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MAHARAS HTRA S TATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (IS O/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2017, Sub. Code : - 17604, , Subject: Design of RCC Structures, , --------------------------------------------------------------------------------------------------------------Important Instructions to examiners:, 1) The answers should be examined by key words and not as word-to-word as given in the model answer, scheme., 2) The model answer and the answer written by candidate may vary but the examiner may try to assess the, understanding level of the candidate., 3) The language errors such as grammatical, spelling errors should not be given more importance. (Not, applicable for subject English and Communication Skills.), 4) While assessing figures, examiner may give credit for principal components indicated in the figure. The, figures drawn by the candidate and those in the model answer may vary. The examiner may give credit, for any equivalent figure drawn., 5) Credits may be given step wise for numerical problems. In some cases, the assumed constant values may, vary and there may be some difference in the candidate’s answers and the model answer., 6) In case of some questions credit may be given by judgment on part of examiner of relevant answer based, on candidate’s understanding., 7) For programming language papers, credit may be given to any other program based on, equivalent concept., ----------------------------------------------------------------------------------------------------------------------------- ---Que., No., Q1, , Sub., Que., A, a), , Model Answers, , Marks, , Attempt any THREE :, , Total, Marks, 12, , Draw the stress block diagram for singly reinforced section., , Ans., , 04, , 04, , (Note: Two marks for sketch and two marks for labeling ), b), Ans., , State any four functions of reinforcement in R.C. sections., Functions of reinforcement in R.C. sections:, 1. In case of slab, beams and wall of water tanks, reinforcement, is mainly provided to carry direct or bending tensile stresses., , Page No 1/ 32

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MAHARAS HTRA S TATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (IS O/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2017, Sub. Code : - 17604, , Subject: Design of RCC Structures, , --------------------------------------------------------------------------------------------------------------Que., No., 1, , Sub., Que., , Model Answers, , Marks, , Total, Marks, , 01, mark, each, (any, four), , 04, , 2. In case of columns the steel is provided to resist the direct, compressive stress as well as bending stresses if any., 3. In case of beams stirrups are provided to resist the diagonal, tension due to shear and hold the main steel in position., 4. The box type mesh of reinforcement is provided to resist, torsion., 5. The steel is provided in the form of rectangular, circular,, lateral ties or spirals to prevent bucking of main bars in, column., 6. The distribution steel is provided to distribute the concentrated, loads and to reduce the effects of temperature and shrinkage, and to hold main bars in position., c), , State two advantages and two disadvantages of prestressed, concrete., , Ans., , Advantages of prestressed concrete., 1. The use of high strength concrete and steel in prestressed, members results in lighter and slender members than is, possible with RC members., 2. In fully prestressed members the member is free from tensile, stresses under working loads, thus whole of the section is, effective., 3. In prestressed members, dead loads may be counter-balanced, by eccentric prestressing., 4. Prestressed concrete member possess better resistance to shear, , 01, mark, each, (any, two), , forces due to effect of compressive stresses presence or, eccentric cable profile., 5. Use of high strength concrete and freedom from cracks,, contribute, , to, , improve, , durability, , under, , aggressive, , environmental conditions., , Page No 2/ 32

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MAHARAS HTRA S TATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (IS O/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2017, Sub. Code : - 17604, , Subject: Design of RCC Structures, , --------------------------------------------------------------------------------------------------------------Que., No., 1, , Sub., Que., , Model Answers, , Marks, , Total, Marks, , 6. Long span structures are possible so that saving in weight is, significant & thus it will be economic., 7. Factory products are possible., 8. Prestressed members are tested before use., 9. Prestressed, , concrete structure deflects appreciably before, , 04, , ultimate failure, thus giving ample warning before collapse., 10. Fatigue strength is better due to small variations in prestressing, steel, recommended to dynamically loaded structures., Disadvantages of Prestressed Concrete, 1. The availability of experienced builders is scanty., 2. Initial equipment cost is very high., 3. Availability of experienced engineers is scanty., 4. Prestressed sections are brittle., 5. Prestressed concrete sections are less fire resistant., d), Ans., , 01, mark, each, (any, two), , State various forms of shear reinforcement in beams., 1. Vertical stirrups, 2. Bent up bars along with stirrups,, , 04, , 04, , 3. Inclined stirrups, e), Ans., , State two ductile detailing provisions in IS 13920., Requirement for longitudinal reinforcement in flexural members:, 1) The top as well as bottom reinforcement shall consist of at, least two bars throughout the member length., 2) The maximum steel ratio on any face at any section, shall not, , 02, marks, exceed Pmax = 0.025, each, (any, 3) The positive steel at a joint face must be at least equal to half two), the negative steel at that face., , 04, , (Note : Any other members ductile detailing provisions should be, considered), , Page No 3/ 32

