STAIR DESIGN
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Stair Design
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Introduction:
Stairs
are means of moving up and down in buildings. A well-planned and designed stair
should provide an easy, quick and safe mode of communication between the
various floors.
Classification of stairs:
1)
Straight stair : the most obvious use of the straight stair is to
form an access to an entrance, porch of portico. Straight stairs cannot be
avoided in places where stair-case hall is long and narrow and the possibility
of any other form of stair may not be practically possible. In this form of
stair all the steps rise in the same direction. If the ascent is steep the straight
flight may broken at an intermediate landing.
2)
Dog-legged stairs: it consists of
two straight flights of steps with abrupt turn between them. Usually, a level
landing is steps with abrupt turn between them. Usually, a level landing is
placed across the two flights at the change of direction. This type of stair is
useful where the width of the stair case hall is just sufficient to accommodate
two widths of the stair.
3)
open-well stairs: It consists of two or more straight flights changed
in such a manner that a clear space, called a “well” occurs between the
backward and the forward flights. If the width of stair case hall is such that
it becomes difficult to accommodate the number of steps in the two flights
without exceeding the maximum allowable limit of steps in each flight, a short
flight of 3 to 6 steps may be provide along the width of the wall. In this sate
the intermediate short flight will have quarter-space landings on its either
side.
4)
Geometrical stair: This is similar to the open-newel stair with the
difference that the open-well between the forward and the backward flights is
curved. In this form of stair, the change in direction is obtained through
winders.
5)
Circular stair: circular stair is commonly provided at the service
purposes. The circular stairs are commonly constructed in R.C.C., iron or
stone. In this from of stair all the steps radiate from a newel post or well
bole in the form of winders.
6)
Bifurcated stair: This type of stair is suitably provided in modern aristocratic
public buildings. In this type of stair the flights are so arranged that there
is a wide flight at the start which is sub-divided into narrow flights at the
mid-landing. The two narrow flights start from either side of the mid-landing.
The stair which will be designed in this chapter
is a doge leg type ….
ØThere are Some
general requirements of a good stair Such as:
Ä Location: It
should be so located that sufficient light and ventilation is ensured on the
stair-way. If possible it should be located centrally so as to be easily
accessible from the different corners of the building.
Ä Width of the stair:
Width of a stair varies with the situation and the purpose for which it is
provided. Obviously in a public building where there is a regular traffic of
people using there stair-way its width should be sufficient while in a
residential building it may be just the minimum. The usually adopted average
value of the stair width for public and residential building is 1.5m. And 1 m
respectively.
Ä Length of light:
For the comfortable ascend of a stair-way the number of steps in a flight
should be restricted to a maximum of 12 and a minimum of 3.
Ä Pitch of stair:
The pitch of long stairs should be made flatter by introducing landings to make
the ascent less tiresome and less dangerous. In general the slope of stair
should never exceed 40° and should not be flatter than 25°.
Ä Head room: The
room or the clear distance between the tread and the soffit of the flight
immediately above it should not be less than 2.14 m.
Ä Materials: The
stair should preferably be constructed of materials which possess
fire-resisting qualities
Ä Landing: the
width of landing should not be less than the width of stair.
Ä Winders: The
introduction of winders in stairs should be avoided as far as possible. They
are liable to be dangerous and involve extra expanse in construction. They are
difficult to carpet and are especially unsuitable for public buildings.
However, where the winders cannot be dispensed with, they should preferably be
provided near the lower end of the flight, thus instead of quarter-space
landing three winders may be used and for a half-space landing five winders or
four radiating risers may be adopted.
Ä Step proportions:
The rise and tread of each step in a stair should be of uniform dimensions
throughout. The ration of the “going”
and the “rise” of a step should be so proportioned as so ensure a
comfortable access of the stair-way
Stair Design Procedure:
Ø Take rise 15 cm.
Ø Take Go 30 cm.
Ø High of floor
= 3m.
