Rectangular Concrete Beam Section Analysis

Beam Deflection and Stress | Civil Engineering Design Resources

RECTANGULAR CONCRETE BEAM/SECTION ANALYSIS
Beam Torsion and Shear
Per ACI 318-05 Code
Input Data:
Reinforcing Yield Strength, fy = ksi
Concrete Comp. Strength, f 'c = ksi
Beam Width, b = in.
Depth to Tension Reinforcing, d = in.
Total Beam Depth, h = in.
Ultimate Design Shear, Vu = kips
Ultimate Design Torsion, Tu = ft-kips
Ultimate Design Axial Load, Pu = kips
Total Stirrup Area, Av+t(used) = in.^2
Closed Stirrup Spacing, s = in.
Edge Distance to Tie/Stirrup, dt = in.
Results:
For Shear:
fVc =
The ultimate shear strength provided by the concrete, 'fVc', is calculated as follows:
For shear and flexure only (no axial load):
fVc = 2*f*(f'c*1000)^(1/2)*b*d
For shear with axial compression:
fVc = 2*f*(1+Pu/(2*Ag)*(f'c*1000)^(1/2)*b*d
For shear with axial tension:
fVc = 2*f*(1+Pu/(0.5*Ag)*(f'c*1000)^(1/2)*b*d
where: f = 0.75
Ag = b*h
Note: for members such as one-way slabs,
footings, and walls, where minimum shear
reinforcing is not required, if fVc >= Vu
then the section is considered adequate.
For beams, when Vu > fVc/2 , then a
minimum area of shear reinforcement is
required.
kips
fVs =
The ultimate shear strength provided by the shear reinforcing in beams, 'fVs', is calculated as follows:
fVs = f*fy*d*Av/s2
where: f = 0.75

The required amount of shear strength to be provided by the shear reinforcing in beams is:
fVs(req'd) = Vu-fVc >= 0.
The required amount of shear strength to be provided must be <= fVs(max).
kips
fVn = fVc+fVs =
fVs(max) =
The maximum allowable ultimate shear strength to be provided by the shear reinforcing in beams, 'fVs(max)', is calculated as follows:
fVs(max) <= 8*f*(f'c*1000)^(1/2)*b*d/1000
where:
f = 0.75
Av(prov) = in.^2?? = Av+t(used)
Av(req'd) =
The required area of shear reinforcing, 'Av(req'd)', perpendicular to axis of the beam is calculated as follows:
Av(req'd) = fVs*s/(f*fy*d) >= Av(min)
where:
f = 0.75
Note: Av(req'd) = area of two legs of a closed stirrup.
Av(min) =
The minimum area of shear reinforcing, 'Av(min)', to be provided for beams is calculated as follows:
Av(min) = 75*SQRT(f'c*1000)*b*s/fy
but not less than: 50*b*s/(fy*1000)
Note: Av(min) = area of both legs of a closed stirrup.
s(max) =
The maximum allowable shear reinforcing (closed stirrup) spacing, 's(max)', shall not exceed d/2, nor 24". However, when fVs > 4*f*(f'c*1000)^(1/2)*b*d, then the maximum spacing given above shall be reduced by one-half.
(Note: f = 0.75)
For Torsion:
Tu(limit) =
The effects of torsion may be neglected when the factored torsional moment, 'Tu', is less than 'Tu(limit)', which is calculated as follows:
Tu(limit) = f*(f'c*1000)^(1/2)*(Acp^2/Pcp)/12000
where:
f = 0.75
Acp = b*h
Pcp = 2*(b+h)
xo = b-(2*dt)
yo = h-(2*dt)
Tu(max) =
The maximum allowable torsion (with or without shear), 'Tu(max)', is calculated as follows:
Tu(max) = (((fVc*1000/(b*d)+f*8*(f'c*1000)^(1/2))^2-(Vu*1000/(b*d))^2)*(1.7^2*Aoh^4/Ph^2))/12000
where:
f = 0.75
Acp = b*h
Pcp = 2*(b+h)
xo = b-(2*dt)
yo = h-(2*dt)
Ph = 2*(xo+yo)
Aoh = xo*yo
Ao = o.85*Aoh
At(prov) = in.^2?? = (Av+t(used)-Av(req'd))/2
At(req'd) =
The required area of transverse torsion reinforcing, 'At', perpendicular to axis of the beam is calculated as follows:
At = Tu*12*s/(f*2*Ao*fy) >= At(min)
where:
f = 0.75
Aoh = xo*yo
Ao = 0.85*Aoh
Note: At = area of only one leg of a closed stirrup.
At(prov) = (Av+t(used)-Av(req'd))/2
At(min) =
The minimum area of transverse torsion reinforcing, 'At(min)', perpendicular to axis of the beam is calculated as follows:
At(min) = (50*b*s/(fy*1000)-Av(req'd))/2
Note: At(min) = area of only one leg of a closed stirrup.
At(prov) = (Av+t(used)-Av(req'd))/2
Al(req'd) =
The required area of longitudinal torsion reinforcing, 'Al(req'd)', parallel to axis of the beam is calculated as follows:
Al = At(req'd)/s*Ph >= Al(min)
where:
xo = b-(2*dt)
yo = h-(2*dt)
Ph = 2*(xo+yo)
Al(min) =
The minimum area of longitudinal torsion reinforcing, 'Al(min)', parallel to axis of the beam is calculated as follows:
Al(min) = 5*(f'c*1000)^(1/2)*Acp/(fy*1000)-At(req'd)/s*Ph
where:
Acp = b*h
xo = b-(2*dt)
yo = h-(2*dt)
Ph = 2*(xo+yo)
Total (Av+t) =
The required total area of transverse shear and torsion reinforcing, 'Av+t', perpendicular to axis of the beam is calculated as follows:
Av+t = Av+2*At >= Av+t(min)
Note: Av = area of two legs of a closed stirrup.
At = area of only one leg of a closed stirrup.
Total (Av+t)(min) =
The required total area of transverse shear and torsion reinforcing, 'Av+t(min)', perpendicular to axis of the beam is calculated as follows:
Av+t(min) = 50*b*s/(fy*1000)
Note: Av = area of two legs of a closed stirrup.
At = area of only one leg of a closed stirrup.

Contribute Article Spider Optimizer

© Copyright 2000 - 2017, by Engineers Edge, LLC www.engineersedge.com
All rights reserved
Disclaimer | Feedback | Advertising | Contact

Spider Optimizer

Home
Engineering Book Store
Engineering Forum
Excel App. Downloads
Online Books & Manuals
Engineering News
Engineering Videos
Engineering Calculators
Engineering Toolbox
Engineering Jobs
GD&T Training Geometric Dimensioning Tolerancing
DFM DFA Training
Training Online Engineering
Advertising Center


Copyright Notice

Publishing Program