Detailed Solutionsof GATE-2014 Morning Session Civil Engineering Write us at
[email protected] | Phone: 011-45124612, 9958995830 www.madeeasy.in Expert Opinion D B. Singh (Ex. IES) CMD, MADE EASY Group ear Students, The Questions of GATE 2014 are based on fundamental and basic concepts of the syllabus. There is no ambiguity and misprint noticed till now, however, it is an observation based on students feedback. The level and standard of GATE 2014 questions are relatively easier than the exam of GATE 2013. There are 3 important observations made by me about GATE 2014 exam. 1. The GATE 2014 exam is conducted in two seating i.e. morning session and afternoon session. The question papers of both seatings are different. The difficulty level of questions are nearly same and due care has been taken to balance both the papers, however small differences are certainly there. The morning session paper seems to be little easier by 2 to 5%, however, it varies on the perception of person to person also. The average marks of both the papers should be equated and necessary scaling criteria should be adopted for this purpose. 2. The GATE 2014 cut-off will be nearly same as that of last year, perhaps it may be little lesser than that of GATE 2013. GATE-2013 cutoff was 33 marks. Though the paper of GATE 2013 was tougher and number of students were less, 6 marks questions were wrongly framed and hence, these 6 marks were awarded to all the candidates, which was certainly a kind of bonus. 3. Therefore expected cut-off for GATE 2014 may be between 30 to 34 marks (General category). It may be noted that the following formulae is used to evaluate GATE cut-off marks. Total Marks obtained by all the candidates GATE Cutoff = Total number of candidates GATE cutoff < | 25 Marks The topper’s marks in GATE 2013 was nearly 83 in which 6 marks of bonus to all are included. In my opinion topper’s marks in GATE 2014 will be between 80 to 85. Disclaimer Dear Students, MADE EASY has taken due care in collecting the data and questions. Since questions are submitted by students and are memory based, therefore the chances of error can not be ruled out. Therefore MADE EASY takes no responsibility for the errors which might have incurred. If any error or discrepancy is recorded then students are requested to inform us at:
[email protected] Super Talent Batches announcing Civil Engineering Super Talent Batches at Kalu Sarai premise of Delhi st Batch : Commencing from May 20th Morning Batch nd Batch : Commencing from June 15th Evening Batch • Top 2000 Rank in GATE • B.Tech from IIT • Appeared in IES or 3 PSUs Interview • B.Tech from NIT with 65% marks (Any of the following) Eligibility • B.Tech from Private Engineering college with 70% marks • Better Teaching Environment Benefits • Extra teaching hours • In-depth coverage of subjects Kalu Sarai.I (Civil Engineering) One Mark Questions Q.1 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 1 For a saturated cohesive soil. a tri-axial test yields the angle of interval friction (φ) as zero.com | Visit: www. (c) Put x= (b) 0 (d) ∞ 1 as x→∞ ⇒ h→0 h 1 1 + sin h h + x sin x lim lim 1 = h→ 0 x →∞ x h Corporate Office: 44-A/1.GATE-2014 Exam Solutions Civil Engineering (Morning Session) Section . New Delhi-16 | Email : madeeasydelhi@gmail. The conducted test is (a) Consolidated Drained (CD) test (b) Consolidated undrain (CU) test (c) Unconfined compression (UC) test (d) Unconsolidated undrain (UU) test (d) Ans.3 x + sin x lim is equal to x →∞ x (a) –∞ (c) 1 Ans. (a) (b) Q (d) S Q. Q.madeeasy.2 The possible location of shear centre of the channel section shown below is P Q R S (a) P (c) R Ans.in . 6 Vs (b) V > 0. (a) Ans. The applied shear force is V and shear capacity of the section is Vs.4 A conventional flow duration curve is a plot between (a) Flow and % time flow is exceeded (b) Duration of flooding and ground level elevation (c) Duration of water supply in a city and proportion of area recurring supply exceeding this duration.2. For such a section. high shear force (as per IS 800-2007) is defined as (a) V > 0.8 Vs (d) V > 0 (a) Clause 9.1 IS 800:2007.5 A steel section is subjected to a combination of shear and bending action.6 In reservoir with an uncontrolled spillway the peak of the plotted outflow hydrograph (a) Lies outside the plotted inflow hydrograph. Kalu Sarai.7 Vs (c) V > 0. (b) Lies on the recession limb of the plotted inflow hydrograph. Ans.GATE-2014 Exam Solutions Civil Engineering (Morning Session) 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 2 1 sin h =1 lim 1 + = h→ 0 1 h Q. Q. Q. (d) is higher than peak of the plotted inflow hydrograph.madeeasy. (c) Lies on the peak of the inflow hydrograph.com | Visit: www.in . Inflow hydrograph Recession limb Outflow hydrograph Corporate Office: 44-A/1. (b) Ans. New Delhi-16 | Email : madeeasydelhi@gmail. (d) Flow rate and duration of time taken to empty of a reservoir at that flow rate. 9 The following statements are related to temperature stress developed in concrete pavement slab with four edge (without any restrain) P : The temperature stress will be zero during both day and night time if the pavement slab is considered weight less. (c) The SI unit of kinamatic viscosity is m2/sec. ∴ dimension of kinamatic viscosity [ν] = L2/T. Q.madeeasy.8 The dimension for kinematic viscosity is (a) (c) L MT (b) (d) L T2 ML T L2 T Ans. y = 0 (b) has a slope +1 (c) is parallel to x-axis (d) has a slope of –1 (c) y = 5x2 + 3 dy = 10 x dx 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 3 Ans.GATE-2014 Exam Solutions Civil Engineering (Morning Session) Q. R : The temperature stress will be compressive at the bottom of the slab during day time if the self weight of the pavement slab is considered. : The temperature stress will be compressive at the bottom of the slab during night time if the self weight of the pavement slab is considered.7 If y = 5x2 + 3 than the tangent at x = 0 and y = 3 (a) passes through x = 0.in .3) = 10 × 0 = 0 Tangent (0. Kalu Sarai. 3) x ⇒ tangent is parallel to x-axis.com | Visit: www. Corporate Office: 44-A/1. New Delhi-16 | Email : madeeasydelhi@gmail. Q. ⇒ y dy dx (0. Q. 5 = 4 4 ⇒ V = 2.madeeasy. New Delhi-16 | Email : madeeasydelhi@gmail. If velocity at the section of 15 cm dia portion of the pipe is 2. Q. Sol.com | Visit: www. the amount of rainfall in August is more than that in January.in . (ii) Every year.GATE-2014 Exam Solutions Civil Engineering (Morning Session) The true statement(s) is(are) (a) P only (c) P and Q only Ans.5 × 9 = 22. π 2 π (5) × V (15)2 × 2.5 m /sec. (c) 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 4 (b) Q only (d) P and R only The temperature stress will be tensile at the bottom of the slab during the day time if the self weight of the pavement slab is considered. Corporate Office: 44-A/1.5 m/s Q. Kalu Sarai.11 The monthly rainfall chart based on 50 years of rainfall in Agra is shown in the following figure which of the following are true ? (K percentile is the value such that K % of data fall below that value) Rainfall (mm) 800 700 650 600 500 400 300 200 100 650 600 400 200 100 50 Feb Mar Apr May June July Aug Sep Oct Nov Dec (i) On average it rains more in July than in Dec. the velocity of fluid (in m/s) at section falling in 5 cm portion of the pipe is _________.10 An incompressible homogeneous fluid flowing steadily in a variable dia pipe having the large and small dia as 15 cm and 5 cm respectively. 1 (b) 0. (a) Q.0 and minimum value is 0. Kalu Sarai.in .13 The minimum value of 15 minutes peak hour factor on a section of a road is (a) 0.GATE-2014 Exam Solutions Civil Engineering (Morning Session) 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 5 (iii) July rainfall can be estimated with better confidence than Feb.7 – 0.com | Visit: www.12 The potable water is prepared from turbid surface water by adopting the foil treatment square. (a) Turbid surface water → Coagulation → Flocculation → Sedimentation → Filtration → Disinfection → Storage and supply (b) Turbid surface water → Disinfection → Flocculation → Sedimentation → Filteration → Coagulation → Storage and supply (a) Turbid surface water → Filteration → Sedimentation→ Disinfection → Flocculation → Coagulation (a) Turbid surface water → Sedimentation → Flocculation → Coagulation → Disinfection → Filteration Ans.madeeasy. (b) Q. V = Peak hourly volume in hr. peak hr factor is used for traffic intersection design PHF = (V/4) V15 Ans.25 Normal range is 0. (iv) In Aug. Maximum value is 1. rainfall.) V15 = Maximum 15 minimum volume within the peak hr. (a) (i) and (ii) (b) (i) and (iii) (c) (ii) and (iii) (d) (iii) and (iv) Ans.25 Corporate Office: 44-A/1. (veh.98 = 0. there is at least 500 mm of rainfall. New Delhi-16 | Email :
