/ / W ./- \...\..S/. i x ju L \sixu. o/xiyjx[a-.xj; wnere, x - Case 2- If NA lies below the flange, with variable stress in the flange, Df<xu<7/3 Df

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1 T and L beams Breadth of flange, bf For T-beams, bf = lo/6 + bw + 6Df For L-beams, bf = lo/12 + bw + 3Df For Isolated beams: For T-beams, bf= bw + lo/[(lo/bw)+4] For L-beams, bf = bw +0.5 lo//[(lo/bw) +4] lo is the distance between points of contraflexure Within the flange Xu<Df Position of Neutral axis Varying stresses in the flange Df<Xu<7/3Df Case 1 Case 2 Case 1> If NA lies with in the flange, xu < Df Qnn^s -y./- / / fy /Es Below the flange Determination of NA, xu: [bf x xu ]x[0.36xfck] = [Ast]x[0.87fy] Moment of Resistance, MR: MR = C x jd = [bfxxux0.36xfck]x[d-.42xxu] MR =T x jd = [Astx0.87xry]x[d-.42xxU] Unit IV Uniform stresses in the flange Xu>7/3 Df Case 3 Column. Footings and T-beam l Page R V R K Prasad, K D K College of Engineering, Nagpur Case 3:- If NA lies below the flange, with uniform stress in the flange, xu>7/3 Df fy/Es Determination of NA, xu: [bw x xu ]x[0.36xfck] + [(bf-bw)xdf]x[0.446xfck] = [Ast]x[0.87fy] Moment of Resistance, MR: MR = Cxjd = Cl xjdl+c2xjd2 = [bwxxux0.36xfck]x[d-.42xxu] +[(bf-bw)xdf x 0.446xfck] ]x[d-.df/2] -r,. :j _ r A,...A o-t.-^.n-.tj TT,.»,i. _ cl(0.42xu)+c2(d//2) i x ju L \sixu. o/xiyjx[a-.xj; wnere, x - Case 2- If NA lies below the flange, with variable stress in the flange, norm Df<xu<7/3 Df \l Df Yf \...\..S/. / / W fy /Es Determination of NA, xu: [bw x xu ]x[0.36xfck] + [(bf-bw)xyf]x[0.446xfck] = [Ast]x[0.87fy] MR = Cxjd = Cl xjdl +C2xjd2 = [bwxxux0.36xfck]x[d-.42xxu] +[(bf-bw)xyf x 0.446xfck] ]x[d-.yf/2] Where, Yf= 0.15 xu Df T x jd = [Astx0.87xfy]x[d-.x] ; = C1+C2

2 Question 1: Calculate the depth of neutral axis and ultimate moment of resistance of T beam section of M20 & Fe 415 for the following data. i) Flange width = 900 mm, ii) Flange thickness = 100 mm, iii) Web width = 300 mm iv) Effective depth = 500 mm v) Effective cover = 50 mm vi) Area of tension reinforcement = 3-25<j> bf=900 Df=lOO Question 2: Calculate the depth of neutral axis and ultimate moment of resistance of T beam section of M20 & Fe 415 for the following data. i) Flange width = 900 mm, ii) Flange thickness = 100 mm, iii) Web width = 300 mm iv) Effective depth = 500 mm v) Effective cover = 50 mm vi) Area of tension reinforcement (f> < > bf=900 Df=100 d=500 d=500 dc= <J> dc=50 -or1 3-20i > +4-25< > bw=300 bw=300 Area of steel = 3 x = mm2 Limiting NA, xu, max =0.48 d = 240 mm Depth of Neutral Axis, xu : Assuming the NA lies within the flange, xu < Df, i.e.; Case 1 [bf x xu ]x[0.36xfck] = [Ast]x[0.87fy] 900 x xu x 0.36 x 20 = x 0.87 x 415 xu = mm < Df [ UR & Assumption of <Df is Correct ] Moment of Resistance = C xjd or Txjd C x jd = [bfxxux0.36xfck]x[d-.42xxu] =(900x x0.36x20]x[ x ] =247.5 knm T x jd = Astx0.87xfy]x[d-.42xxu] = [ x0.87x415]x[ x82.051]= Area of steel = 3 x x = mm2 Limiting NA, xu, max =0.48 d = 240 mm Depth of Neutral Axis, xu : Assuming the NA lies within the flange, xu < Df, i.e.; Case 1 [bf x xu ]x[0.36xfck] = [Ast]x[0.87fy] 900 x xu x 0.36 x 20 = x 0.87 x 415 xu = mm > Df [ Assumption of xu < Df is Wrong] Assuming the NA lies below the flange, with flange subjected to uniform stresses. i.e.; xu >7/3 Df, i.e.: Case 3 [bw x xu ]x[0.36xfck] + [(bf-bw)xdf]x[0.446xfck] = [Ast]x[0.87fy] [300 x xu ]x[0.36x20] + 600x100]*[0.446x20] = [ ]x[0.87x415] ' xu = mm > [ xu > 7/3 Df is correct & UR 2 Page R V R K Prasad, K D K College of Engineering, Nagpur

