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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (ISO/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2018, Sub. Code: 17604, Subject: Design of R.C.C. Structure, --------------------------------------------------------------------------------------------------------------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. Sub., Total, Model Answers, Marks, No. Que., Marks, Q. 1, Attempt any THREE:, 12, (a), Ans., , (b), , State the meaning of partial safety factors for material strength, and loads., Partial safety factor for material strength: It is a strength reduction, factor (greater than unity) when applied to the characteristic strength, gives a strength known as design strength., Partial safety factor for load: It is a load enhancing factor (greater, than unity) which when multiplied to characteristic load gives a load, known as design load for which structure is to be designed., , 2, 4, 2, , Draw a neat sketch showing strain diagram and stress diagram, for a singly reinforced balanced section., , Ans., , 4, , 4, , Page 1 of 30

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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (ISO/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2018, Sub. Code: 17604, Subject: Design of R.C.C. Structure, --------------------------------------------------------------------------------------------------------------Que., No., Q. 1, , Sub., Que., (c), Ans., , (d), Ans., , (e), Ans., , Model Answers, , Marks, , Total, Marks, , State the functions of vertical stirrups provided in the beam., Functions of vertical stirrups provided in the beam are as follows:, i., To prevent sudden failure., ii., To prevent premature failure if the bond between main steel 1 each, and concrete is lost., iii. To act as tie for holding the beam reinforcement., iv., To confine the concrete., , 4, , State the meaning of magnitude of earthquake and intensity of, earthquake., Magnitude of earthquake: It is a measure of the amount of energy, released. It is quantitative measure of the actual size or strength of the, earthquake and it is much more precise measure than intensity., Intensity of earthquake: It is an evaluation of the severity of the, ground motion at a location and it is represented by a numerical value., , 4, , Differentiate between pre-tensioning and post-tensioning., Pre-tensioning, Post-tensioning, 1.Tendons, are, stretched 1.Tendons are stretched, after, before concreting., setting of concrete., 2. It requires heavy abutments 2. Cables are anchored with the, for anchoring steel wires., help of jacks. Cement grout is, forced under pressure to fill the, space in the ducts around, tendons., 3. Losses are about 18%., 3. Losses are about 15%., 4. This method is suitable for 4. This method is suitable for, mass production of small precast large members., members., e.g. Railway sleepers., e.g. Bridge construction., , (B), , Attempt any ONE:, , (a), , A R. C. beam 230 mm x 400 mm deep effective is reinforced with, 1.02 % steel of grade Fe 415. Determine the position of NA and, ultimate moment of resistance of the beam if fck = 25 N/mm2., Given:, b = 230 mm, d = 400 mm, Ast = 1.02 %, fck = 25 N/mm2, fy = 415 N/mm2, , 2, , 2, , 1 each, , 4, , 6, , To find:, Xu = ?, Mu = ?, , Page 2 of 30

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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (ISO/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2018, Sub. Code: 17604, Subject: Design of R.C.C. Structure, --------------------------------------------------------------------------------------------------------------Que., No., Q. 1, , Sub., Que., , Model Answers, Solution:, Ast, %p t =, ×100, b×d, Ast, 1.02=, ×100, 230×400, Ast =938.4 mm 2, , Xu =, , 0.87f y A st, 0.36f ck b, , =, , Marks, , Total, Marks, , 1, , 0.87×415×938.4, 0.36×25×230, , 1, 1, , X u =163.676 mm, , 6, , X umax =0.48d=0.48×400=192 mm, X u =163.676 mm < X umax =192 mm, , 1, , Hence, section is under reinforced,, M u =0.87f y .A st .(d-0.42X u ), , 1, , M u =0.87×415×938.4  400-(0.42)×(163.676) , M u =112.233×106 N-mm, , 1, , M u =112.233 kN-m, , (b), , Ans., , A R. C. slab, 120 mm thick effective, has a simply supported, effective span of 3.2 m. It is reinforced with 12 mm diameter bars, at a spacing of 100 mm. Calculate the safe load (including self, weight) the slab can carry if fck = 20 N/mm2 and fy = 415 N/mm2., Given:, l, = 3. 2 m = 3200 mm, d = 120 mm, ǿ = 12 mm, Spacing of bars = 100 mm, fck = 20 N/mm2, fy = 415 N/mm2, , To find:, w =?, , Solution:, , Page 3 of 30

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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (ISO/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2018, Sub. Code: 17604, Subject: Design of R.C.C. Structure, --------------------------------------------------------------------------------------------------------------Que., No., Q. 1, , Sub., Que., , Model Answers, , Assume 1 m width of slab., b = 1000 mm, Area of one 12 mm φ bar, π, A φ = ×122 =113.097 mm 2, 4, A, Spacing (S) = φ ×b, Ast, A, 113.097, Ast= φ ×b=, ×1000=1130.97 mm 2, S, 100, 0.87f y Ast 0.87×415×1130.97, Xu =, =, 0.36f ck b, 0.36×20×1000, X u =56.713 mm, , Marks, , Total, Marks, , 1, , 1, 1, , X umax =0.48d=0.48×120=57.6 mm, X u =56.713 mm < X umax =57.6 mm, Hence, section is under reinforced,, M u =0.87f y .A st .(d-0.42X u ), , 1, 6, , M u =0.87×415×1130.97 120-(0.42)×(56.713), M u =39.274×106 N-mm, , 1, , M u =39.274 kN-m, w d ×l2, 8, w d ×3.22, 39.274 =, 8, w d =30.683kN/m, , But, M u =, , But, w d = w×γ f, w=, , w d 30.683, =, γf, 1.5, , 1, , w = 20.455 kN/m, Q. 2, , Attempt any TWO:, (a), , 16, , A 3 m wide passage, supported on 230 mm thick side walls, carries, a superimposed load of 3.75 kN/m2 including floor finish. Design, the suitable slab using M20 concrete and Fe 415 steel. Use 8 # and, 6 ǿ bars. Take MF = 1.4. Apply the checks for maximum area of, reinforcement, minimum area of reinforcement and spacing. Do, not apply checks for shear and bond. Sketch the cross-section. Use, effective cover – 20 mm., , Page 4 of 30