[email protected] (c) 15 min.2 (c) 0. veh.25 (d) 0. the area ratio is (a) 2 D2 0 − Di D2 i (b) 2 D2 i − D0 D2 i (c) Ans. Mercury (b) Calcium.14 Some of the non-toxic metal normally found in natural water are (a) Arsenic. Mangnese. copper (d) Iron.16 The degree of static indeterminacy of a rigid jointed frame PQR supported as shown is y 145° S R EI 90° Q EI P x (a) 0 (c) 2 Ans. (a) 2 D2 0 − Di D2 0 (d) 2 D2 i − D0 D2 0 Q. (a) Ds = DSe + Dsi (b) 1 (d) 3 = (re – 3) + 3C – rr = (4 – 3) + 3 × 0 – 1 = 0 Corporate Office: 44-A/1. Lead.in . If outer diameter and inner dia of sample are D0 and Di respecgively. curomium. Silver (c) Cadmium.GATE-2014 Exam Solutions Civil Engineering (Morning Session) Q.com | Visit: www. New Delhi-16 | Email : madeeasydelhi@gmail. Q.madeeasy. Kalu Sarai. Magnesium (d) 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 6 Ans. Sodium.15 The degree of disturbances of a sample collected by sampler is expressed by a term called the area ratio. in . New Delhi-16 | Email : madeeasydelhi@gmail. (c) 2M p L 6M p L (b) (d) 4M p L 8M p L W L/2 Wu θ Mp θ Mp θ Mp L/2 θ From principal of virtual work − M pθ − M pθ − M pθ + Wu L θ = 0 2 ⇒ Wu = 6M p L Corporate Office: 44-A/1.GATE-2014 Exam Solutions Civil Engineering (Morning Session) Q.madeeasy.com | Visit: www. Q.18 The ultimate collapse load (Wu) in terms of plastic moment Mp by kinematic approach for a propped cantilever of length L with W acting at its mid span as shown in fig would be W L/2 L/2 (a) (c) Ans. Kalu Sarai.17 The action of negative friction on the pile is to (a) Increase the ultimate load on the pile (b) Reduce the allowable load on the pile (c) Maintain the working load on the pile (d) Reduce the settlement (b) 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 7 Ans. R – 2.19 Match the following: Group I P.madeeasy. Stedia hair (a) P – 3. Q – 1. R – 2. Kalu Sarai. Q – 4. (a) Sum of eigen values = trace of matrix = 215 + 150 + 550 = 915 Q.21 The probability density function of evaporation E on any day during a year in watershed is given by 1 0 ≤ E ≤ mm/day f (E) = 5 Otherwise 0 The probability that E lies in between 2 and 4 mm/day in a day in watershed is (in decimal) Sol. (d) P – 3. S – 1 (c) P – 1. Theodolite 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 8 Q. Alidade Q. Arrow R. R – 1. S – 4 (b) P – 2.20 The sum of eigen value matrix [M] is When 215 650 795 [M] = 655 150 835 485 355 550 (b) 1355 (d) 2180 (a) 915 (c) 1640 Ans. Plant table surveying 4. 1 0 ≤ E ≤ mm/day f (E) = 5 Otherwise 0 Corporate Office: 44-A/1. S – 4 Ans. New Delhi-16 | Email : madeeasydelhi@gmail. R – 3. Chain Survey 2. S – 3 (d) P – 3.GATE-2014 Exam Solutions Civil Engineering (Morning Session) Q. Q – 2. Levelling 3. Bubble tube S. Q – 2. Q – 1.com | Visit: www.in . R – 4. S – 4 Group II 1. 658 = F cos θ F cos θ from option (a) = 56.389 kN and θ = 28.28° (c) F = 9. Corporate Office: 44-A/1. (a) For no swinging ∑Fhorizontal = 0 90 kN 45° 30° 100 kN θ x 40 kN F ⇒ 90 cos 30° = 40 cos 45° + F cos θ 49.22 A box of weight 100 kN shown in the figure to be lifted without swinging.28° = 49. the magnitude and direction (θ) of force F w. New Delhi-16 | Email :
[email protected] Exam Solutions Civil Engineering (Morning Session) 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 9 P (2 < E < 4) = ∫ f (E) dE = 2 4 ∫ 5 dE = 5 [ E]2 2 4 1 1 4 = 1 2 (4 − 2) = 5 = 0.t.1414° (d) F = –9.055 kN and θ = –1.23 The amount of CO2 generated in kg while completely oxidizing one kg of CH4 is ________. Kalu Sarai. If all the forces are coplanar.com | Visit: www.389 cos 28.in .r.055 kN and θ = 1. x axis is ________.madeeasy.1414° Ans.389 kN and θ = –28.28° (b) F = –56.658 kN Q. y 90 kN 45° 30° 100 kN θ x 40 kN F (a) F = 56.4 5 Q. New Delhi-16 . | Email :
[email protected] fck = 0.26 Three rigid bucket are of identical height and base area. W2. Sol. Permissible bearing stress = 0.25 3 2 1 Given J = 2 4 2 and K = 1 2 6 1 2 then product KT JK is _______. K = −1 1 2 6 3 2 1 1 JK = [1 2 − 1] 2 4 2 2 1 2 6 −1 1 = [ 6 8 − 1 ] 2 = 6 + 16 + 1 = 23 −1 KT Two Marks Questions Q.madeeasy. 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 10 CH 4 + 2O 2 → CO2 + 2H 2O 16g 44g ⇒ ⇒ 16 g of CH4 when completely oxidized leads to 44 g of CO2 1 kg of CH4 when completely oxidized leads to 44 × 1 = 2. −1 Sol.GATE-2014 Exam Solutions Civil Engineering (Morning Session) Sol.. the bearing strength of concrete (N/mm2) in LSM of design as per IS 456: 2000 is ... W3 respectively...24 While designing for a steel column of Fe250 grade the base plate resting on a concrete pedestal of M20 grade. 1 3 2 1 2 2 4 2 J= . Kalu Sarai. Which of the following option are correct.in Corporate Office: 44-A/1.45 × 20 = 9 N/mm2 Q.75 kg CO 2 16 Q.com | Visit: www... Further assume that each of these buckets have negligible mass and are full of water. The weight of water in these bucket are denoted by W1. madeeasy. W 2 > W1 > W3 F1 = F2 = F3 Force on the base in each case will be equal to = γ w h A Q. d 2y dx y P M L x EI y N P 2 + py = 0 EI the mid-span deflection of a member shown in figure is [a is the amplitude constant for y] (a) y = (c) y = 1 2π x 1 − a cos P L a sin n π x L (b) y = (d) y ′ = 1 2π x 1 − a sin P L a cos n π x L Corporate Office: 44-A/1. New Delhi-16 | Email :
[email protected] | Visit: www. (d) Bucket → identical height → identical base area h h h h A 1 A 2 A 3 ⇒ Hence.GATE-2014 Exam Solutions Civil Engineering (Morning Session) 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 11 h h h (a) W2 = W1 = W3 and F2 > F1 > F2 (b) W2 > W1 > W3 and F2 > F1 > F3 (c) W2 = W1 = W3 and F1 = F2 = F3 (d) W2 > W1 > W3 and F2 = F1 = F3 Ans.27 If the following equation establishes equilibrium in slightly bent position.in . Kalu Sarai. (a) 230 (b) 10 (d) 16 450 4 × 12φ τuc = 0. y = 0 ⇒ b= 0 at x = L.36 N/mm2 [For M20] for τuc = 0. and τuc = 0.48 N/mm2 [For M20] for Factored SF = 45 kN = Vu A st × 100 = 0. the diameter (mm) of Fe 500 steel 2 legged stirrups to be used at spacing of 325 mm should be (a) 8 (c) 12 Ans.com | Visit: www. y = 0 ⇒ 0 = sin mL ⇒ mL = n π ⇒ ∴ m= nπ L nπx L y = asin Q. The grade of concrete is M 20. is reinforced with 4 bars of 12 mm diameter. For a factored shear force of 45 kN. New Delhi-16 | Email :
[email protected]. Kalu Sarai.25 bd A st × 100 = 0. the ultimate shear strength τ uc = 0.25.in .28 A rectangular beam of 230 mm width and effective depth = 450 mm.GATE-2014 Exam Solutions Civil Engineering (Morning Session) Ans. grade of steel is Fe 500. Given that for M 20 grade of concrete.5 bd Corporate Office: 44-A/1. (c) d2 y dx 2 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 12 = − P ×y EI = –m2y ∴ Solution of above differential equation is y = a sin mx + b cos mx at x = 0.36 N/mm 2 for steel percentage of = 0.madeeasy.48 N/mm2 for steel percentage = 0. Assume the dilution coefficient n = 1.GATE-2014 Exam Solutions Civil Engineering (Morning Session) 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 13 We have to calculate the dia of Fe 500 2-L egged stirrup to be used at a spacing of 325 mm c/c τv = Vu 45 × 1000 = = 0.5 km long pipe of diameter 45 cm. tCn = K where t = time required to kill all organism Ans.25 ) = 0. Kalu Sarai.27 (a) In the disinfection process we have the relationship.12 × (0.26 mm adopt φ = 8 mm Q.4 (c) 38 (d) 23. Corporate Office: 44-A/1.5 (b) 4.madeeasy.87 × 415 = 82.36 + Since ⇒ ⇒ 0.4 × 325 × 230 ( φ )2 = 0.87 f y Since we limit fy to 415 N/mm2 hence.in . The chlorine at the rate of 32 kg/d is applied at the entry of this pipe so that disinfected water is obtained at the exit.45 N/mm 2 0.814 mm 2 4 φ = 7.437 % % tensile steel = 230 × 450 4× τc = 0.437 − 0.com | Visit: www.87 f bS v y Asv = 0.4 A sv = 0. New Delhi-16 | Email : madeeasydelhi@gmail. These is a proposal to increase the flow through the pipe to 22 MLD from 16 mLD.4 × (S v )( b ) 0.4348 N/mm 2 bd 230 × 450 π (12)2 4 × 100 = 0. Asv = 2 × π 0.25 τv – τc < 0 Min shear reinforcement is required Min shear reinforcement is given by 0. The minimum amount of chlorine (in kg per day) to be applied to achieve the same degree of disinfection for the enhanced flow is (a) 60.29 16 MLD of water is flowing through a 2. 015 Q = 1 m3/s Normal depth of flow between 0.84 (d) 1. The minimum width of throat (in m) that is possible at a given section while ensuring that the prevailing normal depth does not exceed along the reach upstream of the concentration is approximately.76 m and 0. Q = 1 m3/sec given that normal depth of flow ranges between 0.04 (b) 0.8 m.com | Visit: www. where W1 = weight of disinfectant per day .GATE-2014 Exam Solutions Civil Engineering (Morning Session) c = concentration of disinfectant n = dilution coefficient k = constant ⇒ n n = t2 C2 t1 C1 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 14 in our case n = 1 ⇒ t 1C 1 = t 2 C 2 L t1 = v 1 L = length of pipe . New Delhi-16 | Email :