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MAHARAS HTRA S TATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (IS O/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2017, Sub. Code : - 17604, , Subject: Design of RCC Structures, , --------------------------------------------------------------------------------------------------------------Que., No., 1, , Sub., Que., (B), a), , Model Answers, , Marks, , Solve any ONE:, , Total, Marks, 06, , A beam 230mm x 450mm effective size carries a factored B. M. of, 150 kN-m if concrete M20 and Steel grade Fe 500 are used, find, area of steel., , Ans., , M u =150 kN-m = 150 106 N-mm, M umax  (0.133) f ck bd 2,  (0.133)  20  230  (450) 2,  123.889 106 N-mm, , 02, , M umax = 123.889kN  m, Since M u  M u max  123.889kN  m, Section is over reinforced. Hence desiging balanced section, , 02, , For Fe 500, % Pt max = 0.038f ck = 0.038  20 = 0.76%, A, But, % Pt max = st 100, bd, A st, 0.76 , 100, 230  450, A st  786.6 mm 2, , b), , Ans., , 06, , 02, , Find moment of resistance if steel provided is 6 bars of 12 mm, diameter in a beam 300 mm x 500 mm effective. Concrete M20, and. Steel Fe 500 are used., , Ast  6 , Xu , , , ,  (12) 2  678.584 mm 2, , 4, 0.87 f y Ast, 0.36 f ck b, , , , 0.87  500  678.584, 0.36  20  300, , X u  136.659 mm, X u max  0.46d  0.46  500  230 mm, X u  136.659 mm < X u max  230 mm, , 02, 02, , Hence, section is under reinforced,, , 06, , M u  0.87 f y . Ast .(d  0.42 X u ), M u  0.87  500  678.584 500  (0.42)  (136.659), M u  130.649 106 N  mm, M u  130.649 kN  m, , 02, , Page No 4/ 32

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MAHARAS HTRA S TATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (IS O/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2017, Sub. Code : - 17604, , Subject: Design of RCC Structures, , --------------------------------------------------------------------------------------------------------------Que., No., 2., , Sub., Que., , Model Answers, , Marks, , Solve any TWO:, a), , Ans., , Total, Marks, 16, , Design a one-way slab with following data span = 5.0 m, Live load, = 4.5 kN/m2 , Floor finish = 1 kN/m2 . Concrete M 20 and steel Fe, 415, M.F. = 1.4. Sketch c/s of slab showing reinforcement details., , Given:, l =2m, , LL=4.5 kN/m 2, , FF = 1 kN/m 2, , MF =1.4, , f ck =20N/mm 2 , f y =415N/mm 2, , Step (1), Span, 5000, d=, =, =178.571 mm, 20×MF 20×1.4, Assuming, 10mm  bars and cover of 20 mm, , 10, D = d+c+ =178.571+20+ =203.571 mm, 2, 2, Provide, D =210 mm, 10, d = 210 - 20 = 185 mm, 2, Step (2), Effective span, le  l +d=5000+185=5185 mm=5.185m, , 02, , 01, , Step (3), Load & B M calculation, i) D.L. of slab = 0.210×1×1×25 = 5.25 kN/m, ii) L.L. of slab = 4.5×1×1= 4.5 kN/m, i) F.F. of slab = 1×1×1×25 = 1.0 kN/m, Total load = 10.75 kN/m, Factored load (w d )=1.5×w, , 01, , =1.5×10.75, = 16.125 kN/m, w  l  16.125×(5.185) 2, BM = Mu = d e =, 8, 8, 2, , BM = Mu=54.188kN-m, , 01, , Page No 5/ 32

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MAHARAS HTRA S TATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (IS O/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2017, Sub. Code : - 17604, , Subject: Design of RCC Structures, , --------------------------------------------------------------------------------------------------------------Que., No., 2, , Sub., Que., , Model Answers, , Marks, , Step (4), Check for depth, , Total, Marks, 08, , Mu max =M u, 0.138f ck b  d reqd  =54.188×106, 2, , 0.138×20×1000×  d reqd  =54.188×106, 2, ,  d  =140.118mm<d=185mm, , ........Ok, , reqd, , Step (5), Main steel and its spacing, A st =, , 0.5f ck, fy, , , 4.6×Mu×106, 1- 1f ck bd 2, , , ,  bd, , , 0.5×20 , 4.6×54.188×106 , A st =, 1- 1 ×1000×185, 415 , 20×1000×(185) 2 , A st =903.164mm 2, Spacing of bar Min. of, , 01, , π, 2, 10 , 4, =86.960mm, 903.164, , 1000×, , a), , 1000×A, Sx =, =, A st, , b), , Sx =3d=3×185=555mm, , c), , Sx =300mm, Sx =85mm c/c, , Provide 10 mm  bars @ 85mm c/c along the shorter span, , Step 6), Distribution steel and its spacing, 0.12, 0.12, bD=, ×1000×210=252mm 2, 100, 100, Assuming, 8mm φ bars,, Spacing of bars is equal to min. of, π 2, 1000×  8 , 1000×Ad, 4, a), Sy =, =, =199.466mm, Ast d, 252, Astd =, , b), , Sy =5d=5×185=925mm, , c), , Sy =450mm, , 01, , Sy =195mm c/c, Provide 8 mm  bars @ 195 mm c/c along the longer span, , Page No 6/ 32