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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (ISO/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2018, Sub. Code: 17604, Subject: Design of R.C.C. Structure, --------------------------------------------------------------------------------------------------------------Que., No., Q. 2, , Sub., Que., Ans., , Model Answers, Given:, l, = 3 m = 3000 mm, ts = 230 mm, LL + FF = 3.75 kN/m2, ǿ x = 8 mm, ǿ y = 6 mm, MF = 1.4, C, = 20 mm, fck = 20 N/mm2, fy = 415 N/mm2, , Marks, , Total, Marks, , To find:, D =?, Ast in both direction = ?, , Solution:, , Step (1), Span, 3000, d=, =, =107.143 mm, 20×MF 20×1.4, φ, 8, D = d+c+ x =107.143+20+ =131.143 mm, 2, 2, Provide, D =140 mm, 4, d = 140 - 20 - = 116 mm, 2, Step (2), Effective span, Min. of (a) & (b), a) le =l+ d = 3000 + 116 = 3116 mm = 3.116 m, , 1, , 1, , b) le =l+ t s = 3000 + 230 = 3230 mm = 3.230m, le =3.116 m, Step (3), Load & B M calculation, i) D.L. of slab, = 0.140×1×1×25 = 3.5 kN/m, ii) L.L. + FF of slab = 3.75×1×1, = 3.75 kN/m, Total load = 7.25 kN/m, Factored load (w d )=1.5×w, , 1, , =1.5×7.25, = 10.875 kN/m, w  l  10.875×(3.116) 2, BM = Mu= d e =, 8, 8, BM = Mu=13.199 kN-m, 2, , 1, , Page 5 of 30

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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (ISO/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2018, Sub. Code: 17604, Subject: Design of R.C.C. Structure, --------------------------------------------------------------------------------------------------------------Que., No., Q. 2, , Sub., Que., , Model Answers, , Marks, , Total, Marks, , Step (4), Check for depth, Mu max =M u, 0.138f ck b  d reqd  =13.199×106, 2, , 0.138×20×1000×  d reqd  =13.199×106, 2, ,  d  =69.153 mm < d=116 mm, reqd, , ........Ok, , 1, , Step (5), Maximum area of reinforcement, Ast max = 0.04×b×D =0.04×1000×140=5600 mm 2, Minimum area of reinforcement, 0.12, 0.12, Ast min =, bD=, ×1000×140=168 mm 2, 100, 100, Step (6), Main steel and its spacing, , 4.6×Mu×106 , 1- 1 bd, f ck bd 2, , , 0.5×20 , 4.6×13.199×106, A st =, 1- 1415 , 20×1000×(116) 2, A st =335.433mm 2, A st =, , 1, , 0.5f ck, fy, , ,  ×1000×116, , , Spacing of bar Min. of, π 2, 8, 4, =149.853 mm, 335.433, , 1000×, , a), , 1000×Aφ x, Sx =, =, A st, , b), , Sx =3d=3×116=348 mm, , c), , Sx =300mm, Sx =140 mm c/c, , Provide 8 mm φ bars @ 140 mm c/c along the shorter span, , 1, , Step (7), Distribution steel and its spacing, 0.15, 0.15, A std =, bD=, ×1000×140=210 mm 2, 100, 100, , Page 6 of 30

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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (ISO/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2018, Sub. Code: 17604, Subject: Design of R.C.C. Structure, --------------------------------------------------------------------------------------------------------------Que., No., Q. 2, , Sub., Que., , Model Answers, , Marks, , Spacing of bars is equal to min. of, π 2, 1000×Aφ y 1000× 4  6 , a), Sy =, =, = 134.640 mm, Ast d, 210, b), , Sy =5d=5×116=580 mm, , c), , Sy =450mm, , Total, Marks, , 8, , Sy =130 mm c/c, Provide 6 mm φ bars @ 130 mm c/c along the longer span, , (b), , Ans., , 1, , Design a two way slab for panel of effective size 5.6 m x 4 m, simply supported on all four sides. It carries a live load of, 3.5 kN/m2 and a floor finish of 1 kN/m2. Use M20 concrete, Fe 500, steel, MF of 1.2, 10 # bars and effective cover of 20 mm. Take αx =, 0.099 and αy = 0.051. Do not apply check for shear and bond., Draw the cross section along shorter span., To find:, Given:, D =?, lx = 4 m = 4000 mm, Ast in both direction = ?, ly = 5.6 m = 5600 mm, LL = 3.5 kN/m2, LL = 1 kN/m2, ǿ x = 10 mm, MF = 1.2, C, = 20 mm, αx = 0.099 and αy = 0.051, fck = 20 N/mm2, fy = 500 N/mm2, Page 7 of 30

