SBCE – CE
Design of Steel Structures
15. 15.1.
PRACTICE PROBLEMS
PROBLEMS FROM CHAPTER 5
15.1.1. Problem 1 Calculate the ultimate moment carrying capacity of a rectangular beam of M30 concrete having a width of 250mm and depth 400mm and having an effective cover of 50mm. It is reinforced with Fe250 steel of area 1800mm2 on the tension face. Solution Effective depth, d = Overall depth – Effective cover = 400 – 50 = 350mm Step 1: Determination of Neutral axis 0.87
0.36
.
which gives
.
.
145
.
Step 2: Check for type of section (Cl.38 of IS 456:2000) xu,max for Fe250 = 0.53d = 0.53*350 = 185.5mm > xu. Hence Under reinforced section. Step 3: Computation of Ultimate moment carrying capacity . .
Ultimate Moment carrying capacity, Mu = T*Lever arm = 0.87
250
1800
350
0.42
0.42
0.87
145 = 113.18x106Nmm = 113.18kNm.
15.1.2. Example 2 Calculate the moment carrying capacity of a reinforced section with width 250mm and effective depth 400mm having a tensile reinforcement 0f 3600mm2. Assume grade 20 concrete and 415 grade steel. Solution Step 1: Determination of Neutral axis 0.87
0.36
.
which gives
.
.
722
.
Step 2: Check for type of section (Cl.38 of IS 456:2000) xu,max for Fe415 = 0.48d = 0.48*400 = 192mm < xu. Hence Over reinforced section. Limit x to xu,max. Step 3: Computation of Ultimate moment carrying capacity . . 0.36 Ultimate Moment carrying capacity, 6 250 192 400 0.42 192 = 110.37x10 Nmm = 110.37kNm.
0.42
0.36
20
15.1.3. Example 3 A singly reinforced rectangular section is of breadth 150mm and depth 350mm. The tension steel consists of 3Nos HYSD bars of 16mm dia. The stirrups are of mild steel 8mm dia. Cover to steel is 25mm. Assuming M20 concrete, determine the moment capacity of the section. Solution Compiled By: Vaisakh G.
PP1
SBCE – CE
Design of Steel Structures
Effective depth = Overall depth – clear cover – dia. of stirrups – dia. of main bar/2. i.e., d = 350-25-816/2 = 309mm Step 1: Determination of Neutral axis 0.87
0.36
.
.
which gives
201.65
.
.
x
= 0.87fyAst/0.36fckb = 201.65mm Step 2: Check for type of section (Cl.38 of IS 456:2000) xu,max for Fe415 = 0.48d = 0.48*309 = 148.32mm < xu. Hence Over reinforced section. Limit x to xu,max. Step 3: Computation of Ultimate moment carrying capacity . . 0.36 Ultimate Moment carrying capacity, 250 192 400 0.42 192 0.36 20 150 148.32 309 =39.52x106Nmm = 39.52kNm.
0.42 0.36 0.42 148.32
20
15.1.4. Example 4 A rectangular beam 200x400mm is to be constructed in moderate environment. Find the reinforcing steel required to resist a bending moment of 25kNm due to working loads. The diameter of shear steel is 10mm. Assume M20 concrete and Fe415 steel. Solution Ultimate bending moment = 1.5*25 = 37.5kNm. Nominal Cover = 30mm for moderate exposure (Table 16 of IS 456 – 2000) Diameter of Main bar = 16mm (Assume) Effective depth = Overall depth – clear cover – dia. of stirrups – dia. of main bar/2. i.e., d = 400-3010-16/2 = 352mm Step 1: Determination of xu,max (Cl.38 of IS 456:2000) xu,max = 0.48d for Fe415 = 0.48*352 = 168.96mm Step 2: Estimation of reinforcing steel required 0.87
0.36
Number of bars required =
.
which gives .
. .
. .
= 673.87mm2
= 3.35nos, say 4#16dia.
15.1.5. Example 5 Design a RCC beam with a simply supported effective span of 5.5m. The beam is required to support live and superimposed loads of 15kN/m and 10kN/m respectively. Use M20 concrete and Fe415 steel. Solution Compiled By: Vaisakh G.
PP2
SBCE – CE
Design of Steel Structures
fck = 20N/mm2; fy = 415N/mm2; wl = 15kN/m; ws = 10kN/m; leff = 5.5m For estimation of depth, assume D = leff /12 = 5500/12 = 458.33mm say D = 500mm Assume b = D/2 = 500/2 = 250mm Self weight of the beam = wg = 0.25*0.5*1*25 = 3.125kN/m Dead load = wd = wg + ws = 3.125 + 10 = 13.125kN/m Assuming partial safety factors 1.5 for DL and LL, 1.5
8
1.5
1.5
8
13.125 5.5 8
1.5
15
5.5 8
159.52
Assuming the beam as a balanced section, Mu = Mu,lim. Hence, we will obtain xu = xu,max. So, ,
0.36
0.42 20 250
,
0.36
159.52 10 0.42 0.48
,
0.48
Which gives, d = 480.88mm Assuming 30mm cover 16mm dia bars and 8mm stirrups, D = 480.88 + 30 + 8 + 16/2 = 526.88, say 530mm. Effective depth, d = 530 – 30 – 8 – 16/2 = 484mm. 0.36
0.42
159.52
0.36
10
20
250
484
0.42
Which gives, xu = 231.54 ; xu,max = 0.48d = 0.48*484 = 232.32mm; xu < xu,max , So Under reinforced section. 0.87
0.42
159.52
0.87
10
415
484
0.42
231.54
Which gives Ast = 1142.39mm2 .
Number of bars required = Ast/Area of one bar, 250
2
30
2
8
6
16
5 .
Assuming 20mm dia bars, 250
2
3.63 30
2
6
15.6
20,
.
4 8
3
5.68
4
20
31.33
20,
.
15.1.6. Example 6 A rectangular beam is to be simply supported on two walls of 125mm width with a clear span of 6m. The characteristic live load is 12kN/m. fck = 20N/mm2 and fy = 415N/mm2. Design the suitable section for the beam and determine the necessary tension steel. 15.2.
PROBLEMS FROM CHAPTER 6
15.2.1. Example 1
Compiled By: Vaisakh G.
PP3
SBCE – CE
Design of Steel Structures
Design a simply supported beam of effective span 8m subjected to imposed loads of 35 kN/m. The beam dimensions and other data are: b = 300mm, D = 700mm, M20 concrete, Fe415 steel. Minimum overall depth = leff/12 = 8000/12 = 666.66, Assume 700mm. [cl.23.2.1 of IS456:2000] Dead load of the beam, wd = 0.3 x 0.7 x 25 = 5.25 kN/m Imposed loads (given) wl = 35.00 kN/m .
Factored bending moment,
= 482.96kNm.
Assuming d' = 70 mm, d = 700 - 70 = 630 mm; xu,max = 0.48d = 0.48*630 = 302.4 mm [cl.38.1 of IS456:2000] Step 1: Determination of Mu,lim and Ast,lim ,
Using,
0.36
0.42
,
0.87
,
= 328.55kNm
,
0.42
,
, we obtain Ast1 = 1809.14mm2
Step 2: Determination of Mu2, Asc, Ast2 and Ast Mu2 = Mu – Mu,lim = 426.96 – 328.55 = 154.41kNm Here, d'/d = 70/630 = 0.11. From Table 6.3.2, by linear interpolation, we get, fsc = 350.8kNm. From
, Asc = 806.52mm2
Equating C = T or
0.87
, Ast2 = 763.69mm2
Ast = Ast2 + Ast1 = 1809.14 + 763.69 = 2572.83mm2 Step 3: Check for minimum and maximum tension and compression steel. In compression: Minimum Asc = 0.2% of GA = GA =
300
300
700= 420mm2, Maximum Asc = 4%of
700 = 8400mm2. Thus, 420mm2 < 806.517mm2 < 8400mm2. Hence, O.K.
In tension: Minimum Ast = 300
.
.
.
= 387.1mm2, Maximum Asc = 4%of GA =
700 = 8400mm2. Thus, 387.1mm2 < 2572.834mm2 < 8400mm2. Hence, O.K.
