Detailed Solution Civil Engineering
SET - C
1. Which of the following statements are correct Sol. in respect of temperature effect on a long RE carrying three hinged arch? HE E 1. No stresses are produced in a three- 200 N hinged arch due to temperature change 2m alone. C A 2. There is a decrease in horizontal thrust 2m due to rise in temperature. HD 3. There is an increase in horizontal thrust D B due to rise temperature. RD (a) 1 and 2 only (b) 1 and 3 only 4m 4m Taking moment about D (c) 2 only (d) 3 only 200 × 8 = H × 4 Ans. (a) E 1600 2. Consider the frame as shown in the figure H = = 400 N E 4
E 3. The load system in the figure moves from left H 200 N E to right on a girder of span 10 m. 2 m C 150 kN 90 – 120 kN 2 m 70 kN 60 kN D 4 m 4 m
0.5 m 0.5 m 1 m The magnitude of the horizontal support A B reaction at E is (a) 400 kNIES(b) 300MASTER kN 10 m (c) 250 kN (d) 200 kN The maximum bending moment for the girder Ans. (a) is nearly (a) 820 kNm (b) 847 kNm (c) 874 kNm (d) 890 kNm Ans. (d)
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Sol. 80 200 200 80 150 or 120 60 70 2m 2m
0.5m 0.5m 1m A 4m 12m 16m 10 m 12 A 10 16 16 Resultant location of load is given by 150 + 120 60 R 70 2m 2m 10m – x –2 16 (70 0) (150 1) (60 1.5) (120 2) ILD for SF at (A) x = –4 400 16 = 1.2 m When 200 kN load is leading : Max. BM will be taken as below 150 kN Max. (–)ve SF when 200 kN is just to the left of sec. (A)
120 150 4 2 60 R=400 70 VA = 200 80 16 16 960 = = –60 kN 16 4.1m 0.5m 0.3m 1m 3.9m 400 4.9 When 80 kN load is leading : 0.1m 0.1m 10 Max. (–)ve SF when 80 kN is just to the left = 196 kN of (A) BM below 150 kN is 196 × 4.9 – 70 ×1 = 890.4 kNm. 4 2 720 V = 80 200 = = –45 kN A 16 16 16 4. Two wheel loads 80 kN and 20 kN respectively spaced 2 m apart, move on a girder of span When 200 kN is just to the left of (A) 16 m. Any wheel load can lead the other. The maximum negative shear force at a section 4 10 V = 200 80 = 0 IES MASTERA 4 m from the left end will be 16 16 (a) –50 kN IES(b) –60 kN MASTER 5. The maximum possible span for a cable (c) –70 (d) –80 kN supported at the ends at the same level Ans. (b) (assuming it to be in a parabolic profile) Sol. 1 allowing a central dip of of the span with 10 permissible stress of 150 N/mm2 (where the steel weighs 78,000 N/m3) will be nearly
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(a) 1270 m (b) 1330 m 6. A three-hinged arch has a span of 30 m and (c) 1388 m (d) 1450 m a rise of 10 m. The arch carries UDL of 60 kN/m on the left half of its span. It also carries Ans. (c) two concentrated loads of 160 kN and 100 Sol. Let the maximum horizontal span be kN at 5 m and 10 m from the right end. metres. The horizontal thrust will be nearly Dip of the cable = h = metres (a) 446 kN (b) 436 kN 10 (c) 428 kN (d) 418 kN 8 h2 Length of the cable = L = Ans. (c) 3 Sol. 2 8 h 60 kN/m 1 100 kN = 3 = L 160 kN 8 1 308 = 1 = 10m 3 100 300 Let the area of the cable be A mm2 H Weight of the cable = W 30m 308 A 78000 N; R = 2 300 1000 R × 15 – H × 10 – 60 × 15 × 7.5 = 0 W = 0.08008 AN 1.5R – H – 675 = 0 ...(1) Each vertical reaction R × 30 – 60 × 15 × 22.5 – 100 × 10 – 160 × 5 = 0 W = V = ; R = 735 kN 2 H = 427.5 kN W Horizontal reaction = H = 8h 7. An unstable vibratory motion due to combined W 5 bending and torsion which occurs in flexible = (10) = W plate like structures is called 8 4 (a) Galloping (b) Ovalling 2 2 Max. tension = Tmax = VH (c) Flutter (d) Oscillation 2 2 W 5W Ans. (c) = = 1.35 W 2 IES 4 MASTERSol. Flutter is a oscillatory motion that results from the coupling of aerodynamic forces with the = 1.35 × 0.08008 AN elastic deformation of a structre. It is often the results of combined bending and torsion and Tmax Maximum stress = = f effects plate like structures. A max Examples: sign boards and suspension 1.35 0.08008 = 150 bridges decks. = 1387.5 m
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8. A propped cantilever beam of span l and x 12t constant plastic moment capacity Mp carries 30° a concentrated load at mid-span. The load at W collapse will be b e a P 12t 30° 2 M 4 M (a) p (b) p l l
Effective width (We) = a + 24t (If, x > 12t) 6 Mp 8 Mp (c) (d) Effective width (W ) = b (If, x < 12t) l l e Ans. (c) 10. A wind brace is to be provided between two columns spaced at 5m, at an inclination of M 30° with the horizontal, to resist a tension of Sol. P L/2 PL/2 320 kN developed by a wind force. The effective area required will be nearly (considering 150 N/m2 as a relevant factor) (a) 1670 mm2 (b) 1640 mm2 M M P P (c) 1600 mm2 (d) 1570 mm2 Ans. (c) L P = 3MP 2 11. A beam column for non-sway column in a building frame is subjected to a factored axial 6M P = P load of 50 kN, factored moment at bottom of L column of 45 kNm. For ISHB 200, the values 2 45.1, 9. A steel plate is subjected to tension. The are A = 4750 mm , y h = 200 mm, b
tensile force is applied over a width ‘a’ whereas = 200 mm, bf = 9 mm and the effective length the gross width of the plate ‘b’. The dispersion is 0.8 L. Its buckling load will be of the force from the point of application is at (a) 910 kN (b) 930 kN about 30° with the axis and extends to a maximum width of 12 times the thickness t of (c) 950 kN (d) 980 kN the plate. The effective width which into action Ans. (c) will be Sol. Pe = equivalent axial load (a) 2a + 12t (b) a + 12t 2M P = P z , where d = depth of beam (c) a + 24t IES(d) 2a + 24tMASTERe d Ans. (c) IES MASTER45000 = 500 + 2 × = 950 kN Sol. 200
12. Which of the following assumptions are correct for ideal beam behaviour? 1. The compression flange of the beam is restrained from moving laterally 2. The tension flange of the beam is restrained from moving laterally.
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Detailed Solution Civil Engineering
3. Any form of local buckling is prevented. (a) 2 and 3 only (b) 1 and 3 only (c) 1 only (d) 3 only Ans. (b) Sol. Two important assumptions are made for ideal beam behaviour. Fig. (a) (i) Compression flange of beam is restrained Vierendeel girders : When the chords are laterally. parallel as shown in figure below, they are (ii) Local buckling of elements is prevented. called as Vierendeel girders. In a Vierendeel girder, the loading is carried by a combination 13. In which one of the following industrial roofing of pure flexure and flexure due to shear contexts, is the loading carried by the induced by the relative deformation between combination of pure flexure and flexure due the ends of the top and bottom chord to shear induced by the relative deformation members, similar to that found in castellated between the ends of the top and bottom chord beams. members? (a) Vierendeel girders (b) Scissors girders (c) Lenticular girders
(d) Mansard girders Fig. (b) Ans. (a) Lenticular truss : The lens shaped truss is Sol. Mansard truss : When straight members are called as lenticular truss, as shown in figure used for top chords, they are called Mansard below has been used in some bridges in USA. trusses. Any web system may be used, since Here again the web members will carry zero the web stresses are normally small. If, for forces and hence may be removed and the example, a parabolic upper chord is used and joints may be made rigid as shown in figure the truss is subjected to a uniform load, there below. will be no stress in the web members, uniform tension throughout the lower chord, and compression in the upper chord (forces in upper chord members will be slightly higher than the lower chord members due to their slope). In other words, the behaviour will be Fig. (c) similar to that of a tied arch of the same rib form. The diagonals,IES though carryingMASTER zero Scissors truss : The scissors truss is shown forces, are provided for stability under varying below, is used infrequently in steel construction loading conditions, if the joints are pin- but is included to show as efficient form for connected. Both bowstring and mansard use in short spans. In these types of trusses, trusses are used for longer spans. both the chords are sloped.