ØNumber of rise
= 300 /15 = 20 , use two flight, each flight has 10 rise
ØNumber of Go =
number of rise – 1
= 10 – 1
= 9
-we
are taken the horizontal projection to design the stairs:
Then:
Horizontal
Length of each flight = 9 * 0.3 = 2.7m
Ln
= 1/2 of land's + length of flight
= 0.5 + 0.75+ 2.7 = 3.95
The
minimum Thickness of flight slab (h) based on code for simply supported
H = Ln / 20
= 3950 / 20 = 197.5 mm Ü use (h) = 200mm
¯
Load of flight:
ØTile = (20/10) 20 (9.81)
= 0.785 KN/m
ØPlaster = (20/10)
21 (9.81) = 0.412 KN/m
ØMortar = 0.03
(24)= 0.72 KN/m
ØOwn weight of
each step = 0.304 (0.3)(24) (1) = 2.2 KN/0.3
= 7.3 KN/m
Total
dead loads
= 0.785 + 0.412 + 0.72 + 7.3= 9.767 KN/m
Live load of
stair = 400 Kg/m = 3.924 KN/m
Total
load (Wu) =
1.4 * DL + 1.7 * LL
= 1.4 (9.767) + 1.7 (3.924)
= 20.34 KN/m
¯Take length of
flight = 1/2 of land's + length of flight
= 2.7 +0.5+0.75 = 3.95 m
¯Moment (Mu) = (Wu * L2 )/ 8 = 20.34 * (3.95)2
/ 8 = 39.67 KN.m
¯Shear force (Vu) = Wu * L/ 2 = 20.34 * 3.95 / 2 = 40.175 KN
Design for main reinforcement :
Mu = f * fc * b *
d2 w ( 1 – 0.59 w )
39.67*
106 = 0.9 * 25 * 1000 * (175)2
w ( 1 – 0.59 w )
w ( 1 – 0.59 w ) = 0.057571 Ü w = 0.0588
r = w (fc / fy ) = 0.0588 ( 25
/ 414) = 0.0035507 > rmin. = 0.00338
As = r * b *d = 0.0035507
(1000) (175) = 621.4 mm
(Use 5F14/m = 769
mm2)
For short direction:
Use steel for shrinkage
& temperature
As = 0.0018 * b * h
= 0.0018 * 1000 * 200 = 360 mm2
(Use
5F10/m
= 393 mm2)
Design for Shear reinforcement :
f Vc = 0.85 * (fc )0.5
* b * d / 6
= 0.85 *(25)0.5 * 1000 *
175 / 6
= 123.96 KN
Vu = 31.11 KN < f Vc , so that use minimum reinforcement
S = 3 * Av * fyv / b =
3 *157 * 280 / 1000 = 132 mm
Use 1F10/20 cm for each step
along the step
¯
Design of Landing:
t ( min) > L
/ 20
length of
landing = 2.6 m
t = 260 / 20 =
13 cm Ü use h (t) = 20 cm
ØLoads:
ØOwn weight of
landing = 1* 0.2 * 24 = 4.8 KN/m
ØTile
=
(20/10) * 20 * 9.81 * 1 = 0.785 KN/m
ØPlaster = (2/1)
* 21 * 9.81 * 1 = 0.412 KN/m
ØMortar = 0.03 *
24 * 1 = 0.72 KN/m
Total dead
loads = 6.717 KN/m
Live load for
land of stair = 400 Kg = 400 * 1 * 9.81 = 3.923 KN/m
(Wu) = 1.4 Dl + 1.7 LL
= 1.4 (6.717) +
1.7(3.923)
= 16.1 KN/m
From the flight = 40.175 KN/m
Total loads = 16.1 + 40.175= 56.275 KN/m
Moment (Mu) = (Wu * L2 )/ 8 = 56.27 * (2.7)2
/ 8 = 51.3 KN.m
Shear force (Vu) = Wu * L/ 2 = 56.27 * 2.7 / 2 = 76 KN
Design for main reinforcement
Mu = f * fc * b *
d2 w (1 – 0.59 w )
51.3 * 106 =
0.9 * 25 * 1000 * (175)2 w (1 – 0.59 w )
w ( 1 – 0.59 w ) =0.07445 w = 0.0753
r = w (fc / fy ) = 0.0753 ( 25
/ 414) = 0.004547 > rmin. = 0.00338
As = r * b *d = 0.004547(1000)
(175) = 795.8 mm2
(Use
6F14/m
= 923 mm2)
For
short direction:
Use steel for shrinkage
& temperature
As = 0.0018 * b * h
= 0.0018 * 1000 * 200 = 360 mm2
(Use 5F10/m = 393 mm2)
Roof slab design(for stair
case):
The
first step is to determine wither the slab is one-way solid slab or two –way
solid slab ,depend on the ratio between the long direction (L) and the short
direction(B).
