PROJECT REPORT
Design of floor system supported on beams and columns
Submitted By:
Submitted To:
Table of Contents
1 Objectives:…………………………………………………………………………………………………..1
2 Introduction:………………………………………………………………………………………………..1
(a) Design approach:……………………………………………………………………………………..1
(b) Plan of Building:………………………………………………………………………………………1
(c) Load Combinations:…………………………………………………………………………………3
4 Calculations:………………………………………………………………………………………………..3
(a ) Self-Weight of Beam:………………………………………………………………………………4
(b) Self-Load of Slab :……………………………………………………………………………………5
( d ) Effective Flange Width:…………………………………………………………………………..6
( e ) Relative stiffness of beam:……………………………………………………………………….8
( f ) Transform Section:………………………………………………………………………………….9
( g) Calculation of Deflection:………………………………………………………………………..9
5 Free body diagrams:……………………………………………………………………………………….10
6 Results:………………………………………………………………………………………………………11
7 Discussion & Conclusion:……………………………………………………………………………12
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1 Objectives:
The major purpose of this project is to design and analyze reinforced concrete frame
system with different types of dead and live loads and calculate the reinforcement in different
members and design their dimensions according to the deflection control system.
2 Introduction:
Concrete structures are commonly used in now a days. Engineers majorly focused on
increasing trend of concrete construction, analysis and design process. Reinforced concrete
structures are built due to the different features of reinforcement. Reinforcement is an
important component of reinforced concrete. It is usually formed from ridged carbon steel ,
the ridges give frictional adhesion to the concrete. Reinforcement bar is used because
although concrete is very strong in compression it is virtually without strength in tension. To
compensate for this, it cast into it to carry the tensile loads on a structure. Since concrete is a
brittle material and is strong in compression. It is weak in tension, so steel is used inside
concrete for strengthening and reinforcing the tensile strength of concrete.
Reinforced concrete floor systems provides us an economical solution to a wide
variety of situations. Numerous types of non-prestressed and prestressed floor systems are
available to satisfy virtually any span and loading condition. Selecting the most effective
system for a given set of constraints can be vital to achieving overall economy, especially for
low- and mid-rise buildings and for buildings subjected to relatively low lateral forces where
the cost of the lateral-force-resisting system is minimum.
3 Project Description:
(a) Design approach:
The American Concrete Institute (ACI 14) design code is followed for design
calculations and LRFD design procedure is followed for design purpose and ASCE LRFD
load combinations is used for factored load and moment calculations.
(b) Plan of Building:
The plan of building consists of a regular beam and column arrangement and
reinforced concrete floor . The small squares indicate the square column. The structure is a
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plane frame having 69 x 69 ft span in x-direction and z-direction and story height of 10 ft in
y-direction. There are fix support on the bases of column at base level. The building has three
regular bays of 23 ft in x and z direction. The building has total 9 grids. The columns spacing
and stories height given in table given below;
Table 1: Frame Plan and Story Data

The beams, columns and other members are fabricated from concrete having fc’ = 4000 psi
and reinforcement yield strength is fy = 60000 psi.
Figure 1: Plan of Frame Building
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Figure 2: Cross Section of Beam
(c) Load Combinations:
The load combination for given structure is given below;

Table 2: Service Floor Load Details

4 Calculations:
fc ‘ = 4,000 psi
f y = 60,000 psi
Dead load=30 psf
Live load=50 psf
w
u = 1.2DL+1.6LL
=1.2(30) + 1.6(50)
Total factored load= wu =116.00 psf
R =
l
x
l
y
=
23
23 = 1.00
Thus it shows that it is two way slab.
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Load Intensity = 116.00 × 11.5 =1334.00lb/ft
For Beam A2–B2
Load Intensity = 2 ×1334.00 =2668.00 lbs/ft
Reaction =
0.5× 2668× 23
2 =15341.00 lbs
(a ) Self-Weight of Beam:
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A = 23× 12 7 + 11144 ×14
= 14.4858 ¿2
(b) Self-Load of Slab : wSD =
7
12
× 150=87.50 psf
Factored = 1.2 ×87.5=105.00 psf
After the distribution of load in both triangles
= 105 ×11.5 (2)=2415.00lb/ ft
So, Total design load for Beam A2–B2 is,
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= 2415.00 + 2668.00 = 4167.00 lb/ft
So, load on Column from one side is = 1845.00 lb
=
160.41667 × 23
2
× 4×1.2=7380lbs ×1.2=8854.99lbs
The Factored load on Column is,
=
0.5× 5083× 23
2
× 4+8855=125763.89lbs
pu = 125.76376 kips
( c ) For Equivalent UDL on Beam:
=
23
×23×(116+105)+1.2×160.41666=3581.1667 lb
ft
Design the Steel Reinforcement for Positive Moment in Beam A2–B2.
( d ) Effective Flange Width:
i. 16 hf+bw=16×7+14=126 inch
ii. L/4 =
23× 12
4
=60inch