3 Moment of Resistance = C x jd or T x jd Cxjd = Cl xjdl -f C2xjd2 = [bwxxux0.36xfck]x[d-,42xxu] +[(bf-bw)xdf x 0.446xfck] ]x[d-.df/2] = [300x82.05 Ix0.36x20]x[ x ] + [600x100x0.446x20] x[ /2] = knm T x jd = [Astx0.87xfy]x[d-.x] = [ x0.87x415]x[ ]=, * - Cl(Q.42*u)+c2(D/72) _ 514(99.945)+535.2(50) wnere, x ~ Question 3: Calculate the depth of neutral axis and ultimate moment of resistance of T beam section of M20& Fe415 for the following data. i) Flange width ~ 900 mm, ii) Flange thickness = 100 mm, iii) Web width = 300 mm iv) Effective depth = 500 mm v) Effective cover = 50 mm vi) Area of tension reinforcement = 2-20<j» <j> d= 50 D < r i dc=50 bf=900 * K/ u bw= <t> / +4-25( ) Area of steel = 2 x x = mm2 Df=100 Limiting NA, xu, max =0.48 d = 240 mm Depth of Neutral Axis, xu : Assuming the NA lies within the flange, xu < Df, i.e.; Case 1 [bf x xu ]x[0.36xfck] = [Ast]x[0.87fy] 900 xxux 0.36x20 = x 0.87x415 xu= 144,4. mm>df [ Assumption of xu < Df is Wrong] Assuming the NA lies below the flange, with flange subjected to uniform stresses, i.e.; xu>7/3 Df, i.e.; Case 3 [bw x xu ]x[0.36xfck] + [(bf-bw)xdf]x[0.446xfck] = [Ast]x[0.87fy] [300 x xu ]x[0.36x20] + [600xlOO]x[0.446x20] = [ ]x[0.87x415] xu = mm < [ Assumption of>7/3 Df is Wrong] Assuming the NA lies below the flange, with flange subjected to variable stresses, i.e.; Df < xu<7/3df, i.e.; Case 2 [bw x xu ]x[0.36xfck] + [(bf-bw)xyf]x[0.446xfck] = [Ast]x[0.87fy] - Where, Yf =0.15 xu Df [300 x xu ]x[0.36x20] + [600 (0.15xu+65)]x[0.446x20] = [ ]x[0.87x415] xu= mm -> 100<xu < [ Assumption of Df < xu <7/3 Df is Correct, UR] Moment of Resistance = C x jd or T x jd Cxjd = Cl xjdl +C2xjd2 = [bwxxux0.36xfck]x[d-.42xxu] +[(bf-bw)xyf x 0.446xfck] ]x[d-.yf/2] Yf= mm <Df - [300x82.051x0.36x20]x[ x ] + [600x x0.446x20] x[ /2] = knm T x jd = [Astx0.87xfy]x[d-.x] = [ x0.87x415]x[ ]=... cl(oa2xu')+c2(yf/2) (99.945) (47.382) _ Where, x - - C1+C _ DO. obi 3 Page R V R K Prasad, K D K College of Engineering, Nagpur

4 Doubly reinforced T-beams Question 4: Calculate the depth of neutral axis and ultimate moment of resistance of T beam section of M20& Fe415 for the following data, i) Flange width = 600 mm, ii) Flange thickness = 100 mm, iii) Web width = 250 mm iv) Effective depth = 550 mm v) Effective cover dc = 48 mm vi) Effective cover d'= 43 mm vii) Area of tension reinforcement = 3-20$ viii) Area of compression reinforcement = 2-10 d'=43 bf=600 Q Df=100 Question 5: Calculate the depth of neutral axis and ultimate moment of resistance of T beam section of M20& Fe415 for the following data. i) Flange width = 600 mm, ii) Flange thickness = 100 mm, iii) Web width = 250 mm iv) Effective depth = 550 mm v) Effective cover dc = 48 mm vi) Effective cover d'= 44 mm vii) Area of tension reinforcement = 5-20<J> viii) Area of compression reinforcement = 2-12 d'=44 bf=600 Df=100 d=550 dc=48 -o bw-250 Area of Tensile reinforcement = 3x314 = 942 mm2 Area of compression reinforcement = 2 x 78.5 = 157 mm2 Limiting NA, xu, max =0.48 d = 264 mm Depth of Neutral Axis, xu : Assuming the NA lies within the flange, xu < Df, i.e.; Case 1 And further assuming fsc = 0.87 ft & fee = fck bfxuq36fcl( + Asc(fsc - /cc) = Ast(Q.87fy) 600 x xu x 0.36 x (0.87x x20) = 942 x 0.87 x 415 xu = mm < Df [ UR & Assumption of <Df is Correct ] Moment of Resistance = C x jd - Cc x jd 1 + Cs x jd2 C x jd = [bfxxux0.36xfck]x[d-.42xxu] +[ Asc (fsc-fcc)] [d-d'] =[600x x0.36x20]x[ x ] + + [157 x(0.87x x 20)] [550-43) knm d=550 dc=48 -<J bw= <(> Area of Tensile reinforcement = 5 x 314 = 1570 mm2 Area of compression reinforcement = 2 x 113 = 226 mm2 Limiting NA, xu, max =0.48 d = 264 mm Depth of Neutral Axis, xu : Assuming fsc = 0.87 ft & fee = fck Assuming the NA lies within the flange, xu < Df, i.e.; Case I bfxuq36fck 4- Asc(fsc - fcc} = Ast(G.87fy) 600 xxux 0.36x (0.87x x20) = 1570x0.87x415 xu = mm > Df [ Assumption of xu < Df is Wrong] Assuming the NA lies below the flange, with flange subjected to uniform stresses, i.e.; xu>7/3 Df, i.e.; Case 3 bwxuq.36fck + (bf - Asc(fsc - fcc) = Ast(0.87fy) [250 x xu ]x[0.36x20] + [350xlOO]x[0.446x20] + 226x(0.87x x = [1570]x[0.87x415]. xu = mm < [ xu > 7/3 Df is Wrong ] 4 Page R V R K Prasad, K D K College of Engineering, Nagpur

