CIVIL-ENGINEERING-PAPER-I-QP-ESEM-25-110825

Engineering Services (Main) Examination - 2025 NESP-O-CVLE CIVIL ENGINEERING PAPER—I Time Allowed : Three Hours Maximum Marks : 300 QUESTION PAPER SPECIFIC INSTRUCTIONS Please read each of the following instructions carefully before attempting questions There are EIGHT questions divided in TWO Sections. Candidate has to attempt FIVE questions in all. Question Nos. 1 and 5 are compulsory and out of the remaining, any THREE are to be attempted choosing at least ONE question from each Section. The number of marks carried by a question/part is indicated against it. Wherever any assumptions are made for answering a question, they must be clearly indicated. Diagrams/Figures, wherever required, shall be drawn in the space provided for answering the question itself. Unless otherwise mentioned, symbols and notations have their usual standard meanings. Attempts of questions shall be counted in sequential order. Unless struck off, attempt of a question shall be counted even if attempted partly. Any page or portion of the page left blank in the Question-cum-Answer Booklet must be clearly struck off. Answers must be written in ENGLISH only. NESP-O-CVLE/40 1 [P#t-0. SECTION—A 1. (a) (i) Write the qualities of good timber and the factors affecting the strength of timber. 6 (ii) Write short notes on any three of the following : (1) Brick buttresses (2) Brick corbel (3) Brick coping (4) Thresholds (5) Brick jambs (6) Racking back 2x3=6 (b) Draw the bending moment and shear force diagram for the beam shown below : 10 kN/m Hinge (e 2m $ 2m 12 (c) With the help of a sketch, briefly explain the major shear transfer mechanisms in a reinforced concrete beam, having shear reinforcement. Indicate the internal forces acting at a flexural-shear crack. 12 (ad) A pedestal in the shape of a frustum of a cone is made of concrete having a specific weight of 24 kN/m°. Determine the average normal stress acting in the pedestal at its base : - o2 x 12 (e) A four-wheel tractor weighing 18000 kg has weight distribution between the front and the rear wheels of 40 percent and 60 percent respectively. It is operating on a level haul road whose rolling resistance is 45 kg/ton. What is the maximum rimpull of the tractor if the coefficient of traction between the road surface and the tyre is 0°65? 12) NESP-O-CVLE/40 2 (b) (c) <e> x’ 140 mm 0332m i: i<—— 400 mm > <-60 mm «——— 600 mm 6 ———— A short hollow cast iron column having an external diameter of 600 mm and inside diameter 400 mm was cast in a factory. On inspection, it was found that the bore is eccentric as shown in the figure above. If the column carries a load of 2000 kN along the axis of the bore, calculate the extreme intensities of stresses induced in the section. Explain the following terms : () Autogenous shrinkage (i) Bogue compounds (iii) Case-hardening (iv) Elastomers (v) Guniting (vi) Scoriaceous aggregate (vii) Self-desiccation (viii) Shingling (ix) Puddling 20 (x) Wet rot 2x10=20 Design a suitable double-angle discontinuous strut in a steel truss to carry a working axial compressive load of 200 KN. The effective length of the strut is 2:12 m. Use a gusset plate of 20 mm thick. Assume column buckling class c. fy =250 MPa, f,, =400 MPa, yp =1-25. Relevant portion of the code books is enclosed. 20 NESP-O-CVLE/40 3 [ P.O: TABLE ||| ROLLED STEEL EQUAL ANGLES * DIMENSIONS AND PROPERTIES ( Continued ) x Designation Size Thickness Sectional Weight per Centre of Distance of Area Metre Gravity Extreme Fibre AxB t a = , Cre = €+» ണന ಗಾಗಾ mm em? kg നേ cm 158 7070 70 - 70 6-0 6-77 53 1-89 511 60 8:06 63 1-94 5-06 8-0 10:58 8.3 2:02 4-98 10-0 13-02 10-2 2:10 4:90 157575 75x75 50 7-27 57 202 5-48 60 8-66 68 2-06 5.44 8.0 11-38 8- 214 5-36 10-0 14:02 110 222 5:28 ISA 8080 x 80 6-0 9-29 78 2:18 5:82 8.0 12.21 9-6 2:27 5:13 10-0 15-05 118 234 5:66 12-0 17.81 14-0 2-42 5:58 ISA 9090 × 60 10-47 8.2 2-42 6-58 8.0 13-79 10-8 2:51 6:4 10-0 17-03 134 2.59 6-41 12-0 20-19 15-8 2:66 634 ISA 100100 100 100 6.0 11-67 9.2 2:67 7-33 8.0. 15-39 121 276 7-24 10-0 19-03 14-9 3:84 716 1240 22-59 17-7 292 7-08 154 110110 110 110 8-0 17.02 13.4 3-0 8-00 (0-0 21-06 165 3-08 7-92 12-0 25-02 19-6 3:16 7-84 15-0 30-81 24:2 3:27 773 15% 120130 130 × 130 BO 20-22 9 3-50 9-50 10-0 25-06 19-7 3-58 9:42 12-0 29-82 23-4 3-66 9.34 15-0 36-81 26-9 3:78 9.22 1SA 150150 150 × 150 10-0 29-03 22:8 4-06 10-94 12-0 34-59 27-2 4.14 10:86 15-0 4278 33-6 4:26 10.74 18-0 50-79 39-9 438 10:62 15% 200200 200 x 200 12:0 46-61 366 5-36 14:64 15-0 57-80 45-4 5:49 145! 18-0 68-81 54-0 5-61 1429 250 93-80 73-6 5.88 14:12 NESP-O-CVLE/40 4 Cyy Y Moments of Inertia TABLE Ill ROLLED STEEL EQUAL ANGLES DIMENSIONS AND PROPERTIES ( Continued ) Radii of Gyration તના મે ಅರಯದ /:±×=/;+ 1 hoy ‌ 31-4 49.8 1275 36.8 58-8 148 47-4 755 19-3 572 90.7 23:7 387 619 155 457 731 18-4 59-0 94:1 24-0 714 1133 29-4 560 896 22:5 725 115.6 29-4 87-7 139-5 36-0 101-9 161-4 42-4 80:1 1281 32-0 104-2 1664 42-0 1267 201-9 51-6 147-9 2349 609 111-3 178-1 445 145-4 231-8 58.4 177-0 282-2 718 207-0 329-3 847 195-0 311-7 78:2 238-4 380-5 96-3 279-6 4453 113-8 337.4 535-4 139-3 328-3 525-1 13114 402:7 643-4 162-1 473.8 735-9 191-8 574-6 914-2 235-0 622.4 995.4 249.4 735-4 1174-8 296-0 8968 1 429-7 363-8 1048-9 1 6682 429-5 1 788-9 2 862-0 715-9 2197-7 3511-8 883-7 25887 41308 1046:5 3 436-3 $ 460-9 14118 NESP-O-CVLE/40 frr=Syy cm 2:15 2-14 2:12 2:10 231 2:30 2:28 226 2.46 2-44 241 2:39 2:77 2:75 273 2:7! 