[email protected] A rectangular channel flow have bed slope of 0.in .24 Ans. Corporate Office: 44-A/1. equal to (assume negligible loss) (a) 0.madeeasy.8 m. Q1 = discharge per Q1 LA W2 LA W1 × × = Q2 Q2 Q1 Q1 ⇒ W2 = Q2 22 2 × W1 = × 32 kg / day = 60. Kalu Sarai. V1 = velocity of flow t1 = t1 = L Q1 / A LA Q1 C1 = day ⇒ W1 .64 (c) 1.0001 width = 3 m coefficient n = 0. (b) n = 0.5 kg/day 2 16 Q1 2 Q.76 m to 0. GATE-2014 Exam Solutions Civil Engineering (Morning Session) 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 15 If prevailing normal depth of flow is not exceeded, there must not be chocking of the section or there must be just chocking. Thus the width of the section should be such that for the prevailing specific energy there should be critical flow at the contracted section i.e. 3 q2 2 g Q 2 B min 3 2 g 1/3 = EC = Einitial 1/3 ⇒ = Einitial Let is now calculate Einitial Q= 1 AR 2/3 S1/2 0 2 1 3y 3y ) ( 3 + 2y 0.015 2/3 ⇒ ⇒ ⇒ 1= (0.0001)1/2 y = 0.78 m Einitial = y + q2 2gy 2 1 3 2 = 0.78 + 1/3 2 × 9.81 × (0.78) 2 = 0.7893 m ⇒ Q 2 B min 3 2 g 2/3 = 0.7893 ⇒ (1) 3 2 g1/3 ( Bmin )2/3 = 0.7893 Bmin = 0.836 m Q.31 A levelling is carried out to established the reduced level (RL) of point R with respect to the bench mark (BM) at P. The staff reading taken are given below | Email :
[email protected] | Visit: www.madeeasy.in Corporate Office: 44-A/1, Kalu Sarai, New Delhi-16 GATE-2014 Exam Solutions Civil Engineering (Morning Session) 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 16 Staff station P Q R IS FS RL − − 1.655 − −0.95 − −1.5 − – − 0.75 − BS If RL of P is + 100 m, then RL (m) of R is (a) 103.355 (b) 103.155 (c) 101.455 (d) 100.355 Ans. (c) HI = RL + BS and Staff station P Q R BS RL = HI – FS IS FS RL HI RL − − 101.655 1.655 − 100 −0.95 − −1.5 − 102.205 103.155 – − 0.75 − − 101.455 ∴ RL of R = 101.455 m Q.32 A given cohensionless soil has emax = 0.85, emin = 0.5. In the field, the soil is compacted to a mass density of 1800 kg/m3 at water content of 8%. Take the mass density of water as 1000 kg/m3 and GS = 2.7. (a) 56.43 (c) 62.87 (b) 60.25 (d) 65.41 e max = 0.85 emin = 0.5 ρfield = 1800 kg/m3 at water content = 8% ρw = 1000 kg/m3 G S = 2.7 Relative density, ID = ? ρ= ⇒ ⇒ ⇒ 1 + e= Ans. (d) Gρ w (1 + w ) 1+e = 2.7 × 1000 (1.08 ) 1+e 2.7 × 1000 × 1.08 1800 e = 0.62 e max − e × 100 ID = e max − e min | Email :
[email protected] | Visit: www.madeeasy.in Corporate Office: 44-A/1, Kalu Sarai, New Delhi-16 GATE-2014 Exam Solutions Civil Engineering (Morning Session) 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 17 = 0.85 − 0.62 × 100 0.85 − 0.5 = 65.714 % Q.33 Then tension and shear force (both in kN) at both joints as shown below are respectively 4 5 3 Pu = 250 kN (a) 30.3 and 20 (c) 33.33 and 20 Ans. (d) (b) 30.33 and 25 (d) 33.33 and 25 θ 3 Pu 5 Pu cos θ = 4 Pu 5 Pu cos θ = Pu tanθ = cos θ = sinθ = Tension in each bolt = 4Pu 5×6 3 4 4 5 3 5 4 × 250 = 33.33 kN 5×6 Shear in each bolt = 3Pu 3 × 250 = 5×6 5×6 = 25 kN Corporate Office: 44-A/1, Kalu Sarai, New Delhi-16 | Email :
[email protected] | Visit: www.madeeasy.in the Carbonate and Non-carbonate hardness concentration (in mg/l as CaCO3) respectively are. NaHCO3 absent] −3 = 3.5 × 10 × 50g as CaCO3 l = 175 mg/l as CaCO3 Non carbonate hardness = total hardness-carbonate hardness Total hardness = 5 × 50 mg/l as CaCO3 [total hardness is due to Ca2 Mg2+] = 250 mg/l as CaCO3 ⇒ NCH = 250 – 175 = 75 mg/l as CaCO3 + and Q. (c) (b) 175 and 75 (d) 50 and 200 Carbonate hardness = 3. R-1 (d) P-1. R-2 (c) P-3. Q-3.e. Load Settlement J K L (a) P-1. No apparent heaving of soil around the footing.34 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 18 For a sample of water with the ionic composition shown below. Curve L Group-II 1.35 Group-I P. Kalu Sarai. Q-1. R-3 Corporate Office: 44-A/1.5 7 (a) 200 and 500 (c) 75 and 175 Ans. Curve K R. Rankine passive zone develops imperfectly 3. R-2 (b) P-3. Well defined slip surface extends to ground surface. Q-2. 2.madeeasy. 0 meq/l meq/l Ca 2+ 4 Mg 2+ 5 Na 2– 7 + HCO3 – SO4 3. Q-2.5 × 10–3 g-eq [if NCH is present sodium alkalinity will be absent i. New Delhi-16 | Email :
[email protected] | Visit: www.in . Curve J Q.GATE-2014 Exam Solutions Civil Engineering (Morning Session) Q. S) are shown below the correct.25 0.37 In a simply supported beam of length L four influence line diagram for shear at a section located at a distance of L/4 from the left support marked (P. (d) L → General shear failure K → Local shear failure J → Punching shear failure 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 19 Q.madeeasy. L/4 3L/4 Corporate Office: 44-A/1. L/4 0. After impact the jet split symmetrically in a plane parallel to the plane of the plate.5 3L/4 S.0028 m 2 Force on plate = ( ρ w a V ) V = ρ w aV 2 = 1000 × 0.in . R. L/4 3L/4 Q.6 P. The force of impact (in N) of the jet on the plate is (a) 90 (c) 70 (b) 80 (d) 60 Ans. L/4 3L/4 0. Q.0028 m2 hits a fixed vertical plate with a velocity of 5 m/s. (c) 5 m/s area of jet = 0.0028 × (5)2 = 70 N Q.75 0. New Delhi-16 | Email : madeeasydelhi@gmail. ILD is 0. Kalu Sarai.36 A horizontal jet of water with its cross section area 0.6 R.com | Visit: www.GATE-2014 Exam Solutions Civil Engineering (Morning Session) Ans.5 0. 25 X 1 unit 3L/4 B 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 20 (b) Q (d) S ILD for SF at X-X i.5.3 (b) 44. New Delhi-16 | Email : madeeasydelhi@gmail. Use γsat = 18 kN/m3 and γw = 10 kN/m3. Firm strate lies below the slope and it is assumed that water table may occasionally rise to the surface.com | Visit: www.75 + – 0. Kalu Sarai.in . (a) X A L/4 1 unit 0.. with seepage taking place parallel to the slope.3 (c) 12. Assuming a potential failure surface parallel to the slope would be (a) 45. maximum slope angle (in degree) to ensure the factor of safety 1.GATE-2014 Exam Solutions Civil Engineering (Morning Session) (a) P (c) R Ans. (d) C′ = 0 φ = 34° γsat = 18 kN/m γw = 10 kN/m 3 3 β Corporate Office: 44-A/1.3 Ans.38 A long slope is formed in a soil with shear strength parameter C′ = 0. option (a) Q. φ′ = 34°.e.madeeasy.7 (d) 11. 39 For the truss shown below. R 3m P Q 4m 4m Sol. PQ is short by 3 mm We have to find out vertical displacement of joint R in mm ∆ R = ∑u (λ) Let in apply unit load at R as shown below R uPR R P θ u PQ Q 1/2 1/2 Corporate Office: 44-A/1.30° Q.5 β = 11.com | Visit: www. tan φ tan β = γ 1. the member PQ is short by 3 mm.GATE-2014 Exam Solutions Civil Engineering (Morning Session) 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 21 γ sub . Kalu Sarai.5 sat ⇒ tan β = ⇒ (18 − 10) × tan 34° 18 × 1. New Delhi-16 | Email : madeeasydelhi@gmail. tan φ FOS = γ sat tan β ⇒ γ sub tan φ .in . The magnitude of the vertical displacement of joint R in mm is ________. tan β = γ sat FOS γ sub .madeeasy. madeeasy. g = 9.25 m3/s.41 A straight 100 m long raw water gravity main is to carry water from intake to the jackwell of a water treatment plant. New Delhi-16 | Email :
[email protected] .01.com | Visit: www. what is the sum. The largest angle of the triangle is twice its smallest angle. in degrees of the second largest angle of the triangle and the largest angle of the quadrilateral. The ratio between the angle of the quadrilateral is 3 : 4 : 5 : 6. Corporate Office: 44-A/1.75 m/s. Allowable velocity through main is 0. Assume f = 0.40 The smallest angle of a triangle is equal to two third of the smallest angle of a quadrilateral. The required flow of water is 0. Sol. The minimum gradient (in cm/100 length) required to be given to this main so that water flow without any difficulty should be ________ . Kalu Sarai.GATE-2014 Exam Solutions Civil Engineering (Morning Session) 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 22 uPR sin θ = 1 2 uPQ + uPR cos θ = 0 u P Q = –uPR cos θ = − uPQ = − 1 . cos θ 2 sin θ −1 × 4 / 3 −2 1 = cot θ = 2 3 2 −2 ( −3) = 2 mm upwards 3 ∆ R = u PQ × λ PQ = Q. θ2 θ3 4α 6α 18α = 360 α = 20° 3α 5α θ1 Largest angle of quadrilateral = 120° Smallest angle of quadrilateral = 60° ⇒ Smallest angle of triangle = 2 × (2 × 20) = 40° 3 Largest angle of triangle = 2 × 40 = 80° ⇒ Third angle of triangle = 60° = 120° + 60° = 180° ⇒ Sum of largest angle of quadrilateral and second largest angle of triangle Q. GATE-2014 Exam Solutions Civil Engineering (Morning Session) Sol.01 × 100 × ( 0.81 Q 0.75 ) = hf = m 2gd 2 × 9.reaction time for a vehicle travelling at 90 km/h.044 m = 4. Sol.6515 m 2 flv 2 0.madeeasy.com | Visit: www.81 m/s2) is ________ seconds.4 cm = 100 m l Q. Kalu Sarai. Corporate Office: 44-A/1. effective depth = 500 mm.6515 = 0.85 × 230 π (12 )2 4 = 1. given the coefficient of longitudinal friction of 0.01 g = 9.42 For a beam cross section W = 230 mm.35 and the stopping sight distance of 170 m (assume g = 9. hf 4.75 3 4 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 23 ⇒ ⇒ d = 0. New Delhi-16 | Email :