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MAHARAS HTRA S TATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (IS O/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2017, Sub. Code : - 17604, , Subject: Design of RCC Structures, , --------------------------------------------------------------------------------------------------------------Que., No., 2, , Sub., Que., , Model Answers, , Marks, , Total, Marks, , 01, , (Note: Answer may vary on assumption of cover diameter of the bar), b), , The effective dimensions of a slab panel are 4 m x 7 m. it carries, super imposed loads of 4 kN / sqm. Design a suitable slab using., M20 and Fe 415 steel. Take M.F. = 1.25,, , αx = 0.113 and αy =, , 0.037. Find total depth D factored BM and reinforcement details, using suitable bars. Sketch the c/s of slab along shorter span, showing reinforcement details., Ans., , Given: lx =4m, , LL=4 kN/m 2, , MF=1.25, , f ck =20N/mm 2 ,, ,  x  0.113, ,  y  0.037, , FF = 1 kN/m 2, f y = 415N/mm 2, , Step (1), Slab thickness,, LL= 4 kN/m 2 > 3 kN/m 2 and, lx = 4m > 3.5 m, Span, 4000, =, =160 mm, 20×MF 20×1.4, Assuming, 10mm  bars and cover of 15 mm, , d=, , , , 10, =180 mm, 2, 2,  Provide, D = 180 mm d = 160 mm, D=d+c+, , = 160 + 15 +, , 02, , Page No 7/ 32

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MAHARAS HTRA S TATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (IS O/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2017, Sub. Code : - 17604, , Subject: Design of RCC Structures, , --------------------------------------------------------------------------------------------------------------Que., No., 2, , Sub., Que., , Model Answers, , Marks, , Total, Marks, , Step (2), Effective span, lx  lxe = 4 m = 4000 mm, , 01, , l y  l ye = 7 m = 7000 mm, (As given diamensions are effective diamentions of slab panel), (3)Load & B M calculation, i) D.L. of slab = 0.180 × 1 × 1 × 25 = 4.5 kN/m, ii) L.L. of slab = 4×1×1= 4 kN/m, i) F.F. of slab = 1×1×1 = 1.0 kN/m, Total load = 9.5 kN/m, Factored load (wd )=1.5×w, , 01, , =1.5×9.5, = 14.25kN/m, BM calculations,, Mu x = x .wd .  lxe  =(0.113 14.25  (4) 2 ), 2, , 08, 01, , Mu x = 25.764kN-m, Mu y =  y .wd .  lxe  =(0.037 14.25  (4) 2 ), 2, , Mu x =8.436kN-m, , Step (4), Check for depth, Mu max =M ux, 0.138f ck b  d reqd  = 25.764×106, 2, ,  d  =96.616mm < d=160mm, , ........Ok, , reqd, , Step (5), Main steel and its spacing, In X direction, A stx =, , 0.5f ck, fy, , , 4.6×Mu×106, 11, f ck bd 2, , , ,  bd, , , 0.5×20 , 4.6×25.764×106 , 11,  ×1000×160, 415 , 20×1000×(160) 2 , A st = 475.541 mm 2, A st =, , 01, , Page No 8/ 32

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MAHARAS HTRA S TATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (IS O/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2017, Sub. Code : - 17604, , Subject: Design of RCC Structures, , --------------------------------------------------------------------------------------------------------------Que., No., 2, , Sub., Que., , Model Answers, , Marks, , Total, Marks, , Spacing of bar Min. of, π, 2, 10 , 4, = 165.158mm, 475.541, , 1000×, , a), , 1000×A, Sx =, =, A st, , b), , Sx = 3d = 3×160 = 480mm, , c), , Sx = 300mm, Sx = 165mm c/c, , Provide 10 mm  bars @ 165 mm c/c, In Y direction, d '  d    160  10  150mm, A sty =, , 0.5f ck, fy, , , 4.6×Mu×106, 1- 1f ck bd 2, , , ,  bd, , , 0.5×20 , 4.6×8.436×106 , A st =, 1- 1 ×1000×150, 415 , 20×1000×(150) 2 , A st = 138.158 mm 2, , 01, , 0.12,  1000  180  216mm 2, 100, A sty =216mm 2, Ast min , , Spacing of bar Min. of, π, 2, 10 , 4, = 363.61mm, 216, , 1000×, , a), , 1000×A, Sy =, =, A sty, , b), , Sy = 3d = 3×150 = 450mm, , c), , Sy = 300mm, Sy = 300mm c/c, , Provide 10 mm  bars @ 300 mm c/c, , 01, (, Note: Answer may vary on assumption of cover diameter of the bar), , Page No 9/ 32

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MAHARAS HTRA S TATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (IS O/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2017, Sub. Code : - 17604, , Subject: Design of RCC Structures, , --------------------------------------------------------------------------------------------------------------Que., No., 2, , Sub., Que., c), , Model Answers, , Marks, , Total, Marks, , Design a cantilever chajja with following data: Span = 1.50 m,, width = 2.0 m, L.L.= kN/m2 . Floor finish = 0.5 kN/m2 , support, lintel = 230 x 300 mm concrete M 20, Fe 415 steel, sketch the c/s of, chajja. Showing steel details., , Ans., , Given :, Span= l =1.5m=1500mm, Width= 2m=2000mm, LL=1.5kN/m 2 , FF=0.5kN/m 2, Support = 230  300 mm, f ck  20N/mm 2 , f y  415N/mm 2, Step 1), Slab thickness, Span, 7  M .F ., Assume, M.F.1.4, cover=15 mm and   10mm, , d, , d, , 1500,  153.06mm, 7  1.4, , D  d c, , , ,  153.06  15 , , 2, provide, D=180mm,, , 10,  173.06mm, 2, , 01, , 10, =160mm, 2, D=180mm, d=160mm, , d=180-15Step (2), , Effective span, le  l , , d,  1580mm  1.58m, 2, , 01, , Step 3), Load cal. and BM, , i) D.L. of slab =0.180 11 25  4.5kN / m, ii) L.L. of slab =1.5 11  1.5kN / m, iii) F.F. pf slab =0.50 11  0.5kN / m, , 01, , Total laod (w) = 6.5 kN/m', Factored load w d  1.5  6.5  9.75kN / m, BM  M u, , wd  le 2, , , , 2, Check for depth ,, , 9.75 1.5802, ,  12.169kN  m, 2, , 01, , Mu max = M ux, , Page No 10/ 32