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dc For minimum cost, 0 ds 2 –k1/s + 2k2s + k3 = 0 2 k1/s + 2k2s + k3s = 0
Fig. (d) –l + 2p + r = 0 l = 2p + r 14. Bearing stiffeners are provided (a) At the ends of plate girders 16. A welded plate girder of span 25 m is laterally restrained throughout its length. It has to carry (b) At the ends of plate girder and on both a load of 80 kN/m over the whole span besides faces of the wen its weight. If K = 200 and fy = 250 MPa, the (c) At the ends of plate girder and only on thickness of web will be nearly one face of the web (a) 10 mm (b) 14 mm (d) At the points of concentrated loads, to (c) 16 mm (d) 20 mm protect the web from the direct compressive loads. Ans. (a) Ans. (d) Sol. Span (l) = 25 m Sol. Bearing stiffners are provided at joints of Factored applied udl = 1.5 × 80 = 120 kN/m concentrated loads, to protect the web from Total factored applied load (W) = 120 × 25 the direct compressive loads. = 3000 kN 15. If the cost of purlins/unit area is p and the W Let self weight of the girder = kNm cost of roof covering/unit area is r, then cost 200 of trusses/unit area l for an economical 3000 spacing = 15 kNm 200 (a) p + r (b) 2p + r Total uniform factored load = 120 + 15 (c) p + 2r (d) 2p + 2r = 135 kNm Ans. (b) 135 252 Sol. Cost of truss is inversely proportional to the Maximum bending moment (M) = spacing of truss, 8 = 105.46.875 kNm l k / s 1 IES MASTEROptimum thickness of web, Cost of purlins is directly proportional to the 1/3 square of spacing of trusses M 10546.875 106 tw 2 2 IES MASTER 2 fy k P = k2s 250 200 Cost of roof covering is directly proportional = 10.17 mm 10 mm to the spacing of trusses 17. A proper cantilever ABCD is loaded as shown r = k s 3 in figure. If it is of uniform cross-section, the Total cost = l + p + r collapse load of the beam will be nearly 2 C = k1/s + k2s + k3s
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A W W/8
A D B C h MP MP
L/2 L/2 L/3 B C b MP MP (a) 6.5 (b) 5.6 The shape factor will be nearly L L (a) 2.3 (b) 3.2 M M (c) 4.7 P (d) 3.8 P (c) 4.1 (d) 5.0 L L Ans. (a) Ans. (a) Sol. Shope factor of triangular section is 2.34 Sol. Collapse mech-1 19. Fatigue in RCC beams will not be a problem W W/8 if the number of cycles is less than L/2 L/2 L/3 A D (a) 20,000 (b) 25,000 B C (c) 30,000 (d) 35,000 MP Ans. (a) WL 24 M = MW P 8 3 P L 20. The desired characteristic strength of a mix is 20 N/mm2. The standard deviation is 4 N/ Collapse mech-2 mm2 for 150 mm size of concrete cubes; and K = 1.645. The average strength of the cubes W W/8 will be nearly L/2 B L/2 L/3 2 2 A (a) 38.2 N/mm (b) 32.4 N/mm C D (c) 26.6 N/mm2 (d) 22.8 N/mm2 M P Ans. (c) MP MP Sol. Mean strength or average strength of cube
L WL fm = fck + K W – = 3M 2 8 3 P 2 Given fck = 20 N/mm 11 K = 1.645 WL = 3M 24 IESP MASTER = 4 N/mm2 M fm = 20 + 1.645 × 4 W = 6.545 P L = 26.58 N/mm2 26.6 N/mm2 18. Consider a triangular section with base b and height h as shown in the figure. 21. A circular column is subjected to an un- factored load of 1600 kN. The effective length of the column is 3.5 m, the concrete is M 25,
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ASC and the value of g = 2% for Fe 415 Ag 2.5m steel. The design diagram of the column will be nearly 80 N (a) 446 mm (b) 432 mm 2.5m (c) 424 mm (d) 41 mm Ans. (a) Sol. From the options, it is clear that the column is a short colum. 80(5)3 = = 10 × 10–3 48 EI 3 leff 3.5 10 As all the options are > 80(5)3 12 12 EI = 48 10 103 = 291.67 mm 2 Hence we can consider the column as a short EA Critical load = axially loaded column 2
P = 0.4 f A + 0.67 f A 2 2 u ck c y sc E EImin = 2 = 2 Asc = 2% of Ag = 0.02 Ag eff eff 1.5 × 1600 × 103 = 0.4×25 (A – 0.02 A ) g g rmin + 0.67 × 415 × 0.02 A g 2 3 = EI 2400 × 10 = 9.8 Ag + 5.561 Ag 2 eff 3 2400 10 2 2 3 A = 156239.82 mm 80(5) g 15.361 = = 8224.67 N (5)2 48 10 103 D2 = 156239.82 4 23. The recommended imposed load on staircase in residential buildings as per IS 875 is D = 446 mm (a) 5.0 kN/m2 (b) 3.0 kN/m2 22. A strut is made of a circular bar, 5 m long and (c) 1.5 kN/m2 (d) 1.3 kN/m2 pin-jointed at both ends. When freely supported the bar gives a mid-span deflection Ans. (b) of 10 mm underIES a load of 80 N at MASTERthe centre. Sol. The recommended imposed load on stair case in residential building as per IS 875 The critical load will be IES MASTER 2 (IS 875 (part 2), clause 3.1) is 3 kN/m . (a) 8485 N (b) 8340 N (c) 8225 N (d) 8110 N 24. A 230 mm brick masonry wall is to be provided with a reinforced concrete footing on site Ans. (c) having soil with safe bearing capacity of 125 Sol. kN/m2, unit weight of 17.5 kN/m3 and angle of shearing resistance of 30°. The depth of
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Detailed Solution Civil Engineering
footing will be nearly (a) 0.8 m (b) 0.7 m 1 d = 3t (c) 0.6 m (d) 0.5 m 1 280 Ans. (a) 2 1 q 1 sin = 350 Sol. Depth of footing = 3 140 1 sin 1 280
2 = 140mm 125 1 sin30 = 17.5 1 sin30 xb Balance moment = 0.5bxb c d 3 2 125 1 125 d = C 17.5 3 157.5 x = 0.793 m b d x d b 0.8 m 3
25. A rectangular beam 200 mm wide has an T effective depth of 350 mm. It is subjected to 140 a bending moment of 24,000 Nm. The = 0.5 200 140 5 350 N.mm permissible stresses are c = 5 N/mm2, t = 3 140 N/mm2; and m is 18. The required area = 21233.33Nm of tensile reinforced will be MORbalance < B.M (a) 688 mm2 (b) 778 mm2 Section needs to be designed as a over 2 2 (c) 864 mm (d) 954 mm reinforced Ans. (b) Depth of Neutral axis Sol. b = 200mm x d = 350mm 0.5bxc d = 24000Nm 3 M = 24000Nm x C = 5N/mm2 0.5 200 x 5 350 3 (Permissible stress in concrete in bending = 24000×103N.mm compression) 175000x – 166.67x2 – 24000 × 103 = 0 t = 140N/mm2 IES MASTER x = 887.78mm (it is out of section and (Permissible stress in steel) shall be discarded) m = 18 (Modular ratio) x = 162.10mm It is a WSM based quations, 1 d Balance depth of N.A = 3 1 St 280
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Detailed Solution Civil Engineering
modulus of elasticity will have low creep and cbc (C) shrinkage. x These aggregate will generally not allow the change in volume due to higher rigidity. Also, as cement paste and water-cement ratio is responsible for creep and shrinkage, keeping minimum water-cement ratio, larger fst (Strain diagram) aggregate size will lead to less shrinkage. m (For over reinforced section) (strain diagram) 27. During earthquakes, the corner and edge columns may be subjected to c fst = (a) Uniaxial bending x m d x (b) Biaxial bending 5 fst = (c) Combined biaxial bending and torsion 162.20 18 350 162.20 (d) Combined biaxial bending and tension fst = 104.20 N/mm2 Ans. (d) MOR from tension side Sol. During earthquake structre can behave as a x 3 cantilever, inducing the tensile force in the 24000 × 10 = fst A st d 3 columns lying on the side of application of lateral load while inducing compressive force 162.20 = 104.20Ast 350 on the columns on the opposite side. 3 Also, biaxial bending will be there in the corner A = 778.30mm2 778mm2 st and edge columns. 26. Which of the following statements are correct 28. The minimum number of bars required in a with reference to ensuring minimum shrinkage rectangular column for an earthquake resistant of prestressed concrete? design, is 1. The water-cement ratio and proportion of (a) 4 (b) 6 cement paste should be kept minimum to reduce shrinkage. (c) 8 (d) 10 2. Aggregates of larger size, well graded for Ans. (c) minimum void, need a smaller amount of 29. The permissible of allowable compressive cement paste, and attendant shrinkage stress f of brick masonry does not depend will be smaller.IES MASTERac on 3. Harder and denser aggregates of low (a) Type of strength of bricks water absorptionsIES and high modulusMASTER of elasticity will exhibit small shrinkage. (b) Efflorescence of bricks (a) 1 and 2 only (b) 1 and 3 only (c) Strength of mortar (c) 2 and 3 only (d) 1, 2 and 3 (d) Slenderness ratio Ans. (d) Ans. (b) Sol. Concrete with aggregates that are hard, dense Sol. “Efflorescence” is a defect which is caused and have low water absorption and high due to presence of excessive soluble salts in
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brick earth. These salts after getting dissolved Resisting moment about toe in water, appear in form of fine whitish crystals 5 1.5 (5 1.5) on the exposed brick surface. This defect will = 264 308 2 result in ugly appearance. But it doesn’t affect 2 3 the strength of brick masonry. = 1840.67 kNm Eccentricity about centroid 30. A masonry dam 8 m high 1.5 m wide at the 5 = x top and 5 m wide at the base retains water to 2 a depth of 7.5 m, the water face of the dam (MR M 0 ) 1840.67 689.765 being vertical. If the weight of water is 9.81 x = kN/m3, weight of masonry is 22 kN/m3, the Vertical load (308 264) maximum intensity of stress developed at the = 2.01 m base will be nearly. 5 e = 2.01 0.49 m (a) 196 kN/m2 (b) 182 kN/m2 2 (c) 160 kN/m2 (d) 148 kN/m2 W 6e Stress at toe = 1 Ans. (b) b b Sol. 572 6 0.49 = 1 5 5 = 181.80 kN/m2 182 kN/m2 1.5 m 31. Consider the following data Root zone depth = 2 m Existing water content = 5% Dry density of soil = 15 kN/m3 3 8.0 m Water applied to their soil = 500 m 7.5 m Water loss due to evaporation and deep percolation = 10% Area of plot = 1000 m2 5 m The field capacity of the soil will be nearly. (a) 16.8 % (b) 17.7 % Weight of rectangular portion = 1.5 × 8×22 (c) 18.8 % (d) 19.7 % = 264 kN Ans. (d) Weight of triangularIES portion MASTERSol. Root zone depth, D = 2 m
= 0.5 × (5 – 1.5) × 8 × 22 3 d = 15 kN/m = 308 kN = 9.81 Overturning moment about toe w 3 7.5 Water applied to the soil = 500 m = 0.5 9.81 7.52 3 Water loss due to evaporation and deep = 689.765 kN-m
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percolation = 10% foundation. Amount of water lost = 50 m3 2. The difference in weir crest and downstream river bed may not exceed 3m. Total water stored = 450 m3 3. When water passes over it, the Area of plot = 1000 m2 longitudinal location of the formation of 450 hydraulic jump is variable. Depth of water stored, d = 0.45 m 1000 This weir is of the type. Existing water content may be assumed (a) Vertical drop weir as M0 = 5% (b) Masonry or concrete sloping weir d d = D (FC M 0 ) (c) Dry stone slope weir w (d) Parabolic weir 15 0.45 = 2 (F 0.05) Ans. (b) 9.81 C Sol. Masonary on concrete sloping weir. FC = 0.1971 Weir of this type are of recent origin. They F = 19.71% C are suitable for soft sandy foundation, and 32. Consider the following data for irrigation water: are generally used where the difference in weir crest and downstream river bed is limited Milli - equivalent to 3m, when water passes over such a weir, Concentration hydraulic jump is formed on the sloping glacis. per litre 1 Na 24 34. Consider the following data while designing an expansion transition for a canal by Mitra’s 2 Ca 3.6 method: 3 Mg 2 Length of flume = 16 m The solidum-Absorption ratio (SAR) is nearly Width of throat = 9 m (a) 13.1 (b) 14.3 Width of canal = 15 m (c) 15.5 (d) 16.7 If Bx is the width at any distance x from the Ans. (b) flumed section, the values of BX at x = 8 m and at x = 16 m are nearly [Na ] Sol. SAR = (a) 10.8 m and 15 m [Ca2 ] [Mg 2 ] (b) 11.3 m and 15 m IES2 MASTER (c) 10.8 m and 13 m 24 = IES MASTER(d) 11.3 m and 13 m 3.6 2 2 Ans. (b) SAR = 14.34 Sol. According to Mitra, channel at any section, at a distance x from the flumed section is given 33. Consider the following statements with respect by, to weir under discussion: BBLn f f 1. Its design corresponds to soft sandy Bx = Lf B n (B n B f )x
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where, Bn = 15 m, width of canal (c) Train the flow along a certain course
Bf = 9 m, width of throat (d) Reduce the silting in the river bed
Lf = 16 m, length of flume Ans. (c)
Bx = width at any distance x from flumed Sol. Groynes are the embankment type structures, section constructed transverse to the river flow, at x = 8 m extending from the bank into the river. They are constructed to protect the bank from which 15 9 16 they are extended, by deflecting the current B = 11.25 11.3 m x 16 15 (15 9) away from the bank i.e, to train the flow along at x = 16 m a certain course. 37. Which one of the following compounds of 15 9 16 15 m nitrogen, when in excessive amounts in water, Bx = 16 15 (15 9) 16 contributes to the illness known as infant methemoglobinemia? (a) Ammoniacal nitrogen 35. Consider the following data for a drain: (b) Albuminoid nitrogen L = 50 m a = 10 m, b = 10.3 m, and (c) Nitrite k = 1 × 10–5 m/s (d) Nitrate If the drains carry 1% of average annual rainfall in 24 hrs, the average annual rainfall Ans. (d) for which this system has been designed will Sol. Presence of nitrates in too much amount be causes a disease called methemoglobinemia (a) 78 cm (b) 84 cm (also known as blue baby disease). Children suffering from this disease may vomit, their (c) 90 cm (d) 96 cm skin colour may become dark and may die in Ans. (b) extreme cases.
