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A Rectangular Beam Has B

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11.) A rectangular beam has b = 300 mm and d =490 mm . Concrete compressive strength f’c = 27.6 MPa and steel yield strength fy = 276 MPa . calculate the required tension steel area if the factor moment Mu is (a) 20 Knm (b) 140 Kn-m (c) 485 Kn-m (d) 620 Kn-m. SOLUTION: Solve for ρ max and Mu max : ' ¿ Ρb ¿ .85 f c β 1 600 fy (600+ fy) pb 0.85 (27.6)(0.85)(600) 276(600+276) Pb Ρmax= 0.75 ρb = 0.0495 ρmax = .75(0.0495) Ρmax=0.0371 ωmax ¿ ¿ pmax fy f 'c ωmax 0.0371( 276) 27.6 ωmax = 0.371 Rn max = f’c ω(1-0.59ω) 0.59(.371)] Rn maax =27.6( .371)[1Rn max =8.001 MPa Mn max = Rn max b d2 (490) Mn max = 8.001(300) 2 Mn max = 576.279 x 106 n-mm Mn max = 576.279 kN-mm Mu max =Ø Mn max Mu max =0.90 x 576.279 Mu max =518.65 kN-m A) Mu = 20 kN-m < Mu max (singly reinforced) Mu = Ø Rn b d2 20 X 106 = .90 Rn ( 300)(490)2 Rn = 0.309 MPa ρ ¿ 0.85 f ' c fy 1- 1- 2 Rn 0.85 f ' c 0.85 (27.6) 276 ¿ ρ 1- 1– 2(0.309) 0.85(27.6) ρ =0.00113 < ρ min f'c ¿√ 4 fy ρmin ρ min = 1.4 fy if f’c > 31.36 MPa otherwise ρ min = 1.4 fy = .005072 As =ρ b d As = 0.005072(300)(490) As = 746 mm2 b) Mu = 140 kN-m < Mu max (singly reinforced) Mu = Ø Rn b d2 140 x 106 = .90 Rn (300) (490) Rn = 2.16 MPa ρ ρ= ¿ 0.85 f ' c fy ¿ 0.85 (27.6) 276 1- 2 Rn 0.85 f ' c 1- 1- 1– 2(2.16) 0.85(27.6) ρ = 0.00822 As = ρ b d ρ= 0.00822 > ρ min As= 0.00822(300)(490) As = 1,209 mm2 c) Mu =485 kN –m < Mu max (Singly Reinforce) Mu =Ø Rn b d2 485 x106 Rn (300)(490)2 ρ ¿ 0.85 (f ' c ) fy 1- 1– 2 Rn 0.85(f ' c ) ρ= ¿ 0.85 (27.6) 276 1- 1– 2(7.48) 0.85(27.6) ρ =0.03384> ρ min As = ρ b d As =0.03384(300)(490) As = 4,975 mm2 d) Mu = 620 kN –m > Mu max The beam will be doubly reinforced. 12.) A rectangular beam has b = 310 mm and d=460 mm. The beam will be design to carry a service dead load of 230 kN-m and service live load of 190 kN-m .Compression reinforcement if necessary will have its centroid 70 mm from extreme concrete fiber . Determine the required steel area . Use f’c = 30 MPa and fy = 415 MPa. SOLUTION: β= 0.85 Mu = 1.4 MD + 1.7 ML Mu =1.4( 230) +1.7(190) Mu = 645 kN-m Solve for ØMn max : NOTE: for rectangular beams Cmax = .75 cb C max = 0.75 600 d 600 fy = 203.94 mm A =β1 C max = 173.35 mm Mn max =0.85 f’c ab [d – a/2 ] Mn max = 0.85 (30) (173.35)(310) [460- 173.35/2] Mn max = 511.58 kN-m ØMn max = 0.90 (511.58) ØMn max = 460.42 kN-m Since Mu = 645 kN-m > Ø Mn max , Compression steel is necessary b A’s C’s=A's f’s a Cc =0.85f’c a b c = d-d’ d d- a/2 A’s + As As1 As2 T2= As2 Fy Mn1 = Mn max = 511.58 kN-m Mn2 = Mu Ø - Mn1 Mn2 = 645 0.90 - 511.58 Mn2 = 205.088 C = Cmax = 203.94 mm A =173.35 mm Tension steel: T1 = Cc As1 fy = 0.85 f’c a b As1 (415)= 0.85(30)((173.35) (310) As1 = 3,302 mm2 Mn2 = T2 (d –d’) 205.088x 106=As2(415)(460- 70) As2 = 1267 mm2 AS = As1 +As2 As= 3302 + 1267 As = 4569 mm2 Compression Steel: F’s 600 c−d ' c f’s =600 203.94−70 203.94 f’s =394.06 MPa A1 T = 100 mm c bf = 1200 a C1 A1 z A2 C2 y1 d=500mm y2 As Bw = 280 A2 = Ac max – A1 T A2 = 121,759 – 120,000 A2 = 1759 mm2 A2 = bw z 1759= 280 z Z = 6.28 mm Y1 = d- t /2 y1 = 500 – 100/2 Y1= 450 mm Y2 = d – t – z/2 y2 = 500 – 100 – 6.28/2 y 2= 396.86 mm Mn max = C1 y1 + C2 y2 Mn max = 0.85 f'c(A1y1 +A2y2) Mn max -= 0.85(21)(120,00 x 450 +1759 x396.86) Mn m,ax = 976. 36 kN-m Ø Mn max = 0.90 (976.36) Ø Mn max = 878.72 kN –m Since Mu = 1080 kN-m > Ø Mn max , the compression reinforcement must be provided. Bf = 1200 a` A1 C’s A’s d’ = 70 C1 A’s A2 z C2 d= 500mm 500 mm 430 y1 y2 d-d’ As d’ = 70 As1 As2 T2 Bw = 280 mm As1 = Asmax Mn1 = Mn max Mn2 A=t+z a = 100 + 6.28 a= 106.28 mm c =a/ β1 c = 106.28/0.85 c = 125.04 f’s = 600 c−d ' c f’s = 600 125.04−70 125.01 f’s = 264.1 MPa < fy Mn1 = Mn max = 976.36kN-m As1 = Asmax T1 = C1 +C2 As1 fy = 0.85f’c (A1+A2) As1(415)= 0.85(21)(120,000 +1759) As1 = 5237 mm2 Mn2 = Mn – Mn1 Mn2 = Mu Ø Mn2 = 1080 0,90 - Mn1 - 976.36 Mn2 = 223.64 kN-m Mn2 = T2 (d-d’) Mn2 = As2fy (d-d’) 223.64 x 106 =As2(415)(500- 70) As2 = 1253 mm2 Tension steel area ,As = As1+ As2 = 6,490 mm2 Compression steel area: C’s =T2 A’s f’s = As2 fy A’s(264.1)= 1253(415) A’s =1969 mm2 1.2 m 0.1m 1.2 m 0.47 m 0.28 m Dead load = weight of concrete: Area = 1.2 (0.1) + .28(0.47) = .2516 m3 Wc = Yc x area Wc = 23.5 (0.2516) Wc = 5.9126 kN/m dead load Wu A L= 9 m B WuL2/24 - WuL2/12 L= 9m C WuL2/24 - WuL2/12 - WuL2/12 Maximum positive moment ( at midspan) Wu L2 Mu= 24 Wu (9)2 1,080 = 24 Wu = 320 kN/m Wu = 1.4 Wd + 1.7 Wl 1.7 Wl 320 = 1.4(5.9126) + WL =183 .37 kN/m LIVE LOAD