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MAHARAS HTRA S TATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (IS O/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2017, Sub. Code : - 17604, , Subject: Design of RCC Structures, , --------------------------------------------------------------------------------------------------------------Que., No., 2, , Sub., Que., , Model Answers, , Marks, , Total, Marks, , 0.138f ck b  d reqd  = 12×106, 2, , 0.138  20 1000   d reqd   12.169 106, 2, ,  d  = 66.400 mm < d=160mm, reqd, , ........Ok, , 08, , Step (5), Main steel and its spacing, , 4.6×Mu×106 , 1- 1 bd, f ck bd 2, , , 0.5×20 , 4.6×12.169 106, A st =, 1- 1415 , 20×1000×(160) 2, A st = 216.857 mm 2, A st =, , 0.5f ck, fy, , ,  ×1000×160, , , 0.12, A st min =, 1000 180  216mm 2, 100, Spacing of bar Min. of, 1000×A, =, A st, , 02, , π, 2, 10 , 4, = 362.173mm, 216.857, , 1000×, , a), , Sx =, , b), , Sx = 3d = 3×160 = 480mm, , c), , Sx = 300mm, Sx = 300mm c/c, , Provide 10 mm  bars @ 300 mm c/c, Step 6), 0.12, 1000 180  216mm 2, 100, Assuming, 8 mm  bars, Spacing of bar Min. of, A st y =A st min =, , π 2, 8, 4, = 232.710mm, 216, , 1000×, , a), , 1000×A, Sy =, =, A st, , b), , Sy = 5d = 3×160 = 800mm, , c), , Sy = 450mm, , 01, , Sy = 230mm c/c, Provide 8 mm  bars @ 230 mm c/c, , Page No 11/ 32

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MAHARAS HTRA S TATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (IS O/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2017, Sub. Code : - 17604, , Subject: Design of RCC Structures, , --------------------------------------------------------------------------------------------------------------Que., No., 2, , Sub., Que., , Model Answers, , Marks, , Total, Marks, , 01, , (Note: Answer may vary on assumption of cover diameter of the bar), , 3., , 16, , Attempt any FOUR :, a), , Find the moment of resistance (M r) of fec (T) beam with following, data: Df = 120 mm, bf = 1500 mm, bw = 300mm d = 450 mm, Asf =, 2200 mm2 , concrete M25, steel Fe 500., , Ans., , Given :, bf = 1500mm, Df = 120mm, bw = 300mm, d =450mm, Ast = 2200 mm2, To find M u, Step 1 ), To find Xu=? (Assume Xu<Df), , 01, Xu=70.88 mm < Df =120 mm …….ok, , Page No 12/ 32

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MAHARAS HTRA S TATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (IS O/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2017, Sub. Code : - 17604, , Subject: Design of RCC Structures, , --------------------------------------------------------------------------------------------------------------Que., No., 3, , Sub., Que., , Model Answers, Step 2), To find Xumax, Xumax = 0.46 X d, = 0.46 X 450, Xumax =207 mm, As, Xu<Xumax ,, So, beam is under reinforced., , Marks, , Total, Marks, , 01, , 04, , 01, , Step 3), To find Mu=?, Mu = Tu x Zu, = 0.87 x fy x Ast ( d -0.42 Xu), = 0.87 x 500 x 2200 ( 450 -0.42 x 70.88), = 402.16 x 106 N-mm, , 01, , Mu = 402.16 KN-m, State the conditions of formation of flanged beams & state, b), Ans., , effective flange width for T & L beam., Is code recommends the following two provisions for beam spanning, parallel to slab to act as a flanged beam,, 1) Transverse reinforcement (perpendicular to beam) is required to be, provided at the top in flanged portion for a length equal to L/4 on each, side of beam. L= Span of slab., , 01, , 2) Transverse reinforcement > 60% of main reinforcement at mid span, of slab., , 01, , 3) The slab shall cast integrally with the web or the web and slab shall, be effectively bonded together in any other manner., 04, The effective width of the flange may be taken as following in no case, greater than the width of the web plus half the sum of the clear, distance to the adjacent beam on the either side., , 01, , a) For T beam, bf =, , + bw +6 Df, , 01, , b) For L beam, bf =, , + bw+ 3Df, Page No 13/ 32

