ACI Beam Ledge

04/17/2023 at 17:30:51 Of 7 document.xls

document.xls - Rev. 1.0 / 7 04/17/2023

DESIGN CALCULATION SHEET

Project No. Date

------- 17-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)

N21

a/d < 1 according to Aci - 11.9.1

M48

0.7*(0.85*f'c*w*L)

M52

4 f SQRT(f'c)(W+2L+2d f)*d f

M56

Vu =< f . Av . fy .S/ss

M58

V =< Av .(0.5 fy) . (W+3a)/ss

document.xls - Rev. 1.0 / 7 04/17/2023



Project No. Date

------ 17-Apr-23


-- 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


= 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


= F 20 @ 150



Total Thickness (h)




a /d=

Width of bearing plate / pad (W)

Distance between Ledge Loads (S)

Distance to Ledge end (C)



















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)

N21


M48

0.7*(0.85*f'c*w*L)

M52

4 f SQRT(f'c)(W+L+d f)*d f

M56


M58

V =< Av .(0.5 fy) . (W+3a)/ss



Project No. Date

---------


-- 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


= 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


= F 11 @ 200



Total Thickness (h)




a /d=

Unit Width of bearing plate / pad (W)

Unit distance of distributed Loads (S)



















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)

N21


M48

0.7*(0.85*f'c*w*L)

M52

4 f SQRT(f'c)(W)*d f

M56


M58

V =< Av .(0.5 fy) . (W)/ss

ACI Beam Ledge

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