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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (ISO/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2018, Sub. Code: 17604, Subject: Design of R.C.C. Structure, --------------------------------------------------------------------------------------------------------------Que., No., Q. 2, , Sub., Que., , Model Answers, , Marks, , Total, Marks, , Step (1), Slab thickness,, as l x =4 m > 3.5 m and LL = 3.5 kN/m 2 > 3kN/m 2 and Fe 500is used., Span, 4000, =, =166.667 mm, 20×MF 20×1.2, φ, 10, D = d+c+ x =166.667+20+ =191.667 mm, 2, 2, Provide, D =200 mm, 10, d = 200 - 20 = 175 mm, 2, Step (2), Effective span, l x =l xe = lx + d = 4000 +175 = 4175 mm = 4.175 m, d=, , 1, , 1, , l y =l ye = l y + d = 5600+175 = 5775 mm = 5.775 m, Step (3) Load & B M calculation, i) D.L. of slab = 0.2 × 1 × 1 × 25 = 5.0 kN/m, ii) L.L. of slab = 3.5×1×1, = 3.5 kN/m, i) F.F. of slab = 1×1×1, = 1.0 kN/m, Total load = 9.5 kN/m, Factored load (w d )=1.5×w, , 1, , =1.5×9.5, = 14.25 kN/m, BM calculations,, Mu x =α x .w d .  l xe  =(0.099×14.25×(4.175) 2 ), 2, , Mu x = 24.590 kN-m, Mu y = α y .w d .  l xe  =(0.051×14.25×(4.175) 2 ), 2, , 1, , Mu y =12.667 kN-m, Step (4), Check for depth, Mu max =M ux, 0.133f ck b  d reqd  = 24.590×106, 2, ,  d  = 96.148 mm < d =175 mm, reqd, , ........Ok, , 1, , Page 8 of 30

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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (ISO/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2018, Sub. Code: 17604, Subject: Design of R.C.C. Structure, --------------------------------------------------------------------------------------------------------------Que., No., Q. 2, , Sub., Que., , Model Answers, , Marks, , Total, Marks, , Step (5), Main steel and its spacing, In X direction, , 4.6×Mux×106 , 11,  bd, f ck bd 2, , , 0.5×20 , 4.6×24.590×106 , A st =, 1- 1 ×1000×175, 500 , 20×1000×(175)2 , A st = 339.665 mm 2, A stx =, , 0.5f ck, fy, , 1½, , Spacing of bar Min. of, 1000×Aφ x, =, A st, , π, 2, 10 , 4, = 231.227 mm, 339.665, , 1000×, , a), , Sx =, , b), , Sx = 3d = 3×175 = 525 mm, , c), , Sx = 300mm, Sx = 230 mm c/c, , Provide 10 mm φ bars @ 230 mm c/c, In Y direction, , 8, , d ' =d-φ x =175-10=165 mm, 0.5f ck , 4.6×Muy×106 , A sty =, 1- 1 bd', f y , f ck bd'2, , 0.5×20 , 4.6×12.667×106 , A sty =, 1- 1 ×1000×165, 500 , 20×1000×(165)2 , A sty = 181.565 mm 2, , 1½, , 0.12, ×1000×175=210 mm 2, 100, A sty =181.565 mm 2 > Astmin =210 mm 2, A stmin =, , A sty =210 mm 2, Using 8 mm dia.bar, Spacing of bar Min. of, 1000×Aφ y, , π 2, 8, 4, = 239.359 mm, 210, , 1000×, , a), , Sy =, , b), , Sy = 3d' = 3×165 = 495 mm, , c), , Sy = 300mm, , A sty, , =, , Sy = 230mm c/c, Provide 8 mm φ bars @ 230 mm c/c, , Page 9 of 30

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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (ISO/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2018, Sub. Code: 17604, Subject: Design of R.C.C. Structure, --------------------------------------------------------------------------------------------------------------Que., No., Q. 2, , Sub., Que., , Model Answers, , Marks, , Total, Marks, , 4, , 4, , (c)-(i) Draw the cross-section of a dog-legged staircase showing, reinforcement details., , Fig. Dog legged staircase, (Note: 3 marks for sketch and 1 marks for labeling.), , c)-(ii) A cantilever slab of effective span 1.0 m carries a superimposed, load of 1.5 kN/m2 including floor finish. Calculate the depth and, area of reinforcement. Use M20 concrete and mild steel. Take, MF = 1.55., , Page 10 of 30

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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (ISO/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2018, Sub. Code: 17604, Subject: Design of R.C.C. Structure, --------------------------------------------------------------------------------------------------------------Que., No., Q. 2, , Sub., Que., Ans., , Model Answers, Given:, le = 1 m = 1000 mm, LL + FF = 1.5 kN/m2, MF = 1.55, fck = 20 N/mm2, fy = 250 N/mm2, , Marks, , Total, Marks, , To find:, D =?, Ast in both direction = ?, , Solution:, , Step 1), Slab thickness, Span, d=, 7×M.F., Assume, Cover =15 mm and φ x =10 mm, 1000, =92.165 mm, 7×1.55, φ, 10, D=d+c+ x =92.165+15+ =112.165 mm, 2, 2, provide, D=120mm,, 10, d=120-15- =100 mm, 2, D=120 mm, d=100 mm, d=, , Step (2), Effective span, 100, le =1000 +, =1050 mm=1.05 m, 2, Step 3), Load cal. and BM, i) D.L. of slab, =0.120×1×1×25=3.0 kN/m, ii) L.L.+ F.F. of slab =1.5×1×1, =1.5 kN/m, Total laod (w), = 4.5kN / m, Factored load w d =1.5×4.5=6.75 kN/m, BM=M u =, ,  wd  le, 2, , 2, , =, , 1, , 1, , 6.75×1.052, =3.72 kN-m, 2, , Page 11 of 30