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MAHARAS HTRA S TATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (IS O/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2017, Sub. Code : - 17604, , Subject: Design of RCC Structures, , --------------------------------------------------------------------------------------------------------------Que., No., 3, , Sub., Que., , Model Answers, , Marks, , Total, Marks, , where,, bf = effective width of flange, lo = distance between points of zero moment in the beam, bw = breath of web, Df = thickness of flange, b = actual width of flange., c), , Define development length & state factors affecting development, length., , Ans., , Development Length:, , Development length is defined as the length of the bar required on, either side of the section to develop the required stress in steel at that, section through bond., OR, , 02, , Development length is an embedded length of the bar required to, develop the design stress in reinforcement at the critical section, Factors affecting development length:, 1. Grade of steel., , ½, mark, each, , 2. Grade of concrete., 3. Diameter of bar., 4. Design bond stress., d), , Diameter of a steel bar is 20 mm, steel Fe 415 grade and design, bond stress is 1.2 MPa, , for plain bars in tension, calculate the, , development length for bar inn compression., Ans., , Given:, Φ = 20mm, fck = 20 N/mm2, fy = 415N/mm2, Bond stress, τbd =1.2 N/mm2, , Page No 14/ 32

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MAHARAS HTRA S TATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (IS O/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2017, Sub. Code : - 17604, , Subject: Design of RCC Structures, , --------------------------------------------------------------------------------------------------------------Que., No., 3, , Sub., Que., , Model Answers, , Marks, , Total, Marks, , Development length for bar in compression, 02, For Fe 500 steel value of τbd shall be increased by 60% and for, 04, , bar in compression, the value of bond stress for bar in tension shall be, increased by 25%, 02, Ld = 752.1875 mm, , e), , Design a rectangular column with following data:, Factored load = 3500 kN, concrete M 20, Steel Fe 415,, , Ans., , Unsupported length = 4.0 m. Assume 1 % steel., (Note: Answer may vary according to shape of column assumed), Given:, Pu = 3500 kN, L = 4 m = 4000mm, Fck = 20 N/mm2, Fy = 415 N/mm2, Step 1), To find Size of column, Pu = ( 0.4fck Ac ) + ( 0.67 fy Asc), Assume 1% of steel in column, Area of steel, Asc = 0.01 Ag, Area of concrete Ac = Ag- Asc, Ac = 0.99Ag, 3500x103 =(0.4x20x0.99Ag)+(0.67x415x0.01Ag ), Ag = 327.087x103 mm2, , 01, , Assume b = 400 mm, , Page No 15/ 32

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MAHARAS HTRA S TATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (IS O/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2017, Sub. Code : - 17604, , Subject: Design of RCC Structures, , --------------------------------------------------------------------------------------------------------------Que., No., 3, , Sub., Que., , Model Answers, , Marks, , Total, Marks, , 01, , 04, , (Students may assume any other value of ‘b’ according the ‘d’ will, changed), D=, , =, , = 817.71 mm = 820 mm, , Step 2), , Check for minimum eccentricity, e min = L/500 + D/30 OR 20mm whichever is greater, e min, , = 4000/500 + 820/30, = 35.33 mm OR 20mm whichever is greater, , e min =35.33 mm, e max = 0.05D, 0.05D = 0.05 X 820 =41 mm, e min 35.33mm < e max 41 mm .............ok for minimum eccentricity., Provide size of column = 400 mm x 820 mm, A sc = 0.01 Ag, 01, , = 0.01 X 400 X 820, A sc = 3282 mm 2, Provide 8 bars of 20 mm  bar and 2 bars of 25 mm , A scprod . = 8x314.15 + 2 x 490.87 = 3494.94 mm 2 > 3282 mm 2 .. ok, %Pt =, %Pt min, , 3494.24, 100  1.06%, 400  820, = 0.18%<%Pt =1.06<%Pt, , max, ,  6%, , Step 3)Lateral Ties, Diameter of ties = ¼ X diameter of longitudinal steel bar., = ¼ X 25, = 6.25 mm, So, provide 8 mm dia. lateral ties., Pitch should not be greater than, i) Least lateral dimensions of column i.e. 400mm., ii) 16 x dia. of longitudinal steel =16 x ϕ =16 x 20 = 320 mm, iii) 300mm, , 01, , (Select minimum of above values), Therefore, provide lateral ties 8mm ϕ @ 300mm c/c., , Page No 16/ 32

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MAHARAS HTRA S TATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (IS O/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2017, Sub. Code : - 17604, , Subject: Design of RCC Structures, , --------------------------------------------------------------------------------------------------------------Que., No., Q4, , Sub., Que., A), a), Ans., , Model Answers, , Marks, , Attempt any THREE :, , Total, Marks, 12, , State methods of prestressing and explain in brief any one., i)Pre-Tensioning, 1.Hoyer system, ii)Post-Tensioning, 1. Freyssinet system, 2. Magnel system, , 02, , 3. Leonhardt system, 4. Lee-McCall system, 5. Gifford-Udall system, , 04, , Hoyer system:, This system is generally used for mass production. The end abutments, are kept sufficient distance apart, and several members are cast in a, single line. The shuttering is provided at the sides and between the, , 02, , members. This system is also called the Long Line Method. The, following figure is a schematic representation of the Hoyer system., The end abutments have to be sufficiently stiff and have good, foundations., , Schematic representation of Hoyer system, , (Note: Any one of above method should be considered), , Page No 17/ 32

