QP-ESEM-26-220626-CIVIL-ENGINEERING-PAPER-I

Engineering Services (Main 6 (M laaitser DETACHABLE Examination, 2026 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. Questions No. 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. 1 SNCS-O-CVLE 1.(a) 1.(b) 1.(c) 1.(d) 1.(e) (nisM) asotvis2 aninssrignd AcE KHON’ AB, a vertical post of a crane as shown in figure, is hinged at A and B. The horizontal arm CD is hinged to AB at C and supported by the strut EF which is hinged at E and F. Draw the bending moment diagram for AB and CD. Va 7 3m 18kN , | + oF D kK——— 4m ah 3m ——>| 3m +O 3m _ B 12 Explain Fibre Reinforced Plastics (FRPs). Draw a comparison between different type of FRPs based on their composition, engineering properties and their civil engineering applications. 12 There are two circular beams of same length, same material and of same weight. One of these is solid with diameter D and the other one is hollow with outer diameter D, and inner diameter D;. Show that the ratio of section modulus of solid beam to that of the hollow beam is (Pes 12 a) + i What are the different types of failure of black bolts ? Explain with sketches. 12 A short column of a 4 m effective height is subjected to an ultimate load of 1500 KN. The column is of 400 mm diameter. The column is reinforced by 6 bars of 20 mm of Fe-415 grade. If M-30 concrete is used, obtain the ultimate moment carrying capacity of the column. The nominal cover to the longitudinal bar is 40 mm. 6-20mm¢ (Circular cross-section of 400mm dia) f;_—___} 12 400mm SNCS-O-CVLE 2 2.8) 2.൪൫) _ ગા 56 COMPRESSION WITH BENDING - Circular Section 11 7 २58 1, «હોડ N/mm? | 6/0 = 0-10 1-0 રડ ‘ভৰ ৰ ০০ 0-8 ت8‎ = న్‌ छो [ | خش کے‎ 7९7 ट ?® 8 £= SAY < 06 ଏ હ St os ےل ایاپ‎ ୪୪ ৮ Erie 2 کم‎ வாண்ட்‌ 4 ® aN لاٹ ریا‎ प्नं य 23 (६ ୯ ५ RAN کاچ ا یہ‎ பண்ட 1 ९.५: ન ୦ ଏ शि 12 ਢੇ ® تا‎ ۱9۸۷5۱۱ 92 [1 + 111 01 711 [7 ۲ 7 ப 74! 4 > > / 7ھ‎ 1 ആ 0-08" 0 ૦-૬ 0-20 0-25 દય Determine the horizontal displacement of support D of the frame shown in figure. Take उ = 200 GPa and I = 300x10~ 114. B 0 | ୩ 60 2m E 2m | A D 20 शि Sm + ದ ન the classification of building lime based on IS 712 : 1973 specifically mentioning different classes, their names and uses in construction. Also enlist the field tests to be carried out on building lime as recommended by IS 1624 : 1986. 10 3 SNCS-O-CVLE ee ee ee 2.(b)(ii) What is cold weather concreting ? Discuss the effects of cold weather on concreting. Explain in brief, various measures recommended by IS 7861 (Part-II) : 1981 to deal with the problems of cold weather concreting. 10 2.(c) A T-shaped retaining wall for a height 4m above ground level is as shown in figure below. The angle of repose of earth is 30° and density 18 KN/m?. The safe bearing capacity of soil is 100 KN/m? and coefficient of friction between concrete and soil is 0-55. Apply all the stability checks. Also design the stem if M-25 concrete and Fe-415 bars are used. Show details of reinforcement in stem. Consider nominal cover to steel as 75 mm. 200 (All dimensions vd are in mm) aoe y= 18kN/m3 4000 $= 30° GL 7 PEN 1200 500 L 1500 p 500 1 1000 | 20 4 4 4 4 3.