Table of Content
S.No.
Title
Table of contents
List of Figures
List of Tables
Acknowledgement
1
Introduction
1.1
General information
1.2
Layout of the report
2
Slab design
2.1
Flat plate design
2.2
Flat slab design
2.3
Slab with interior beam
3
Beam and Column design
3.1
General information
3.2
Edge beams
3.3
Interior beams
3.4
Corner columns
3.5
Edge column
3.6
Interior column
4
Staircase design
4.1
General information
`4.2
Staircase design
5
Shear wall design
5.1
General information
5.2
Shear wall design for staircase
5.3
Shear wall design for elevator
6
Foundation design
6.1
General information
6.2
Footing under column
6.3
Footing under edge column
6.4
Footing under interior column
6.5
Combined footing under two adjacent corner edge
columns
6.6
Staircase foundation
6.7
Shear wall foundation
Page
No
1
1
4
6
19
29
35
35
36
49
66
69
74
79
79
79
93
93
93
99
109
109
109
113
117
120
127
132
i
7
7.1
7.2
8
9
145
145
158
179
180
Retaining wall
General information
Green engineering
Concrete material estimation
References
List of figures
Figure No.
Content
Figure 1.1 Structural feature of the office building
Page no.
1
Figure 1.2
2
The top view of the office building project using
25’x25’ panels
Figure 2.1 The four types of slab for the office building
Figure 2.2 The position of the use of the expansion joint on
each floors
Figure 2.3 The expansion joint (Source by Design Handbook:
section 4 from
Figure 2.4 Panel assignment
Figure 2.5 Bars arrangement
Figure 2.6 Shear check for flat slab with drop panel
Figure 3.1 Panel load share that goes to an Edge Beam
Figure 3.2 Glazing load dispersion diagram
Figure 3.3 Loaded Frame on Axis ((A1-A2-A3-A4- A5)
Figure 3.4
Loading and shear force diagram for beam A1-A2
of the flat plate floor
Figure 3.5 Loading diagram (axis 1B-2B-3B-4B-5B) for the
purpose of calculating additional moments due to
self weight of beam
Figure 3.6 Loading diagram (axis 1B-2B-3B-4B-5B) for the
purpose of calculating shear in internal beams due
to loads from slab
Figure 3.7 Corner column attachment and section of beams
attached to it
Figure 3.8 Load transfer to an edge column
Figure 3.9 Edge Column attachment to slabs/ beams in both
directions
Figure 3.10 Interior Column attachment to slabs/ beams
4
5
6
6
11
20
37
38
40
47
50
50
67
70
71
75
ii
Figure 4.1
Figure 4.2
Figure 5.1
Figure 6.1
Figure 6.2
Figure 6.3
Figure 6.4
Figure 6.5
Figure 6.6
Figure 6.7
Figure 6.8
Figure 6.9
Figure 6.10
Figure 6.11
Figure 6.12
Figure 6.13
Figure 6.14
Figure 6.15
Figure 6.16
Figure 6.17
Figure 6.18
Figure 7.1
Figure 7.2
Figure 7.3
Figure 7.4
Figure 7.5
Figure 8.1
Figure 8.2
Figure 8.3
Figure 8.4
Figure 8.5
Building Plan View
Staircase Plan View
Shear force and bending moment due to wind load
Footing load elements
Corner Footing Bearing Pressure distribution
Corner Footing Critical Sections: a) two way shear
, b) one way shear and c) bending moment
Edge footing Bearing Pressure distribution
Edge Footing Critical Sections: a) two way shear ,
b) one way shear and c) bending moment
Interior footing Bearing Pressure distribution
Interior Footing Critical Sections: a) two way
shear , b) one way shear and c) bending moment
Common footing Bearing Pressure distribution
Common Footing Critical Sections: a) two way
shear , b) one way shear and c) bending moment
Load at typical floor on wall
Critical section for shear
Critical section for shear
Plan of shear wall & vertical forces acting on
footing
Plan of footing integrated with beams 1, 2, 3
showed in red hatch
load & reactions on strip A
load & reactions on Beam 1
load & reactions on Beam 2
load & reactions on Beam 3
Example of Retaining Wall
Drainage system in retaining wall
Free body of soil & pressure
Footing Details
Critical section for shear in toe
Increases in Environmental Regulation
Graphs on Emission in Industrial
Typical Unitized Curtain Wall Systems
Typical Unitized Curtain Wall System Assembled
product
Typical Unitized Curtain Wall System Air Vapor
Barrier
79
81
94
109
110
110,112
114
114,115,116
118
118,119
121
123,124
127
129
130
132
134
134
137
139
141
145
146
147
151
155
163
164
169
169
171
iii
Figure 8.6
Figure 8.7
Figure 8.8
Figure 8.9
Figure 8.10
Figure 8.11
Load Transfers
Glass Panel Sizes at different floors
Components of Roofing System
Positive Drainage for healthy roots
Fixing Method
Scheme representation of Sewage Treatment Plant
174
175
177
178
178
180
List of Tables
Table 2.1
Table 2.2
Table 3.1
Table 3.2
Table 3.3
Table 3.4
Table 3.5
Table 3.6
Table 3.7
Table 3.8
Table 3.9
Table 3.10
Table 4.1
Table 5.1
Table 7.1
Table 8.1
Table 8.2
Description of different types of panel
Minimum thickness of slabs without interior beams
(ACI table 9.5c)
Summary of loading on Edge Beams
Design Moments from SAP analysis output for frame
on Axis ((A1-A2-A3- A4- A5)
Design Moments for internal beams on axis 1B-2B3B-4B-5B
Design Shear forces for internal beams on axis 1B2B-3B-4B-5B
Corner column loading from frame analysis
Design Loads Summary and Reinforcement provide
for a Typical Corner
Column
Acting design loads before magnification for edge
column
Design Loads Summary and Reinforcement provided
for a Typical Edge Column
Acting Column Axial loads for an interior column
Design Loads Summary and Reinforcement provided
for a Typical interior Column
Detailing of reinforcement
Specification of elevators
Vertical forces and moment acting at the toe of the
footing
Materials considered in green engineering
Glass weight
7
10
39
40
51
51
66
69
71
74
75
77
88
102
149
161
169
iv
Acknowledgment
Our deepest gratitude goes to Prof Riyad S. Aboutah for his continuous and constructive
advise and follow up. His successive advisories and comments were the pillars in our
every step during the design process of the building and its comments. We are thankful
to him for the fact that he has inspired and helped us know about the ACI Code, the
concepts and ways of analysis and design of Shear walls, staircases, beams, column,
foundations and retaining walls.
We are also thankful to our peers who helped us in one way or another in bringing this
project into the shape it has assumed now.
v
1. Introduction
This is a report for a design project of a three-story office building which is located in the
city of Syracuse, New York. The main structure of the office building project is wholly
made of reinforced concrete. The height of the building is 58 feet, and story heights of
each floor are as specified in Figure 1.1 Each panel size of the building is 25’x 25’.
Figure 1.1: Structural feature of the office building
In addition, the building is designed for fire safety purpose by having two staircases
situated at both corners of the building. For the convenient usage, the building is also
comprised of two elevators and two freight elevator in the middle part of the building.
Moreover, the outside parts of property include the parking lot, and retaining wall with
the height of 10 feet run through the edge area of the property. This is illustrated in the
figure 1.2 in the next page
1
Figure 1.2: The top view of the office building project using 25’x25’ panels
The flooring system used in the design of the building is fashioned in such a way that the
design project will give an opportunity to design different slab types. In line to this, slabs
of flat plate type on the top floor, slab of flat slab type on the middle floor and slabs with
interior beams on the first floor are used. On the top of the building along the peripheries
there is a 1-ft high parapet wall. At the edges of the panels of staircase and elevators,
shear wall will be incorporated for the design from the roof slab through ground slab. The
thick square lines at the panel edges represent shear wall to be used. On the roof slab, a
mechanical room is placed next to the two elevators in the same panel.
Furthermore, as green engineering is becoming a big concern for the recent world
environment, the design the building intends to include the aspects of the use of green
roof at the top of the building (roof slab), glass panels placed outside on all sides of the
building and Sewage Treatment Plant which gives the clean water as required for clean
environment. In Building Sewage Treatment plant is fixed at the parking lot as it is a
2
modular system .Apart from that few finishes items are explained like waterproofing,
Paints were its advantages to the environment are explained
Throughout this project design a concrete strength capacity of 6000 psi and the steel
strength capacity of 60000 psi are used. The design procedures are pertinent to those
specified in the ACI-318 code and international building code design.
Nonetheless to mention, aesthetics as an integral part of a building design, though not
thoroughly dealt, has been tried to be incorporated. For this aspect, glass walling has been
used as façade for the external part of the building. Material use estimation with a focus
on concrete usage is also a part of the design project.
The report is organized in such a manner that it will have flow when read. The second
chapter is devoted to the design of slabs. Chapter three presents the design of beams and
columns followed by design of stair case and shear wall in chapter four. In chapter six
designs of foundations for the building and the shear walls is addressed. The green
engineering and aesthetics part is included in chapter eight proceeded by retaining wall in
chapter six. At the end comes the Concrete estimation followed by references.
3
2. Slabs Design
2.1
General information
This office building uses the different type of design on each slab. Therefore, four types
of slab will be used as specified by a plan requirement. These four slabs are designed
using two-way slab systems, direct design method. A flat plate is used for a design for a
roof slab which has supporting beams running over along the edge of the roof slab. Next,
a flat slab is implemented for a design for the second floor. It also has supporting beams
running over along the edge of the floor like the way it is done on the flat plate. However,
they are different in that the flat slab makes a use of drop panels on the top of columns
supporting the flat slab, resulting in a use of a less slab thickness than that of flat plate.
For the first floor, the slab with interior beams is used. In this case, supporting beams at
four sides of each panel squarely connect all the columns below the slab. Finally, the
ground slab is designed to be used on the first floor of the building. The illustration of the
slab design is shown below.
Figure 2.1: The four types of slab for the office building
4
Moreover, expansion joint is provided for this building. Due to the temperatureproduced change, a potential impact is likely to occur to the office building. The
utilization of the expansion joint brings a benefit for the office building in terms of the
structural integrity and building serviceability. Before the concrete is placed, the
expansion joint will be constructed by installing pre-formed or pre-molded
elastic/resilient material. The thickness of such material to be incorporated at edge of
slabs is half inch with the wideness of each slab types. Once the material has been
inserted, the brass or bronze cover plate will be attached. The feature below shows the
use of expansion joint at the interior edges of the office building on each floors and a
feature example of this usage.
Figure 2.2: The position of the use of the expansion joint on each floors
5
Figure 2.3:
The expansion joint (Source by Design Handbook: section 4 from
http://www.copper.org/homepage.html)
2.2
Flat plate
Figure 2.4: Panel Assignment
6
The slab has been divided into panels (as indicated) for the purpose of design
using the Direct Design Method as adopted by ACI Code. The axis coordinates have also
been assigned to the top view of the building plan in order to show the direction of the
particular strips to be considered in the calculation using excel sheet. The particular
description of the panels are as detailed in the next page.
Table 2.1: Description of different types of panel
Panel 1

An exterior ( but not on the corner) panel in 
An exterior corner panel, both of its exterior
which its exterior side is support on edge beam,
edges are supported on edge beam
which runs over the exterior columns

Panel 2

Fulfills the requirements stated for direct design
Fulfills the requirements stated for direct design
method as per ACI Code
method as per ACI Code
Panel 3
Panel 4

An interior panel


Fulfills the requirements stated for direct design
method as per ACI Code
An exterior panel like with support conditions
like Panel 1

Does not meet, the requirement for the presence
of at least three continuous panels
direction
7
each
Preliminary design of minimum column dimension
In order to obtain a value of (ln) , the preliminary column dimension must be determined.

The typical column considered is an interior column with the following tributary
area.

Assume the depth of slab to be 8 inches.

Loads on this tributary area include snow load, live load, dead load, rain load.

Rain load, RL
=
62.5 psf

Snow load, SL
=
46.2 psf

Roof Live load, Lr =
12.0 psf

Dead load
=
150 lb/ft3 *(8/12) ft = 100 psf

Topping load
=
20 psf

Live load
=
50 psf

Finishing load
=
20 psf
Load combination (preliminary)
This formula below is used to determine the combination of loads (ACI code 9.2)
U
=
1.2(DL+ FL + TL) +1.6*(LL+HL) + 0.5[max (Lr, SL,RL)]
=
1.2(100 + 0 + 20) + 1.6(0 + 0) + 0.5*62.5
=
175.25 psf
=
1.2(100 + 20 +0) + 1.6(50 + 0) + 0.5 (0)
=
224 psf
For flat plate
U
For flat slab
U
For slab with beams
8
U
=
1.2(100 + 20 +0) + 1.6(50 + 0) + 0.5 (0)
=
224 psf
Load on Column
Pu
=
0.80 [0.85
(Ag -Ast) +
Ast]
=
specified constant by code = 0.7 for tied columns
=
6,000 psi
=
60,000 psi
Assume Ast is equal to 1.5% of Ag = 0.015 Ag
For flat plate,
P
=
175.25 * (25’*25’)
Pu
=
1.5* P = 1.5 * 109.53 = 164.3 kips
164.3 =
= 109.53 kips
0.7 *0.8 [0.85 * (6,000/1,000) * (Ag -0.015Ag) + (60,000/1,000) *
0.015Ag]
Ag
=
50 in2
Required column size = 7” x 7”
For flat slab,
Pu
368.7 =
=
164.3 + 1.5[218 * (25’*25’)] =
368.7 kips
0.7 *0.8 [0.85 * (6,000/1,000) * (Ag -0.015Ag) + (60,000/1,000) *
0.015Ag]
Ag
=
111.15 in2
Column size = 11” x 11”
For slab with beams,
Pu
573.07 =
=
368.7 + 1.5[190.25 * (25’*25’)]
=
573.07 kips
0.7 *0.8 [0.85 * (6,000/1,000) * (Ag -0.015Ag) + (60,000/1,000) *
0.015Ag]
Ag
=
172.76 in2
Column size = 14” x 14”
9
As a result, the column size to be used is 14”x 14” in the design for flat plate, flat slab,
and slab with beams.
Depth requirements
For the specified design concrete strength, as per ACI (table 9.5c) the minimum
thickness required for flat slab and flat plate is
Table 2.2: MINIMUM THICKNESS OF SLABS WITHOUT INTERIOR BEAMS (ACI
table 9.5c)
fy, psi
Without drop panels
Exterior panels
With drop panels
Interior
Exterior panels
Interior
panels
panels
With edge
Without
With edge
Without
beams
edge
beams
edge
beams
60000
ln/33
ln/30
beams
ln/33
ln/36
ln/33
ln/36
Minimum slab depth requirement for flat plate
hmin
=
ln/33 ;
ln
=
25-(14’/12’) = 23.833’
=
23.833/33
=
0.722’ = 8.667”
As a result, we try hmin = 9 inches.
New design load for roof flat plate
DL
= 150 * (9/12) = 112.5 psf
Ud
= 175.25 + 1.2 (112.5 – 100) = 190.25 psf
With the green roof load, an additional load of 30 psf is included.
Ud
= 190.25 + 1.2 (30)
= 226.25 psf
Static Moment for all panels of flat plate
Mo
= (Ud l1 ln2)/8
(ACI code 13-4)
= 226.25 * 25 * [25-(14/12)2] / (8*1000) = 401.613 ft-k
10
Consider bar arrangement as follows
Figure 2.5: Bars arrangement
dx= depth for bars spanning in x-direction
dy = depth for bar spanning in y-direction
For an assumed j=0.925, compute trial area As, at a section of maximum moment, in this
case (first interior negative column strip) with Mu = 210.85 ft-kips
Shear check for flat plate
Bars # 5 are used for reinforcement.
dx = 9-3/4-(0.625/2) = 7.9375”
dy = 7.9375 – 0.625 = 7.3125”
dav = (7.9375+7.3125)/2 = 7.625”
b0 = (2) (14” + 7.625”) + (2) (14”+7.625”) = 86.5”
Vu
ØVc
=
[(25’*25’)-{(14”+7.625”)/12}2]*(226.25/1000)
=
140.827 kips
=
=
Ø4
*b0 * d
(0.85) * 4 *
=
173.704 kips > 140.827 kips OK
(ACI code 11-35)
* (86.5) * (7.625)/1000
11
12
13
14
15
16
17
18
2.3
Flat Slab
Depth requirement
From table 2.2, minimum slab depth requirement for flat slab
hmin
=
ln/36 ;
ln
=
25-(14’/12’) = 23.833’
=
23.833/36
=
0.662’ = 7.944”
As a result, we try hmin = 8 inches.
This thickness is the same as one assumed before. Therefore, Ud remains the same as one
calculated from the preliminary design for a column dimension.
Ud
=
224 psf
Static Moment for all panels of flat slab
Mo
= (Ud l1 ln2)/8
= 226.25 * 25 * [25-(14/12)2] / (8*1000) = 401.613 ft-k
Note:

Height of drop panel >= ¼ of slab thickness

Width of drop panel extension on each side >=1/6 of span length measured
center-to-center of supports

 f for each beam is considered the smallest value 0.8
19
Shear check for flat slab with drop panel
Figure 2.6: Shear check for flat slab with drop panel
There are 2 critical sections for slab with drop panel (dash line)
h
=
8”
hd
≥
h/4
8”/4
=
2”
Therefore, the thickness of drop panel = 8” + 2”
=
10”
;
hd
=
Assume bar #5 to be used for reinforcement.
Determine d1
d1
=
10” –
– 0.625”
=
8.625”
Determine d2 by using
dx
=
8”-3/4-(0.625”/2)
=
6.9375”
dy
=
6.9375” – 0.625”
=
6.3125”
d2
=
(6.9375”+6.3125”)/2 =
L/6
=
25’/6 =
6.625”
4.167’
Consequently, use 4.2’ for a length of drop panel from the center of the column.
20
Check shear at the edge of drop panel
b0
Vu
ØVc
=
2[(4.2’*12”) + (6.625”)] + 2[(4.2’*12”) + (6.625”)]
=
228.1”
=
[25’ * 25’ – {4.2’+ (6.625”/12”)} 2] * (224/1000)
=
134.942 kips
=
(0.85) 4
=
(0.85) 4
=
397.984 kips > 134.942 kips OK
b0 d
* 228.1” * 6.625”
Check punching shear at the column
Since the drop panel is thicker than the thickness of flat slab, the adjustment must be
implemented.Therefore, The new U for the section over the drop panel
Upanel =
b0
Vu
224 psf + 1.5 * 150 psf * [(10”-8”)/12]
=
261.5 psf
=
2(14”+8.625”) + 2(14”+8.625”)
=
90.5”
=
[(25’*25’ – 4.2’*4.2)*224/1000’] + [4.2’*4.2’ –
{(14”+8.625”)/12”}2]261.5/1000
Vc
=
139.732 kips
=
(0.85) 4
=
205.571 kips > 139.732 kips OK
* 90.5 * 8.625/1000
For a flat slab, the slab thickness to be used for design is 8”.
Ud = 224 psf
Static moment = 397.619 ft-k
21
22
23
24
25
26
27
28
2.3
Slabs with interior beams between columns
In order direct design method to be applicable the following condition shall be satisfied
(ACI 318 section 13.6.1.6)
0.2≤
where
Step 1 : Slab thickness and beam size:

If
>1 for all beams all the shear is transferred to the columns by beams, making it
unnecessary to check shear while selecting the slab thickness
Assume all beams to have sizes of 14 in width and 20 in depth
The calculated α for both the edge and interior beam respectively are 1.86 and 2.1
respectively and this satisfy the above condition  no shear failure checking required
for slabs