Step 4: Selection of bar diameter and numbers. Asc: Provide 2-20 T + 2-12 T (= 628 + 226 = 854 mm2) Ast: Provide 4-25 T + 2-20 T (= 1963 + 628 = 2591 mm2) It may be noted that Ast is provided in two layers in order to provide adequate space for concreting around reinforcement. Also the centroid of the tensile bars is at 70 mm from bottom. 15.2.2. Example 2 Determine the ultimate moment capacity of the doubly reinforced beam of b = 350mm, d' = 60mm, d = 600mm, Ast = 2945mm2 (6-25 T), Asc = 1256mm2 (4-20 T), using M20 and Fe415. Compiled By: Vaisakh G.
PP4
SBCE – CE
Design of Steel Structures
Step 1: To check if the beam is under-reinforced or over-reinforced.
0.00379. 0.002
.
,
xu,max = 0.48d = 0.48*600 = 288 mm [cl.38.1 of IS456:2000].
,
Yield strain of Fe 415 = .
.
= 0.0038 > 0.00379. Hence, the beam is over-reinforced.
Step 2: To determine Mu,lim and Ast,lim ,
0.36
0.42
,
,
= 347.67 kNm.
Equating compression to tension, 0.36
0.87f A
,
,
, Ast,lim = 2016mm2
Step 3: To determine Ast2 and Asc Ast2 = Ast - Ast,lim = 2945 - 2016 = 929mm2. The required Asc will have the compression force equal to the tensile force as given by 929mm2 of Ast2. By equating the compression and tension due to Asc and 0.87
Ast2,
.For fsc let us calculate εsc: 2
0.0035 1
,
= 0.002771.
2
From Table Table 9, fsc = 351.90N/mm . So Asc = 977.96mm . Step 4: To determine Mu2, Mu and Ast = 181.12kNm. Mu = Mu,lim + Mu2 = 347.67 + 181.12 = 528.79kN. Therefore, with Ast = Ast,lim + Ast2 = 2016 + 929 = 2945 mm2 the required Asc = 977.956mm2 (much less than the provided 1256mm2). Hence, O.k. 15.2.3. Example 3 Design a doubly reinforced beam to resist Mu = 375kNm when b = 250mm, d = 500mm, d' = 75mm, fck = 30N/mm2 and fy = 500N/mm2. 15.2.4. Example 4 Determine the moment of resistance of the doubly reinforced beam with b = 300mm, d = 600mm, d' = 90mm, fck = 30N/mm2, fy = 500N/mm2, Asc = 2236mm2 (2-32 T + 2-20 T), and Ast = 4021mm2 (4-32 T + 4-16 T). 15.2.5. Example 5 Design a simply supported beam of effective span 8m subjected to imposed loads of 35kN/m. The beam dimensions and other data are: b = 300mm, D = 700mm, M 20 concrete, Fe 415 steel. 15.2.6. Example 6 Determine the ultimate moment capacity of the doubly reinforced beam of b = 350mm, d' = 60mm, d = 600mm, Ast = 2945mm2 (6-25 T), Asc = 1256 mm2 (4-20 T), using M20 and Fe415. 15.3.
PROBLEMS FROM CHAPTER 7
15.3.1. Example 1 Determine the moment of resistance of the T-beam. Given data: bf = 1000 mm, Df = 100 mm, bw = 300mm, cover = 50mm, d = 450mm and Ast = 1963 mm2 (4- 25 T). Use M 20 and Fe 415. Compiled By: Vaisakh G.
PP5
SBCE – CE
Design of Steel Structures
Step 1: To determine the depth of the neutral axis xu Assuming xu in the flange and equating total compressive and tensile forces from the expressions of C and T as the T-beam can be treated as rectangular beam of width bf and effective depth d, we get 0.87fyAst = 0.36fckbfxu and on solving, xu = 98.44mm < 100mm (Df). So, the assumption of xu in the flange is correct. xu, max for the balanced rectangular beam = 0.48d = 0.48x450 = 216mm, xu Df. The ratio Df/d is computed. Df/d = 100/450 = 0.222 > 0.2 Hence, it is a problem of case discussed in sec. 7.5.2.2 Step 2: Computations of yf , C and T First, we have to compute yf and then find C, T and Mu, respectively. yf = 0.15 xu,max + 0.65 Df = 0.15 x 216 + 0.65 x 100 = 97.4 mm. C = 0.36fckbwxu,max + 0.45fck(bf - bw)yf = 0.36x20x300x216+0.45x20x(1000-300)x97.4 = 1,080.18kN. T = 0.87 fy Ast = 0.87 x 415 x Ast. Equating C and T, we have, Ast = 2991.77mm2. Provide 4-28 T + 3-16 T = 3,066 mm2. Step 3: Computation of Mu Mu = 0.36(xu, max /d){1 - 0.42( xu, max/d)} fck bw d2 + 0.45 fck(bf - bw) yf(d - yf /2) = 413.87 kNm. 15.3.3. Example 3 Determine the moment of resistance of the beam when Ast = 2,591 mm2 (4- 25 T and 2- 20 T). bf = 1,000 mm, Df = 100 mm, bw = 300 mm, cover = 50 mm and d = 450 mm. Use M 20 and Fe 415. Step 1: To determine xu Assuming xu to be in the flange and the beam is under-reinforced and C = T, we have 0.36fckbfxu=0.87fyAst or xu = 129.93mm > 100mm (Df). Since xu > Df, the neutral axis is in web. Here, Df/d = 100/450 = 0.222 > 0.2. So, we have to substitute the term yf, assuming Df / xu > 0.43 in the equation of C = T. Accordingly, we get: 0.36 fck bw xu + 0.45 fck (bf - bw) yf = 0.87 fy Ast. or xu = 169.398 mm < 216 mm (xu,max = 0.48d = 216 mm). So, the section is under-reinforced. Step 2: To determine Mu Df /xu = 100/169.398 = 0.590 > 0.43. This is the problem of case as detailed in sec.7.5.3.2. The corresponding equations for yf, C, T and Mu, can be employed. We have yf = 0.15 xu + 0.65 Df = 0.15 x 169.398 + 0.65 x 100 = 90.409mm. Mu = 0.36(xu /d){1 - 0.42( xu /d)} fck bw d2 + 0.45 fck(bf - bw) yf (d - yf /2) = 369.18 kNm. Compiled By: Vaisakh G.
PP6
SBCE – CE
Design of Steel Structures
15.3.4. Example 4 Determine the moment of resistance of the flanged beam with Ast = 4,825 mm2 (6- 32 T). bf = 1000 mm, Df = 100 mm, bw = 300 mm, cover = 50 mm and d = 450 mm. Use M 20 and Fe 415. Step 1: To determine xu Assuming xu to be in the flange and the beam is under-reinforced and C = T, we have 0.36fckbfxu=0.87fyAst or xu = 241.95 mm > 100mm (Df). Here, Df/d = 100/450 = 0.222 > 0.2. So, we have to determine yf by yf = 0.15 xu + 0.65 Df and equating C and T we get, 0.36 fckbwxu + 0.45fck(bf bw)yf = 0.87 fy Ast, or xu = 429.17mm. xu,max = 0.48xd = 216mm. Since xu > xu,max, the beam is overreinforced. Accordingly. xu = xu, max = 216 mm. Step 2: To determine Mu This problem belongs to case explained in sec. 7.5.4.2. So, we can determine Mu from Mu = 0.36(xu, 2 max /d){1 - 0.42(xu, max /d)} fck bw d + 0.45 fck(bf - bw) yf (d - yf /2) where yf = 0.15xu,max+0.65Df = 97.4mm. Employing the value of yf=97.4mm, we get Mu=413.87kNm. 15.3.5. Example 5 Design the simply supported flanged beam, given the following: Df = 100 mm, D = 750 mm, bw = 350 mm, spacing of beams = 4000 mm c/c, effective span = 12 m, cover = 90 mm, d = 660 mm and imposed loads = 5 kN/m2. Fe 415 and M 20 are used. Step 1: Computation of factored bending moment Weight of slab per m2 = 0.1 x 1 x 1 x 25 = 2.5 kN/m2. So, Weight of slab per m = 4 x 2.5 = 10.00 kN/m. Dead loads of web part of the beam = 0.35 x 0.65 x 1 x 25 = 5.6875 kN/m. Imposed loads = 4 x 5 = 20 kN/m. Factored Bending moment = 1.5wdl2/8 + 1.5wll2/8 = 1.5x15.68x122/8 + 1.5x20x122/8 = 963.56kNm Step 2: Computation of Mu,lim Effective width of flange =(lo/6) + bw+ 6 Df = (12000/6) + 350 + 600 = 2,950 mm. xu,max = 0.48 d = 0.48 x 660 = 316.80 mm. This shows that the neutral axis is in the web of this beam. Df /d = 100/660 = 0.1515 < 0.2, and Df /xu = 100/316.8 = 0.316 < 0.43. The expression of Mu,lim is obtained is as follows: Mu,lim = 0.36(xu,max/d){1 - 0.42 (xu,max /d)} fck bw d2 + 0.45 fck (bf - bw) Df (d - Df /2) = 0.36x0.48x{1–0.42x0.48}x20x350x650x650+0.45x20x(2950–350)x100x(660–50) = 1,835.43kNm. The design moment Mu = 963.5625 kNm is less than Mu,lim. Hence, one under-reinforced beam can be designed. Step 3: Determination of xu Since the design moment Mu is almost 50% of Mu,lim, let us assume the neutral axis to be in the flange. The area of steel is to be calculated from the moment equation, when steel is ensured to reach the design stress fd = 0.87x415 = 361.05 N/mm2. It is worth mentioning that the term b is here replaced by bf as the T- beam is treated as a rectangular beam when the neutral axis is in the flange. Hence the moment capacity of the section can be expressed as Mu = 0.36fckbfxu(d-0.42xu). xu is obtained from this expression. xu = 71.98mm. Step 4: Determination of Ast Equating compression and tension, we obtain, 0.87fyAst = 0.36fckbfxu. From this we obtain, Compiled By: Vaisakh G.