(a) A bracket plate is welded to the flange of a column of ISMB 600 as shown in figure below. Determine the load carrying capacity of the joint for 10mm fillet weld. Assume E250 grade steel and shop welding. Use limit state method. y,,,= 1-25. P it p1S01 Imm |, 200mm 4 TT YL 300mm ALIN 20 SNCS-O-CVLE 4 3.(b) Determine the member forces in the truss as shown in the figure. The members in tension have a cross-sectional area a and the members in compression have a cross- sectional area 2a. Take E = constant. 50kN 30° 30° oO 30 30° c & & 3.(c) The network of a certain project is shown in figure given below, with the estimated duration of various activities. Determine the following : 20 (i) Earliest event time and latest event time. (ii) Earliest and latest start and finish times of each activity. (iii) Total and free floats for each activity. (iv) Critical path for the network. 20 5 SNCS-O-CVLE tree |] The floor plan of a building is as shown in figure below. The floor load consists of a live load of 5 KN/m~ and dead load of 2 kN/m (excluding self weight). Design the slab thickness and reinforcement required at the various critical sections. Use Fe-500 steel and M-20 concrete. Assume that the structure is subjected to mild exposure conditions. 20 --.-:--.:--.-. „ -. |, | T T | | | Interior, Beams =>, | nterior, Beams 1 | Edge ~ | KN ‎ | | | 2 Edge - - !| | Beam [~ + | | of Beam | | 1 | | | | | || -_ ‎ ~... 3500 3500 300 | 3500 | | 1 1 1 7 (All dimensions are in mm) Table 12 Bending Moment Coefficients (Clause 22.5.1) Type of Load Span Moments Support Moments -----------------.. =... ચારા Near Middle At Middle At Support At Other ` of End Span of Interior Next to the Interior Span 800 Support Supports (1) (3) (4) ) Dead load and imposed + + _ 1. _ 1 load (fixed) 12 16 10 12 sed imposed load (not 7 a + 1 1 fixed) 10 12 ‎ NOTE — For obtaining the bending moment, the coefficient shall be multiplied by the total design load and effective span. Table 13 Shear for Coefficients (Clauses 22.5.1 and 22.5.2) Type of Load At End At Support Next to the At All Other Support End Support Interior Supports Outer Side 11055510 (1) (2) ˆ ) (4) Dead 1020 and imposed 0.4 9.6 0.55 0.5 load (fixed) Imposed load (not 0.45 0.6 0.6 0.6 fixed) NOTE — For obtaining the shear force, the coefficient shall be multiptied by the total design load. SNCS-O-CVLE 6 22.6 Critical Sections for Moment and Shear 22.6.1 For‘monolithic construction, the moments computed at the face of the’supports shall be used in the design.of the members at. those sections. For non- monolithic construction the design of the member shall be done keeping in view 22.2. 22.6.2. Critical Section for Shear The shears computed at the face of the support shall be used in the design of the member at that section except as in 22.6.2.1. 22.6.2.1 When the reaction in the direction of the applied shear introduces compression into the end region of the member, sections located at a distance less than d from the face of the support may be _ designed for the same shear as that computed at distance d (see Fig. 2). NOTE—The above clauses are applicable for beams generally carrying uniformly distributed load or where the principal load is located farther than 2d from the face of the support. 