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MAHARAS HTRA S TATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (IS O/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2017, Sub. Code : - 17604, , Subject: Design of RCC Structures, , --------------------------------------------------------------------------------------------------------------Que., No., 4., , Sub., Que., b), , Ans., , Model Answers, , Marks, , Total, Marks, , Calculate load carrying capacity of column 300 mm in diameter, reinforced with 4- 16mm ɸ and 6-12mm ɸ bars use M20 concrete, and Fe 415 steel., Given data :, Size of column = 230 x 230mm, Asc = 4x (π/ 4 ) x (16) 2 + 6 x (π/4) x (12)2, = 804.24+678.58 mm2, =1482.82 mm2, fck = 20 N/mm2, fy = 415 N/mm2, To find, load carrying capacity of column, P, Step 1 :, Gross area = Ag = (π/4) x 3002, , 01, 2, , Ag = 70685.83 mm, Step 2 :, , Area of steel, Asc = 1482.82 mm2, , 01, , Step 3 :, , 04, , Area of concrete, Ac =Ag – Asc, Ac = 69203 mm2, , 01, , Step 4 :, Ultimate load carrying capacity, Pu, Pu = [ 0.4 x fck x Ac ] + [ 0.67 x fy x Asc ], = [ 0.4 x 20 x 69203] + [ 0.67 x 415 x 1482.82 ], , 01, , Pu = 965.92 x 103 N = 965.92 kN, , Page No 18/ 32

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MAHARAS HTRA S TATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (IS O/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2017, Sub. Code : - 17604, , Subject: Design of RCC Structures, , --------------------------------------------------------------------------------------------------------------Que., No., , Sub., Que., c), , Model Answers, , Marks, , Total, Marks, , Define:, (i) Characteristic strength and, (ii) Characteristic load., , Ans., , i) Characteristic Strength, Characteristic Strength means that value of the strength of the material, below which not more than 5% of the test results are expected to fall., , 02, 04, , ii) Characteristic Load, Characteristic Load means that value of load which has 95%, , 02, , probability of not being exceeded during the life of the structure., , d), Ans., , State four situations where doubly reinforced section are, preferred., 1) When the singly reinforced beams need considerable depth to, resist large bending moment, it becomes necessary to provide, doubly reinforced section., 2) When the size of rectangular beam cross-section is limited, because of architectural reasons or practical reasons then it, becomes necessary., 3) When the sections are subjected to reversal of bending, , 01, mark, each, (any, four), , 04, , moment., 4) When it is required to reduce the long-term deflection, it, becomes necessary to provide doubly reinforced section., 5) When a beam is continuous overall several supports; the beam, is subjected to alternate sagging also it becomes necessary to, provide doubly reinforced section., , Page No 19/ 32

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MAHARAS HTRA S TATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (IS O/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2017, Sub. Code : - 17604, , Subject: Design of RCC Structures, , --------------------------------------------------------------------------------------------------------------Que., No., 4, , Sub., Que., b), , Ans., , Model Answers, , Marks, , Total, Marks, , Fine the moment of resistance of a beam 230mm x 450mm deep, reinforced with 4-16mm diameter bars in tension zone and 212mm diameter bars in compression zone. Assume effective cover, of 40mm., Given data :b = 230 mm, D = 450 mm, d = D - d’= 450 - 40 = 410 mm, d’ = 40 mm, Asc = 2x (π/4)x122 = 226.19mm2, Ast = 4x (π/4)x162 = 804.24 mm2, fck = 20 N/mm2, fy = 415 N/mm2, Step 1 – To find Xu max, Xumax = 0.48 d………for Fe415, = 0.48 x 410, , 01, , = 196.8 mm, Step 2 - To find actual Xu ,, , Fcc= 0.45fck = 0.45x20 = 9 N/mm2, fsc is depend on d’/d ratio, , 01, , d'/d, , .05, , 0.10, , 0.15, , 0.20, , fsc, , 355, , 352, , 342, , 329, , (Note : Given data is insufficient any suitable value of ‘fsc’ assumed, should be considered), , Page No 22/ 32

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MAHARAS HTRA S TATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (IS O/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2017, Sub. Code : - 17604, , Subject: Design of RCC Structures, , --------------------------------------------------------------------------------------------------------------Que., No., 4, , Sub., Que., , Model Answers, , Marks, , Total, Marks, 06, , fsc = 352.15 N/mm2, , 01, 2, , 214.97 mm, Ast1 = Ast- Ast2, = 804.24-214.97, = 589.27 mm2, , 01, , Step 3 Find Xu, , 01, Xu < Xu max the section is Under reinforced, Step 4 Moment of Resistance, , M u  0.87  f y  Ast   d  0.42 X u    f sc  f cc   Asc  d  d ', M u  0.87  415  589.27   410  0.42 128.47    352.15  9   226.19  410  40 , , 01, , M u  104.468 106 N  mm, M u  104.468kN  m, 5., , 16, a), , Ans., , Attempt any TWO:, A doubly reinforced beam section 230mm x 40mm effective, carries a factored moment of 175 kN-m. fine the area of steel., Required if M 20concrte and Fe 500 are used. Assume d’=50mm, and σsc=353 N/mm2 ., Given :, b  230mm, , d  450mm (effective), , Md  175kN  m, M 20 Fe500 Fck  20 N / mm 2, fy  500 N / mm, Step 1), To find x umax, xu max  0.46d,  0.46  450,  207 mm, , 01, , Page No 23/ 32