Corresponding average α values for a corner panel and interior panel are then
The slab thickness will be determined by the thickness of the interior panel
But not less than
Therefore the assumed h= 7 inches can be comfortably selected as the thickness of the
slab with interior beams running between columns
29
Loading , based on h= depth of slab 7 in considered
U= 1.2 (150*7/12+15) +1.6*(50) =203 Psf
Design static bending moment, Mo
The design static moments for edge strips and interior strips are

Edge strip
L2= 12.5+ (4/12) = 12.83 ft ln= 25 ft 

Interior strip
l2= 25 ft ln= 25 ft 
Reinforcement Calculation

Assume bar # 5 is used , the depths, d, in the orthogonal direction are
dx= depth for bars spanning in x-direction
dy = depth for bar spanning in y-direction

For an assumed j=0.925, compute trial area As, at a section of maximum moment,
in this case (first interior negative middle strip) with Mu=-64.43 ft-kips per 12. Ft
length
30
=2.915 in2

Compute a and a/d and check ρ≤ 0.75 ρb
For ρ= 0.75 ρb from a table a/d= 0.444, the value obtained above is much less than this
hence the slab thickness is safe against bending.
Maximum spacing of shrinkage reinforcement in slabs
18in
Smax=  
i.e Smax=18 in
 2h
31
32
33
34
3. Beam and Column Design
3.0 General Information
The beam and column design is done as per the requirements of the ACI code and
particular procedures has been adopted from the book “Reinforced Concrete: Mechanics
and Design” by James G. Macgregor, second edition.
The beams designed include the edge beams in all floor and the interior beams in the first
floor. In designing the columns, however, the regularity of the building both in plan and
elevation has been considered in selecting representative columns. The panel dimensions
are all the same between columns. This condition of the building integrity has made it
easy in selecting representative columns, in which only these representative columns are
designed and duplicated every where in the building where the category falls into.
Accordingly, each column of the building will fall in any one of the following three
categories. Moreover, as the frames are braced by shear walls as indicated in plan, the
frame is considered as braced against side sway.
Category one: A corner column
This column is found in corner of each of the building units separated by expansion joint.
This column is subjected to a biaxial bending which is attributed to the bending actions
coming from the frames in each direction but balanced in opposite directions.
Category two: An edge column
This column is found on the periphery of the building units and is subjected to a small
amount of bending in one direction and a substantial amount of bending in the other
direction (that accounts to the bending due to an interior frame that joins the edge column
and which not balanced on the other side)
Category three: An interior column
Pertaining to the uniformity in panel dimensions of the panels surrounding an interior
column, the unbalanced moments that are transferred to an interior column is
35
insignificant. Thus, this column will be designed for an axial loading and minimum
moments that will take care of accidental eccentricities.
3.1 Edge Beam Design
The Edge beams designed include those beams on axes namely: Axis (A1- A2-A3-A4A5), Axis (5E-5D-5C-5B-5A), Axis (1A-1B-1C-1D, Axis (B6-B7-B8-B9-B10), and Axis
(6B-6C). The choice of axes for design covers all the edges on all the three floors and the
design have been made separately, for each floor as the loading is different.
Flat plate Edge beam Design
The flat plate edge beams are subjected to the following loads

Share of the Bending moments in the edge column strips of the slab analysis by
direct design method

Weight of the stem of the beam

Weight of the 1ft high parapet wall

Loads due to the glass wall claddings
For easiness of analysis, the bending moments from the column strips has been
alternatively considered by just analyzing the beam by subjecting to a load share of the
slab weight and floor loading of each panel adjacent to the beam. The load share to each
beam from a particular panel is as shown in figure 3.1 below. This approach of loading
the beam results in a slightly magnified moment as compared to the moments computed
using the direct design method of the ACI code.
36
Where
w= 12.5 ft * Pu ( panel )
w = 12.5 ft * 226.25 lb / ft 2 *
1k
 2.82k / ft
lb
Figure 3.1 Panel load share that goes to an Edge Beam
The thickness of parapet wall (assuming made of concrete) has been decided based on a
guideline obtained from (www.carryduff.designes.co.uk/ technical/limiting-height-ofchimney-and-parapet-walls.html). As per this guideline;
For 1ft high parapet wall the thickness of wall t= 6 in

Factored Dead load of parapet wall
= 1.2*12in *6in *

150lb 1 ft 2
1k
k
*
*
 0.09
3
2
ft
144in 1000lb
ft
Factored Dead weight of the beam stem/web (on the assumption that the total
depth of the beam is 16 in and its width is 12 in). For instance, the thickness of the
flat plate is 9 in, which has been already included in the floor loading, the
corresponding beam stem load will be
150lb 1 ft 2
1k
k
= 1.2*12in *7in * 3 *
*
 0.11
2
ft
144in 1000lb
ft
Similarly for the edge beams in other floor the factored beam stem loads will be
o on the edge beams on the floor with flat slab ( thickness of slab = 8 in)
= 0.11
k 8
k
*  0.125
ft 7
ft
o on the edge beams of the floor slab with beams ( thickness of slab = 7 in)
= 0.11
k 9
k
*  0.141
ft 7
ft
37
The Glass curtain wall is assumed to be made from a 10mm (3/8 in) thick insulating glass
unit paneled into vertical mullions which are 5ft apart on center to center and a horizontal
mullion fitted into the vertical mullions to divide the storey glazing into Vision area and
the Spandrel area. The glass curtain wall is hanging from the floor edges beams with
vertical mullions attached to the edge beams. Therefore, the weight of the glass curtain
wall which consists of: the glass itself, vertical and horizontal mullion, capping,
connection to fix it in to the floor edge beams will be transferred into the edge beams as
concentrated loads. However the following approximation has been made to simplify the
load calculations attributed to the difficulty we face in understanding the details of the
connections and finding out the detailed sizes of the vertical and horizontal mullions plus
the capping. Approximations made
o The weights of the connections, horizontal and vertical mullions and
capping are assumed to amount as 50% of the glass weight
o The load transferred to the edge beams is taken as distributed allover the
length rather than concentrated at the points of connections.
The glass loading and its accessories of a given panel along the story height is carried
towards to the top and bottom floor edge beams equally. A diagrammatic representation
of the load transfer is as indicated below.
Figure 3.2 Glazing load dispersion diagram
Load due to glass curtain wall

Weight of a 10mm (3/8 in) glass is 5psf (glass information bulletin)

Weight of connections, mullions, capping = 0.5*5= 2.5 psf

Total weight of glass curtain wall = 7.5 psf
38
Accordingly, the loads due to curtain wall that will go to the edges beams as uniformly
distributed loads will be as shown below, which is also clearly indicated in Table 3.1 that
shows the summary of loading on edge beams
Floor level
Flat plate
Flat slab
Floor with beams
Grade beams
Factored load to beams due to glazing
1.2( 7.5*16/2) =72lb/ft = 0.072k/ft
1.2(7.5*(16/2+16/2)= 144 lb/ft = 0.144 k/ft
1.2(7.5*(16/2+26/2)=189 lb/ft =0.189 k/ft
1.2*(7.5*26/2)=117 lb/ft =0.117 k/ft
Table 3.1 Summary of loading on Edge Beams
Floor level
Due to
parapet wall
Udl- k/ft
Flat plate
0.09
Flat slab
0
Floor with beams 0
Grade beams
0
Factored Design loads
Due to self weight
of beam stem/web
Udl- k/ft
0.11
0.125
0.141
0.251
Due to glass
curtain walls
Udl – k/ft
0.072
0.144
0.189
0.117
Weight from slabs
( triangular)
w (k/ft)
2.82
2.79
2.61
0
In soliciting out the design actions (bending moment and shear forces) for the edge beams
a frame analysis using SAP 2000 is done to analyze the frame consisting of edge beams
from top to bottom. Frame on axis (A1- A2-A3-A4-A5), one of the frame types analyzed,
with full loading is as shown in figure 3.3 below.
In addition the edge beams will be designed for torsion moments which are the negative
exterior column strip moments of the column strips spanning into the edge beams from
the slab analysis using the directed design method.
The output of the frames analyzed for the beams has been hand picked. The outputs taken
are the negative moments at the supports and the span moments plus the shear forces at
each end of the beam. A sample of the outputs selected for the frame on axis (A1-A2-A3A4- A5) is shown in Table 3.2.
39
Figure 3.3 Loaded Frame on Axis ((A1-A2-A3-A4- A5)
Table 3.2 Design Moments from SAP analysis output for frame on Axis ((A1-A2-A3-A4- A5)
Moments (kips-ft)
Beams
@level
Flat plate
Flat Slab
Slab
w/beams
Ground
A1
support
66.75
90.53
74.42
A1-A2
span
76.01
65.43
68.6
A2
support
118.74
110.00
111.71
A2-A3
span
59.64
60.2
57.58
A3
support
103.49
105.26
100.73
A3-A4
span
59.64
60.2
57.88
A4
support
118.74
110
111.71
A4-A5
span
76.01
65.43
68.6
A5
support
66.75
90.53
74.42
18.21
9.77
19.5
9.45
19.29
9.45
19.5
9.77
18.21
40
Reinforcement calculations
Checking assumed depth of beam for moment and shear capacity.
Maximum moment every where in the beams is Mu = 118.79 kips-ft i.e. at the Flat plate
edges beam at support A2
Assuming no compression reinforcement, the minimum depth required is given by
Mu 
 knbd 2
12000
d 
1200M u
; b(in) M u (k  ft )
 kn b
Assume the values of the steel ratio as follow
  0.5b ; b  0.0377 for f c'  6000 psi and f y  60000 psi
   0.019
for   0.019  kn  911 j  0.888
 d  12.76in
The required over all depth of beam D for the width b= 12 in is then
D= 12.76 in +1.5in (cover) +.375in ( dtirrup)+.5 ( half bar diameter# 8) = 15.135 in
 D= 16 in used is ok!
Checking shear capacity
Maximum shear in the beams is Vmax = 22. 18 kips. This shall not be greater than the
maximum allowed shear Vu , which is calculated based on the concrete strength , steel
strength and the depth of the cross-section
Vu= (Vc+Vs) = 2 fc/ bwd  8 f c/ bwd  106.65 kips is greater than the
maximum applied shear so depth is satisfactory and can proceed with design
of
reinforcements.
Longitudinal Reinforcement Calculations
a) Support/ Negative moments (compression on web) – section is rectangular
As 
M u (k  ft )*12000
;
 f y jd
d  16  2.5  13.5in Mu,max = 118.79 k-ft
b= 12 in
and assume j = 0.875
 As= 2.24 in2
41
Corresponding a = 2.2 in
Since compression is on the web the beam behaves as rectangular with b= bw= 12 in and
its ductile failure can easily checked that ≤b by checking if a/d ≤ 0.75(ab/d)
a/d = 0.163 ≤ 0.333 ok ( failure is ductile failure)
Using the above value of a re-compute As, this will result in a slightly more
reinforcement than required. Thus,
As 
M u (k  ft )*12000
 0.018M u
 f y (d  a / 2)
This formula is used to calculate the reinforcements for all support moments.
b) Span/ Positive moments
Mu,max = 76.01 k-ft
Assume that compression zone rectangular and like with in the depth of the slab for the
section of the beam to be considered (that accounts part of the slab as an effective flange)
shown below.
Assuming j = 0.95
M (k  ft ) *12000
As  u
 1.31in 2
 f y jd
Then
a
As f y
0.85* f c'b
 0.815 in
a< h , therefore the assumption is correct
compression zone lies on the flange
Since compression zone lies all in all on the flange and section of beam beam is
rectangular with b= 19 in it can easily checked that ≤b by checking if a/d ≤
0.75(ab/d)
a/d = 0.06 ≤ 0.333 ok ( failure is ductile failure)
Therefore, for the positive moment regions, As can be calculated from
42
As 
M u (k  ft )*12000
 0.017 M u
 f y (d  a / 2)
This formula is used to calculate the reinforcement in span of the edge beams. All the rest
of the calculations are done in an excel sheet presented below
Flat Plate Edge beams
Longitudinal Reinforcement
bw(in)= 12
Beam (bw=12 in; d=14.5in)
Support Moment
A1
A2
-66.18
59.8
2.1528
1.2939
0.54
0.54
1.1912 1.2939
-66.18
76.56
1.854
2.1528
1.01062
0
1.01062
0.54
0.54
0.54
2.1528 1.01062
1.854
1.01062
0.54
A5
-119.6
59.8
1.1912
Min. reinf
A4
-103
76.56
Req'd reinf.(in2), span
Reinf Provide
A3
-119.6
Span Moment
Req'd reinf.(in2), supp
d(in)= 13.5
1.19124
1.293864
0.54
0.54
0.54
2.1528 1.293864
1.19124
Bar # used
7
7
8
7
8
7
8
7
7
area of bar
0.6
0.6
0.79
0.6
0.79
0.6
0.79
0.6
0.6
1.9854 2.1564 2.7250633 1.68437 2.346835 1.684367 2.725063
2.15644
1.9854
#bars req'd
bars used
2#7
2#7+1#6 3#8
bw(in)= 12
Beam (bw=12 in; d=14.5in)
Support Moment
1B
-66.02
Reinf Provide
-119.8
1.1884
Req'd reinf.(in2), span
2#8 +1#7
2#7+1#6
-102.8
2.1564
0.54
1.1884 1.2934
1D
-113.06
60.2
1.2934
0.54
2#7
1C
76.53
Min. reinf
2#8 + 1 #6
d(in)= 13.5
1A
Span Moment
Req'd reinf.(in2), supp
2#7
61.22
1.8504
1.01738
0.54
0.54
2.1564 1.01738
2.03508
0 1.034618
0.54
0.54
0.54
1.8504 1.034618
2.03508
Bar # used
7
7
8
7
8
7
8
area of bar
0.6
0.6
0.79
0.6
0.79
0.6
0.79
#bars req'd
bars used
1.9806 2.1556 2.7296203 1.69563 2.342278 1.724363 2.576051
2#7
2#7+1#6 3#8
2#7
2#8 + 1 #6
2#7
2#8 +1#7
43
2#7
bw(in)= 12
Beam (bw=12 in; d=14.5in)
Support Moment
6B
6C
58.44
125.83
Span Moment
Req'd reinf.(in2), supp
72
1.0636
Req'd reinf.(in2), span
2.290106
1.2456
Min. reinf
0.54
Reinf Provide
d(in)= 13.5
0.54
0.54
1.0636 1.2456
2.290106
Bar # used
7
7
8
area of bar
0.6
0.6
0.79
#bars req'd
1.7727
2.076 2.8988684
# bars used
2#7
2#7 +1#6 3#8
Beam (bw=12 in; d=14.5in)
5E
5D
Support Moment
-64.08
-115.04
Span Moment
Req'd reinf.(in2), supp
1.1534
2.07072
0
1.1534 1.2498
57.52
1.79496
0.98071
0
0
-66.42
76.56
2.13228
0 0.972088
0
2.07072 0.98071
5A
-118.46
58.03
1.2498
Min. reinf
5B
-99.72
73.95
Req'd reinf.(in2), span
Reinf Provide
5C
0
1.293864
0
1.79496 0.972088
1.19556
0
0
0
2.13228 1.293864
1.19556
Bar # used
7
7
8
7
8
7
8
7
7
area of bar
0.6
0.6
0.79
0.6
0.79
0.6
0.79
0.6
0.6
1.9224 2.0829 2.6211646 1.63451 2.272101 1.620147 2.699089
2.15644
1.9926
#bars req'd
bars used
2#7
2#7+1#6 2#8 +1#7
2#7
2#8 + 1 #6
2#7
2#8 +1#7
2#7+1#6
2#7
.
Flat Slab Edge beams
Longitudinal Reinforcement
bw(in)= 12
Beam (bw=12 in; d=14.5in)
Support Moment
A1
Req'd reinf.(in2), supp
Req'd reinf.(in2), span
A2
-83.53
Span Moment
d(in)= 13.5
A3
-99.74
59.39
1.5035
-96
55.2
1.79532
1.0037
A5
-99.74
-83.53
55.2
1.728
0.93288
A4
0
59.39
1.79532
0.93288
1.50354
1.003691
44
Min. reinf
Reinf Provide
0.54
0.54
0.54
1.5035 1.0037
0.54
0.54
0.54
1.79532 0.93288
1.728
0.93288
0.54
0.54
0.54
1.79532 1.003691
1.50354
Bar # used
7
7
8
7
8
7
8
7
7
area of bar
0.6
0.6
0.79
0.6
0.79
0.6
0.79
0.6
0.6
2.5059 1.6728
2.272557
1.5548 2.272557 1.672818
2.5059
#bars req'd
bars used
2#7+1#6 2#7
2#8 +1#6
bw(in)= 12
Beam (bw=12 in; d=14.5in)
Support Moment
-83.41
-99.9
59.37
1.7982
1.0034
0.54
2#7
2#8 +1#6
2 #7
1C
1D
-94.84
-101.2
55.45
1.5014
Min. reinf
2#8+1#6
1B
Req'd reinf.(in2), span
Reinf Provide
2#7
1.70712
1.8216
0.93711
0
0.93119
0.54
0.54
0.54
0.54
1.7982 0.93711
1.70712
0.93119
1.8216
0.54
0.54
1.5014 1.0034
55.1
Bar # used
7
7
8
7
8
7
8
area of bar
0.6
0.6
0.79
0.6
0.79
0.6
0.79
#bars req'd
bars used
2.5023 1.6723 2.2762025 1.56184 2.160911 1.551983 2.305823
2#7+1#6 2#7
2#8 +1#6
bw(in)= 12
Beam (bw=12 in; d=14.5in)
Support Moment
Reinf Provide
77.74
2#8 +1#6
59.82
1.4149
1.921192
1.0349
0.54
0.54
0.54
1.4149 1.0349
1.921192
7
7
8
area of bar
0.6
0.6
0.79
# bars used
2#7
105.56
Bar # used
#bars req'd
2#8+1#6
6C
Req'd reinf.(in2), span
Min. reinf
2#7
d(in)= 13.5
6B
Span Moment
Req'd reinf.(in2), supp
2#7+1#6
d(in)= 13.5
1A
Span Moment
Req'd reinf.(in2), supp
1.5548 2.187342
2.3581 1.7248 2.4318886
2#7+1#6 2#7
2#8 +1#6
45
Edge beams on floor slab with interior beams
Longitudinal Reinforcement
bw(in)= 12
Beam (bw=12 in; d=14.5in)
Support Moment
A1
A2
-69.34
Span Moment
Req'd reinf.(in2), supp
A3
-104.5
0.54
0.54
1.2481 1.0955
64.82
1.6938
0.91784
0.54
-69.34
54.31
1.881
1.0955
A5
-104.5
54.31
1.2481
Min. reinf
A4
-94.1
64.82
Req'd reinf.(in2), span
Reinf Provide
d(in)= 13.5
1.881
0 0.917839
0.54
0.54
1.881 0.91784
1.24812
1.095458
0.54
0.54
1.6938 0.917839
0.54
0.54
1.881 1.095458
1.24812
Bar # used
7
7
8
7
8
7
8
7
7
area of bar
0.6
0.6
0.79
0.6
0.79
0.6
0.79
0.6
0.6
2.0802 1.8258 2.3810127 1.52973 2.144051 1.529732 2.381013 1.825763
2.0802
#bars req'd
bars used
2#7
2#7
2#8 +1#6
bw(in)= 12
Beam (bw=12 in; d=14.5in)
1B
-69.4
Reinf Provide
1.881
1.0951
0.54
2 #7
-99.8
55.13
1.6812
0.92156
0.54
0.54
1.2492 1.0951
0.54
1.881 0.92156
1.7964
0 0.931697
0.54
0.54
0.54
1.6812 0.931697
1.7964
Bar # used
7
7
8
7
8
7
8
area of bar
0.6
0.6
0.79
0.6
0.79
0.6
0.79
#bars req'd
bars used
2.082 1.8252 2.3810127 1.53593 2.128101 1.552828 2.273924
2#7+1#6 2#7
2#8 +1#6
bw(in)= 12
Beam (bw=12 in; d=14.5in)
Support Moment
6B
2#7
2#8+1#6
2#7
2#8 +1#6
d(in)= 13.5
6C
64.97
Span Moment
Req'd reinf.(in2), supp
2#7+1#6
1D
54.53
1.2492
Min. reinf
2#8 +1#6
-93.4
64.8
Req'd reinf.(in2), span
2#7
1C
-104.5
Span Moment
Req'd reinf.(in2), supp
2#8+1#6
d(in)= 13.5
1A
Support Moment
2#7
111.56
63.92
1.1825
2.030392
46
Req'd reinf.(in2), span
1.1058
Min. reinf
0.54
Reinf Provide
0.54
0.54
1.1825 1.1058
2.030392
Bar # used
7
7
8
area of bar
0.6
0.6
0.79
#bars req'd
1.9708
# bars used
2#7
1.843 2.5701165
2#7
2#8 +1#7
Design for Shear and Torsion
For the edge beams the analysis results showed that the shear forces at critical sections
are almost the same for all the edge beams in a floor with a slight difference of only
2kips. Therefore all edge beams are designed for the highest shear forces value in the
beams in each floor. The load and shear force diagram for the edge beam with the larger
shear, which is found on the flat plate floor, is as depicted in figure 3.4.
Figure 3.4 loading and shear force diagram for beam A1- A2 of the flat plate floor
Design shear at d distance from face of support is
Vd = 23.1 -.5*(.248*20.5/12+.27)*20.5/12
= 22.5 kips
Design for shear
Vn 
Vd