PP7
SBCE – CE
Design of Steel Structures
Ast = 4234.72mm2. Minimum Ast = (0.85/fy)bwd = (0.85/415)x350x660 = 473.13 mm2. Maximum Ast = 0.04bwD = 0.04x350x660 = 9,240 mm2. Hence, Ast = 4,237.41 mm2 is o.k. Provide 6-28T + 2-20 T to have total Ast = 4,322 mm2. 15.3.6. Example 6 Design a beam in place of the beam of the previous example if the imposed loads are increased to 12kN/m2. Other data are: Df = 100 mm, bw = 350 mm, spacing of beams = 4000 mm c/c, effective span = 12 m simply supported and cover = 90 mm. Use Fe 415 and M 20, bf = 2,950 mm. Step 1: Computation of factored bending moment Weight of slab per m2 = 0.1 x 1 x 1 x 25 = 2.5 kN/m2. So, Weight of slab per m = 4 x 2.5 = 10.00 kN/m. Dead loads of web part of the beam = 0.35 x 0.65 x 1 x 25 = 5.6875 kN/m. Imposed loads = 4 x 12 = 48 kN/m. Factored Bending moment = 1.5wdl2/8 + 1.5wll2/8 = 1.5x15.68x122/8 + 1.5x48x122/8 = 1719.56kNm Step 2: Computation of Mu,lim Effective width of flange =(lo/6) + bw+ 6 Df = (12000/6) + 350 + 600 = 2,950 mm. xu,max = 0.48 d = 0.48 x 660 = 316.80 mm. This shows that the neutral axis is in the web of this beam. Df /d = 100/660 = 0.1515 < 0.2, and Df /xu = 100/316.8 = 0.316 < 0.43. The expression of Mu,lim is obtained is as follows: Mu,lim = 0.36(xu,max/d){1 - 0.42 (xu,max /d)} fck bw d2 + 0.45 fck (bf - bw) Df (d - Df /2) = 0.36x0.48x{1–0.42x0.48}x20x350x650x650+0.45x20x(2950–350)x100x(660–50) = 1,835.43kNm. The factored moment of this problem (1,719.5625 kNm) is close to the value of Mu,lim of the section. Step 3: Determination of d Assuming Df /d < 0.2, we have, Mu = 0.36xu,maxfckbw{d - 0.42xu,max}+0.45fck(bf–bw)Df(d - Df/2). Solving the above equation, we get d = 624.09 mm, giving total depth = 624.09 + 90 = 715 mm (say). Since the dead load of the beam is reduced due to decreasing the depth of the beam, the revised loads are calculated as follows: Dead loads (revised) = 0.615 (0.35) (25) = 5.38125 kN/m. Factored Bending moment = 1.5wdl2/8 + 1.5wll2/8 = 1.5x15.38x122/8 + 1.5x48x122/8 = 1711.29kNm Step 4: Determination of Ast xu = xu,max = 0.48x625 = 300mm. Equating T and C, we have 0.87fyAst = 0.36xu,maxbwfck + 0.45fck(bf – bw)Df or Ast = 8574.98mm2. Maximum Ast = 0.04bD = 0.04x350x715 = 10,010.00 mm2. Minimum Ast = (0.85/fy)bwd = (0.85/415)x350x625 = 448.05 mm2.Hence, Ast = 8,574.98 mm2 is o.k. So, provide 836 T + 2-18 T, Ast provided = 8,651 mm2. Step 5: Determination of xu Using Ast = 8,651 mm2 in the expression of T = C, we have 0.87fyAst = 0.36xubwfck + 0.45fck(bf-bw)Df or xu = 310.89 >xu,max. So, Ast provided is reduced to 8-36 + 2-16 = 8,545 mm2. Accordingly by using expression, 0.87fyAst = 0.36xu,maxbwfck + 0.45fck(bf – bw)Df we obtain xu = 295.7mm < xu,max. Step 6: Checking of Mu Df /d = 100/625 = 0.16 < 0.2; Df /xu = 100/215.7 = 0.33 < 0.43. Hence Mu can be obtained from Mu = 0.36fckbf xu{1 - 0.42xu} + 0.45fck(bf - bw)Df(d - Df/2) = 1,718.68 kNm > (Mu)design. Hence o.k. 15.3.7. Example 7
Compiled By: Vaisakh G.
PP8
SBCE – CE
Design of Steel Structures
Determine the tensile reinforcement Ast of the flanged beam when the imposed loads = 12 kN/m2. All other parameters are the same as those of the previous case. Df = 100 mm, D = 750 mm, bw = 350 mm, spacing of beams = 4000 mm c/c, effective span = 12 m, simply supported, cover = 90 mm and d = 660 mm. Use Fe 415 and M 20. Step 1: Computation of factored bending moment Weight of slab per m2 = 0.1 x 1 x 1 x 25 = 2.5 kN/m2. So, Weight of slab per m = 4 x 2.5 = 10.00 kN/m. Dead loads of web part of the beam = 0.35 x 0.65 x 1 x 25 = 5.6875 kN/m. Imposed loads = 4 x 12 = 48 kN/m. Factored Bending moment = 1.5wdl2/8 + 1.5wll2/8 = 1.5x15.68x122/8 + 1.5x48x122/8 = 1719.56kNm Step 2: Computation of Mu,lim Effective width of flange =(lo/6) + bw+ 6 Df = (12000/6) + 350 + 600 = 2,950 mm. xu,max = 0.48 d = 0.48 x 660 = 316.80 mm. This shows that the neutral axis is in the web of this beam. Df /d = 100/660 = 0.1515 < 0.2, and Df /xu = 100/316.8 = 0.316 < 0.43. The expression of Mu,lim is obtained is as follows: Mu,lim = 0.36(xu,max/d){1 - 0.42 (xu,max /d)} fck bw d2 + 0.45 fck (bf - bw) Df (d - Df /2) = 0.36x0.48x{1–0.42x0.48}x20x350x650x650+0.45x20x(2950–350)x100x(660–50) = 1,835.43kNm. The design moment Mu = 1719.56 kNm is less than Mu,lim. Hence, one under-reinforced beam can be designed. Step 3: Determination of xu Assuming xu to be in the flange, we have considering b = bf, Mu = 0.36fckbf xu{d - 0.42xu}, xu = 134.1>100mm. So, let us assume that the neutral axis is in the web and Df/xu < 0.43, we have Mu = 0.36fckbwxu{d - 0.42xu} + 0.45fck(bf - bw)Df(d - Df /2). Substituting the value of Mu = 1,719.56kNm in the above equation and simplifying, xu2 - 1571.43xu + 276042 = 0 or xu = 201.5 mm. Df/xu = 100/201.5 = 0.496 > 0.43.So, we have to use yf in the equation to find Mu. Thus, we have Mu = 0.36fckbwxu{d - 0.42xu} + 0.45fck(bf - bw)yf(d - Df /2) where, yf = (0.15xu + 0.65Df). Solving the above expression, we obtain xu = 213.63 mm. Df /xu = 100/213.63 = 0.468 > 0.43.Hence, o.k. Step 4: Determination of Ast Equating C = T, we have 0.87fyAst = 0.36fckbwxu + 0.45fck(bf – bw)yf where, yf = 0.15xu + 0.65Df. Here, using xu = 213.63 mm, Df = 100 mm, we get yf = 0.15x213.63 + 0.65x100 = 97.04 mm. On solving, Ast = 7780.32mm2. Minimum Ast = (0.85/fy)bwd = 0.85x350x660/415 = 473.13 mm2. Maximum Ast = 0.04bwD = 0.04x350x750 = 10,500 mm2. Hence, Ast = 7,780.32 mm2 is o.k. Provide 6-36 T + 3-28 T (= 7,954 mm2). Step 5: Checking of xu and Mu using Ast = 7,954 mm2 From T = C, we have 0.87fyAst = 0.36fckbwxu + 0.45fck(bf – bw)yf, where, yf = 0.15xu + 0.65Df or xu = 224.01 mm; Df /xu = 100/224.01 = 0.446 > 0.43. Accordingly, we have Mu = 0.36fckbwxu{d - 0.42xu} + 0.45fck(bf - bw)yf(d - yf /2) = 1,779.439 kNm > 1,719.5625 kNm Hence, o.k. 15.3.8. Example 8 Design the flanged beam, given in following: Df = 100 mm, D = 675 mm, bw = 350 mm, spacing of beams = 4000 mm c/c, effective span = 12 m simply supported, cover = 90 mm, d = 585 mm and imposed loads = 12 kN/m2. Use Fe 415 and M 20. Step 1: Computation of factored bending moment Compiled By: Vaisakh G.
