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note : enter data in cell ma
DesignofanCorbelasperAci318-95 ACI 318-08-
Ref codes
Building Code Requirements for Structural Concrete (ACI 318-08) and Commentary
only Bracing plate and Anchor to be welded to reinforcement.
As primary reinforcement.
:Shear Reinforcement for Slabs Slabs - Reported by Joint ACI-ASCE Committee Committee 421 : Recommendations for Design of SlabColumn Connections in Monolithic Reinforced Concrete Structures
av
Vu
Data : Strength of concrete fc
40
Mpa
5802
ps i
Strength of reinforement fy
415
Mpa
60191
psi
Service Dead Load Vsdl
45.00
Kips
200
Kn
Service Live Load Vsll
75.00
Kips
334
Kn
0.00
Kips
0.00
Kn
400
mm
15.75
in
Factored Horizontal force Nucenter 0 if data not available
Corbel W idth bw Solu So luti tion on
:
Thee ver Th verti tica call fac facto tore redd loa loadd to to be be car carri ried ed is
����� ������������������ Vu = 1.4 * ���� ������� 45.00 ������� + ���1.7 * = 190.5 ��������� Kips ��� ������ = 847.34 ������� ��� ���������� �� Vu 190.5�� 254.00 Vn = = = � � ������ φ 0.75 ���������������������� ������������������ ������������������ 1) W idth of Bearing plate(W b) = 250 mm ���������� vu <= ¢(Pab) = ¢(0.85fc'Al) where Al = Area of bearing ����������
Al =
Vu / (Ø*(0.85*fc'))
=
847.34 0.70
= .: Bearing length =
Al / W b
.: Bearing plate size =
250
x
cm
1
=
Ah(horizontal stirrups or ties)
0.75
75.00
(11.8.3.1)
Kn Kips
=
=
1129.79
9.84
in
1401.68
in2
Kn
ok
)
*
40
mm2
142.41
1
1000
0.85
35603
=
d h
φ
*
*
d 5 . 0
=
.:
use Bearing length =
200
200
mm
ok
mm
2) Determine shear span 'a' with 25 mm max clearance at beam end.Beam reaction is assumed at third
point of bearing plate, to simulate rotation of supported Girder and triangular distribution of stress under bearing pad. av > 2 * bl /3 or
=
Av = bl / 2 + 25
us e av =
135
133 = mm
mm 125
=
=
5.25
mm
=
5.31
in
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in 4.92
in
maximum
ok
3) In absence absence of roller or low-friction low-friction support support pad pad , a horizontal tensile tensile force of Nuc .:
=
0.2
*
1.60
*
=
0.2
*
192.00
=
=
38.40
Adopt Nuc <
0.2fc'bwd
75.00 ) Kips
=
170.81
-
254.00
=
d
=
254.00
=
0.2
*
13.90
170.81
Kn
Kn
5802
*
15.75 *
d
in
=
353.08
mm
*
5802
=
433.92
(480 + 0.08fc')bwd (
480
+
15.75 * <
=
(clause - 11.8.3.2)
(11.8.3.2.1) <
45.00 + 38.4
Kips
For normal weight concrete . Check Vn
(
1600 bw d
al s o
d
=
17.08
254.00
=
1600
in *
15.75 *
d
=
10.08
0.5 d
>=
Length of bearing plate
= ( Therefore max
d
=
19.69
Adopt
d
=
20.00
Cover from c/l of r/f to top =
0.08
50.0
=
in mm
in
200
in
= =
)
*
d
1.97
= +
50
256.05
mm
50
mm
+ )
500
mm
508
mm
in
mm
d
*
2
=
( 25 mm clearance from both sides )
500
mm
=
19.69
ok assumed value , since bar dia is not known
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in
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Calculation Sheet Project : Subject : CALCULATION
Total depth
h
=
21.97
in
=
558
mm
If a 45° slope is used then the corbel depth outside of the bearing area will be This is Check (11.8.1-a)
>=
0. 5 d
=
av / d < 1
10.00 5.31
=
in
=
=
0.266
10.00
254
<
in
=
254
mm
mm
1.00
ok
20.00
The total shear friction is found from equation Avf
Vu
=
Where ff is the friction factor. For monolithic construction - for normal concrete ff = 1.40
Ø * ff * fy 190.5
=
0.75
*
1000
*
1.40
=
*
The bending moment to be resisted is Mu
=
190.5
Mu
*
1088.09
=
3.01
in2
=
1944.66
mm2
60191 =
5.31 =
Vu *
+
90.67
38.40 kip-ft
=
av + *
Nuc *
(
(h-d)
21.97
122.94
-
20.00
)
Kn.m
12 Find Af using conventional flexural design method or coservatively use jud = 0.85d Af = Mu / ( ¢ * fy * ju * d )
90.67
=
0.75 The Tensi;e force Nuc =
38.40
*
k i ps
Note No te : for al alll des desig ignn cal calcu cula latio tions ns ¢ = 0. 0.75 75
12000 60191
=
*
0.85
*
1.418
As
=
>= Af +
in2
=
(11. (1 1.8.3 8.3.1) .1)
914.73
mm2
20.00
requires an additional steel area.
1000 = 0.851 in2 = 548.79 60191 Thus from the above two equation, the total steel area at top of the bracket must not be less than An
Nuc Ø fy
*
=
38.40 0.75 *
*
An >=
1.418
+
0.851
=
2.268
2 *
3.01
+
0.851
5802
^
i n2
=
mm2
1464
mm2
nor less than As
>=
2
Avf +
An
=
3
=
2.860
in2
=
1845
mm2
3
nor less than Asmin >=
3*
√fc' bw d
=
3*
0.5
fy =
1.196
*
15.75 *
20.00
60191 i n2
=
771
mm2
nor less than Asmin >=
200 bw d / fy
200
=
*
15.75 *
20.00
=
1.047
in2
=
675
mm2
60191 Therefore adopt
Ass A
=
provide
mm
dia
20
2.860 6
i n2
=
nos bars
1845
mm2
area =
2.9217
in2
=
1885 mm2
Close Clo sedd hoo hoopp ste steel el / st stirr irrup upss or tie tiess Ah Ah hav havin ingg a tot total al are areaa no nott less less tha thann 0.5 0.5(A (Ass - An An)) mus mustt be pr prov ovide ided. d.
ok ( 11.8 11.8.4) .4)
Thus Ah
=
0.5
provide 10 mm Spacing between bars =
*
(
dia
2.860 9
1.67
-
0.851
nos bars in
)
area = =
42.33
=
1.005
1.0956
i n2
i n2 =
=
648
706.86 mm2
mm2 ok
mm
11.8.4 - Total area, Ah, of closed stirrups or ties parallel to primary tension reinforcement shall not be less than 0.5(Asc- An). Distribute Ah uniformly within (2/3)d adjacent to primary tension reinforcement.
Prep By : A B Quadri- Abq Consultants - 9959010210 -
[email protected]/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-India.-www.abqconsultants.com Prepared by :
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Calculation Sheet Project : Subject : CALCULATION
WidthofCorbel=400mm 250wideX200long 2 270 70
X25thk X 25 thk
135
) t h g i e h l a t o T (
6/20Ø
9/10Ø 1
4 7 5
d e r i u q e r ' d ' n i m 8 0 5
4 0 3
1
5/10Ø
Concretegrade=40Mpa SteelGrade(PrimaryandSecondary)=415 Mpa
0
270
Prep By : A B Quadri- Abq Consultants - 9959010210 -
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5y 7y 7yy
Last Change 4/6/2014-----12:25:12 PM
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