere Peber eet 26 CONTACT 94864 31610, 85248 7077, $2521 KO010