[email protected] The perception . Q = 0.25 m3/s Allowable velocity = 0. the number of reinforcement bars of 12 mm diameter required to satisfy minimum tension reinforcement requirement specified by IS-456-2000 (assume grade of steel is Fe500) is __________.81 × 0.25 1 2 πd 2 = = m = V 0.4 cm ⇒ Minimum gradient = Hence answer is 4.75 m/s f = 0.in .729 = 2 bars Q.4. A st min bd = 0.85 fy 0.85 × 230 × 500 mm 2 500 A st πd 4 2 A st = n= = 0. 7 K. K = 50 C = 70 × 50 – 0.1510 sec.madeeasy.278 v ) ( + 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 24 2gf ⇒ ⇒ 170 = tr = 0. where v is in km/hr and k is in veh/km. ⇒ ⇒ Capacity. 2 × 9.265 = e 1 + 5 + 3 2 6 Q.44 A traffic office impose on an average 5 number of penalties daily on traffic violators. Kalu Sarai. Assume that the number of penalties on different day is independent and follows a Poisson distribution.com | Visit: www.35 (0. SSD = 0.278 × 90 × tr + 3.45 The speed-density (v-k) relationship on a single lane road with unidirectional flow is v = 70 – 0.4 K = 0 dK Ans.278 vt r 2 0. dC = 70 – 1.7(50)2 = 1750 veh/hr Corporate Office: 44-A/1. Sol. New Delhi-16 | Email :
[email protected] Exam Solutions Civil Engineering (Morning Session) Sol.81 × 0.7 K2 Now.in . The probability that there will be less than 4 penalties in a day is _________. The capacity of the road (veh/hr) is (a) Capacity = Velocity × Density ⇒ C = V × K = 70 K – 0.278 × 90)2 Q. Mean λ = 5 P (x < 4) = p (x = 0) + p (x = 1) + p (x = 2) + p (x = 3) = e −5s 0 e −5 51 e −5 52 e −5 53 + + + 0! 1! 1! 3! 25 125 118 −5 + = e −5 = 0. y and z show variation of the distance covered by the particles in (cm) with time (t) (in second).com | Visit: www. Kalu Sarai. The magnitude of the acceleration of the particle (in cm/s2) at t = 0 is __________. Sol. y = –3e-2t and z = 2 sin (5t). New Delhi-16 | Email :
[email protected] . 20 kN 2 mm 1m 2m Corporate Office: 44-A/1. where x.47 For a cantilever beam of a span 3m as shown a concentrated load of 20 kN applied to the free end causes a vertical displacement of 2 mm at a section located at a distance of 1 m from the fixed end (with no other load on the beam) the maximum vertical displacement in the same (in mm) is ___________.46 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 25 A particle moves along a curve whose parametric equation are x = t3 + 2t.madeeasy. x = t3 + 2t y = –3 e–2t z = 2 sin (5t) dx = 3 t2 + 2 dt ⇒ ax = d 2x dt 2 = 6t dy = – 3 e–2t × (–2) = 6 e–2t dt ⇒ ⇒ ⇒ ay = d 2y dt 2 = −12 e −2t dz = –10 × 5 sin (5t) = –50 sin5t dt az = d 2z dt 2 = −50 sin 5t ˆ ˆ + a yˆ j + azk a = axi ˆ ˆ − 12 ˆ j + 0k a at t = 0 = 0 i a = − 12 ˆ j ⇒ Magnitude of acceleration at t = 0 = 12 cm/s2 Q.GATE-2014 Exam Solutions Civil Engineering (Morning Session) Q. 5 × 12 + 5 = 92 sec.com | Visit: www.in Corporate Office: 44-A/1. Optimum cycle length.3 and 0.2.75 Total lost time in a cycle L = 4 × 3 = 12 sec.49 Mathematical idealization of a crane has three bar with their vertices arranged as shown with load of 80 kN hanging vertically. The optimum cycle length (in sec.75 Q. | Email :
[email protected] An isolated three-phase traffic signal is designed by webster’s method.GATE-2014 Exam Solutions Civil Engineering (Morning Session) Sol. Sol. 0. C0 = 1. The critical flow ratio for three phase are 0.5L + 5 1−y 1. The coordinate of the vertices are given in parenthesis. 20 kN 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 26 1m ∆12 2m 10 kN 1m 2m ∆21 From Betti’s law ⇒ ⇒ P1 × ∆12 = P2 × ∆21 10 × 2 = 20 × ∆21 ∆21 = 1 mm Q. Kalu Sarai. y = y1 + y2 + y3 = 0.25 respectively and lost time per phase is 4 second.3 + 0. New Delhi-16 . 1 − 0.2 + 0. Sum of the flow. The force in member QR is ______.) is ________.madeeasy.25 = 0. 4) 22.13° R(3. cos φ = 17 17 Corporate Office: 44-A/1.0 Q(1.0) Sol.madeeasy.GATE-2014 Exam Solutions Civil Engineering (Morning Session) 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 27 P(0.84° φ 80 kN 4m θ 10 4 . (i) 1 4 .8 4° y x 10 4 . P(0.4) 80 kN 22 .0 Q(1...(i) From eq.com | Visit: www.in .13° R(3.0) 3° 55. Kalu Sarai.0) 3° 55. New Delhi-16 | Email :
[email protected]) VQ 1m 3m VR VQ + VR = 80 ΣMR = 0 ⇒ ⇒ 80 × 3 = VQ × 2 VQ = 120 kN V R = –40 kN tan φ = ⇒ sinφ = tanθ = ⇒ cos θ = 3 4 4 5 1 4 . com | Visit: www.50 The flow net constructed for a dam is shown in the figure below. 50m 6.03° ΣF y = 0 ⇒ ⇒ ⇒ ⇒ FPQ cosα = +VQ = 0 FPQcos14. the quantity of flow (in cm3/sec) under the dam per m is _________.03° – 90° = 14.2 m | Email : madeeasydelhi@gmail. Kalu Sarai. New Delhi-16 17.GATE-2014 Exam Solutions Civil Engineering (Morning Session) Consider joint Q F PQ α F QR VQ 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 28 α = 104.3 m 1.986 kN ≈ 30 kNm Q.03° + 120 = 0 FPQ = –123. Q = KH Nf Nd Corporate Office: 44-A/1.4 m Sol. Quantity of flow.6 m 9.in . Taking coefficient of permeability as 3.6897 kN ΣF x = 0 FPQ sinα = FQR FQR = –29.8 × 10–6 m/s.madeeasy. If thickness of the clay specimen is 25 mm.3 × 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 29 3 10 = 7.8 MPa Ans. e = 0. the value of coefficient of volume compressibility is ________ × 10–4 m2/kN. the effective curvature of the section (in per mm) is (a) 2 × 10–6 (b) 3 × 10–6 –6 (c) 4 × 10 (d) 5 × 10–6 (c) B Xu = 58 mm d 5. e0 = 1.1 = 7. Given. Assuming Linear elastic behavior of the concrete.182 × 10–6 m3/s/m = 7.619 × 10–4 m2/kN Q.3 m −6 Q = 3. ∆e av = ∆σ mv = 1 + e0 1 + e0 = (1. the stress at extreme fibre in compression is 5.9) 125 × 2.51 The full data are given for laboratory sample σ′0 = 175 kPa. The depth of Neutral Axis in the section is 58 mm and grade of concrete is M25.GATE-2014 Exam Solutions Civil Engineering (Morning Session) Here.182 × 10–6 × 106 cm3/s/m Q = 7. Kalu Sarai.8 × 10 × 6. New Delhi-16 | Email :
[email protected]. Sol.52 The reinforced concrete section.1 σ′0 + ∆σ′0 = 300 kPa.com | Visit: www.9. of flow channels = 3 Nd = No. and ∴ Nf = No.8 MPa.8 × 10–6 m/s H = 6.in . of equipotential drops = 10 K = 3.1 − 0. M25 conrature Corporate Office: 44-A/1.182 cm3/s/m Q. Kalu Sarai. the coefficient of discharge of venturimeter (correct upto 2 decimal) is _________.8 m/s2.GATE-2014 Exam Solutions Civil Engineering (Morning Session) Modulus of elasticity of concrete E = 5000 25 = 25000 N/mm2 ∴ σ E M = = y R I σ 5. The pipe carries incompressible fluid at steady rate of 30 l/s. The difference of pressure head measured in terms of the moving fluid in between the enlarged and the throat of the vent is observed to be 2.81 Q = Cd A1 A 2 2 − A2 A1 2 2gh Cd = Q A1 A 2 2 − A2 A1 2 2gh Corporate Office: 44-A/1.8 1 = = = Ey EX u 58 × 25000 R 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 30 ⇒ = 4 × 10–6 Per mm ⇒ Curvature = 4 × 10–6 per mm Q.5 cm φ 1 2 Q = 30 l/s P1 P + Z1 − 2 + Z 2 = 2.com | Visit: www.5 cm at the throat and 15 cm at the enlarged end is installed in a horizontal pipeline of 15 cm diameter.madeeasy.45 m = h w w g = 9.45 m.8 5. Taking the g = 9. Sol.in . New Delhi-16 | Email :
[email protected] A venturimeter having diameter of 7. 1 2 15 cm φ 7. The yearly Traffic growth rate is 7.2 8.989 N = 15 365 × 5000 (1.075) − 1 × 4.2 8. Calculation of vehicle damage factor W W W W W V1 1 + V2 2 + V3 3 + V4 4 + V5 5 Ws W3 Ws Ws Ws V1 + V2 + V3 + V4 + V5 4 y 4 4 4 VDF = where Ws = standard axle load = 80 kN = 8. Axle load (tones) Frequency of traffic (f) 18 14 10 8 6 10 20 35 15 20 The design period of the road is 15 years.95 Q.2 ⇒ VDF = 10 + 20 + 35 + 15 + 20 = 4.075)2 4 × 2 × 9.GATE-2014 Exam Solutions Civil Engineering (Morning Session) 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 31 = 30 × 30 −3 m 3 /s π (0.2 8.5% the load safety factor (LSF) is 1.806 MSA Corporate Office: 44-A/1.989 0.15 0.54 A traffic surveying conducted on a road yield an average daily traffic count of 5000 vehicle.075 − ( ) ( ) Cd = 0.45 2 2 0. The axle load distribution on the same road is given in the following table.madeeasy.2 8.com | Visit: www.3.2 tonn 18 14 10 8 6 10 + 20 + 35 + 15 × + 20 × 8. the design traffic (In million standard axle load MSA) is _______ .in . New Delhi-16 | Email :
[email protected] × 2. Sol.075 4 4 4 4 4 ⇒ = 237. Kalu Sarai.15)2 × (0. If the vehicle damage factor (VDF) is calculated from the above data. Kalu Sarai. assume the head loss due to friction only and the Darcy-weisbach friction factor to be same. (c) Sajan decided to give into Rajan’s request to work with him. (d) Corporate Office: 44-A/1.com | Visit: www.56 Rajan was not happy that Sajan decided to do the project on his own on observing his unhappiness. (b) Rajan and Sajan were formed into a group against their wishes.II (General Aptitude) One Mark Questions Q. The velocity ratio between bigger and smaller branched pipe is _________. From a section the pipe divides into two horizontal parallel pipes of diameter (d1 and d2) that run. Which one of the statement below is logically valid and can be inferred from the above sentences? (a) Rajan has decided to work only in group. For both the pipes. (d) Rajan had believed that Sajan and he would be working together. for a distance of L each and then again join back to a pipe of the original size.madeeasy. d1 C Q1 B A Q2 d2 D L d 1 = 4d2 2 flv1 flv 2 2 = 2gd1 2gd 2 ⇒ 2 2 V1 V2 = d1 d2 2 V1 2 V2 = d1 =4 d2 v1 = 2 v2 Section .55 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 32 An incompressible fluid is flown at steady rate in horizontal pipe.in . Sol. New Delhi-16 | Email :
