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DESIGN CALCULATION SHEET
Project No. Date
------- 19-Apr-23
Sheet No. Computed by
-- of -- --
Subject ------------------------------Checked by Approved by
-- _Building Design of Beam Ledge
DESIGN OF BEAM LEDGE ACCORDING TO ACI 318 - 99Beam-B1 Ledge
Material Properties
= 350
= 4200
Section Properties
= 300.00 Cm
= 27.00 Cm
= 10.00 Cm
12.00 Cm 0.37 < 1
= 11.50 Cm
= 122.00 Cm
Straining Actions
= 10.50 t
= 2.10 t (Min. = 0.20 Vu = 2.10 t) ACI - 11.9.3.4
= 71.50 Cm
= 6.99 m. t ACI - 11.9.3
Effective width for Shear =(W+4a) or S = 51.50 Cm
= 66.48 t ACI - 11.9.3.2.1
= 7.03 71.50 Cm
= 0.59 71.50 Cm ACI - 11.9.3.4
= 2.10 51.50 Cm ACI - 11.7.4.1, Eqn. (11-25)
= 0.1066 ACI - 11.9.3.5 & 11.9.5
= 0.0492 ACI - 11.9.4
0.1558 @ F 17 @ 150 mm
Check of bearing
= 11.50 Cm
= 27.54 t > Vu O.K.
Check of punching
Section
= 40.30 t > Vu O.K.
= 150 mm
= 0.362
Verifying the serviceability1.205 (governs) Plan
= F 12 @ 150
Concrete Cylinder Strength (f 'C) Kg/Cm2
Steel Yield Stress (fY) Kg/Cm2
Total Thickness (h)
Depth of Section (d)
Load distance to face of Beam (a)
Load distance to center of Stirrup's Leg (a f)=
a /d=
Width of bearing plate / pad (W)
Distance between Ledge Loads (S)
Ultimate Vertical Force (Vu)
Ultimate Normal Force (Nuc)
Effective width for M & N=(W+5a f )or S
Ultimate Moment (MU)=Vu.af+Nuc(h-d)
Shear Strength ( f Vn)
Main Reinf. (F1) Calculations
Reinforcement for Flexure (A f) Cm2 / effective width of
Reinforcement for Normal force (A n) Cm2 / effective width of
Reinforcement for Shear Friction (Avf) Cm2 / effective width of
Main Reinforcement (A s) Cm2 / Cm
Total Area of Stirrups (A h) Cm2 / Cm
Provide (A s+A h) = Cm2 / Cm
Bearing plate length ( L )
Bearing Strength (f Pnb)= f.0.85.f'c.W.L
Assume d f = dPunching Strength (f Pnp)
Check Required Area for Hanger Reinf. (F2)
Spacing between Vertical hangers(ss)
A v = Vu . Spacing (ss) / ( f fy.S) Cm2
A v = (Vu/1.5) . (SS) / ( 0.5 . Fy).(W+3a) = Cm2
A v
d
NucVu
af
Av (F2)
As + Ah (F1)
4 F1 + Cover
h
af
a
d f
d f / 2
d f /2
L
d f / 2L
W
d f /2
d f /2
As + Ah (F1)
Av(F2)
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DESIGN CALCULATION SHEET
Project No. Date
------ 19-Apr-23
Sheet No. Computed by
-- of -- --
Subject ---------------------Checked by Approved by
-- _Building Design of Ledge-End
DESIGN OF BEAM LEDGE ACCORDING TO ACI 318 - 99Beam-B1 Ledge-end
Material Properties
= 350
= 4200
Section Properties
= 35.00 Cm
= 30.00 Cm
= 10.00 Cm
16.00 Cm 0.33 < 1
= 10.00 Cm
= 150.00 Cm
= 15.00 Cm
Straining Actions
= 19.00 t
= 3.80 t (Min. = 0.20 Vu = 3.80 t) ACI - 11.9.3.4
= 30.00 Cm
= 3.23 m. t ACI - 11.9.3
Effective width for Shear =(W+4a) or S = 30.00 Cm
= 43.03 t ACI - 11.9.3.2.1
= 3.19 30.00 Cm Min. to ACI-10.5
= 1.06 30.00 Cm ACI - 11.9.3.4
= 3.80 30.00 Cm ACI - 11.7.4.1, Eqn. (11-25)
= 0.1417 ACI - 11.9.3.5 & 11.9.5
= 0.0531 ACI - 11.9.4
0.1949 @ F 19 @ 150 mm
Check of bearing
= 15.00 Cm
= 31.24 t > Vu O.K.
Check of punching
Section
= 27.83 t > Vu O.K.