3-09 3-07 3-05 3:03 3-38 3-36 3:34 3:31 4-03 4-01 3-99 3-95 4-63 4-61 4-58 4:54 6:20 617 613 6-05 7 cm 271 2-70 2-67 2:64 2.92 291 288 284 3:11 3-08 3:04 301 3:50 3:47 3.44 3:41 3-91 3:88 385 3:82 4:28 425 4:22 4:17 5:10 507 5-03 49% 5:86 5-83 5-78 5.73 7.84 7-79 7.75 7-63 cm 1-36 1-36 1-35 1:39 1:46 146 1:49 1-45 1-56 1-55 1:55 1:54 1-75 1:75 1:74 1-74 1-95 1:95 1-94 1:94 2:14 2:14 2:13 2:13 2:55 2:54 2:54 2:53 2:93 2-93 2-92 2-91 3-92 3-91 3-88 Section Z,,=2,, 6.1 34 9.5 11:7 7 4 110 13-5 96 12-6 155 18:3 12:2 150 19-8 23:3 15-2 20-0 247 29:2 24-4 30:1 35-7 437 345 427 50-7 62:3 59 677 83-5 98:7 422-2 15114 179-9 243-3 Modulus of Radius at Root 4 mm 70 70 8-5 85 10-0 10-0 12-0 150 Toe 45 45 55 క్స్‌ 6-0 60 8-0 10:0 Radius at Product of Inertia hy ‌ 18-4 217 23 33-3 22-8 27-0 348 417 33-0 427 51-4 59-2 47.2 615 745 865 65-7 85-8 104-4 121-6 1155] 1406 1645 197-0 194-2 238-3 279-9 337-8 368-2 435-0 5291 616-0 1058-9 1 301-2 1530-5 20157 Designation 154 7020 ISA 7575 ISA 8080 ISA 9090 ISA 100100 ISA 110110 ISA 130130 154 150150 154 200200 [ TS 800 : 2007 10.3.1.3 In the calculation of thread length, allowance should be made for tolerance and thread run off. 10.3.2 A bolt subjected to a factored shear force (Voy) shall satisfy the condition Va = Vay where V,, is the design strength of the bolt taken as the smaller of the value as governed by shear, V4, (see 10.3.3) and bearing, Vigp (see 10.3.4). 10.3.3 Shear Capacity of Bolt The design strength of the bolt, V4. as governed shear strength is given by: Vass = Vaso / Yon where V. =nominal shear capacity of a bolt, calculated as follows: Visa Fun Aytn, Ay) where Jf, = ultimate tensile strength of a bolt; n, = number of shear planes with threads intercepting the shear plane; n, = number of shear planes without threads intercepting the shear plane; Ag = nominal plain shank area of the bolt; and A,, = net shear area of the bolt at threads, may be taken as the area corresponding to root diameter at the thread. 10.3.3.1 Long joints When the length of the joint, /; of a splice or end connection in a compression or tension element containing more than two bolts (that is the distance between the first and last rows of bolts in the joint, measured in the direction of the load transfer) exceeds 15d in the direction of load, the nominal shear capacity (see 10.3.2), V,, shall be reduced by the factor B,,, given by: B,, = 1.075 - t;/(200 d) but 0.75 < B, < 1.0 = 1.075 - 0.005(1; /d) where d = Nominal diameter of the fastener. NOTE — This provision does not apply when the distribution of shear over the length of joint is uniform, as in the connection of web of a section to the flanges. 10.3.3.2 Large grip lengths When the grip length, /, (equal to the total thickness of NESP-O-CVLE/40 the connected plates) exceeds 5 times the diameter, d of the bolts, the design shear capacity shall be reduced by a factor B,,, given by: By, = 8 d/G d+ 1) = 8 (341, Md) B,, shall not be more than B, given in 10.3.3.1. The grip length, J, shall in no case be greater than 8d. 10.3.3.3 Packing plates The design shear capacity of bolts carrying shear through a packing plate in excess of 6 mm shall be decreased by a factor, B,, given by: Bye = (1 - 0.0125 t,) where t,, = thickness of the thicker packing, in mm. 10.3.4 Bearing Capacity of the Bolt The design bearing strength of a bolt on any plate, Vio as governed by bearing is given by: Veen = Vagw / Yb where Vip» = nominal bearing strength of a bolt =25k,dtf, where €.2P fu i : -0.25, 5 AO; k, is smaller of 3d,’ 3d, iy e, p =end and pitch distances of the fastener along bearing direction; d, = diameter of the hole; Su» f, = ultimate tensile stress of the bolt and the ultimate tensile stress of the plate, respectively; d = nominal diameter of the bolt; and t = summation of the thicknesses of the connected plates experiencing bearing Stress in the same direction, or if the bolts are countersunk, the thickness of the plate minus one half of the depth of countersinking. The bearing resistance (in the direction normal to the slots in slotted holes) of bolts in holes other than standard clearance holes may be reduced by multiplying the bearing resistance obtained as above, Vp by the factors given below: a) Over size and short slotted holes — 0.7, and b) Long slotted holes -0.5. NOTE — The block shear of the edge distance due to bearing force may be checked as given in 6.4, : 2007 IS 800 ৬ 992 $$6 हाइट ଟନ୍‌ 25 15 0୪ ண કટ (ण्ट $ કટ દાટ छ ୨୮ ୨୭ 65 ಕಾ 05 تا‎ 9 ୧୯୪ કટ 2 35 1 0૮૮ शष्ट ୯୯୨ 992 ১৮% ୮୨ 29 ष्ट 09८ 6୧ ൧൦ ୨୯୭ छट 00६ 68 ୫86 ८6८ €൦ ୨୮୧୪ ୮୪ ଅଃ (ट 68८ 882 9८ چیم‎ ஸஹ 282 082 6୧ 82 १८ 0६८ 9૮૬ शट श्ट ૮૮૪ 1૮૬ ઠાદ ୫୮୮ ୬୮୮ काह কাছেও દાદ 01 80 590 €0 5805 ଅଃ 10 ண 022 96 1718 ಟಟ 19 কে 6 ୭ ୮ ୭୫୭ مور‎ 00 44 1 10 - ப வ தில்‌ ರ್ಟ ਨਨ 017 066 ୫୫୭୮ 986 ೫8: 895 18૬ 60 ୮୮ 5૮૬ ೯0% 05 : ੪੩੬. ૦59૬ દક 19६ 6% ८५६ ୮୪ ୮୭୮ 002 ணை Ste tty Ot 819 ண ૪૪ 11% 60» 90 घण 666 ൧6 ೯6 षहः 06% 8% ൭89 061 દા 17 6%. 9% દા 19 8% 950 வல்‌ அண 66 லை ள்‌ न नत ठत 081 ୨ ண ण्ट ൧5 $ $ ண 19 शलो धह ளை 18% ୭ ୦୫ . [2 ୭ ୨୪୨୧ 0/1 ८8 وہ‎ ୨୫୭ 975 275 695 59 195 ୦୮୭ 65૬ 86 کیج‎ 98 ૬૬ 95 95 1૭6 91: 115 091 699 659 19 99 19 ୯୨୭ न 051 ےہ‎ 12 9६८ ഞാ ے۶‎ ୫୩୮୬ 21८ 20 00 ୮୦୨ 989 ८७ (9 29 959 09 ৮৮৪ ୨୮୨ 609 01 61% ମଧ୍ଯ 9% 6૮ ౦0 ೫18 ൧08 00% 26 ६४८ ౯౬౯ ८9८ 00 £৮% ७६८ 8೭ 61८ 01 00 01 ୨୪୫ 896 696 6৮৪ ୫6 06 1૮6 116 106 688 968 ୯୨୫ 91% ൧28 4/8 218 90 ೫6 2೬ 021 ६।। 2೭11 111 011 801 201 901 ೫01 ६0 0 001 6८6 ८.५6 9৮6 ६६6 026 506 068 ६८४ 011 হো ౬ 6% ध्व டைன ணை ઠા! ୨। शा ટા டை ൧0 6501 ੧01 20 00 ೯೬6 001 9५1 ات‎ ટકા 691 ୨୮୮ চা trl 01 (81 pel 161 ೭೭1 ا4ت‎ 121 611 911 ೪11 111 801 06 ୨୫ દા 0 9८1 ౬ ഒ $9 ଅ୨ یو‎ ౪ हा छा ஸ்‌! ୨୯ ୮୮ 06 21 டை ण्ट | 08 £ਟ शट દાટ श्ट 20૮ ८6 ౭61 ¿/୫। 281 9 01 ౪ 951 ઠા भा ೫! फा 98 (ध | 0 £ ৫ படல்‌ ୩୬୪୩ க ୬୪୭ ୫୪ ୧୪୧୭ 610 ୬ હા 161 থে £ 891 છા 851 छा 8 দে| છ କଃ ட ટ ୭ دھیوں‎ ୩୭ ୧୫୦ ୭ ୧୭ হত তোত 10 ହୋ છા ୫୬୮ ୮ ୧୨୮ 641 દડા! 