[email protected] Exam Solutions Civil Engineering (Morning Session) Q. Ans. sajan explained to Rajan that he preferred to work independently. madeeasy.GATE-2014 Exam Solutions Civil Engineering (Morning Session) Q. Total cost for 100 tonne = 8000 × 100 + 50000 = 850000 Cost per tonne = 850000 = Rs.25 Corporate Office: 44-A/1. 2 percent of the people of country Y are taller than 6 ft.com | Visit: www.59 A student is required to demonstrate a high level of comprehension for the subject.57 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 33 A boundary has a fixed daily cost of Rs 50. (a) Q. 8000 Q. especially in the social sciences. 100 Q. New Delhi-16 | Email : madeeasydelhi@gmail. Two Marks Questions Q. OYEIX (a) ALRVX (b) EPVZB (c) ITZDF (d) OYEIX (d) Ans. EPVZB.5 (c) 1.61 One percent of the people of country X are taller than 6 ft. What is the cost of production in Rs. The word closes in meaning to comprehension is (a) Understanding (b) Meaning (c) Concentration (d) Stability Ans.5 (d) 1. (a) Vice (b) Virtues (c) Choices (d) Strength (b) Ans. There are thrice as many people in country X as in country Y.60 Find the odd one in the following group ALRVX.000 whenever it operates and variable cost of Rs. Sol. Q. What % of people are taller than 6 ft? (a) 3 (b) 2.in . one of his biggest _______ was his ability to forgive. Taking both countries together. ITZDF. Kalu Sarai. per tonne for a daily production of 100 tonnes. where Q is the daily production in tonnes. 8500/tonne.58 Choose the most appropriate word from the option given below to complete the following sentences. y1 = 1. Y) the vertices of a triangle have x1.62 With reference to the conventional certesion (X. the area of triangle is (a) (c) 3 2 4 5 y (b) (d) 3 4 5 2 Ans.in .com | Visit: www.2) a C(4.25 Q. x2. New Delhi-16 | Email : madeeasydelhi@gmail. 0.madeeasy. y2 = 2. y3 = 4. 3. Kalu Sarai. and x3. (a) Area of triangle is A= p ( p − a )( p − b )( p − c ) B(2.3) a+b+c when P = 2 c b a= b= c = p = ( 4 − 2 )2 + ( 2 − 1)2 ( 4 − 1 ) 2 + (3 ) 2 (2 − 1)2 + (2 )2 = 5 A(1. 2.GATE-2014 Exam Solutions Civil Engineering (Morning Session) Ans.0) x =3 2 = 5 5 + 5 +3 2 3 = 5+ 2 2 A= 3 2 3 3 3 3 − 3 2 5 + − 3 2 5 + − 5 5+ 5+ 2 2 2 2 = Corporate Office: 44-A/1. (d) Let the population of county Y is P ∴ Population of country X is 3P 3P × 1 P × 2 + 100 100 % of people taken than 6 ft = 3P + P 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 34 = 1. in | Phone: 011-45124612.in . 9958995830 www.madeeasy.Detailed Solutions of GATE-2014 Evening Session Civil Engineering Write us at info@madeeasy. 6 marks questions were wrongly framed and hence. There is no ambiguity and misprint noticed till now. MADE EASY has taken due care in collecting the data and questions. 1. morning session and afternoon session. In my opinion topper’s marks in GATE 2014 will be between 80 to 85. The GATE 2014 cut-off will be nearly same as that of last year. perhaps it may be little lesser than that of GATE 2013. therefore the chances of error can not be ruled out. it is an observation based on students feedback. The Questions of GATE 2014 are based on fundamental and basic concepts of the syllabus. If any error or discrepancy is recorded then students are requested to inform us at: madeeasydelhi@gmail. There are 3 important observations made by me about GATE 2014 exam. It may be noted that the following formulae is used to evaluate GATE cut-off marks. MADE EASY Group ear Students. 3. The morning session paper seems to be little easier by 2 to 5%. however small differences are certainly there. The GATE 2014 exam is conducted in two seating i. these 6 marks were awarded to all the candidates. Though the paper of GATE 2013 was tougher and number of students were less. Since questions are submitted by students and are memory based. The level and standard of GATE 2014 questions are relatively easier than the exam of GATE 2013. which was certainly a kind of bonus. Singh (Ex. however. it varies on the perception of person to person also. Total Marks obtained by all the candidates GATE Cutoff = Total number of candidates GATE cutoff < | 25 Marks The topper’s marks in GATE 2013 was nearly 83 in which 6 marks of bonus to all are included. The difficulty level of questions are nearly same and due care has been taken to balance both the papers. Therefore expected cut-off for GATE 2014 may be between 30 to 34 marks (General category). Therefore MADE EASY takes no responsibility for the errors which might have incurred.Expert Opinion D B. however. The average marks of both the papers should be equated and necessary scaling criteria should be adopted for this purpose. 2. The question papers of both seatings are different. Disclaimer Dear Students.com .e. IES) CMD. GATE-2013 cutoff was 33 marks. Tech from IIT • Appeared in IES or 3 PSUs Interview • B.Tech from NIT with 65% marks (Any of the following) Eligibility • B.Tech from Private Engineering college with 70% marks • Better Teaching Environment Benefits • Extra teaching hours • In-depth coverage of subjects .Super Talent Batches announcing Civil Engineering Super Talent Batches at Kalu Sarai premise of Delhi st Batch : Commencing from May 20th Morning Batch nd Batch : Commencing from June 15th Evening Batch • Top 2000 Rank in GATE • B. 7%. w = 20% then the degree of saturation is __________. 0 1 2 3 Sol.2 The determinant of matrix is __________.1 Sol. Kalu Sarai. n = 40%.GATE-2014 Exam Solutions Civil Engineering (Evening Session) Section .71 × 20 = 81. 1 0 3 0 2 3 0 1 3 0 1 2 ∆ = 0 1 2 3 1 0 3 0 2 3 0 1 3 0 1 2 R 4 → R 4 – R2 – R 3 0 1 ∆ = 1 0 2 3 3 0 2 3 0 1 0 −3 −2 1 R 4 → R4 + 3R1 ∆ = 0 1 2 0 0 1 2 0 1 0 3 0 2 3 3 0 0 1 4 10 R 3 → R3 – 3R1 ∆ = 1 2 3 0 3 0 0 −6 −8 0 4 10 | Email :
[email protected] | Visit: www.madeeasy.in Corporate Office: 44-A/1.67 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 1 If Gs = 2. New Delhi-16 .3% 0.67 1−n 2. Se = wG and ⇒ e = S = n = 0.I (Civil Engineering) One Mark Questions Q. Q. madeeasy. r e = 4 Number of reaction released. Kalu Sarai.3 The static indeterminacy of two span continuous beam with internal hinge is __________. Internal Hinge Number of member.GATE-2014 Exam Solutions Civil Engineering (Evening Session) Interchanging column 1 and column 2 and taking transpose 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 2 1 0 − 2 ∆ = 3 0 0 0 1 2 0 3 −6 4 0 −8 10 1 2 0 = −1 × 3 −6 4 0 −8 10 = –1 × {1(–60 + 32) + 2(0 – 30)} = –(–28 – 60) = 88 Q. T2(≠ 0) are constants having dimension of time. rr = 1 Degree of static indeterminacy. v=− and w = 0 along x.in . New Delhi-16 | Email :
[email protected] A plane blow leak velocity component u = x y . m = 4 j = 5 Number of external reaction. y T1 T2 and z direction respectively where T1(≠ 0). Internal Hinge Sol. D s = 3 m + re – 3j – rr = 3 × 4 + 4 – 3 × 5 – 1 = 0 Q. Number of joint.com | Visit: www. The given blow is incompressible if Corporate Office: 44-A/1. (a) Group-I Curve J Curve K Curve L P-1. Stress J K L Strain P. R-2 1. T2 2 (b) T1 = − (d) T1 = T2 (d) For a flow to exist ⇒ ⇒ ⇒ ∂u ∂v + = 0 ∂x ∂y 1 1 − = 0 T1 T2 T 1 = T2 Q. Q-3. R-2 (c) P-2. (b) Corporate Office: 44-A/1.GATE-2014 Exam Solutions Civil Engineering (Evening Session) T2 2 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 3 (a) T1 = –T2 (c) T1 = Ans.825σ (c) fck + 0. Q-2.64σ (d) fck + 0. New Delhi-16 | Email : madeeasydelhi@gmail. R-1 Q. Match the stress-strain curves with corresponding materials. Kalu Sarai. Q. R-1 Ans. (d) (d) P-3.com | Visit: www. While Group-II gives the list of materials.in . R. Q-1.50σ (b) fck + 1.725σ Ans.madeeasy. Q-3. 2. 