5 Assuming the NA lies below the flange, with flange subjected to variable stresses, i.e.; Df < xu < 7/3 Df, i.e.; Case 2 bwxuq36fck + (bf - Asc(fsc - /«) = Ast(0.87fy) Where, Yf = 0.15 xu Df [250 x xu ]x[0.36x20] + [350 (0.15xu+65)]x[0.446x20] + 223(0.87x x20) = 1570 x 0.87 x 415 xu= mm -» 100<xu < [ Assumption of Df < xu <7/3 Df is Correct, UR] Moment of Resistance = C x jd = Ccl x jdl +Cc2 x jd2+ Cs x jd3 = [bwxxux0.36xfck]x[d-.42xxu] +[(bf-bw)xyf x 0.446xfck] ]x[d-.yf/2] +[Asc(fsc-fcc)][d-d'] Yf= 83.82mm <Df = [250x x0.36x20]x[ x ] + [350x83.82 x0.446x20] x[ ] + [226 x(0.87x x 20)] [450-44] = knm Question 6: Calculate the depth of neutral axis and ultimate moment of resistance of T beam section of M20 & Fe 415 for the following data. i) Flange width = 600 mm, ii) Flange thickness = 100 mm, iii) Web width = 250 mm iv) Effective depth = 550 mm v) Effective cover dc = 48 mm vi) Effective cover d'= 44 mm vii) Area of tension reinforcement = 5-20(j> viii) Area of compression reinforcement = 2-12 d'=46 bf= Df=100 Area of Tensile steel = 4 x = 2463 mm2 Area of Comp. steel = 2 x 201 = 402 mm2 Limiting NA, xu, max =0.48 d = 264 mm Depth of Neutral Axis, xu : Assuming the NA lies within the flange, xu < Df, i.e.; Case 1 Jyx«0.36/cfc + Asc(fsc - /«) = A«(0.87/y) 600 xxux0.36x (0.87x x20) =2463x0.87x415 xu=173. mm> Df Wrong Assuming the NA lies below the flange, with flange subjected to uniform stresses, i.e.; xu >7/3 Df, i.e.; Case 3 bwxuq36fck + (bf - bw)dfu6fck + Asc(fsc - fcc] = [300 x xu ]x[0.36x20] + [350 xloo] x[0.446x20]+ 402(0.87x x20) =[2463]x[0.87x415] xu = mm -> Assumption of Df < xu <7/3 Df is Correct, UR] Moment of Resistance = C x jd = Ccl xjdl +Cc2 x jd2 + Cs x jd3 = [bwxxux0.36xfck]x[d-.42xxu] +[(bf-bw)xdf x 0.446xfck] ]x[d-.df/2] +[Asc(fsc-fcc)][d-d'] = [250x x0.36x20]x[ x ] + [350x100 x0.446x20] x[ /2] + [402x(0.87x x20)][550-46] = knm d= $ Analysis of L beams are similar to that of T-beams dc-52 bw=250 5 Page R V R k Prasad, K D K College of Engineering, Nagpur

6 Short and Slender Compression Members Short column: A compression member may be considered as short when both the slenderness ratios Ix /D and ly/b are less than 12: Unsupported Length : The unsupported length, L, of a compression member shall be taken as the clear distance between end restraints. Effective Length of Compression Members: In the absence of more exact analysis, effective length 1, of columns may be obtained as described below: Degree of End Restraint of Symbol / Compression Members Diagram Effectively held in position at both ends, but not restrained against rotation Effectively held in position at both ends, restrained against rotation at only one end Effectively held in position and restrained against rotation in both ends Effectively held in position and restrained against rotation at one end but not held in position nor restrained against rotation at the other end 5c 7^r M 7 7^ > '- V r Theoretical Value of Effective Length l.ool 0.7 L 0.5 L 2.00 L Recommended Value of Effective Length l.ool 0.8 L 0.65 L 2.00 L [Note: In cases of practical difficulties in placing and compacting of concrete and lapping of reinforcement, a maximum percentage of 4 is recommended.] Minimum number of longitudinal bars: 4 in rectangular and 6 in circular columns. Minimum diameter of bar in longitudinal steel : Shall not be less than 12 mm Maximum Spacing of longitudinal bars: The distance between two bars, measured along the periphery of the column shall not exceed 300 mm. Transverse reinforcement: Diameter: The dia of the polygonal links or lateral ties shall be not less than 1/4 of the dia of the largest longitudinal bar, and in no case less than 6 mm. Pitch of lateral ties: The pitch of transverse reinforcement shall be not more than the least of the following distances: (i) The least lateral dimension of the compression members; (ii) 16 times the smallest dia of the longitudinal reinforcement bar to belied; and (iii) 300 mm. Pitch of helical reinforcement: It shall be not more than 75 mm, nor more than 1/6 of the core diameter of the column, nor less than 25 mm, nor less than 3 (j)t of the steel bar forming the helix. Arrangement of transverse reinforcement: If the longitudinal bars are not spaced more than 75 mm on either side, transverse reinforcement need only to go round comer and alternate bars for the purpose of providing effective lateral supports Minimum Eccentricity All columns shall be designed for minimum eccentricity, equal to the L/ 500 plus lateral dimensions/30, subject to a minimum of 20 mm. Longitudinal reinforcement: The cross-sectional area of longitudinal reinforcement, shall be not less than 0.8 percent nor more than 6 percent of the gross cross-sectional area of the column. 6 Page R V R K Prasad, K D K College of Engineering, Nagpur If the longitudinal bars spaced at a distance of not exceeding 48 times the diameter of the tie are effectively tied in two directions, additional longitudinal bars in between these bars need to be tied in one direction by open ties.