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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (ISO/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2018, Sub. Code: 17604, Subject: Design of R.C.C. Structure, --------------------------------------------------------------------------------------------------------------Que., No., Q. 2, , Sub., Que., , Model Answers, , Marks, , Total, Marks, , Step 4), Check for depth ,, Mu max = M ux, 0.149f ck b  d reqd  = 3.72×106, 2, , 0.149×20×1000×  d reqd  =3.72×106, 2, ,  d  = 35.33 mm < d=100 mm, reqd, , ........Ok, , Step (5), Main steel and its spacing, , 4.6×Mu×106 , 11,  bd, f ck bd 2, , , 0.5×20 , 4.6×3.72×106, A st =, 1- 1250 , 20×1000×(100)2, A st = 174.945 mm 2, A st =, , 0.5f ck, fy, , ,  ×1000×100, , , 1, , 0.15, ×1000×100=225 mm 2, 100, A st = 174.945 mm 2 < Ast min =225 mm 2, A st min =, , Hence, A st = 225 mm 2, Spacing of bar Min. of, π, 2, 10 , 4, = 349.06 mm, 225, , 1000×, , a), , 1000×Aφ x, Sx =, =, A st, , b), , Sx = 3d = 3×100 = 300 mm, , c), , Sx = 300 mm, , 4, , Sx = 300 mm c/c, Provide 10 mm φ bars @ 300 mm c/c, Step 6), 0.15, ×1000×150=225mm 2, 100, Assuming, 8 mm φ bars, Spacing of bar Min. of, π 2, 1000×Aφ y 1000× 4  8 , a), Sy =, =, = 223.402 mm, A st, 225, Ast y =Ast min =, , b), , Sy = 5d = 5×100 = 500 mm, , c), , Sy = 450 mm, , 1, , Sy = 220 mm c/c, Provide 8 mm φ bars @ 220 mm c/c, , Page 12 of 30

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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (ISO/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2018, Sub. Code: 17604, Subject: Design of R.C.C. Structure, --------------------------------------------------------------------------------------------------------------Que., No., Q. 2, , Sub., Que., , Q. 3, , Model Answers, , Marks, , 16, , Attempt any FOUR:, (a), , Ans., , State the necessary conditions for the beam to act as a flanged, beam., Following are the situations where a flanged RCC section is, preferred :, i. When slab and beam are to be casted together., ii. When main reinforcement of the slab is to be kept parallel to, the beam, transverse reinforcement is not less than 60% of the, main reinforcement at mid span of the slab., , Total, Marks, , 4, , 4, , Page 13 of 30

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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (ISO/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2018, Sub. Code: 17604, Subject: Design of R.C.C. Structure, --------------------------------------------------------------------------------------------------------------Que., No., Q. 3, , Sub., Que., (b), Ans., , (c), , Ans., , Model Answers, Write the expressions for effective flange width of T and L beams., State the meaning of each term., Expressions for effective flange width :, i. For T beam, l, bf = 0 +b w +6Df, 6, ii. For L beam, l, b f  0  bw  3D f, 12, 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., , State when and how minimum shear reinforcement is provided., Write the expression for minimum shear reinforcement giving the, meaning of terms involved., If Nominal shear stress (ζv) < Design shear strength of concrete (ζc),, minimum shear reinforcement should be provided., It is provided in form of stirrup., , Marks, , Total, Marks, , 1½, , 1½, , 4, , 1, , 1, , 1, , Expression for minimum shear reinforcement:, , Asv,  0.4/ 0.87fy, (b x Sv), Where,, Asv = total cross section area of stirrups legs effective in shear, Sv = stirrups spacing along the length of the member, b = breadth of beam or web of flanged beam, fy= characteristic strength of stirrup reinforcement in N/mm2 which, shall not be taken greater than 415N/ mm2., (d), , 4, 1, , 1, , A 16 mm diameter bar of grade Fe 500 is used for resisting, compression. Calculate the development length if the design bond, stress is 1.2 N/mm2 for plain bars in tension., , Page 14 of 30

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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (ISO/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2018, Sub. Code: 17604, Subject: Design of R.C.C. Structure, --------------------------------------------------------------------------------------------------------------Que., No., Q. 3, , Sub., Que., Ans., , Model Answers, 0.87×fy×φ, 4×τ bd ', ,  τ bd '=1.6×1.25×τ bd ---- for deformed bar, , , , the value of τ bd increased by 60% and, , , 0.87×500×16 , =, for bar in compression τ bd shall be increased , 4×1.6×1.25×1.2 , ,  by 25 %., , , , , , Ld = 725 mm, , Ans., , Total, Marks, , Given data: φ = 16 mm, fy = 500 N/mm 2 ,τ bd =1.2N/mm 2 , bar is in compression, Ld =, , (e), , Marks, , Write IS specifications for longitudinal and transverse, reinforcement of an axially loaded short column., IS specifications for longitudinal reinforcement of an axially, loaded short column:, i. Minimum diameter of bar in column = 12 mm, ii. Minimum number of bars in circular column = 6 Nos, iii. Cover of the column = 40 mm, iv. Minimum and maximum steel in column, Max % of steel = 6 % of gross cross sectional area of column, Min % of steel = 0.8 % of gross cross sectional area of column, IS specifications for transverse reinforcement of an axially loaded, short column:, i. IS specification for diameter of lateral ties: The diameter of the link, should be maximum of the following:, a) The diameter of the links should be at least one fourth of the, largest diameter of the longitudinal steel., b) In any case the links should not be less than 6mm in diameter., ii. IS specification for pitch: The spacing of the link should not exceed, the least of the followinga) The least lateral dimension of column., b) Sixteen times the diameter of the smallest longitudinal bar., c) 300 mm, , 1, , 2, , 4, , 1, , 2, , 4, , 2, , Page 15 of 30