22.5
 26.4kips
0.85
47
Check if stirrups are required as per ACI 11.5.5
Vc
2 fcbw d
25.1kips
Vn
26.4kips
is greater than
Vc
2
12.54
Therefore stirrups are required
Check selected stirrups for anchorage and maximum spacing
Try No. 3 double leg stirrups with fy = 60000psi in which case Av= 0.22 in2 . Since the
bar used for stirrups is less No. 6 ACI recommends that the stirrups be anchored by a 900
hook around a bar.As per ACI 11.4.5, the maximum spacing shall be the smaller of the
following
Smax


0.5d  6.75in


d / 4  3.375 if Vn
 6 f c' bw d  75.3 kips (not to beused )
 f
 y Av ,min  60000*0.22  14.7 in
50*12
 50bw
 Smax = 6.75 in
Compute the required stirrup spacing to resist the design shear forces
For vertical stirrups
s
Av f y d
Vn  Vc

.22*60000*13.5
 137in
(26.4  25.1)*1000
 use double leg # 3 bars at a spacing s=6.75 in all over the beam. As this is the
maximum shear, the same spacing of stirrup is to be used all over the edge beams.
Torsion Consideration at Edge Beams
The torsional constant of the section is given by
x
x
x
2
y  max{(92 *19 7 2 *12), (162 *12  7 2 *9)}
2
y  max{2127,3198}
2
y 3198in3
48
The torsion moment that acts on the edge beam is equal to the negative moment in the
column strips of the floor system that span into the edge beams. That is the moments at
the exterior supports of the flood system shown in the tables in the slab design section.
For Flat Plate floor edge beams, the maximum exterior negative column strip moment is
114.89 k-ft per 12.5 ft strip
 Td =114.89/12.5 = 9.19 k-ft
Check whether torsion reinforcement is required or not
 fc/  x2 y  .85* 6000 *3198/12000  17.54k  ft
Td = 9.19 is less than to 17.54 k-ft , therefore no torsion reinforcement is required .
Thus provide reinforcement for the edge beams as per the specification stated in the
tables below for main reinforcement and provide #3 bar double-legged stirrup @ 6.75 in
center-to-center
3.2 Interior beams
Interior beams are only present at the second floor of the building. The loads to which
these beams are subjected include
o Self weight of the beam
o Floor loads and floor weight
In the analysis of direct design method, the moments from the slabs carried by an
adjacent beam to a column strip of the slab has been calculated during slab analysis.
Therefore the moment due to floor weight and floor loading are picked from the table of
moments done for slab analysis. However, the beams shear forces and additional moment
due to its unit weight need to be calculated in a different manner. For which two different
frame analyses has been done for each axis of beams, one for additional moment
calculation and the other for shear calculation(assumption all the shear is transferred to
the column through shear force in beams). Figures 3.5 and 3.6 show the frame loadings
for axis (axis 1B-2B-3B-4B-5B) for this purpose.
49
Figure 3.5 Loading diagram (axis 1B-2B-3B-4B-5B) for the purpose of calculating
additional moments due to self weight of beam
Figure 3.6 loading diagram (axis 1B-2B-3B-4B-5B) for the purpose of calculating
shear in internal beams due to loads from slab
However, the shear calculated using the diagram in figure 3.6 need to be adjusted by
multiplying by
1l2
l1
if
1l2
l1
 1.0 as per ACI 13.6. Therefore the total shear force
used in design for the design of the beam will be the shear force obtained from figure 3.5
50
and the adjusted shear from figure 3.6 . For an interior beam with the beam the value of
1l2
l1
is calculated as follows
3
0.7165* 25
E b a
 0.72
 f  cb   f  1l2 l 
1
25
Ecs l  h 
b,ft
h,ft
a
1 0.583333
f
L,slab
1.5
a/h
b/h
1.333333 2.285714 1.714286
α
25
0.716501
The design moments and shear forces for the internal beams on axis 1B-2B-3B-4B-5B
are as indicated in Table 3.3 and Table 3.4 respectively.
Table 3.3 Design Moments for internal beams on axis 1B-2B-3B-4B-5B
Interior
beams
From slab
Due to own
weight
Md, Design
Moment
1B
support
34.75
4.95
1B-2B
span
96.7
4.41
2B
support
118.78
8.21
Moments (kips-ft)
2B-3B 3B
3B-4B
span
support span
59.39 110.3
59.39
3.52
7.14
3.52
4B
support
118.78
8.21
4B-5B
span
96.72
4.41
5B
support
34.75
4.95
39.7
101.1
127
62.91
127
101.1
39.7
117.44
62.91
Table 3.4Design Shear forces for internal beams on axis 1B-2B-3B-4B-5B
Interior beams
From slab ( 1)
Adjusted shear
from slab (1)*0.72
Due to own
weight
Vd , design shear
1B
support
31.2
22.46
2B support
Left
right
37.55
34.82
27
25.1
3B support
left
right
33.93
33.93
24.43
24.43
4B support
left
right
34.87
37.55
25.1
27
5B
support
31.2
22.46
1.62
1.88
1.77
1.73
1.73
1.77
1.88
1.62
24.08
28.88
26.87
26.16
26.16
26.87
28.88
24.08
51
There will no be torsion load in internal beams as the panels on both sides of the beams
are having the same loading and the same dimensions. Therefore the interior beams will
be designed for shear and bending only.
Reinforcement calculations
Checking assumed depth of beam for moment and shear capacity.
Maximum moment every where in the interior beams is Mu = 118.78 kips-ft i.e. at the
support 2B
Assuming no compression reinforcement, the minimum depth required is given by
Mu 
 knbd 2
12000
d 
1200M u
; b(in) M u (k  ft )
 kn b
Assume the values of the steel ratio as follow
  0.5b ; b  0.0377 for f c'  6000 psi and f y  60000 psi
   0.019
for   0.019  kn  911 j  0.888
 d  12.76in
The required over all depth of beam D for the width b= 12 inches is then
D= 12.76 in +1.5in (cover) +.375in ( dtirrup)+.5 ( half bar diameter# 8) = 15.135 in
 D= 16 in used is ok!
Checking shear capacity
Maximum shear in the beams is Vmax = 28. 88 kips. This shall not be greater than the
maximum allowed shear Vu , which is calculated based on the concrete strength , steel
strength and the depth of the cross-section
Vu= (Vc+Vs) = 2 fc/ bwd  8 f c/ bwd  106.65 kips is greater than the
maximum applied shear so depth is satisfactory and can proceed with design
of
reinforcements.
52
Longitudinal Reinforcement Calculations
a) Support/ Negative moments (compression on web) – section is rectangular
As 
M u (k  ft )*12000
;
 f y jd
d  16  2.5  13.5in Mu,max = 118.789 k-ft b= 12 in
and assume j = 0.875
 As= 2.24 in2
Corresponding a = 2.2 in
Since compression is on the web and beam section is rectangular with b= bw= 12 inches
and it can easily checked that ≤b by checking if a/d ≤ 0.75(ab/d)
a/d = 0.163 ≤ 0.333 ok ( failure is ductile failure)
Re-compute As
As 
M u (k  ft )*12000
 0.018M u
 f y (d  a / 2)
This formula is used to calculate the reinforcements for all support moments.
b) Span/ Positive moments
Mu,max = 101.1 k-ft
Assume that compression zone rectangular and lies within the depth of the slab for the
section of the beam to be considered (that accounts part of the slab as an effective flange)
shown below.
Assuming j = 0.95
M (k  ft ) *12000
As  u
 1.743 in 2
 f y jd
Then
a
As f y
 0.686 in
0.85* f c'b
a< h , therefore the assumption is correct
compression zone lies on the flange
Since compression zone lies all in all on the flange and beamsecion is rectangular with
b= 19 inches and it can easily checked that ≤b by checking if a/d ≤ 0.75(ab/d)
a/d = 0.051 ≤ 0.333 ok ( failure is ductile)
53
Therefore, for the positive moment regions, As can be calculated from
As 
M u (k  ft )*12000
 0.0169M u
 f y (d  a / 2)
This formula is used to calculate the reinforcement in the span of the interior beams. All
the rest of the calculations are done in an excel sheet presented below
Interior Beams
Beam (bw=12 in; d=14.5in)
Support Moment
Span Moment
Req'd reinf.(in2), supp
Req'd reinf.(in2), span
Min. reinf
Reinf Provide
Bar # used
area of bar
#bars req'd
bars used
Beam (bw=12 in; d=14.5in)
Support Moment
Span Moment
Req'd reinf.(in2), supp
Req'd reinf.(in2), span
Min. reinf
Reinf Provide
Bar # used
area of bar
#bars req'd
bars used
Beam (bw=12 in; d=14.5in)
Support Moment
Span Moment
Req'd reinf.(in2), supp
Req'd reinf.(in2), span
Min. reinf
Reinf Provide
Bar # used
area of bar
#bars req'd
bars used
bw(in)= 12
d(in)= 13.5
B1
B2
B3
-41.75
-130.5
-120.5
103.03
64.3
0.752
2.349
2.169
1.7412
1.08667
0
0.54
0.54
0.54
0.54
0.54
0.752 1.7412 2.349 1.08667 2.169
7
7
8
7
8
0.6
0.6
0.79
0.6
0.79
1.253 2.902 2.9734 1.81112 2.7456
2#6
3#7
bw(in)= 12
8B
-41.75
3#8
2#7
d(in)= 13.5
8C
-130.5
96.72
0.752
1.6346
1.08667
0.54
0.54
0.54
0.54
0.752 1.6346 2.349 1.08667
7
7
8
7
0.6
0.6
0.79
0.6
1.253 2.7243 2.9734 1.81112
2#6
2#7 +1#6 3#8
2#7
64.3
8D
-120.5
8E
-120.5
0.54
2.174
7
0.6
3.624
3#8
2#7
3#8
2#7
3#7
64.3
3#8
2#6
8F
-131
64.3
2.169
8G
-42
96.75
2.349
0.75
1.09
1.0867
1.6351
0.54
0.54
0.54 0.54
0.54 0.54
1.09
2.169 1.0867 2.349 1.6351 0.75
7
8
7
8
7
7
0.6
0.79
0.6 0.79
0.6 0.6
1.81 2.74557 1.8111 2.973 2.7251 1.25
2#8+1#7 2#7
d(in)= 13.5
D3
D4
-120.8
-120.8
64.12
64.12
2.1744
2.1744
1.0836
1.08363
0
0.54
0.54
0.54
0.54
1.0836 2.1744 1.08363 2.1744
7
8
7
8
0.6
0.79
0.6
0.79
1.806 2.7524 1.80605 2.7524
0.752
1.09
1.7412
0.54
0.54
0.54 0.54
1.09
2.349 1.7412 0.752
7
8
7
7
0.6
0.79
0.6
0.6
1.81 2.97342 2.902 1.253
bw(in)= 12
D2
-120.8
2.174
103.03
3#8
2.169
0
0.54
2.169
8
0.79
2.7456
B5
-41.8
2.349
2#8+1#7 2#7
64.3
2.349
B4
-130.5
3#8
2#7 +1#6 3#8
2#7 +1#6 2#6
D5
-120.8
64.1
2.1744
1.08
0.54
0.54
1.08 2.1744
7
8
0.6
0.79
1.81 2.75241
2#7
3#8
54
Beam (bw=12 in; d=14.5in)
Support Moment
Span Moment
Req'd reinf.(in2), supp
Req'd reinf.(in2), span
Min. reinf
Reinf Provide
Bar # used
area of bar
#bars req'd
bars used
Beam (bw=12 in; d=14.5in)
Support Moment
Span Moment
Req'd reinf.(in2), supp
Req'd reinf.(in2), span
Min. reinf
Reinf Provide
Bar # used
area of bar
#bars req'd
bars used
Beam (bw=12 in; d=14.5in)
Support Moment
Span Moment
Req'd reinf.(in2), supp
Req'd reinf.(in2), span
Min. reinf
Reinf Provide
Bar # used
area of bar
#bars req'd
bars used
bw(in)= 12
C1
-41.75
d(in)= 13.5
C2
-130.5
103.3
0.752
C3
-120.5
64.3
2.349
1.7458
1.08667
0.54
0.54
0.54
0.54
0.752 1.7458 2.349 1.08667
7
7
8
7
0.6
0.6
0.79
0.6
1.253 2.9096 2.9734 1.81112
2#6
3#7
3#8
bw(in)= 12
D7
-135.1
2#7
64.3
2.169
2.169
1.09
1.0867
0.54
0.54
0.54 0.54
1.09
2.169 1.0867 2.169
7
8
7
7
0.6
0.79
0.6
0.6
1.81 2.74557 1.8111 3.615
2#8+1#7
2#7
2#8+1#7
D9
-135.1
55
2.4318
0.9295
0.9295
0.54
0.54
0.54
0.54
2.432 0.9295 2.4318 0.9295
7
7
8
7
0.6
0.6
0.79
0.6
4.053 1.5492 3.0782 1.54917
3#7 +1#8 2#7
C5
-121
64.3
2.169
0
0.54
2.169
8
0.79
2.7456
2#8+1#7 2#7
d(in)= 13.5
D8
-135.1
55
2.432
C4
-120.5
3#7 +1#8 2#7
2.4318
0
0.54
2.4318
8
0.79
3.0782
3#7 +1#8
bw(in)= 12
d(in)= 13.5
7B
7C
-7
-198.8
111.25
0.126
3.5784
1.8801
0.54
0.54
0.54
0.54 1.8801 3.5784
7
7
8
0.6
0.6
0.79
0.9 3.1335 4.5296
2#6
2#8 +1#6 4#8 +1#6
Design for Shear
For the interior beams, the analysis results showed that the shear forces at critical sections
are almost the same for all the interior beams in the floor with a slight difference of only
2kips. Therefore all interior beams in the floor are designed for the highest shear force
55
values of 28.88 kips (shear at 2B to the left) at all interior supports and 24.1 kps (shear at
1B) at exterior support.
Vd = is taken to be the maximum shear at support as it not too large compared to shear at
d-distance as seen in the design shear force calculation for edge beams
= 28.88 kips
Design for shear
Vn 

Vd


28.88
 33.97kips
0.85
Check if stirrups are required as per ACI 11.5.5
Vc  2 fc/ bwd  2* 6000 *12*13.5/1000  25.1kips
Vn =33.97 kps is greater than Vc/2 = 12.55 kips, therefore stirrups are required.
Check selected stirrups for anchorage and maximum spacing
Try No. 3 double leg stirrups with fy = 60000psi in which case Av= 0.22 in2 . Since the
bar used for stirrups is less No. 6 ACI recommends that the stirrups be anchored by a 900
hook around a bar. As per ACI 11.4.5, the maximum spacing shall be the smaller of the
following
Smax


0.5d  6.75in


d / 4  3.375 if Vn
 6 f c' bw d  75.3 kips (not to beused )
 f
 y Av ,min  60000*0.22  14.7 in
50*12
 50bw
 Smax = 6.75 in
Compute the required stirrup spacing to resist the design shear forces
For vertical stirrups
s
Av f y d
Vn  Vc

.22*60000*13.5
 20in
(33.97  25.1)*1000
 Use double leg # 3 bars at a spacing s=6.75 in all over the beam as governed by the
maximum stirrup spaced for a given depth d. As this is the maximum shear, the same
spacing of stirrup is to be used all over the interior beams.
56
3.5 Beam Reinforcement Detailing/ Drawings
57
FLAT PLATE EDGE BEAMS
FLAT SLAB EDGE BEAMS
58
59
60
EDGE BEAMS ON SLAB WITH INTERIOR BEAMS
INTERIOR BEAMS
61
INTERIOR BEAMS
62
63
64
Notes:
1. Cover to reinforcements 1.5 inches
2. Concrete 6000 psi and steel 60000psi are used
3. all top reinforcements labeled 2#4 are hanging bars
65
3.6 Corner Column (A1)
Column dimensions: 14” x 14”
Table 3.5 shows corner column loads from frame analysis (frame
loading)
Table 3.5 Corner column loading from frame analysis
Column
level
Third story
Second
story
First story
Foundation
column
Axial load
P ( kips)
Moment,
ft)
Bottom
66.75
49.75
My (k-
37.82
77.9
Moment,
Mx ( k-ft)
bottom top
66.75
-49.28
49.75
-41.26
115.82
124.92
24.26
4.88
24.26
4.88
-13.33
2.13
-13.33
2.13
top
-49.28
-41.26
Consider the third story column
Moment
M 2b  66.75 k  ft and M1b  49.28 k  ft . These are the same for both directions.
Axial force
Pu = 37.82 kips + self weight of columns
14 in*14in*16ft* 150 lb/ft 3
= 37.82 +
(12 in/ ft) 2 *1000lb / k
 37.82  3.27  41.1 kips
Check for Slenderness
ACI 10.10 considers a braced (non-sway)
Column is short if
klu
M
 34  12( 1b )  40
r
M 2b
16in
 14.6 ft
12in / ft
r  0.3* column dim ension  0.3*14in  4.2in
lu  16  depth of beam  16 ft 
a) Column attachments
66
b) Top and bottom beam sections
Figure 3.7 Corner column attachment and section of beams attached to it
These sections are used in determining the slenderness of the column as per ACI 10.10
 Ic 
Ic 
14*143 in 4 1 ft
14*143 in 4 1 ft
3 


16.67
in

2*
 33.34in3
 
 
l
12*16
ft
12
in
l
12*16
ft
12
in
c top
c bot


 Ib 
163 *12in 4  93 *13in 4  13*9**3.52 in 4 1 ft
 21.06in3
  
25 ft
12in
 lb top
 Ib 
163 *12in 4  83 *13in 4  13* 8**42 in 4 1 ft

 21.05in3
 
l
25
ft
12
in
b bot

t 
 I
 I
c
b
/ lc top
/ lb top

  I c / lc bot  33.74in3  1.602
16.67in3

0.791


b
21.06in3
  Ib / lb  21.05in3
bot
0.7  0.05(t  b )  0.82  1.0
k
 k  0.82
 0.85  0.05min  0.88  1.0
klu 0.82*14.67 *12in
M