7 Axial load carrying capacity of column with zero eccentricity: Puz = 0.45 fck x Ac fy x Asc Short Axially Loaded Members in Compression: When the minimum eccentricity does not exceed 0.05 times the lateral dimension, the members may be designed by the following equation: Pu = 0.4 fck x Ac fy x Asc Compression Members with Helical Reinforcement : The strength of compression members with helical reinforcement satisfying the requirement of (*) shall be taken as 1.05 times the strength of similar member with lateral ties. The Requirement of helical reinforcement (*) : The ratio of the volume of helical reinforcement to the volume of the core shall not be less than Members with Axial Load and Uniaxial Bending: A member subjected to axial force and uniaxial Bending shall be designed by using charts in SP-16 (Design aids for reinforced concrete to IS 456) Question 1: Determine the ultimate load carrying capacity of a short column of 500 x 500 mm reinforced with 8 bars of 28((>. Consider concrete of grade M25 and reinforcement of Grade Fe 415 Solution Area of steel, Asc = 8 x = 4926mm2 i) Ultimate concentric axial load capacity Puz = 0.45 fck Ac fy Asc = 0.45 fck Ag + Asc( 0.75 fy fck) = 0.45 x 25 x(500x500) (0.75x x25) = 4290 JcN ii) Ultimate axial load capacity when the minimum eccentricity does not exceed times the lateral dimension Pu = 0.4 fck Ac fy Asc = 0.4 fck Ag + Asc( 0.67 fy fck) = 0.4 x 25 x(500x500) (0.67x x25) 3820 kn (89%) Nominal cover: for longitudinal reinforcing bar > 40 mm Incase of columns of minimum dimension of 200 mm or undr, whose reinforcing bars do not exceed 1 2mm, a nominal cover of 25mm may be used. Assumptions: In addition to the flexure, f) The maximum compressive strain in concrete in axial compression is taken as g) The max. compressive strain at the highly compressed fibre in concrete subjected to axial compression and bending and when there is no tension on the section shall be times the strain at least compressed fibre. 7 Page R V R K P ra sa d, K D K College of Engineering, Nagpur

8 Question 2: Design square column to carry superimposed load of 1700 kn. Unsupported length of column is 3.6m. Both ends are hinged. M25,Fe 415 Solution: 1) Assuming the column is short and ex < 0.05 D & ey < O'.OS b 2) Pu= 1.5 x 1700 = 2550 kn Assuming 1% steel Pu = 0.4 fck A + Asc( 0.67 fy fck) 2550 x 103 = 0.4 x 25 x(a) + (0.01A)(0.67x x25) A = 201,096 mm2; b = D= mm 3) Longitudinal reinforcement Pu = 0.4 fck A + Asc( 0.67 ry fck) 2550 x 103 = 0.4 x 25 x(450x450) + Asc(0.67x x25) Asc= 1959 mm2 Provide 1959mm2 Try 8 Nos Asc= 1959/8 = 244 $ = 17.65» 20 X Provide 4-20<J> Asc bal = x314=703mm2 Try 4 bars asc = 703/4= 175 < >= «16 Asc ^4-20$ $ r\ \y Provide 450 mm mj Select No. of bars Try 8 Nos asc= 1959/8 = 244 $=17.65*20 Provide 8-20$ Asc = 8x314 =2512 Asc =450x450* 1/1 00 = 2025 mm2 < > Asc =0.0 1* =20 10 /\] w Select dia. Try 1 6 mm<j> No of bars = 1959/201 =9.74 Provide 12-16$ Asc =12x201 = 2412 Provide 4-20$ $; Provided Asc = 4x x201 = 2060 mm2 4) Transverse reinforcement (lateral tie / link) Diameter, $t i) $m/4 = 20/4 = 5 mm ii) 6 mm V Spacing (pitch) i) least lateral dimension.b = 450 mm ii) 16 $m = 16 x 16 =256 mm V iii) 300 mm Provide 6 mm lateral ties at a pitch of 250 mm c/c 5) Initial assumption Checks i) le = 1.0 x L = 1.0 x 3600 = 3600 mm le/ b = 3600/450 = 8 < 12 -> short column -» OK ii) emin = L/500 +D/30 subjected to a minimum of 20 mm emin = (3600/500) +(450/30) = 22.2mm 0.05 D = 450/20 = 22.5 mm emin < 0.05 D -> OK 8 Page R V R K Prasad, K D K College of Engineering, Nagpur

9 Question 3: Design circular column with circular ties to carry superimposed load of 1600 kn. Unsupported length of column is 3.6m. Both ends are hinged. UseM25&Fe415 Solution: 1) Assuming the column is short and ex < 0.05 D & ey < 0.05.b 2) Pu = 1.5xl600 = 2400kN Assuming 1% steel Pu = 0.4 fck A + Asc( 0.67 fy fck) 2400x I03 =0.4x25 x(a) + (0.01A)(0.67x x25) A = 189,267 mm2; Dia= 491 mm Provide 500 mm 3) Longitudinal reinforcement Pu = 0.4 fck A + Asc( 0.67 fy fck) 2400 x 103 = 0.4 x 25 X(TI 2502) + Asc(0.67x x25) Asc = 1628 mm2 Provide 1628 mm2 Try 8 Nos Asc = 1628/8 = <(i=16.1 «20 X Provide 4-20<fr Asc bal = x mm2 Try 4 bars asc = 372/4 = 93 > =10.9* 12 O R O R Asc= /100 = 1893mm2 Try 8 Nos Asc =1628/8 = > = 16.1»20 Provide 8-20< > Asc = 8x314 =2512 5) Initial assumption Checks i) le= 1.0xL= 1.0x3600 = 3600 mm le/ b = 3600/500 = 7.2 < 12 -» short column -» OK ii) emin = L/500 +D/30 subjected to a minimum of 20 mm emin = (3600/500) +(500/30) = mm 0.05 D = 500/20 = 25 mm emin < 0.05 D -> OK Provide 4-20< > <(>; Provided Asc = 4x x113 = 1708mm2 4) Transverse reinforcement (lateral tie / link) Diameter, <f)t i) ((>m/4 = 20/4 = 5 mm i ii)6mm \g (pitch) i) leasi lateral dimension,b = 500 mm ii) 16<J>m = 16 x 12 =192 mm V iii) 300 mm Provide 6 mm lateral ties at a pitch of 175 mm c/c 9 Page R V R K Prasad, K D K College of Engineering, Nagpur