06 25 હા একি ய ୭ છઠ ഇ, ୫୨୪ ಟ್ಟ .. ਲਏ છે ୬୪୮ ଲୌ ୫୫ 161 ആ ઠા 01 છા ৮ ୭୫୮୮ ~ ଏମି કેદ એ সহে હાદ ಟ್ಟ பலத்‌ પદ હદ. ୭୭୮ এচ 11% "ट છઠા 981 068 ய 0 BSH, দে ಹಚು પદ. કાઇ પાદ ಯಯ કાઈ ക. 90 . પિટ 0000 ಯೂ 0 000 હા છા. છા 4 16 ৮৮ % ୦୮ ರ್ಜ پیم‎ $£ ઠટદ 60६ હિ છટ ୩୮ દદ દાદ 0 છાટ કો ଆଃ 0} 0୩ 05 00 05% 0৮৮ 00 ४७६ 0૬ ৮৮ ઉદ 00६ 082 092 0 ୦୪୯ 0୮ 02 07 00८ 4 (ea) 4 ജട PITA ૪7૪ (UTE ୨୫୩୪/୦) 3 ୫୭୪[୦ Surpyong 10102 40} (egw) ۳/ 55૩૩૩૬ 31553400100 നടാ (2)6 31081 1.1.0. NESP-O-CVLE/40 3. (a) For the network shown below, the time estimates (in days) for each activity are mentioned. Determine the probability of completing the project in 35 days. Given : Standard normal distribution function Z P% (% probability) 0:8 78:51 0-9 81:59 1-0 84-13 11 86-43 1:2 88-49 (b) Design a short reinforced concrete column subjected to a working axial load of 1400 kN and service moments of 60 kN-m and 40 kN-m about its major and minor axes respectively. The least cross-sectional dimension of the column shall be 300 mm. Adopt limit state design. Use M 30 concrete and Fe 500 grade steel. The effective concrete cover to longitudinal reinforcement is 60 mm. Sketch the reinforcement details. Relevant portions of IS 456 and SP 16 are enclosed. NESP-O-CVLE/40 8 NESP-O-CVLE/40 39.3 Short Axially Loaded Members in Compression The member shall be designed by considering the assumptions given in 39.1 and the minimum eccentricity. When the minimum eccentricity as per 25.4 does not exceed 0.05 times the lateral dimension, the members may be designed by the following equation: P, = O4f,.A,+0.67f A, where P, = axial load on the member, J, = characteristic compressive strength of the concrete, = Area of concrete, = characteristic strength of the compression reinforcement, and area of longitudinal reinforcement for columns, 39.4 Compression Members with Helical Reinforcement The strength of compression members with helical reinforcement satisfying the requirement of 39.4.1 shall be taken as 1.05 times the strength of similar member with lateral ties. 39.4.1 The ratio of the volume of helical reinforcement to the volume of the core shall not be less than 0.36 (AJA-1) f,/f, gross area of the section, 5 area of the core of the helically reinforced column measured to the outside diameter of the helix, : characteristic compressive strength of the concrete, and characteristic strength of the helical reinforcement but not exceeding 415 Nimm?, 39.5 Members Subjected to Combined Axial Load and Uniaxial Bending A member subjected to axial force and uniaxial bending shall be designed on the basis of 39.1 and 39.2. NOTE— The design of member subject to combined axial load and uniaxial bending will involve lengthy calculation by trial and error. in order to overcome these difficulties interaction diagrams may be used. These have been prepared and published by BIS in ‘SP ; 16 Design sids for reinforced concrete to IS 456°. IS 456 ; 2000 39.6 Members Subjected to Combined Axial Load and Biaxial Bending The resistance of a member subjected to axial force and biaxial bending shall be obtained on the basis of assumptions given in 39.1 and 39.2 with neutral axis so chosen as to satisfy the equilibrium of load and Moments about two axes. Alternatively such members may be designed by the following equation: Fa +h 's10 Ma Muy, where MM, = moments about x and y axes due to design loads, May Mays = maximum uniaxial moment capacity for an axial load of P., bending about x and y axes respectively, and , is related to PJP where P,=0.45f,.A,+0.75f.A, For values of P/P,, = 0.2 to 0,8, the values of a, vary tinearly from 1.0 to 2.0. For values less than 0.2, a, is 1.0; for values greater than 0.8, a, is 2.0. 39.7 Slender Compression Members The design of slender compression members (see 28.1.1) shall be based on the forces and the moments determined from an analysis of the structure, including the effect of deflections on moments and forces. When the effect of deflections are not taken into account in the analysis, additional moment given in 39.7.1 shall be taken into account in the appropriate direction. 39.7.1 The additional moments M,, and M,, shall be calculated by the following formulse: é - feof 2000 | b ‘ # axial load on the member, effective length in respect of the major axis, effective length in respect of the minor axis, depth of the cross-section at right angles to the major axis, and b width of the member. For design of section, 39.5 or 39.6 as appropriate shall apply. ~ iw] " u [P.T.O. 500 Chart 48 COMPRESSION WITH BENDING—Rectangular Section—Reinforcement Distributed Equally on Four Sides 14 13 1:2 11 10 P,, /f,4,bD NESP-O-CVLE/40 |fy= 500 N/mm?] d’/D = 0:10 Ag= pbD/100 +a) +44’ AXIS OF BENDING cov 03 0-4 02 Mau/fcxbD 2 10 Compression Members 1.4 ط,/٤ںیطاح‎ 500 Chart 50 COMPRESSION WITH BENDING—Rectangular Section—Reinforcement Distributed Equally on Four Sides T T T = 500 N/mm? | 641/2 = 0:20 | As= pbD/100 € ® = Hee. ঢালে ಗಹ $ a ‌ AXIS OF BENDING se" of 6. “ < 4 0878 4. i | 3 04 நம்ம? Compression Members NESP-O-CVLE/40 ] [ P.T.O. 4.0 Chart 63 VALUES OF a for COMPRESSION MEMBERS ७ [ಗೆ 2; نو‎ کچ‎ < | ಆ > ਨਕ ഛി ५५०० 4 4.1 2 3-0 ய்‌ 3 [11 તવ ଞ୍ଚ ଛି नि aS “| va ¢ 2-0 గ ய [4 ea 9 2 [1] ਨੂ [1717 | | (3 a 1 2 ਛੱ ۴ छ 1-0 ర్‌ Fs ಸ್ಟಾ ۶2 न्य 7127) | 11] ध (| ¬ IH 0.0 દિ | | oA - ജ്ജ | +| ୪ | \ | 10 ছা 81 | + "६ [ = £ + ப 5 | ਹੇ... ଏ 8 = 18 ا‎ [1] ৰ لا 1 ات‎ < ਇੱ = 3 થા 5 ५ ಪ್ರಾಣಾ = कक एक कुछ कक ಅಲ ° = == ಕಣಣ 125] 5 ٦ 20 ریت‎ eee ~ sf "39 మై 4 44 یں‎ NY 25 ಜೆ ২ NESP-O-CVLE/40 Design Aids for Reinforced Concrete 12 () (a) (b) (c) For the continuous beam shown below, draw the bending moment diagram using Clapeyron’s theorem (three-moment equation). The support C sinks by lcm. Take E =200 GPa, I=10000cm? : Osim 20 kN 20 kN/m 10 kN/m ¥ 5 kN/m Hinge 10m A of B Cc + it ee b> yan ft Gi &®> ya Ac aia alia At main Ha 2m 2m 2m 2m 2m 1m 20 Five wheel loads of 90 kN, 90 kN, 90 kN, 50 kN and 50 kN magnitudes spaced 3m, 2 m, 1-5 m and 2 m apart respectively cross a simply supported girder of 20 m span from right to left with 90 kN load leading. Calculate the maximum positive and negative shear force at the centre of the span and absolute maximum value of bending moment that occurs anywhere in the girder. 