4k 2 2 38. Consider the following data regarding a Sol. S = (b a ) q theoretical profile of a dam:
4 105 (10.3 2 10) 2 Permissible value of compressible stress 50 = q 350 tonnes/m2 q = 4.872 × 10–6 m3/s Specific gravity of concrete s = 2.4 1 (S 1) 100 Uplift coefficient c = 0.6m q = IES8.64 104 MASTER The value of 1 = 0.841m 84 cm
36. The purpose of constructing a ‘Groyne’ is to The height and base width will be nearly (a) Expand a river channel to improve its (a) 125m and 63m depth (b) 175m and 63m (b) Encourage meandering (c) 125m and 93m
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(d) 175m and 93m in a city of one lakh population. The factor of Ans. (c) safety is taken as 1.5, detention time as 4 h and overflow rate as 20,000 litres/day/m2. The 2 Sol. = 350 tonnes/m area of 3m deep plain sedimentation tank as
GC = 2.4 per surface loading consideration will be = 1 tonnes/m3 (a) 1025m2 (b) 1075m2 = wH(G c C 1) (c) 1125m2 350 = 1 × H (2.4 – 0.6 + 1) (d) 1175m2 H = 125 m Ans. (c) H Sol. Given, B = GC C Water demand = 150litres/head/day 125 Population = 1 lakh B = 93 m 2.4 0.6 Factor of safety = 1.5 39. Chlorine usage in the treatment of 25,000 m3/ Detention time = 4hrs day of water has been 9 kg/day. The residual Overflow rate = 20,000litres/day/m2 chlorine after 10 minutes contact is 0.2 mg/ . Depth of tank = 3m The chlorine demand of water would be nearly Water flow = 150 × 105 × 1.5 (a) 0.28mg / = 22.5MLD (b) 0.22mg / Q Area of tank = Overflow rate (c) 0.16mg / 22.5 106 (d) 0.12mg / = 20,000 Ans. (c) = 1125m2 Sol. Given, Flow = 25000m3/day 41. The rain intensity over 54 hectares of land is Chlorine use = 9kg/day 50 mm/h, 30% of area consists of roof surface Residual chlorine = 0.2 mg / with runoff rate as 0.9, 30% is open field with runoff rate of 0.2 and remaining 40% is road 9kg / day network with runoff rate of 0.4. The storm Chlorine doseIES= 3 MASTER 25000m / day water flow will be nearly (a) 2.6m3/s (b) 3.7m3/s IES6 MASTER 9 10 mg = (c) 4.8m3/s (d) 5.9m3/s 25000 103 Ans. (b) = 0.36mg / Sol. Chlorine demand = 0.36 0.2mg / = 0.16mg /
40. The demand of water is 150 litres/head/day
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Ans. (c) CACACA1 1 2 2 3 3 Ceq = Sol. As, sludge density index AAA1 2 3 100 = 0.9 × 0.3 + 0.2 × 0.3 = + 0.4×0.4 Sludge Volume index m / gm = 0.49 100 = 2000mg 2g Q = C i A 176 2 3 50 10 4 = 0.49 × 54 10 = 1.14g / m 3600 = 3.675 m3/sec 44. Which one of the following gases is the 3.7 m3/sec. principal by-product of anaerobic decomposition of the organic content in waste 42. Critical dissolved oxygen (D.O.) deficit occurs water ? in which one of the following zones of pollution (a) Carbon monoxide of oxygen sag curve in case of self-purification of natural streams ? (b) Ammonia (a) Zone of recovery (c) Hydrogen sulphide (b) Zone of active decomposition (d) Methane (c) Zone of degradation Ans. (d) (d) Zone of clear water Sol. Methane is the principal by-product of anaerobic decomposition of the organic Ans. (b) content in waste water (around 60%). Sol. 45. Consider the following statements with Zone of Zone of degradation recovery reference to the mixing of industrial waste water with domestic waste water : Clearer Clearer DO water water 1. The industrial waste water can be mixed Saturation with domestic water when it has higher BOD. D.O D .O sa 2. The industrial waste water can be mixed g cu 40% of D.O rve with domestic water when the pH value of Zone of active industrial waste water is highly alkaline. decomposition Saturation Which of the above statements is/are correct? 43. The MLSS concentrationIES in an aerationMASTER tank (a) 1 only (b) 2 only as 2000 mg/l and the sludge volume after 30 minutes of settling in a 1000 ml graduated (c) Both 1 and 2 (d) Neither 1 nor 2 cylinder is 176ml. The value of sludge density Ans. (d) index (SDI) will be nearly Sol. Limit for disposal of industrial waste in sewer (a) 3.34g / m (b) 2.22g / m line for BOD is 500mg/l and for pH is 5.5 – 9.0 according to BIS standard. (c) 1.14g / m (d) 0.26g / m
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And industrial waste compatible with domestic sewage can be mixed with domestic sewer ELR t (inversion) h
hence neither of statements is true. g
Inversion i condition e 46. The waste water from a factory having a pH H ALR of 10, contains KOH only. For waste water Temp. discharge is 80m3/day. The total quantity of KOH per day will be nearly 48. A thermal power plant burns coal at the rate (a) 4.5 kg/day (b) 5.4 kg/day of 8t/h. The coal has sulphur content of 4.5%.
(c) 6.3 kg/day (d) 7.2 kg/day The rate of emission of SO2 will be Ans. (*) (a) 180 g/s (b) 200 g/s Sol. pH = 10 (c) 220 g/s (d) 240 g/s pOH = 14 – 10 = 4 Ans. (b) [OH–] = 10–pOH Sol. Rate of coal Burning = 8t/hr = 8000kg/hr [OH–] = 10–4 mole/litre Sulphur content in coal = 4.5% Molecular weight of KOH = 39 + 16 + 1 Sulphur (S) present in coal after burning will be converted into SO (sulphur di-oxide) = 56gm 2 Rate of burning of sulphur present in coal Total quantity of KOH per day = discharge × concentration of KOH 415 = 8000kg/hr = 80 × 103 × 10–4 1000 = 8 mol/day = 360kg/hr = 8 × 56 gm/day As the sulphur converted into sulphur di-oxide, moles of sulphur present in coal will be equal = 0.448 kg/day to SO2 released
47. Fanning type of plume behaviour takes place S + O2 SO2 when Moles of sulphur converting into (a) Super-adiabatic lapse rate prevails with 360kg / hr light to moderate wind speed SO = 2 Molecular weight (b) Extreme inversion conditions exist in the presence of light wind 360 103 = / hr (c) There existsIES a strong super-adiabatic MASTER 32 lapse rate above a surface of inversion = 11250 moles/hr (d) Plume is caught between two inversion IES MASTERRate of emission of SO2 = Rate of burning of layers sulphur × molecular weight of SO2 Ans. (b) = 11250 × 64gm per hour Sol. Fanning plume behaviour takes place when 11250 64 extreme inversion conditions exist in the = gm / sec presence of light wind. 3600 = 200gm/sec
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49. The property of clays by virtue of which they 35 22 regain, if left alone for a time, a part of the 0.42 W 0.3 W strength lost due to remoulding at unaltered S S moisture content, is known as WS = 0.0969 = 9.69% (a) Thixotropy (b) Sensitivity VV L d 100 (c) Consistency (d) Activity Vd Shrinkage ratio (SR) = Ans. (a) WWLS Sol. The property of soil due to which loss of 35 = 1.083 1.1 strength on remoulding can be regained if left (42 9.69) undisturbed for sometime is known as “Thixotropy”. 51. The ratio of a given volume change in a soil expressed as percentage of the dry volume, Increase in strength with passage of time is to the corresponding change in water content due to tendency of clay soil to regain their is called chemical equilibrium with the reorientation of water molecule in adsorbed layer. (a) Specific gravity of soil solids (b) Mass-specific gravity of soils (c) Shrinkage ratio of soils 50. The plastic limit and liquid limit of a soil are 30% and 42% respectively. The percentage (d) Density ratio of soils volume change from liquid limit to dry state is Ans. (c) 35% of the dry volume. Similarly the Sol. percentage volume change from plastic limit to dry state is 22% of the dry volume. The VV1 2 shrinkage ratio will be nearly 100 Vd (a) 4.2 (b) 3.1 S.R. = (w1 w 2 ) (c) 2.2 (d) 1.1 Here, V – V = change in volume of soil Ans. (d) 1 2 V = dry volume of soil Sol. d