 34.3 and 34  12( 1b )  42.86  40
r
4.2in
M 2b

klu
 34.3  40
r
67
Therefore, column is not slender and treated as short column.
Check for minimum moment requirements
For braced columns emin= (0.6+0.03h)= 0.6+0.03*14= 1.02 in
Pu*emin= 41.1*1.02/12= 3.49 k-ft
Actual moments are greater than this minimum moment and hence the column is
designed for actual moments.
Compute magnified moments
Magnified moment Mc, is given by
M c   b M 2b where  b 
Cm
1  ( Pu /  Pc )
  0.75; Cm  0.6  0.4 M 1b M  0.304
2b
Pc 
 EI
2
 klu 
2
;
EI 
0.4 Ec I g
1   dns
(design stage) ;
Ec  57000 f c'  4.41*106 psi I g  144 /12  3201in 4
 dns 
factored axial dead load
 0.7 appx.
factored servive axial load
 2 4.41*106 *3201lb.in 2
1kips
 3303kips
(.82*16*12in) *(1  0.7) 1000lb
Cm
0.304
b 

0.309  1   b  1
1  ( Pu /  Pc ) 1  (41.1/(0.75*3303))
Pc 
2
*
Therefore no magnification of moments is required due to curvature. And the design
moment will be
Mc = M2b = 66.75 k-ft in both directions
The reinforcement required for the factored design actions of Pu=41.1 kips and Mx=My=
66.75 k-ft is 8 # 8 bars. The Design for reinforcement is made using STAAD etc of
STAAD Pro using the option of ACI Code.
A similar procedure is followed for the columns in other stories and the results are as
indicated in table 3.6
68
Table 3.6 Design Loads Summary and Reinforcement provide for a Typical Corner
Column
Column@
level
Column
type
Design actions
P (kips)
Mx(k-ft)
Second st.
short
84.44
first
slender
foundation short
Magnified actions
Reinforcement
My(k-ft)
Mx(k-ft)
My(k-ft)
provided
49.75
49.75
49.75
49.75
4#8 bars
127.67
24.26
24.26
24.26
24.26
4#8 bars
141.39
4.88
4.88
4.88
4.88
4#8 bars
Tie design
As lateral forces taken by the shear wall and hence the shear forces in column are
insignificant. For this reason no shear is considered in the design of the columns and the
resulting tie requirement from the design software is # 3 bars ties spaced 14 inches on
center to center.
3.5 Edge Column (B1)
Column dimensions: 14” x 14”
The loading to each column of the edge column includes the loadings from the edge
frame that runs from the top of the building down to the foundation, analyzed using SAP
2000 and from the adjacent slabs through shear and moment transfer. Consider the edge
located at B1 of the top storey, the tributary area for shear and strip for moment transfer
as indicated in figures 3.8 (a) and 3.8(b)
The column loadings in the y- direction can be taken from the frame analysis done earlier
for the purpose of the design of edge beams. However, the column loading (moment and
shear) in the x- directions has to be obtained from the above two diagrams as per the ACI
code of shear and moment transfer to columns from slabs. The moment in the x-direction
will be the moment at the exterior support of column strip of the strip shown in figure
3.7(b). And the shear will be approximately the total flood load in the tributary area
shown in figure 3.7(a).
As this also includes the floor load that has been already
considered in the frame loads of the edge frame only half of the load over the area shall
be considered and added to the axial load obtained from the frame analysis.
69
a) Tributary area for shear transfer
b) strip for moment transfer
Figure 3.8 Load transfer to an edge column
The column loadings in the y- direction can be taken from the frame analysis done earlier
for the purpose of the design of edge beams. However, the column loading (moment and
shear) in the x- directions has to be obtained from the above two diagrams as per the ACI
code of shear and moment transfer to columns from slabs. The moment in the x-direction
will be the moment at the exterior support of column strip of the strip shown in figure
3.8(b). And the shear will be approximately the total flood load in the tributary area
shown in figure 3.8(a).
As this also includes the floor load that has been already
considered in the frame loads of the edge frame only half of the load over the area shall
be considered and added to the axial load obtained from the frame analysis.
Load from frame loading ( Y direction)
P = 44.42 kips M 1by= -3.74kips
M2by = 7.41 kips
Load transfer from slab
P = 0.5* 226.25 psf * 12.5ft*25ft * 1k/1000lb= 35.35 kips
M1bx = 114.89 k-ft (taken from flat plate)
M2bx = -0.5 *113.75 =56.87 k-ft (this is the moment transferred to the bottom of the third
story column B1 from the flat slab and it is shared between this column and the column
70
below it. The share of the moments is based on their corresponding stiffness which in this
case is the same and each will take half of the moment.
Total column loading
P= 44.42+33.35+ self weight of column= 77.77+3.27=81.04 kips
M 1by= -3.74kips
M2by = 7.41 kips M1bx = 114.89 k-ft M2bx = -56.87 k-ft
A similar procedure is followed for the columns in the other stories and the edge column
(B1) loadings are as indicated in table 3.7.
Table 3.7 acting design loads before magnification for edge column
Column
level
Axial load
P ( kips)
81.04
163.65
Moment,
My ( k-ft)
bottom top
-3.74
7.41
-1.96
4.12
Moment, Mx (kft) (from slab)
Bottom
top
-56.87
114.89
-28.48
56.87
third story,
Second
story
First story
Foundation
column
244.14
255
-1.02
-0.15
-28.48
0
2.35
0.48
28.48
0
For details of the calculation on magnification and slenderness of the columns, consider
the third story column B1. Figure 3.9 shows its attachment / framing to the building
elements.
Figure 3.9 Edge Column attachments to slabs/ beams in both directions
71
Check for Slenderness
ACI 10.10 considers a braced (non-sway)
Column is short if
16in
 14.6 ft
12in / ft
9in
klu
M 1b
l

16

depth
of
slab

16
ft

 15.25 ft
 34  12(
)  40 ux
12in / ft
r
M 2b
r  0.3* column dim ension  0.3*14in  4.2in
luy  16  depth of beam  16 ft 
Cross sections of beams are the same sections used in the checking of slenderness for the
corner column of the third story. The slab x-section to be used for the x-direction are
shown below
Y-direction
 Ic 
3
    16.67in
 lc top
 Ic 
3
    2*16.67  33.34in
 lc bot
 Ib 
3
3
    2*21.06in  42.12in
 lb top
 Ib 
3
3
    2*21.05in  42.1in
 lb bot
t 
 I
 I
c
b
/ lc top
/ lb top

16.67in3
 0.396
42.12in3
X-direction
 Ic 
3
    16.67in
l
c top

 Ic 
3
3
    2*16.67in  33.34in
 lc bot
 Ib 
93 in3 *150in /12 1 ft

 30.375in 4
 
25 ft
12in
 lb top
 Ib 
83 in3 *150in 4 /12 1 ft

 21.33in3
 
l
25
ft
12
in
b bot

t 
 I
 I
c
b
/ lc top
/ lb top

16.67in3
 0.55
30.375in3
72
Y-direction
b 
 I
 I
c
b
/ lc bot
/ lb bot

X-direction
33.74in3
 0.801
42.10in3
0.7  0.05(t  b )  0.76  1.0
k
 0.85  0.05 min  0.87  1.0
 k  0.76
kluy
0.76*14.67 *12in
 31.8
r
4.2in
M
and 34  12( 1b )  40.06  40
M 2b


kluy
r
b 
c
b
/ lc bot
/ lb bot

33.74in3
 1.58
21.33in3
0.7  0.05(t  b )  0.81  1.0
k
 0.85  0.05min  0.88  1.0
 k  0.81
klux 0.81*15.25*12in

 35.29
r
4.2in
M
and 34  12( 1b )  39.94  40
M 2b

 31.3  40
 I
 I
klux
 35.29  39.94
r
Therefore, column is not slender in both directions.
Check for minimum moment requirements
For braced columns emin= (0.6+0.03h)= 0.6+0.03*14= 1.02 in
Pu*emin= 81.04*1.02/12= 7.14 k-ft
Actual max end moments are greater than this minimum moment and hence the column is
designed for actual moments.
Compute magnified moments
Magnified moment Mc, is given by M c   b M 2b where  b 
Y-direction
X-direction
Cmy  0.6  0.4
 by 

M 1by
M 2by
 0.4
Cmy
0.4
  by  1
(0.75*3303)
Cmy  0.6  0.4
 bx 
1  ( Pu /  Pc )
1  81.04
Cm
1  ( Pu /  Pc )
 0.41
M 1by
M 2by
 0.402
Cmy
1  ( Pu /  Pc )
0.402
 0.41
1  81.04
(0.75*3303)
  bx  1

73
Therefore no magnification of moments is required due to curvature. And the design
moments will be
Mcy = M2by = 7.43 k-ft and Mcx=M2bx= 114.89 k-ft
The reinforcement required for the factored design actions of Pu=81.04 kips and
Mx=7.43 k- ft, My= 114.89 k-ft is 6 # 8 bars. The Design for reinforcement is made
using STAAD etc of STAAD Pro using the option of ACI Code.
A similar procedure is followed for the columns in other stories and the results are as
summarized in table 3.8
Table 3.8 Design Loads Summary and Reinforcement provided for a Typical Edge
Column
Column@
level
Column
type
Analysis actions
P (kips)
Mx(k-ft)
Second st.
short
163.65
first
slender
foundation short
Magnified actions
Reinforcement
My(k-ft)
Mx(k-ft)
My(k-ft)
provided
4.12
56.87
13.9
56.87
4#8 bars
244.14
2.35
28.48
28.66
28.66
4#8 bars
255
.48
0
21.65
21.65
4#8 bars
Tie design
As lateral forces taken by the shear wall and hence the shear forces in column are
insignificant. For this reason no shear is considered in the design of the columns and the
resulting tie requirement from the design software is # 3 bars ties spaced 14 inches on
center to center.
3.6 Interior Column
Column dimensions: 14” x 14”
The regularity of the building units and the equity of the panel dimension both in the xand y- directions will result in an almost zero unbalanced moment to be transferred to the
columns. Therefore, the interior columns will be designed for an axial force (which is
determined based on the tributary area of the floor that is carried by a specific internal
column plus the self weight of the column and internal beams, if present) plus the
74
minimum moment that account for minimum accidental eccentricity. The summary of
the axial loads that act on each column in respective story is as shown in table 3.9
Table 3.9 Acting Column Axial loads for an interior column
Column Load from
level
top story
kips
Third
00
story
second 144.67
First
287.93
ground 425.2
Load from floor- kips
226.25*25*25/1000=141.4
224*25*25/1000=140
209*25*25/1000=130.63
0
Load due
to beam if
any (kips)
0
Self weight
of column
-kips
3.27
Total axial
force
kips
144.67
0
1.33
1.33
3.27
5.31
1.02
287.93
425.2
427.55
Consider the column at the bottom story. Its attachment to the building elements is as
indicated in figure 3.9 in both directions.
Figure 3.10 Interior Column attachments to slabs/ beams
Check for Slenderness
ACI 10.10 considers a braced (non-sway) Column is short if
klu
M
 34  12( 1b )  40
r
M 2b
16in
=24.67ft
12in/ft
r=0.3*column dimension=0.3*14in=4.2in
lu  26-depth of beam=26ft-
75
 Ic 
Ic 
14*143 in 4 1 ft
14*143 in 4 1 ft 14*143 in 4 1 ft
3 

16.67


26.93
in


 63.61in3
 
 
l
12*
26
ft
12
in
l
12*
26
ft
12
in
12*5
ft
12
in
c top
c bot


 Ib 
163 *12in 4 /12  73 *18in 4 /12  18*7 **4.52 in 4 1 ft
 47.75in3
    2*
25 ft
12in
 lb top
 Ib 
163 *12in 4 /12 1 ft

2*
 27.3in3
 
l
25
ft
12
in
b bot

t 
 I
 I
c
b
/ lc top
/ lb top

  I c / lc bot  63.61in3  2.33
26.93in3

0.564


b
47.75in3
  Ib / lb  27.3in3
bot
0.7  0.05(t  b )  0.845  1.0
k
 k  0.845
 0.85  0.05min  0.88  1.0
klu 0.845* 24.67 *12in
M

 59.56 and 34  12( 1b )  22  40
r
4.2in
M 2b
klu
 59.26  22
r
Therefore, column is slender in both directions.

Minimum moment requirements
For braced columns emin= (0.6+0.03h)= 0.6+0.03*14= 1.02 in
Pu*emin= 425.2*1.02/12= 36.14 k-ft
These are the moments to be considered for design as governed by minimum eccentricity.
Compute magnified moments
Magnified moment Mc, is given by M c   b M 2b where  b 
Cm
1  ( Pu /  Pc )
and since
klu
 59.26 is less 100 the limit set by ACI 10.11 then the un-cracked gross section of
r
the column can be used to calculate EI for the computation of Pc.
76
M c   b M 2b where  b 
Cm
1  ( Pu /  Pc )
  0.75; Cm  0.6  0.4 M 1b M  1
2b
Pc 
 EI
2
 klu 
2
;
EI 
0.4 Ec I g
1   dns
(design stage) ;
Ec  57000 f c'  4.41*106 psi I g  144 /12  3201in 4
 dns 
factored axial dead load
 0.7 appx.
factored servive axial load
 2 4.41*106 *3201lb.in 2
1kips
 1179kips
(.845* 26*12in) *(1  0.7) 1000lb
Cm
1
b 