L / B = 6 / 2.8=2.14 > 2
Ä Design slab as one way solid slab
¯ Determine the
thickness of slab:
The min thickness for the simply supported slab(t) = L/20
h or
(t) = 280 / 20 = 14 cm use h = 15 cm
ØLoads:
ØOwn weight of
slab = 0.15 * 2.8 * 1 * 24 = 10.08 KN/m
Ølive loads = 2 KN/m
Wu = 1.4*(10.08)+1.7( 2 ) = 17.512 KN/m
Moment (Mu) = (Wu * L2 )/ 8 = 17.512 * (2.8)2 / 8 = 17.162 KN.m
Design for main reinforcement
Mu = f * fc * b *
d2 w ( 1 – 0.59 w )
17.162 * 106
= 0.9 * 25 * 1000 * (125)2 w ( 1 – 0.59 w )
w ( 1 – 0.59 w ) = 0.488164 Ü w = 0.04935
r = w (fc / fy ) = 0.04935( 25/
414) = 0.00298 < rmin. = 0.00338
As = 0.00338* b * d =
0.00338 * 1000 * 125 = 422.5 mm2
(Use
4 F12/m
= 452 mm2)
For
long direction:
Use steel for shrinkage
& temperature
As = 0.0018 * b * h
= 0.0018 * 1000 * 150 = 270 mm2
(Use
3F12/m
= 339 mm2)
Roof slab design(for elevator):
The
first step is to determine wither the slab is one-way solid slab or two –way
solid slab ,depend on the ratio between the long direction (L) and the short
direction(B).
L / B = 2.2 / 2 =1.1 < 2
Ä Design slab as Two way solid slab
¯Determine the thickness
of slab:
The min thickness for the simply supported slab(t) = L/20
h or
(t) = 2.2 / 20 = 0.11 cm
use h = 15 cm
ØLoads:
ØOwn weight of
slab = 0.15 * 1 * 24 = 3.6 KN/m
Ølive
loads = 0
Total load =
dead load only
Wu = 1.4 DL =
1.4*(3.6) = 5.04 KN/m
Ws = 5.04*1.14/1+1.14 = 3 KN/m
Wl = 5.04/1+1.14 = 2.04 KN/m
Ms = Ws*l*l/8 = 1.5KN.m
Ml = Wl*l*l/8 = 1.23
Design for main reinforcement
Mu = f * fc * b * d2
w ( 1 – 0.59 w )
1.5 * 106
= 0.9 * 25 * 1000 * (125)2 w
( 1 – 0.59 w )
w ( 1 – 0.59 w ) =0.0004195
w=0.0004267
Ü
Where r =w* fc / fy
r=0.0004267*(25/414)
=0.0000258
r < rmin. = 0.00338
So
That Use r Minimum
As = 0.00338* b
* d = 0.00338 * 1000 * 125 = 422.5 mm2
(Use
4F12/m = 452 mm2)
For
Second direction:
Mu = f * fc * b *
d2 w ( 1 – 0.59 w )
1.23 * 106 =
0.9 * 25 * 1000 * (125)2 w ( 1 – 0.59 w )
w ( 1 – 0.59 w )=0.0004012
r < rmin. = 0.00338
So That Use r Minimum
As = 0.00338* b *
d = 0.00338 * 1000 * 125 = 422.5 mm2
(Use
4F12/m
= 452 mm2)
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Walls Design
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WALLS
are generally utilized to
provide lateral support for an earth fill, embankment, or some other material,
and/or to support vertical loads. Some of the more common types of walls are
described later. One of the primary purposes for these walls is to maintain a
difference in the elevation of the ground surface on each side of the wall.
The earth whose ground surface is at the
higher elevation is commonly called the backfill, and the wall is said to
retain this backfill.
All the walls have applications in either building or bridge project.