w1 – w2 = change in water content 52. A masonry dam is founded on pervious sand V L D a factor of safety of 4 is required against B boiling. For the sand, n = 45% and G = 2.65. V C B P The maximum permissible upward hydraulic O gradient will be nearly VS A IES MASTER(a) 0.18 (b) 0.23 (c) 0.28 (d) 0.33 WS WP WL VVVV Ans. (b) L d P d Sol. G = 2.65 WWWWLSPS S VVVV n = 45% = 0.45 L d 100 P d 100 V V d S WWWWLS PS
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n 0.45 0.45 C = 170.38 × (0.11 + 0.0037 × 30) e 0.82 u 1 n 1 0.45 0.55 = 37.65 37.7 kN/m2 G 1 54. A 6m high retaining wall with a vertical back Critical hydraulic gradient (i ) = s Cr 1 e has a backfill of silty sand with a slope of 10° for the backfill. With values of KR = 760 kg/ 1.65 2 2 = 0.906 m /m and KV = 100kg/m /m, the total active 1.82 earth pressure will approximately
iCr (a) 128kN/m (b) 134kN/m FOS = ipossible (c) 138kN/m (d) 142kN/m 0.906 i = 0.2265 0.23 possible 4 Ans. (c) 2 Sol. kH = 760 kg/m /m 53. The representative liquid limit and plastic limit k = 100 kg/m2/m values of a saturated consolidated clay deposit V are 60% and 30%, respectively. The saturated By Peck, Hanson and Thornburn unit weight of the soil is 19 kN/m3. The water 12 1 2 table is at 8 m below ground level. At a depth PHH K H 760 6 of 10m from the ground surface, the undrained 2 2 shear strength of the soil will be nearly = 13680 kg/m = 136.8 kN/m (a) 37.7 kN/m2 (b) 33.5 kN/m2 12 1 2 2 2 P K H 100 6 (c) 29.3 kN/m (d) 25.1 kN/m VV2 2 Ans. (a) = 1800 kg/m = 18 kN/m Sol. Given : WL = 60% P P2 P 2 136.82 18 2 WP = 30% AHV 3 = 138 kNm sat 19 kN / m 55. The vertical stress at any point at a radial distance r and at depth z as determined by 8m using Boussinesq’s influence factor KB and Westergaard’s influence factor KW would be 2m r IES MASTERalmost same for ratios equal to or greater z For clayey soil, Cu IES MASTERthan C u 0.11 0.037 I (%) (a) 2.0 (b) 1.8 p z (c) 1.5 (d) 1.2 2 z = 8 × 19 + 2(19 – 9.81) = 170.38 kN/m Ans. (c)
Ip = 60 – 30 = 30% Sol. Boussinesq’s influence factor = kB
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1 Df 3 2 1, d d 1 = r B c q 2 1 z For strip footing Westergaard’s influence factor = kW sc = sq = 1 1 = For 0, Nc 5.14, Nq 1,N 0 2 3/2 r 1 2 z ( 1 0.5) N qu = CNc + t sub q r q = 40 × 5.14 + (20 + (20 – 9.81)×0.5) × 1 When 1.5, k k u z WB = 230.695 kN/m2 231 kN/m2 Hence, QBoussinesq QWestergaard (Assuming t sat for clay) 56. A strip footing 2m in width, with its base at a depth of 1.5m below ground surface, rests on 57. The settlement due to secondary compression 3 is predominant in a saturated clay soil with sat 20kN / m ; c u (a) Granular soils 2 2 = 40kN/m ; u 0; c’ = 10kN/m ; and (b) Inorganic clays ' 20 . The natural water table is at 1m (c) Organic clays depth below ground level. As per IS : 6403 – 1981, the ultimate bearing capacity of this (d) Very fine sand and silts footing will be Ans. (c)
(taking the relevant N as 5.14) Sol. Seondary compression is caused by creep, c viscous behaviour of the clay-water system, 2 2 (a) 327kN/m (b) 285kN/m compression of organic matter etc. 2 2 (c) 253kN/m (d) 231kN/m In organic clays due to high content of organic Ans. (d) matter, secondary consolidation is predominant. Sol. 58. A raft foundation 10m wide and 12m long is to be constructed in a clayey soil having shear strength of 12kN/m2. Unit weight of soil is Dw = 1m 16kN/m3. the ground surface carries a Df = 2 1.5m surcharge of 20kN/m ; the factor of safety is 1.2 and the value of Nc = 5.7. The safe depth of foundation will be nearly IESB = 2m MASTER As per IS6403:1981 (a) 8.2m (b) 7.3m (c) 6.4m (d) 5.5m qu = scdcic CNc + sqdqiq qNq For footing we consider end of construction Ans. (d) stability, Sol.
2 C = Cu = 40 kN/m , u 0
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k, depend upon the pile material.
60. An excavation is made with a vertical face in Heaving H 2 Qf a clay soil which has Cu = 50 kN/m , q q 3 t 18kN / m and sn = 0.261. The maximum depth of a stable excavation will be nearly (a) 10.6m (b) 12.4m (c) 14.2m (d) 16.0m 2 2 Given : q = 20 kN/m , u C u 12 kN / m , Ans. (a) = 16 kN/m3, B = 10 m, L = 12 m C Sol. S = Bearing capacity of soil for rectangular raft n FH footing in cohesive soil is given by : c t As it is asked to calculate maximum depth of B Qf 1 0.3 CNc D f q excavation, FOS = F = 1 L c 50 10 0.261 = = 1 0.3 12 5.7 16Df 20 1 18 H 12 H = 10.64 10.6 m = 105.5 + 16D 61. Reconnaissance survey for determining Base failure will occur when Qf = 0 feasibility and estimation of scheme falls under 105.5 Hence, D = 6.59 the classification based on the 16 (a) Nature of the field of survey (minus sign indicates that it is excavation) (b) Object of surveying D 6.59 Safe depth = = 5.49 m (c) Instruments used FOS 1.2 (d) Method employed 59. The skin frictional resistance of a pile driven Ans. (b) in sand does not depend on Sol. During reconnaissance survey, surveyor (a) Lateral earth pressure coefficient should first of all thoroughly examine the (b) Angle of friction between pile and soil ground to ascertain as how the work can be (c) Pile material arranged in the best possible manner. (d) Total stressIES analysis MASTERIt falls under classification based on purpose or object of surveying (for which surveying is Ans. (d) IES MASTERconducted). Sol. Skin frictional resistance for a driven pile in On the basis of object of survey the sand classification can be as given below. (1) Control survey Qf K av tan Asurface (2) Hand survey Where, k = lateral earth pressure coefficient (3) Topographic survey = angle of friction between pile and soil (4) Engineering survey
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(a) Reconnaissance survey tan (90 ) = cot (b) Preliminary survey 63. In plane surveying where a graduated staff is (c) Location survey observed either with horizontal line of sight or (5) Route survey inclined line of sight, the effect of refraction is to (6) Construction survey (a) Increase the staff reading (7) Astronomic survey (b) Decrease the staff reading (8) Mine survey (c) Neither increase nor decrease the staff 62. A survey line BAC crosses a river, A and C reading being on the near and distant banks (d) Duplicate the staff reading respectively. Standing at D, a point 50m measured perpendicularly to AB from A, the Ans. (b) bearings of C and B are 320° and 230° Sol. Refraction is the phenomenon of light rays respectively, AB being 25m. The width of the deviating from a straight line as they pass river will be through different layers of air of different (a) 80m (b) 90m densities. (c) 100m (d) 110m Due to refraction ray of light bend towards centre of earth. Hence refraction make reading Ans. (c) lower than what it should be with a horizontal Sol. line of sight.