10 Question 4: Design circular column with HELICAL ties to carry superimposed load of 1600 kn. Unsupported length of column is 3.6m. Both ends are hinged. UseM25&Fe4l5 Solution: 1) Assuming the column is short and ex ^ 0.05 D & ey ^ 0/05 b 2) Pu= 1.5 x 1600 = 2400 kn Assuming 1% steel Pu = 1.05 { 0.4 fck A + Asc( 0.67 fy fck)} 2400 x 103 =1.05 { 0.4 x 25 x(a) + (0.01A)(0.67x x25)} A = 180,250 mm2; Dia= 4791 mm Provide 480 mm 3) Longitudinal reinforcement Asc = 180,250 /100 = 1803 mm2 Provide 1803 mm2 Try 8 bars; asc = 1803/8 = 225 4>=16.9 Provide 8-20$.. 4) Transverse reinforcement (lateral tie / link) - -: - Diameter, 4>t i) <}tm/4 = 20/4 = 5 mm ii) 6 mm V Pitch (spacing) * «= Diaofcore,Dc= 480-2x50 +2x6 = 392 mm Minimum pitch i) 25 mm V ii)3<t»t =3x6 =18 mm Minimum pitch i)dc/6 = 392/6 = V ii) 75 mm Vcore 7T62X = > Saf 7T3922x50 6) Initial assumption Checks i) le/ b = (1.0x3600)/500 = 7.2 < 12 -> short column -> OK ii) emin = L/500 +D/30 subjected to a minimum of 20 mm emin = (3600/500)+(480/30) = 23.2 mm < 24 [480/20] -» OK Provide 6 mm helical ties at a pitch of 50 mm c/c 5) check for helical reinforcement Vḣel.ties 7r >0.36 fy TT = Considering one pitch length = 50 mm Volume of core,vcore = nr2h = n 1962 x 50 = 6,034,370 Length of circular tie = 2nr = 2n x 386 = mm Length of helical tie = V SO Page R V R K Prasad, K D K College of Engineering, Nagpur

11 Columns subjected to axial compression and uniaxial bending M = Pxe. M The loads are rarely axial, for several reasons like bad workmanship, uncertainty in location of application of loads etc. Minimum Eccentricity All columns shall be designed for minimum eccentricity, equal to the IV 500 plus lateral dimensions/30, subject to a minimum of 20 mm. IT IS CERTAIN THAT EVERY COLUMN MUST BE DESIGNED F BENDING As eccentricity increases, BM increases and axial load capacity reduces Various combinations of load and eccentricity can be easily determined by using the monograms given by BIS in SP-16 A typical interaction curves (Chart 27 ~ Chart 62) as shown in figure. Question 5: Determine the ultimate load carrying capacity of a short column of 400 x 500 mm reinforced with 10-16<t>. Take M25 & Fe 415 Solution: It is a short column and assuming emin < O.OStimes the lateral dimension Asc= 10x201 =2010mm2 Pu = 0.4 fck Ac ry Asc = 0.4 fck A + Asc( 0.67 fy fck) = 0.4 x 25 x(400x500) (0.67x x25) = kn Question 6: Determine the maximum eccentricity about major axis, at which an ultimate load 1800 kn can be applied on a short column of 400 x 500 mm reinforced with 10-16$. Take M25 & Fe 415 Solution: Column is subjected to axial load and uniaxial bending Asc= 10x201 =20IOmm2 Fe 415 type of steel and reinforcement distributed equally on four faces with dvd = 50/500= 0.1 -> Chart 44 of SP-16 p = 100 Asc /(bd) = 100 x 2010 /(400x500) = 1 % p/fck = 1/25 = » Diagram 3rd from left Pu/fck b D = 1800,000/(25x400x500) = 0.36 From interaction Diagram Mu/fck b D2 = Mu = 0.07 (25x 400 x 5002) = 175 knm [Units] e = Mu/Pu = 175/1800 = m = 97 mm 11 P a g e R V R K Prasad, K D K College of Engineering, Nagpur

12 Question 7: Determine the ultimate load a short column of 400 x 500 mm reinforced with <J> can carry with an eccentricity of 1 25mm about major axis. Take M25 & Fe 415 Solution: Column is subjected to axial load and uniaxial bending dyd = 50/500 = 0.1 Asc = 10x201 =2010mm2 p= 100 Asc/(bD)= 100 x 20107(400x500)= 1% p/fck = 1/25 = > Chart 44 - Diagram 3rd from left dy Pu/fckbD D Draw a line on interaction chart with the given slope; the intersection point with the corresponding interaction diagram is the ultimate load it carry with 125 mm eccentricity Pu/fckbD = 0.33; Pu = 0.33 (25x400x500) = 1650 kn Question 8: Determine the reinforcement required in a short column of 400 x 500 mm, if it has to carry an axial load of 1400 kn with an eccentricity of 95mm about major axis. Take M25 & Fe Solution: Pu=1.5xP=1.5x 1400 = 2100 kn Pu/fck b D xl03/(25x400x500) = 0.42 Mu = Pux e = 2100x knm Mu/fck b D2 = x!06/(25x400x500-) = 0.08 d Yd = 50/500= 0.1 from Chart 44 Pu/fck b D = 0.42 & Mu/fck b D2 = > p/fck = 0.06 /. p = 0.06x25= 1.5% Asc = (1.5/100) * (400 x 500) mm2 Try 20 mm <(»; no of bars = 3000/3 1 4 = 9.55 say 1 0 bars Transverse reinforcement (lateral tie/ link) Diameter, <j)t i) <J>m/4 = 20/4 = 5 mm ii) 6 mm V Spacing (pitch) i) least lateral dimension,b = 400 mm ii) 16 (ftm =16x20 =320 mm iii) 300 mm V Provide 6 mm lateral ties at a pitch of 250 mm c/c 0 0 fl 12 (P a g e R V R K P r a s a d, K D K C o l l e g e of E n g i n e e r i n g, N a g p u r