PP9
SBCE – C CE
Desig gn of Steel Sttructures
5 = 10.00 Weight of slab per m2 = 0.1 x 1 x 1 x 25 = 2.5 kN/m2. So, Weight of slab per m = 4 x 2.5 kN/m. Dead D loads of web part of the beam = 0.35 x 0.57 75 x 1 x 25 = 5.032 kN/m m. Imposed loads l =4 2 2 x 12 = 48 kN/m. Factored Bending B mooment = 1..5wdl /8 + 1.5wll /8 = 1.5x15.03x x122/8 + 1.5x48xx122/8 = 17011.87kNm. Step 2: Determinatio D on of Mu,lim Assuminng the neutrral axis to bee in the webb, Df/xu < 0.43 0 and Df/dd = 100 / 5885 = 0.17 < 0.2, we considerr Mu,lim = 0.336fckbwxu,max{1 – 0.42xu,mmax} + 0.45fck Nm. Since c (bf – bw)Df(d – Df/2) = 1,582.4kN factoredd Mu > Mu,lim, the beam iss designed ass doubly rein nforced. Mu2 = Mu - Mu,limm = 1701.87 - 1582.4 = 119.477 kNm Step 3: Determinatio D on of area off steel Ast,lim is obtained equuating T=C,00.87fyAst,lim = 0.36bwxu,maaxfck + 0.45fckk(bf - bw)Df, Ast,lim = 8440 0.98mm2 w fsc = 3553 N/mm2 fo or d'/d = 0.1, fcc = 0.446fcck = 0.446 x2 20 = 8.92 From Mu2 = (fsc – fccc)Asc(d-d’) where 2 N/mm , Mu2 = 119.447kNmm, d' = 58.5 mm, d = 585 mm m. Using the above a valuess in the expression of 7fyAst2 = (fsc – fcc)Asc, Ast22 = 628.48 mm m 2 Mu2, we get Asc = 6559.63mm2. By equating C2 to T2, 0.87 6 = 9,,069.46 mm m2, Maximum m Ast = 0.0 04bwD = Total Ast = Ast,lim + Ast2 = 8,440.98 + 628.48 2 0.04x350x675 = 9,4450 mm andd minimum Ast = (0.85/ffy)bwd = (0.885/415)x350xx585 = 419..37 mm2. Hence, Ast = 9, 069.446 mm2 is o.k. Provide 8-36 8 T + 3-20 0 T (= 9,0855 mm2) for Ast and 1-20 + 2-16 (= 716 mm m2 for Asc. Step 4: To T check forr xu and Mu Assuminng xu in the web w and Df/xxu < 0.43 andd using T = C, C we have 0.87f 0 0 (bf yAst = 0.36bwxufck + 0.45 bw)fckDf + Asc(fsc - fcc). This givves xu = 275..33 mm. xu,m max = 0.48d = 0.48x585 = 280.8mm. So, xu < umptions, xu,max, Df/xu = 100//275.33 = 0.363 < 0.433 and Df/d = 100/585 = 0.17 < 0.2. The assu thereforee, are correctt. So, Mu cann be obtainedd with additiional momennt due to com mpression steeel, as Mu = 0.36 bw xu fck (d - 0.42 xu) + 0.45 0 (bf - bw) fck Df (d - Df /2) + Asc (f ( sc - fcc) (d - d') = 1,707.23 kNm. Factoredd moment = 1,701.87 kN Nm < 1,707.23 kNm. Hen nce, o.k. 15.4.
PROBLEM MS FROM CHAPTER C 8
15.4.1. Example 1: mply supporrted beam off effective sppan 8 m who ose crossDetermiine the shearr reinforcemeent of the sim section is i shown in Figure F 61. Faactored shearr force is 250 0 kN. Use M 20 and Fe 4415.
Figu ure 61 Examp ple 1
Here, Asts =2-25T+2--20T gives thhe percentagee of tensile reinforcemen r nt. Compiled B By: Vaisakh G.
1.43. PP10
SBCE – C CE
Desig gn of Steel Sttructures
S 456:2000 (Table 10), τc = 0.67 + 0.036 = 0.7066 N/mm2 (by linear interp polation). From Taable 19 of IS = 2.22N/mm m2. and τcmax = 2.8 N/mm m2 from Tablee 20 of IS 4556:2000(Tabble 11). Hencce, τc < τv S shear reinnforcement is i needed forr the shear fo orce (Cl. 40.44 of IS 456:22000). < τcmax. So, Vus = Vu – τc b d = 250 2 – 0.706 x 250 x 4500 x 10-3 = 170.575 kN. Providing 8 m mm, 2-legged d vertical svv 2 stirrups, we have A = 2 x 50 = 100 mm . Hence, H spacin ng of the stirrrups as obtaiined from Cll. 40.4 of IS 456:22000.
.
.
= 95.25mm, say 95m mm.
For 10 mm, 2 legged vertical stirrups, s (Asvv = 157 mm m2), spacing, Sv = 149.54mm2. Acco ording to mum spacing of the stirrup ps = 0.75d = 0.75x450 = 337.5mm = 300 mm cl.26.5.11.5 of IS 4566, the maxim (say). Minimum M shear reinforcem ment (cl. 26.55.1.6 of IS 456) is obtainned from
. .
. So Asv, min =
40.16mm m2. So, we seelect 10 mm,, 2 legged stiirrups @ 145 5 mm c/c. 15.4.2. Example 2 Design the t bending and shear reinforcement of the tapered cantileverr beam of wiidth b = 300 mm and as show wn in Figure 62 using M 20 and Fe 415 4 (i) witho out any curtaailment of beending reinfo orcement and (ii) redesign thee bending and shear reinfforcement iff some of thee bars are cuurtailed at secction 2-2 of the beeam.
Figu ure 62 Examp ple 2
Here, Mu = wl2/2 = 75x3.52/2 = 459.375kN Nm. Mu/bd2 = 6.125/mm2. Ast = 2-255T + 2-20T gives g the percentaage of tensilee reinforcem ment, p = 1000As/bd = 1.4 43. Reinforccing bars of 3-28T + 1-16T give 2048 mm m2. Asc at secction 2-2 = 0.686x400x3 0 300)/100 = 82 23.2mm2. Reeinforcing baars of 2-20T + 2-12T 2 give 8544 mm . Thouugh it is betteer to use 4-288T as Ast and d 2-20T + 2-16 as Asc with proper currtailment from the practical aspects a of construction, c here the bars are seleected to have areas closse to the requirem ments for the academic innterest only. Figure 63 6 shows the reinforcemeent at sec. 2-22. Case (i): No curtailm ment (all baars of bendinng reinforcem ment are conntinued): Bennding momeent Mu at 2 = 234.3775 kNm. Sheear force Vu at a section 2-2 2 = 187.5 kN N. Effective ddepth d at section 2-2 section 2-2 = 450–550=400mm, and a Width b = 300mm tanβ=(550–2 t 200)/3500=0..1. Clause 400.1.1 of IS 456 4 gives 1 1.074 / Here, thhe negative siign is used as a the bendinng moment in ncreases num merically in tthe same dirrection as the effecctive depth inncreases. Compiled B By: Vaisakh G.