20 The initial cost of an equipment is ¥ 1,100, salvage value is € 100, life of the equipment is 5 years. The rate of interest for sinking fund is 8% . Calculate the yearly depreciation and book value at the end of each year by straight line method, declining balance method, sum of years digital method and sinking fund method. Present the value in tabular form. 20 A single angle ISA 100 mm x75 mm x10 mm is connected to a gusset plate of 12 mm thick with five numbers of 16 mm diameter bolts. Determine its tensile capacity if the gusset plate is connected to the 100 mm leg. The cross-sectional area of the angle is 1650 mm7?, fu =410MPa, fy =250 MPa, Ym, =1-1, Ym =1-25. 12 mm thick gusset plate ISA 100 eae 10 Se 100 mm ===> Fi Boze 40 50 50 50 40 \ mmmm mm mm mm 75 mm Relevant portion of the IS 800 : 2007 is enclosed. 20 NESP-O-CVLE/40 13 Peo. TS 800 : 2007 5.6.1.1 Where the deflection due to the combination of dead load and live load is likely to be excessive, consideration should be given to pre-camber the beams, trusses and girders. The value of desired camber shall be specified in design drawing. Generally, for spans greater than 25 m, a camber approximately equal to the deflection due to dead loads plus half the live load may be used. The deflection of a member shall be calculated without considering the impact factor or dynamic effect of the loads on deflection. Roofs, which are very flexible, shall be designed to withstand any additional load that is likely to occur as a result of ponding of water or accumulation of snow or ice. 5.6.2 Vibration Suitable provisions in the design shall be made for the dynamic effects of live loads, impact loads and vibration due to machinery operating loads. In severe cases possibility of resonance, fatigue or unacceptable vibrations shall be investigated. Unusually flexible structures (generally the height to effective width of lateral load resistance system exceeding 5:1) shall be investigated for lateral vibration under dynamic wind loads. Structures subjected to large number of cycles of loading shall be designed against fatigue failure, as specified in Section 13. Floor vibration effect shall be considered using specialist literature (see Annex C) $.6.3 Durability Factors that affect the durability of the buildings, under conditions relevant to their intended life, are listed below: Environment, Degree of exposure, Shape of the member and the structural detail, Protective measure, and Ease of maintenance. a) b) c) d) e) 5.6.3.1 The durability of steel structures shall be ensured by following recommendations in Section 15. Specialist literature may be referred to for more detailed and additional information in design for durability. 5.6.4 Fire Resistance Fire resistance of a steel member is a function of its mass, its geometry, the actions to which it is subjected, its structural support condition, fire protection measures adopted and the fire to which it is exposed. Design provisions to resist fire are briefly discussed in Section 16. Specialist literature may be referred to for more detailed information in design of fire resistance of steel structures. NESP-O-CVLE/40 14 SECTION 6 DESIGN OF TENSION MEMBERS 6.1 Tension Members Tension members are linear members in which axial forces act to cause elongation (stretch). Such members can sustain loads upto the ultimate load, at which stage they may fail by rupture at a critical section. However, if the gross area of the member yields over a major portion of its length before the rupture load is reached, the member may become non-functional due to excessive elongation. Plates and other rolled sections in tension may also fail by block shear of end bolted regions (see 6.4.1). The factored design tension 7, in the members shal] satisfy the following requirement: T<T, where T, = design strength of the member. The design strength of a member under axial tension, T,, is the lowest of the design strength due to yielding of gross section, 7,,. rupture strength of critical section, Ty, and block shear Ty, given in 6.2, 6.3 and 6.4, respectively. 6.2 Design Strength Due to Yielding of Gross Section The design strength of members under axial tension, T,, as governed by yielding of gross section, is given by Tyg = Af, Novo where hy = yield stress of the material, A, = gross area of cross-section, and Yoo = partial safety factor for failure in tension by yielding (see Table 5). 