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MAHARAS HTRA S TATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (IS O/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2017, Sub. Code : - 17604, , Subject: Design of RCC Structures, , --------------------------------------------------------------------------------------------------------------Que., No., 5, , Sub., Que., b), , Model Answers, , Marks, , Total, Marks, , A beam 250mm x 415 mm effective depth is reinforced with 4 bars, of 16mm dia of grad Fe 415. The shear force of the support is 90, kN. Design the shear reinforcement. Use M 20 concrete and 6 mm, dia. vertical stirrups of Fe 415 steel., , Ans., , %Pt., , 0.5, , 0.75, , τc (MPa), , 0.48, , 0.56, , Given data:, b=250mm, , f y =415N/mm 2, , d=415mm, , f ck =20N/mm 2, , 4 bars 16mm φ, V=90kN, π, Ast=4× ×162 =804.25mm 2, 4, Stirrups 6 mm φ, Step 1), To calculate factored shear force, Vu=V×1.5, =90×1.5, =135kN, Step 2), To calculate shear stress(τ v ), Vu 135×103, =, =1.3N/mm 2, bd 230×415, τ cmax =2.8N/mm 2, ........for M 20 concrete, , 01, , τv =, , 01, , Step, 3), Calculate pt%, Ast, τ c, Ast, 804.25,  100 ,  100  0.78, bd, 250  415, pt  0.78, pt, c, pt , , 0.75, , 0.56, , 1.0, , 0.62, , 01, , (Note: Given data is insufficient. Any suitable value of τ c assumed, should be considered ), Page No 25/ 32

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MAHARAS HTRA S TATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (IS O/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2017, Sub. Code : - 17604, , Subject: Design of RCC Structures, , --------------------------------------------------------------------------------------------------------------Que., No., 5, , Sub., Que., , Model Answers, , Marks, ,  c1  0.56, pt2  1.0,  c 2  0.62, pt  0.78, then  c, (   ),  c   c1  c 2 c1  ( pt  pt1 ), , Total, Marks, , pt1  0.75, , 08, , ( pt2  pt1 ), ,  c  0.56 , , (0.62  0.56),  (0.78  0.75), (1.0  0.75), ,  c  0.57 N / mm 2,  v  1.3N / mm 2   c  0.57 N / mm 2, Shear reinforcement, Step, , 01, , is required, , 4), , Shear force for which shear reinforcement is required, Vus=Vu-τ c bd, , 01, , =135×103 -0.57×230×415, =80.593kN, =80593N, , 01, , As bent up bars are not provided, Vusv=Vus=80593N, Provide 6mm  two legged vertical stirrups, Step, , 5), , Spacing(Sv)=, , 0.87f y Asv.d, , Vusv, π, 0.87×415×2× ×62 ×415, 4, Sv=, 80593, Sv=105.13  100mm c/c, , Svmin , Svmin, , 0.87 f y Asv, , 01, , 0.87  415  2 , , , 0.4b,  221.92  220mm, , Svmax  0.75d or, , 300, , Svmax  0.75  415, , or, , Svmax  311.25 or, , 300, , 0.4  230, , , 4, ,  62, , 01, , 300, , Provide 6mm two legged vertical stirrups @ 100mmc/c., , Page No 26/ 32

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MAHARAS HTRA S TATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (IS O/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2017, Sub. Code : - 17604, , Subject: Design of RCC Structures, , --------------------------------------------------------------------------------------------------------------Que., No., 5, , Sub., Que., c), , Model Answers, , Marks, , Total, Marks, , Design on R.C. column footing with following data., Size of column = 400mm x 400mm, Safe bearing capacity = 1200 kN/m2, Concrete M20 and steel Fe 415 is used., Calculate depth of footing from B.M criteria., No shear check is required., , Ans., , GivenSize of column – 400mm X 400 mm, Safe bearing capacity of soil = 200 kN/m2, Load on column is 1200 kN, fck = 20 N/mm2, fy= 415 N/mm2, Step 1 :, Ultimate S.B.C ( qu )= 2 X 200= 400 kN/m2, , 01, , Step 2 :, Size of footingWU = W X ‫ﻵ‬f = 1200 X 1.5 = 1800 kN, Af = 1.05 X WU / qu, , 01, , = 1.05 X 1800/400, =4.725 m2, L=B=, , AF = 4.725 = 2.173m = say 2.20m, , Adopt footing of size 2.20m X 2.20m, , 01, , Step 3 :, Upward soil pressure, p = Wu /(L X B) = 1800/(2.2 X 2.2) = 371.90 KN/m2, , 01, , Page No 27/ 32

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MAHARAS HTRA S TATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (IS O/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2017, Sub. Code : - 17604, , Subject: Design of RCC Structures, , --------------------------------------------------------------------------------------------------------------Que., No., 5, , Sub., Que., , Model Answers, , Marks, , Total, Marks, , Step 4 :, Depth for flexure, , 01, Let X1 = Y1 = projection beyond column, = (2.2-0.4) / 2 = 0.9, Mx = My = 1 × X1 × p × ( X1 / 2), Mx = My = 1 X 0.9 X 371 X( 0.9/2), , 08, , = 150.62 KNm, , d reqd  Mx / q. fck .b, dreqd  (150.62 106 / 0.138  20 1000), , = 233.61 mm say 240 mm., Adopt cover of 80 mm, 01, , D= 240+80= 320 mm, Step 5 :, Ast , , 0.5 f ck , 4.6  M u , 1  1 ,  bd, fy , f ck bd 2 , , Ast , , 0.5  20 , 4.6 150.62 106, 1  1 , 415 , 20 1000  2402, , ,  1000  240, , , Ast  2132.120mm 2, , Using 16 mm diameter, Spacing, Sx = Sy = 1000 x Aø/ Ast, = 1000 X (π/4) X 162 / 2132.120, = 94.30 mm say 90 mm c/c, , 01, , Provide 16 mm ø @ 90 mm c/c both way, , Page No 28/ 32

