Analysis of Doubly Reinforced Beam (Investigation) [PDF]

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Analysis of Doubly Reinforced Beam (Investigation)



Problem 1 (Compression Steel Yields) Determine the permissible ultimate moment capacity of the beam as shown in the figure. Use f’c = 20.7 MPa, fy = 345 MPa. Gillesania, p. 116



4 - 36 mm ∅



600 mm



2 - 28 mm ∅



60 mm



540 mm



350 mm



𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧: π 36 2 4 𝐀𝐬 = 𝟒, 𝟎𝟕𝟏. 𝟓 𝐦𝐦𝟐 As = 4



As − A′ s ρ − ρ′ = bd 4,071.5 − 1,231.5 ρ − ρ′ = 350(600) 𝛒 − 𝛒′ = 𝟎. 𝟎𝟏𝟑𝟓𝟐𝟒



𝐒𝐢𝐧𝐜𝐞



π 28 2 4 𝐀′𝐬 = 𝟏, 𝟐𝟑𝟏. 𝟓 𝐦𝐦𝟐 A′s = 2



0.85f ′ c β1 600 d′ ρ1 = dfy 600 − fy 0.85(20.7) (0.85) 600 60 ρ1 = (600)(345) 600 − 345 𝛒𝟏 = 𝟎. 𝟎𝟏𝟎𝟐



𝛒 − 𝛒′ > 𝛒𝟏 , 𝐜𝐨𝐦𝐩𝐫𝐞𝐬𝐬𝐢𝐨𝐧 𝐬𝐭𝐞𝐞𝐥 𝐲𝐢𝐞𝐥𝐝𝐬



Based from the free body diagram:



T = C1 + C2 As fy = 0.85f′c ab + A′s f′s 4,071.5 (345) = 0.85(20.7)(0.85c)(350) + 1231.5 345 𝐜 = 𝟏𝟖𝟕. 𝟏𝟖 𝐦𝐦 c − d′ c 187.18 − 60 f′s = 600 187.18 f′s = 407.67 MPa > 345 MPa f′s = 600



Actual strain of tension steel d−c ϵs = ϵ c c 600 − 187.18 ϵs = (0.003) 187.18 𝛜𝐬 = 𝟎. 𝟎𝟎𝟔𝟔𝟏𝟔



𝐂𝐨𝐦𝐩𝐫𝐞𝐬𝐬𝐢𝐨𝐧 𝐬𝐭𝐞𝐞𝐥 𝐲𝐢𝐞𝐥𝐝𝐬 f′s ≥ fy



𝐂𝐚𝐥𝐜𝐮𝐥𝐚𝐭𝐞 𝐚, ∅, 𝐚𝐧𝐝 𝐀𝐬𝟏 a = β1 c a = (0.85)(187.18) 𝐚 = 𝟏𝟓𝟗. 𝟏𝟎 𝐦𝐦



ϵy =



fy Es



345 ϵy = 200,000 ϵy = 0.001725 T𝐞𝐧𝐬𝐢𝐨𝐧 𝐬𝐭𝐞𝐞𝐥 𝐲𝐢𝐞𝐥𝐝𝐬 ϵs > ϵy



ϵs > 0.005 , ∅ = 𝟎. 𝟗𝟎



C2 = T2 A′s fy = As2 fy 𝐀′𝐬 = 𝐀𝐬𝟐



As1 = As − As2 As1 = 4,071.5 − 1,231.5 𝐀𝐬𝟏 = 𝟐, 𝟖𝟒𝟎 𝐦𝐦𝟐



Ultimate moment capacity: Mu = ∅ M𝑢1 + M𝑢2 a Mu = ∅ T1 d − + T2 d − d′ 2 a Mu = ∅ As1 fy d − + As2 fy d − d′ 2 159.10 Mu = 0.9 2,840 345 600 − + 1,231.5 (345) 600 − 60 2 Mu = 665,429,000 N ∙ mm 𝐌𝐮 = 𝟔𝟔𝟓. 𝟒𝟑 𝐤𝐍 ∙ 𝐦