3. (b) Group-II Cement paste Coarse aggregate Concrete P-3.6 As per IS : 456-2000 in design of concrete target mean strength is taken as (a) fck + 0.5 Groups-I contains representative stress-strain curves as shown in figure. Flexural strength = 0.in .11 Sol. = e∫ k2dt Q.7 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 4 Modulus of elasticity of concrete is calculated (as per IS 456 : 2000) by (a) Secent modulus (c) Initial tangent modulus (b) Tangent modulus (d) None of these Ans. The flexural tensile strength of M 25 grade of concrete in N/mm2. (a) Q.5 N/mm2 Q. Polar moment of inertia (Ip) in cm4 at a rectangular section having width b = 2 cm and depth d = 6 cm is ___________ . (a) Q.GATE-2014 Exam Solutions Civil Engineering (Evening Session) Q.9 Survey which is conducted for geological features like river. cities etc.madeeasy. is denoted as (a) Land survey (c) Engineering survey (b) Geological survey (d) Topographical survey Ans. Kalu Sarai.10 The integrating factor for the differential equation −k t (a) e 1 −k t (b) e 2 k t (d) e 2 dP + k 2P = k1L 0e− k tt is dt (c) ek1t Ans. I p = Ix + Iy = = bd 2 b + d2 12 bd3 db3 + 12 12 ( ) = 2×6 2 2 + 62 12 ( ) = 40 cm4 Corporate Office: 44-A/1. (d) I. New Delhi-16 | Email : madeeasydelhi@gmail. as per IS:456-2000. natural resources.com | Visit: www.8 Sol. is ________.7 fck = 3.P. Polar moment of inertia. building. 12 Sol. The probability of obtaining a “TAIL’ when the coin is tossed again is (a) 0 (c) 4 5 (b) (d) 1 2 1 5 Ans. then the density (in veh/km) of stream is ___________.73 (WL – 20) (c) Ip = 0. Kalu Sarai.madeeasy.70 (WL – 10) Ans.15 The contact pressure for a rigid footing resting on clay at the centre and the edges are respectively (a) maximum and zero (c) zero and maximum (b) maximum and minimum (d) minimum and maximum Corporate Office: 44-A/1. Q.14 As per ISSCS (IS : 1498 -1970) an expression of A-line is (a) Ip = 0. (a) Q.73 (WL – 10) (b) Ip = 0.70 (WL – 20) (d) Ip = 0.in . New Delhi-16 | Email : madeeasydelhi@gmail. Capacity = ⇒ Density = 1000 × V = V × density S 1000 = 20 veh/km S 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 5 The average spacing between vehicles in a traffic stream is 50 m.13 A fair (unbiased) coin was tossed 4-times in a succession and resulted in the following outcomes (I) H (II) H (III) H (IV) H.GATE-2014 Exam Solutions Civil Engineering (Evening Session) Q.com | Visit: www. {T}] = 2 n(E) = {(T)} = 1 ∴ P(E) = 1 2 Q. (b) P(E) = n ( E) n (S ) n(s) = [{H}. madeeasy.in . Kalu Sarai.5 m q = 4/2.16 The clay mineral primarly governing the swelling behaviour of black cotton soil is (a) Halloysite (c) Kaolinite (b) Illite (d) Montmorillonite Ans.com | Visit: www.639 m g Corporate Office: 44-A/1.18 A rectangular channel of 2. (a) Two Marks Questions Q. Q.5 m3/s/m = 1. Considering that acceleration due to gravity as 9. (a) Aerobic heterotrops (c) Autotrops (b) Anaerobic heterotrops (d) Phototrops Ans. B = 2.5 m width is carrying a discharge of 4 m3/s.6 m3/s/m yc3 = Sol. then velocity of flow (in m/s) corresponding to the critical depth (at which the specific energy is minimum) is ________ .81 m/s2. (d) Q.GATE-2014 Exam Solutions Civil Engineering (Evening Session) Ans.17 Dominating micro-organisms in Active Sludge Reactor process. Q = 4 m3/s. (d) 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 6 Stress distribution for cohesive soil Minimum at centre and maximum at edge. New Delhi-16 | Email : madeeasydelhi@gmail. ⇒ q2 ⇒ yc = 0. Y R = 0. Y0 = = Yw = 90 mg/l = 90 gm/m3 Q R = 12 m3/sec (2 × 90) + (12 × 5) 2 + 12 180 + 60 240 = = 17.143 1 − (10 ) = 15.25/day.921 .434 ×0.639 = 2g 2 V = 0..1085 Y t = Y0 1 − (10 ) Area of river = 50 m2 Flow of river = 12 m3/sec Stream velocity = Time taken.in .504 m/s Q.1085× 9. Assuming the de-oxygenation rate constant. ∴ From eq. (i) t = 14 = 0.68 (c) 15. Kalu Sarai.5 mg/l = 5 gm/m3 BOD of mixture. Cross-section area of the river is 50 m2.(i) Y t = 17.19 A waste water stream (flow = 2 m3/s. ultimate BOD = 5 mg/l).28m / s −0.28 m / sec 50 10 km = 9.921 days 0.143 gm/m3 14 14 − k Dt kD = 0.GATE-2014 Exam Solutions Civil Engineering (Evening Session) 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 7 At critical depth velocity head = ⇒ ⇒ yc 2 V2 0. k′ = 0. The BOD (in mg/l) of the river water 10 km downstream of the mixing point is (a) 1. New Delhi-16 | Email :
[email protected] = 0.70 mg/lit Corporate Office: 44-A/1. Flow of river.37 Ans. (c) Flow of waste water stream. Qw = 2 m3/sec Ultimate BOD.434 K = 0. ultimate BOD = 90 mg/l) is joining a small river (flow = 12 m3/s. Both water streams get mixed up instantaneously.70 gm/m3 = 15. Ultimate BOD of river.madeeasy.639 × g = 2.com | Visit: www.63 (d) 1..46 (b) 2. madeeasy. New Delhi-16 | Email : madeeasydelhi@gmail. L/2 σ L/2 R = From properties of circle h αT L L × 2 2 (Considering ‘δ’ very small so neglect δ2) (2R – δ)δ = ⇒ 36°C δ = L2 L2αT = 8R 8h 72°C 1.5 m Corporate Office: 44-A/1. Kalu Sarai.50 × 10–5/°C.5 m Sol.com | Visit: www. The temperature at the top and the bottom surfaces of the beam are 36°C and 72°C respectively.in 250 mm . the vertical deflection of the beam (in mm) at its mid span due to temperature gradient is _________. 36°C 250 mm 72°C 1. Considering coefficient of thermal expansion (α) as 1.5 m 1.20 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 8 The beam of an overall depth 250 mm (shown below) is used in a building subjected to two different thermal environments.GATE-2014 Exam Solutions Civil Engineering (Evening Session) Q.5 m 1. Kalu Sarai.22 Match the List-I (Soil exploration) with List-II (Parameters of subsoil strength characteristic) and select the correct answer from the codes given below: List-I P. Skin resistance (fc) 4. S-1 List-II 1. (a) xα − 1 α→0 α lim 0 0 form eα ln x − 1 α eα ln x ln x 1 Use L-Hospital Rule α→ 0 lim lim α→ 0 = ln x Q. Undrained cohesion (Cu) (b) P-1. Vane shear test (VST) (a) P-1. R-4.com | Visit: www.madeeasy. ⇒ α = 1.50 × 10−5 × (72° − 36°) × 32 8 × 250 × 10−3 ( ) = 2. Static cone penetration (SCPT) R.50 × 10–5/°C δ = αTL2 8h 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 9 = 1. Menard’s method (Em) 2. Q-1. R-2.43 mm Q. Q-3. S-3 Ans. S-4 (d) P-4. R-2. R-3. New Delhi-16 | Email :
[email protected] The expression lim (a) log x (c) x log x xα − 1 is equal to α→0 α (b) 0 (d) ∞ Ans. (a) Corporate Office: 44-A/1. Standard penetration (SPT) S. Number of blows (N) 3. Q-2. S-4 (c) P-2. Q-3. Pressure meter test (PMT) Q.in .GATE-2014 Exam Solutions Civil Engineering (Evening Session) Given. 75 and 25 (c) 1500.204 × 10–3 m/s Vs = ⇒ BL = Q BL 0. The minimum surface area (in m2) required for this settling tank to remove particles of size 0.204 × 10−3 = 31. New Delhi-16 (b) 1250. 300 and 100 | Email :
[email protected] m2 Q. specific gravity of particles = 2.0105 × 10−2 × 102 2 = = 3.in .GATE-2014 Exam Solutions Civil Engineering (Evening Session) Q.10 m3/s.24 The values of axial stress (σ) in kN/m2.2 m) 3m Q 50 kN (a) 1000. Kalu Sarai.06 mm and above with 100% efficiency is ________.81 m/s2.81 × 36 × 10 −4 18 × 1. bending moment (M) in kNm and shear force (V) in kN acting at point P for the arrangement shown in figure are respectively Cable Frictionless Pulley P Beam (0. Sol.madeeasy.2 m × 0.10 mm and below is flowing into a settling tank at 0.1 3.0105 × 10–2 cm2/s.com | Visit: www. Vs = ( G s − 1) g d 2 18ν = (2.65 × 9. 225 and 75 Corporate Office: 44-A/1. Assume g = 9.23 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 10 The suspension of sand like particles in water with particles of diameter 0.0105 × 10 −2 1. and kinematic viscosity of water = 1.65 − 1) × 9.81 × (6 × 10−2 ) 18 × 1.65. 150 and 50 (d) 1750. New Delhi-16 | Email : madeeasydelhi@gmail. (c) Distribution Factor Joint Member BA B BP PB P PE PC CP C CD RS I 6 4I 8 4I 8 I 6 4I 8 4I 8 I 6 (b) 172 (d) 178 TRS 2 I 3 D.in . Kalu Sarai. the magnitude of the bending moment (in kNm) at P (Preferably using the moment distribution method) is 24 kN/m B C 4Ic Ic P Ic E 4Ic Ic 6m A 8m D 8m (a) 170 (c) 176 Ans. (b) Loading after removing the cable 50 kN 50 kN 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 11 50 kN Axial stress = 50 = 1250 kN/m2 0.GATE-2014 Exam Solutions Civil Engineering (Evening Session) Ans.madeeasy.2 × 0.F.2 Bending moment = 50 × 3 = 150 kNm Q. 1 4 3 4 3 7 1 7 3 7 3 4 1 4 7 I 6 2 I 3 Corporate Office: 44-A/1.25 Considering the symmetry of a rigid frame as shown.com | Visit: www. com | Visit: www. | Email :