6.3 Design Strength Due to Rupture of Critical Section 6.3.1 Plates The design strength in tension of a plate, 7,, as governed by rupture of net cross-sectional area, A,, at the holes is given by Tan = 0.9 Ag fa! Your where Ym! = partial safety factor for failure at ultimate stress (see Table 5), f, = ultimate stress of the material, and A, = net effective area of the member given by, NESP-O-CVLE/40 -| ne + , Bi] t b,t =width and thickness of the plate, respectively, where d, = diameter of the bolt hole (2 mm in addition to the diameter of the hole, in case the directly punched holes), & = gauge length between the bolt holes, as shown in Fig. 5, Pp, = staggered’pitch length between line of bolt holes, as shown in Fig. 5, n = number of bolt holes in the critical section, and i = subscript for summation of all the inclined ‘legs. ae a t Pen a Pas oes b S — aa ated Fic. 5 Plates with Botts HoLes IN TENSION 6.3.2 Threaded Rods The design strength of threaded rods in tension, T,,, as governed by rupture is given by Tan = 0.9 Ag fs! Your where A, = net root area at the threaded section. 6.3.3 Single Angles The rupture strength of an angle connected through one leg is affected by shear lag. The design strength, T,,, a8 governed by rupture at net section is given by: Ty, =0.9 Age fa! You + BA, £, mo where B = 1.4-0.076 (wt) Ff.) (B/L.) < FY m0 Yoar) > 07 where w = outstand leg width, b, = shear lag width, as shown in Fig. 6, and 15 TS 800 : 2007 L , = length of the end connection, that is the distance between the outermost bolts in the end joint measured along the load direction or length of the weld along the load direction. For preliminary sizing, the rupture strength of net section may be approximately taken as: Ty = Ay fa! Yu where 0.6 for one or two bolts, 0.7 for three bolts and 0.8 for four or more bolts along the length in the end connection or equivalent weld length; = net area of the total cross-section; = net area of the connected leg; R u = gross area of the outstanding leg; and = thickness of the leg. by=w bs=wew;-t Fic. 6 ANGLES WITH SINGLE LEG CONNECTIONS 6.3.4 Other Section The rupture strength, 7,, of the double angles, channels, I-sections and other rolled steel sections, connected by one or more elements to an end gusset is also governed by shear lag effects. The design tensile strength of such sections as governed by tearing of net section may also be calculated using equation in 6.3.3, where f is calculated based on the shear lag distance, b, taken from the farthest edge of the outstanding leg to the nearest bolt/weld line in the connected leg of the cross-section. 6.4 Design Strength Due to Block Shear The strength as governed by block shear at an end connection of plates and angles is calculated as given in 6.4.1. 6.4.1 Bolted Connections The block shear strength, Ty, of connection shall be taken as the smaller of, Ta = [Avg LB Ymo) + 0.9Ain fa Ym | or Tay = (0.9A vy f, (V3 Your) + Arg Sy Yoo ) ER:T.O. 1S 800 : 2007 where A\gvAy,= minimum gross and net area in shear along bolt line parallel to external force, respectively (1-2 and 3-4 as shown in Fig. 7A and 1-2 as shown in Fig. 7B), A, A,, = minimum gross and net area in tension from the bolt hole to the toe of the angle, end bolt line, perpendicular to the line of force, respectively (2-3 as shown in Fig. 7B), and Sf, = ultimate and yield stress of the material, : respectively. 6.4.2 Welded Connection The block shear strength, Ty, shall be checked for welded end connections by taking an appropriate section in the member around the end weld, which can shear off as a block. SECTION 7 DESIGN OF COMPRESSION MEMBERS 7.1 Design Strength 7.1.1 Common hot rolled and built-up steel members used for carrying axial compression, usually fail by flexural buckling. The buckling strength of these members is affected by residual stresses, initial bow and accidental eccentricities of load. To account for all these factors, the strength of members subjected to axial compression is defined by buckling class a, b, c, or d as given Table 7. 7.1.2 The design compressive strength P,, of a member is given by: P< Py where Py = Acfog ~-e <4 ee 7A Plate where A, = effective sectional area as defined in 7.3.2, and Sig = design compressive stress, obtained as per 7.1.2.1. 7.1.2.1 The design compressive stress, Sia, Of axially loaded compression members shall be calculated using the following equation: faa fc "leoee SON, IN ted Veo where @ = O5[1+a(A-0.2)+ A] 2 = non-dimensional effective slenderness ratio 2 = Vi/fe= yf, (KY) fee wE = Euler buckling stress = 2 Loc B (KA) where KL/r = effective slenderness ratio or ratio of effective length, KL to appropriate radius of gyration, r; &® = imperfection factor given in Table 7; X — = Stress reduction factor (see Table 8) for different buckling class, slenderness ratio and yield stress 1 = [o+(@-a)"] Amo = partial safety factor for material strength. 1 2h (ti aah leer 4. r) 3} Cnr | — Pn aetge ee: mae Eaba aise 7B Angle Fic. 7 Bock SHEAR FAILure NESP-O-CVLE/40 16 5. (a) i) (ii) SECTION—B What is smart concrete? Write the key features and benefits of smart concrete. 6 What is self-compacting concrete? How is it obtained? Explain the advantage and disadvantage of it. 6 (b) A vibration test is conducted on the model of a tank. A cable attached to the tank induces a force of 20 kN horizontally and pulls the tank by 5 mm. The cable is cut and the resulting vibration is recorded. At the end of four complete cycles, the time elapsed is 2 seconds and the amplitude is 0-5 mm. Compute the damping ratio, natural period of vibration (undamped), stiffness and damping coefficient. Also find the number of cycles required for the displacement amplitude to decrease to (35) of the initial amplitude. 12 (c) Explain the following terms, with sketches, pertaining to the masonry walls : ) (i) (iti) (w) ) (i (i) Solid wall with piers Cavity wall Faced wall Veneered wall 12 List the factors that affect strength of a steel column. 4 Briefly explain the possible modes of failure of axially loaded steel column. 