Example of Bar Cutoff

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  Example of Bar Cutoff

A floor system consists of single span T-beams 8 ft on centers, supported by 12 in masonry walls spaced

at 25 ft between inside faces. The general arrangement is shown in below. A 5-inch monolithic slab to

be used in heavy storage warehouse. Determine the reinforcement configuration and the cutoff points.

Check the provisions of ACI 318 for bar cutoff. f’c = 4,000 psi (normal weight) fy = 60,000 psi

  8.0' b

  8.0' 8.0' 26' Typical

  1.0' 1.0' ?

  5" 18" 22"

  26'-0" See ACI 8.7 for definition

  See ACI 8.9 for definition of span

  of span length

  length

  12"

  !

  Dead Load Dead Load

  5 3 Weight of slab = ( ft )(7 )(150 / ft lb ft ) 440 lb ft / =

  12

  12

  22 3 Weight of beam = ( ft )( ft )((150 / lb ft ) 275 lb ft / =

  12

  12 440 275 715 /

  w lb ft

  = + =

  D

  1.2 860 /

  w lb ft

  =

  D Live Load w

  250 Referring to Table of 1.1 in your notes, for Storage Warehouse – Heavy, psf L 2

  =

  w (250 / lb ft )(8 ) ft 2, 000 / lb ft L = =

  1.6 w 3, 200 lb ft /

  L =

  Find Flange Width

  26 12 ×

  ↔ / 4

  78 inches Controls b = 48 inches

  L = =

  4

  80

  12

  92

  h inches b f w = = + 8 12

  • 16

  96 Centerline spacing = inches × = Determine Factored Load w

  1.2 w 1.6 w 860 3, 200 4060 lb ft / = 4.06 kips/ft

  • u = D L = =

  Determine Factored Moment

  1 2 M w l u u =

  8

  1 2 M

  ft-kips u = = (4.06)(26) 343

  8 Design the T-beam Use a trial and error procedure. First, assume for the first trial that the stress block depth will be equal to the slab thickness (a = 5 inches):

  M 343 12 ×

  76.2 2

  u A

  =

  4.92 in

  = = =

  s φ f d ( − a / 2) 0.9 60(18 × −

5 / 2) 18 5 / −

  2

  y 4.92 ×

  60 A f s y

  4.92 × 0.226

  1.11 5 → .

  a h i nches ok

  = = = = < = ' f

  0.85 0.85 × 4 ×

  78

  f b c

  The stress block depth is less than the slab thickness; therefore, the beam will act as a rectangular beam and the rectangular beam equations are valid. Adjust trial

  M

  76.2 2

  u A =

4.37 in

  = = =

  s φ f d ( − a / 2 )

  1 8 −

  1.1 1 / 2 y

  A f s y 4. 73 0.226

  0.99

  a

  = ' = × =

  0.85

  f b c

  Next trial

  M

  76.2 2

  u A =

4.35 in

s = = = f d ( − a / 2 )

  1 8 −

  0.9 9 / 2 φ y 2 Close enough to previous iteration of 4.37 in . Stop here.

  = Since 2 2

  T A f

  4.71

  23.16 s

  A in in

  = ≤ , we satisfy the ACI code and we will have tension failure.

  78" c '

  0.85 c

  f s s y

  = u ε

  = ×

  0.004 s t

  ε ε

  = = 18"

  d =

d c

  − a c C max 2

  ? s

  A in

  • × →

  s y c s A f f A in

  1.5" clear 1.5" clear

  → =

  Use 6- #8 bars As = 4.71 in2 Check ACI for Maximum Steel:

  Using similar triangles: 0.007

  7.71 0.004 0.004

  18 u

  c c c inches d c c

  ε =

  → =

  − − 1

  23.16

  0.85

  7.71 6.65 a c inches

  β

  = = ×

  =

  [ ] max ' max 2

  0.85 78 5 12 1.56

  =

  1.5" clear d = 18+0.5+.5 d = 18 in

  4- #8 2- #8 6- #8 d =19 in

  As=3.14 in2 As=1.57 in2 As=4.71 in2 CASE 1 CASE 2 CASE 3 2 bars 4 bars

  6 bars As = 1.57 As = 3.14

  As = 4.71 d = 19 in d = 18 in d = 18 in clear spacing = 2.63 in clear spacing =6.25 in clear spacing = 6.25 in