 2.08  1   b  2.08
1  ( Pu /  Pc ) 1  (425.2 /(0.75*1179))
Pc 
2
*
Therefore magnification of moments is required due to curvature. And the design
moments will be
Mc = M2by = M2bx=2.08* 36.14 = 75.17 k-ft
The reinforcement required for the factored design actions of Pu=425.2k and Mx=My=
75.17 k- ft is 8 #8bars. The Design for reinforcement is made using STAAD etc of
STAAD Pro using the option of ACI Code. A similar procedure is followed for the
columns in other stories and the results are as tabulated in table 3.10.
Table 3.10 Design Loads Summary and Reinforcement provided for a Typical
Interior Column
Column@
level
Column
type
Design actions
P (kips)
Mx(k-ft)
Third stor.
short
144.67
Second st.
short
first
slender
foundation short
Magnified actions
Reinforcement
My(k-ft)
Mx(k-ft)
My(k-ft)
provided
0
0
13.1
13.1
4#8 bars
287.93
0
0
27.68
27.68
4#8 bars
425.2
0
0
75.17
75.17
8#8
427.5
0
0
44
44
8#8 bars
77
3.8 Column Reinforcement Detailing /Drawings
Note:
 Cover to reinforcement 1.5 inches
 Depth of foundation column considered is 5 feet
 Concrete 6000psi and steel 60000psi are used
 Reinforcements of the foundation column run into the footing as dowels for
anchorage purpose
78
4 Staircase Design
4.1 General Information
Design of Staircase here it is adopted by the Cantilever Stairs
Cantilever Stairs most commonly on fire escape stairs where vertical wall offers
Protection in addition to its structural use. These may be adapted to many forms full
flight, half flight or semi-spiral .Because the span of the cantilever is the same for each
case the waist thickness is no different for full flight than for half flight .
The Main reinforcement is placed in the top of the tread this beign the tension zone and it
is tied back into the far face of the supporting wall for a full tension lap
4.2 Staircase Design
Fig4.1 Building Plan View
79
Flight From First Floor to Second Floor
Functional Planning
Riser =7 inches (Refer Page 212 IBC)
Tread =11 inches (Refer Page 212 IBC)
No. Of Riser =
Floor to Floor Height = 192/7 =28
Riser
Riser = 192/28= 6.86 inches
No. of Riser in each flight = 28/2 =14 nos
No. of Tread in each flight =14-1 =13 nos
Going = 13 x 11 =143 Inches
Passage = ((24 x12) -Going )/2 = 145/2 = 72.5 inches
As from page 213 IBC the width of Landing shall not be less than the width of Stairway
they Serve that is 145 inches> 143 inches So Assumption is correct
Waist Slab = 125mm (4.925 inches)
Density of Concrete (ρ) = 150 Psf
Wall Thickness =10 inches (From page 242 ACI Code)
Steel Grade Fy = 60000 Psi
Concrete Grade Fc= 5000 Psi
Cosφ =6.86/12.96
Φ= 58.04
θ= 180-90-58.04= 31.96
d1/6.86 = Sin 58.04
d1 =5.82’
d2 = 4.925’
D=d1+d2 = 10.75’
80
Fig4.2 Staircase Plan View
Consider Span AB
81
Load Calculation
Total Load =Live Load + Dead Load
= (L.L+ Floor to Floor Finish + Self Weight of Waist Slab + Weight of Step)
Live Load (Wl ) = 100 Psf (Refer page 286 IBC)
Dead Load (Wd ) =( Floor finish+ Self weight of Waist Slab+ Wt. Of Step)
Floor Finish= 23 Psf (Refer Page 247 ASCE)
Self Weight of Waist Slab= (150 x 4.925/12 x Cos31.96) =72.56 lb/ft
Weight of Step= (150 x 141 x 6.86 /2 x 144 ) = 503.78 lb/ft
Wd = 23+72.56+503.78 = 600 lb/ft
Factored Load ( Wu ) = 1.4Wd+1.7Wl = 1.4 x 600 + 1.7 x 100
= 1010 lb/ft
For Cantilever = 0.8 Wu= 808 lb/ft
Ma = Wl2 /2 = (808 x 11.752/ 2) = 55.78 Kips-Ft
Tension is on the top of the beam
Start a= 2 inch in then
d=D-a = 10.75 -2 =8.75”
Steel Area = Ast =Mu/ φ Fy (d-0.5a)
= 55.78 x 12 / 0.9 x 60 x 7.75
= 1.6 inch2
a= Ast Fy/0.85 Fc’b = (1.6 x 60) /(0.85 x 5 x 12) = 1.9
Improved Steel Area = Ast= ( 56.768 x 12) / (0.9 x 60 x 7.8)
= 1.617 inch2
Steel Reinforcing Ratio ρ =Ast/bd = (1.62) / (12 x 8.75) = 1/65
Minimum Steel Reinforcement ρmin = Max.
=
ρ > ρmin
=
……………………….OK
82
ρmax = 0.6375 x 0.85 x
= 0.6375 x 0.85 x
x(
)
x
)
ρmax =
ρmax > ρ …………………………OK
Detailing of Reinforcement
Ast = 0.6 inch2 # 7 Dia of bar =0.875 inch
To find Spacing
n x 0.6 =1.617
no. of spacing = 2.695
Spacing=
=
= 4.45”
#7 @4.45”
The Main Reinforcement is placed in the Top of the tread this being the Tension Zone
and it is tied back into the far face of the supporting wall for a full tension lap
Shrinkage & Temperature Reinforcement Cracking in grade 60 Slabs
Area of Shrinkage = 0.0018 x 8.75 x 12 = 0.189 in2/ft
By using ASTM Standards
#3
Ast for one bar= 0.11 inch2
n x 0.11 = 0.189
Where, n= no.of Spacing
n = 1.72
Spacing =
= 7’
Therefore Shrinkage & Temperature bar can be placed at #3 @ 7”
Development Length
Development length required
Ld=
α = Location Factor =1
83
β = Coating Factor =1
λ = light weight agg. Concrete factor = 1
γ = Bar size factor = 1
= 42.64”
Ld =
Ld =
+ la
la = Max.(12
db, d)
db= Bar diameter = 8.75”
d= Effective depth= 0.875”
Mu=56.768 Kips-Ft
la = Max.(12
0.875, 8.75)
= Max.(10.5,8.75) =10.5”
a= 1.9”
Mn=1.62
= 758.16
Ld=
Ld=
+ 10.5 = 113.194”
113.194” > 42.64”
Consider Span CD
84
Load Calculation
Total Load =Live Load + Dead Load
= (L.L+ Floor to Floor Finish + Self Weight of Waist Slab + Weight of Step)
Live Load ( Wl )= 100 Psf (Refer page 286 IBC)
Dead Load ( Wd ) =( Floor finish+ Self weight of Waist Slab+ Wt. Of
Step)
Floor Finish= 23 Psf (Refer Page 247 ASCE)
Self Weight of Waist Slab= (150 x 4.925/12 x Cos31.96) =72.56 lb/ft
Wd = 23+72.56 = 95.56 lb/ft
Factored Load ( Wu ) = 1.4Wd+1.7Wl = 1.4 x 95.56 + 1.7 x 100
= 303.786 lb/ft
For Cantilever = 0.8 Wu= 243 lb/ft
Ma = Wl2 /2 = (243 x 6.052/ 2) = 4.43 Kips-Ft
Tension is on the top of the beam
Start a= 2 inch in then
d=D-a = 10.75 -2 =8.75”
Steel Area = Ast =Mu/ φ Fy (d-0.5a)
= 4.43 x 12 / 0.9 x 60 x 1.178
= 0.84 inch2
a= Ast Fy/0.85 Fc’b = (0.83 x 60) /(0.85 x 5 x 12) = 0.98”
Improved Steel Area = Ast= ( 4.43 x 12) / (0.9 x 60 x 1,68)
= 0.58 inch2
Steel Reinforcing Ratio ρ =Ast/bd = (0.58) / (12 x 2.178) = 1/44.86
Minimum Steel Reinforcement ρmin = Max.
=
ρ > ρmin
=
……………………….OK
ρmax = 0.6375 x 0.85 x
x(
)
85
= 0.6375 x 0.85 x
x
)
ρmax =
ρmax > ρ …………………………OK
Detailing of Reinforcement
Ast = 0.2 inch2 # 4 Dia of bar =0.5 inch
To find Spacing
n x 0.2 =0.52
no. of spacing = 2.6
Spacing=
=
= 4.62”
#7 @4.62”
The Main Reinforcement is placed in the Top this being the Tension Zone and it is tied
back into the far face of the supporting wall for a full tension lap
Shrinkage & Temperature Reinforcement Cracking in grade 60 Slabs
Area of Shrinkage = 0.0018 x 2.178 x 12 = 0.047 in2/ft
By using ASTM Standards
#3
Ast for one bar= 0.11 inch2
n x 0.11 = 0.189
Where, n= no.of Spacing
n = 1.72
Spacing =
= 7’
Therefore Shrinkage & Temperature bar can be placed at #3 @ 7”
Development Length
Development length required
Ld=
α = Location Factor =1
β = Coating Factor =1
86
λ = light weight agg. Concrete factor = 1
γ = Bar size factor = 1
= 42.64”
Ld =
Ld =
+ la
la = Max.(12
db, d)
db= Bar diameter = 0.5”
d= Effective depth= 2,178”
Mn=53.92 Kips-Ft
la = Max.(12
0.5, 2.178)
= Max.(6,2.178) =6”
a= 0.9”
Mn=0.52
= 53.92
Ld=
+ 6 = 58.69”
Ld=
58.96” > 42.64”…………………OK
Flight from Ground Floor to First Floor
Functional Planning
Riser =7 inches (Refer Page 212 IBC)
Tread =11 inches (Refer Page 212 IBC)
No. Of Riser =
Floor to Floor Height = 312/7 =45
Riser
Riser = 312/45= 6.86 inches
Let us Assume 3 Flight
Flight No 1= 17 Riser
No. of Riser in each flight = 28/2 =14 nos
87
No. of Tread in each flight =14-1 =13 nos
Going = 13 x 11 =143 Inches
Design of Staircase remains same for Staircase from Ground Floor to First Floor
Table 4.1 Detailing of Reinforcement
Description
ASTM Standard Size
Location
Main Reinforcement
Along AB Span
#7@4.45”
In Tread Tension Zone
Along CD Span
#4@4.62”
In Mid landing Span
Shrinkage & Temperature
#3@7”
In Tread & Waist Slab in
Reinforcement
1
both direction
Shrinkage Cracking and Temperature Reinforcement is provided to minimize the
cracking and tie the structure together and achieve structural integrity
2
Tension is on top of tread so main reinforcement is provided
3
Development length is provided is equal to length of the bar required to develop the
required stress in steel
88
89
90
91
92
5 Shear Wall Design
5.1 General Information
Horizontal forces acting on reinforced concrete structure are resisted by different system.
Frames are commonly used for low rise construction. Such frames may also rely on the
contribution of infill walls and partitions of lateral load resistance, especially if wind is
the only lateral force consideration. When large load must be resisted, such as
earthquake, or very high wind loads, shear walls are often used. In such buildings, shear
walls may be constructed between column lines or may be incorporated into stair wells,
elevator shafts or utility shafts. We have incorporated shear walls near elevators and
staircase.
The foundation must be designed to resist he shear and moment at the base of the wall.
The reinforcement at the base must be careful detailed so that the forces can be
transferred between the wall and the foundation.
Here we have provided minimum Ast to control cracking and also the Shear reinforcement
in both the longitudinal & transverse direction and generally it is placed in both faces of
the wall. The longitudinal (Vertical) bars in the wall serve also as flexural reinforcement.
5.2 Shear Wall Design for Staircase
Wind Load Calculation
With reference ASCE Wind load calculation Page 27 ASCE we can find chapter on wind
load calculation
Basic Wind Speed V is determined for Syracuse Region from page 37 ASCE were V= 40
m/s
Ps=
=Adjustment Factor for Building Height
I= Importance Factor
Ps30=Simplified design wind Pressure for exposure
With reference fig 6.2 From page 43 ASCE
λ= 1.22
Ps30 here from table we can find the value of
93
A =12.8 Psf
B=-6.7 Psf
C= 8.5 Psf
D=-4 Psf
For Ps30=12.8+8.5 =21.3 Psf
Ps=1.22 x 1 x 21.3 =25.986 Psf
Therefore considering Ps=26 Psf
Fig 5.1 Shear Force & Bending Moment Due to Wind Load
Design of Shear wall follows the same approach as that for beams.
The provision for Shear wall design is contained in ACI 11.10 Special provision for wall.
Design of Horizontal section for Shear in the plane of wall is based on
Vu ≤ øVn …………….From Code ACI equation 11.1
Vn =Vc+Vs………… From Code ACI equation 11.2
94
Check Maximum Shear Strength Permitted
φVn= φ 10√fc’x hw x d
Where d=0.8 x lw =0.8 x 24 x 10 = 192 inch
Where lw is the Length of the Wall
φVn= 0.75 x 10 x √5000 x 12 x 192/1000
φVn=1221.88 Kips
Calculate Shear Strength Provided by Vc
Critical Section for Shear is
Therefore
=
= 12 feet
Vc= 3.3√fc’ hw d + Nu d/4 lw
Where Nu is the factored Axial Tensile Force
Slab Reaction coming on to the wall = 3.0375 Kips
Where Nu=3037.5 + 10
1
Therefore, Nu= 4537.5 lb
Nu=4.537 Kips
Vc=
Vc= 538.36 Kips
Minimum Reinforcement of Wall
Vertical Reinforcement = 0.0012
= 0.144 inch2
As per ASTM Standards Reinforcing Bars
Nominal Diameter =0.375 inch
Nominal Area =0.11 inch2
Ast=0.1728/0.11 =1.57
Spacing =12/1.57 = 7 inch
Therefore provide #3 @ 7 inch
Minimum Horizontal Steel
Ast= 0.0020 x b x d
95
Ast=0.002 x 12 x (12-0.5-0.25)
Ast= 0.27 inch2
Provide # 3 @ 10 inches
Providing Shrinkage Cracking Steel
The minimum length of a lap splice when fy= 60 ksi is
Lap = 0.0005xdbxfy
= 0.0005x0.375x60 = 11.25in < Min lap (12in)
Provide lap = 12in
Development length in compression (ACI Code, section 12.2)
ld = 0.04Abfy/fc
= 0.04x0.31x60000/(6000) =10in< 12in
Provided ld=12 in (min development length)
As each wall is governed by the Minimum Area of Reinforcement .Therefore for each
wall reinforcement & detailing remains the same.
96
97
98
5.3 Shear Wall Design for Elevator
1. Design of machine room slab
Loads acting on slab are:
 Passenger load
 Lift load
= 2205 lb
= 4500 lb
As we are providing two passenger elevators, the total load per ft² over roof from
lifts is
= 2*(2205+4500)/25² = 13410 psf
Roof live load
= 100 psf
 Finish load
= 15psf
 Self wt. of slab: Assume depth of slab ‘d’=10”
Self wt. of slab = 150x1’x1’x(10/12)’ = 125 psf
 Factored load(Wu) = 1.4D.L+1.7L.L
= 1.4(125+22+15)+1.7(100)= 400 lb = 0.4 k
Analyze the slab as a unit strip (b=12) &designed it as a beam:

Factored Moment
Mu = Wu*l²
8
Mu = 0.4x25² = 28.8k-ft
8

Checking shear capacity
Maximum shear in the beams is Vu= (5x1.4=7kips) This shall not be greater than the
maximum allowed shear, which is calculated based on the concrete strength, steel
strength and the depth of the cross section
Vn= 0.9x10xx12x7.5 = 69.71 kips > 7 kips
1000
Unit strip is safe against shear & we can proceed furthur.

Area of steel
As =
Mu
fy(d-a/2)
=
28.8x12
0.9x60x (7.5-2/2)
= 0.984in2
99
a = As.fy = 0.984x60 = 1.157
0.85fc’.b
0.85x5x12
Improvised Steel area =
Mu
=
28.8x12
fy(d-a/2 0.9x60(7.5-1.157/2)
= 0.924 in²
Provide #7 @7” (As = 1.02 in²) in both directions
Development length in compression (ACI Code, section 12.2)
ld = 0.04Abfy/
= 0.04x0.0.875x60000/(6000) =23.3in
Provided ld =24 in
2. Design of Shear wall (above roof level)
 Axial compression force due to slab
= 3.125k
 Axial compression force due self wt. of wall = 150x10’x1’x1’
= 1500 lb
Maximum shear strength permitted
Vu  Vn
where Vn = Vc+Vs
Vc=3.3sqrt(fc’)h.d+Nu.d
4Lw
Vc = 3.3sqrt(6000)x10x0.8x12 + 4625x0.8x12
1000
4x12x1000
Vc = 24.54+ 3.7
Vc= 0.85x28.24 = 24k
Vn = 10sqrt(6000)x10x(0.8x12) = 74.36k
1000
Vn = 0.85x74.36 = 63.20k
Vu = 0.260x1.4 = 0.365k
Since Vu<<ØVc therefore no reinforcement is required but as per ACI code, their should
be some minimum reinforcement provided to wall.
The minimum shear reinforcement is provided by ACI wall section 14.3.2

Mini. Horizontal steel:
Ast = 0.0020.b.d’= 0.0020x12x[10-0.75(cover)-0.25(half dia of bar)]
Ast = 0.216 in²
100
Provide #4@ 10”
The minimum length of a lap splice when fy= 60 ksi is (ACI code section 12.15&12.16)
Lap = 0.0005xdbxfy
= 0.0005x0.5x60 = 115in
Provide lap = 15in
Mini vertical steel: Ast = 0.0012.b.d’
= 0.0012x12x (10-0.5-0.75)
= 0.1296 in²
Provide #3 @9 (Ast =0.1467 in²)
The minimum length of a lap splice when fy= 60 ksi is
Lap = 0.0005xdbxfy
= 0.0005x0.375x60 = 11.25in < Min lap(12in)
Provide lap = 12in
3. Design of Shear wall (@base level)
 Axial compression force due to slabs =(5 + 3.125=8.125k)
 Axial compression force due self wt. of wall = 150x68.41’x1’x1’
= 102615lb = 10.26 kip
 Reactions from slabs = 8 kip (due to all slabs)
Maximum shear strength permitted
Vu  Vn
where Vn = Vc+Vs
Vc=3.3sqrt(fc’)h.d + Nu.d
4Lw
Where, Nu= 8.125+10.26+8 = 26.385 kips
Lw =12in
d = 0.8Lw=0.8x12 in
h = 10in
Vc = 3.3sqrt (6000) x10x0.8x12 + 102615x0.8x12
1000
Vc = 24.53 +20.523
4x12x1000
Vc = 0.85x45.053 = 36.05k
101
Vn = 10sqrt(6000)x10x(0.8x12) = 74.36 kips
1000
Vn = 0.85x74.36 = 93.206kips
Vu = 1.768x1.4 = 2.475 kips
Since Vu<<ØVc therefore no reinforcement is required but as per ACI code the their
should be some minimum reinforcement provided to wall.
The minimum shear reinforcement is provided by ACI wall section 14.3.2
Mini. Horizontal steel: Ast = 0.0020.b.d’
= 0.0020x12x9 = 0.216 in2
Provide #4@ 10”
The minimum length of a lap splice when fy= 60 ksi is
Lap = 0.0005xdbxfy= 0.0005x0.5x60 = 115in
Provide lap = 15in
Development length in compression (ACI Code, section 12.2)
ld = 0.04Abfy/fc = 0.04x0.20x60000/(6000) =16in
Mini vertical steel:
Ast = 0.0012.b.d’ = 0.0012x12x9.25= 0.1296 in²
Provide #3 @9 (Ast =0.1467 in²)
As the shear wall is all around the elevator only one part of shear wall was analyzed and
designed and has resulted in a minimum steel at both the two extreme levels (i.e. one at
the roof level and one at the foundation level). Therefore, the same minimum steel
reinforcement is provided at all parts of shear wall as per ACI code.
Also, the freight elevator shear wall is similar to the passenger elevator in terms of
loading and shape, therefore we have provided the same reinforcement to it also as per
ACI code.
Specification of passenger and freight elevators installed
Type
Capacity(kg) Door
Car
dimension
Passenger
1000
2S
5.9’x4.92’
elevator
Freight
1000
3S
7.22’x7.4’
elevator
Speed(m/s)
1.5
1.3
102
103
Detaailing of shear wall and slab connection
104
Slab1 – Machine room slab @roof level
Slab2 – Machine room roof slab
105
106
Beams detail incorporated with shear wall footing (page 111,112)
107
Footing joint with shear wall
108
6. Foundations
6.1 General Information
The foundation chosen for the building are of four types
1. Isolated footing under corner column
2. Isolated footing under edge columns
3. Isolated footing under interior columns
4. Common footing for columns which are at the junction of the different building
units i.e. at points where we have the expansion joint.
6.2 Isolated footing under the corner column
Load from superstructure
P = 141.39 kips
Mx=My=4.88 kips
Figure 6.1 Footing load elements
Assuming the following data
Bearing capacity of soil = 6000psf = 6 ksf
First floor loading is 100 psf = 0.1 psf
Density of soil = 120 lb/ft3  weight of soil = 120*5= 600psf =0.6 ksf
Thickness of footing =2* column size= 28in
Net allowable soil pressure, qn
qn  6.0  footpad  soil  floor  floorload
 6.0  [
28
*0.15  5*0.12  0.5*0.15  0.1]
12
 4.875 ksf
Preliminary area of footing required, Aplim
109
Aplim 
141.39kips
 29 ft 2 , then try a square footing of 6ft by 6ft and 28 inch thick.
4.875ksf
Acting net pressure calculation at corners of the footing
qact 
P Mx y M yx
(6 ft ) 4


but I x  I y 
 108 ft 4
A
Ix
Iy
12
Where
P= 141.39 + 1.2*(6ft*6ft)*(0.1+0.6+(28/12+6/12)*.15)=190 kips
Mx=My=4.88 kips and A=36 ft2
At corner A
qact 
190 4.88*3 4.88*3


=5.55 ksf
36
108
108
At corner B
qact 
190 4.88*3 4.88*3


=5.28 ksf
36
108
108
At corner C
qact 
190 4.88*3 4.88*3


=5.01 ksf
36
108
108
At corner D
qact 
190 4.88*3 4.88*3


=5.28 ksf
36
108
108
Figure 6.2 Corner footing Bearing Pressure distribution
Check the thickness for two way shear
Assume # 9 bars are used for footing reinforcement
dav  28" (3"cover)  (bar diameter)=28"-3"-1.128"=23.87 in . The shear perimeter is
shown in dashed lines.
110
Average pressure under the footing, qav
5.55  5.28  5.28  5.01
qav 
 5.28 ksf
4
Acting shear, Vu

Vu  5.28*(36 ft 2  37.8
Figure 6.3 a) Corner two way shear
Critical Section
ft )  137.7 kips
12  
2
Length of critical shear perimeter
bo = 4*37.8” =151.2 in
As per ACI , the resistance shear is given by
 (2  4 ) f ' b d
c o av
c


Vc  min  ( s d av b ) f c' bo d av
o


 4 f c' bo d av

 s = 40 for column at center of footing
 c =long side of column/short side of column =1
1421.59kips

Vc  min  1965.6kips  Vc  947.7kips
 947.7kips

Vc is about seven times of Vu  reduce the thickness assumed take h = 14 in
d av  14" (3"cover)  (bar diameter)=14"-3"-1.128"=9.87 in
bo  4*(14  9.78)  95.12in
14  9.87 2
Vu  5.28*(36  (
) )  169kips
12
379kips

Vc  min  381kips  Vc  248kips  Vu  169kips
248kips

Therefore take thickness D of footing as 14 inches.
111
Check one-way shear
qav 
 5.55  5.28
 5.415ksf
2
Vu  5.415*6*(3  (7  9.87) /12)  51.79kips
Vc   2 f c' bd  .85* 2* 6000 *6*9.87 /12
 93.57kips  Vu
 Footing is ok against one way shear
Figure 6.3 b) Corner footing one way shear critical section
Reinforcement Design
qav 
 5.55  5.28
2
 5.415ksf
acting moment, M u  5.415*[6*
(29 /12) 2
]
2
 94.87 k  ft
Assume that j= 0.9
Mu
94.87*12000
As =
=
=2.376 in 2
 f y jd av .9*60000*(.9*9.87)
Minimum As as per ACI 10.5 =0.0018bh
=0.0018*72*14
 1.81in 2
 Minimum As doesn't govern.
Figure 6.3 c) Corner footing critical section for bending
Maximum spacing as per ACI 7.12 is 18 inches
Try 6 # 6 bars @ 12 inches center to center As = 2.64 in2
112
Re-compute  M n
a
As f y
'
c
0.85* f b
since

2.64*60000
 0.431in
0.85*6000*72
0.75ab
a
0.431

 0.044  0.333 
d av 9.78
d
0.9* 2.64*60000*(9.87  .431/ 2)
 114.7k  ft
12000
Therefore the reinforcement provided is enough and
  M n   As f y (d av  a ) 
2
Mn  Mu
thus, provide 6#6 bars @ 12 in center to center both ways.
6.3 Isolated footing under the Edge column B1
Load from superstructure
P = 255 kips
Mx= 0 , My=21.65 kips
Assuming the following data
Bearing capacity of soil = 6000psf = 6 ksf
First floor loading is 100 psf = 0.1 psf
Density of soil = 120 lb/ft3  weight of soil = 120*5= 600psf =0.6 ksf
Thickness of footing =2* column size= 28in
Net allowable soil pressure, qn
qn  6.0  footpad  soil  floor  floorload
 6.0  [
28
*0.15  5*0.12  0.5*0.15  0.1]
12
 4.875 ksf
Preliminary area of footing required, Aplim
Aplim 
255kips
 52.3 ft 2 , then try a square footing of 8 ft by 8 ft and 28 inch thick.
4.875ksf
Acting net pressure calculation at corners of the footing
113
qact 
P Mx y M yx
(8 ft ) 4


but I x  I y 
 341.33 ft 4
A
Ix
Iy
12
Where
P= 255 + 1.2*(8ft*8ft)*(0.1+0.6+(28/12+6/12)*.15)=341.5kips
Mx= 0 My=21.65 kips and A=64 ft2
At corners A and B
qact 
341.5 21.56* 4