[email protected] An effluent at a flow rate of 2670 m3/d from a sewage treatment plant is to be disinfected.in Corporate Office: 44-A/1.145t where Nt = number of micro organism surviving at time t (in min) and N0 = number of microorganism. The laboratory data at disinfected studies with a chlorine dosage of 15 mg/l yield the model Nt = N0e–0.madeeasy. Kalu Sarai. New Delhi-16 .GATE-2014 Exam Solutions Civil Engineering (Evening Session) Fixed End Moment M AB = M BA = M PE = M EP = M CD = M DC = 0 M BP 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 12 24 × (8) = −128 kNm = − 12 2 M PB = +128 kNm M PC = –128 kNm MCP = 128 kNm Distribution Table 1 4 A 0 3 4 B 0 –128 32 16 32 96 –32 128 0 48 3 7 3 7 P –128 0 –48 128 3 4 1 4 C 0 0 D 1 7 P 0 0 –16 0 E 0 –96 –32 0 Final 16 32 –32 176 –176 32 –16 0 0 88 kNm + 32 kNm 32 kNm – – 176 kNm – – + 88 kNm 32 kNm P 32 kNm – + 16 kNm + 16 kNm Q. GATE-2014 Exam Solutions Civil Engineering (Evening Session) 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 13 present initially (at t = 0).83 N π Q. Force = ρaV2 Q = aV = 15 litre/sec ⇒ Force = ρ Q a 2 Sol.27 A horizontal nozzle of 30 mm diameter discharges a study Jet of water into the atmosphere at a rate of 15 litres/second. (50 m3) Q. (b) 2k 3 q = v × k (b) 2400 (d) 9600 v = 80 − Capacity. The volume of disinfection unit (in m3) required to achieve a 98%. The diameter of inlet to the nozzle is 100 mm. The capacity (in 3 veh/hr) of this section of the highway would be (a) 1200 (c) 4800 Ans. is _______ .in Corporate Office: 44-A/1. Sol. the force exerted by the Jet (in N) on the plate is_________ . New Delhi-16 .28 On a section of a highway the speed density relations is linear and is given 2 by ν = 80 − k .madeeasy. (given) = 100 × 152 kg 3 π × 302 m 4 × 10 −6 m6 s2 × 1 10−6 m2 = 100 × 1 N = 31. Neglecting air friction and considering density of water as 1000 kg/m3.com | Visit: www. Kalu Sarai. Kill of M. where ν is in km/hr and k in vehicle/km.O. = 80k − For q to be maximum ⇒ dq = 0 dk 2k 2 3 4k dq =0 = 80 − 3 dk | Email : madeeasydelhi@gmail. The Jet impinges normal to a flat stationary plate held close to the nozzle end. 0 10 100 = = 1.com | Visit: www. k = 60 q = 80 × 60 − = 4800 − = 2400 2 (60 )2 3 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 14 2 × 3600 3 Q.GATE-2014 Exam Solutions Civil Engineering (Evening Session) ⇒ Maximum capacity.30 0 4 4 6 The rank of the matrix −2 14 8 18 is _______ − − 14 14 0 10 Sol. Given. New Delhi-16 | Email : madeeasydelhi@gmail. 0 4 4 6 −2 14 8 18 14 −14 0 −10 R 3 → R3 – 2R1 + R2 Corporate Office: 44-A/1.3 gm/cm3 1. the depth of the irrigation water (in mm) required for irrigating the crop is __________ .madeeasy.0 gm/cm3 γ d ( Fc − w ) γw 1. 70 cm 28% 18% 1.3 × (70 × 10 ) × = 91 mm Q.3 × (70 × 10)(28% − 18%) 1. If the densities of the soil and water are 1. FC = Existing moisture content. γ = Density of water. γw = Depth of irrigation water required.29 Irrigation water is to be provided to a crop in a field to bring the moisture content of the soil from the existing 18% to the field capacity of the soil at 28%. Root zone depth. The effective root zone of the crop is 70 cm. d = Field capacity. dw = Sol. w = Density of soil. Kalu Sarai.3 g/cm3 and 1 gm/cm3 respectively.in . 31 An infinitely long slope is made up of a C-φ soil having the properties cohesion (C) = 20 kPa and dry unit weight (γd) = 16 kN/m3.madeeasy. As the given slope is in dry condition.21° Q. 60 vehicles over took the student (Assume the number of vehicle overtaken by the student is zero) during the ride and 45 vehicles while the student stopped.in . New Delhi-16 | Email : madeeasydelhi@gmail. The angle of inclination and critical length of slope are 40° and 5 m respectively.5 (c) 40 (b) 12 (d) 60 Ans. Kalu Sarai. (d) Velocity of student = (40 − 15) min 5km = 12km/hr Corporate Office: 44-A/1.com | Visit: www. The student stopped for 15 minutes during this ride.6782 φ ≥ 59. therefore. factory of safety should be more than 2. −2 14 ∴ Rank is 2. Sol.32 A student riding a bicycle on a 5 km one way street takes 40 minutes to reach home. Q. The speed of vehicle stream on that road (in km/hr) is (a) 7.GATE-2014 Exam Solutions Civil Engineering (Evening Session) 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 15 6 0 4 4 −2 14 8 18 14 − 2 (6) + ( −2) −14 − 2 (0) + (14 ) 0 − 2 (4 ) + 8 −10 − 2 (4 ) + (18 ) 6 0 4 4 −2 14 8 18 0 0 0 0 6 0 Determinant of matrix is not zero. the angle of internal friction of soil (in degree) is ________ . To maintain the limiting equilibrium. Fs = ⇒ ⇒ ⇒ tan φ ≥ 2 and β = 40° tan b tanφ ≥ 2tan (40°) tanφ ≥ 1. The sample includes 5% of bitumen (by total wt.t.441.r. The void filled with bitumen (VFB) in the Marshall sample (in %) is ________. student moving = Relative speed of vehicle w.10. h.564 = × 100 VMB 15. the maximum and average velocities are respectively Corporate Office: 44-A/1. The theoretical maximum specific gravity of mix is 2. Kalu Sarai.t. is given by the expression u=− 2 h2dP y 1 − 4 h 8µdx In this equation the y = 0 axis lies equidistant between the plates at a distance h/2 from the two plates.74% Q.594 = 67. New Delhi-16 | Email :
[email protected] In a Marshall sample.594% VFB = Vb × 100 10. Sol.com | Visit: www.03% 2.madeeasy.324 wb 5 = 2.324 × = 10.564 Gb 1.324 × 100 = 5. the parabolic velocity distribution profile of fully developed laminar flow in x-direction between two parallel plates. of mix) of specific gravity 1.546 respectively. P is the pressure variable and µ is the dynamic viscosity term. Vv = G t − Gm × 100 Gm 2. y) plane.441 − 2. Stationary and identical plates that are separated by distance.GATE-2014 Exam Solutions Civil Engineering (Evening Session) 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 16 Vehicle/min Relative speed of vehicle w. the bulk specific gravity of mix and aggregates are 2.1 = Vb = Gm VMB = Vv + Vb = 15.34 With reference to a standard Cartesian (x.324 and 2. student standing Vehicle/min ⇒ ⇒ 60 / 25 45 /15 = x − 12 x−0 x = 60 km/hr Q.in .r. madeeasy.in Corporate Office: 44-A/1. (a) 3 h2dP and Uavg = Umax 8 8µdx y h h/2 x Velocity expression for a laminar flow between two parallel plates is U = − End condition. ⇒ 2 h2 dP y − 1 4 8 µ dx h U = Umax at y = 0 Umax = − h2 dP 8 µ dx dQ = Area × Velocity 2 h2 dP y − 1 4 ( dy × 1) dQ = − dx h 8µ ⇒ Q = − h2 dP h/2 4y2 − 1 dy 8µ dx ∫− h/2 h2 h/2 h2 dP 4y3 − y = − 8 µ dx 3h2 = − − h/2 h2 dP h h 4 h3 h3 − − − − − dx 2 2 3h2 8µ 8 8 | Email : madeeasydelhi@gmail. Kalu Sarai. New Delhi-16 . ⇒ Discharge.com | Visit: www.GATE-2014 Exam Solutions Civil Engineering (Evening Session) 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 17 (a) Umax = − h2dP 2 and Uavg = Umax 8µdx 3 (b) Umax = h2dP 2 and Uavg = Umax 8µdx 3 3 h2dP and Uavg = Umax 8 8µdx (c) Umax = − (d) Umax = Ans. 150 (d) 0.015 Ans. assuming that the energy losses are negligible.35 A venturimeter having a throat diameter of 0.A1 A 2 rgh 2 − A2 A1 2 Corporate Office: 44-A/1. Pressure difference.2 m d = 0.1 m is used to estimate the flow rate of a horizontal pipe having a dia of 0.GATE-2014 Exam Solutions Civil Engineering (Evening Session) 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 18 = − ∵ h3 dP 12 µ dx Q = AV − h3 dP = (h × 1) × Uavg 12µ dx Uavg = − − ⇒ h2 dP 12 µ dx Uavg Umax = h2 dP 8 2 12µ dx = = 2 h dP 12 3 − dx 8µ ∴ Uavg = 2 U max 3 Q. 1 D = 0.in .1 m D = 0.500 (c) 0. Kalu Sarai.01 m 1 2 Coefficient of discharge. (c) Diameter of throat. The flow rate (in m3/s) through the pipe is approximately equal to (a) 0. Discharge.com | Visit: www.2 m for an observed pressure difference of 2 m of water head and coefficient of discharge equal to unity.madeeasy.2 m P1 − P2 = 2 m = h w 2 d = 0. CD = 1 Q = CD . Diameter of pipe.050 (b) 0. New Delhi-16 | Email : madeeasydelhi@gmail. Kalu Sarai.madeeasy.1) 2 × 9.2 ) (0.GATE-2014 Exam Solutions Civil Engineering (Evening Session) 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 19 A1 = A2 = π (0. (c) (d) Here degree of static indeterminacy = 0 ∴ Number of plastic hinges required for mechanical = Ds + 1 = 0 + 1 = 1 P/2 P θ θ Mp θ Mp Corporate Office: 44-A/1.81 × 2 4 Q = 4 4 π 0.050 m3/sec Q.0508 ≈ 0.in .36 A prismatic beam (shown) has plastic moment capacity of Mp. then the collapse lead P of the beam is P P/2 L/2 L/2 L/3 (a) 2M p L 6M p L (b) 4M p L 8M p L (c) Ans.2)2 4 π (0.com | Visit: www.2 ) − (0. New Delhi-16 | Email :