  60 s)

  40 p ki ( u

  1

  2

  3

  4

  5

  

6

  7

  8

  9

  10

  11

  12

  13 Distance From Support (ft) 400 350 300 250 s) p

  200

  • ki ft ( u

  150 M

  100

  50

  1

  2

  3

  4

  5

  6

  7

  8

  9

  10

  11

  12

  13 Distance From Support (ft) Note: Code allows discontinuing 2/3 of longitudinal bars for simple spans. Therefore, let’s cut 4 bars. Capacity of section after 4 bars are discontinued:

  1.57

  60 A f s y ×

  0.355

  a inches

  = ' = =

  0.85

  0.85

  4

  78

  f b c × × a M (2 bars ) M A f d ( − ) u n s y = φ = φ

  2

  0.35

  5

  1 M (2 bars ) u = × × − × = −

  0.9

  1.57 60( 19 ) 133 ft kips

  2

  12 Capacity of section after 2 bars are discontinued:

  3.14

  60 A f s y ×

  0.71

  a inches

  = = = '

  0.85

  0.85

  4

  78

  f b c × × a M (4 bars ) M A f d ( − ) u = φ n = φ s y

  2

  0.7

  1

  1 M (4 bars )

  0.9

  3.14 60( 18 ) 250 ft kips u = × × − × = −

  2

  12

  4.06 / w kip ft

  = u

  Find the location where the moment is equal

  M bars

  to (2 ) : u

  1 2 M x x

  x

  52.7 (4.06) = −

  2 2 M x x

  52.7

  2.03

  52.78

  = − 2

  kips M bars x x u (2 )

  52.7 2.03 133 = − = 2 2

  52.78 52.78 − 4 133 × ×

  2.03 ±

  2.03 − 52.78 133

  2.8

  x x x ft

  = → = = + 2 ×

  2.03 M bars Find the location where the moment is equal to (4 ) : 2 u

  M bars x x u = − = (4 )

  52.7 2.03 250 2 2

  52.78 52.78 − 4 × 250 ×

  2.03 ±

  2.03 − 52.78 250

  6.3

  x x x ft

  • 2

  = → = =

  2.03 ×

  = = <

  2 1.5 3 / 8 0.5 2.375

  3

  9 0.22 60, 000

  0.33 1500 1500 9 3 tr tr yt tr

  A in n s in A f k sn

  = =

  = ×

  = = = × ×

  1 (3.63) 1.8

  in control c in

  = = = = 2

  ⎧ =

  ← ⎪

  = ⎨ ⎪

  1.8

  0.33

  2.13

  2.5

  1.00 tr b

  

c k

ok

d

  0.22

  1.0 t e s ψ ψ ψ λ

  Note: Clear bar spacing is equal to: 1.5" clear

  8

  6- #8 As=4.71 in2 2- #8 As=1.57 in2 d = 18 in d = 18+0.5+.5 d =19 in

  6 bars As = 4.71 d = 18 in

  2 bars As = 1.57 d = 19 in

  4- #8 As=3.14 in2 4 bars As = 3.14 d = 18 in

  CASE 1 CASE 2 CASE 3 clear spacing = 2.63 in clear spacing =6.25 in clear spacing = 6.25 in center to center spacing = 3.63 in center to center spacing = 7.25 in center to center spacing = 6.25 in

  Note: Clear bar spacing is equal to: ( )