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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (ISO/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2018, Sub. Code: 17604, Subject: Design of R.C.C. Structure, --------------------------------------------------------------------------------------------------------------Que., No., Q. 4, , Sub., Que., (A), (a), Ans., , Model Answers, Attempt any THREE:, Define characteristic strength and characteristic load., i. characteristic strength:, Characteristic strength of a material is the value of the material, below which not more than 5% of test results are expected to fail., ii. Characteristic load:, Characteristic load is that value of load which has 95% probability, of not being exceeded during the service life time of the structure., , (b), Ans., , Marks, , Total, Marks, 12, , 2, 4, , 2, , Why doubly reinforced beam is provided? Write the expression, for its moment of resistance if Xu < Xu max., Conditions where doubly reinforced section is provided are as, follows:, i) When the applied moment exceeds the moment resisting capacity, of a singly reinforced beam., 1 each, ii) When the dimension b and d of the section are restricted due to, (any, architectural, structural or constructional purposes., three), iii) When the sections are subjected to reversal of bending moment., e.g. piles, underground water tank etc., iv) In continuous T-beam where the portion of beam over middle, support has to be designed as doubly reinforced., v) When the beams are subjected to eccentric loading, shocks or, impact loads., , 4, , Expression for moment of resistance for Doubly reinforced beam, if Xu < Xu max, Mu = Mu1 + Mu 2, Mu = (Tu1 ×a1 ) + (Cu 2 ×a 2 ), , 1, , Mu = 0.87×f y ×Ast1  d-0.42×x u1   +  f sc -f cc  A sc  d-d'  , , (c), Ans., , Enlist the losses in prestressed concrete. Explain any one in brief., Losses in prestressed concrete:, i., Due to elastic shortening of concrete, ii., Due to creep of concrete, iii. Due to shrinkage of concrete, iv., Due to creep in steel, v., Due to frictional loss, vi., Due to slip at anchorages, , ½, each, (any, four), , Page 16 of 30

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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (ISO/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2018, Sub. Code: 17604, Subject: Design of R.C.C. Structure, --------------------------------------------------------------------------------------------------------------Que., No., Q. 4, , Sub., Que., , Model Answers, , Marks, , i., Due to elastic shortening of concrete: As the prestress is, transferred to concrete, the member shortens and prestressing steel, also gets shortened along with it, resulting in loss of prestressed in, steel., OR, 2 each, ii. Due to creep of concrete: Creep is a plastic deformation under (any, constant stress. Concrete under the action of constant stress continues one), to deform with time, causing loss of prestress., OR, iii. Due to shrinkage of concrete: During the process of drying and, hardening, concrete undergoes contraction reducing the prestressing, force., OR, iv. Due to creep in steel - The loss of prestress due to creep of steel, is the product of modulus of elasticity of steel and creep strain of steel., OR, v., Due to frictional loss: It takes place only in post-tensioning, system sue to relative movement between the tendon and the wall of, the duct., OR, vi. Due to slip at anchorages: The loss of prestress due to slip is due, to slipping of wires during anchoring., (d), , Ans., , Total, Marks, , 4, , A square column of side 425 mm is reinforced with 8 bars of 20, mm diameter of grade Fe 500. If the grade of concrete is M25,, calculate the safe load the column can carry., , Step 1, Gross area, A g = 425×425, = 180625 mm 2, Step 2, 2, π, Area of steel  A sc  = 8×   ×  20 , 4, = 2513.274 mm 2, Step 3, , Area of concrete  A C  = Ag - Asc, , 2, , = 180625 - 2513.274, = 178111.726 mm 2, , Page 17 of 30

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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (ISO/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2018, Sub. Code: 17604, Subject: Design of R.C.C. Structure, --------------------------------------------------------------------------------------------------------------Que., No., Q. 4, , Sub., Que., , Model Answers, , Marks, , Total, Marks, , 1, , 4, , Step 4, Ultimate load carrying capacity (Pu ), , Pu =  0.4×fck×A C  +  0.67×fy×A sc , =  0.4×25×178111.726  +  0.67×500×2513.274 , , = 2623064.05 N, = 2623.06 kN, Safe load carrying capacity (P), Pu 2623.06, P=, =, γf, 1.5, P =1748.707 kN, , 1, , 6, (B), , Attempt any ONE:, , (a), , A doubly reinforced beam of size 250 mm x 400 mm is reinforced, with 3, 20 # bars in tension and 2, 16 # bars in compression each, at an effective cover of 40 mm. Calculate the ultimate moment of, resistance if fck = 20 MPa, fy = 415 MPa and fsc = 353 MPa., , Ans., , Given:, b, = 250 mm, D, = 400 mm, C, = 40 mm, d, = D – C = 360 mm, π, Ast = 3× ×202 =942.477 mm 2, 4, π, Asc = 2× ×162 =402.123mm2, 4, fsc = 353 N/mm2, fck = 20 N/mm2, fy = 415 N/mm2, , To find:, Mu = ?, , Solution:, , Page 18 of 30