13 Isolated Column Footings for axial loads (IS 1904) Footings shall be designed to sustain the applied loads and the induced reactions and to ensure that any settlement which may occur shall be as nearly uniform as possible, and the safe bearing capacity of soil is not exceeded. The thickness of edge shall not be less than 150 mm Critical sections for design: for BM is at the face of column for SF = at d from face of column ( designing as wide beam) = at d/2 from the periphery of column (two-way) Permissible shear stress = k 0.25 Vfck (For two-way) Where k = 0.5 +pc < 1 ; p = L/B && pc = B/L Reinforcement in central band: In two-way reinforced rectangular footings, for reinforcement in the shorter direction, Ast required in the central band = 2/(p+l) Ast,tot Transfer of loads: The compressive stress at base of column < k 0.45 fck Where, k = V(A1/A2) but not greater than 2 For calculating A1 consider dispersion of 1 in 2 R i V: Rj K! i pi i r a! i S] a! d! Question 10: Design a pad footing for column 400x600mm carrying an axial load of 1400 kn. SBC. of soil is 220 kn/m2 and unit weight is 17 kn/m3. Use M 20 concrete and Fe 415 steel. Draw a neat reinforcement sketch Solution: 1). Dimensions of footing:- Load on column = 1400 kn Selfwt (10%) =140kN Total load =1540 kn Area of foundation = 1540/220 =7 m2 Try 2.5 x 2.8 m Offset, longer direction = ( )/2 =l.lm Offset shorter direction = ( ) = 1.05m 2). Factored Net soil pressures Kj Net upward pressure = 14007(2.5x2.8) = 200 kn/m2 D; Factored soil pressure = 1.5 x 200 = 300 kn/m2 KJ G EJ N: i aj si pi ui 1.1 3). Check for Bending i) In longer direction - considering unit width of footing Critical face of column, section 1-1 BM = 1.0x[300 xl.!2/2 ]= knm d = V(M/Qb) = V(181.5xl06/2.76xlOOO) -256 Provide, d = 440 mm & D = /2 = 500 mm Ast = fck fy i 4.6xl81.5xl xlOOOx Mu 1 fck b d2 bd V 1000x440 = 1212mm2 Ast, min = 85/415=0.205 % -> x 1000x440 =902 mm2 Spacing of 16< > bars = 1000x201/ say 150mmc/c 13 Page R V R K Prasad, K D K College of Engineering, Nagpur

14 ii) In shorter direction - considering unit width of footing Critical face of column, section 2-2 BM= 1.0x[300 xl.052/2 ]= knm d = V(M/Qb) = V( 165 Ax 106/2.76x 1000) = 244 < 440 Ast = x16 5.4x106 20xlOOOx x440 = 1098mm2 Spacing of 12< > bars = 1 OOOx 113/1098 = 102 say 100 mm c/c ii). Two way shear (punching shear) Critical section the periphery of column 600+d= 1140;400 + d = 940 Perimeter of critical section, bo = 2( ) = 4160 mm Max SF, Vu= (2.5x xo.94) x 300 = kn Shear stress, TV =1778,500/(4160x540) = 0.79 N/m2 < ( kx0.25vfck ) m B 1 ', T/ / / 1.05 > / / I \/ / / 1 1m 4. Check for shear in longer direction : i) One way shear (wide beam) Critical section d from face of column Offset i.e.; = 0.66 m from free end Max SF, Vu = (1 xo.66) 300 = 198 kn Shear stress, TV = 198,0007(1000x440) = 0.45 N/m2 Percentage of steel, p = 100 Ast/bd = 100 x 1212 /(1000x440) =0.275% Linear interpolation, y = y± + A* For M20; p TC 0.25% 0.36 Mpa 0.50% 0.48 MPa For 0.275% ; TC = (0.12/0.25)0.025 = MPa TV > T c unsafe Increase depth such that the shear stresses are within permissible limit = 198,000/( loooxd) -> d = 532 Increase effective depth to 540 mm,.'. D = /2 =60o mm 7. Development length Ld = <t>(0.87fy)/4tbd = < >0.87x415/(4xl.6xl.2) = For 12 mm-* 564 mm < ( x16) For 16mm -> 752 mm < ( x16) Safe Safe 8. Transfer of load at base of column For Al -> *0.54 = 2.76; *0.54 =2.56 > 2.5 V(A1/A2) = V(2.76x2.5)/(0.6x0.4) =5.4 > 2 Permissible bearing pressure, =2 x 0.45 x 20=18 N/mm2 Max. stress in column = 1.5x 1400x107400x600 = 8.75 N/mm2 < 18 Safe 14 Page R V R K Prasad, K D K College of Engineering, Nagpur

