Example of Bar Cutoff
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 psi8.0' 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
okd
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
l ≤ inches d =
- 52.78
2.54 2.54 12 31 116
l ft × inches ≤ inches → ok 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 "