8 (e) If the vertical shearing force acting on the thin-walled channel section shown in the figure below is 2000 N, compute and show the shear flow, and determine the shear centre : 100 mm 25 mm 7 K- 250mm 25 mm——] 25 mm 100 mm 12 NESP-O-CVLE/40 Lz Ee He 6. (a) The culvert shown below is of constant section throughout and the top beam is subjected to a central concentrated load of 25 kN. Assuming that the base pressure is uniform throughout, analyze the box culvert. Draw the bending moment diagram using moment distribution method : 25 kN #5 m5 m— B é c 10m APPR TR VAD 2:5 kN/m 20 (b) A rectangular beam 300 mm x600 mm is prestressed with parabolic cables having cross-sectional area 1200mm?. The parabolic cables have an eccentricity of 150mm at mid-span and zero eccentricity at the ends. The beam is simply supported over a span of 12 m and the initial prestress in the cables is 1100 MPa. Estimate the deflection of the beam due to initial prestress plus self-weight of the beam. Assume losses of prestress as 18%. The beam is subjected to a live load of 20 kN/m over its entire span. Estimate the final deflection. Also derive only the expression for deflection due to prestress of the parabolic cable. The unit weight of concrete is 25 kN/m*. Assume E, =206 kN/mm? and E, =35 kN/mm?. 20 (c) () An element in a stressed material has tensile stress of 400 MN/m? anda compressive stress of 300 MN/m? acting on two mutually perpendicular planes and equal shear stresses of 80 MN/m? on these planes. Find the principal stresses and position of the principal planes. Find also the maximum shearing stress. Solve using analytical method. 10 (i) A simply supported beam 150 mm wide and 220 mm deep is 4m long and carries a load of 20 kN at mid-span. The load is inclined at an angle of 30° to the vertical. The line of action is passing through the centroid of the section. Find the locations and magnitudes of maximum tensile and compressive stresses set up due to bending. 10 NESP-O-CVLE/40 18 T. (a) () What is ferrocement and fiber-reinforced concrete? Write their advantages and disadvantages. 8 (ii) Describe, in short, the various methods of proportioning concrete. 6 (ii) Write, in short, the various advantages of RC structures over other masonry structures. 6 (b) Asimply supported reinforced concrete (RC) slab, having a ciear span of 3 m, is supported only on two opposite sides on brick walls of 230 mm thick. If the live load on the slab is 3 kN/m? and floor finish being 1 kN/m?, design the RC slab as per limit state method. Use M 30 concrete and Fe 500 grade steel. Also check for deflection. Sketch the reinforcement details. Assume clear cover to the reinforcement as 20 mm. Relevant portion of the IS 456 is enclosed. 20 NESP-O-CVLE/40 19 [P-0. {S 456 : 2000 ta me (a) te FEEDEw ae 4 (ec) + (b) Fic. 2. Typica, Support Conprrions Por LocaTinc Factorep SHEAR Force but in no case greater than the breadth of the web plus _ structure or finishes or partitions. The deflection shall half the sum of the clea: distances to the adjacent beams generally be limited to the following: on either side, a) a) For T-beams, b, -2 +b, + 6D, b) For L-beams, 6, = a +b, + 3D, The final deflection due to all loads including the effects of temperature, creep and shrinkage and measured from the as-cast level of the supports of floors, roofs and all other horizontal members, should not normally exceed span/250. Z ’ 5 b) The deflection including the effects of c) For isolated beams, the effective flange width temperature, creep and shrinkage occurring after shall be obtained as below but in no case greater erection of partitions and the application of than the actual width: finishes should not normally exceed span/350 1 or 20 mm whichever is less. T~beam,b, <— +b, : CaS ‘s a) +4 23.2.1 The vertical defiection limits may. generally be b assumed to be satisfied provided that the span to depth ratios are not greater than the values obtained as below: L=beam,b, = OSA, 4b: a) Basic values of pen to effective depth ratios 1, Pei for spans up to 10 m: Cantilever 7 where Simply supported 20 b, = effective width of flange, Continuous 26 1, = distance between points of zero moments b) For spans above 10 m, the values in (a) may be in the beam, b, = breadth of the web, D, = thickness of flange, and 5 = actual width of the flange. c) NOTE — For continuous beams and fremes, ‘I, may be assumed as 0.7 times the effective span. 23.2 Control of Deflection The deflection of a structure or part thereof shall not adversely affect the appearance or efficiency of the 4) NESP-O-CVLE/40 20 multiplied by 10/span in metres, except for cantilever in which case deflection calculations should be made. Depending on the area and the stress of steel for tension reinforcement, the values in (a) or (b) shall be modified by multiplying with the modification factor obtained as per Fig. 4. Depending on the area of compression reinforcement, the value of span to depth ratio be further modified by muttiplying with the modification factor obtained as per Fig. 5. SECTION XX 19 456 : 2000 Fic. 3 TRANSVERSE REINFORCEMENT IN FLANGE OF T-BEAM WHEN Main REINFORCEMENT OF ) For flanged beams, the values of (a) or (b) be Stas 15 PARALLEL 70 THE BEAM modified as per Fig. 6 and the reinforcement percentage for use in Fig. 4 and 5 should be based NESP-O-CVLE/40 MODIFICATION FACTOR on area of section equal to 8, . 