Chapter 6. Compression Reinforcement - Flexural Members If a beam cross section is limited because of architectural or other considerations, it may happen that the concrete cannot develop the compression force required to resist the give bending moment. In this case, reinforcing is added in the compression zone, resulting in a so-called doubly reinforced beam, i.e., one with compression as well as tension reinforcement. Compression reinforced is also used to improve serviceability, improve long term deflections, and to provide support for stirrups throughout the beam. 6.1. Reading Assignment: Text Section 5.7; ACI 318, Sections: 10.3.4, 10.3.3, and 7.11.1 6.2. Strength Calculations 0.85f c′ Á u = 0.003 A’s h d A bs d’ d-c b b Á s′ cb a b = β 1c b Cs Cc T bs h-c b Ás = Áy strains stresses forces From geometry we can find the strain in compression steel at failure as: d′ Á s′ = 0.003 c − c CIVL 4135 (6.1) 118 Compression Reinforcement 6.3. Nominal Resisting Moment When Compression Steel Yields d’ Á u = 0.003 A’s h c Áy 0.85f c′ a 0.85f c′ Cs = d As d-c Cc + A s′f y Ts T s = (A s − A s′)f y > Áy b a A s′f y Cc Case I Case II Doubly Reinforced Rectangular Beam Total resisting moment can be considered as sum of: 1. Moment from corresponding areas of tension and compression steel 2. The moment of some portion of the tension steel acting with concrete. M n = (A s − A s′) f y (d − β 1c ) + A s′ f y (d − d′) 2 (6.2) and from equilibrium: (6.3) 0.85f c′ ab = (A s − A s′)f y Solve for “a”: a= CIVL 4135 A s − A s′ f 0.85f c′ b y (6.4) 119 Compression Reinforcement 6.4. Compression Steel below Yield Stress (strain compatibility check). Whether or not the compression steel will have yielded at failure can be determined as follows: d’ 0.85f c′ Á u = 0.003 A’s c Á s′ = Á y a Cs Cc d h As lim d-c Ts ≥ Áy b From geometry: Áu c = Á′ s c − d′ if compression steel yield (6.5) Á′ s = Á y then: Áu c Áy = c − d ′ → c = Áu d′ − Áy Áu (6.6) Equilibrium for case II: − A′ s)f y = 0.85 × (β 1c) b f ′ c (A lim s (6.7) Substitute for “c” from Eq. (6.6) and (6.7) and divide both sides by “bd” gives: (A lim − A′ s)f y s = 0.85 × β 1 × b × f ′ c bd Áu Áu d′ − Áy A lim f′ A ′s s = + 0.85 × β 1 × c bd bd fy Áu Áu d′ − Áy d or à lim = à ′ s + 0.85 × β 1 × Ã actual > à lim if if f ′c 87, 000 d′ × fy 87, 000 − f y d then compression steel will yield f′ As − A ′s 87, 000 d′ ≥ 0.85 × β 1 × c × bd fy 87, 000 − f y d CIVL 4135 1 bd 120 (6.8) (6.9) (6.10) this is common for shallow beams using high strength steel then compression steel will yield Compression Reinforcement 6.5. Example of analysis of a reinforced concrete section having compression reinforcement. Determine the nominal moment, Mn , and the ultimate moment capacity, Mu , of the reinforced concrete section shown below. 2.5” A’s= 3.8 in2 22.2 f c′ = 5, 000 psi f y = 60, 000 psi As= 7.62 in2 12” Solution Mn can be calculated if we assume some conditions for compression steel. Assume that compression steel yields: C c = 0.85f ′ c β 1 cb = 0.85 × (5 ksi) × (0.80) × c × (12) = 40.8c C s = A s′f y = 3.8 × (60ksi) = 228 kips T s = (7.62 in 2) × ( 60 ksi) = 457 kips Equilibrium: Cs + Cc = Ts solve for c: c = 457 − 228 = 5.6 in 40.8 d’ Á u = 0.003 0.85f c′ check assumption Á′ s c c − d′ c 5.6 − 2.5 = 0.003 = 0.0017 5.6 Á′ s = 0.003 Á′ s = 0.0017 < d d-c Ts Áy fy 60 = = 0.00207 Es 29, 000 wrong assumption This means the compression steel does not yield. Therefore, our initial assumption was wrong. We need to make a new assumption. CIVL 4135 121 Compression Reinforcement Assume f ’s < fy C s = A′ sf ′ s = A′ s Á′ s E s c − 2.5 c − 