=5.59 ksf
64
341.33
At corner C and D
qact 
341.5 21.56* 4

=5.08 ksf
64
341.33
Check the thickness for two way shear
Figure 6.4 Edge footing Bearing Pressure distribution
Assume # 9 bars are used for footing reinforcement
dav  28" (3"cover)  (bar diameter)=28"-3"-1.128"=23.87 in . The shear perimeter is
shown in dashed lines.
Figure 6.5 a) Edge footing two way shear
Critical Section
Average pressure under the footing, qav
5.59  5.59  5.08  5.08
qav 
 5.34 ksf
4
Acting shear, Vu

Vu  5.34*(64 ft 2  37.8
 s = 40 for column at center of footing
ft )  288.77 kips
12  
2
Length of critical shear perimeter
bo = 4*37.8” =151.2 in
As per ACI, the resistance shear is given by
 (2  4 ) f ' b d
c o av
c


Vc  min  ( s d av b ) f c' bo d av
o


 4 f c' bo d av
114

 c =long side of column/short side of column =1
1421.59kips

Vc  min  1965.6kips  Vc  947.7kips
 947.7kips

Vc is about three times of Vu  reduce the thickness assumed take h = 16 in
d av  16" (3"cover)  (bar diameter)=16"-3"-1.128"=11.87 in
bo  4*(14  11.87)  103.48in
14  11.87 2
Vu  5.34*(64  (
) )  317kips
12
 483.81kips

Vc  min  486.1kips  Vc  322.55kips  Vu  317kips
322.55kips

Therefore take thickness D of footing as 16 inches.
Check one-way shear
qav  5.34ksf
Vu  5.34*8*(4  (7  11.87) /12)  103.7 kips
Vc   2 f c' bd  .85* 2* 6000 *8*11.87 /12
 150.04kips  Vu
 Footing is ok against one way shear
Figure 6.5 b) Edge footing one way shear critical section
115
Reinforcement Design
qav 
 5.55  5.28
2
 5.415ksf
(41/12) 2
]
2
 249.35k  ft
acting moment, M u  5.34*[8*
Assume that j= 0.9
Mu
249.35*12000
As =
=
=5.187 in 2
 f y jd av .9*60000*(.9*11.87)
Minimum As as per ACI 10.5 =0.0018bh
=0.0018*96*16
 2.76in 2
 Minimum As doesn't govern.
Figure 6.5 c) Edge footing critical section for bending
Maximum spacing as per ACI 7.12 is 18 inches
Try 8 # 8 bars @ 12 inches center to center As = 6.32 in2
Re-compute  M n
a
As f y
'
c
0.85* f b
since

6.32*60000
 0.775in
0.85*6000*96
0.75ab
a
0.775

 0.066  0.333 
d av 11.78
d
0.9*6.32*60000*(11.87  .775 / 2)
 326.5k  ft
12000
Therefore the reinforcement provided is enough and
  M n   As f y (d av  a ) 
2
Mn  Mu
thus, provide 8#8 bars @ 13.5 inches center to center both ways.
116
6.4 Isolated footing under an interior column
Load from superstructure
P = 427.5 kips
Mx= 0 , My=0
Assuming the following data
Bearing capacity of soil = 6000psf = 6 ksf
First floor loading is 100 psf = 0.1 psf
Density of soil = 120 lb/ft3  weight of soil = 120*5= 600psf =0.6 ksf
Thickness of footing =2* column size= 28in
Net allowable soil pressure, qn
qn  6.0  footpad  soil  floor  floorload
 6.0  [
28
*0.15  5*0.12  0.5*0.15  0.1]
12
 4.875 ksf
Preliminary area of footing required, Aplim
Aplim 
427.5kips
 87.7 ft 2 , then try a square footing of 10.5 ft by 10.5 ft and 28 inch thick.
4.875ksf
Acting net pressure calculation at corners of the footing
qact 
P Mx y M yx
(8 ft ) 4


but I x  I y 
 341.33 ft 4
A
Ix
Iy
12
Where
P= 427.5 + 1.2*(10.5ft*10.5ft)*(0.1+0.6+(28/12+6/12)*.15)=576.34 kips
Mx= 0 My=0 and A=110.25 ft2
117
At corners A ,B,C and D
qact 
562.5
=5.63 ksf
100
Check the thickness for two way shear
Figure 6.6 Interior footing Bearing Pressure distribution
Assume # 9 bars are used for footing reinforcement
dav  28" (3"cover)  (bar diameter)=28"-3"-1.128"=23.87 in . The shear perimeter is
shown in dashed lines.
Figure 6.7 a) Interior footing two way shear
Critical Section
Average pressure under the footing, qav
qav  5.63 ksf
Acting shear, Vu

Vu  5.63*(100 ft 2  37.8
ft )  506.93 kips
12  
2
Length of critical shear perimeter
bo = 4*37.8” =151.2 in
As per ACI , the resistance shear is given by
 (2  4 ) f ' b d
c o av
c


Vc  min  ( s d av b ) f c' bo d av
o

 s = 40 for column at center of footing

 4 f c' bo d av

 c =long side of column/short side of column =1
1421.59kips

Vc  min  1965.6kips  Vc  947.7kips
 947.7kips

118
Vc is about two times of Vu  reduce the thickness assumed take h = 21 in
d av  21" (3"cover)  (bar diameter)=21"-3"-1.128"=16.87 in
bo  4*(14  15.87)  123.48in
14  16.87 2
Vu  5.63*(100  (
) )  525.74kips
12
822.92kips

Vc  min 749.52kips  Vc  548.61kips  Vu  525.74kips
 548.61kips

Therefore take thickness D of footing as 21 inches.
Check one-way shear
qav  5.63ksf
Vu  5.63*10*(5  (7  16.87) /12)  169.5kips
Vc   2 f c' bd  .85* 2* 6000 *10*16.87 /12
 271.2kips  Vu
 Footing is ok against one way shear
Figure 6.7 b) Interior footing one way shear
critical section
Reinforcement Design
qav  5.63ksf
(53 /12) 2
]
2
 549.12k  ft
acting moment, M u  5.63*[10*
Assume that j= 0.9
Mu
549.12*12000
As =
=
=8.04 in 2
 f y jd av .9*60000*(.9*16.87)
Minimum As as per ACI 10.5 =0.0018bh
=0.0018*120*21
 4.54in 2
 Minimum As doesn't govern.
Figure 6.7 b) Interior footing critical section for bending
119
Maximum spacing as per ACI 7.12 is 18 inches
Try 9 # 9 bars @ 12 inches center to center As = 9 in2
Re-compute  M n
a
As f y
'
c
0.85* f b
since

9*60000
 0.882in
0.85*6000*120
0.75ab
a
0.882

 0.053  0.333 
d av 16.78
d
0.9*9*60000*(16.87  .882 / 2)
 665.37k  ft
12000
Therefore the reinforcement provided is enough and
  M n   As f y (d av  a ) 
2
Mn  Mu
thus, provide 9#9 bars @ 15 inches center to center both ways.
6.5 Common footing under the two Edge columns meeting at the
expansion joint
The two footings for the edge columns that come together at the expansion joint are
considered as a one and simplified to a single column of dimension 14” x 28 “ ( x and y
directions dimensions respectively). This is valid because the expansion joint size
between the columns is almost zero compared to the individual footings. Moreover, as
the two columns come together the unbalanced moment to at the footing will be zero.
Load from superstructure
P =510 kips
Mx= 0 , My=0
Assuming the following data
Bearing capacity of soil = 6000psf = 6 ksf
120
First floor loading is 100 psf = 0.1 psf
Density of soil = 120 lb/ft3  weight of soil = 120*5= 600psf =0.6 ksf
Thickness of footing =2* column size= 28in
Net allowable soil pressure, qn
qn  6.0  footpad  soil  floor  floorload
 6.0  [
28
*0.15  5*0.12  0.5*0.15  0.1]
12
 4.875 ksf
Preliminary area of footing required, Aplim
Aplim 
510kips
 104.6 ft 2 , try a square footing of 10.5 ft by 10.5 ft and 28 inch thick.
4.875ksf
Acting net pressure calculation at corners of the footing
qact 
P
A
Where P= 510 + 1.2*(10.5 ft*10.5ft)*(0.1+0.6+(28/12+6/12)*.15)=586.32 kips and
A=110.25 ft2
qact 
586.32
 5.31ksf
110.25
thickness for two way shear
Figure 6.8 Common footing Bearing Pressure distribution
121
Check the thickness for two way shear
Assume # 9 bars are used for footing reinforcement
dav  28" (3"cover)  (bar diameter)=28"-3"-1.128"=23.87 in . The shear perimeter is
shown in dashed lines.
Figure 6.9 a) Common footing two way shear
Average pressure under the footing, qav
qav  5.31ksf
Acting shear, Vu
Vu  5.31*(110.25 ft 2  37.8 * 51.8
ft 2
12
12
 582.1 kips
Length of critical shear perimeter
bo = 2*37.8”+2*51.8 =179.2 in
As per ACI , the resistance shear is given by
 (2  4 ) f ' b d
c o av
c


Vc  min  ( s d av b ) f c' bo d av
o


 4 f c' bo d av

 s = 40 for column at center of footing

Critical Section
 
 c =long side of column/short side of column =1
2820.74kips

Vc  min  1500.6kips  Vc  1128.3kips
 1128.3kips

Vc is about two times of Vu  reduce the thickness assumed take h = 20 in
d av  20" (3"cover)  (bar diameter)=20"-3"-1.128"=15.87 in
bo  2*(14  15.87  28  15.87)  147.48
Vu  5.31*(110.25  (43.87
 1541kips

Vc  min 663.3kips
616.4kips

12
* 29.87
))  537.1kips
12
 Vc  616.4kips  Vu  537.1kips
Therefore take thickness D of footing as 20 inches.
Check one-way shear
122
The critical section for one way shear lies in the direction where we have the shortest
dimension of the assumed combined column size.( i.e. parallel to the longer dimension)
qav  5.31ksf
Vu  5.31*10.5*(5.25  (7  15.87) /12)  186.45kips
Vc   2 f c' bd  .85* 2* 6000 *10.5*15.87 /12
 1828.56kips  Vu
 Footing is ok against one way shear
Figure 6.9 b) Common footing one way shear Critical Section
Reinforcement Design
For uniformity of the reinforcement in both directions, the most critical section (i.e.
parallel to the longer dimension of the column) is considered.
qav  5.31ksf
(56 /12) 2
acting moment, M u  5.31*[10.5*
]
2
 607.11k  ft
Assume that j= 0.9
Mu
607.11*12000
As =
=
=9.45 in 2
 f y jd av .9*60000*(.9*15.87)
Minimum As as per ACI 10.5 =0.0018bh
=0.0018*126*20
 4.56in 2
 Minimum As doesn't govern.
Figure 6.9 c) Common footing two way shear critical section
123
Maximum spacing as per ACI 7.12 is 18 inches
Try 10 # 9 bars @ 12 inches center to center As = 10 in2
Re-compute  M n
a
As f y
'
c
0.85* f b
since

10*60000
 0.934in
0.85*6000*126
0.75ab
a
0.934

 0.059  0.333 
d av 15.78
d
0.9*10*60000*(15.87  .934 / 2)
 693.1k  ft
12000
Therefore the reinforcement provided is enough and
  M n   As f y (d av  a ) 
2
Mn  Mu
thus, provide 10#9 bars @ 14 in center to center both ways.
Development lengths and splices
Tension bars
As per ACI 318 section 12.2

For # 6 bar and smaller

For # 7 bar and greater

If the tension reinforcement is a top reinforcement the development length shall be
multiplied by 1.3
As per ACI 318-08 section 12.15, the class of the splice is considered to be Class B and
thus,
Bar splice = 1.3*
the values indicated below are rounded to the nearest inch of
multiple of five.
Bar #
Dev. Length(in)
Bar splice (in)
6
bottom top
25
35
35
45
7
bottom top
35
45
45
55
8
bottom top
40
55
55
70
9
bottom top
45
60
60
75
124
6.6 Foundation Reinforcement Detailing/ Drawings
125
Note:
 Cover to reinforcement 3 inches
 Depth of foundation column considered is 5 feet
 Concrete 6000psi and steel 60000psi are used
 The dowels that run from the foundation column into the
footing are of the same size and and equal in number to the
reinforcements of the foundation column
 Lean concrete may be placed under footing for smoothening of
the surface
126
6.6 Shear Wall Foundation (At Staircase)
Figure 6.10 Load at typical floor on wall
As we have design for maximum load coming to the wall and we have considered the
case of Load coming from Slab at each floors and load coming from staircase these load
and wind load is transferred at the bottom of the Footing on that basis we have calculated
the Area of footing which is same for each wall
127
Footing are used to transfer loads from the structure to the soil
Load Calculation
Load from Roof =0.243 kip/ft
Load from Flat Plate =5.68 kip/ft
Load from Flat Slab = 5.168 kip/ft
Load from Slab with Beam = 5.168 kip/ft
Load due to wind load= 60.112/25 =2.4 Kips
Load due to Flat Plate from Moment = 195.289/25 =7.83 Kips
Load due to Flat Slab =193.83/25 =7.75 Kips
Load due to Slab and Beam =180.81/25 =7.23 Kips
Load due to moment From Staircase = 55.78 x 4/25 =8.92 Kips
Load from Staircase Treads = 9.5 x 4/25 =1.52 Kip
Load from Staircase slab =0.22 x 2 =0.44 kips
Load due to Shear Wall = Density of Concrete x b x d x h=150 x 10/12 x 68.41 =8.55 kip
First Floor Live Load = 100 lb/ft
Dead Load = 59.37 +1.52 =60.89 Kips…….Adding loads from 1 to 12
Wind Load at the bottom = 0.288 Kips
Area Required =
Bearing Capacity of the Soil =qa= 6 ksf
Assume 2 ft depth of footing
128
Weight of concrete = 150 x 1 x 1 x 2 /1000 =0.3 kips/ft
Weight of Surcharge 120 x 1 x 1 x 5/1000 = 0.6 kips/ft
qn= qa-weight of concrete- weight of surcharge
qn= 6-0.3-0.6 =5.1 kips/ft
=11.84 ft2
Area required =
Try Footing Area =12 ft2
Width =12/2 = 6ft wide
Factored net Pressure qnu=
qnu = 7.142
Check the Shear is significant in a wall footing
d= 24-cover- 0.5 x bar diameter
Cover =3 inch
Therefore, d= 20.5”
Tributary Area
Figure 6.11 Critical section for shear
129
Vu=qnu x (x/12 x1)
Vu=7.142 x (10.5/12 x 1) = 6.24 Kips/ft
Φ Vc = Φ
Φ Vc =
= 24.64 kips/ft
Vu < Φ Vc Hence depth is ok
Therefore a depth as 2 feet and with 6 feet wide
Design the reinforcement
Tributary area for Flexure
Figure 6.12 Critical section for flexure
Mu=7.142
(31/12)2
Mu= 23.84 Kips/ft of length
130
Mu= Φ Mn= Φ Ast Fy j d
Footing are generally very lightly reinforced therefore assume that
J = 0.925
= 0.28 inch2/ft
Ast=
From ACI Section 10.5.3 and 7.12.2
Minimum Ast=0.0018
=0.5184 inch2/ft
Ast=0.0018
Therefore use # 5 Bars
Ast of one bar = 0.31
Spacing = 7 inches
Diameter of one bar= 0.625 inches
Provide #5 bars @ 7”
Development Length
Bar diameter db=0.625 inches
Cover = 3 inch =4 x db=4 x 0.625 =2.5
Bottom Bar x 1
Regular Concrete x 1
Regular Bar x 1
Ld=
db = 42.42 x 0.625………………….. ACI 12.2.4
Ld= 26.52’
As we have,
36inch – cover 3inch =33 inch
Therefore 26.52”< 33” ……………..Hence Safe
Shrinkage & Temperature Reinforcement
With Reference to ACI Code 7.12.2.1
Ast=0.0018
= 0.0018
12
Ast=0.5184 inch2
131
132
Therefore provide #5 @ 7”
As Per ACI maximum Spacing is 18 inches ……….ACI 7.12
Providing Minimum Area of steel at the Compression Zone
Therefore Provide #5 @ 7” in Both direction as it will Avoid Cracking in the Footing
In Shear wall Footing I have considered same Footing because these footing were design
for maximum load coming From the Slab so it is decided in order to keep same Area of
footing for each wall.
6.7 Shear wall Footing
Figure 6.13 Plan of shear wall & vertical forces acting on footing
133
Analyze of Shear wall Footing

Total Load acting on the footing due to Self weight of Shear wall, Reaction from
slabs, elevator & passenger load , machine room is 1935.71 Kips (calculated the
above load reaction shown in the figure 1 with the help of Staad- pro software)

Moment acting on the footing due to wind is 60.84 k-ft (-Mz)

Moment acting on the footing due to slabs – 839.6 k-ft (Mz)

Moment acting on the footing due to slabs – 839.6 k-ft (Mx)
I have calculated the eccentricity of total load acting on the footing due to individual load

ex = -2.32 in

ey = -28.26 in
Base soil pressure
Base soil pressure has been calculated by considering the total load & moments acting on
the footing
+
Where:
P= total load (kips)
A= Area of footing (ft2)
I= moment of inertia of footing
y = distance from centre line of footing to farthest end of footing
=
= 4.548 ksf
Hence the calculated soil base pressure is less than the allowable soil bearing
capacity (6 ksf). Therefore the assumed area is correct and we can proceed
further with the designing.
134
Design
Consider a unit strip of footing in both x & z direction and analyze & design it as a beam
Figure 6.14 Plan of footing integrated with beams 1, 2, 3 showed in red hatch.
Strip in Z- direction
Figure 6.15 load & reactions on strip A
M1=42.46(k-ft)
V1=35.03 k
M2=24.625(k-ft)
V2=13.175 k
135
M3=1.135(k-ft)
1. First design the strip for the maximum moment & shear to get the safe depth of footing
Factored moment Mu= 1.4x42.46 =59.45 k-ft
Assume the values of the steel ratio as follow
=
=
= 0.0377
The depth of the footing has been calculated by
fybd2
Mu=
59.45x12=0.9x0.019x60x12xd2
d=8.5in
9.5in
The overall depth required of strip D for the width b=12in is then
D = 9.5+3(cover) +1(dia of bar) + 0.5(half bar dia) = 14”
Checking shear capacity
Maximum shear in the beams is Vu=49.042k . This shall not be greater than the
maximum allowed shear, which is calculated based on the concrete strength , steel
strength and the depth of the c/s
Vu
Vn (10
bwd)
Vn= 0.9x10x
x12x9.5 = 79.46 kips > 49.042 (Vu=35.03x1.4)
1000
So the depth is satisfactory
Recomputed
59.45x12=0.9
by solving with b=12in & d= 9.5in
60x12x9.52
136
= 0.01692 >
min
=
= 0.0033
Area of steel given by: As= bd
As=0.01692x12x9.5= 1.93 in2
Check for minimum steel
As per ACI code the minimum steel is given by
. bw.d = 0.395in2 < As
As,min=
Provide #9 @ 6”
Development length in Tension (ACI Code, section 12.2)
ld = 0.04Abfy/
= 0.04x1.128x60000/
=35in
Provided ld=35 in
2. Calculate the steel As for moment M2=24.625(k-ft), b=12in, d=9.5in
Factored moment Mu= 1.4x24.625=34.475
Mu=
fybd2
34.475(12) =0.9x x60x12x9.52
= 0.0074 >
min
=
= 0.0033
Area of steel (As) = 0.0094x12x9.5 = 0.844 in2
Provide #6 @ 6 (As= 0.88in2)
Development length in Tension (ACI Code, section 12.2)
ld = 0.04Abfy/
= 0.04x0.75x60000/
=23.24in
Provided ld=23.5 in
137
Beam1
M1= 7.32(k-ft)
V1=87.77 k
M2= 14.65(k-ft)
Figure 6.16 load & reactions on Beam1
1. Compute the dimensions of beam based on maximum value of Mu.
Factored Moment Mu=1.4x14.65 = 20.51(k-ft)
Compute the depth of beam by the following equation, assume width of beam (b) = 8”
Mu=
fybd2
20.51(12) = 0.9x0.019x60xdx19.52
b = 3” (Since the width of wall is 10” so the I have provided the minimum width should
not be less
10”, therefore took b=10”)
Checking shear capacity
138
Maximum shear in the beams is Vu (87.77x1.4=122.9k). This shall not be greater than the
maximum allowed shear, which is calculated based on the concrete strength , steel
strength and the depth of the c/s
Vu
Vn (10
bwd)
x10x10.5 = 79.46 kips < 122.9 …………not safe
Vn= 0.9x10x
1000
Since the beam failed in shear check, therefore calculate ‘b’ using Vu
123x1000=0.9x10x
xbx10.5 => b=18”
for b=18” & d=10.5”
Recomputed
20.5(12) =0.9x x60x18x10.52
= 0.00243 <
min
= 0.00333
Area of steel given by: As= bd
As=0.0033x18x10.5= 0.63 in2
Provide 6no. #3(As=0.66 in2)
2. Calculate the steel As for moment M1=7.35(k-ft), b=10, d=8.4n
Factored moment Mu=1.4x7.35 = 10.248(k-ft)
Mu=
fybd2
10.248(12) =0.9x x60x18x10.52
= 0.0011 <
min
=
= 0.0033
Area of steel (As) = 0.0034x18x10.5 = 0.642 in2
Provide 3no. #6 (As= 0.66in2)
Design for shear
Vb = 35.03-2.274x9.5/12
= 33.323 kips
139
Vn=