Horizontal line of sight C T.V R efra cte d r ay x 320° M.V A 50 m D 25 m 230° 64. A sidereal day is the average time taken by (a) The Earth to move around the sun once B (b) The Moon to move around the Earth once IES MASTER(c) The first point of Aries to cross the same In ABD meridian successively
25 1 (d) The Earth to move around its own axis tan = cot 2 50 2 once x Ans. (c) In DAC 2 50 Sol. Side real day is the time interval between the x tan (90 ) = x 100 m 50
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movement of the first point of aries over the = 100 m; and the size of each photograph is same meridian twice. 20cm × 20cm. The minimum required number of photographs will be 65. In triangulation in order to control the (a) 170 (b) 158 accumulation of errors of length and azimuth subsidiary bases are selected. At certain (c) 146 (d) 134 stations the astronomical observations for Ans. (d) azimuth and longitude are also made. These A stations are called Sol. Number of photograph, N = (a) Transportation stations a (b) Bowditch stations A N = 1cm 1 (c) Universe stations (1 P )l (1 P )w S L s 100 m 10,000 (d) Laplace stations SS Ans. (d) 150 106 Sol. The defect of triangulation is that it tends to N = (1.06) 0.2 (1 0.3) 0.2 accumulate errors of length and azimuth, since 1 1 the length and azimuth of each line is based 104 104 on the length and azimuth of the preceding 6 line. 150 10 N = 8 To control the accumulation of errors, 0.4 0.2 0.7 0.2 10 subsidiary bases are also selected. At certain 15 107 N = stations, astronomical observations for 4 2 7 2 104 azimuth and longitude are also made. These N = 133.393 134 Nos. stations are called Laplace stations. 68. Which one of the following conditions is not 66. A vertical photograph is taken at an altitude correct with respect to the transition curve? of 1200m above mean sea level (a.m.s.l.) of a terrrain lying at a elevation of 80m a.m.s.l. (a) It should be tangential to the straight The local focal length of camera is 15cm. The approaches at the two ends scale of the photograph will be nearly (b) It should meet the circular curve (a) 1 : 8376 (b) 1 : 7467 tangentially (c) 1 : 6558 (d) 1 : 5649 (c) Its curvature will necessarily be non-zero at the point of take-off from the straight Ans. (b) IES MASTERapproaches f 15 102 (d) The rate of increase of curvature along Sol. Scale, S = H hIES 1200 80 MASTERthe transition reach should match with the increases of cant. 1 1 = Ans. (c) 7466.66 7467 Sol. Radius of transition curve should be infinite 67. Aerial photographs are required to be taken when taking off from straight line or meeting to cover an area of 150 km2. The longitudinal the straight line. and side overlaps are to be 60% and 30% respectively. The scale of photograph is 1cm
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1 same radius. If the lines are 12m apart and Curvature = and R = R the maximum distance between tangent points measured parallel to the straights is 48m, then Curvature = 0 the maximum allowable radius will be 69. A circular curve has a long chord of 80m and (a) 51.1m (b) 52.3m a versed sine of 4m. The height and ordinate (c) 53.5m (d) 54.7m at a distance of 30m from the mid-ordinate will be nearly Ans. (a) (a) 3.06m (b) 2.72m Sol. (c) 2.24m (d) 1.76m 0 Railway T2 Ans. (d) Sol. Long chord, L = 80 m 12 R R Versine = Mid-ordinate (M) = Oo = 4m 30m Oo Ox T1 T2 48 L/2 2 2 T1T2 = 48 12 = 49.47m R 12 sin / 2 = 0.2426 49.47 = 28.07° 2 2 L ORRo 12 2 Radius = 2 1 cos 2 2 L R R 4 12 = 2 2 1 cos28.07 2 2 L = 51.01m (R 4) R 2 71. In an old map, a line AB was drawn to a 2 2 (R – 4) = R – (40) magnetic bearing of 5°30’, the magnetic R2 – 8R + 16 = R2 – 402 declination at the time being 1° East. If the present magnetic declination is 8°30’ East, 8R = 1616 the line should be set to a magnetic bearing R = 202 m IES MASTERof x2 (a) 358° (b) 2° OOx o 2R (c) 6 30 (d) 257° 302 O 4 1.77 m Ans. (a) x 2 202
70. Two parallel railway lines are to be connected by a reverse curve, each section having the
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Sol. 73. Consider two cars approaching from the opposite directions at 90km/h and 60km/h. If T.N the reaction time is 2.5s, coefficient of friction M.N is 0.7 and brake efficiency is 50% in both the 1° cases, the minimum sight distance required 5°30 B T.B = 6°30 to avoid a head-on collision will be nearly A (a) 154m (b) 188m (c) 212m (d) 236m Ans. (d) Sol. Min sight distance required Old map 2 2 v1 v2 T.N = v1 t r v2 t r B 2g 2g
T.B = 6°30 2 90 0.278 M.N = 90 0.278 2.5 2° 2 9.81 0.35 8°30 2 60 0.278 + 60 0.278 2.5 2 9.81 0.35
A M.B = 360° – 2° = 235.92 m = 358° 236 m. Note : Co-effn of friction is taken as 0.35 (is 0.7 × 0.5) New map 74. What is the extra widening required (as 72. An unconformity is nearest magnitude) for a pavement of 7m width on a horizontal curve of radius 200m, if (a) A surface of erosion or non-deposition as the longest wheel of vehicle expected on the detected in a sequence of rocks road is 6.5m and the design speed is 65km/ (b) A layer of boulders and pebbles in a h? sequence of rocks (a) 0.3m (b) 0.5m (c) A layer of clay or shale in an igneous (c) 0.7m (d) 0.9m mass IES MASTER Ans. (c) (d) A type of joint especially associated with folded and faultedIES rocks MASTERSol. Ans. (a) nl2 v We = Sol. An unconformity is a buried erosional or non- 2R 2.64 R depositional surface separating two rock
masses or strata of different ages, indicating 2 2 6.5 (65 0.278) that sediment deposition was not continuous. = 2 200 2.64 200
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= 0.6952 Ans. (d) 0.7 m. Sol. Various methods of speed and delay studies are : 75. A vehicle moving at 40km/h speed was 1. Floating car method stopped by applying brake and the length of the skid marks was 12.2m. If the average 2. License plate method or vehicle number skid resistance of the pavement is 0.70, the method brake efficiency of the test vehicle will be 3. Interview technique nearly 4. Elevated observation (a) 80% (b) 74% 5. Photographic technique (c) 68% (d) 62% Ans. (b) 78. Consider the following data with respect to the design of flexible pavement : Sol. Design wheel load = 4200kg v2 µ = 2 braking test 2gl Tyre pressure = 6.0 kg/m Elastic modulus = 150 kg/cm2 40 0.2782 = 2 9.81 12.2 Permissible deflection = 0.25cm = 0.5166 1 (take 1/2 1.77, 1/2 0.564, 0.318 and 0.5166 = 100 braking 0.7 2 9.87 ) = 73.79% The total thickness of flexible pavement for a 74%. single layer elastic theory will be nearly 76. The main drawback of automatic counters- (a) 42cm (b) 47cm cum-classifiers, used for traffic volume studies, (c) 51cm (d) 56cm is that it is not yet possible to classify and record Ans. (c) (a) Vehicle type (b) Axle spacing Sol. (c) Axle load (d) Speed 2 1/2 3P 2 Ans. (a) Total thickness = a 2 E 77. Which one ofIES the following is notMASTER a part of ‘speed and delay’ studies? 2 1/2 3P P (a) Floating car method = 2 E p (b) Vehicle number method
(c) Interview technique 2 1/2 3 4200 4200 = (d) License number method 2 3.14 150 0.25 3.14 6
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51 cm. (d) Wind tends to carry drifting sand in a zigzag way 79. The minimum possible grade that can be Ans. (d) provided in a tunnel and its approaches with providing adequately for proper drainage is 83. Consider the following data for designing a (a) 0.1% (b) 0.2% taxiway for operating Boeing 707 – 320 (c) 0.3% (d) 0.4% aeroplane : Ans. (b) Wheel base = 17.70m
80. The section of the tunnel adopted perfectly in Tread of main loading gear = 6.62m lieu of ease of construction and maintenance in hard rock tunnels, where the risk of roof Turning speed = 40km/h failure or collapse caused by external pressure Coefficient of friction between tyres and from water, or from loose or unstable soil pavement surface = 0.13 conditions on tunnel lining is practically non- existent is The turning radius of the taxiway will be (a) Circular section (a) 98.5m (b) 94.5m (b) Segmental roof section (c) 89.5m (d) 86.5m (c) Horse-shoe section Ans. (a) (d) Egg-shaped section 84. Which one of the following instances of Ans. (a) performance of aircraft is not considered for Sol. According to IRC : SP-91, circular sections determining basic runway length ? are structurally best and are commonly used (a) Normal landing case for underwater tunnels, tunnels through soft ground and for tunnels excavated with TBM. (b) Normal take-off case (c) Engine failure case 81. Which one of the following methods is adopted for tunneling in soft soils? (d) Emergency landing case (a) Pitot tunnel method Ans. (d) (b) Drift method Sol. Basic runway length is determined by 3 cases: (c) Needle beam method (i) Normal landing case (d) Heading and benching method (ii) Normal take off case Ans. (c) IES MASTER(iii) Engine failure case 82. Which one of theIES following features doesMASTER not Directions: Each of the next six (06) times pertain to Littoral drift? consists of two statements, one labeled as ‘Statement (I)’ and the other as ‘Statement (a) It depends on length of wave (II)’. You are to examine the two statements (b) It is the process of erosion of deposition carefully and select the answers to these six by waves items using the codes given below :
(c) Waves caused by prevailing wind, stir up Codes: and move sand particles
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(a) Both Statement (I) and Statement (II) are the area required for maximum mid-span individually true, and Statement (II) is the moment in the slab. correct explanation of Statement (I). Ans. (b) (b) Both Statement (I) and Statement (II) are Sol. Both statements are individually correct. We individually true, but Statement (II) is not provide torsion reinforcement in order to the correct explanation of Statement (I). minimize the crack due to torsion. (c) Statement (I) is true, but Statement (II) is false 89. Statement (I): The inclination of the resultant stress with normal can exceed angle of repose (d) Statement (I) is false, but Statement (II) is (adopting old terminology). true. Statement (II): The ratio of the difference 85. Statement (I): Expansive cement is used in between greatest and least intensities of repair work for opened up joints. pressure to their sum cannot exceed the sine Statement (II): Expansive cement expands of the angle of repose (adopting old while hardening. terminology). Ans. (a) Ans. (d) Sol. Expansive cement is a cement, which when Sol. mixed with water, it will have a tendency to increase in volume significantly while setting. = resultant stress This will be helpful to repair the damaged r concrete surfaces. tan 86. Statement (I): Plastic hinges are developed when stress at every point is equal to yield stress. Statement (II): Plastic hinges are formed at (,) sections subjected to the greatest curvature. angle of repose Ans. (b) 3 1 87. Statement (I): If degree of fixity at supports is lessened, the maximum hogging moment at The inclination of the resultant stress with the ends will decrease. normal (i.e., angle of obliquity) cannot exceed angle of repose, so statement (I) is incorrect Statement (II): If degree of fixity at supports is lessened, the maximum sagging moment 1 greatest pressure intensity at mid-span IES decreases. MASTER least pressure intensity Ans. (c) 3 88. Statement (I): Torsion reinforcement is Now, sin = 1 3 provided at (and near) corners in a two-way 1 3 slab which is simply supported on both edges statement II is correct meeting at the corner. 90. Statement (I): Alum works in slightly alkaline Statement (II): The area of reinforcement in range. each of the layers shall be three-quarters of
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Statement (II): At higher temperatures, 93. A construction equipment has an initial cost viscosity of water (resistance to settling) of Rs. 2,00,000 and salvage value of Rs. decreases and flocs settle better. 50,000 at the end of an economic life of 5 Ans. (b) years. The rate of straight-line depreciation and total depreciation will be Sol. Alum is a coagulant which works efficiently in slightly alkaline range i.e. between 6.5 to 8.5. (a) 0.1 and Rs. 1,50,000 At higher temperature viscosity of water (b) 0.2 and Rs. 1,50,000 decreases and floc formation is better (c) 0.1 and Rs. 1,00,000 (d) 0.2 and Rs. 1,00,000 91. A front-end loader on a given job moves a load of 1.5 m3 of loose soil in one cycle Ans. (b) consisting of loading-lifting-travelling- Sol. Total depreciation = Initial cost – Salvage value unloading-return trip-and-ready for next = 200,000 – 50,000 loading. If each cycle time is 1.2 minutes, the actual output will be = 1,50,000 (a) 75 m3/hour (b) 70 m3/hour Depreciation by straight line method 3 3 CC (c) 65 m /hour (d) 60 m /hour = i s Ans. (a) n 1 1 Sol. Actual output = Volume in one cycle in cum Rate of depreciation = 0.2 × No. of cycle per hour n 5 60 = 1.5 75 m3 1.2 During this we will not consider idle time 94. Consider the following assembly with different because in question it is accounted as ready operations for next loading. B D G 92. Which of the following techniques belong to A ‘Project Time Plan’? E H K 1. Critical path method C F 2. Precedence network analysis Operation Standard time minutes 3. Line of balance technique A 60 4. Linear programme chart IES MASTERB 65 (a) 1, 2 and 3 only (b) 1, 2 and 4 only C 29 (c) 3 and 4 onlyIES(d) 1, 2, 3 andMASTER 4 D 37 Ans. (a) E 28 Sol. Techniques for project time plan are critical F 63 path method, precedence network analysis and line of balance technique. G 36 H 126 Linear programme chart is used for profit maximisation, cost minimisation and resources K 64 allocation.
