Civil MDC

# Computer Aided Design Beam

Load Data A Single span beam
P1DL kips L
P1LL kips a
P2DL kips b
P2LL kips
DL kips / ft.
LL kips / ft.

Dead Weight of Beam = 0.52 kips/ft
total factored Point Load 1 = 0.00 kips
total factored Point Load 2 = 0.00 kips
total factored Uniform Load = 0.72 kips

VA = #DIV/0! kips
VB = #DIV/0! kips
Max. Shear = #DIV/0! kips
Max. ultimate positive moment = #DIV/0! kips.ft
Max. ultimate negative moment = 0 kips.ft
hmin = 0 inches
req’ d = #DIV/0! inches

Determine depth (d) and thickness (h)
Use Beam Depth = #DIV/0! inches Thickness =

Check the beam weight by recalculating the load â€¦â€¦. Passed

Determine As
req’ As = #DIV/0! inches.2

``````Bar No. 5   6   7   8
Area / bar (in2)    0.31    0.44    0.6 0.79
No. of bars*    #DIV/0! #DIV/0! #DIV/0! #DIV/0!
Min. number of 2 steel bars is required
Use the top bar   =         2   No. 14 bar      Use the bottom bar size    =``````

Check Shear
Check if Bd(f’c)1/2 > Vmax / f
The shear reinforcement is #DIV/0!
Bar No. 3 4 5
Area (in2) 0.22 0.4 0.62
Spacing of bars #DIV/0! #DIV/0! #DIV/0!