22.7 Redistribution of Moments Redistribution of moments may be done in accordance with 37.1.1 for limit state method and in accordance with B-1.2 for working stress method. However, where simplified analysis using coefficients is adopted, redistribution of moments shall not be done. IS 456 : 2000 23 BEAMS 23.0 Effective Depth Effective depth of a beam is the distance between the centroid of the area of tension reinforcement and the maximum compression fibre, excluding the thickness of finishing material not placed monolithically with the member and the thickness of any concrete provided to allow for wear. This will not apply to deep beams. 23.1 T-Beams and L-Beams 23.1.1 General A slab which is assumed to act asa compression flange of a T-beam or L-beam shall satisfy the following: . a) The slab shall be cast integrally with the web, or the web and the slab shall be effectively bonded together in any other manner; and- b) Ifthe main reinforcement of the slab is parallel to the beam, transverse reinforcement shall be provided as in Fig. 3; such reinforcement shali not be less than 60 percent of the maia reinforcement at mid span of the slab. 23.1.2 Effective Width of Flange In the absence of more accurate determination, the effective width of flange may be taken as the following ~~ --- 7-7 | | « ] BEAM SECTION XX Fic. 3 TRANSVERSE REINFORCEMENT IN FLANGE OF T-BEAM WHEN MAIN REINFORCEMENT OF SLAB IS PARALLEL TO THE BEAM it SNCS-O-CVLE . e) For flanged beams, the values of (a) or (b) be on area of section equal to 6, . modified as per Fig. 6 and the reinforcement percentage for use in Fig. 4 and 5 should be based. NOTE = Wien: ‎ method given in Annex © may be used. 2-0 | 1 ‌ || | [ | Rw ‎ 720 ` 12 rN ‌ = ‎ ¬ [ ‎ 190 - ¬ 2 ‎ [| - 2- = ae < | r 55290 ‎ | || | 04 4 Note: fg 15 STEEL STRESS OF SERVICE LOADS N/mm? | ' | 0 04 08 12 1-6 20 24 28 30 PERCENTAGE TENSION REINFORCEMENT £ Areaof cross - section of steel required + = 0.58 - - 7 Areaof cross-section of steel provided Fic. 4 Mopirication Factor FoR TENSION REINFORCEMENT 4.(b Determine the plastic moment capacity for the frame of uniform cross-section shown pacity in figure below under applied factored load. [ ] | ‎ 1001- 8 & ail 7777 20 0+41 - 6m SNCS-O-CVLE 8 4.(c) 5.(a) 5.(b) 5.(c) 5.(d) A circle of 60 mm is marked on a mild steel plate prior to the application of stresses on the plate. The plate is stressed as shown in figure. Find the lengths of the major and minor axes of an ellipse formed as a result of the deformation of the circle marked. Take E = 2x10° N/mm? and Poisson’s ratio, p= 0-25 30N/mm a D Cc | 50N/mm* : I I LOIN rat et wt Pies Gin Nps ee 100 N/mm? \ \ ' 2 50N/mm° A [> B v 30N/mm? 20 SECTION ‘B’ Design a bolted connection for double angle member ISA 100*100x10 in a truss connected on either side of a 10 mm thick gusset plate. The length of the member is 2-5m. Use 16 mm dia bolt of grade 4-6. Assume net area at threads is 0-85 times area at shunk. y,,= 1:25, K,=0-6. The factored loads in the member are 175 kN (Compression) and 150 kN (Tension). 