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 - Chapter 6

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, 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 stir- rups 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

  C s

  Á s

  A’ b b

  ′

  s a c

  = β

  C

  1 c b c d’ d h b

  A s b b b d-c

  T h-c s

  Á

s y

  = Á

  b strains stresses forces

  From geometry we can find the strain in compression steel at failure as:

  c

  (6.1) − d′

  Á s

  ′ = 0.003

  c

6.3. Nominal Resisting Moment When Compression Steel Yields

  d’

  0.85f 0.85f

  c c

  ′ ′

  Á u

  = 0.003

  C s

  Á y a A’ a

  A s s y

  C ′f C c c c = d

• + h A d-c

  s A s y

  ′f

  T s

  T s s s y

  = (A − A ′)f

  y

  > Á

  b 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.

  β c

  1

  (6.2)

  M (d ) (d n s s y s y

  = (A − A ′) f − + A ′ f − d′)

  2 and from equilibrium: (6.3)

  0.85f

  c s s y

  ′ ab = (A − A ′)f Solve for “a”:

  A s s

  − A ′ (6.4)

  a f y

  = 0.85f

  c

  ′ b

6.4. Compression Steel below Yield Stress (strain compatibility check).

  bd

  or

  y  d

  ′

  ′

  

  1

  c f y

  Á u

  ′

  

f

  (6.5) (6.6) (6.7) (6.8) (6.9)

  b

  s bd

  ′

  A

  − Á

   Á u

  A lim s bd

  C s

  ′ = Á

  y Á s

  ≥ Á

  T s

  c

  ′

  C c

  A s lim A’ s a h d d’ c d-c

  c 

  (A

  lim s − A′ s

  )f

  y bd

  = 0.85 × β

  1

  × b × f ′

  =

  Substitute for “c” from Eq. (6.6) and (6.7) and divide both sides by “bd” gives:

  = 0.003 0.85f

  Á

  Á u

  then:

  y

  = Á

  s

  ′

  − d ′ if compression steel yield

  =

  c c

  =

  s

  ′

  Á

  Á u

  Whether or not the compression steel will have yielded at failure can be determined as fol- lows: From geometry:

  Á y

  c c

  c

  lim s

  ′

  

1

c ) b f

  = 0.85 × (β

  y

  )f

  s

  − A′

  (A

  − d ′ →

  ′ Equilibrium for case II:

  y d

  − Á

  Á u

  Á u

  =

  c

  y Á u

• 0.85 × β

  then compression steel will yield if this is common for shallow beams using high strength steel

  ′

  if

  A s

  − A ′

  s bd

  ≥ 0.85 × β

  1

  ×

  f

  c f y

  (6.10)

  ×

  

  87, 000 87, 000

  − f

  y  d

  ′

  d

  then compression steel will yield

  1

  ×

  

f

  ′

• 0.85 × β

  y  d

  − f

  87, 000 87, 000

  

  ×

  c f y

  ′

  d à actual > à lim

   Á u

  1

  s

  = Ã ′

  Ã lim

  d

  y  d

  − Á

  Á u

  ×

6.5. Example of analysis of a reinforced concrete section having compression reinforce- ment.

  ′

  Á

  d’

  s

  ′

  Á

  

d

T s

  c

  s

  0.85f

  = 0.003 Á y d-c c

  check assumption Á u

  40.8 = 5.6 in

  − 228

  = 457

  ′

  = 0.003

  Equilibrium:

  = 0.0017 <

  wrong assumption

  0.00207 This means the compression steel does not yield. Therefore, our initial assumption was wrong. We need to make a new assumption.

  60 29, 000 =

  =

  E s

  f y

  s

  c

  ′

  Á

  5.6 = 0.0017

  5.6 − 2.5

  = 0.003

  c

  − d′

  c

  ) × ( 60 ksi) = 457 kips solve for c:

  f c

  A

  22.2 2.5”

  2

  = 3.8 in

  2 A’ s

  = 7.62 in

  s

  Solution

  M n can be calculated if we assume some conditions for compression steel.

  = 60, 000 psi

  f y

  , of the reinforced concrete section shown below.

  u

  , and the ultimate moment capacity, M

  n

  ′ = 5, 000 psi Determine the nominal moment, M

  12”

  Assume that compression steel yields:

  2

  = A

  = (7.62 in

  T s

  = 3.8 × (60ksi) = 228 kips

  y

  ′f

  s

  C s

  C s

  = 0.85 × (5 ksi) × (0.80) × c × (12) = 40.8c

  1 cb

  c β

  = 0.85f ′

  s C c

  = T

  c

• C
• C

• 330 ×

Calculate forces:

  0.284 < 0.375

  = 6.31 in

  check assumption f

  ′

  s

  = 0.003 ×

  6.31 − 2.5

  6.31 × 29, 000 = 52.5 ksi < f

  y

  = 60 ksi assumption o.k.

  check ACI Code requirements for tension failure c d =

  6.31 22.2 =

  We are in the tension-controlled section and satisfy the ACI code requirements.