[email protected]) ( 4 = 0.1)2 4 2 ⇒ 2 2 π 1 × × (0. GATE-2014 Exam Solutions Civil Engineering (Evening Session) From principal of virtual work − M pθ − M pθ + P L P L θ− × θ = 0 2 2 3 PL PL θ− θ = 0 2 6 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 20 ⇒ −2M pθ + ⇒ ⇒ 2Mp = P = PL PL (3 − 1) PL 1 × = = PL 2 6 6 3 6 Mp L Q.. New Delhi-16 | Email : madeeasydelhi@gmail.(i) Corporate Office: 44-A/1. P 2m 160 kN Q 2m S 2m R Sol.madeeasy.. Kalu Sarai.in .37 The axial load (in kN) in the member PQ for the assembly/arrangement shown in figure given below is ________. Free body diagram For principle of superposition VQ Q VQ 160 kN S 2m 2m VQ L3 VQL 160 × 23 160 × 22 + ×2− = 3EI 2 EI 3EI AE .com | Visit: www. Kalu Sarai.GATE-2014 Exam Solutions Civil Engineering (Evening Session) 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 21 Deflections due to axial forces will be very less as compared to bending forces. H = 1 m ⇒ ⇒ ⇒ ⇒ dH = C1 dx H = C1x + C2 C2 = 5 H = C1x + 5 1 = C1 × 10 + 5 10C1 = –4 C1 = − 2 5 2 H = − x +5 5 Corporate Office: 44-A/1. the value of H (in m) at the middle of the strip is __________ .38 Water is flowing at a steady rate through a homogeneous and saturated horizontal soil strip at 10 meter length. So we can neglect the axial deformation. equation of the flow of soil strip is d2 H dx2 = 0 ⇒ ⇒ at x = 0. ∴ From equation (i) VQ 43 160 × 23 160 × 22 + ×2− = 0 3EI 2 EI 3EI VQ × 43 160 × 23 160 × 22 × 2 + = 3 2 3 ⇒ ⇒ VQ = 50 kN Q. New Delhi-16 | Email : madeeasydelhi@gmail. Given.madeeasy. The strip is being subjected to a constant water head (H) of 5 m at the beginning and 1 m at the end.com | Visit: www. Sol. If the governing equation of the flow in the soil strip is d 2H dx 2 = 0 (where x is the distance along the soil strip).in . H = 5 m ⇒ ⇒ at x = 10. 30) = = 2.320 and 0. Kalu Sarai.21) + 0. PQ = ks + C = 100(0.1 m Corporate Office: 44-A/1.40 A tachometer was placed at point P to estimate the horizontal distance PQ and PR. of vehicles = 240 vehicle/km e −240 ×30 3600 P(1.32) + 0.1 = 21. e−λt .2707 240 3600 × 30 1! 1 Q. n λ = no.39 An observer counts 240 veh/hr at a specific highway location. If the stadia multiplication constant = 100 and stadia addition constant = 0. Assume that the vehicle arrival at the location is Poisson distributed.10 m. New Delhi-16 | Email : madeeasydelhi@gmail. Q P R Sol. t) = n! Here.GATE-2014 Exam Solutions Civil Engineering (Evening Session) at x = 5 m H = − 2 ×5+5 5 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 22 H = 3 m Q. ( λt ) P(n.in . the probability of having one vehicle arriving over 30 sec time interval is _______.1 m PR = ks + C = 100(0. the horizontal distance (in m) between the point Q and R is ________. Sol.madeeasy.210 m respectively. The corresponding stadia intercept with the telescope kept horizontal are 0.1 = 32. The ∠QPR is measured to be 61°30′30′′.e–2 = 0.com | Visit: www. 06 m Q.0 MPa Q.41 For the state of stresses (in MPa) shown in the figure below.508° 1030. Kalu Sarai.in .508° = = 1030.GATE-2014 Exam Solutions Civil Engineering (Evening Session) Applying the cosine rule QR = 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 23 PQ2 + PR2 + 2 ( PQ)( PR ) cos θ Where.21 + 696. the maximum shear stress (in MPa) is 4 4 2 2 4 4 Sol. θ = 61°30′30′′ = 61. New Delhi-16 | Email :
[email protected])(21.41 + 445.42 The tension (in kN) in a 10 m long cable shown in figure neglecting its self weight is 3m P S 3m Q Cable y Cable R 120 kN Corporate Office: 44-A/1.41 + 445.1) cos61. τmax = σx − σ y σ1 − σ2 = + τ2 xy 2 2 = 5.21 + 2 (32.2 = 46.madeeasy.com | Visit: www. Kalu Sarai. New Delhi-16 | Email : madeeasydelhi@gmail.. If the cycle length is 60 seconds.in . Corporate Office: 44-A/1. The lost time is given as 4 seconds for each phase. The efficiency gree time of 4th phase is ____________ (in seconds).GATE-2014 Exam Solutions Civil Engineering (Evening Session) (a) 120 (c) 60 Ans. (b) 3m P S 4m θ θ R 120 kN 3m Q 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 24 (b) 75 (d) 45 5m 5m Freebody diagram T S 4m θ θ T 5m 5m 120 kN ⇒ Here. ΣF y = 0 2Tcosθ = 120 cos θ = 2T × 4 5 .madeeasy..com | Visit: www. 187 and 210 veh/hr with saturation flow rate of 1800 veh/ hr/lane for all phases.43 A pre-timed four phase signal has critical lane flow rate for the first three phases as 200.(i) 4 = 120 5 ⇒ T = 120 × 5 = 75 kN 2×4 Q. 5 × 16 + 5 1−y 24 + 5 1− y 29 60 1 − 29 60 − 29 = 60 60 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 25 Saturation flow rate is 1800 veh/hr/lane y2 = y3 = C0 = ⇒ ⇒ ⇒ ⇒ And ⇒ ⇒ 60 = 60 = 1 – y = y = = 0.185 G = = = 15.madeeasy.517 y 4 = 0.in . length of the cycle.GATE-2014 Exam Solutions Civil Engineering (Evening Session) Sol. q3 = 210 veh/hr L = 4 × 4 = 16 sec C 0 = 60 sec y1 = q1 200 = s1 1800 q2 187 = s2 1800 q3 210 = s3 1800 1. Flow rates for the first three phase are given as q1 = 200 veh/hr q2 = 187 veh/hr and Lost time.5L + 5 1− y 1. Now. Kalu Sarai.com | Visit: www. New Delhi-16 | Email :
[email protected] y = y1 + y2 + y3 + y4 0.185 (60 − 16 ) 0.517 = 597 + y4 1800 y4 (C0 − L ) y 0.745 sec Corporate Office: 44-A/1. GATE-2014 Exam Solutions Civil Engineering (Evening Session) Q. δ = 2 2 φ = × 30 = 20° 3 3 Lateral earth pressure coefficient. Qu = ? Vertical effective stress at 20 m. The pile is embedded in a homogeneous sandy stratum where.in . K = 2. the ultimate bearing capacity of the pile is ________ Sol.7 and the bearing capacity factor (Nq) = 25. σ v = 20 × 20 = 400 kN/m2 From 0 to 20 m.7 Bearing capacity factor. unit point bearing resistance and skin friction resistance remain constant at σ v = 400 kN/m2 The ultimate load capacity is given by Qu = qup⋅Ap + qs⋅As = Qup + fs where qup = σ v Nq = 400 × 25 = 10000 kN/m2 and qs = 1 ⋅ σ v ⋅ K ⋅ tanδ 2 Corporate Office: 44-A/1.44 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 26 A single vertical friction pile of dia 500 mm and length 20 m is subjected to a vertical compressive load.com | Visit: www. N q = 25 Ultimate load capacity. 20 m 500 Homogeneous sandy stratum φ = 30° γ d = 20 kN/m3 Wall friction angle. New Delhi-16 | Email : madeeasydelhi@gmail. Kalu Sarai.madeeasy. dry unit weight (γd) = 20 kN/m3 and angle of wall friction (δ) = 2φ/3. angle of internal friction ( φ ) = 30°. Considering the coefficient of lateral earth pressure (k) = 2. 5)2 4 = 1963.38 and 1648.45 The chainage of the intersection point of two straights is 1585.49 (d) 418.88 (b) 218.38 Ans.88 and 218.49 = 8137.49 kN Ultimate load on the pile = 1963.54 kN/m2 Skin friction resistance.49 kN = Qup = q up × π (0.7 tan 20° 2 = 196. (c) V¢ (P.5 × 20 = 6174. If the radius of a circular curves is 600 meter.GATE-2014 Exam Solutions Civil Engineering (Evening Session) 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 27 1 × 400 × 2. Kalu Sarai.88 and 1466.000 × Q.54 × π × 0. New Delhi-16 | Email :
[email protected])2 4 ∴ π (0.49 + 6179.com | Visit: www.madeeasy.I)V ∆ = 40° C T1 (P. f s = 196.in . the tangent distance (in m) and length of the curve (in m) respectively are (a) 418.C) D T 2 (P.08 (c) 218.98 kN = 10.38 and 418.60 meter and the angle of inter section is 140°.T) R ∆/2 ∆/2 R O ∆ = 180° – 140° = 40° Corporate Office: 44-A/1. 47 The population of a new city is 5 million and is growing at a rate of 20% annually. New Delhi-16 | Email : madeeasydelhi@gmail. (a) 3-4 year (c) 5-6 year (b) 4-5 year (d) 6-7 year Corporate Office: 44-A/1.46 A surface water treatment plant operates round the clock with a flaw rate of 35 m3/min. Given data Flow rate. Sol.82 m 180 = Tangent distance (T) is the distance between P-C to P. Alum quantity (kg) required for 30 days = 35 × 103 × 60 × 24 × 30 × 25 × 10–6 = 37.T) ⇒ T = T1V = T2V = OT1 tan = 600 tan 20° = 218.800 kg Section . The water temperature is 15°C and Jar testing indicated and alum dosage of 25 mg/l with flocculation at a Gt value of 4 × 104 producing optimal results. How many year would it take to double at this growth rate.in .I (also the distance from P.madeeasy.I to P.com | Visit: www. Kalu Sarai. The alum quantity required for 30 days (in kg) of operation of the plant is _________ . Q = 35 m3/min Gt = 4 × 104 Alum dosage = 25 mg/lit.GATE-2014 Exam Solutions Civil Engineering (Evening Session) 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 28 Length of the curve = πR ∆ 180° π × 600 × 40 = 418.88 m ∆ ∆ = R tan 2 2 Q.II (General Aptitude) One Mark Questions Q. in . (a) experiences (c) is experiencing (b) has experienced (d) experienced Ans.50 If X is 1 km North east of Y.2 n = 3. Q is 1 km East of P.8 year Q. (a) r Pnew = Pold 1 + 100 20 10 = 5 1 + 100 n n 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 29 ⇒ ⇒ ⇒ ⇒ 2 = (1. (c) Two Marks Questions Q. Ans.madeeasy.com | Visit: www.2)n log 2 = n log1.GATE-2014 Exam Solutions Civil Engineering (Evening Session) Ans. Kalu Sarai.48 A person affected by Alzheimers disease _______ short term memory loss. (b) Women are not citizens of any country. (c) Women solidarity knows no national boundary. I have no country” (a) Women have no country. (a) Q. P is 1 km South of W. Then find the distance between X and Q. New Delhi-16 | Email : madeeasydelhi@gmail. Y is 1 km South east of Z and W is 1 km West of Z. (a) 1 (c) (b) 2 3 (d) 2 Corporate Office: 44-A/1.49 Select the closest in meaning “As a women. (d) Women of all country have equal legal rights. madeeasy. Anshu is youngest in the group. −1 + 2 2 XQ = 1 1 2 + + ( −1) = 3 2 2 2 Q. Shiv is younger than Riyaz. Ans. 2 2 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 Page 30 Y(0. 2 2 1 1 Z− . (c) Statement (1) and statement (2) both are required. (d) Statement (1) and (2) both are insufficient to find.in . 0) 1 1 P −1 − .51 In the group of four children.GATE-2014 Exam Solutions Civil Engineering (Evening Session) Ans. If Som is younger than Riyaz and Shiv is elder than Anshu. 2. Kalu Sarai. (a) Corporate Office: 44-A/1. 2 2 1 1 X . Som is younger than Anshu. −1 + 2 2 1 1 Q− .com | Visit: www. (b) Statement (2) is sufficient to recognize. (c) 1 1 W −1 − . (a) Statement (1) is sufficient to recognize. so the eldest in the group 1. New Delhi-16 | Email : madeeasydelhi@gmail.