15 Question 11: Design a footing for column carrying an ultimate load of 1800 kn The size of column is 300 x 600 mm and S.B.C of Soil 275 kn/m2 Use M20 concrete and Fe 415, Draw reinforcement details. O O P o ooo O o o o o o o 1. Dimensions of footing: T Service load =1800/1.5= 1200 kn Load on column = 1200 kn Selfwt =120kN Q Total load = 1320kN Area of foundation = 1320/275= 4.8 m2 Try 2.05 \5 m Longer side offset = (L-D)/2 = ( )/2 = 0.875m Shorter side offset = (B-b)/2 B = ( X2 = m 2. Factored Soil pressures Factored net soil pressure = 1800/(2.05x2.35) = kn/m2 3. Check for Bending consider unit width i) Bending in longer direction "<-» *,. $ ai 2.8m Critical face of column, BM = x /2 = 143 knm d = V(M/Qb) = V(143xl06/2.76x2000) = 228 Provide d = 490 mm & D = 550 mm fck fy A 4.6xl43xl06 20xlOOOx4902 [shear POV] Mu 1 1 bd 1 * r I L J7 J M b d2 looox^ion Ast, min = 85/415= % -^ x 1000x490 = 1005 mm2 Try 12 mm<t> bars, spacing = 1000 x 113/1005 = 112 mm c/c Provide I2< >@ 110 c/c 15 Page R V R K Prasad, K D K College of Engineering, Nagpur

16 ii) Bending in shorter direction As the offset is same, Provide 12< HOc/c 4. Check for shear i) one way shear : for unit width Critical d from face of column i.e.; = 385 mm Max S?=\x 0.385x = kn - Shear stress, T = 143,8407(1000x490) = 0.29 N/m2 For M20; p TC 0.15% 0.28 Mpa 0.25% 0.36 MPa For 0.205% ; TC = (0.08/0.10)0.055 = MPa TV <TC safe ii). Check for two way shear (punching shear) Critical d/2 from face of column Dimension of critical section =1090 & =790mm Perimeter, bo = 2( ) = 3760 mm Max SF = (2.05x x1. 09) x = kn Shear stress, T= xlo3 7(3760x490) = 0.98 N7m2 < 1.118(0.25^fck ) 5. Development length Ld =(t>(0.87fy)74tbd = For 1 2 mm-» 564 mm (t 0.87x4157(4x1.6x1.2) = 47< ) < ( x 1 2) Safe 6. Transfer of load at base of column The compressive stress at base of column = 1800, x600= 10N7mm2 For A 1 -> *0.49 = 2.26 > 2.05; *0.49 =2.56 > 2.35 V(A17A2) = V(2.05x2.35)7(0.3x0.6) =5.2 > 2;.-. k =2 Permissible bearing pressure, ^2 x 0.45 x 20=18 N7mm2 Safe R V R K P r a s a H K K F. a g n u r )5rr L 10 c) ") (~) fi r) ^ \ ^ ^^ () r u 0 ) f 1 \- U< ) 2.3; ) Sc 3 rr ( s i } r) r) () i-) (u x v 16 Page R V R K Prasad, K D K College of Engineering, Nagpur

17 Question 12: Design a sloped footing for column carrying an ultimate load of 1800 kn The size of column is 300 x 600 mm and S.B.C of Soil 275 kn/m2 Use M20 concrete and Fe 415, Draw reinforcement details. 1. Dimensions of footing: T Service load = 1800/1.5 = 1200 kn Load on column = 1200 kn Self wt = 120 kn Total load = 1320 kn Area of foundation = 1320/275= 4.8 m2 Try 2.05 x 2.35m Longer side offset = (L-D)/2 = ( )/2 = 0.875m Shorter side offset = (B-b)/2 = ( )/2 = 0.875m 2.05 B [ Solve (0.3 +2x)(0.6+2x) = 4.8; ->x«0.875m ] 2. Factored Soil pressures Factored net soil pressure = 1800/(2.05x2.35) = kn/m2 3. Check for Bending consider unit width i) Bending in longer direction Critical face of column, Area of portion (1) = ( )/2 x = m2 CG from face of column, x =(0.3+2x2.05)/( )(0.875/3) =0.546 m BM = (373.6 x 1.028)0.546 = knm d = V(M/Qb) = V( xl06/2.76x300) = 503 mm Provide d = 540 mm & D = 600 mm 0.5 fck fy I 4.6 Mu 1 1 * r fckbd2 bd ii) Bending in shorter direction Area of portion (2) = ( )/2 x = 1.29 m2 CG from face of column, x =(0.6+2x2.35)/( )(0.875/3) =0.524 m BM = (373.6 x 1.29) = knm d = V(M/Qb) = V(252.54x 106/2.76x600) = 390 < 540 mm Ast = I J 4.6x252.54xl06 20x600x x540 = 1426mm2 Ast, min = 85/415= % -> x600x540 =664 mm2 Try 12mm<t>bars,Noofbars = 1426/113 = 13 bars Provide 200 free end (eff. depth = 140) 4. Check for one way shear Critical d (540mm) from face of column i)in longer direction Dimensions of the critical section are bo = /875 * 540 = 1380 mm do = (300/875)540 = 355 mm = 335 mm Area of portion (1') =( )72 x = m2 MaxSF= 0.575x373.6 =214.8 kn Shear stress, T = (1380x355) = 0.44 N/m2 For M20; p TC 0.15% 0.28 Mpa 0.25% 0.36 MPa For 0.205%; TC = (0.08/0.10)0.055 = MPa TV > T c Unsafe Increase depth T = (1380xd) = >d = d required at face of column = ( )/335 x875 = 1029 mm sayd=1040&d = 1100 mm Ast = x xl06 20x300x x540 = 1289mm2 Ast, min «= 85/415= % -^ x300x540 =332 mm2 Try 12 mm<t> bars. No of bars = 1289 // 113 = 12 bars 17 Page R V R K Prasad, K D K College of Engineering, Nagpur ii) In shorter direction bo = /875 * 540 = 1680 mm do = (900/875)540 = 484 mm = 335 mm Area of portion (!')=(! )/2 x = m2

