# 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"

!

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

=

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

= = = = &lt; = ' 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 ×

= = &lt;

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

&gt; ld = 2.54' &gt; ld = 2.54' &gt; 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

&gt; ld = 2.54' &gt; ld = 2.54' &gt; 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 &gt; (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 &gt; (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 "