12 What is preservation of timber ? Explain various methods of preservation of timber. 12 A beam of square section is used with one diagonal horizontal. The beam is subjected to a shear force F at a section. Find the maximum shear stress in the cross-section of the beam and draw the shear stress distribution diagram for the section. 12 A post tensioned beam of span 8 m is pre-stressed by a parabolic cable, concentric at the support and having an eccentricity of 240 mm at midspan. The cross-sectional area of high tensile wire of the cable is 600 mm2. The wires are stressed by using a jack at the left end so that the initial force in the cable at the right end is 260 KN. Calculate the jacking force at the left end and the total losses of stress in wires. 9 SNCS-O-CVLE 5.(e) 6.(a) 6.(b) Assume : Coefficient of curvature effect = 0-55 Coefficient for wave effect = 0-003/m Relaxation of steel stress = 4% Shrinkage of concrete = 0-0002 Creep coefficient = 2:2 Anchorage slip at jacking end = 3 mm E, = 210 kN/mm? and E, = 35 kN/mm? Size of beam = 250x750 mm (bxD) 12 Discuss various safety measures to be taken for workers while working at height. 12 A cantilever has a quadratal shape as shown in figure with a radius of curvature of R. It carries a vertical downward load P at its free end 4, while its other end is fixed. If EI is constant, find horizontal and vertical deflection of the free end A. A 20 B A factory shed is to be constructed using pitched roof trusses supported on R.C. columns. The spacing of the truss is 5 m. Purlins are placed at the nodes (Refer figure below). Design the purlins with circular hollow section for the unfactored loading as follows : Dead load = 180 N/m Live load 920 N/m Wind load = 2080 N/m (suction) Assume E250 grade steel. Apply limit state method of design. Assume sections are compact. Yyn9 = 1:1. Draw also the connection with the rafter (100 mm Nominal bore dia). Purlin ll 3m 16m 77177 T7177 SNCS-O-CVLE 10 6.(c) Given : Nominal Outer dia. Thickness bore dia. (mm) (mm) (mm) 50 60:3 3-6 65 76-1 3-6 80 88-9 4-0 90 101-6 4-0 100 114-3 4-5 Moment of Inertia (cm*) 25-9 54-0 96-4 146-3 234-3 Plastic section Modulus (cm?) 11-6 19-0 28-8 38-1 54-3 20 Design a circular water tank of capacity 350 m3 resting on ground. The base of wall is fixed with base slab. Height of Tank is restricted to 3-4 m including free board of 300mm. The materials used are M-30 concrete and Fe-500 steel. Use IS-3370 (PartIV) method. Adopt working stress design approach. Use wall thickness of 200 mm. Show details of reinforcement. 20 Table Shear at the Base of Cylindrical Wall { Yh? (triangular foad) Shear V = coefficient x4 pH (rectangular load) (M/H (moment at base) HIDt Triangular load, Rectangular load, Triangular or Moment fixed base fixed base rectangular load, at edge hinged base 04 +0.436 +0.755 +0.245 -1.58 08 +0.374 +0.552 +0,234 -1.75 42 +0.339 +0.460- +0.220 ~2.00 1.6 +0.317 +0.407 +0.204 2.28 2.0 +0.299 +0.370 +0.189 (72.57 3.0 +0.262 +0.310 +0.158 -3.18 40 +0.236 +0.271 +0.137 ~3.68 5.0 +0.213 +0.243 +0.121 —4.10 6.0 +0.197 +0.222 +0.110 4.49 8.0 +0.174 +0.193 +0.096 . -5.18 10.0 +0.158 +0.172 +0.087 . 