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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (ISO/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2018, Sub. Code: 17604, Subject: Design of R.C.C. Structure, --------------------------------------------------------------------------------------------------------------Que., No., Q. 4, , Sub., Que., , Model Answers, Step 1 :Find X umax =0.48d, , Marks, , Total, Marks, , ------ for Fe 415, , =0.48×360, X umax =172.8 mm, , 1, , Step 2 : Find Ast 2, f cc =0.45×f ck =0.45×20=9 N/mm 2, Ast 2 =, , (f sc -f cc )×A sc (353-9)×402.123, =, 0.87×f y, 0.87×415, , 1, , Ast 2 =383.133mm 2, Ast1 =Ast- Ast 2 =942.477-383.133=559.344 mm 2, Step 3 : Find Xu1, Xu1 =, , 0.87×f y ×Ast1, 0.36×f ck ×b, , =, , 0.87×415×559.344, =112.195 mm, 0.36×20×250, , Step 4 : Find type of section, As Xu1 =112.195 mm < X umax =172.8 mm, , 1, 6, 1, , Section is under-reinforced., Step 5 : Find Moment of Resistance M u, M u =0.87×f y ×Ast1×  d-0.42Xu1  +  f sc -f cc  ×A sc  d-d'  , , 1, , M u =0.87×415×559.344×  360-0.42×112.195  +  353-9  ×402.123×  360-40  , M u =107.451×106 N-mm, , 1, , M u =107.451kN-m, , (b), , Ans., , Calculate the area of steel reinforcements required for a doubly, reinforced beam 250 mm x 450 mm over all, subjected to ultimate, bending moment of 165 kN-m. Take fck = 20 MPa, fy = 415 MPa,, d’ = 45 mm and fsc = 353 MPa. The effective cover to tension steel, is 45 mm., Given:, b, = 250 mm, D, = 450 mm, C = d’ = 45 mm, d, = D – C = 405 mm, Mu = 165 kNm, fsc = 353 N/mm2, fck = 20 N/mm2, fy = 415 N/mm2, , To find:, Ast = ?, Asc = ?, , Page 19 of 30

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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (ISO/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2018, Sub. Code: 17604, Subject: Design of R.C.C. Structure, --------------------------------------------------------------------------------------------------------------Que., No., Q. 4, , Sub., Que., , Model Answers, , Marks, , Total, Marks, , Solution:, , Step, , 1) To find x umax, , x umax, , =0.48d, , 1, , =0.48×405, =194.4 mm, Step 2) To find M u1, M u1 =M ulim =0.138f ck bd 2, =0.138×20×250×4052, Step, , 1, , =113.177×106 N-mm, 3) To find A st1, , Pt lim =0.048fck=0.048×20=0.96% ---- for M20 Concrete, A st1 =, , Pt lim ×bd 0.96×250×405, =, 100, 100, , 1, , A st1 =972 mm 2, Step, , 6, , 4) Balanced moment of resistance (Mu 2 ), , Mu 2 =Mu-Mu1, =165×106 -113.177×106, , 1, , 6, , =51.823×10 N-mm, Step 5) To find Asc, fcc=0.45fck=0.45×20=9 N/mm 2, fsc=353 N/mm 2, Mu 2 =Asc(fsc-fcc)(d-d'), 51.823×106 =Asc(353-9)×(405-45), Asc=418.467 mm 2, Step, , 1, , 6)To find Ast 2, , Cu 2 =Tu 2, Asc(fsc-fcc)=Ast 2 ×0.87×fy, 418.467×(353-9)=Ast 2 ×0.87×415, Ast 2 =398.706 mm 2, , 1, , \Total Ast=Ast1 +Ast 2, =972+398.706, Ast=1370.706 mm 2, , Page 20 of 30

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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (ISO/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2018, Sub. Code: 17604, Subject: Design of R.C.C. Structure, --------------------------------------------------------------------------------------------------------------Que., No., Q. 5, , Sub., Que., , Model Answers, , Marks, , Attempt any TWO:, (a), , Ans., , Total, Marks, 16, , A simply supported beam of span 4 m carries a superimposed, load of 50 kN/m. The size of beam is limited to 230 mm x 400 mm, effective. Design the beam using concrete M20 and Fe 415 steel., Assume the cover of 40 mm to both reinforcements. Take, fsc = 353 N/mm2 and unit weight of R.C.C. as 25 kN/m3., Given:, b, = 230 mm, d, = 400 mm, C = d’ = 40 mm, w, = 50 kN/m, l, = 4 m = 4000 mm, Ρconc = 25 kN/m3, fsc = 353 N/mm2, fck = 20 N/mm2, fy = 415 N/mm2, , To find:, Ast = ?, Asc = ?, , Solution:, Step 1) To find Mu, D = d + c = 400 + 40 = 440 mm = 0.44 m, Total load acting on beam, Self weigth of beam = (b×D×ρconcrete )=(0.23×0.44×25)=2.53 kN/m, Superimposed load, , = 50 kN/m, Total load (w) = 52.53 kN/m, Factored load (w d ) =1.5×w =1.5×52.53=78.795 kN/m, , w d ×l2 78.795×42, =, =157.59 kNm, 8, 8, Step 2) To find x umax, Mu =, , x umax, , 1, 1, , =0.48d, =0.48×400, =192 mm, , 1, , Page 21 of 30

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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (ISO/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2018, Sub. Code: 17604, Subject: Design of R.C.C. Structure, --------------------------------------------------------------------------------------------------------------Que., No., Q. 5, , Sub., Que., , Model Answers, , Marks, , Total, Marks, , Step 3) To find M u1, M u1 = M ulim =0.138f ck bd 2, =0.138×20×230×4002, =101.568×106 N-mm, M u1 =101.568 kNm < M u =157.59 kNm, , 1, , Hence, Doubly reinforced beam is required., Step 4) To find A st1, Pt lim =0.048fck=0.048×20=0.96% ---- for M20 Concrete, A st1 =, , Pt lim ×bd 0.96×230×400, =, 100, 100, , A st1 =883.2 mm, Step, , 1, , 2, , 5) Balanced moment of resistance (Mu 2 ), , Mu 2 =Mu-Mu1, =157.59×106 -101.568×106, Step, , =56.022×106 N-mm, 6) To find Asc, , 8, , fcc=0.45fck=0.45×20=9 N/mm 2, fsc=353 N/mm 2, Mu 2 =Asc(fsc-fcc)(d-d'), 56.022×106 =Asc(353-9)×(400-40), Asc=452.374 mm, Step, , 1, , 2, , 7)To find Ast 2, , Cu 2 =Tu 2, Asc(fsc-fcc)=Ast 2 ×0.87×fy, 452.374×(353-9)=Ast 2 ×0.87×415, Ast 2 =431.011mm 2, Step, , 1, , 8)Total Ast=Ast1 +Ast 2, =883.2+431.011, Ast=1314.211mm 2, , 1, , Page 22 of 30

