Concrete Design Project
Concrete Design Project
Concrete Design Project
Professor Amde
Submitted: 12/10/18
Erin Nolan
Evan Siegel
Olivia Hancock
Sumaccha Duyn
Table of Contents
1. Project Description 2
2. Building Layout 3
4. References 15
1
1. Project Description
The team is tasked with designing and performing the structural analysis on a six-story
office building in College Park, Maryland. ACI 318 Building Code and Commentary will be
used to design this building. A reinforced concrete frame with a one-way slab and beam floor
system will be used. Determine the amount of steel required at the location of maximum
positive and negative moments for the T-beam B2-B6 on the fifth floor. Also, determine
the column section required for column E4 on the third floor and design the footing for
column E4. Assume that the column locations and framing arrangements obtained in the
preliminary design and shown in the attached figure satisfy both architectural and structural
requirements. Assume normal weight concrete with fc’= 4,000 psi; fy= 60,000 psi. Assume
that shear walls, not shown in the figure, will resist lateral loads.
Floor Loads:
Footings:
2
2. Building Layout
3
4
3. Calculations and Results
3.1 Column Design:
Roof:
Snow 40 324 12960
Roof Mats 18 324 5832
Equip, Ceilings 22 324 7128
Slabs 75 324 24300
Total 50220 65448
Floor:
Live Load 75 324 24300
Equip, Ceilings 20 324 6480
Partitions 35 324 11340
Slabs 75 324 24300
Total 66420 89424
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The factored load was found using equation W= (1.2 × ∑ D) + (1.6 × ∑ L)
To determine the design for column E4 at floor 3 it is important to first determine the required
dimensions of the column at the footing in order to design a column that is flush at all sections.
Once the factored load is determined we need to solve for the moment about the column. To do
so we used equation (13-7) from ACI 318-11 in section 13.6.9.
l2 = 18’
ln = 18’-2(15”/12”)
qLu = 415 psf
qDu = 765 psf
Inputting a factored load of 529.44 kip and moment of 72.35 kip-ft the following column was
designed using SpColumn.
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Now that we have determined the dimensions of the column to be 15” x 15” we can solve for the
steel reinforcement for the column E4 @ the third floor. Once again we will be using equation
13-7 so we need to solve for the following variables once again.
l2 = 18’
ln = 18’-2(15”/12”)
qLu = 265 psf
qDu = 505 psf
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3.2 T-Beam Design
In order to determine the amount of steel required for the T-beam B2-B6, we had to consider the
location of maximum positive and negative moments for on the fifth floor. The T-Beam
was designed using SpBeam. The design code used was ACI 318-11 with ASTM A615
reinforcement.
Design Inputs:
The beam will have five supports with four spans, two exterior and two interior. Each span is 18
feet long as noted in the building plan view. The story height (Ha, Hb) is 12 feet as shown in the
elevation view of the office building. A thickness of 6 inches was used as given by the slab
thickness. The unit density of the concrete is 150 lb/ft 3 .
An initial estimate of 1.5 feet for width left and width right (3 feet total) was used for the flanges
of the T-Beam. This size was selected using ACI 8.12.2, which states that the T-beam flange
shall not exceed one-quarter of the span length of the beam (4.5 feet). In addition, ACI 8.12.3
states that for beams with a slab on one side only, the effective overhanging flange shall not
exceed one-twelfth the span length of the beam (1.5 feet). This is why we selected 1.5 feet to
compensate for both these code requirements.
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The only loads included in the design of this beam were the floor loads assuming that any roof
loads would be supported by a sixth floor beam. These loads were entered as area loads which
were given as preliminary load estimates in the problem statement. Each of these loads was
specified as a dead or live load. Under load combinations, coefficients 1.2, 1.2, and 1.6 were
used for self, dead, and live loads respectively to reflect the equation W= (1.2 × ∑ D) + (1.6 × ∑
L).
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The beam will be experiencing both tension and compression. These moments could have a very
small difference or cancel eachother out.
Design Results:
In order to account for tension and compression on the beam, double reinforcement was used.
SpBeam gave us data for the span’s between nodes. Below is a diagram showing the beams
spans.
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Top Reinforcement Design Results:
Using the results generated above using SpBeam, the following conclusions were made about the
steel reinforcement:
In span 1, there were 6 #3 bars on the left side and 7 #3 bars on the right. In span 2 and 3
there were 7 #3 bars on the left side and on the right side. For span 4 there were 7 #3 bars
on the left side and 6 #3 bars on the right side.
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Bottom Reinforcement Design Results:
Using the results generated above using SpBeam, the following conclusions were made about the
bottom bar steel reinforcement:
In each span there are 4 #3 bars.
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3.3 Footing Design
13
a 0.81
c 1.035
Et 10.3.5 0.067
.067 > .005 therefore tension controlled
Check development of reinforcement 15.6
Ld Eq. (12-1) 12.2.1, 12.2.3, 12.2.4 28.5
28.5 < 49.5 therefore bars fully developed
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4. References:
● ACI Code
● Software:
○ SpBeam
○ SpColumn
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