=
= 39 kips
Check if stirrups are required as per ACI 11.5.5
Vc=2
bwd =17.67 kips
Since Vn>Vc
Therefore stirrups are required
Check selected stirrups for anchorage and maximum spacing
Try No. 3 double leg stirrups with fy = 60000psi in which case Av= 0.22 in2 . Since the
bar used for stirrups is less No. 6 ACI recommends that the stirrups be anchored by a 90 0
hook around a bar.As per ACI 11.4.5, the maximum spacing shall be the smaller of the
following.
Smax
Therefore, Smax = 4.75 in
Compute the required stirrup spacing to resist the design shear forces
For vertical stirrups
S=
=
= 31.87”
Use double leg # 3 bars at a spacing s=4.75” in all over the beam. As this is the maximum
shear, the same spacing of stirrup is to be used all over the beam.
Beam 2
M1=661.5(k-ft)
V1=271.67 k
M2=330.75(k-ft)
V2=170.85 k
140
Figure 6.17 load & reactions on Beam 2
1. Compute the dimensions of beam based on maximum value of Mu.
Factored Moment Mu=1.4x661 = 926.1(k-ft)
Compute the depth of beam by the following equation, assume width of beam (b) = 24”
fybd2
Mu=
926.1(12) = 0.9x0.019x60x24xd2
d = 23.5”
3”(cover) + 0.5”(half of dia.) = 27”
Checking shear capacity
Maximum shear in the beams is Vu(271.67x1.4=380.4k) .
Vu
Vn (10
bwd)
x24x23.5 = 393.2 kips >380.4 kip ………safe
Vn= 0.9x10x
1000
Recomputed
for b=24” & d=23.5”
926.1(12) =0.9x x60x24x23.52
= 0.011 >
min
= 0.00333
Area of steel given by: As= bd
141
As=0.011x24x23.5= 6.64 in2
Provide 5no.s #9 & 3no. #7 (Ast =6.8 in2)
2. Calculate the steel As for moment M2=330.75(k-ft), b=24, d=23.5n
Factored moment Mu=1.4x330.75 = 463.05(k-ft)
fybd2
Mu=
463.05(12) =0.9x x60x24x23.52
= 0.00815 >
min
= 0.0033
=
Area of steel (As) = 0.00815x24x23.5 = 4.6 in2
Provide 7no.s #7 & 1no. #6 (Ast =4.64in2)
Design for shear
Vb = 271.37-35.03x23.5/12
= 202.76 kips
Vn=

=
= 238.55 kips
Check if stirrups are required as per ACI 11.5.5
Vc=2
bwd =87.37kips
Since Vn>Vc
Therefore stirrups are required
Check selected stirrups for anchorage and maximum spacing
Try No. 3 double leg stirrups with fy = 60000psi in which case Av= 0.22 in2 . Since the
bar used for stirrups is less No. 6 ACI recommends that the stirrups be anchored by a 90 0
hook around a bar. As per ACI 11.4.5, the maximum spacing shall be the smaller of the
following
Smax
Therefore, Smax = 11 in
142
Compute the required stirrup spacing to resist the design shear forces
For vertical stirrups
S=
=
= 205.1”
Use double leg # 3 bars at a spacing s=11” in all over the beam. As this is the maximum
shear, the same spacing of stirrup is to be used all over the beam.
Beam 3
M1=185.5(k-ft)
V1=85.64 k
M2=92.75(k-ft)
V2=171.28
Figure (6) load & reactions on Beam (3)
1. Compute the dimensions of beam based on maximum value of Mu.
Factored Moment Mu=1.4x185.5 = 260(k-ft)
Compute the depth of beam by the following equation, assume width of beam (d) = 10.5”
Mu=
fybd2
260(12) = 0.9x0.019x60xbx10.52
b = 32”
Provided b=34” for shear to come safe
Checking shear capacity
143
Maximum shear in the beams is Vu (171.28x1.4=239.8k) .
Vu
Vn (10
bwd)
x34x10.5 = 248.9 kips >239.8 kip ………safe
Vn= 0.9x10x
1000
for b=34” & d=10.5”
Recomputed
260(12) =0.9x x60x34x10.52
= 0.017 >
min
= 0.00333
Area of steel given by: As= bd
As=0.017x32x10.5= 5.714 in2
Provide 5no. #9 & 1no. #8 (As = 5.79 in2)
2. Calculate the steel As for moment M2=92.75(k-ft), b=34in, d=10.5in
Factored moment Mu=1.4x92.75 = 129.85(k-ft)
Mu=
fybd2
129.85(12) =0.9x x60x34x10.52
= 0.00808 >
min
=
= 0.0033
Area of steel (As) = 0.00808x34x10.5 = 2.89 in2
Provide 5no. #7 (As= 0.3in2)
Design for shear
Vd = 171.28-13.17x10.5/12
= 159.75 kips
Vn=

Vc=2
=
= 187.94 kips
Check if stirrups are required as per ACI 11.5.5
bwd =55.30 kips
Since Vn>Vc
Therefore stirrups are required
144
Check selected stirrups for anchorage and maximum spacing
Try No. 3 double leg stirrups with fy = 60000psi in which case Av= 0.22 in2 . Since the
bar used for stirrups is less No. 6 ACI recommends that the stirrups be anchored by a 900
hook around a bar. As per ACI 11.4.5, the maximum spacing shall be the smaller of the
following
Smax
Therefore, Smax = 5.25 in
Compute the required stirrup spacing to resist the design shear forces
For vertical stirrups
S=
=
= 104.5”
Use double leg # 3 bars at a spacing s=5.25” in all over the beam. As this is the
maximum shear, the same spacing of stirrup is to be used all over the beam.
145
7. Cantilever Retaining Wall
7.1 Introduction
Retaining walls are structures that support backfill and allow for a change of grade. For
instance a retaining wall can be used to retain fill along a slope or it can be used to
support a cut into a slope.
Retaining wall structures can be gravity type structures, semi-gravity type structures,
cantilever type structures, and counter fort type structures. Walls might be constructed
from materials such as fieldstone, reinforced concrete, gabions, reinforced earth, steel
and timber. Each of these walls must be designed to resist the external forces applied to
the wall from earth pressure, surcharge load, water, earthquake etc.
Cantilever retaining wall consists of vertical wall, the stem, supported on a footing. The
front and rear end of the footing (or base) are termed the toe and heel, respectively. The
weight of the fill directly over the heel of the foundation stabilizes the wall, that is,
prevents it from sliding outward and overturning .When the lateral thrust is large, some
designers may add the key, a short vertical cantilever, to the base of footing to increase
resistance to sliding
Drainage System
At times, the water pressure together with the soil pressure of a submerged backfill will
produce a much larger forces than a dry backfill. To lower these forces engineers provide
146
drains or weep holes to reduce the buildup of water pressure behind walls. We have
provided the perforated soil pipe surrounded by a graded, stone filter. A filter prevents
inflow of sand or silt that would clog the drain. In our project we have provided a
perforated pipe 8” in diameter lay along the base of the wall and surrounded by gravel.
Fig 7.1 Drainage System in Retaining Wall
Specification:
Provided perforated 8” diameter pipe laid along the base of the wall &surrounded by
gravels (stone filter)
Purpose
To release the hydrostatic pressure.
147
Figure (7.3) free body of soil & pressure
148
Height of wall from G.L = 8’
Live load =100psf
Soil Density (
=120lb/ft³
Allowable soil bearing capacity = 6ksf

Angle of internal friction (
fy = 60 ksf
fc = 4 ksf
1. Using the Rankin equation: Ca = 1-sin/1+sin
= 1-


= 0.33
2. Determinations of the dimensions of the retaining wall:Height of wall (h)
= 8’+3’= 11’
Base thickness
= h/12 to h/10
= 0.92’ to 1.1’
Base length
= 0.5h to 0.7h
= 5.5’ to 7.7’ = 6.6’ (assumed)
Stem thickness
= 0.08h to 0.1h
= 0.9’ to 1.1’ = 10‘(assumed)
Toe projection phase of stem = 0.17’ to 0.125’h
= 1.87’ to 1.375’ = 1.7’ (assumed)
Base analysis on the 1’ thick slice of wall
3. Factor of safety against overturning:Calculate the actual unfactored forces acting on the retaining wall, first those acting to
overturn the wall
P1 = Ca.W
= 0.33x 0.1 = 0.033 kip/ft³
P2 = Ca  h
= 0.33x0.12x11= 0.436 kip/ft³
Ha1 = 0.033x11 = 0.363 kip……………………………….. arm 11/2 = 5.5’
Ha2 = 0.436x11/2 = 2.4 kip………………………………....arm 11/3 = 3.67’
149

Overturning moment about ‘a’ due to horizontal forces:Ma = (0.1815x1.5’) + (2.4x11/3)
= 9 k-ft

Balancing Moment against overturning:-
Forces(kip)
Arm(ft)
Moment(k-ft)
W1= 0.834’x9.8’x0.15 = 1.23
2.12
2.16
W2= 6.6’x1’x0.15 = 1
3.3
3.3
W2= 9.8’x 4.07’x0.12= 4.79
4.57
21.9
W4 = 0.1x4.07 = 0.407
4.57
1.86
W = R= 7.547
Ma= 29.67
Table 7.1 Vertical forces and moment acting at the toe of footing
 F.O.S = against overturning = Ma
R
= 29.67 = 3.29 > 2
9
4. Calculate base pressure:Take moments about the toe end ‘a’ to determine the location of the resultant ‘R’ of the
vertical forces.
x = Ma - h(y) = Balancing Moment - overturning moment
R
R
x = 29.67 - 9 = 2.8’>2.2’ (6.6’/3 = 2.2’)
7.547
 The eccentricity ’e’ = 6.6/2 - 2.8 = 0.5’
The resultant ’R’ acts within the middle third of the base and has an eccentricity e = 0.5’
from the centre of the base.
Assuming compressive stresses act over the entire base, compute the magnitude of the
soil stresses at each end of the footing by direct &bending stresses produced by ‘R’
150
p = R + ( R.e) .c
A
I
where R = 7.547 kips
A = 6.6x1 ft2
c = 6.6/2 ft
I=
ft4
Substituting above terms into above equation,
Pmax = 1.66 ksf
Pmin = 0.62 ksf
Since the maximum value of soil pressure (at the toe) is less than the allowable bearing
capacity of 6 ksf, the soil is not overstressed.
5. Factor of safety against sliding: - Passive pressure is neglected unless a key is
provided

Forces causing sliding = Ha1 + Ha2
= 0.363 + 2.4 = 2.736 kip

Resisting forces = R ………………… (Assume  = coarse grained soil without
silt)
= 0.5x7.545 = 3.77 kip
 F.O.S against sliding = R
Ha
= 3.77/2.676 = 1.36 < 1.5
So, in this case, key should be provided to develop passive pressure larger enough to
resist the excess forces that causes sliding.
In the calculation of the passive pressure ,the top foot of the earth at the toe side is
usually neglected , leaving a height of 2ft. Assume a key depth ‘t’=1.25’ and width ‘b’=
1.25’
Cp
Hp
= 1+sin = 1.5/0.5 = 0.3
1-sin
= ½ Kp(h’+t)²
151
= ½ x 3x 0.12 x (2+1.25)² = 1.9 kip
The sliding occurs on the surfaces ac, cd, df
Figure 7.4 Footing Details
Frictional forces (F) = (R1.1) + (R2.2)
R1= Reaction on AC =
2.12
= 2.69 kip
R2= Reaction on CDF =
5.32
= 4.136 kip
 F = (0.7) x 2.69 + (0.5) x4.136
= 3.951 kip
Hence, the total Resisting force (F+Hp) = 3.951+1.9
= 5.85 kip

F.O.S against sliding = Resisting force
Ha
= 5.85/2.736 = 2 > 1.5.…………………….SAFE
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DESIGN
6. Design of wall (Stem):-Design the stem for shear and moment created by lateral earth
pressures using a load factor of 1.7 .The thickness of the stem is generally controlled by
moment.
The critical section for bending moment is at the bottom of wall,
Height of wall above footing = 11’- 1’ = 10’
Moment at base of stem (Ms):--
Ha1= 0.033x10
= 0.33 kip (Due to surcharge)
Ha2= ½ x 0.396x10
= 1.98 kip (Due to lateral earth pressure)
V= 2.313 kip
Ms = 0.3(10/2) + 1.98(10/3) = 8.24 k-ft

Mu(at the bottom of wall) = 1.7(Ms) = 1.7x8.24 = 14 k-ft

Check for depth:Mu = fybd²
12x14 = 0.9x0.014x12x60xd² (1 - 0.014x60)
1.7x5
d = 4.6”  5”
Hence, the total depth of stem’D’ = 5+2+0.5 = 7.5”< 10”

(what assumed previously)
Check for critical shear:Vc = 2
= 0.85x2x
bd
) x12x7.5
= 11.85 kip
Vu = 1.7(V) = 1.7x2.313 = 3.93 kip
Hence Vu < Vc …………………………Shear capacity is O.K
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
Steel Requirement
 Reinforcement at internal face:o Main Reinforcement:Ru = Mu = 14x1200 = 248.88
bd²
12x(7.5)²
%= 0.0048
 Ast =%bd = 0.0048x12x7.5
Ast, provided = 0.432  0.465 in²
Providing #5@8” [Vertical steel]
Development length in Tension (ACI Code, section 12.2)
ld = 0.04Abfy/
= 0.04x0.31x60000/
=10in< 21in
Provided ld=21 in (min development length)
o Distribution Steel:Minimum Steel = 0.0025bd
= 0.0025x12x7.5 = 0.225
Divided the above steel into internal and external face equally.
 Mini Steel (each face) = 0.1125in²............ [ACI - 14.3.4]
Provide #3 @ 11’
Development length in Tension (ACI Code, section 12.2)
ld = 0.04Abfy/
= 0.04x0.11x60000/
=3.5in< 12in
Provided ld=12 in (min development length)
 Reinforcement at external face:Main reinforcement:Ast,mini = 0.0012xbxd
= 0.0012x 12x7.5
= 0.11in²
 Provide #3 @12”
154
Development length in compression (ACI Code, section 12.3)
ld = 0.02Abfy/
= 0.02x0.11x60000/
=1.8in< 8in
Provided ld=8 in (min development length)
7. Design of Heel
The upward soil pressure is neglected as it will reduce the effect of the backfill &
concrete on the heel. The total load on the heel:
Vu = 1.7(Rw) + 1.4(Wsoil) + 1.4(Wconc.)
= 1.7x0.407+1.4x4.79+1.4x4.07x0.15
= 8.25 kip
Mu (at face of wall) = 8.25x2.035 = 16.8 k-ft
Check for shear (depth required for shear)
Vc = 0.85(2(
) bd
Vc = Vu
0.85(2
)12xd = 1000x8.25
d = 6.4” (effective depth)
Required depth = 6.4”+3”+0.5” = 10” = 10” (assumed)

Steel reinforcement:Ru = Mu = 16.28x12000 = 232.526
bd²
12x8.5²
% = 0.45
Ast = 0.0045x12x8.5
= 0.46in²
 Provide #5 @8”
Development length in Tension (ACI Code, section 12.2)
ld = 0.04Abfy/
= 0.04x0.31x60000/
=9.6in< 21in
Provided ld=21 in (min development length)
155
8. Design of the Toe The toe of the base acts as a cantilever beam subjected to upward
pressure. The toe is subjected to upward pressure from soil &downward pressure due to
self weight of the toe slab. The critical section for shear is at the distance’s’ from the
front face of the stem.

Vu = 1.7(1.6+1.48) x0.825’ - 1.4(1’x0.15’0.835)
2
= 2.34 kip (this is less than Vc)

Mu = 1.7[1.35x1.7² + (1.6-1.35) x1.7 (2/3 x1.7)] -- 1.4[1x0.15x 1.7² ]
2
2
2
= 3.45 (k-ft)

Steel reinforcement:Ru = Mu = 3.45x12000 = 47.75
bd²
[ took mini % = 0.003 from steel table ]
12x8.5²
Ast =
Ast = 0.204in²
 Provide #4 @12”
Development length in Tension (ACI Code, section 12.2)
ld = 0.04Abfy/
= 0.04x0.5x60000/
=15.5in
Provided ld=16 in
Figure 7.5 Critical section for shear in toe
156
157
158
8 Green Engineering
8.1 Introduction
Green engineering is the design, commercialization, and use of processes and products,
which are feasible and economical
Helps in Minimizing Environment aspect

generation of pollution at the source

risk to human health and the environment
Green engineering initially cost more but later on it gets benefited.
As every year USA generates billions tons of waste as the managing waste cost hundred
of billions of dollars as we can see following graph which displays the increase in
environmental regulations from 1870 to present, demonstrates this trend.
Figure 8.1 Increases in Environmental Regulation
Although the number of regulations was on the rise in the mid 60's, the majority of the
regulations (75%) were passed after 1970 following the creation of the U.S. EPA
159
As industry continues to expand and permissible exposure limits continue to decrease,
waste treatment and the associated risk are becoming less cost efficient and more difficult
to handle. The figure below depicts the general trend observed over time of reduced
emission limits in conjunction with increased industrial growth.
Figure 8.2 Graphs on Emission in Industrial
Engineers have used several strategies to minimize the impact of chemical processes on
human health and the environment. By selecting chemicals that have low toxicity,
potential hazards are eliminated and chemical releases are prevented, thereby reducing
chemical exposures. Thus, environmentally conscious design reduces risks.
http://www.epa.gov/oppt/greenengineering/pubs/whats_ge.html
What Makes a Building Green?
A green building, also known as a sustainable building, is a structure that is designed,
built, renovated, operated, or reused in an ecological and resource-efficient manner.
Green buildings are designed to meet certain objectives such as

Protecting occupant health

Using energy, water, and other resources more efficiently and reducing the overall
impact to the environment
What Are the Economic Benefits of Green Buildings?

A green building Initial cost is high but saves through lower operating costs over
the life of the building.
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
Some benefits, such as improving occupant health, comfort, productivity,
reducing pollution and landfill waste are not easily quantified. Consequently,
they are not adequately considered in cost analysis. For this reason, consider
setting aside a small portion of the building budget to cover differential costs
associated with less tangible green building benefits or to cover the cost of
researching and analyzing green building options.