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Desig gn of Steel Sttructures
m 2), we get p = 2.555 %. % Table 19 oof IS 456:200 00 (Table Continuuing 4-28T annd 3-16T barrs (= 3066 mm 2 2 10) givees τc = 0.82 N/mm < τv (= 1.074 N/mm N ). Hen nce, shear reeinforcementt is needed for f shear -3 force obbtained as Vus 00x10 kN = 89.1 kN. Frrom cl.40.4 of o IS 456, u = Vu - τv b d = 1.875 – 0.82x300x40 we havee
.
. wheree Asv = 100 m mm2 for 8 mm, m 2 leggedd vertical stirrrups. This gives g sv =
162.087 mm (fy = 415 4 N/mm2). IS 456, cl.. 26.5.1.6 giives the spaccing consideering minimu um shear .
reinforceement.
.
or sv = 300.8755mm. Hence,, provide 8 mm, m 2 leggeed vertical stirrups @
150 mm m c/c.
Figure 63 SSection 2‐2 off Example 2
Case (ii)): With curtaailment of baars: Clause 26.2.3.2 2 of IS I 456 stipuulates that anny one of th he three connditions is too be satisfied d for the terminattion of flexurral reinforcem ment in tensiion zone. Heere, two of thhe conditionss are discusseed. (a) Conddition (i): urtailment, att section 2-22 Ast = 2048 mm2 (3orr Vus ≥ (1.5 τv – τc) b d. After the cu 28T + 1-16T 1 bars), gives p = 1.71 %. Tablee 19 of IS 456:2000 4 givves τc = 0.7452 N/mm2 when w p= 1.71% (making ( lineer interpolatioon). gives thee spacing off stirrups
.
1.074 /
. So, Vus = 103.896 kN k which
. Using 8 mm m, 2 legged vertical v stirruups (Asv = 10 00 mm2),
we havee sv = 139.005 mm. Hence, provide 8 mm, 2 leggeed vertical stiirrups @ 1300 mm c/c. (b) Conddition (ii): Additionnal stirrup arrea for a disttance of 0.755 d {= 0.75x4 400 = 300 mm} m = 0.4 b s/fy, where spacing s s is not greater g than (d/8βb), wheere βb = cut off bar area/total bar area a = 2048//3066 = 0.67 7. Since, additionnal stirrups are a of loweer diameter, mild steel bars are prreferred withh fy = 250 N/mm2. Maximuum spacing s = d/8βb = 75 7 mm. Excess area = 0.4 0 b s/fy = 36 3 mm2. Provvide 6 mm, 2 legged 2 mild steeel vertical stirrups (56 mm m ) @ 75 mm c/c for a distance of o 300 mm, i.e., five num mbers of stirrups (additional). 15.4.3. Example 3 Design the t flexural and shear reinforcement of the simply supportedd T-beam (Fiigure 64) of effective span 8 m placed @ 4.2 m c/c annd subjected to a total faactored load of 150 kN/m m. Use M 30 0, Fe 415 and SP-16 tables forr the design of o flexural reeinforcement. Compiled B By: Vaisakh G.
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Desig gn of Steel Sttructures
Figu ure 64 Examp ple 3
o flexural reeinforcementt with SP-16: Design of Effective width of flange fl bf = lo/6 / + bw + 6 Df = 2353 mm m > 2100 mm m (breadth of the web plus half the sum of the clear distance to the t adjacent beams on eiither side). So, S bf = 21000 mm. (Mu)faactored = 1200 kN Nm, at the miid-span. (Vu)factored ) = 600 6 kN, at thee support.
F Figure 65 BM & & SF diagram ms of Example 3
The bennding moment and shearr force diaggrams are sh hown in Figuure 65. At thhe mid-span n =0.37. Df /d = 120/6600 = 0.2 annd bf /bw = 2100/300 = 7. 7 Table 58 of o SP-16 (fy = 415 N/mm m2) gives ,
= 0.62 (whenn bf /bw = 7.00 and Df /d = 0.2) > 0.37 7 in this case.. Hence, o.k.. 2
.
2
mm . Providee 7-32T + 1-28T (= 62455 mm ) bars at mid-span and up to seection 5-5 (Figure 66, = 6155m sec. 5-5)). The flexurral reinforcem ment is crankked up and the t reinforceement diagraams are show wn at five sections in Figure 666. o shear reinforcement: Design of
Compiled B By: Vaisakh G.
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Desig gn of Steel Sttructures
f the sectio on 1-1 in six steps. Resullts of all fourr sections The detaails of calcullations are shhown below for are preseented in Table 19. Step 1: Ast at sectionn 1-1 is deterrmined (= 32217 mm2 = 4-32T) from Figure F 65 to calculate p = 1.79%. From Taable 19 of IS S 456:2000 (T Table 10), τc is determined for p = 1..79% as 0.811 N/mm2. Tab ble 20 of IS 456:22000 (Table 11) gives τc,mmax for M 30 = 3.5 N/mm2. Step 2: Vu is 600 kN N at section 1-1 1 from Figgure 65 to geet τv = Vu/(bwd) = 3.33N//mm2. Here, τv is less than τcmaax.
Figgure 66 Reinfo orcement detaails of Examp ple 3
Step 3: The magniitude of sheear force forr which sheear reinforceement is neeeded (say, Vreinf.) is determinned from Figgure 65. For section 1-1,, Vreinf = Vu – τc b d = 4554.2 kN. Thee magnitude of shear force takken by bent up u bar(s) is obtained o as, Vbent = 0.87 fy Asv sinα = 206.5 kN. T This force sh hould not be greater than 0.5 (V Vreinf), whichh, at this secttion, is 227.1 1 kN. The maagnitude of thhe shear forcce for the design of o vertical stirrup = Vus = Vreinf – Vbent = 454.2 – 20 06.5 = 247.77 kN. Table 19 D Design of stirrrups using 10 mm 2 legged vertical stirru ups, (τc,max = 3 3.5 N/mm2)
Step
Values of Secc. 1-1 Sec. 2--2 S 3-3 Sec. Sec. 4-4 4 Ast (mm2) 3,217 4,021 1 4,825 5,62 29 1 p (%) 1.79 2.23 2.68 3.13 3 2 τc (N/mm ) 0 0.81 0.877 7 0.93 0.96 6 Vu (kN) 6 600 525 450 375 5 2 τv (N/mm2) 3.33 2.92 2.50 2.08 8 Vreinf (kN) 4554.2 367.14 4 282.6 202.2 3 Vbent (kN) 206.5* (< <0.5 Vreinf) 206.5(< 18 83.57) 206..5 (< 141.3) 157.24**(< < 101.1) Vus (kN) 2447.7 183.57 7 141.3 101.1 4 sv (mm) 1337.3 185.27 7 240.7 336.4 5 M sv (mm)) Min ≤ 300 ≤ 300 0 ≤ 300 ≤ 30 00 6 Provided sv 1 135 180 240 300 0 * Diameeter of bent up u bar = 32 mm; m ** diam meter of bent up u bar = 28 mm m Compiled B By: Vaisakh G.
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Desig gn of Steel Sttructures
A the diameter d of vertical v stirruup bars as 10 0 mm, 2 leggged (Asv = 1557 mm2). Thee spacing Step 4: Assume of verticcal stirrups sv = 0.87 fy Asv s d/Vus = 137.3 mm c/c. Step 5: Check sv considering minimum m shhear reinforccement from m cl. 26.5.1.66 of IS 456 6 as sv ≤ 0.87fyAsv F cl. 26.5.1.5 2 stipu ulates the maaximum spaccing = 0.75 d on 300 s /0.4bw; sv ≤ 472 mm. Further, mm. Here, the maxim mum spacingg = 300 mm. Step 6: So, S provide 10 1 mm, 2 leggged stirrups @ 135 mm c/c. The ressults of all the sectionns are givenn below: To o avoid sevveral spacinggs for the practical considerration, providde stirrups @ 130 mm c//c for first 1 m, @ 230 mm m for next 1 m and theen @ 300 mm up to t the mid-sppan in a symm metric mannner 15.5.