b
f = 69 inch
Assume Neutral axis is on bottom of Flange,
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A =
A
s f y
0.85f c‘ b=
2×60,000
0.85×4,000×69
= 0.7578 in ¿7‘ ‘=hf
So, Design as singly reinforced beam,
M
u=
w
u L2
14
=
3581.2×232
14 = 135318.19 lb-ft
Mn = Mu / 0.9 = 135318.19/0.9 = 150353.5 lb-ft
R =
M
u
bd2=
150353.5 ×12
69× 152 =116.22 psi
1/m =
0.85f c‘
f y =
0.85× 4
60
=0.0567
p=1/m[1–√1–2mR fy ]=0.05667[1–√(1–2× 17.64 60000 ×116.22]]
p=0.0019716
AS=pbd=0.0019716 ×69×15=2.204¿2
Choose #6 bars
Use 4#6
A
S = 2.2089 ¿2
Design a short column at B2,
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Assume square Tied Column.
pu=15.764 kips
p = 0.04
A
S =0.04 A g
pu=¿ 0.85 f ‘c( Ag–As)+f y As
∅P
n=0.65 ×0.8¿ ]
125.764 = 0.65 ×0.8[0.85× 4( Ag–0.04 Ag)+60(0.04) A g]
A
g=42.69¿2
One Side = 6.53 inch
Choose 9’’ × 9’’ Column
A
g=
24.2538
56.6 = 0.4285 ¿2
3#4 ut one bar must provide on each face so use 4#4
Design of ties, for #3 tie,
Spacing:
i. 48 ×
38
=18∈¿
ii. 16 ×
48
=8∈¿
c) Least Dimension = 8 inch
If we use 14’’ ×14 ‘ ‘ section then minimum 1% must provide the
reinforcement.
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( e ) Relative stiffness of beam:
Locating Centroidal Axis
Y =
14 ×11×5.5+69×7× 14.5
14 ×11+69×7
= 12.3242 inch from bottom
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g=
14×113
12
+14× 11×6.824172+ 69× 2.17582
12
+69× 7×2.17582
= 12983.328 ¿2
= 41.643 k-ft
( f ) Transform Section: Ec=5700√4000
=3604996.533 psi
Ὺ= Es
E
c
¿ 29×106
3604996
=8
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A
s=5¿6=2.2089¿2
n A
s=17.672¿2
I
cr=
(bf–bw)hf 3
12
+(bf–bw)(hf )(x–h2f)2+ b12 w x3+ bw2x2+n As(d–x)2
For ‘x’
(bf–bw ) (hf )(x– h2f ).+ bw2x2 =n As(d–x)
X = 3.759 in ¿hf
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cr=
(69–14)
12
× 73+55×7(3.759–3.5)2 +
14 ×3.7593
12
+
14× 3.7592
2
+17.6715(15–3.759)2
= 3991.76 ¿4
M
a=135.318k–ft
I
e=
41.64263
135.3183 (12983.328) + [1-
41.643
135.323 ] × 3991.759
=4254.043 ¿4
( g) Calculation of Deflection:
℘=
wl L4
384 E
c Ie
w
l=50× 2
3
×23=766.67 lb
ft
℘=
766.67×234× 123
384× 3604996× 4254
=0.0629∈¿
L
360
=
23×12
360
=0.77∈¿0.0629∈¿
So, the slab and beam has proper resistance to deflection
5 Free body diagrams:
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Figure 3: Model of Frame
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Figure 4: Shear Force Diagram
Figure 5: Bending Moment Diagram
6 Results:
The factored area load on slab is 116 psf, and factored 7”slab dead load is 105 psf. It is
the type of two way slab. The load is transferred by two way slab action. The factored dead
load of beam is 2214 lb/ft. The factored load on column is 126 kips. The loading diagram
with analysis (maximum shear, moment and axial response) is shown in table given below,
also the factored load on column is given below;
Table 3: Floor Beam Analysis Details

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7 Discussion & Conclusion:
There concrete frame building was designed and analyzed for given load
combinations. The suggested cross section of beam and column was mentioned in plan
Figure-1 and beam section Figure-2. From the design calculations it is noticed that, in beam
A2-B2 max shear and moments is 34.1 K and 135.4 K-ft respectively. The column B2 was
loaded with point load of 126 kips and. The designed dimensions of the column B2 are 8×8
inch with 4#4 bars and #3@8” ties.
Floor beam and column taken from design results with their maximum shear and
bending moment values for analysis and design. We selected the “ T ” beams for our structure
with reinforcement 4#6 bars .The design result shows that the T-beam has enough section to
resist all types of loads. The design is ok according to the deflection control criteria. It has
good relative stiffness to resist all types of deflections within code limits. The bending
moment resisting capacity is too much the dimensions of the beam are 14” *14”.
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