Beam Ledge Design
2
Sheet : Cont'd : fc' = fy = = x + x = = x + x = 1) = Lb = mm wb = mm Ab = cm2 = = = = = = <= = = 2) The reaction is considered at outer third point of bearing pad. d = a) cover = a = lb*(2/3) + 25 mm = mm d' = = mm Vu = + = x + x = b) Assume mm cover and mm bar stirrups. af = a + cover + bar stps/2 = mm :. Effective width flexure = Ewf = Wb + 5af = mm 3) ØPnb = Ø (0.85fc'Ab) where Ø = = > Vu = 4) for fc = 30 mpa Vn (max) = min( 800 * ew * d' , 0.2 *fc' * ew * d' ) = kn :. ØVn = > Vu = 5) Avf = Vu / (Ø * fy * µ ) = mm2 / per effective width of Ew. = mm = mm2 / m where µ = 6) Check for punching shear Vups = 4 * Ø * fc^0.5 * ( Wb + 2Lb + 2de ) * de where Wb = mm Ø = = kn > Vu = Lb = mm de = mm (assumed) 7) Nu = max ( 0.2Vu , T ) = kn An = Nu / Ø fy = mm2 / per effective width of mm = mm2 / m 8) Mu = Vu * af + Nu * (d - de) = kn.m find Af using conventional flexural design methods. For beam ledges Af = mu / Ø fy jd = mm = mm2 / m use jd = 0.8 and Ø = 0.85. 9) As = max ( 2 * Avf / 3 + An , Af + An, As(min) ) = mm2 / m As (min) = 0.04 * ( fc / fy ) * b *de = mm2 / m 10) Ah = 0.5 ( As - An ) = mm2 / m try mm dia bar ar = mm2 = use Ø @ c/c mm2 / per effective width of Ew. = 0.85 821 (reinforcement within 2/3 d and close top As) 12 125 ok (11.9.5) 1346 400 200 (11.7.4.3) Determine Shear friction reinforcement. 1564 1514 763.08 1.4 ok 165.08 1442 1184 40 kn ok (10.17.1) 763 kn 1.4 363 1.7 150 1.4 * Pdl 1.7 * Pll (9.3.2.4) 0.7 208 For flexure : critical section is at c/l of hanger r/f (Av) 1442 Check concrete bearing strength. 40 4.99525 mpa 410 mm Beam reactions (service load) kn Pll T 763 kn 500 kn kn Depth 65 150 512.92 kn Determine Shear Span and effective widths for both shear and flexure 1.4 363 1.7 150 12.917 mpa Design of Beam Ledge as per ACI 318M- 95 ( 9.2.1) mpa Allowable Bearing Capacity For shear friction 158 Effective width = Ew = Wb + 4a 1033 460 mm mm 363 Pdl Pll Check Bearing pad size : 400 200 725 Bearing Capacity of plate not ok 399.62 kn psi Prepared by : A B Quadri 7/17/2011 7/17/2011 7/17/2011 7/17/2011 Date : note : enter data in blue cells only. Verified by : 30 A B Quadri Project : Subject : 12 113 217 mm max spacing Smax Check effective ledge section for maximum nominal shear transfer strength Vn 1428.00 2228.46 Determine flexural reinforcement : - Af 153 Determine reinforcement to resist direct tension (Nu) : - An ok 1223.81 (11.9.3.2.1) kn 763.08 kn 438 kn 1033 (11.7.4.1) Prep By : A B Quadri- Abq Consultants - 9959010210 - [email protected]/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-India. 500028-www.abqconsultants.com Abq Consultants Abq Consultants Abq Consultants Abq Consultants Job No. : J-1023 J-1023 J-1023 J-1023 1 of 34 Civil & Structural Design Engineers Revision Notes : Calculation sheet Total dead load + live load (service) Total dead load + live load (ultimate) 1.0 363 1.0 150 513 kn Determine Shear reinforcement :- Ah 521 kn 350 kn sw of ledge Total Pdl Pdl 800 Determine primary tension reinforcement :- As 1346 1442 304 460 763 kn 2622 20
Transcript of Beam Ledge Design
Sheet :
Cont'd :
fc' =
fy =
=x + x
=
=x + x =
1) =
Lb = mm wb = mm Ab = cm2 =
= = =
= = <= =
=
2)
The reaction is considered at outer third point of bearing pad.
d =
a) cover =
a = lb*(2/3) + 25 mm = mm d' =
= mm
Vu = +
= x+
x =
b)
Assume mm cover and mm bar stirrups.
af = a + cover + bar stps/2 = mm
:. Effective width flexure = Ewf = Wb + 5af = mm
3)
ØPnb = Ø (0.85fc'Ab) where Ø =
= > Vu =
4)
for fc = 30 mpa Vn (max) = min( 800 * ew * d' , 0.2 *fc' * ew * d' ) = kn
:. ØVn = > Vu =
5)
Avf = Vu / (Ø * fy * µ ) = mm2 / per effective width of Ew. = mm = mm2 / m
where µ =
6) Check for punching shear
Vups = 4 * Ø * fc^0.5 * ( Wb + 2Lb + 2de ) * de where Wb = mm Ø =
= kn > Vu = Lb = mm
de = mm (assumed)
7)
Nu = max ( 0.2Vu , T ) = kn
An = Nu / Ø fy = mm2 / per effective width of mm = mm2 / m
8)
Mu = Vu * af + Nu * (d - de) = kn.m find Af using conventional flexural design methods. For beam ledges
Af = mu / Ø fy jd = mm = mm2 / m use jd = 0.8 and
Ø = 0.85.