PROBLEM MS FROM CHAPTER C 9
15.5.1. Example 1 Determiine the anchhorage lengtth of 4-20T reinforcing bars going into the suupport of thee simply supporteed beam shoown in Figuure 67. The factored sheear force Vu = 280kN, w width of thee column support = 300mm. Use U M 20 conncrete and Fee 415 steel. 2 = 1.92 N/m mm2. Ld = (φσs)/(4τbd). When W σs = τbd for M 20 and Fe 415 (with 60% increaseed) = 1.6x1.2 0.87fy, Ld = 47.01φ. Ld (when σs = fd) ≤ M1/V V + L0. Heree we need to find xu, from m C = T, 0.36fckbxu = 0.87fyAsts , xu = 209.994mm; xu,max = 0.48d = 2440mm.
Figu ure 67 Examp ple 1
Since xu < xu,max ; M1 = 0.87 fy Ast (d – 0.42 xu) = 187.75 54 kNm and V = 280 kN. With the sttipulation of 30 peer cent increaase assumingg that the reinnforcing barss are confineed by a comppressive reaction Ld ≤ 1.3(M1/V V) + L0. Assuming L0 = 0, 0 φ≤18.54m mm. Thereforre, 20 mm diameter bar ddoes not allow w Lo = 0. Determiination of Lo from Ld ≤ 1.3(M1/V) + L0, Minimum m L0 = 68.48mm. So, the bars are exteended by 100 mm m to satisfy thhe requiremennt as shown in Figure 68.
Figure 68 8 Solution of EExample 1 Compiled B By: Vaisakh G.
PP15
SBCE – CE 15.6.
Design of Steel Structures
PROBLEMS FROM CHAPTER 10
15.6.1. Example 1. Determine the reinforcement required of a ring beam of b = 400 mm, d = 650 mm, D = 700 mm and subjected to factored Mu = 200 kNm, factored Tu = 50 kNm and factored Vu = 100 kN. Use M 20 and Fe 415 for the design. Step 1: Check for the depth of the beam. We have the equivalent shear Ve = Vu + 1.6(Tu/b) = 300 kN. the equivalent shear stress τve = Ve/bd = 1.154 N/mm2. From Table 20 of IS 456:2000(Table 11), τc,max = 2.8 N/mm2. Hence, the section does not need any revision. Step 2: Check if shear reinforcement shall be required. Assuming percentage of tensile steel as 0.5, Table 19 of IS 456:2000(Table 10) gives τc = 0.48 N/mm2 < τve < τc,max. So, both longitudinal and transverse reinforcement shall be required. Step 3: Longitudinal tension reinforcement: Me1 = Mu + Mt = Mu + (Tu/1.7){1 +(D/b)} = 280.88 kNm. Proceeding further by the design for moment, we obtain Ast = 1340.56 mm2. Provide 2-25T and 216T = 1383 mm2. This gives percentage of tensile reinforcement = 0.532, for which τc from Table 19 of IS 456:2000(Table 10) is 0.488 N/mm2. The minimum percentage of tension reinforcement = (0.85/fy)(100) = 0.205 and the maximum percentage of tension reinforcement is 4.0. So, 2-25T and 216T bars satisfy the requirements. Step 4: Longitudinal compression reinforcement: Here, in this problem, the numerical value of Mt (80.88 kNm) is less than that of Mu (200 kNm). So, longitudinal compression reinforcement shall not be required. Step 5: Longitudinal side face reinforcement: Side face reinforcement shall be provided as the depth of the beam exceeds 450 mm. Providing 2-10 mm diameter bars (area = 157 mm2) at the mid-depth of the beam and one on each face, the total area required = 120 mm2 < 157 mm2. Hence o.k. Step 6: Transverse reinforcement: Providing two legged, 10 mm diameter stirrups (area = 157 mm2), we have d1 = 700 - 50 - 50 = 600 mm, b1 = 400 - 2(25 + 10 + 12.5) = 305 mm. We have .
.
.
. Using the numerical values of Tu, b1, d1 and Vu, we have 0.87fyAsv/sv =
339.89N/mm. Also we have 0.87fyAsv/sv ≥ 269.6N/mm. So, the first case is governing and we get for 2 legged 10 mm stirrups (Asv = 157 mm2), sv = 0.87x415x157/339.89 = 166.77 mm. Step 7: Check for sv: The two legged 10 mm diameter stirrups for which x1 = 340 mm and y1 = 628.5 mm. The maximum spacing sv should be the least of x1, (x1 + y1)/4 and 300 mm. Here, x1 = 340 mm, (x1 + y1)/4 = 242.12 mm. So, provide 2 legged 10 mm T stirrups @ 160 mm c/c. 15.6.2. Example 2 A reinforced concrete rectangular beam of b = 300 mm, d = 600 mm and D = 650 mm is subjected to factored shear force Vu = 70 kN in one section. Assuming the percentage of tensile reinforcement as 0.5 in that section, determine the factored torsional moment that the section can resist if (a) no additional reinforcement for torsion is provided, (b) maximum steel for torsion is provided in that section, and (c) determine the reinforcement needed for the case (b). Assume M 30 concrete, Fe 500 for longitudinal and Fe 415 for transverse reinforcing steel bars. 15.7.
PROBLEMS FROM CHAPTER 12
Compiled By: Vaisakh G.
PP16
SBCE – C CE
Desig gn of Steel Sttructures
15.7.1. Problem 1 t one-way continuous slab of Figurre 69 subjectted to uniforrmly distribuuted imposed d loads of Design the 2 2 5 kN/m using M 200 and Fe 4155. The load of o floor finish is 1 kN/m . The span ddimensions shown s in the figurre are effective spans. Thhe width of beams b at the support s = 3000 mm.
Figure 69 Slab specified iin Problem 1
Step 1: Selection S of preliminary p depth of slabb The bassic value of span to effeective depth ratio for thee slab havinng simple suppport at the end and continuoous at the inttermediate iss (20+26)/2 = 23 (cl.23.2 2.1 of IS 4566:2000). Moddification facctor with 2 assumedd p = 0.5 annd fs = 240 N/mm N is obbtained as 1.2 from Fig.44 of IS 456::2000. Thereefore, the minimum m effective depth d = 1.2x33000/23= 1008.69 mm. Leet us take thee effective deepth d = 115 5 mm and with 25 mm cover, thhe total deptth D = 140 mm. m D loadss, bending moment and shhear force Step 2: Design Dead loads of slab of o 1 m widthh = 0.14x25 = 3.5 kN/m m. Dead load of floor finiish =1.0 kN//m. Total m. Factored dead load = 1.5x4.5 = 6.75 6 kN/m. Factored liv ve load = Dead looad = 3.5 + 1 = 4.5kN/m 1.5x5.0 = 7.50 kN N/m. Total factored f loaad = 14.25 kN/m. Maxximum mom ments and shear s are T 12 an nd 13 (Page 36 of IS 4556: 2000). Maximum M determinned from thee coefficientts given in Tables positive moment = 14.25x3x3/1 1 2 = 10.68755 kNm/m. Maximum M neggative momeent = 14.25x3 3x3/10 = 12.825 kNm/m. k Maxximum shearr Vu = 14.25xx3x0.6 = 25.65 kN Step 3: Determinatio D on of effectivve and total depths d of slab b Mu,lim=R R,limbd2 wherre R,lim is 2.776N/mm2 froom Table 5.. So, d = √{{12.825x106//2.76x1000} = 68.17 mm. Sinnce, the compputed depth is much lesss than that deetermined inn Step 1, let uus keep D = 140 mm and d = 115 mm. D of slabb for shear foorce Step 4: Depth Table 199 of IS 456:2000 (Tablee 10) gives τc = 0.28 N/m mm2 for the lowest perceentage of steeel in the slab. Furrther for the total depth of 140 mm, let us use th he coefficientt k of cl. 40..2.1.1 of IS 456:2000 4 2 as 1.3 too get τc= k τc = 1.3x0.28 = 0.364 N/m mm . Table 20 2 of IS 456::2000 (Tablee 11) gives τcmax = 2.8 c N/mm2. For this prooblem τv = Vu/ bd = 25.665x103/(115x x1000) = 0.2223 N/mm2. Since, τv<τc< τ c,max, i safe. the effecctive depth d = 115 mm is Step 5: Determinatio D on of areas of steel Mu = 0.87 fy Ast d {1 – (Astfy)/(f ) ckbd)} frrom Cl. G.1.1(b) of IS 456: 4 2000. ((i) For the maximum m m 12825000 = 0.887x415xAstx115{1 x – (A Astx415)/(10000x115x20)} or Ast2negativee bending moment, 5542.16 Ast + 17118871.646 = 0. Solving the quadratic equation, wee have the nnegative Ast = 328.34 mm2. (iii) For the maximum positive beending mom ment, 10687500 = 0.877x415xAstx115x{1 – Compiled B By: Vaisakh G.