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There are 250 working days in a year to (a) Activity float (b) Free float produce 4000 units in a year. The minimum (c) Total float (d) Interfering float number of work stations required will be Ans. (b) (a) 13 (b) 12 Sol. Activity float is range within which the start (c) 11 (d) 10 time may fluctuate without affecting the Ans. (a) completion time of the project Sol. Total float is the maximum time by which an activity can be delayed without affecting
EST Activity name EFT project completion time. Duration Free float is the time by which an activity can
60 B 125 125 D 162 162 G 198 65 37 36 be delayed without affecting earliest start time of succeeding activity 0 A 60 60 89 E 117 28 Interfering float is difference between total float 60 C 89 198 H 324 324 K 388 29 126 64 and free float.
89 F 152 63 97. A crew consisting of two carpenters and one helper can fix 10 m2 of a slab form work in 8 The time required to assemble 1 unit is 388 hours and the hourly labour rate of a carpenter minutes. is Rs. 85 and for a helper is Rs. 69.50. An average hourly rate per worker of the crew Assuming 8 hours of working in a year. will be nearly No. of unit manufactured by 1 working station (a) Rs. 90 (b) Rs. 80 250 8 60 in 250 days will be = 309.27 (c) Rs. 70 (d) Rs. 60 388 Ans. (b) No. of working station required to manufacture Sol. Average hourly rate per worker 4000 units in a year will be 2 Carpenter rate 1 Helper rate = 4000 3 = 12.933 13 309.27 2 85 1 69.5 = 95. Flattening and smoothing the road surface by 3 scrapping is called = 79.83 80Rs. (a) Compaction (b) Consolidation 98. A project with the production cost of Rs. 100 (c) Grading (d) Ditch digging crores, has a 20,000 man-months as direct Ans. (c) IES MASTERlabour, of which 60% is non-productive time. The labour cost as estimated while tendering Sol. A motor grader in road construction is used is 20% of project cost. If 15% of the wastage for cutting, spreading and levelling of material resulting from non-productive time is i.e., flattening and smoothing. eliminated by using improved methods, the 96. The amount of time by which the start of the resulting saving in labour cost will be activity may be delayed without interfering with (a) 14.5% (b) 18.5% the start of any succeeding activity is called
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(c) 22.5% (d) 26.5% 1200 Duration = d = 35.3 days Ans. (c) 40 0.85 Sol. Lobour costed = 20% of project cost 35.3 = 0.2 × 100 = = 5.648 5.5 weeks 5 1.25 = 20 crores Non productive labour time at 60% of labour 100. A systematic measurement and evaluation of cost the way in which an organization manages its health and safety programme against a series = 0.6 × 20 = 12 crores of specific and attainable standards is called Non productive time = 0.15 × 12 (a) Safety inspection (b) Safety audit = 1.8 crores (c) Safety plan (d) Safety committee Saving as percentage of productive work value Ans. (b) 1.8crores Sol. Safety audit-It is systematic measurement and = 100 40% of 20crores evaluation of the way in which an organization = 22.5 manages its health and safety programme against a series of specific and attainable 99. Consider the following data : standards.
Work is carried out by a contractor employing 101. On a construction project, the contractor, on labour with 25% overtime per day an average, employed 100 workers with 50 Working for 5 days a week hours working per weeks. The project lasted Contractor peak manpower is 40 per day for 35 weeks and, during this period, 14 disabling injuries occurred. The injury- Build-up period is 20% frequency rate will be (based on one lakh of Rundown period is 10% man hours worked) Total effort in standard man days is 1200 (a) 5 (b) 6 (c) 7 (d) 8 The duration of work by Trapezoidal manpower distribution pattern will be Ans. (d) (a) 5.5 weeks (b) 6.5 weeks Sol. Injury frequency rate
(c) 7.5 weeks (d) 8.5 weeks No. of injuries 5 = 10 Ans. (a) Total No. of hours worked Sol. Effort in man-daysIES = peak manpower MASTER 5 14 10 = 8 (build up period) 100 50 35 × IES peak level MASTER period 2 102. The graphical representations wherein long duration jobs are broken down to key rundown period + segmental elements, wherein events are 2 shown in chronological order without attention to logical sequencing, and wherein 0.2d 0.1d 1200 = 40 0.7 d interdependencies between the events is not 2 2 highlighted, is referred to as = 40 × 0.85 d
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(a) CPM (b) Milestone chart M (c) GANTT chart (d) PERT Ans. (b) Sol. In milestone chart long duration jobs are broken down to key segmental elements which 10 m are called as events, these are present in G Q chronological order. Interdependencies x P
between events of various jobs is not shown. Wboat
WTotal 103. A ship weighs 127 MN. On filling the ship’s B´
boats on one side with water weighing 600 Wboat = 600kN kN with the mean distance of the boats from Take moment of all force about “M”. the centre line of the ship being 10 m, the angle of displacement of the plumb line is MM = 0 (equilibrium) 2°16'. The metacentric height will be nearly WTotal × x = Wboat × GQ (Take sin2°16' = 0.04, cos2°16' = 0.9992 and 127600x = 600 × 10 tan2°16' = 0.04) x = 4.4022 × 10–2m (a) 1.73 m (b) 1.42 m x = (c) 1.18 m (d) 0.87 m GMP sin GM Ans. (c) x 4.7022 102 GM = Sol. sin 0.04 GM = 1.176
GM 1.18m M 104. For frictionless adiabatic flow of compressive fluid, the Bernoulli’s equation with usual notations is
G P B 2 2 x k p1 v 1 k p2 v 2 (a) z1 z2 h L W W B Total Total k 1 w1 2g k 1 w2 2g B´ Wboat
10m 2 2 k p1 v 1 k p2 v 2 (b) z z Ship k 1 w 2g1 k 1 w 2g 2 IES MASTER 1 2 M = Metacentric p v2 p v2 1 1 z H 2 2 z GM = Metacentric height (c) 1 m 2 w1 2g w2 2g sin = 0.04 (Given) 2 2 Total weight (W ) = 127000 + 6000 = k p1 v 1 p2 v 2 total (d) z1 H m z2 h L 127600kN k 1 w1 2g w2 2g
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Ans. (b) 1. The flow is uniform and pressure Sol. For compressive flows in fluid dynamic in distribution is hydrostatic before and after adiabatic state Bernoulli’s equation is the jump. 2. Losses due to friction on the surface of 2 V k p the bed of the channel are small and gz = constant 2 k 1 hence neglected. Where k is ratio of specific heats of the fluid. 3. The slope of the bed is small, so that the component of the weight of the fluid in p V2 k p V2 k 1 1 2 2 the direction of flow is negligibly small. z1 = z2 k 1 w1 2g k 1 w2 2g 107. Water is to be pumped out a deep well under 105. The phenomenon of generation of lift by a total head of 95 m. A number of identical rotating an object placed in a free stream is pumps of design speed 1000 rpm and specific known as speed 900 rpm with a rated capacity of 150 (a) Coanda effect l/s are available. The number of pumps required will be (b) Magnus effect (a) 1 (b) 3 (c) Scale effect (c) 5 (d) 7 (d) Buoyancy effect Ans. (b) Ans. (b) Sol. Given, Sol. Magnus effect is a phenomenon associated with spinning object moving through a fluid Total head, H = 95m producing lift force on the object N = 1000rpm
106. Which of the following assumptions is/are Ns = 900rpm made in the analysis of hydraulic jump? Q = 150 / S 1. It is assumed that before and after jump NQ formation the flow is essentially two- We know, Ns = 1/4 dimensional and that the pressure Hm distribution is hydrostatic. 1000 150 2. The length of the jump is small so that 900 = 3/4 Hm the losses due to friction on the channel H = 32.5m floor are small and hence neglected. m 3. The channel floor is horizontal or the slope For lefting water to a higher head, pumps are is so gentleIES that the weight componentMASTER of to be installed in series. the water mass comprising the jump is H very high. IES MASTERRequired number of pumps = Hm (a) 1 only (b) 2 only 95 = (c) 3 only (d) 1, 2 and 3 32.5 Ans. (b) = 2.92 3 Sol. Before deriving the expression for the depth of hydraulic jump, the following assumptions 108. Consider the following data from a test on are made. Pelton wheel :
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Head at the base of the nozzle = 32 m (c) Base-land plant Discharge of the nozzle = 0.18 m3/s (d) Run-of-river plant Area of the jet = 7500 mm2 Ans. (d) Power available at the shaft = 44 kW Sol. Run-off river plants are one which does not Mechanical efficiency = 94% store any water and utilises the water as it flows. The power lost in the nozzle will be nearly 110. Two turbo-generators, each of capacity 25,000 (a) 3.9 kW (b) 4.7 kW kW, have been installed at a hydel power (c) 3.5 kW (d) 2.3 kW station. The load of the hydel plant varies from Ans. (b) 15,000 kW to 40,000 kW. The total installed plant capacity and the load factor are nearly Sol. Given, Q = 0.18m3/S (a) 40,000 kW and 68.8% A = 7500mm2 (b) 50,000 kW and 68.8% = 7500 × 10–6m2 (c) 40,000 kW and 62.3% Shaft power = 44kW (d) 50,000 kW and 62.8% x = 94% m Ans. (b) H = 32m Sol. Since two generators each of capacity 25,000 Power at the base of the nozzle = gQH kW have been installed, hence = 9.81 × 0.18 × 32 Total installed plant capacity = 2 × 25000 = 56.5 kW = 50,000kW 1 Average load over Kinetic energy per second of Jet = QV2 2 a certain period Now, load factor = 2 Peak load during 1 0.18 = 0.18 that period 2 7500 106 15000 40000 1000 kW 2 1000 = 40000 = 51.84kW = 0.6875 Power lost in nozzle = 56.5 – 51.84 = 68.8%