  = 457 kips → solve for c →

  φ = 0.9 Á t = 0.005 c d t

  = 0.375

  Á t

  = 0.002

  c d t

  = 0.600

  0.75

  0.90

  0.65 φ φ = 0.75 + (Á t

  − 0.002)(50)

  SPIRAL OTHER Compression Controlled

  c

  c

  Assume f

  2

  ’ s < f y

  C s

  = A′

  s f

  ′

  s

  = A′

  s Á

  ′

  s E s

  = (3.8 in

  ) × (0.003

  − 2.5

  c

  − 2.5

  c

  ) × (29, 000 ksi) = 330

  c

  − 2.5

  c

  Now for equilibrium: C

  s

  c = T s

  40.8c

  c

  Transition Tension Controlled

  C c

  = 40.8 × (6.31 in) = 258 kips 258+200=458

  C

  Equilibrium

  s

  = 3.8 × (52.5ksi) = 200 kips is satisfied

2 T )

  s

  = (7.62 in × ( 60ksi) = 457 kips Take moment about tension reinforcement to determine the nominal moment capacity of the section:

  β c

1 M d (d

  n c s

  = C − + C − d′)

  2

   

  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 n

  = φ M = 0.9 × 9020 = 8118 in − k

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

  2 A’ = 1.2 in s

  21”

  f c

  ′ = 5, 000 psi

  f y

  = 60, 000 psi

  4 No. 10

  2 A = 5.08 in s

  14”

  Solution A s

  5.08

  Ã 0.0173

  =

  bd =

  14 × 21 =

  Ã

  − Ã′ = 0.0173 − 0.0041 = 0.0132

  A s

  ′

  1.2

  Ã 0.0041

  ′ =

  bd =

  14 × 21 =

  Check whether the compression steel has yielded, use Eq. (6.10): ?

  f c 87, 000

  ′

  d

  ′ 0.0132

  0.85 ≥ × β × ×

  1

f d

y 87, 000 y

   − f 

  ? 87, 000

  2.5 0.0132

  ≥ 0.85 × 0.80 × 5 60 ×

  21

   87, 000 

  − 60000 ?

  0.0132 ≥ 0.0217 Therefore, the compression steel does not yield.

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

  2 A’ = ? in s

  22.2”

  f c

  ′ = 5, 000 psi

  Required M n

  = 9020 in − k

  2 A = ? in s

  12”

  Solution

  For singly reinforced section:

  c use 0.375 d = f c c ′ Ã

  = 0.85β

  1 d × f y

  5 ksi

  Ã 0.0213

  = (0.85)(0.80)(0.375) 60 ksi = Maximum A for singly reinforced section then is:

  s1

  2 A

  = Ã × b × d = (0.0213) × (12) × (22.2) = 5.66 in

  s

  1 f y

2 M bd

  1

  n y

  = Ãf − 0.59Ã

   f  c

  ′

  2

  2 M )(60 ksi)(12 in)(22.2 in) 1–0.59(0.0213) n

  = (0.0213 in × 60 = 6409 in.kips

   

  5 M

  n u = φM = 0.9 × 6409 = 5747 in.kips

2 Moment which must be resisted by additional compression and tension reinforcement

  M

  = M − M

  u u u

  1

  1

  2 M u = 0.9 × 9020 – 5747 = 2365 in.kips

  1 Assuming compression steel yields we will have: M f (d s y s s u = φA′ − d′) = 0.9 × A′ × (60) × (22.2 − 2.5) = 1063.8 × A′

  1

  2

  2365 in–k A 2.23 in

  s s

  = 1063.8 × A′ → ′ = 2365 1063.8 =

  Therefore, the design steel area for tension and compression reinforcement will be:

  2

  8-#9

  A s

  = 5.66 + 2.23 = 7.89 in

  2

  3-#8

  A s

  ′ = 2.23 in

  A s

  ′b

  ≥ 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

  a

  2.5

  = (8.00–2.37)60

  0.85(5)(12) = 6.62 in

  c

  =

  a β

  1

  = 6.62 0.80 =

  8.28 in

  c d =

  8.28 22.2 =

  0.373 < 0.375

  22.2 0.0211

  

  = ? in

  ′

  2 A’ s = ? in

  2

  22.2” 2.5 ”

  12” Check whether the compression steel has yielded, use Eq. (6.10):

  A s

  − A ′

  s bd

  ≥ 0.85 × β

  1

  ×

  f

  c f y

  − 60000

  ×

  

  87, 000 87, 000

  − f

  y  d

  ′

  d

  8 − 2.37

  22.2 × 12 ≥

  0.85 × 0.80 × 5 60 ×

  



  87, 000 87, 000

  Tension controlled see the following page for the rest of the solution done in a speadsheet.