18 MaxSF= 0.675x373.6 = kn Shearstress,-c = /(1680x484) = 0 N/m2 <0.324( 0.205% 5). Check for two way shear (punching shear) Critical d/2 (270) from face of column Dimension of critical section =1 140 & =840mm Perimeter, bo = 2( ) = 3760 mm Effective depth do = (300/875)270 = mm Max SF = (2.05x xu4)x =1442 kn Shear stress, T= 1442 x 103/(3760x447.4) = N/m2 < 1.118(0.25Vfck 6. Development length Ld = <t>(0.87fy)/4tbd = <1> 0.87x4 15/(4x 1.6x1. 2) = For 1 2 mm-» 564 mm < ( x 1 2) Safe 7. Transfer of load at base of column The compressive stress at base of column = 1800,000/300x600= 10N/mm2 For A 1 -» *0.54 = 2.46 > 2.05; *0.54 =2.76 > 2.35 V(Al/A2) = V(2.05x2.35)/(0.3x0.6)=5.2>2;.-. k =2 Permissible bearing pressure, =2 x 0.45 x 20=18 N/mm2 Safe Note: slope is too steep -» depth at free end to be increased R V R K P r a s a d K D K C E Ml a >.05n in 0 nonnoooooo 0 n o U LT 0 n Shear msy bre checked by TV = bd,;* 2.35m 18 Page R V R K Prasad, K D K College of Engineering, Nagpur

19 Winter 14 Q(7)[14M]: Q(8)[ 14M]: Summer 15 Q (7a) [ 6M ] :Determine the moment of resistance of T-beam for the following data bf = 1000 mm; Df = 120 mm, bw = 300mm, Ast = 6 x 25, d = 600mm, M20andFe415. Q (7b) [ 7M ]: Design a RCC column of rectangular section having unsupported length of 3m subjected to axial compressive load of 1200 kn. Use M20 and Fe 415. One dimension is restricted to 400 mm. Q (8) [ 13M ]: Design a rectangular pad footing for a column size of 300 x 500 mm with compressive load of 1000 kn. Use M20 and Fe 415. Density of soil is 21 kn/m3 and SBC of soil is 150 kn/m2. Give the necessary checks as per IS 456 and draw a neat sketch. Winter 15 Q (7) [ 13M ] : Design a short axially loaded circular column with helical rings. The diameter of column is 400 mm and axial service load on column is 800 kn. Use M 20 concrete and Fe 415 steel. Draw a neat reinforcement sketch. Q (8) [ 13M ] : Design a pad footing for column 400x600mm carrying an axial load of 1000 kn. S. B. C. of soil is 200 kn/rm. Use M 20 concrete and Fe 415 steel. Draw a neat reinforcement sketch Summer 16 Q (7) [ 13M ]: Design a short column of reinforced concrete to carry an axial land of 1150KN. Width of the column is restructed to 300mm. Also Design a rectangular footing of uniform thickness for an axial lended column of 1150 KN. Safe bearing capacity of soil is 200 kn/m2. Use M20 concrete and Fe 415 grade of steel. Q (8a) [ 6M ] : A simply supported T beam has flange width of 2400 mm and flange thickness of 120 mm. The effective span of beam is 3.6m. The effective depth of beam is 580 mm and its width 300 mm. It is reinforced with 8-20 mm diameter, Fe 415 grade steel. Determine the moment of resistances of the section. Use M 20 grade of concrete. Q (8b) [ 7M ] : Design a circular column of diameter 400 mm subjected to a load of 1200 KN. The column is having spiral ties. The column is 3 m long and is effectively held in position at both ends but not restrained against rotation. Use M25 concrete and Fe 415 steel. Winter 16 Q (7) [ 13M ]: Design the reinforcement in a circular column of diameter restricted to 450 mm with helical reinforcement of 8mm diameter to support a factored load of 1500 kn. The column has unsupported length of 3.5m and is braced against sideway. Also design a rectangular footing of uniform thickness for an axial loaded column of 1500 KN safe bearing capacity of soil is kn m Use M20 concrete and Fe 415 grade of steel Q (8a) [ 7M ] : Calculate the depth of neutral axis and ultimate moment of resistance of T beam section; for the following data. Flange width = 800 mm, Flange thickness = 150mm, Web width = 300mm Effective depth = 420mm Area of tension reinforcement = mm Assume = M 25 grade concrete and Fe 415 grade of steel Q (8b) [ 6M ]: Design a circular column of diameter 500mm subjected to a load of 1350 KN. The column is having spiral ties. The column is 3.5m long and is effectively held in position at both ends but not restrained against rotation. Use M20 concrete and Fe 415 steel Summer 17 Q (7a) [ 6M ] : Determine the moment of resistance of T-beam for the following data using M20 and Fe 415 bf = 1200 mm; Df = 110 mm, bw = 300mm, Ast = 6 x 20, d = 600mm, Q (7b) [ 7M ]: Design a short rectangular RCC column of rectangular section having unsupported length of 3m subjected to axial compressive load of 700 kn. Use M20 and Fe415. One dimension is restricted to 350 mm. Q (8) [ 13M ]: Design a rectangular slopped footing for a column size of 400 x 400 mm carrying an axial load of 600 kn. Use M20 and Fe 415. SBC of soil is 200 kn/m2. Give the necessary checks as per IS 456 and draw a neat sketch. 19 (P a g e R V R K Prasad, K D K College of Engineering, Nagpur