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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (ISO/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2018, Sub. Code: 17604, Subject: Design of R.C.C. Structure, --------------------------------------------------------------------------------------------------------------Que., No., Q. 5, , Sub., Que., (b), , Ans., , Model Answers, , Marks, , Total, Marks, , A beam 230 mm x 450 mm deep effective is reinforced with 4 – 16, # bars of grade Fe 415. The beam is subjected to a factored shear, force of 147 kN. Design the shear reinforcement. Use two legged, vertical stirrups of 8 # bars. Take ζuc = 0.57 N/mm2., Given:, b, = 230 mm, d, = 450 mm, , Ast = 4   (16) 2  804.248 mm 2, 4, Vu, = 147 kN, ϕ, = 8 mm diameter 2 legged, ζuc, = 0.57 N/mm2, fy = 415 N/mm2, , To find:, Spacing of stirrups = ?, , Solution :, , Step1) Nominalshear stress, Vu 147×103, =, =1.42N/mm 2, b×d 230×450, Step 2)Shear strength of concrete, ςv =, , ς uc =0.57 N/mm 2 <ς v =1.42N/mm 2, Shear reinforcement is required., Step 3)Shear force for which shear reinforcement is required, Vus =Vu - (ς uc ×b×d)=(147×103 )-(0.57×230×450), Vus =88.005 kN, Step 4)Shear force to be resisted by verticalstirrups, Assuming bentup bars are not provided., Shear force to be resisted by verticalstirrups, Vusv =Vus =88.005 kN, , 1, , 1, , 1, , 1, , Step 5)Spacing of stirrups, π, Asv=2× ×82 =100.53mm 2, 4, , Page 23 of 30

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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (ISO/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2018, Sub. Code: 17604, Subject: Design of R.C.C. Structure, --------------------------------------------------------------------------------------------------------------Que., No., Q. 5, , Sub., Que., , Model Answers, , Marks, , Total, Marks, , 3, , 8, , Spacing of stirrups = Min.of following 0.87×f y ×Asv×d 0.87×415×100.53×450, a)Sv=, =, =185.596 mm, Vusv, 88.005×103, 0.87×f y ×Asv 0.87×415×100.53, =, =394.525 mm, 0.4×b, 0.4×230, c)Sv=0.75×d=0.75×450=337.5 mm, d)Sv=300 mm, Hence,Sv=180 mm, b)Sv=, , 1, , Provide8 mm dia. 2 legged vertical stirrups at 180 mm c/c, (c), , Ans., , Design a square column to carry an axial load of 1500 kN. The, unsupported length of the column is 3.5 m. Use M20 concrete and, 1 % Fe 500 steel for longitudinal reinforcement. Use MS bar for, lateral ties. Apply the check for minimum eccentricity., Given:, l = lo = 3.5 m = 3500 mm, P, = 1500 kN, Asc = 1 % of Ag, fck = 20 N/mm2, fy = 500 N/mm2 for main steel, fy = 415 N/mm2 for transverse, steel, , To find:, Size of column = ?, Main steel = ?, Transverse steel = ?, , Solution:, Step 1) Factored axial load, Pu = 1.5×P = 1.5×1500 =2250 kN, Step 2) Size of column, 1, Asc =, ×Ag=(0.01)Ag, 100, Ac =Ag-Asc=(1)Ag-(0.01)Ag=(0.99)Ag, Using formulaPu, = (0.4×f ck ×Ac)+(0.67×f y ×Asc), , 1, , 1, , 1, , 3, , 2250×10 = (0.4×20×0.99×Ag)+(0.67×500×0.01×Ag), Ag =199.645×103 mm 2, For square column, b = Ag = 199.645×103 =446.816 mm, Provide column of size 450 mm×450 mm, , 1, , Page 24 of 30

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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (ISO/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2018, Sub. Code: 17604, Subject: Design of R.C.C. Structure, --------------------------------------------------------------------------------------------------------------Que., No., Q. 5, , Sub., Que., , Model Answers, , Marks, , Total, Marks, , Step 3) Check for eccentricity, e max =0.05×b=0.05×450=22.5 mm, b ,  l, e min =  o +  or 20 mm whichever is larger,  500 30 ,  3500 450 , =, +,  or 20 mm whichever is larger,  500 30 , e min =22 mm < e max =22.5 mm, , 1, , Step 4) Main Steel, Asc=0.01×Ag=0.01×4502 =2025 mm 2, Providing, 25 mm dia.bars, Asc 2025, No.of bars =, =, =4.12, Aφ π ×252, 4, Provide 6 bars of 25 mm dia. as main steel, , Step 5) Transversesteeli.e.links, 1, Dia.of link = ×φ or 6 mm whichever is greater, 4, 1, Dia.of link = ×25or 6 mm whichever is greater, 4, Dia.of link = 6.25 mm or 6 mm whichever is greater, Provide8 mm dia.links, Spacing of links = Minimum of below, a)S= b =450 mm, b)S=16×φ=16×25=400 mm, c)S=300 mm, S=300 mm, , 1½, , 1½, , Provide8 mm dia.links at 300 mm c/c, , Page 25 of 30