2-0 N 1-6 ‎ N [ ‎ [ ` 720 12 AAS IN NY rt ‎ 190 ¬ 08 (~ 15.29 = 04 Note: 1g IS STEEL STRESS OF SERVICE: LOADS N/mm? 04 08 1⁄2 16 20 24 28 30 Fig. 4 11001701710 Factor FoR TENSION REINFORCEMENT PERCENTAGE TENSION REINFORCEMENT f= 058 / Areaof cross-section of 9) required " Ageaof cross-section of steel provided 21 NOTE—When deflections are required to be calculated, the given in Annex © may be used. 976 NESP-O-CVLE/40 IS 456 : 2000 used the horizontal distance between bars of a group may be reduced to two-thirds the nominal maximum size of the coarse aggregate, provided that sufficient space is left between groups of bars to enable the vibrator to be immersed. Where there are two or more rows of bars, the bars shall be vertically in line and the minimum vertical distance between the bars shall be 15 mm, two-thirds the nominal maximum size of aggregate or the maximum size of bars, whichever is greater. c) 26.3.3 Maximum Distance Between Bars in Tension Unless the calculation of crack widths shows that a greater spacing is acceptable, the following rules shall be applied to flexural members in normal internal or external conditions of exposure. a) Beams — The horizontal distance between parallel reinforcement bars, or groups, near the tension face of a beam shall not be greater than the value given in Table 15 depending on the amount of redistribution carried out in analysis and the characteristic strength of the reinforcement. b) Slabs 1!) The horizontal distance between parallel main reinforcement bars shall not be more than three limes the effective depth of solid slab or 300 mm whichever is smaller. 26.4 Nominal Cover to Reinforcement 26.4.1 Nominal Cover Nominal cover is the design depth of concrete cover to all steel reinforcements, including links. It is the dimension uscd in design and indicated in the drawings. It shall be not less than the diameter of the bar. 26.4.2 Nominal Cover to Meet Durability Requirement Minimum values for the nominal cover of normal- weight aggregate concrete which should be provided to all reinforcement, including links depending on the condition of exposure described in 8.2.3 shall be as given in Table 16. 26.4.2.1 However for a longitudinal reinforcing bar in a column nominal cover shall in any case not be less than 40 mm, or less than the diameter of such bar. In the case of columns of minimum dimension of 200 mm or under, whose reinforcing bars do not exceed {2 mm, a nominal cover of 25 mm may be used. 26.4.2,2 For footings minimum cover shall be 50 mm. 26.4.3. Nominal Cover to Meet Specified Period of Fire Resistance Minimum values of nominal cover of normal-weight aggregate concrete to be provided to all reinforcement including links to meet specified period of fire resistance shall be given in Table 16A. 26.5 Requirements of Reinforcement for Structural Members 2) The horizontal distance between parallel 26.5.1 Beams reinforcement bars provided against : : shrinkage and temperature shall not be more 26:5-1-1 Tension reinforcement than five times the effective depth of a solid a) Minimum reinforcement—The minimum area of slab or 450 mm whichever is smaller. tension reinforcement shall be not less than that Table 15 Clear Distance Between Bars (Clause 26.3.3) 1; Percentage Redistribution to or from Section Considered -# -15 | 0 | +15 +30 Clear Distance Between Bars Nim? mm mm mm mm mm 280 215 260 300 300 300 415 125 155 180 210 235 500 105 130 150 175 195 NOTE — The spacings given in the table are not applicable to members subjected to particularly aggressive environments unless in the calculation of the moment of resistance, f, has been limited to 300 N/mm in limit state design and ¢, limited to 165 N/mm? in working stress design. 22 ANNEX G IS 456 : 2000 (Clause 38.1) MOMENTS OF RESISTANCE FOR RECTANGULAR AND T-SECTIONS G-0 The moments of resistance of rectangular and T-aections based on the assumptions of 38.1 are given in this annex. G-1 RECTANGULAR SECTIONS G-1,1 Sections Without Compression Reinforcement The moment of resistance of rectangular sections without compression reinforcement should be obtained as follows ; 8) Determine the depth of netutral axis from the following equation : aw 0.87 4 Ay d 0.36 fy bd b) If the value of x /d is less than the limiting value (see Note below 38.1), calculate the moment of resistance by the following expression : M, = 0.87 f, Ay af i) c) If the value of x/d is equal to the limiting value, the moment of resistance of the section is given by the following expression : : Mo sim = 0.36 7use. (: ~0.42 ume at 7m d) Ifx,/d is greater than the limiting value, the section should be redesigned. In the above equations, 4, = depth of neutral axis, d = effective depth, Bf = characteristic strength of reinforce- ment, A, = _ area of tension reinforcement, 4 = characteristic compressive strength of concrete, b = width of the compression face, M, im = limiting moment of resistance of a section without compression reinforcement, and pk limiting value of x, from 39.1. G-1.2 Section with Compression Reinforcement Where the ultimate moment of resistance of section a NESP-O-CVLE/40 23 exceeds the limiting value, M, .. compression teinforcement may be obtained from the following equation ; M, * My im Sg Ag (dd) where My +My qe @ are same as in Gel.1, f= design stress in compression reinforce- ment corresponding to a strain of Xu, max where 4, mx the limiting value of x, from 38.1, A, = _aseaofcompression reinforcement, and d’ == depth of compression reinforcement from compression face. The total area of tension reinforcement shall be obtained from the following equation : A,=A,, + Ay; where A, = area of the total tensile reinforcement, A,, = area of the tensile reinforcement for a singly reinforced section for M, ,,.. and Ay, = A, f,,/ 0.87 f,. G-2 FLANGED SECTION G-21 For x,< D,, the moment of resistance may be calculated from the equation