5.81 12.0 +0.145 +0.158 +0.079 +6.38 414.0 +0.135 +0.147 +0.073 6.88 16.0 +0.127 +0.137 +0.068 -7.36 Notes: 1. , Density of the liquid 2. Positive sign indicates shear acting inward. 11 SNCS-O-CVLE Table Hoop Tension in Cylindrical Walls Fixed at Base and Free at Top (IS: 3370-1967, Part IV) Hoop tension T = (Coefficient) x (,H “HEIDE Coefficients at point. 0.0H 0.1H 0.2H 0.3H 0.4H 0.5H 0.6H 0.7H 0.8H 0.9H 04 +0.149 +0.134 +0.120 +0.101 +0.082 +0.066 +0.049 +0.029 +0.014 +0.004 08 +0.263 +0.239 +0.215 +0.190 +0.160 +0.130 +0.096 +0.063 +0.034 +0.010 4.2 40,283 40.271 +0.254 +0.234 +0.209 +0.180 +0.142 +0.099 +0.054 +0.016 1.6 +0.265 +0.268 +0.268 +0.266 40.250 +0.226 +0.185 +0.134 +0.075 +0.023 2.0 +0.234 +0.251 +0.273 +0.285 +0.285 40.274 +0.232 +0.172 +0.104 +0.031 3.0 +0.134 40.203 +0.267 +0.322 +0.357 +0.362 +0.330 +0.262 +0.157 +0.052 4.0 +0.067 +0.164 +0.256 +0.339 +0.403 +0429 +0.409 +0.334 +0.210 +0.073 5.0 +0.025 +0.137 +0.245 +0.346 +0.428 +0.477 +0.469 +0.398 +0.259° +0.092 . 6.0 +0.018 +0.119 +0.234 +0.344 +0.441 +0.504 +0.514 +0.447 +0.301 +0.112 8.0- -0.011 +0.104 +0.218 +0.335 +0.443 +0.534 +0.575 +0.530 +0.381 +0.151 10.0 -0.011 +0.098 +0.208 +0.323 +0.437 +0.542 +0608 +0.589 +0.440 +0.179 12:0 -0.005 +0.097 +0.202 +0.312 +0.429 +0.543 .+0.628 +0.633 +0.494 +0.211 14.0 -0.002 +0.098 +0.200 +0.306 +0.420 +0.539 +0.639 +0.666 +0.541 +0.241 16.0. -0.000 +0.099 +0.199 +0.304 +0412 +0.531 +0.641 +0.687 +0.582 +0.265 “Note: Positive sign indicates tension. Table Tension in Circular Ring Wall, Hinged Base, Free Top and Subjected to Triangula Load Tension T = Coefficient x y,Hr H?/Dt Coefficients at point 0.0H O1H 02H 03H 04H 05H 06H 0.7H 08H 0.9H 0.4 +0.474 +0.440 +0.395 +0.352 +0.308 +0.264 +0.215 +0.165 +0.111 +0.057 0.8 +0.423 +0.402 +0.381 +0358 40.330 +0.297 +0.249 +0.202 +0.145 +0.076 42 +0350 +0.355 +0.361 +0.362 +0.358 +0.342 +0309 +0.256 +0.186 +0.098 41.6 +0.271 +0.303 +0.341 +0369 +0.385 +0.385 +0.362 +0.314 +0.233 +0.124 2.0 +0.205 +0.260 +0.321 +0.373 +0.411 +0.434 +0.419 +0.369 +0.280 +0.151 3.0 +0.074 +0.179 +0.281 +0.375 +0.449 +0.506 +0.519 +0.479 +0.375 +0.210 4.0 +0.017 +0.137 +0.253 +0.367 +0469 +0.545 +0.579 +0.553 +0.447 +0.256 5.0 -0.008 +0.114 +0.235 +0.356 +0469 +0.562 +0617 +0.606 +0.503 +0.294 6.0 --0.011 +0.103 +0.223 +0.343 +0463 +0.566 +0.639 +0.643 +0.547 +0.327 8.0 -0.015 +0.096 +0.208 +0.324 +0.443 +0.564 +0.661 +0697 +0.621 +0.386 40.0 —0.008 +0.095 +0.200 +0.311 +0.428 +0.552 +0.666 +0.730 +0.678 +0.433 12.0 ~0.002 +0.097 +0.197 +0.302 +0.417 +0.541 +0.664 +0.750 +0.720 +0.477 14.0 0.000 +0.098 +0.197 +0.299 +0.408 0.531 +0.659 +0.761 +0.752 +0.513 46.0 +0.002 +0.100 +0.198 +0.299 +0.403 +0.521 +0.650 +0.764 +0.776 +0.536 Notes: 1. y, is the density of the liquid. 2. Positive sign indicates tension. SNCS-O-CVLE 12 Table Bending Moment in Cylindrical Walls Fixed at Basé and Free at Top (IS: 3370-1967, Part IV) Moment M = (Coefficient) x y,H® H?