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MAHARAS HTRA S TATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (IS O/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2017, Sub. Code : - 17604, , Subject: Design of RCC Structures, , --------------------------------------------------------------------------------------------------------------Que., No., 5, , Sub., Que., , Model Answers, , Marks, , Total, Marks, , Step 6 :, Development lengthLd = (0.87 fy x ø ) / (4 τbd ), = (0.87 X 415 X 16 ) / (4 X 1.2 X 1.6), = 752.187 mm say 760 mm, This length is available from face of column., Provide 350mm depth near the face of column and reduce depth of footing, , 150mm at the edge., , 01, , Page No 29/ 32

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MAHARAS HTRA S TATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (IS O/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2017, Sub. Code : - 17604, , Subject: Design of RCC Structures, , --------------------------------------------------------------------------------------------------------------Que., No., 6., , Sub., Que., , Model Answers, , Marks, , Attempt any FOUR :, a), , Differentiate under reinforced and over reinforced section with, reference to area of steel. Depth of NA moment of resistance., Under reinforced, section, , Ans., , Area Of, Steel, Neutral, Axis, Moment, of, resistance, b), Ans., , Total, Marks, 16, , Less compared to over, reinforced section., , Over reinforced Section, , More compared to under, reinforced section., 04, , Ast < Ast max., , Ast > Ast max., , Xu < Xumax, , Xu > Xumax, , Mu = Tu.z, , Mu = Cu.z, , =0.87 fy Ast(d-0.42xu), , =0.36 fck b xu(d-0.42xu), , 04, , Write IS specification of minimum eccentricity and transverse, reinforcement for an axially loaded column., IS specification for emin . & transverse steel, Minimum eccentricity: emin .= Lo/500 + D/30 Or 20 mm whichever is greeter, Lo= unsupported Length, , 02, , D=lateral dimension, Transverse reinforcement: I ) Pitch - The spacing of the link should not exceed the least of the, following., 1. The least lateral dimension of column., , 04, 01, , 2. Sixteen times the diameter of the smallest longitudinal bar., 3. 300mm, , Page No 30/ 32

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MAHARAS HTRA S TATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (IS O/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2017, Sub. Code : - 17604, , Subject: Design of RCC Structures, , --------------------------------------------------------------------------------------------------------------Que., No., 6, , Sub., Que., , Model Answers, , Marks, , Total, Marks, , II ) Diameter of link, 1. The diameter of the link should be at least one fourth of the, longitudinal steel., 2. In any case the links should not be less than 6mm in, diameter., , 01, , 3. Largest diameter of condition 1 and 2., c), Ans., , d), Ans., , What is minimum and maximum percentage of tension steel that, should be provided in flanged beam as per IS specification., As per IS 456:2000 Clause No. 26.5.1.1, Min. % of Tension Steel, Max. % of Tension Steel, It shall be not less than It shall not exceed (0.04bD), AS/bd = 0.85 / fy, Hence, As = (0.85 x b x d) / fy, Where, b = Breadth of web of Flanged beam, d = Effective Depth, , 04, , 04, , Find limiting moment of resistance (Mu) of a T beam with, following data, bf= 1500 mm, bw =230mm, Asf=2200 mm2 ,, concrete M 20 & Fe 415 steel., , Given , bf  1500mm, bw  230mm, d  730mm, Df  120mm, Asf  2200mm 2, M 20 and Fe415, Step 1) To find Xumax, Xumax = 0.48d, =0.48  730, , 01, , =350.4mm, Consider Ast=Asf, Xu=, , 0.87 fy Ast 0.87 X 415 X 2200, , 0.36 fck bf, 0.36 X 20 X 1500, , Xu  73.54mm, Xu  Df, Neutral axis within the flange., , 01, 04, 01, , Page No 31/ 32

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MAHARAS HTRA S TATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (IS O/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2017, Sub. Code : - 17604, , Subject: Design of RCC Structures, , --------------------------------------------------------------------------------------------------------------Que., No., 6, , Sub., Que., , Model Answers, , Marks, , Mu= 0.87f y . Ast  d  0.42 X u , , Total, Marks, , Mu= 0.87  415  2200   730  0.42  73.54 , 01, , Mu  555.313 106 N  mm, Mu  555.313kN  m, e), Ans., , Calculate the area of longitudinal, steel for short circular column, of dia. 300mm with eff. Length 5.0m to carry a factored load of, 1000kN. Use M20 concrete & Fe 500 steel., Given :, , D  300mm, Leff  5.0m  5000mm, , Pu  1000kN ., , M20 And Fe-500, For circular column, Ag=, , , , Ag , , 4, , , ,  D2, , 01, ,  3002, , 4, Ag  70.69 103 mm 2, Ag  Asc  Ac, , 01, , Ac  Ag  Asc, , 01, , Ac  (70.69 103  Asc), Pu  0.45 fckAc  0.67 fyAsc, , 04, , 1000 10  0.45  20  (70.69 10  Asc)  0.67  500  Asc, 3, , 3, , 01, , Asc  1115.92mm 2, , Page No 32/ 32