2.5 = (3.8 in 2) × (0.003 ) × (29, 000 ksi) = 330 c c Now for equilibrium: 40.8c + 330 × Cs + Cc = Ts c − 2.5 = 457 kips c → solve for c c = 6.31 in → check assumption f ′ s = 0.003 × 6.31 − 2.5 × 29, 000 = 52.5 ksi < f y = 60 ksi 6.31 assumption o.k. check ACI Code requirements for tension failure c = 6.31 = 0.284 < 0.375 22.2 d 0.90 φ 0.75 0.65 We are in the tension-controlled section and satisfy the ACI code requirements. φ = 0.75 + (Á t − 0.002)(50) SPIRAL OTHER Compression Transition Controlled CIVL 4135 φ = 0.9 Tension Controlled Á t = 0.002 Á t = 0.005 c = 0.600 dt c = 0.375 dt 122 Compression Reinforcement Calculate forces: C c = 40.8 × (6.31 in) = 258 kips C s = 3.8 × (52.5ksi) = 200 kips 258+200=458 T s = (7.62 in 2) × ( 60ksi) = 457 kips Equilibrium is satisfied Take moment about tension reinforcement to determine the nominal moment capacity of the section: Mn = Cc d − β 1c 2 + C (d − d′) s Nominal moment capacity is: M n = (258 kips) × (22.2 − 0.80 × 6.31) + 200(22.2 − 2.5) 2 = 5080 + 3940 = 9020 in − kips Ultimate moment capacity is: M u = φ M n = 0.9 × 9020 = 8118 in − k CIVL 4135 123 Compression Reinforcement 6.6. Example of analysis of a doubly reinforced concrete beam for flexure Determine whether the compression steel yield at failure. 2.5” 2 No. 7 21” f c′ = 5, 000 psi f y = 60, 000 psi A’s= 1.2 in2 4 No. 10 As= 5.08 in2 14” Solution à = As = 5.08 = 0.0173 14 × 21 bd à − Ã′ = 0.0173 − 0.0041 = 0.0132 A′ s 1.2 Ã′ = = = 0.0041 14 × 21 bd Check whether the compression steel has yielded, use Eq. (6.10): ? 0.0132 ≥ 0.85 × β 1 × f ′c 87, 000 d′ × fy 87, 000 − f y d ? 87, 000 2.5 0.0132 ≥ 0.85 × 0.80 × 5 × 60 87, 000 − 60000 21 ? 0.0132 ≥ 0.0217 Therefore, the compression steel does not yield. CIVL 4135 124 Compression Reinforcement 6.7. Example: Design of a member to satisfy a nominal moment capacity. Assume we have the same size beam as Section 6.6. example and wish to satisfy the same nominal conditions: 2.5 ” f y = 60, 000 psi Required A’s = ? in 2 22.2” f c′ = 5, 000 psi M n = 9020 in − k As = ? in 2 12” Solution For singly reinforced section: use c = 0.375 d f′ à = 0.85β 1 c × c d fy 5 ksi = 0.0213 60 ksi Maximum As1 for singly reinforced section then is: à = (0.85)(0.80)(0.375) A s1 = Ã × b × d = (0.0213) × (12) × (22.2) = 5.66 in 2 fy M n = Ãf y bd 2 1 − 0.59à f c′ M n = (0.0213 in 2)(60 ksi)(12 in)(22.2 in) 2 1–0.59(0.0213) × 60 = 6409 in.kips 5 M u2 = φM n = 0.9 × 6409 = 5747 in.kips Moment which must be resisted by additional compression and tension reinforcement M u1 = M u1 − M u2 M u1 = 0.9 × 9020 – 5747 = 2365 in.kips Assuming compression steel yields we will have: M u1 = φA′ s f y (d − d′) = 0.9 × A′ s × (60) × (22.2 − 2.5) = 1063.8 × A′ s 2365 in–k = 1063.8 × A′ s A′ s = → 2365 = 2.23 in 2 1063.8 Therefore, the design steel area for tension and compression reinforcement will be: A s = 5.66 + 2.23 = 7.89 in 2 A′ s = 2.23 CIVL 4135 8-#9 3-#8 in 2 125 Compression Reinforcement 2.5 ” A’s = ? in 2 22.2” As = ? in 2 12” Check whether the compression steel has yielded, use Eq. (6.10): f′ As − A ′s 87, 000 d′ ≥ 0.85 × β 1 × c × bd fy 87, 000 − f y d 87, 000 8 − 2.37 ≥ 0.85 × 0.80 × 5 × 2.5 22.2 × 12 22.2 60 87, 000 − 60000 0.0211 ≥ 0.206 Therefore the compression steel yields at failure Check to make sure that the final design will fall under “tension-controlled” a= (A s − A s′)f y 0.85f c′b a= (8.00–2.37)60 = 6.62 in 0.85(5)(12) c = a = 6.62 = 8.28 in 0.80 β1 c = 8.28 = 0.373 < 0.375 22.2 d Tension controlled see the following page for the rest of the solution done in a speadsheet. CIVL 4135 126 Compression Reinforcement