/Dt Coefficients at point 0.1H 0.2H 03H 04H 0.5H 0.6H 0.7H 0.8H 0.9H 1.0H 04 +0.0005 +0.0014 +0.0021 +0.0007 -0.0042 -0.0150 -0.0302 -0.0529 -0.0816 -0.1205 0.8 +0.0011. +0.0037 +0.0063 +0.0080 +0.0070 +0,0023 -0.0068 ~0.0224 -0.0465 -0.0795 1.2 40.0012 +0.0042 +0.0077 +0.0103 +0.0112 +0.0090 +0.0022 -0.0108 -0.0311 -0.0602 16 +0.0011 +0.0041 +0.0075 +0,0107 +0.0121° +0.0111 +0.0058 ~0.0051 -0.0232 -0.0505 20 +0.0010 +0.0035 +0.0068 +0.0099 +0.0120 +0.0115 +0.0075 ~0.0021 -0.0185 -0.0436 3.0 +0.0006 +0.0024 +0.0047 +0.0071 +0.0090 +0.0097, +0.0077 +0.0012 -0.0119 -0.0333 4.0 +0.0003 +0.0015 +0.0028 +0,0047 +0.0066 +0.0077 +0.0069 +0.0023 -0.0080 -0.0268 5.0 +0.0002 +0.0008 +0.0016 +0.0029 +0,0046 +0,0059 +0.0059 +0.0028 -0.0058 -0.0222 6.0 +0.0001 +0.0003 +0.0008 +0,0019 +0.0032 +0.0046 +0.0051 +0.0029 -0.0041 -0.0187 8.0 0.0000 +0.0001 +0.0002 +0.0008 +0,0016 +0.0028 +0.0038 +0.0029 -0.0022 -0.0146 10.0 0.0000 +0.0000 +0,0001. +0.0004 +0.0007 +0.0019 +0.0029 +0,0028 -0.0012 0.0122 12.0 0.0000 ~0.0001 +0.0001 +0.0002 +0.0003 +0.0013 +0.0023 +0,0026 -0.0005 -0.0104 140 0.0000 0.0000 0.0000 0.0000 0.0001 +0.0008 +0.0019 +0,0023 ~0.0001 -0.0090 16.0 0.0000 0,0000 -0.0001 -0.0002 -0.0001 +0:0004 +0.0013 +0,0019 +0.0001 -0:0079: Note: Positive sign indicates tension at the outside face. 7.(a)(i) 7.(a)(ii) 7.(b) A beam of 200 mm x 200 mm section is 4-0 m long and is simply supported at its ends. The beam fails under bending when a point load of 200 KN is applied at the centre of the beam. Find the intensity of the uniformly distributed load which will cause failure of 5-0 m long cantilever of same material having cross-section of 400 mm wide and 500 mm deep. 10 Determine the output of an angle dozer with a blade width=3-35 m, blade height = 0-855 m, forward speed = 9-4 kmph on a haul road 90 m long consisting of earth and gravel (swell factor = 0-83). The efficiency of dozer is 80% and its working time is 50 min/hr. The job conditions are good and management conditions are fair (Factor for these conditions may be taken as 0-69). 10 An unstiffened suspension cable carries a uniformly distributed load of 65 kN/m over a span of 30m as shown in the figure. The suspension cable is supported on frictionless rollers, fixed to the piers. The anchor cables are inclined at 30° to horizontal. One pier is 4-5 m below the other and the maximum dip of the cable at the lowest point (C) is 3m below the lower pier. Calculate (i) the maximum and minimum tension in the cable. (ii) the horizontal and vertical forces at support A. 13 SNCS-O-CVLE cable | (ల उपमा ആ 20 7.(6) Design a combined footing with a strap beam for two reinforced concrete columns 400 mm x 400 mm size spaced 5 m apart and each supporting a factored load of 1000 KN. Assume the safe bearing capacity of soil as 100 KN/m?. Maximum length of footing is restricted to 8-0 M. Adopt M-25 concrete and Fe-500 Bars. Nominal cover for foundation shall be considered as 50 mm. Load factor considered is 1-5. Show details of reinforcement. 