Example: tourism industry

# Chapter 6. Compression Reinforcement - Flexural …

Chapter 6. Compression Reinforcement - Flexural members If a beam cross section is limited because of architectural or other considerations, it may hap- pen that the concrete cannot develop the Compression force required to resist the give bending mo- ment. In this case, reinforcing is added in the Compression zone, resulting in a so-called doubly rein- forced beam, , 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 stir- rups throughout the beam. Reading Assignment: Text Section ; ACI 318, Sections: , , and Strength Calculations u = c . Cs A's s . a b = 1c b Cc d' cb h d A bs d-c b h-c b T bs b s = y strains stresses forces From geometry we can find the strain in Compression steel at failure as: s = c . c d ( ). CIVL 4135 118 Compression Reinforcement Nominal Resisting Moment When Compression Steel Yields d'. u = c c . Cs A's y a a c Cc A s f y Cc d = +.

CIVL 4135 118 Compression Reinforcement Chapter 6. Compression Reinforcement - Flexural Members Ifabeamcrosssectionislimitedbecauseofarchitecturalorotherconsiderations,itmayhap-

### Information

Domain:

Source:

Link to this page:

Please notify us if you found a problem with this document:

### Transcription of Chapter 6. Compression Reinforcement - Flexural …

1 Chapter 6. Compression Reinforcement - Flexural members If a beam cross section is limited because of architectural or other considerations, it may hap- pen that the concrete cannot develop the Compression force required to resist the give bending mo- ment. In this case, reinforcing is added in the Compression zone, resulting in a so-called doubly rein- forced beam, , 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 stir- rups throughout the beam. Reading Assignment: Text Section ; ACI 318, Sections: , , and Strength Calculations u = c . Cs A's s . a b = 1c b Cc d' cb h d A bs d-c b h-c b T bs b s = y strains stresses forces From geometry we can find the strain in Compression steel at failure as: s = c . c d ( ). CIVL 4135 118 Compression Reinforcement Nominal Resisting Moment When Compression Steel Yields d'. u = c c . Cs A's y a a c Cc A s f y Cc d = +.

2 H As d-c Ts A s f y T s = (A s A s )f y b > y 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. 1c ( ). M n = (A s A s ) f y (d ) + A s f y (d d ). 2. and from equilibrium: c ab = (A s A s )f y ( ). Solve for a : A s A s ( ). a= f c b y CIVL 4135 119 Compression Reinforcement Compression Steel below Yield Stress (strain compatibility check). Whether or not the Compression steel will have yielded at failure can be determined as fol- lows: d'. u = c . Cs A's s = y a c Cc d h As lim d-c Ts b y From geometry: u c ( ). =. s c d . if Compression steel yield s = y then: u c u y = c d c = d ( ). u y Equilibrium for case II: (A lim s A s)f y = ( 1c) b f c ( ). Substitute for c from Eq. ( ) and ( ) and divide both sides by bd gives: (A lim s A s)f y bd = 1 b f c u u y d 1. bd ( ). or A lim s bd =.

3 A s bd f . + 1 c fy u u d . y d ( ). lim = s + 1 . f c fy . 87, 000 d . 87, 000 f y d ( ). this is common for shallow if actual > lim then Compression steel will yield beams using high strength steel if As A s bd f . 1 c . fy 87, 000. d . 87, 000 f y d then Compression steel will yield CIVL 4135 120 Compression Reinforcement Example of analysis of a reinforced concrete section having Compression reinforce- ment. Determine the nominal moment, Mn , and the ultimate moment capacity, Mu , of the reinforced concrete section shown below.. A's= in2. f c = 5, 000 psi f y = 60, 000 psi As= in2. 12 . Solution Mn can be calculated if we assume some conditions for Compression steel. Assume that Compression steel yields: C c = c 1 cb = (5 ksi) ( ) c (12) = C s = A s f y = (60ksi) = 228 kips T s = ( in 2) ( 60 ksi) = 457 kips Equilibrium: Cs + Cc = Ts solve for c: 457 228. c = = in d'. c u = check assumption c s c d d s = c Ts d-c = = y fy 60. s = < = = wrong assumption Es 29, 000. This means the Compression steel does not yield.

4 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 s E s c c = ( in 2) ( c ) (29, 000 ksi) = 330 c Now for equilibrium: Cs + Cc = Ts c + 330 c = 457 kips solve for c c = in check assumption f s = 29, 000 = ksi < f y = 60 ksi assumption check ACI Code requirements for tension failure c = = < We are in the tension-controlled section and satisfy d the ACI code requirements. = = + ( t )(50). SPIRAL.. OTHER. Compression Tension Transition Controlled Controlled t = t = c = c = dt dt CIVL 4135 122 Compression Reinforcement Calculate forces: C c = ( in) = 258 kips 258+200=458. C s = ( ) = 200 kips Equilibrium is satisfied T s = ( in 2) ( 60ksi) = 457 kips Take moment about tension Reinforcement to determine the nominal moment capacity of the section: Mn = Cc d 1c 2. + C (d d ). s Nominal moment capacity is: M n = (258 kips) ( ) + 200( ). 2. = 5080 + 3940 = 9020 in kips Ultimate moment capacity is: M u = M n = 9020 = 8118 in k CIVL 4135 123 Compression Reinforcement Example of analysis of a doubly reinforced concrete beam for flexure Determine whether the Compression steel yield at failure.

5 2 No. 7. A's= in2. f c = 5, 000 psi 21 . f y = 60, 000 psi 4 No. 10. As= in2. 14 . Solution As = = = bd 14 21. = = A s = = = bd 14 21. Check whether the Compression steel has yielded, use Eq. ( ): ? 1 . f c fy 87, 000 d . 87, 000 f y d . ? 5 . 60.. 87, 000 87, 000 60000 21.. ? Therefore, the Compression steel does not yield. CIVL 4135 124 Compression Reinforcement Example: Design of a member to satisfy a nominal moment capacity. Assume we have the same size beam as Section example and wish to satisfy the same nominal conditions: . f y = 60, 000 psi A's = ? in 2. f c = 5, 000 psi Required M n = 9020 in k As = ? in 2. Solution 12 . For singly reinforced section: use c = d f . = 1 c c d fy 5 ksi = ( )( )( ) = 60 ksi Maximum As1 for singly reinforced section then is: A s1 = b d = ( ) (12) ( ) = in 2.. M n = f y bd 2 1 . fy f c .. M n = ( in 2)(60 ksi)(12 in)( in) 2 1 ( ) 60 = 6409 5. M u2 = M n = 6409 = 5747 Moment which must be resisted by additional Compression and tension Reinforcement M u1 = M u1 M u2.

6 M u1 = 9020 5747 = 2365 Assuming Compression steel yields we will have: M u1 = A s f y (d d ) = A s (60) ( ) = A s 2365 in k = A s A s = 2365 = in 2. Therefore, the design steel area for tension and Compression Reinforcement will be: A s = + = in 2 8-#9. A s = in 2 3-#8. CIVL 4135 125 Compression Reinforcement . A's = ? in 2. As = ? in 2. 12 . Check whether the Compression steel has yielded, use Eq. ( ): As A s bd f . 1 c . fy . 87, 000 d . 87, 000 f y d . 8 5 . 12 60.. 87, 000. 87, 000 60000.. Therefore the Compression steel yields at failure Check to make sure that the final design will fall under tension-controlled . (A s A s )f y a=. c b ( )60. a= = in (5)(12). c = a = = in 1 c = = < d Tension controlled see the following page for the rest of the solution done in a speadsheet. CIVL 4135 126 Compression Reinforcement