= 4.66kW 111. An airfoil is a streamlined body as shown in the figure below. Because of the streamlining 109. A certain hydropowerIES plant utilizes MASTER the flow of the body, the seperation occurs only at the as it occurs, without any provision for storage. extreme rear of the body, resulting in It is premised that a defined minimum dry weather flow is available. Such a plant is l classified as V (a) Diverted-flow plant (b) Pooled storage plant (a) A very high pressure drag
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(b) A small wave and consequently small 113. Which of the following are components parts pressure drag for an oil pressure governor in modern (c) A moderate pressure drag turbines? (d) No pressure drag 1. Servomotor, known as relay cylinder Ans. (b) 2. Oil sump Sol. As Airfoil is a streamlined body, so separation 3. Oil pump which is driven by belt connected of boundary layer occur only at the extreme to turbine main shaft rear of the body. So due to small wake size 4. Draft tube at back, pressure difference between front and (a) 1, 2 and 3 only (b) 1, 2 and 4 only back reduces. So, form drag (pressure drag) is comparatively very small in airfoil. (c) 1, 3 and 4 only (d) 2, 3 and 4 only Ans. (a) 112. A plate 0.025 mm distant from a fixed plate moves at 60 cm/s and requires a force of 0.2 Sol. The main components of an oil pressure kgf/m2 to maintain this speed. The dynamic governor are : viscosity of the fluid between the plates will (i) the servomotor or relay cylinder be nearly (ii) The distribution valve or control valve –10 2 (a) 9.2 × 10 kgfs/cm (iii) Actuator or pendulum –10 2 (b) 8.3 × 10 kgfs/cm (iv) Oil pump –10 2 (c) 7.4 × 10 kgfs/cm (v) Gear pump which runs by tapping power (d) 6.5 × 10–10 kgfs/cm2 from the power shaft by belt drive. Ans. (b) (vi) A pipe system communicating with the Sol. control valve servometer and the pump Pendulum or actulator 60cm/s 0.025mm Fly ball Lever Sleeve Fluid of dynamic Fulcrum From A viscosity ‘ ’ turbine shaft Distribution valve = 0.2 kg.f/m2 Gear B pump du Towards = turbine dy guiding IES MASTERServomotor or mechanism Oil sump kg.f 600mm / sec relay cylinder 0.2 = m2 IES0.025mm MASTERFig. Oil pressure governor
6 kg.fs 114. A double-acting reciprocating pump having = 8.33 10 2 m2 piston area 0.1 m has a stroke 0.30 m long. The pump is discharging 2.4 m3 of water per 10 2 = 8.33 10 kgfs / cm minute at 45 rpm through a height of 10 m. The slip of the pump and power required to drive the pump will be nearly
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(a) 0.005 m3/s and 4.8 kW 116. Which one of the following points should be (b) 0.003 m3/s and 4.8 kW kept in mind while selecting the site for a rain gauge station? (c) 0.005 m3/s and 4.4 kW (a) The site where a rain gauge is set up 3 (d) 0.003 m /s and 4.4 kW should be close to a meteorological Ans. (c) observatory. Sol. Given, (b) The rain gauge should be on the top of a Piston Area, A = 0.1m2 hill. Stroke length, L = 0.3m (c) A fence, if erected to protect the rain gauge from cattle etc. should be located 3 Actual discharge, Qact = 2.4m /min within twice of the height of the fence. 3 = 0.04m /s (d) The distance between the rain gauge and Speed, N = 45rpm the nearest object should be atleast twice the height of the object. H = 10m Ans. (d) 2ALN Q = th 60 117. Which of the following statements relates to a retarding reservoir? 2 0.1 0.3 45 = 60 1. There are no gates at the outlets and hence the possibility of human error in = 0.045m3/s reservoir operation is eliminated. Slip = Q – Q th act 2. The high cost of gate installation and also = 0.045 – 0.04 its operation is saved. = 0.005m3/s 3. An automatic regulation may cause coincidence of floor crest farther Theoretical power required = gQ .H th downstream where two or more channels = 9.81 × 0.045 × 10 taking off from retarding reservoirs join = 4.4kW together. (a) 1, 2 and 3 (b) 1 and 2 only 115. In intensity-duration analysis of Sherman, the intensity of rainfall i is represented as (c) 1 and 3 only (d) 2 and 3 only
n Ans. (a) bn a (a) (b) (t a) (t b)n 118. The coefficient of transmissibility T for a confined aquifer can be determined by a (a t)n a pumping-out test together with other relevant (c) IES(d) MASTERn b (t b) observations. The applicable formula is (where Ans. (d) Q = Discharge, and S = Difference in drawdowns in two wells) Sol. Sherman equation is given as : Q Q a (a) (b) i = n 2.72 S 1.72 S t b
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1. Broken-brick aggregate is obtained by Q Q (c) S (d) S crushing waste bricker, and it has a 2.72 2.72 density varying between 1000 kg/m3 – Ans. (a) 1200 kg/m2. 2. Such aggregate is usable in concrete for 119. The volume of water below the minimum pool foundation in light buildings, floorings and level in a reservoir is known as walkways. (a) Useful storage 3. Such aggregate may also be used in light (b) Surcharge storage weight reinforced concrete floors. (c) Dead storage (a) 1 and 2 only (b) 2 and 3 only (d) Bank storage (c) 1 and 3 only (d) 1, 2 and 3 Ans. (c) Ans. (b) 120. Depending upon the source from which the Sol. water is drawn, flow irrigation can be sub- Broken–brick aggregate is obtained by divided into crushing waste brick and has a density 1. River canal irrigation varying between 1600 – 2000 kg/m3 2. Reservoir or tank irrigation It is used in concrete in light-weight reinforced concrete floors. 3. Combined storage and lift irrigation It is used in concrete for foundation in 4. Combined storage and diversion irrigation light building flooring and walkways. Which of the above designations are relevant? 122. In handling air-entraining admixtures the (a) 1, 2 and 3 only (b) 1, 2 and 4 only beneficial amount of entrained air depends (c) 1, 3 and 4 only (d) 2, 3 and 4 only upon certain factors like Ans. (b) 1. Type and quantity of air-entraining agent Sol. Flow irrigation system can be further classified 2. Water-cement ratio of the mix on the basis of source of water from which 3. Strength of aggregates the flow irrigation canal take off. 4. Extent of compaction of concrete (a) Direct irrigation system or river canal (a) 1, 2 and 3 only (b) 1, 2 and 4 only irrigation weir or barrage (c) 1, 3 and 4 only (d) 1, 2, 3 and 4 (b) Storage irrigation system eg reservoir or tank irrigation.IES MASTERAns. (b) (c) Combined storage and diversion irrigation Sol. The beneficial amount of entrained air Dam or IESriver and a diversion MASTER weir or depends on barrage or the river at a suitable place on (i) Type and quantity of air entraining agent the down stream of dam to divert water (ii) Water cement ratio of mix into the canal. (iii) Mixing time 121. Which of the following statements are wholly (iv) extent of compaction of concrete correct regarding broken-brick aggregate useable in concrete? (v) Type of cement
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123. Which one of the following statements is not free lime in cement in presence of water, correct with respect to fly ash? improve durability of concrete at later stage, (a) As part replacement of cement in the but doesn’t contribute to early age strength of range of 15%-30%, fly ash reduces the composite cement. strength in the initial period, but once the 125. Hydration of which compound is responsible Pozzolanic process sets in, higher for increase in strength of cement in later age? strength can be obtained. (a) Tri-calcium Aluminate (C A) (b) Fly ash as a part replacement of sand 3 has a beneficial effect on strength even (b) Tetra-calcium Aluminoferrite (C4AF) at early age. (c) Tri-calcium Silicate (C3S) (c) Fly ash as a part replacement of sand is (d) Di-calcium Silicate (C2S) economical. Ans. (d) (d) A simultaneous replacement of cement Sol. C S hydrates and hardens slowly and takes a and fine aggregates enables the strength 2 long time to add to the strength. Generally, at a specified age to be equalled after one year contribution to the strength and depending upon the water content. hardness of cement is predominately due to Ans. (c) C2S. Sol. Flyash is costlier than sand. Hence using this 126. The creep strain of cement attains its terminal in the place of sand will not be economical. values by Flyash will only contribute to strength only (a) 1 year (b) 2 years when the pozzolanic reaction sets in. (c) 5 years (d) 6 months 124. Which one of the following statements is not Ans. (c) correct with respect to the properties of cement? Sol. Creep is a time dependent phenomenon and creep strain decreases with time. Although it (a) Highly reactive Pozzolanas enhance the doesn’t stop for a long period, but creep strain early age strength of the composite at 5-years are taken as terminal values. cement (b) Pozzolanic activity refines pore structure 127. Which of the following methods will help in which decreases electrolytic resistance of reducing segregation in concrete? concrete. 1. Not using vibrator to spread the concrete (c) The expansion due to alkali-silica reaction 2. Reducing the continued vibration can be controlled by replacement of as 3. Improving the cohesion of a lean dry mix high as 60% of OPC with high-calcium through addition of a further small quantity Pozzolana. IES MASTERof water. (d) Such high amounts of replacement (a) 1, 2 and 3 (b) 1 and 2 only cements result in higher accelerated carbonation depths compared to pure use (c) 1 and 3 only (d) 2 and 3 only of OPC only. Ans. (d) Ans. (a) Sol. Segregation can be reduced by preventing Sol. Pozzolanas don’t have any cementitious over-vibration, using good design mixes, properties as such. But when they react with avoiding dropping concrete from heights, avoiding excess water etc.