20 TABLE 3 FLEXURE — REINFORCEMENT PERCENTAGE, 2 FOR SINGLY REINFORCED SECTIONS நீடி = 25 N/mm? , fy Niomt ‎ 4 — 0 Nim 50258 40 50 | mm 20 2590 48 500 030 916 091 004 003 0000 | 255 1415 1358 0818 0708 0672 035 009 0085 0062 | 28 148 1390 087 974 040 0195 0188 0113 0098 0-094 265 1482 1422 0-857 0741 0711 045 0220 0211 0127 110 0106 270 1515 1-455 0876 0758 04 050 0245 26 ০10 ০13 018 | 275 159 147 ୧୮୨୫ 075 0144 055 0271 0260 0156 0135 0130 280 1584 1520 0916 0192 076 060 0296 0284 171 0148 142 2:85 1618 1554 0809 977 085 0321 "309 0186 0161 0154 290 1653. 1587 0956 0-827 0794 070 034 0333 0201 0114 0167 295 1-689 1621 0844 0811 675 9373 0358 0216 0186 0179 3-00 1724 1-655 0997 0862 01 080 0399 0383 0231 0199 0191 305 1760 1690 1018 0880 085 0425 0408 0246 0212 0204 3-10 1797 1725 1039 0-898 0863 990 . 0451 0433 26 0: 0216 315 1834 1-760 1-061 0917 095 0477 0458 0276 0239 0229 3:20 1871 1-796 1-082, 0936 0898 100 0483 0291 7 242 3-25 1909 1832 1104 0954 0916 105 0-307 0264 0255 330 1-947 148 1126 0973 0935 110 0557 535 0322 0279 0267 332 1-962 1-884 1135 0-981 0942 ris 0561 0338 0292 0-280 334 1978 1899 1144 0989 120 0611 0587 0353 0306 0293 336 1993 1914 1153 125 0638 0613 0369 0319 9: 38 2009 1929 1162 130 0666 0639 35 0333 340 2025 1944 1171 135 0693 "666. 0401 0347 0333 342 2040 1959 1180 140 0721 0417 344 2056 1974 1189 145 0749 0719 0433 0374 03: 346 2 1-989 150 0777 0746 0449 0388 0373 348 2088 2005 155 5 0-773 0466 3 0387 350 2104 2020 160 0834 0 0-482 0417 0-400 3-52 2120 2036 165 62 0-828 0499 0431 0414 3:54 2131 2051 170 0891 0515 0446 3.56 2153 2067 175 0420 0883 0532 0460 0442 338 170 2083 189 0949 911 : 0475 0456 360 2186 2099 18 శ్లో ९ 056 049 0470 | 362 2203 2115 190 1009 0968 0504 0484 3:64 2219 2131 195 1038 0997 1 0519 366 2236 2147 39 198 1026 ‎ 0334 53 | 368 223 2163 205 1099 105 065 0549 0527 | 370 2270 ಟ್ಟಿ 210 1129 1084 0-653 9565 3-72 2287 2196 215 1160 1114 0-671 0580 0357 3-74 2304 220 1191 1143 0-689 0596 225 122 1173 0-707 611 0587 230 1-254 1-204 0-725 0602 235 1285 1-234 0743 0643 9.617 240 1317 1-265 0762 0-659 0-632 245 1350 1-296 9781 0675 0648 250 132 137 98. 98 807% — Blanks indicate inadmissible reinforcement percentage (see Table 15). SNCS-O-CVLE 14 ୩୮୫୨ ~~ ~~~ 8४838 2४४४४ 83535 3333 ৪ 61 SHEAR — DESIGN SHEAR STRENGTH OF CONCRETE, = N/mm* 25 30 033 33 039 040 045 0:45 49 050 53 : 56 057 059 06 98 064 066 0:66 0:69 070 71 72 72 074 074 076 0-76 78 077 080 0-79 ! 3 082 3 086 4 687 6 08 0:87 0:88 91‎ 9 % 093 091 094 0-92 095 092 096 A built up column of effective length 5m consists of two 151/0 250 (back to back). The sections are laced. Find the spacing of the sections for maximum axial compressive load capacity. Also design the single lacing system using bars connected to the channels by weld (shop). Draw the lacing system also. For ISMC 250, A = 38-67 cm’, I, = 3817 05, Iy, 219 , Cy, = 23 cm. Assume lacing bars are at 50° with the vertical. Given : ‎ 40 50 | 60 | 70 | | % | 100 | 110 | 120 0 (MPa)} 198 | 183 | 168 | 152 | 136 | 121 107 95 84 20 Discuss the important features of common post tensioned systems in prestressed concrete. 10 Calculate the moment of resistance of a T-beam as shown in figure given below. The grade of concrete is M-25 and Grade of steel is Fe-500. . 