Chapter 6. Compression Reinforement - Flexural Members If a beam ross setion is limite beause of arhitetural or other onsierations, it may happen that the onrete annot evelop the ompression fore require to resist the give bening moment. In this ase, reinforing is ae in the ompression zone, resulting in a so-alle oubly reinfore beam, i.e., one with ompression as well as tension reinforement. Compression reinfore is also use to improve servieability, improve long term efletions, an to provie support for stirrups throughout the beam. 6.1. Reaing Assignment: Text Setion 5.7; ACI 318-95, Setions: 10.3.4, 10.3.3, an 7.11.1 6.2. Strength Calulations A s b h Á u = 0.003 0.85f Á s a b = β 1 b C C s A b s - b h- b T b s b Á s = Á y strains stresses fores From geometry we an fin the strain in ompression steel at failure as: Á s = 0.003 (6.1) 120 Compression Reinforement
6.3. Nominal Resisting Moment When Compression Steel Yiels Á u = 0.003 0.85f 0.85f C s A Á y s a C = A s a C h + A s - T s A s T s = (A s A s ) b > Á y Case I Case II Doubly Reinfore Retangular Beam Total resisting moment an be onsiere as sum of: 1. Moment from orresponing areas of tension an ompression steel 2. The moment of some portion of the tension steel ating with onrete. M n = (A s A s ) ( β 1 2 ) + A s ( ) (6.2) an from equilibrium: 0.85f ab = (A s A s ) (6.3) Solve for a : a = A s A s 0.85f b (6.4) 121 Compression Reinforement
6.4. Compression Steel below Yiel Stress (strain ompatibility hek). Whether or not the ompression steel will have yiele at failure an be etermine as follows: h Á u = 0.003 0.85f C s A s Á s =Á y a C A s lim - T s b From geometry: Á y Á u = Á s (6.5) if ompression steel yiel Á s = Á y then: Á u Á = y Á = u Á u Á y (6.6) Equilibrium for ase II: (A lim s A s ) = 0.85 (β 1 ) bf (6.7) Substitute for from Eq. (6.6) an (6.7) an ivie both sies by b gives: (A lim s A s ) b Á = 0.85 β 1 b f u 1 Á u Á y b (6.8) or A lim s b = A s b + 0.85 β 1 f Á u Á u Á y (6.9) à lim = à s + 0.85 β 1 f 87, 000 87, 000 (6.10) if if à atual > à lim then ompression steel will yiel A s A s b 0.85 β 1 f 87, 000 87, 000 this is ommon for shallow beams using high strength steel then ompression steel will yiel 122 Compression Reinforement
6.5. Example of analysis of a reinfore onrete setion having ompression reinforement. Determine the nominal moment, M n, an the ultimate moment apaity, M u, of the reinfore onrete setion shown below. 2.5 f = 5, 000 psi = 60, 000 psi 22.2 A s =3.8in 2 A s =7.62in 2 Solution 12 M n an be alulate if we assume some onitions for ompression steel. Assume that ompression steel yiels: C = 0.85f β 1 b = 0.85 (5 ksi) (0.80) (12) = 40.8 C s = A s = 3.8 (60ksi) = 228 kips T s = (7.62 in 2 ) ( 60 ksi) = 457 kips Equilibrium: C s + C = T s solve for : = hek assumption 457 228 40.8 = 5.6 in 0.85f Á u = 0.003 Á s Á s = 0.003 5.6 2.5 = 0.003 = 0.0017 5.6 T s Á y - Á s = 0.0017 < E s = 60 29, 000 = 0.00207 wrong assumption This means the ompression steel oes not yiel. Therefore, our initial assumption was wrong. We nee to make a new assumption. 123 Compression Reinforement