given in G-1.1. G-2.2 The limiting value of the moment of resistance of the section may be obtained by the following equation when the ratio D, / d does not exceed 0.2 : M, no6 a8 -0.42 “anes } fe, by? $0.45 fig (b; —by) Dy (4-2) where M, %, mas @ and f, are same as in G-1.1, b, = breadth of the compression face/flange, b, = breadth of the web, and D, = thickness of the flange. [ P.T.O. (c) In a material, the principal stresses are 60 MN/m?, 48MN/m? and -36 MN/m?. Calculate : () Total strain energy (i) Volumetric strain energy (ii) Shear strain energy (iv) Factor of safety on total strain energy criterion if the material yields at 120 MN/m? Take E =200 GN/m?2 and 1 =0-3. m 8. (a) () A three-hinged stiffening girder of suspension bridge of span 130m is subjected to two point loads of 450 kN and 600 KN at distances 25 m from left support and 40 m from right support respectively. The dip of the cable is 12 m. Determine : (1) Maximum tension in the cable (2) Shear force and bending moment values for girder at 40 m from the left support (i) A parabolic arch has a span of 20 m and is supported at different levels such that the crown C is 12 m from left support A and 8m from right support B. The right support is higher than the left support by 2 m and the crown is higher by 1-5 m with respect to right support. The arch is hinged at the two supports and at the crown. Find the bending moment in the arch at a section Q, 4-5 m from the left support. (b) A rectangular cantilever reinforced concrete (RC) beam of 300 mm x600 mm cross-section has an effective span of 3m. It is subjected to a dead load (self-weight plus floor finishes) of 35 kN/m and a live load of 1-5kN/m at service state. The beam is reinforced with four rebars (reinforcing bars) of 25mm diameter in tension zone and two rebars of 20mm diameter in compression zone. Assume effective cover to both tension and compression reinforcement as 50mm. Use M25 concrete and Fe500 grade steel. Estimate only the ‘initial plus creep’ deflection due to permanent loads. Creep coefficient = 1-6, E, = 5000) fc. . Flexural strength of concrete f,, =0-7/fx,- Relevant portion of the IS 456 is enclosed. NESP-O-CVLE/40 24 20 10 10 20 IS 456 : 2000 ANNEX C (Clauses 22.3.2, 23.2.1 and 42.1) CALCULATION OF DEFLECTION C-1 TOTAL DEFLECTION C-1.1 The total deflection shall be taken as the sum of the short-term deflection determined in accordance with C-2 and the long-term deflection, in accordance with C-3 and C-4. C-2 SHORT-TERM DEFLECTION C-2.1 The short-term deflection may be calculated by the usual methods for elastic deflections using the short-term modulus of elasticity of concrete, E, and an effective moment of inertia /,, given by the following equation: Legg = z 3 but by H 12-2. Md For continuous beams, deflection shall be calculated using the values of /, I, and M, modified by the following equation: xX, =h [4] +(1- &)X, where X, = modified value of X, X,,X, = values of X at the supports, X, = value of X at mid span, k, = coefficient given in Table 25, and X = value of /,, /, or M, as appropriate. C-3 DEFLECTION DUE TO SHRINKAGE C-3.1 The deflection due to shrinkage a, may be d}b computed from the following equation: I Slag SIy a, =k, YP where where : k, is a constant depending upon the suppor. {, = moment of inertia of the cracked section, ” conditions, Sort 0.5 for cantil M, = cracking moment, equal to SE where saa eae ' t 0.125 for simply supported members, J, is the modulus of rupture of concrete, 0.086 for members continuous at one end, /,, is the moment of inertia of the gross and section about the centroidal axis, : neglecting the reinforcement, and y, is the 0.063 for fully continuous members. distance from centroidal axis of gross $s Sates " ato k 2 section, neglecting the reinforcement, to Vos HaRRGaEEn aeette F, D extreme fibre in tension, where €,, is the ultimate shrinkage strain of concrete M = maximum moment under service loads, (see 6.2.4), pee el k= 0.72% AF < 1.0 for0.25<P-P.<1.0 x = depth of neutral axis, IP d = effective depth, b, = breadth of web, and =0.65 x apts 1.0 for P - P.2 1.0 b = breadth of compression face. A Table 25 Values of Coefficient, k, (Clause C-2.1) im 0.5 of less 06 07 08 09 10 ia 12 3 14 cA 0) 0.03 0.08 0.16 0.30 0.50 073 091 0.97 10 NOTE — k, is given by Nica M, +i, My, +My, where M,,M, = suppor moments, and M,,,M,, = fixed end moments. NESP-O-CVLE/40 25 [P0: where p= ee and P, = Ae and D is the total depth of the section, and / is the length of span. C-4 DEFLECTION DUE TO CREEP C-4.1 The creep deflection due to permanent loads a, (perm) 4. (perm) 7 NESP-O-CVLE/40 4, ice (perm) — Qi cresm may be obtained from the following equation: 26 where 4 cc perm) 1 (perm) IS 456 : 2000 = initial plus creep deflection due to permanent loads obtained using an elastic analysis with an effective modulus of elasticity, at @ being the creep coefficient, and = short-term deflection due to permanent load using E. . (c) () Describe the different types of contract in brief. How is a tender document prepared? 10 (i) Estimate the number of carriers, if the data for the project are the following : NESP-O-CVLE/40 Quantity of material to be handled = 5000000 m? Capacity of the loaders to be engaged = 2-3 m° Capacity of bottom dampers = 30 m? Project to be completed in two shifts in 5 years with yearly working hours = 2000 hr Job and management factor = 0-70 Operating efficiency = 0-85 Bucket fill factor = 0-85 Swell factor = 0-90 Cycle time for loader = 0-50 minute Lead distance = 6 km Speed during empty haul @ 25 km/hr and loaded haul @ 20 km/hr 10 kkk 27 26BS—2064