= 150 mm
= 2.661
Verifying the serviceability3.016 (governs) Plan
= F 20 @ 150
Concrete Cylinder Strength (f 'C) Kg/Cm2
Steel Yield Stress (fY) Kg/Cm2
Total Thickness (h)
Depth of Section (d)
Load distance to face of Beam (a)
Load distance to center of Stirrup's Leg (a f)=
a /d=
Width of bearing plate / pad (W)
Distance between Ledge Loads (S)
Distance to Ledge end (C)
Ultimate Vertical Force (Vu)
Ultimate Normal Force (Nuc)
Effective width for M & N=(W+5a f )or S
Ultimate Moment (MU)=Vu.af+Nuc(h-d)
Shear Strength ( f Vn)
Main Reinf. (F1) Calculations
Reinforcement for Flexure (A f) Cm2 / effective width of
Reinforcement for Normal force (A n) Cm2 / effective width of
Reinforcement for Shear Friction (Avf) Cm2 / effective width of
Main Reinforcement (A s) Cm2 / Cm
Total Area of Stirrups (A h) Cm2 / Cm
Provide (A s+A h) = Cm2 / Cm
Bearing plate length ( L )
Bearing Strength (f Pnb)= f.0.85.f'c.W.L
Assume d f = dPunching Strength (f Pnp)
Check Required Area for Hanger Reinf. (F2)
Spacing between Vertical hangers(ss)
A v = Vu . Spacing (ss) / ( f fy.S) Cm2
A v = (Vu/1.5) . (SS) / ( 0.5 . Fy).(W+3a) = Cm2
A v
d
NucVu
af
Av (F2)
As + Ah (F1)
4 F1 + Cover
h
af
a
Ledge End
d f
d f / 2
d f /2
L
d f / 2L
W
d f /2
d f /2
As + Ah (F1)
Av(F2)
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DESIGN CALCULATION SHEET
Project No. Date
---------
Sheet No. Computed by
-- of -- --
Subject -------------Checked by Approved by
_Building Design of Beam Ledge
DESIGN OF BEAM LEDGE ACCORDING TO ACI 318 - 99Beam-B2 Ledge
Material Properties
= 280
= 3600
Section Properties
= 20.00 Cm
= 200.00 Cm
= 12.00 Cm
17.00 Cm 0.06 < 1
= 250.00 Cm
= 250.00 Cm
Straining Actions
= 31.00 t
= 0.00 t (Min. = 0.20 Vu = 6.20 t) ACI - 11.9.3.4
= 250.00 Cm
= -5.89 m. t ACI - 11.9.3
Effective width for Shear =(W+4a) or S = 250.00 Cm
= 2380.00 t ACI - 11.9.3.2.1
= -1.21 250.00 Cm Min. to ACI-10.5
= 2.03 250.00 Cm ACI - 11.9.3.4
= 7.24 250.00 Cm ACI - 11.7.4.1, Eqn. (11-25)
= 0.6222 ACI - 11.9.3.5 & 11.9.5
= 0.3071 ACI - 11.9.4
0.9293 @ F 49 @ 200 mm Big Bar Diameter
Check of bearing
= 10.00 Cm
= 416.50 t > Vu O.K.
Check of punching
Section
= 2021.49 t > Vu O.K.
= 200 mm
= 0.810
Verifying the serviceability0.919 (governs) Plan
= F 11 @ 200
Concrete Cylinder Strength (f 'C) Kg/Cm2
Steel Yield Stress (fY) Kg/Cm2
Total Thickness (h)
Depth of Section (d)
Load distance to face of Beam (a)
Load distance to center of Stirrup's Leg (a f)=
a /d=
Unit Width of bearing plate / pad (W)
Unit distance of distributed Loads (S)
Ultimate Vertical Force (Vu)
Ultimate Normal Force (Nuc)
Effective width for M & N=(W+5a f )or S
Ultimate Moment (MU)=Vu.af+Nuc(h-d)
Shear Strength ( f Vn)
Main Reinf. (F1) Calculations
Reinforcement for Flexure (A f) Cm2 / effective width of
Reinforcement for Normal force (A n) Cm2 / effective width of
Reinforcement for Shear Friction (Avf) Cm2 / effective width of
Main Reinforcement (A s) Cm2 / Cm
Total Area of Stirrups (A h) Cm2 / Cm
Provide (A s+A h) = Cm2 / Cm
Bearing plate length ( L )
Bearing Strength (f Pnb)= f.0.85.f'c.W.L
Assume d f = dPunching Strength (f Pnp)
Check Required Area for Hanger Reinf. (F2)
Spacing between Vertical hangers(ss)
A v = Vu . Spacing (ss) / ( f fy.S) Cm2
A v = (Vu/1.5) . (SS) / ( 0.5 . Fy). (W) = Cm2
A v
d
NucVu
afAv (F2)
As + Ah (F1)
4 F1 + Cover
h
a
d f
d f / 2
d f /2
L
W
As + Ah (F1)
Av(F2)
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