Even with a tight budget, many green building measures can be incorporated with
minimal or zero increased up-front costs and they can yield enormous savings
. What Are the Elements of Green Buildings?

Energy efficiency

Materials efficiency

Water efficiency
Energy efficiency
Passive design strategies can dramatically affect building energy performance. These
measures include building shape and orientation, passive solar design, and the use of
natural lighting. Develop strategies to provide natural lighting. Studies have shown that it
has a positive impact on productivity and well being.
Install high-efficiency lighting systems with advanced lighting controls. Include motion
sensors tied to dimmable lighting controls. Task lighting reduces general overhead light
levels. Use a properly sized and energy-efficient heat/cooling system in conjunction with
a thermally efficient building shell. Maximize light colors for roofing and wall finish
materials; and use minimal glass on east and west exposures. Minimize the electric loads
from lighting, equipment, and appliances. Consider alternative energy sources such as
photovoltaic’s and fuel cells that are now available in new products and
applications. Renewable energy sources provide a great symbol of emerging technologies
for the future.
161
Materials Efficiency
Selection of sustainable materials by evaluating several characteristics such as reused and
recycled content, zero or low off gassing of harmful air emissions, zero or low toxicity,
sustainably harvested materials, high recyclability, durability, longevity, and local
production. Such products promote resource conservation and efficiency. Using
recycled-content products also helps develop markets for recycled materials that are being
diverted from California's landfills, as mandated by the Integrated Waste Management
Act. Use dimensional planning and other material efficiency strategies. These strategies
reduce the amount of building materials needed and cut construction costs.
Water Efficiency
Design for dual plumbing to use recycled water for toilet flushing or a gray water system
that recovers
rainwater or other non potable water for site irrigation. Minimize
wastewater by using ultra low-flush toilets, low-flow shower heads, and other water
conserving fixtures. http://www.ciwmb.ca.gov/greenBuilding/Basics.htm
8.2 What material we have consider in Commercial Building Project
Table 8.1 Materials considered in green engineering
Materials
Function
Application
Glazing Curtain Wall
Weather protection &
Glass on all exterior surface
System
Insulation
Roof Garden
Plantation & Aesthetics
Sewage Treatment Plant
To Generate Methane as an Drainage Treatment of
on roof
energy
Building
Paints
Environmental Friendly
All interior portion
Lighting
Less Energy Consumption
Both Interior & Exterior
Water Proofing
Water proof structure
For Concrete & Masonary
162
Glazing Curtain Wall System:
The glass and aluminum curtain wall is found in city centres on many new buildings and
it is quite popular as a cladding and exterior wall on all types of commercial, industrial,
institutional and residential buildings. The curtain wall is characterized with colored
vision and spandrel glass areas, a grid of aluminum caps and most recently with metal or
stone spandrel covers It is also combined with other types of cladding systems such as
precast, brick or stone to create attractive and durable building facades.
The curtain wall comprises a complete cladding and exterior wall system with the
exception of the indoor finishes. It is generally assembled from aluminum frames, vision
glass and spandrel glass (or metal or stone) panels to enclose a building from grade to the
roof.
Function
 Airtight and weather resistant
 Air leakage control
 Rain Penetration Control
 Heat Loss
 Condensation Control
 Fire Safety
System Adopted for Fixing Glass System
The Unitized Curtain Wall
A glass and aluminum curtain wall fabricated and installed as a panel system is referred
toas a unitized curtain wall system. A unitized curtain wall will have the same
components as a stick built curtain wall system. It will comprise aluminum mullions, an
IGU and a spandrel panel mounted in a prefabricated aluminum frame. However, instead
of assembling the glass and aluminum curtain wall in the field, most of the system
components are assembled in a plant under controlled working conditions. This promotes
quality assembly and allows for fabrication lead-time and rapid closure of the building.
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Infiltration allowance
Air infiltration is the air which passes through the curtain wall from the exterior to the
interior of the building. The air is infiltrated through the gaskets, through joints which are
opened as we know allowance rate is very low as per US specification as joints should be
between horizontal & vertical mullions, through weep holes, and through imperfect
sealing air can easily come. As per The American Architectural Manufacturers
Association (AAMA) is the main body how looks after the specification related to Glass
and which has sets the acceptable levels of infiltration through a curtain wall. This limit is
expressed (in America) in cubic feet per minutes per square foot of wall area at a given
Test pressure. As per standards it is followed practice less than 0.6 CFM/sq ft as
acceptable
http://en.wikipedia.org/wiki/Curtain_wall#Fire_safety
Components Involved are:
 Mullions vertical Frame & rails horizontal mullions
 Vision Glass, insulation
 Hardware components – Anchors, Aluminum connector, Settings blocks, Corner
blocks, Pressure plates, caps, gaskets
Fixing System
The unitized system is assembled on the building as panels. The mullions and rails are
fabricated as half sections instead of tubular sections, which mate at assembly ime to
form the curtain wall system. The panels are installed in shingle fashion, starting from the
bottom of the building and going around each floor and up the building. While the
unitized system offers many advantages with respect to quality assembly and speed of on
building closure, there is one design concern with respect to installed performance and
durability. Installers should ensure that the air seals are properly installed between the
164
split mullions. Nevertheless, the unitized system is now as popular as the stick system
according one
Figure 8.3 Typical Unitized Curtain Wall System
Figure 8.4 Typical Unitized Curtain Wall System Assembled product
165
Function –
1. Air leakage control
To perform satisfactorily, a cladding an exterior wall system must meet several
performance requirements. These include air leakage control (the air barrier function),
vapour diffusion control (the vapour barrier function – not the same as the air barrier
function), heat loss/gain control (insulation and thermal breaks), rain and melt water
penetration control(the rain screen principle). In addition, the curtain wall must
accommodate various differential movements. The most critical of the performance
requirements is air leakage control. The leakage of air through a curtain wall system in
winter may result in excessive ice buildup on aluminum caps, at parapets or soffits. The
ice can grow to become a safety hazard to persons and property below. Air leakage also
causes condensation in glazing cavities to the Detriment of the IGUs and it can corrode
back pans and fasteners.
The leakage of air at the glass aluminum joint is minimized with either a wet or dry seal.
In a unitized system, the mullions are split and therefore include an additional air barrier
joint between the half mullions. This joint is usually hidden and inaccessible once
assembled.
Fig. 8 – Air
166
Figure 8.5 Typical Unitized Curtain Wall System Air Vapour Barrier
167
The air leakage rate through a curtain wall fabricated for the United States market is
limited to 0.3 litres/sec*metre2 at 75 Pa (0.06 cfm per ft2 at 1.57 lbs/ft2) air pressure
difference.
Location
2. Rain Penetration Control
Like any wall system, the curtain wall system must prevent the penetration of rain or melt
water to the inside of the building. Being constructed of glass, aluminum, steel, fibrous
insulation and sealants, the components have no ability to absorb and release even the
most incidental amounts of water penetration. While some types of insulation can absorb
moisture, very little moisture enters the back pan area. Also, because most of the
materials comprising the curtain wall are corrosion resistant, water does not damage the
system materials, except the seals of the IGUs if they remain wetted for long periods of
time.
To enhance the rain penetration control of a glass and aluminum curtain wall system, the
rain screen principle is applied. The rain screen principle incorporates various features to
control:
• Direct entry of rain or melt water,
• Capillary action,
• Surface and cavity drainage,
• Pressure equalization of the glazing cavities.
In a stick built system, resistance to the direct entry of rain and melt water penetration is
provided by the outside seal or gasket at the glass-to-cap joints. However, should a minor
Amount of rain or melt water penetrate through the head, jamb or sill gasket or seal of the
Vision glass, it is channeled sideways and downwards via the jamb cavity to the sill
glazing cavity below. The rainwater is then diverted horizontally by the corner blocks the
168
drain holes in the pressure plates and into the curtain wall snap caps to drain to the
outside.
3 Fire Safety
Fire stopping at the perimeter slab edge which is a gap between the floors and the back
pan of curtain wall is essential slow the passage of fire and combustion gases between
floors. Spandrel areas must have non combustible insulation at the interior face of the
curtain wall. As per some building code requires the mullion is rapped at in heat
insulation near the ceiling to prevent the mullions from melting and spreading the fire to
the floor above. It is important to note that the fire stop at the perimeter slab edge is
considered a continuation of the fire resistance rating of the floor slab. The curtain wall
itself. However, is not ordinarily required to have a rating .To avoid Fire and Smoke
Migrations beyond engaged compartment. As curtain wall by it is nature prevents the Fire
hence fire safety is maintained using curtain wall
http://en.wikipedia.org/wiki/Curtain_wall#Fire_safety
Vendor
www.aamanet.org or call or write AAMA at; Tel. (847)303-5664; Fax (847) 303-5774
1827 Walden Office Square, Suite 104 Schaumburg, Illinois, 60173.
Analysis of curtain wall system
All exterior walls, of whatever materials, are subject to, and must withstand the ravaging
effects of nature. These nature forces are sunlight, temperature, water, wind and gravity.
Except for gravity, the intensity and relative significance of these forces vary some what
169
from one region to another, but all of them are been considered, and their effects
provided for, in all location
Approximate Weight of Architectural Flat Glass
The table below provides approximate weights of architectural flat glass by thickness
designations as published by North American manufacturers:
Table 8.2 Glass weight
Glass Thickness
2.5
3
4
5
6
8
10
12
16
19
1.2
1.6
2
2.4
3
4
5
6.4
8.1
9.8
Designation (mm)
Approximate weight
lb/ft2
2945 SW Wanamaker Drive, Suite A Topeka, KS 66614-5321 (785) 271-0208 Fax: (785)
271-0166 www.glasswebsite.com
Weight consideration for design in project
Curtain Wall is a term used to describe a building façade which does not carry any dead
load from the building other than its own dead load, and to transfer horizontal loads
(wind loads) applied on the curtain wall. These Loads, are transferred to the main
building Connection through hardware connection at floors or columns of the building .a
curtain wall is designed to resist air and water infiltration, wind force acting on the
building. Hence we can see Fig 8.6 how load is transferred floor to floor as mentioned
below complete load calculation with some approximation the panel size and connection
we have calculated the load transferred into beams
170
Figure 8.6 Load Transfer
Load due to glass curtain wall
 Weight of a 10mm (3/8 in) glass is 5psf (glass information bulletin)
 Weight of connections, mullions, capping = 0.5*5= 2.5 psf
 Total weight of glass curtain wall = 7.5 psf
Specification of Glass
We have consider Panels for Glass as 10mm thick
For Ground Floor to First Floor we have 26 feet height. Therefore we have divided the
height of the floor into two parts . 5 Panels .Below Figure represents the Glass panel
Sizes for Each Floor we have provided
171
Figure 8.7 Glass Panel Sizes at different floors
Roof Garden
Aesthetic Benefits
• Green roofs are much better looking than asphalt or tar.
• Natural views create more productive, healthy, happy, creative, relaxed people.
• Green roofs expand the usefulness of buildings via patios, gardens and vistas.
How it is fixed?
It is a modular system
172
Figure 8.8 Components of Roofing System

LiveRoof module is filled to the top of Soil Elevator with LiveRoof engineered
soil.

LiveRoof Plants are grown to maturity approximately 1 inch above the LiveRoof
module.

Placed directly into the roof as it is modular system it can be replaced easily
Why Use LiveRoof System for Green Roof?

It is an instant Green roof

Saves Money rather than farming on the roof

Drainage pattern is good please refer the figure below

Easily changeable
173
Figure 8.9 Positive Drainage for healthy roots
Figure 8.10 Fixing Method
174
Advantages
The Live Roof® system is superior in design and implementation to both conventional
green roofs and to other modular green roof systems. Specific benefits include a better
aesthetic presentation and instantly functional evaporative cooling (instantly green vs.
brown and no multi-year grow in period), substantially lower maintenance, elimination of
wind and rain erosion, simple installation, expertly engineered soil and plant selections,
durable indefinite-lasting components
Specification
SIZE: Module: 1’ x 2’ x 3” (soil height approximately 4”) Soil fills soil elevator, plants
and soil
obscure container edges
MATERIAL: 100% post-industrial recycled polypropylene 100 mil. thick walls
WATER DISPERSAL: Approximately 7 gal. per min. per lineal foot.
COLOR: Black
WEIGHT VEGETATED (fully saturated): Approx. 27-29 lbs./sq. ft
DRAINAGE: Positive drain holes, placed at lowest point in container. Avoids root rot
associated with water reservoirs. Special Hi-Flow option available
SOIL MEDIA: It is recommend proprietary LiveRoof® specified engineered soil,
enhanced German FLL specifications, 93+% (by dry weight) inorganic content for
minimal shrinkage/decomposition
ACCEPTABLE PROTECTIVE UNDERLYING MATERIALS:
Modules are typically placed directly upon a heavy duty (HDPE, Polypropylene, TPO,
EPDM or recyclable PVC) slip sheet/root barrier of 45-60 militre thickness with
effectively bonded seams. This is placed as an additional protective barrier above roof
waterproofing membranes. Alternatively low profile drain boards work well and
manufacturers of cold fluid applied reinforced urethane membranes typically warrant
their systems for use in conjunction with LiveRoof® system.
IRRIGATION SYSTEM: Generally not needed, but recommended for backup during
prolonged hot dry windy weather patterns. Simple overhead system, spigot/hose/sprinkler
175
or drip irrigation systems are inexpensive and effective insurance. Irrigation requirements
are dependent on plant selection. If lightweight module is used, irrigation will be
essential
EDGE TREATMENTS: Co engineered Permaloc® aluminum L-shaped unaffixed
edging recommended.
PLANTS: See LiveRoof®.net for grower in your region, for specific recommendations.
MAXIMUM STACKABLE HEIGHT: LiveRoof® modules are designed to be stacked on
their sides so as to protect the Soil Elevator TM holders.http://liveroof.com/
Sewage Treatment Plant
Fig 8.11 Scheme representation of Sewage Treatment Plant
176
Sewage Treatment Plant is basically clean the sewage comes into it is done basically as
show in scheme representation here by Bioelectrolytic Process here main process is run
by Electro tank were electrolysis takes place. Electrolysis is a technique involving the
passing of electricity through the effluent. The electric current distabilize dissolved
colloidal particles and alters the charge on suspended particles permitting their separation
Benefit of Electrolytic Treatment System
1 Non Chemical Treatment as per green engineering specification
2 No Aeration and no MLSS maintained
3Result in pathogen free disinfected water
4Toxicity easily handled
5Sludge is pathogen free, Easy to dispose & non hazardous to the environment
6 High Performance to cost ratio
7 Noiseless system
Advantages

It generates Methane which can be used as a Source of Energy. We can use the
piping to send to appropriate location

It is an Custom make and modular in size

Maintenance and Operation cost is economical

It maintains the BOD & COD level of Water is obtained which give the clean
water as per environmental department required
Location: It is fixed in parking lot and very little space is occupied as it is a compact
system
Quantity of water treated is 300 KLD (Approximation as per size of Project)
Paints
Consideration for most interior paints is their potential impact on occupant health. To
this end, very-low-VOC or zero-VOC paints are generally preferred, though some
177
chemically sensitive people find that even these can be difficult to tolerate. Note that
liquid carriers (including water and exempt VOCs) are usually excluded when stating
VOC levels, which are expressed as grams of VOC per liter of VOC-plus-paint-solids.
Most zero-VOC paints still use colorant systems that contain VOCs, so custom-coloring
will increase emissions. Some paints are made from minimally processed plants and
minerals; while these products may contain relatively high levels of VOCs, they may not
be as troublesome as the compounds released from petrochemical-based paints. Products
that are less than 50 grams of VOCs per liter is recommended
Sherwin-Williams has increased the quality and depth of its paint inventory with the
introduction of two new industrial paints with low levels of volatile organic compounds
(VOCs). The Pro Industrial 0 VOC waterborne acrylic formula and the alkyd (oil-based)
Enamel 100 are designed for interior or exterior use in healthcare, commercial, industrial,
or other applications where paint will be exposed to high traffic, dirt, and moisture.
These Paints are useful for Interior & Exterior wall painting so that it has special
properties which are helpful human environment as it has low volatile organic
compounds (VOCs)
Lighting
Fluorescent lamps have long been preferable to incandescent lighting, relative to energy
efficiency. New developments with fluorescent technology, including the high-efficacy
T5 lamps, have pushed the energy efficiency envelope further. Recently, attention has
also been paid to the mercury content of fluorescents and the consequences of mercury
releases into the environment. As with all resource use and pollution issues, reduction is
the best way to limit the problem. Even with low-mercury lamps, however, recycling of
old lamps remains a high priority
As these kinds of lamps is provided in the passage of the building and staircase and
common area in the building as it saves lot of energy and gives the same output hence we
save a good amount of energy as
178
Waterproofing
Waterproof Coatings for Concrete and Masonry.While most conventional concrete
sealers and waterproofing agents have very high VOCs, some products react with the
concrete or otherwise provide a seal without resorting to solvents that emit VOCs. Green
Products contains zero VOCs, either nominally or actually.
.Product for Waterproofing: Aquafin-IC Crystalline Waterproofing
Company: Aquafin, Inc.
Aquafin-IC is a penetrating, inorganic, cementitious material used to permanently
waterproof and protect new or existing structurally sound concrete
Application is on Concrete structure or wall were we can apply waterproofing component
and increase its life span were no water leakage takes place
http://www.buildinggreen.com/menus/drillBC.cfm?BuilderCategoryI
179
9. Material (Concrete) usage Estimation
Components
Quantity in (ft3)
Slabs
75000
Beams
6973
Columns
5488
Staircase
1750
Shear Wall with Staircase
5667
Shear Wall with Elevator(2)
Footing for Shear Wall with Staircase
Footing for Shear Wall with Elevator
(2)
Footing Under Column
Retaining wall
11861.54
1200
2434.86
7232
15688.52
Total Concrete Quantity = 133294.9 Ft3
180
References
Books

ACI Code

International building Code –IBC Load calculation & Specification

ASCE- American Society of Civil Engineering

Christian Meyer, Design of Concrete Structures, Prentice Hall, New Jersey

Christian Meyer, Design of Concrete Structures, Prentice Hall, New Jersey

Brian Boughton, “Reinforced Concrete Detailer Manual”-Detailing of Staircase ,
Granada Publication, London, 3rd Edition

Phil M Ferguson, John E. Breen, James Jirsa, “Reinforced Concrete
Fundamental”, John Wiley & sons Publication, New York

Kenneth Leet, Dionisio Bernal, “Reinforced Concrete Design” ,The McGraw-Hill
Companies, INC, New York, 3rd Edition

M.Nadim Hassoun, “Design of Reinforced Concrete Structures”, PWS
Engineering, Boston

James G Macgregor ,“Reinforced Concrete Mechanics and Design”, (2nd edition)
by Prentice hall, A Simon and Schuster company, Upper saddle river ,NJ – 07458

Jack C.McCormac, “Design of Reinforced Concrete” (4th edition),AddisonWesley, California.
181
Sites


Expansion joint http://www.copper.org/homepage.html) 6
Elevator Load and Specification, Mitsubishi Company,
www.mitsubishielevator.com/

Green Engineering-Introduction
http://www.epa.gov/oppt/greenengineering/pubs/whats_ge.html

Green Building http://www.ciwmb.ca.gov/greenBuilding/Basics.htm

Glass
Vendor: AAMA at; Tel. (847)303-5664; Fax (847) 303-5774
1827 Walden Office Square, Suite 104 Schaumburg, Illinois, 60173.
www.aamanet.org
http://en.wikipedia.org/wiki/Curtain_wall#Fire_safety
www.glasswebsite.com

Green engineering Other Component
http://www.buildinggreen.com/menus/drillBC.cfm?BuilderCategoryID=23

Aesthetics –Green roof Live roof System http://liveroof.com/

Sewage Treatment Plant, www.superklean.in
182
183