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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (ISO/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2018, Sub. Code: 17604, Subject: Design of R.C.C. Structure, --------------------------------------------------------------------------------------------------------------Que., No., Q. 6, , Sub., Que., , Model Answers, , Marks, , Attempt any FOUR:, (a), Ans., , Total, Marks, 16, , Differentiate between balanced and under-reinforced sections., Under reinforced, section, Strain in concrete and Strain in steel steel, steel reaches to its reaches to its maximum, maximum value at value first., same time., Equals to required for Less, compared, to, balanced section., balanced section., Ast = Ast max., Ast < Ast max., Xu = Xumax, Xu < Xumax, Balanced section, , i) Strain, , ii) Area, of Steel, , iii), Neutral, Axis, iv), Moment, Mu = Mumax, of, = qmax. fck. b . d2, resistance, (b), Ans., , 1 each, , 4, , 4, , 4, , Mu = Tu.z, =0.87 fy Ast(d-0.42xu), , Draw the sketch showing the cross-section of an isolated square, slopped footing. Show all the reinforcement details., , (Note : 3 marks for sketch and 1 for labeling)., , Page 26 of 30

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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (ISO/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2018, Sub. Code: 17604, Subject: Design of R.C.C. Structure, --------------------------------------------------------------------------------------------------------------Que., No., Q. 6, , Sub., Que., (c), , Ans., , Model Answers, , Marks, , Total, Marks, , A column of size 400 mm x 400 mm carries an axial load of 1500, kN. Calculate the size and depth for B.M. of a square pad footing, using M20 and Fe 500. The safe bearing capacity of soil is, 350 kN/m2., To find:, Size of footing = ?, Main steel = ?, , Given:, b = 400 mm, P, = 1500 kN, SBC = 350 kN/m2, fck = 20 N/mm2, fy = 500 N/mm2, Solution:, , Step 1, Ultimate S.B.C (q u )=2×350, = 700 kN/m 2, Step 2, Size of footing, Assuming 5% as self wt.of footing, Area of footing  A f  =, , 1.05×Pu  = 1.05×(1.5×1500) , qu, , = 3.375 m, L=, , 700, , 2, , 1, , Af, , = 3.375, = 1.837 m »1.9 m, Adopt size 1.9 m×1.9 m, , 1, , Page 27 of 30

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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (ISO/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2018, Sub. Code: 17604, Subject: Design of R.C.C. Structure, --------------------------------------------------------------------------------------------------------------Que., No., Q. 6, , Sub., Que., , Model Answers, , Marks, , Total, Marks, , Step 3, Upword soil pressure  p , p=, , Pu, 1.5×1500, =, = 623.268 kN/m 2, L×B, 1.9×1.9,   , , x1, 0.75, = 1×0.75×623.268×, 2, 2, = 175.294 kN-m, , M x = M y = 1×x1×p×, , Mx, d req =, =,  0.133×f ¶ ×b , , 6, , 175.294×10,  0.133×20×1000 , , 1, , = 256.709 mm»260mm, adopt cover of 50 mm, D = d + 50 = 260 + 50 = 310 mm, Provide, D = 310 mm and d = 260 mm, Step 5, , ,  4.6×M  , 0.5×fck , ux,   ×bd, Ast x =Ast y =, × 1- 1- , 2 , , , fy,   fck×bd   , , =, , ,  4.6×175.294×106  , 0.5×20 ,   ×1000×260, × 1- 1- ,   20×1000×2602   , 500 , ,  , , , = 1896.524 mm 2, using 16mm diameter, , 4, , 1, , π, 1000× ×162, 4, Sx = S y =, =, Ast, 1896.524, = 106.016 mm»100mm c/c, , 1000×Aφ , , Provide 16 mm φ @ 100 mm c/c both way, , (d), , Calculate the ultimate moment of resistance of a T-beam having –, flange width 1250 mm, thickness of slab – 115 mm, effective depth, – 600 mm, width of web – 300 mm and tension reinforcement, consisting of 4 bars of 25 mm diameter of grade Fe 500. The grade, of concrete is M20., , Page 28 of 30

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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (ISO/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2018, Sub. Code: 17604, Subject: Design of R.C.C. Structure, --------------------------------------------------------------------------------------------------------------Que., No., Q. 6, , Sub., Que., Ans., , Model Answers, , Marks, , Total, Marks, , Given :, bf = 1250 mm, Df = 115 mm, bw = 300 mm, d, = 600 mm, π, Ast = 4× ×252 =1963.495 mm 2, 4, fck = 20 N/mm2, fy = 500 N/mm2, To find: Mu =?, Solution:, , Step 1, Find x u, 0.36  fck  b f  x u = 0.87  fy  Ast, 0.36  20 1250  x u = 0.87  500 1963.495, , 1, , x u = 94.902 mm < Df =115 mm, Step 2, Find x umax, x umax = 0.46  d for Fe 500, = 0.46  600, = 276 mm,  As x u < x umax section is under reinforced, , 1, , 4, , Step 3, Find M u, M u =Tu  a, , = 0.87  fy  Ast   d - 0.42  x u , , 1, , = 0.87  500 1963.495   600  0.42  94.902 , = 478.428 106 N-mm, Mu = 478.428 kN-m, , 1, , Page 29 of 30

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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION, (Autonomous), , (ISO/IEC - 27001 - 2005 Certified), , Model Answer: Summer 2018, Sub. Code: 17604, Subject: Design of R.C.C. Structure, --------------------------------------------------------------------------------------------------------------Que., No., Q. 6, , Sub., Que., (e), , Model Answers, , Marks, , Total, Marks, , 4, , 4, , Draw the cross-section, strain diagram and stress diagram for a, singly reinforced T beam with the neutral axis within the flange., , Ans., , Page 30 of 30