PP17
SBCE – C CE
Desig gn of Steel Sttructures
Ast+1426559.705 = 0, Soolving the quuadratic equaation, we (Astx4155)/(1000x1155x20)} or Asts 2 -5542.16A 2 have thee positive Asts = 270.615 mm . Distriibution steell bars along longer span ly. Distributtion steel area = Minimum M steeel area = 0.112x1000x1400/100 = 168 mm2. Since,, both positivve and negattive areas of steel are a higher thhan the minim mum area, we w provide: (aa) For negatiive steel: 10 mm diameteer bars @ 2 230 mm m c/c for whicch Ast = 341 mm giving ps = 0.2965. (b) For posiitive steel: 8 mm diameteer bars @ 180 mm m c/c for which Ast = 2779 mm2 givinng ps = 0.24 426 (c) For distribution steel: Provid de 8 mm 2 diameterr bars @ 2500 mm c/c for which Ast (m minimum) = 201 mm . Step 6: Selection S of diameter andd spacing of reinforcing bars b The diam meter and sppacing alreaady selected in step 5 fo or main and distribution bars are ch hecked as follows: For main bars (cl. 26.3.3.b.1 of IS 456:2000), the t maximum m spacing iss the lesser of o 3d and m i.e., 300 mm m. For distriibution bars (cl. 26.3.3.b b.2 of IS 4566:2000), the maximum sp pacing is 300 mm the lesseer of 5d or 450 mm i.ee., 450 mm. Provided spacings, theerefore, satissfy the requiirements. Maximuum diameter of the bars (cl. ( 26.5.2.2 of IS 456:2000) shall noot exceed 1440/8 = 17 mm m is also satisfiedd with the baar diameters selected herre. Figure 70 0 presents thhe detailing of the reinfo orcement bars. Thhe abbreviatiion B1 to B33 and T1 to T4 are the bottom and top bars, reespectively which w are shown inn Figure 49 for f a typical one-way slaab. The aboove design annd detailing assume abseence of supp port along shhort edges. W When supports along short eddges exist annd there is eventual e clam mping top reinforcemen r nt would be necessary at a shorter supportss also.
Figure 70 Reinforcemen nt detailing fo or the Slab of Problem 1
15.7.2. Problem 2 t cantileveer panel of thhe one-way slab shown in n Figure 71 subjected s to uuniformly diistributed Design the imposedd loads 5 kN N/m2 using M 20 and Fee 415. The load l of floorr finish is 0..75 kN/m2. The T span dimensioons shown inn the figure are a effective spans. The width w of the support is 3000 mm.
Figure 71 Slab specified iin Problem 2
Step 1: Selection S of preliminary p depth of slabb Compiled B By: Vaisakh G.
PP18
SBCE – C CE
Desig gn of Steel Sttructures
Basic vaalue of span to depth ratiio (cl. 23.2.11 of IS 456) = 7. Modificcation factor = 1.18 (see Problem 15.7.1). Minimum efffective deptth = 1850/7xx1.18 = 223.9 97mm. Assum me d = 225m mm and D = 250mm. 2 Step 2: Design D loadss, bending moment and shhear force Factoredd dead loadss = 1.5x0.25xx25 = 9.375 kN/m. Facttored load off floor finishh = 1.5x0.75 5 = 1.125 kN/m. Factored F live loads = 1.5xx5 = 7.5 kN//m. Total facctored loads = 18.0 kN/m m. Maximum negative momentt = 18.0x1.855x1.85x0.5 = 30.8025 kN Nm/m. Maxim mum shear foorce = 18.0xx1.85 = 33.3 kN/m. k Step 3: Determinatio D on of effectivve and total depths d of slab b d = √{30.8025x106/22.76x103}= 105.64 mm, considering g the value of R = 2.76 N N/mm2 from Table 5. This deppth is less thaan assumed depth d of slabb in Step 1. Hence, H assum me d = 225 m mm and D = 250 2 mm. Step 4: Depth D of slabb for shear foorce Using thhe value of k = 1.1 (cl. 40.2.1.1 of IS S 456:2000) for f the slab of o 250 mm ddepth. From Table 19 of IS 456:2000 (Tabble 10), we haave τc = 1.1xx0.28 = 0.308 8 N/mm2. Taable 20 of IS 456:2000 (T Table 11) 2 2 gives τcmmax= 2.8 N/m mm . τv=Vu/bbd = 33.3/2225 = 0.148 N/mm N Heree, the depth oof the slab is i safe in shear as τv<τc<τcmax D on of areas of steel (usingg table of SP-16) Step 5: Determinatio Table 444 of SP16 gives g 10 mm m diameter bars b @ 200 mm c/c cann resist 31.433 kNm/m > 30.8025 kNm/m.. Fifty per ceent of the barrs should be curtailed at a distance off larger of Ld or 0.5lx. Tab ble 65 of SP-16 gives Ld of 100 mm bars = 470 mm andd 0.5lx = 0.5x x1850 = 925 mm from thhe face of thee column. The curttailment distaance from thhe centre linee of beam = 925 9 + 150 = 1075, say 11100 mm. The aboove, howeverr, is not adm missible as thhe spacing off bars after the t curtailmeent exceeds 300 3 mm. So, we provide p 10 mm m @ 300 c/c c and 8 mm m @ 300 c/c.. The momennt of resistannce of this seet is 34.3 kNm/m > 30.8025 kNm/m k (see Table 44 of SP-16). Figure 72 pressents the dettailing of reiinforcing bars of this t problem..
Figure 72 Reiinforcement d detailing for SSlab specified d in problem 2 2
15.8.
PROBLEM MS FROM CHAPTER C 1 12
15.8.1. Problem 1 t slab paneel 1 of Fig.133.6.1 subjectted to factoreed live load of o 8 kN/m2 in addition to o its dead Design the load usiing M20 andd Fe415. Thhe load of floor f finish is i 1 kN/m2. The spans shown in fiigure are effectivee spans. The corners of thhe slab are prrevented from m lifting. Compiled B By: Vaisakh G.
PP19
SBCE – C CE
Desig gn of Steel Sttructures
Figure 73 3 Figure for p problem 1
Step 1: Selection S of preliminary p depth of slabb The spann to depth raatio with Fe 415 4 is taken from cl. 24..1, Note 2 off IS 456:20000 as 0.8 (35 + 40) / 2 = 30. Thhis gives the minimum efffective depthh d = 4000/3 30 = 133.33 mm, m say 1355 mm. The to otal depth D is thuss 160 mm. Step 2: Design D loadss, bending moments and shear s forces Dead loaad of slab (11 m width) = 0.16x25 = 4.0 4 kN/m2. Dead D load off floor finish (given) = 1.0 kN/m2. 2 Factoredd dead load = 1.5x5 = 7.55 kN/m . Facctored live lo oad (given) = 8.0 kN/m2. Total factoreed load = 15.5 kN/m2 Table 20 Maximum m bending moments of Problem 1
For
Sho ort span Lon ng span αx Mx (kNm/m)) αy My (kNm/m)) Negative moment at conntinuous edge 0.075 18.6 0.047 11.66 Positivee moment at mid-span 0.056 13.89 0.035 8.68 The coeefficients of bending b mom ments and thhe bending moments m Mx and My perr unit width (positive and negative) are deetermined ass per cl. D-1.1 and Tablee 26 of IS 4556 for the caase 4, “Two adjacent M edges diiscontinuous” and presennted in Table 13.5.3. The ly / lx for this problem is 6/4 = 1.5. Maximum shear forrce in either direction is determined d a Vu = w lx/2 as 2 = 15.5 x 4//) = 31 kN/m m Step 3: Determinatio D on/checking of the effectiive depth and d total depthh of slab Using thhe higher vallue of the maaximum bendding momentts in x and y directions fr from Table 20 0, we get 2 6 3 2 from Mu,lim = R,lim bd or d = √[18.6x10 /(2.76x10 )] = 82.09 8 mm, where w 2.76 N N/mm is the value of u R,lim takken from Tabble 5. Since, this effectivee depth is less than 135 mm m assumedd in Step 1, we w retain d = 135 mm and D = 160 mm. D of slabb for shear foorce Step 4: Depth Table 199 of IS 456:22000 (Table 10) gives thee value of τc = 0.28 N/mm m2 when the lowest perceentage of steel is provided p in the slab. Hoowever, this value needss to be modified by multtiplying with h k of cl. 40.2.1.1 of IS 456:2000. The vallue of k for the t total deptth of slab as 160 mm is 11.28. So, thee value of τc is 1.228x0.28 = 0.33584 N/mm2. Table 20 of o IS 456:20 000 (Table 11) gives τcmmax = 2.8 N/m mm2. The Compiled B By: Vaisakh G.