1000 1 _ ~ 6 bars of 2510011 300 (All Dimensions are in mm) 15 10 SNCS-O-CVLE IS 456 : 2000 ANNEX G (Clause 38.1) MOMENTS OF RESISTANCE FOR RECTANGULAR AND T-SECTIONS G-0 The moments of resistance of rectangular and ‘T-sections 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 : a) Determine the depth of netutral axis from the 0.003 5 (1-6) % , max where: = «the limiting value of x, from 38.1, 4 = areaof compression reinforcement, and d’ = depth of compression reinforcement from compression face. The total area of tension reinforcement shall be obtained from the following equation following equation : `. 4„=4„,+4, where , ± = 4, = of the total tensile reinforcement, 136 74 8, Aq, = of the tensile reinforcement for ) If the value of x,/d is less than the limiting singly reinforced section for M, value (see Note below 38.1), calculate the and moment of resistance by the following 10.87 expression : Aga = Ate! 0.871 Aah = 0.87 - My, ty An {i ©) | If the value’ 2/ is equal to the limiting value, the moment of resistance of the section is given by the following expression : ce = = 0) Ifx,/dis greater than the limiting value, the section should be redesigned. ) -2 FLANGED SECTION G-2.1 For x,< 20 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 06 obtained by the following equation when the ratio D, 1d does not exceed 0.2 : x 4. = 0.36 = ( -0.42 d 0-0 7 bd? + 0.45 fig (bp - ) 0) (4-3) In the above equations, =, x, = depth of neutral axis, bere’ : d = effective depth, × ‎ as in G-1.1, 23 = characteristic strength of reinforce- £ = breadth of the compression face/flange, [1 8, = breadth of the web, and A, = area of tension reinforcement, 2 of the flange. fg = characteristic compressive strength ( thickness of concrete,’ £ G-2.2.1 When the ratio D,/d exceeds 0.2, the moment ‌ 957 of resistance of the section may be calculated by the M, = = limiting moment of resistance of following equation : a section without compression reinforcement, and = limiting value of x, from 39.1. G-1.2 Section with Compression Reinforcement ‘Where the ultimate moment of resistance of section exceeds the limiting value, compression reinforcement may be obtained from the following ose [1 042. ‎ 032 +045.60%-) -} where y, = (0.15 x, + 0.65 0), but not greater than equation and the other symbols are same as in G-1.1 M, = یزحیے ں۸‎ ర (డగ డీ) (क య” < G-2.3 For x, > ±, > D, the moment of resistance may be calculated by the equations given in 6-22 M,.M, dare same as in 6.11 when ‎ does not exceed 0.43 and G-2.2.1 when f= design stress in compression reinforce- ment corresponding to a strain of D/x, exceeds 0.43; in both cases substituting , ‎ 9. 8.(c) A hollow marine propeller shaft turning at 120 rpm is required to propel a vessel at 16 m/s at the expense of 6500 kW of shaft power. The efficiency of the propeller is 70%. The ratio of the outer diameter to inner diameter of the shaft is 2. If the direct stress due to thrust is not to exceed 10 N/mm‘, calculate the following : (i) Outer and inner diameter of the shaft (ii) Maximum shearing stress due to torque 20 SNCS-O-CVLE 16