9)
As = max ( 2 * Avf / 3 + An , Af + An, As(min) ) = mm2 / m As (min) = 0.04 * ( fc / fy ) * b *de
= mm2 / m
10) Ah = 0.5 ( As - An ) = mm2 / m
try mm dia bar ar = mm2 = use Ø @ c/c
mm2 / per effective width of Ew. =
0.85
821
(reinforcement within 2/3 d and close top As)
12 125 ok
(11.9.5)1346
400
200
(11.7.4.3)
Determine Shear friction reinforcement.
1564 1514
763.08
1.4
ok
165.08
14421184
40
kn ok
(10.17.1)
763 kn
1.4 363 1.7 150
1.4 * Pdl 1.7 * Pll
(9.3.2.4)
0.7
208
For flexure : critical section is at c/l of hanger r/f (Av)
1442
Check concrete bearing strength.
40
4.99525
mpa410
mm
Beam reactions (service load)
kn
Pll
T
763 kn
500
kn
kn
Depth
65
150512.92 kn
Determine Shear Span and effective widths for both shear and flexure
1.4 363 1.7 150
12.917
mpaDesign of Beam Ledge as per ACI 318M-
95
( 9.2.1)
mpaAllowable Bearing Capacity
For shear friction
158
Effective width = Ew = Wb + 4a 1033
460
mm
mm
363
Pdl Pll
Check Bearing pad size :
400200
725
Bearing Capacity of plate not ok399.62 kn
psi
Prepared by :
A B Quadri 7/17/20117/17/20117/17/20117/17/2011
Date :
note : enter data in blue cells only.
Verified by :
30
A B Quadri
Project :
Subject :
12 113 217 mmmax spacing Smax
Check effective ledge section for maximum nominal shear transfer strength Vn
1428.00
2228.46
Determine flexural reinforcement : - Af
153
Determine reinforcement to resist direct tension (Nu) : - An
ok
1223.81
(11.9.3.2.1)kn 763.08 kn
438
kn
1033
(11.7.4.1)
Prep By : A B Quadri- Abq Consultants - 9959010210 - [email protected]/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-India. 500028-www.abqconsultants.com
Abq Consultants Abq Consultants Abq Consultants Abq Consultants Job No. : J-1023J-1023J-1023J-10231 of 34
Civil & Structural Design EngineersRevision Notes :
Calculation sheet
Total dead load + live load (service)
Total dead load + live load (ultimate)
1.0 363 1.0 150 513 kn
Determine Shear reinforcement :- Ah 521
kn
350 kn
sw of ledge
Total Pdl
Pdl800
Determine primary tension reinforcement :- As
1346
1442 304
460
763 kn
2622
20
Sheet :
Cont'd :
11) As = mm2 / m
try mm dia bar ar = mm2 max spacing Smax = mm use Ø @ c/c
12) Av = Vu * s / ( Ø * fy * S ) = mm2 For s = c/c and S = c/c
For serviceability : Av = ( V / 0.5 * fy ) * ( s / ( Wb + 3 * af )) = mm2 (governs) :. Av = mm2
s = stirrup spacing
S = Support Spacing
use Ø @ c/c
Bw- beam width
Bearing pad
x x
Main Reinforcement
Ø
Horizontal Reinforcement
Ø
Vertical Reinforcement
Ø
Reactions on Main Beam due to Ledge load
=
Kn
Kn.m
Kn
Kn.m
182
ok
Check required area of hanger reinforcement : 125 1500
M (dl+ll - ultimate) = 209
Bw -width of beam
not ok20
Prepared by : Date :
Abq Consultants Abq Consultants Abq Consultants Abq Consultants Job No. : J-1023J-1023J-1023J-1023
201
Final Ledge reinforcement 125
16
note : sufficient stirrups for combined shear and torsion must be provided for
global effects in the ledge beam :
125
Determine final size and spacing of ledge reinforcement 1346
Verified by :Civil & Structural Design Engineers
Revision Notes :
A B Quadri
Project :
Subject :
149 16
357 357
Calculation sheet
Beam Ledge dimension and reinforcement datails
20 125 c/c@
12 @ 125
12 @ 100
125
500
125
250 min
200 L 400 W t
Thickness of bearing pad as per
manufacturers recommendations.
1 of 34
A B Quadri 7/17/20117/17/20117/17/20117/17/2011
230
513V (dl+ll - service) =
V (dl+ll - ultimate) = 763
M (dl+ll - service) = 140
1033
Prep By : A B Quadri- Abq Consultants - 9959010210 - [email protected]/2 Royal Residency, Besides Amba talkies, Mehdipatnam , Hyderabad-India. 500028-www.abqconsultants.com
max1033