PP20
SBCE – C CE
Desig gn of Steel Sttructures
m2. Since, τv < τc < τcmax, tthe effectivee depth of computeed shear stresss τv = Vu/bdd = 31/135 = 0.229 N/mm the slab as 135 mm and a the total depth as 1600 mm are saffe. Step 5: Determinatio D on of areas of steel The resppective areass of steel in middle and edge strips are to be deetermined em mploying Equ uation of Step 5 of o the designn of one wayy slab. How wever, the areeas of steel computed c frrom the equaation and those obbtained from the tables off SP-16 are in i good agreeement. Accoordingly, the areas of steeel for this problem m are computted from thee respective Tables 40 and a 41 of SP P-16 and presented in Table T 21. Table 400 of SP-16 is for the effe fective depth of 150 mm,, while Tablee 41 of SP-116 is for the effective depth off 175 mm. The T followinng results arre, thereforee, interpolateed values obbtained from m the two tables off SP-16. Table 21 Rein nforcing bars of Problem 1 1
Particu ulars Taable N No. 40,41
Sh hort span lx Nm/m) Dia.&spacing Mx(kN
Long sspan ly TableNo.. My(kNm//m) Dia.& &spacing
Top steeel for 18.68 >18.6 10 mm @ 200 40,41 12.31>11.66 8 mm m @ 200 mm c/c negattive mm m c/c mom ment Bottom m steel >13.89 8 mm @ 170 40,41 9.20 >8.668 8 mm m @ 250 40,41 14.38> mm c/c for possitive mm m c/c mom ment The minnimum steel is determinned from thee stipulation of cl. 26.5..2.1 of IS 4556:2000 and d is As = (0.12/1000)x1000x1660 = 192 mm m2 and 8 mm m bars @ 250 mm c/c (= 201 mm2) is acceptaable. It is worth mentioning m thhat the areas of steel as shown in Tab ble 21 are moore than the minimum am mount of steel.
Figgure 74 Reinfo orcement dettails of section n 1‐1 of panell 1 for problem m 1
S of diameters annd spacings of o reinforcing g bars Step 6: Selection The advvantages of using u the tabbles of SP-166 are that thee obtained values v satisfyy the requireements of diameterrs of bars and a spacingss. However, they are ch hecked as reeady referencce here. Neeedless to mentionn that this steep may be om mitted in succh a situation n. Maximum diameter alllowed, as giv ven in cl. 26.5.2.2 of IS 456:20000, is 160/88 = 20 mm, which w is morre that the diaameters usedd here. The maximum m spacing of main barrs, as given in cl. 26.3.3(1) of IS 45 56:2000, is thhe lesser of 3x135 and 300 3 mm. a satisfiedd for all the bars. b The maxximum spaciing of minim mum steel (distribution baars) is the This is also lesser off 5x135and 450 4 mm. This is also satissfied. Compiled B By: Vaisakh G.
PP21
SBCE – C CE
Desig gn of Steel Sttructures
Figgure 75 Reinfo orcement dettails of section n 2‐2 of panell 1 for problem m 1
D on of torsionaal reinforcem ment Step 7: Determinatio
Figgure 76 Reinfo orcement dettails of cornerr C1
Figgure 77 Reinfo orcement dettails of cornerr C2
Torsionaal reinforcingg bars are deetermined foor the three different d typees of cornerss as explained earlier. The lenggth of torsionnal strip is 4000/5 4 = 8000 mm and thee bars are to be providedd in four layeers. Each layer wiill have 0.75 times the stteel used forr the maximu um positive moment. Thhe C1 type of corners will havve the full am mount of torrsional steel while C2 ty ype of corneers will havee half of thee amount Compiled B By: Vaisakh G. PP22
SBCE – C CE
Desig gn of Steel Sttructures
d any torsional steel. The results are presented p providedd in C1 type. The C3 typpe of corners do not need in Tablee 22 and Figuure 76, Figurre 77 and Figgure 78.
Figure e 78 Torsion R Reinforcemen nt bars of corn ner C3 Table 22 Torsional reinforcemen nt bars of problem 1
Type Dimensiions along Bar diametter & spacin ng No. of bars b along Cl. no. of IS S 456 x (mm) y (mm) x y C1 800 800 8 mm @ 200 2 mm c/c 5 5 D-1.8 C2 800 1600 8 mm @ 250 2 mm c/c 5 8 D-1.9 C2 1600 800 8 mm @ 2250 mm c/c 8 5 D-1.9 15.8.2. Problem 2 Design a two-way siimply supported slab of Figure 79, not n having addequate provision to resisst torsion at corneers and to preevent the corrners from liifting. The factored f live load is 6 kN N/m2 and thee load of the floorr finish in 1 kN/m2. The spans show wn in the figu ure are effecttive spans. U Use M 20 and d Fe 415. The widdth of the suppport is 300 mm. m
Figu ure 79 Proble em 2
Step 1: Selection S of preliminary p depth of slabb As per cl.24.1, c Notee 2 of IS 4566:2000, the sppan to effecttive depth rattio = 0.8x35 = 28. The minimum m effectivee depth = d = 4200/28 = 150 mm andd, therefore, D = 175 mm. Step 2: Design D loadss, bending moments and shear s forces Compiled B By: Vaisakh G.
PP23
SBCE – C CE
Desig gn of Steel Sttructures
o the slab = 1.5x0.175x25 = 6.5625 kN/m2. Facttored load off floor finish h = 1.5x1 Factoredd dead load of 2 2 2 = 1.5 kN N/m . Factoreed live loads = 6.0 kN/m . Total facto ored loads = 14.0625 kN//m . For this slab ly/lx = 5880//4200 = 1.4, Table 27 of IS 456:22000 gives αx = 0.099 and a αy = 0.051. Mx = αxwlx2 = 0.099x14.0625x4.2xx4.2 = 24.558 kNm/m. My = αywlx2 = 0.051x14..0625x4.2x4.2 = 12.651 kNm/m. N/m. Vu = 0.55 wlx = 0.5x114.0625x4.2 = 29.531 kN Step 3: Determinatio D on/checking of the effectiive depth and d total depthh of slab d = √{244.558x103/2..76} = 94.328 mm < 150 mm. So, wee keep d = 1550 mm and D = 175 mm. Step 4: Depth D of slabb for shear foorces Table 199 of IS 456:22000 (Table 10) gives τc = 0.28 N/mm m2. Clause 40.2.1.1 4 of IS S 456:2000 gives g k= 2 1.25 forr D = 175 mm. m So, τc = 1.25x0.28 = 0.35 N/mm m . Table 20 of IS 456:20000 (Table 11) 1 gives 2 3 2 τcmax = 2.8 2 N/mm . For F this problem τv = Vu/bbd = 29.531x x10 /(150x1000) = 0.19668 N/mm . Since S τv < τc < τcmaxx, the depth is i safe. Step 5: Determinatio D on of areas of steel The positive steel inn the two direections and thhe minimum m steel are furrnished below in Table 23. 2 These are the results r obtainned from the use of Tablee 41 of SP-16. However,, the followinng values can also be obtainedd from the ussual design procedures p deetailed in thee earlier sectiions. Table 23 Rein nforcing bars of Problem 2 2
Partiiculars
Positivve steel Minimuum steel = 0.12xx1750 = 2100 mm2
SP Table No. 41 96
Short span lx Mx D & spaciing Dia. (kkNm/m) 226.40 > 24.558 M steel Min.
8 mm @ 10 00 mm c/c (50 03 mm2) 6 mm @ 12 20 mm c/c (236 mm2) > 210 0 mm2
SP Tablee No. 41 96
Long span ly My Dia. & spacing (kNm/m m) 12.93 > 12.6511
m @ 120 6 mm mm c/c c (236 mm m 2) 6 mm m @ 120 mm c/c c (236 mm2) > 210 mm m 2
Min. steeel
Figure 80 De etailing of barrs of problem m 2, corners no ot held down (Section 1‐1 of Figure 79)
Figure 81 De etailing of barrs of problem m 2, corners no ot held down (Section 2‐2 of Figure 79)
Compiled B By: Vaisakh G.
PP24