  1

  3

  12 2 no. of bars 2 1.5 no. of bars in one row - 1

  1.0

  8

  8 ⎡ ⎤

  ⎛ ⎞ ⎛ ⎞ =

  − − × − ⎜ ⎟ ⎜ ⎟

  ⎢ ⎥ ⎝ ⎠ ⎝ ⎠

  ⎣ ⎦

  Determine the development length

  1.0

  1.0

  • = ⎩

  ⎛ ⎞ ⎜ ⎟

  f

  3 ⎛ 3 60, 000 1 1 1 1 ⎞ y ψ ψ ψ λ t e s × × × ⎜ ⎟

  1

  33

  l d in d b = = × = ' ⎜ ⎟

  ⎜ ⎟

  40 40 4, 000

  2.13

  c k

  ⎛ ⎞

  f tr c

  ⎝ ⎠

  • ⎜ ⎟

  ⎜ ⎟ ⎜ ⎟

  d b

  ⎝ ⎠ ⎝ ⎠

  l d = = 33 in required 2.75 ft A s

  4.35

  2.75

  2.75

  2.54

  l ft d provided = × = × = A s

  4.71

  • ki p s)
  • ki p s)

  50 100 150 200 250 300 350 400

  11

  12

  13 Distance From Support (ft) M u ( ft

  CL 2.8' 6.7' 1.5'

  3" 12"

  One Layer 2-#8 2-#8

  1.5' 3.5' 5'

  11.7' 8.2' 2-#8

   > ld = 2.54' > ld = 2.54' > ld = 2.54' 3.05'

  1.3'

  1

  9

  2

  3

  4

  5

  6

  7

  8

  9

  10

  11

  12

  10

  8

  Extend bars:

  3" 12"

  12 12 1.00

  12

  1

  18

  1.5 b

  d inches ft d inches ft controls

  = ×

  = = ⎧ ⎨

  = = ←

  ⎩

  CL 2.8' 6.7' 1.5'

  One Layer 2-#8 2-#8

  7

  1.5' 3.5' 5'

  11.7' 8.2' 2-#8

   > ld = 2.54' > ld = 2.54' > ld = 2.54' 3.05'

  1.3'

  50 100 150 200 250 300 350 400

  1

  2

  3

  4

  5

  6

  13 Distance From Support (ft) M u ( ft

  Check Zero Moment: M n

  1.3

  l l

  • d a

  V u 3"

  343

  M u 381 . M ft kips n = = =

  0.9 φ 381 12

  ×

  1.3 3.00 116

  linches d =

  • 52.78

  2.54 2.54 12 31 116

  l ft × inchesinchesok d = = = 12"

  This is to ensure that the continued steel is of sufficiently small diameter and the required anchorage requirement of the ACI code is satisfied.

  Check for shear Complication (ACI 12.10.5) '

  = = =

  V

  2 f b d 2 4,000 12 18 27.3 kips c c w × ×

  A f d v y (0.22) 60 18 × ×

  = = =

  V s 26.4 kips s

  9 = = = + +

  V φ u ( s ) ( )

  V c V 0.75 27.3 26.4

  40.3 kips

  4.06 k/ft Vu(x = 1.3) = 52.78 - 4.06 x 1.3 = 47.5 kips Vu(x = 1.3) = 47.5 kips > (2/3) x 40.3 = 26.9

  Vu x

  Vu(x = 4.8) = 52.78 - 4.06 x 4.8 = 33.3 kips 52.78 kips

  Vu(x = 4.8) = 33.3 kips > (2/3) x 40.3 = 26.9 Need additional reiforcements at both cutoff points.

  Check for Shear Complications (ACI12.10.5), Continued

  3

  2-#8

13'

  1.5' 11.7' 8.2'

  One Layer 2-#8 2-#8

  = × =

  d inches

  4

  4

  13.5

  18

  3

  ⎝ ⎠ Provide additional reinforcement for a length of (3/4)d/

  (0.22) 60, 000

  ⎛ ⎞ ×⎜ ⎟

  × = = = ←

  β = = =

  A f b s in b d s in controls

  6 v y w w d

  8

  8

  2

  18 6.7 use 6 inches

  18.33 60 60 12

  13.5 " 13.5 "

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