Question

design a 12" wide beam with a simple span of 28′ to carry
P_{D} = 2000lb & P_{L} = 2000lb at the center
of the span f’_{c} = 4,000lb/in^{2} f_{y} =
60,000lb/in^{2}

factor loads

estimate height

depth=h-2.75"

Estimate reinforcement – A_{s} =
M_{u}/4d_{est}

Check minimum reinforcing
requirements

A_{smin} = (200/f_{y})bh

A_{smin} = 3(√f’_{c}/f_{y})bh

Select Reinforcing Bars

Answer #1

design a 12" wide beam
with a simple span of 28′ to carry PD = 2000lb &
PL = 2000lb at the center of the span
factor loads 1.2D+1.6L
Calculate Mu = PuL/4
f’c = 4,000lb/in2 fy =
60,000lb/in2
estimate height
depth = h - 2.75
Estimate reinforcement – As =
Mu/4dest
Check minimum reinforcing requirements
Asmin = (200/fy)bh
Asmin = 3(√f’c/fy)bh
Select reinforcing
bars

Design for flexure: a simple span beam 26’ long to carry PD =
1000# and PL = 2000# at the center of the span.
f’c = 4,000psi
fy = 60,000psi
(show all work)

Design for flexure a simple span beam 26’ long to carry PD =
1000# and PL = 2000# at the center of the span. f’c = 4,000psi and
fy = 60,000psi. show all work, not in milimeters

Please select the reinforcement spacing for the entire span of a
simply supported beam on a 22 ft span. Use the detailed method to
calculate Vc. Do not use a shear envelope diagram use the regular
shear force diagram. Use b = 16 in., d = 27 in., the main bending
reinforcement consists of 4 #9 bars. The unfactored dead and live
loads are 2.6 Kip/ft and 2.5 kip/ft, respectively, f’c = 4500 psi,
and fy = 60,000 psi. Use...

Design a simple-span one-way slab to carry a uniformly
distributed live load of 450 psf. this is the only load to be
supported other than the self weight of the slab. the span is 7-ft,
center-to-center of the supports. Assume f'c = 4000 psi, fy = 60000
psi. Design per foot of width.

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