Assume f s < C s = A s f s = A s Á s E s = (3.8 in 2 ) (0.003 2.5 ) (29, 000 ksi) = 330 2.5 Now for equilibrium: C s + C = T s 40.8 + 330 2.5 = 457 kips solve for = 6.31 in hek assumption 6.31 2.5 f s = 0.003 6.31 29, 000 = 52.5 ksi < = 60 ksi assumption o.k. hek ACI Coe requirements for tension failure = 6.31 = 0.284 < 0.375 22.2 We are in the tension-ontrolle setion an satisfy the ACI oe requirements. φ = 0.9 0.90 φ = 0.57 + 67Á t φ 0.70 0.65 SPIRAL OTHER φ = 0.48 + 83Á t Compression Controlle Transition Tension Controlle Á t = 0.002 t = 0.600 Á t = 0.005 t = 0.375 124 Compression Reinforement
Calulate fores: C = 40.8 (6.31 in) = 258 kips C s = 3.8 (52.5ksi) = 200 kips T s = (7.62 in 2 ) ( 60ksi) = 457 kips 258+200=458 Equilibrium is satisfie Take moment about tension reinforement to etermine the nominal moment apaity of the setion: M n = C β 1 2 + C s ( ) Nominal moment apaity is: M n = (258 kips) (22.2 0.80 6.31) + 200(22.2 2.5) 2 = 5080 + 3940 = 9020 in kips Ultimate moment apaity is: M u = φ M n = 0.9 9020 = 8118 in k 125 Compression Reinforement
6.6. Example of analysis of a oubly reinfore onrete beam for flexure Determine whether the ompression steel yiel at failure. 2.5 f = 5, 000 psi 21 2No.7 A s = 1.2 in 2 = 60, 000 psi 4 No. 10 A s = 5.08 in 2 Solution à = A s b = 14 5.08 21 = 0.0173 à = A s b = 1.2 14 21.5 = 0.0041 14 à à =0.0173 0.0041 = 0.0132 Chek whether the ompression steel has yiele, use Eq. (6.10):? 0.0132 0.85 β 1 f 87, 000 87, 000? 0.0132 0.85 0.80 5 87, 000 2.5 60 87, 000 60000 21? 0.0132 0.0217 Therefore, the ompression steel oes not yiel. 126 Compression Reinforement
6.7. Example: Design of a member to satisfy a nominal moment apaity. Assume we have the same size beam as Setion 6.6. example an wish to satisfy the same nominal onitions: = 60, 000 psi f = 5, 000 psi 22.2 2.5 A s =? in 2 Require M n = 9020 in k A s =? in 2 Solution 12 For singly reinfore setion: use = 0.375 à = 0.85β 1 f à = (0.85)(0.80)(0.375) 5 ksi 60 ksi = 0.0213 Maximum A s1 for singly reinfore setion then is: A s1 = à b = (0.0213) (12) (22.2) = 5.66 in 2 M n = à b 2 1 0.59à f y f M n = (0.0213 in 2 )(60 ksi)(12 in)(22.2 in) 2 1 0.59(0.0213) 60 5 = 6409 in.kips M u2 = φm n = 0.9 6409 = 5747 in.kips Moment whih must be resiste by aitional ompression an tension reinforement M u1 = M u1 M u2 M u1 = 0.9 9020 5747 = 2365 in.kips Assuming ompression steel yiels we will have: M u1 = φa s ( ) = 0.9 A s (60) (22.2 2.5) = 1063.8 A s 2365 in k = 1063.8 A s A s = 2365 1063.8 = 2.23 in2 Therefore, the esign steel area for tension an ompression reinforement will be: A s = 5.66 + 2.23 = 7.89 in 2 A s = 2.23 in 2 8-#9 3-#8 127 Compression Reinforement
2.5 22.2 A s =? in 2 A s =? in 2 12 Chek whether the ompression steel has yiele, use Eq. (6.10): A s A s b 0.85 β 1 f 87, 000 87, 000 8 2.37 0.85 0.80 5 87, 000 2.5 22.2 12 60 87, 000 60000 22.2 0.0211 0.206 Therefore the ompression steel yiels at failure Chek to make sure that the final esign will fall uner tension-ontrolle a = (A s A s ) 0.85f b a = (8.00 2.37)60 = 6.62 in 0.85(5)(12) = a = 6.62 = 8.28 in β 1 0.80 = 8.28 22.2 = 0.373 < 0.375 Tension ontrolle see the following page for the rest of the solution one in a speasheet. 128 Compression Reinforement