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But vibration is an excellent method for V (1 ) compaction so it shouldn’t be avoided. (a) 2nL (1 2 )(1 ) 128. On an average, in a 125 mm slump, the concrete may lose about (in first one hour) V (1 ) (b) (a) 15 mm of slump (b) 25 mm of slump 2nL (1 2 )(1 ) (c) 40 mm of slump (d) 50 mm of slump V2 (1 ) Ans. (d) (c) 2nL (1 2 )(1 ) Sol. The loss of workability varies with the type of cement, the concrete mix proportions, the 2 V2 (1 ) initial workability and the temperature of the (d) concrete. 2nL (1 2 )(1 ) On an average a 125 mm slump concrete where may lose about 50 mm slump in the 1st one V is pulse velocity, in mm/s, hour. n is resonant frequency of longitudinal 129. Permeability in concrete is studied towards vibration, in Hz, providing for, or guarding against, which of L is distance between transducers, in mm. the following features? Ans. (*) 1. The penetration by materials in solution 2 may adversely affect the durability of 2nl (1 )(1 2 ) concrete; moreover, aggressive liquids Sol. V 1 ‘attack’ the concrete. [IS 13311(part-I : 1992)] 2. In case of reinforced concrete, ingress of 2 moisture and air will result in corrosion of v 1 steel leading to an increase in volume of 2nl (1 )(1 2 ) steel, resulting in cracking and spelling of the concrete cover. 131. Which one of the following methods/ 3. The moisture penetration depends on techniques will be used for placing of concrete permeability and if the concrete can in dewatered ‘Caissons or Coffer’ dams? become saturated with water it is less (a) Tremie method vulnerable to frost action. (b) Placing in bags (a) 1, 2 and 3 (b) 1 and 2 only (c) Prepacked concrete (c) 1 and 3 onlyIES(d) 2 and MASTER 3 only (d) In-the-dry practice Ans. (b) IES MASTERAns. (d) Sol. In permeable conctete, ingress of water leads Sol. The placing of concrete in dewatered caissons concrete susceptible to chemical attack, frost or coffer dams follows the normal in-the-dry- action, rusting of steel reinforcements. practice If pores are saturated with water, the concrete will be more vulnerable to frost action. 132. The minimum cement content (kg/m3) for a pre-specified strength of concrete (using 130. Poisson’s ratio of concrete µ can be standard notations) premised on ‘free water- determined using the formula cement ratio’ will be as
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(a) 5 mm and 120 MN/m2 CW (a) 1 (b) 9 mm and 120 MN/m2 1000SC 1000 (c) 5 mm and 150 MN/m2 Water content (d) 9 mm and 150 MN/m2 (b) Water Cement ratio Ans. (b)
(c) Water content × Water cement ratio Sol. Free expansion = T 100F = 15000 mm × 12 × 10–6 × (65 – 15) (d) CF = 9 mm
Ans. (b) Temperature stress = ET Sol. The minimum cement content for a pre- = 200 × 109 N/m2 × 12 × 10–6 × 50 specified strength of concrete premised on = 1.2 × 108 N/m2 “free-water-cement ratio” will be as = 120 MN/m2. Water content Water cement ratio 135. A bar of uniform rectangular section of area A is subjected to an axial tensile load P; its 133. A bar specimen of 36 mm diameter is Young’s modulus is E and its Poisson’s ratio subjected to a pull of 90 kN during a tension 1 is . Its volumetric strain, e is test. The extension on a gauge length of 200 m v mm is measured to be 0.089 mm and the P 3 P 2 change in diameter to be 0.0046 mm. The (a) 1 (b) 1 AE AE Poisson’s ratio will be m m (a) 0.287 (b) 0.265 P 2 P 1 (c) 1 (d) 1 AE AE (c) 0.253 (d) 0.241 m 2m Ans. (c) Ans. (a) Sol. lateral strain Sol. µ = A longitudinal strain P P 1 E, µ 0.0046 m 36 = = 0.287 ( )(1 2µ) 0.089 Volumetric strain (e ) = x y z V E 200 IES MASTERP x = 134. A steel rod 15 m long is at a temperature of A 15°C. The values of = 12 × 10–6/°C and E y = 0 = 200 GN/m 2 are adopted. When the temperature is raised to 65°C, what is the z = 0 free expansion of the length; and if this P 2 expansion of the rod is fully prevented, what 1 eV = is the temperature stress produced? AE m
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136. The normal stresses on two mutually Ans. (d) 2 perpendicular planes are 140 N/mm (Tensile) Sol. and 70 N/mm2 (Tensile). If the maximum shear stress is 45 N/mm2, the shear stress on these 60 planes will be nearly (a) 20.9 N/mm2 (b) 24.6 N/mm2 (c) 28.3 N/mm2 (d) 32.0 N/mm2 Ans. (c) 120 Sol.
70 = y
xy 2 2 x y x y 1/2 = xy 2 2 xy 140 = x = 90 (30)2 (30) 2
= 90 2 30 = 90 ± 42.426 2 = 132.426, 47.574 = x y 2 max xy Both principal stress are tensile. 2
2 138. At a point in a material, the stresses acting 140 70 2 45 = xy on two planes at right angles to each other 2 are :
2 2 2 = xy = 28.284 N/mm (45) (35) z = 120 MPa and y = –200 MPa and zy = –80 MPa. 137. The normal stresses on the two mutually perpendicular planes at a point are 120 MPa The maximum shear stress on the element (Tensile) and 60 MPa (Tensile). If the shear will be nearly stress across these planes is 30 MPa, the (a) 142 MPa (b) 155 MPa principal stresses will be nearly (c) 167 MPa (d) 179 MPa (a) 124 MPaIES (Tensile) and MASTER 24 MPa (Compressive) Ans. (d) (b) 132 MPa IES (Tensile) and 24MASTER MPa 2 y z () 2 (Compressive) Sol. max = yz 2 (c) 124 MPa (Tensile) and 48 MPa (Tensile) 2 (d) 132 MPa (Tensile) and 48 MPa (Tensile) 320 2 = (80) 2
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Detailed Solution Civil Engineering
2 2 2 = (160) (80) = 702 10 3
= 80 5 = 72.11 MPa = 178.885 MPa 140. The change in shearing force between two points on the beam is equal to the area of 139. The principal stresses in the wall of a container are 40 MN/mm2 and 80 MN/mm2. The normal (a) Loading diagram between the two points makes an angle of 30° with a direction of (b) Shear force diagram between the two maximum principal stress. The resultant points stresses (in magnitude) in the plane will be (c) Bending moment diagram between the nearly two points
2 40 MN/mm (d) M/EI diagram between the two points Ans. (a)
141. Which one of the following statements °
2 0 2 80 MN/mm 3 80 MN/mm specifies shear flow? 30° (a) Flow of shear force along the beam (b) It is the product of the shear stress at any level and the corresponding width b (of the section) 2 40 MN/mm (c) Unbalanced force on any side of given 2 2 (a) 84 MN/m (b) 72 MN/m section divided by area of section 2 2 (c) 64 MN/m (d) 58 MN/m (d) The deformation at any level due to Ans. (b) sudden variation in shear stress Sol. Ans. (b) Normal to the plane
142. Which one of the following statements is (60+Rcos60°, Rsin60°) () correct for the rotating shafts transmitting 60° power? (40, 0) (60, 0) (80, 0) (a) Lower the frequency of shaft lower will be the torque (b) Higher the frequency of shaft lower will 80 40 R = 20MN mm2 be the torque 2 IES MASTER(c) Frequency of the shaft does not influence = 60 20cos 60 the torque = 70MPa (d) Higher the frequency of shaft higher will be the torque 3 = 20 10 3 2 Ans. (b)
2 2 Sol. Resultant Stress = r T = Power = T 2 f
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4 2 Power G = 8.16 × 10 N/mm T = 2 f P Gd4 Stress of spring = As, f , T 64R3 n
143. The maximum shear stress induced in a solid 8.16 104 N/mm 2 (10) 4 mm 4 = circular shaft of diameter 15 cm, when the 64(50mm)3 15 shaft transmits 150 kW power at 180 rpm, will be = 6.8 N/mm (a) 16 N/mm2 (b) 14 N/mm2 145. The shear force diagram of a beam is shown (c) 12 N/mm2 (d) 10 N/mm2 in the figure Ans. (c) Sol. + 800N + 1000N Tr Td 2 16T Max shear stress = 800N – – J d3 d4 1200N 32 P T = The total of the vertically downward loads on the beam is 16P = max d3 (a) 2600 N (b) 2000 N 16 150 103 (c) 2400 N (d) 2800 N = 3 180 2 Ans. (d) 0.15 60 Sol. = 12 106 N m 2 800 N 2000 N 1000 N
= 12N mm2
144. A closely coiled helical spring made of 10 mm diameter steel wire has 15 coils of 100 mm 1600 N 2200 N mean diameter. The spring is subjected to an Total Force = 3800 N axial load of 100 N. For a modulus of rigidity of 8.16 × 104 N/mm2, the stiffness of the spring 146. A beam of triangular cross-section is subjected will be nearlyIES MASTERto a shear force of 50 kN. The base width of (a) 5.9 N/mm (b) 6.8 N/mm the section is 250 mm and the height is 200 mm. The beam is placed with its base (c) 7.7 N/mm (d) 8.8 N/mm IES MASTERhorizontal. The shear stress at neutral axis Ans. (b) will be nearly Sol. d = 10 mm (a) 2.2 N/mm2 (b) 2.7 N/mm2 n = 1 (c) 3.2 N/mm2 (d) 3.7 N/mm2 D = 100 mm P = 100 N
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Ans. (b) 2 w 2000 28 Sol. 2 150 8 100 250 mm 6
N w 21 N/mm A 3 V 200 mm = max 2N mm2 max 2 bd w 2000 3 2 4 = 2 = 2 100 150 N.A. 3 av w 20 N/mm 50 103 N 4 w should be min of the w from the above two 1 3 = 250 20 criteria. 2 = 2.67 MPa wmax = 20 KN.m
147. A timber beam, 100 mm wide and 150 mm 148. A 1.5 m long column has a circular cross- deep, supports a UDL over a span of 2 m. If section of 50 mm diameter. Consider the safe stresses are not to exceed 28 MPa Rankine’s formula with values of fd = 560 N/ in bending and 2 MPa in shear, the maximum 1 load that the beam can support is mm2, for pinned ends and factor of 1600 (a) 16 kN/m (b) 20 kN/m safety of 3. If one end of the column is fixed (c) 24 kN/m (d) 28 kN/m and the other end is free, the safe load will be Ans. (b) (a) 9948 N (b) 9906 N Sol. (c) 9864 N (d) 9822 N w KN/m Ans. (b)
l = 2m f A F.O.S. Sol. P = C 1 a 2 b=100 mm
fC A F.O.S. d = 150 mm- = 2 l 1 rmin wl2 Mmax = 1 8 for pinned end = IES MASTER1600 wl V = max 2 for one end fixed other free = 4 2 Mmax 560 50 3 = 28N mm2 max bd2 4 P = 2 6 4 1.5 1000 1 1600 50 4
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I d4 d 150. The maximum shear stress across a circular r = min mm A d2 4 section is 64 4 4 = 9905.92 N (a) Average shear stress 3 149. A continuous beam with uniform flexural rigidity is shown in the figure. 3 (b) Average shear stress 2 40 kN 5 A C (c) Average shear stress B 4 2m 2m 6m 9 The moment at B is (d) Average shear stress 5 (a) 18 kNm (b) 16 kNm Ans. (a) (c) 14 kNm (d) 12 kNm Ans. (d) Sol. 40 KN
A B C 2m 2m 6m
EI = Constant Joint Member Member Stiffness DF 3EI 9EI 9 BBA 0.6 4 12 15 3EI 6EI 6 BC 0.4 6 12 15 A B C 0.6 0.4 40 4 40 4 0 0 8 8 40 4 IES40 2 MASTER 8 8 0 30 0 0 –18IES–12 MASTER 0 12 –12 0
MBA = 12 KNm
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