They do not necessarily behave in an
identical manner under load.
but still serve the same basic
function of providing lateral support for a mass of earth or other material
which is at a higher elevation behind the wall than the earth or other material
in front of the wall.and the walls could be classified based on its functions;
Types of Walls:
1) Cantilever
wall:
Is the most common
type of retaining structure and generally is used for walls in the range 10-25
ft in height, it derives its name from the fact that its individual part (toe,
heel. And stem) behave as, and are designed as, cantilever beams. Aside from its
stability, the capacity of the wall is a function of the strength of its
individual parts
2) Basement
or foundation wall:
This type may
act as a cantilever retaining wall. However, the first floor may provide an
additional horizontal reaction similar to the basement floor slab, thereby
making the wall act as a vertical beam. This wall would then be designed as a
simply supported member spanning the first floor and the basement floor slab.
3) Bearing
wall or (Bridge abutment):
This type may
exist with or without lateral loads. “The American Standard Building Code
Requirement for Masonry” defines a bearing wall as a wall that supports any vertical load in addition to its
own weight. Depending on the
magnitudes of the vertical and lateral load, the wall may
have to be designed for combined bending and axial compression.
4) Gravity
wall:
Depends mostly
on its own weight for stability. It is usually made of plain concrete and is
used for walls up to approximately 10 ft in height. The semi gravity wall is a
modification of the gravity wall in which a small amount of reinforced steel is
introduced. This, in effect, reduces the massiveness of the wall.
Ü In these project
the Bearing wall it will classified according to its load value carried
by the wall, bearing wall; which carry building stone faces, and shear wall;
which carry the stairs and some load from the slab.
O In
this Project the bearing wall was used ,which carry building stone faces, and shear wall; which carry the
stairs and some load from the slab.
Design Procedure:
According to (ACI-99 code), The minimum Thickness of shear wall is
190 mm, and walls with thickness more than 25cm shall have reinforcement for
each direction placed in two layers parallel with faces of the wall.
Firstly the thickness of the wall
must be assumed according to the load, which carries,and the thickness must not
be less than 150 mm for any wall
Ä The design Procedure for
shear wall :
The axial load strength as the following equation according
to ACI-99 code:
k = 1.0 (for
unrestrained walls against rotation at both ends)
= 0.7
Ag: area of one meter strip
of bearing wall(m²).
f`c: concrete compression
strength.
Pu is
the max factor load subjected to the wall.
Ø And
for steel reinforcement :
• For vertical steel, according to ACI-02 code,
the minimum ratio of vertical reinforcement area to gross concrete area shall
be: 0.0025 .
• For
horizontal steel, according to ACI-02 code, the minimum ratio of horizontal
reinforcement area to gross concrete area shall be: 0.0015.
Design calculation:
¯ Total height
of the shear wall:
H= Number of stories *height of
story
=3 * 4 =12 m
The min thickness of wall according to ACI(14.5.3.1)
T =Lc/25=3/25=0.12m
¯ Take thickness
of shear wall = 20 cm = 0.2 m
- Own weight /M.R = 0.2 * 1 *
12 * 24 * 1.4 = 80.64 KN/ M.R
-
Load from stair = 228.7*4*2 = 914.8 KN / Parameter
= 914.8/
(6+6+2.8+2.8+1.9+1.9+1.8+1.8 )
= 36.6 KN/ M.R
- Load
from slab = wl/2 ----> 13.48*3/2 = 40.4 KN/ M.R
- Load from Beam= 68.9*4 =
275.6 KN / Parameter
= 275.6/
(7+7+2.6+2.6+1.8+1.8+1.6+1.6) ------>10.6 KN/M.R
¯ TOTAL LOAD =
80.64+36.6 +40.4 +10.6
= 168.24
KN/M.R
Φ Pnw
=0.55* Φ *fC`*Ag*[1-(KLC/32t)2]
-Φ:0.7
-k: relative stiffness.(K =1)
-Lc: height of the shear wall
between two supports, height of one floor (m).
-t:thickness of the bearing wall(m).
Φ Pnw = 0.55* Φ *fC`*Ag*[1-(KLC/32t) 2]
= 0.55* 0.7
*25*1000* 0.2* [1-(1*3/32*0.2) 2]
= 1502 KN/M.R
¯ Φ Pnw (1502KN/M.R) >>load in the shear wall
(168.24 KN/M.R)
ÄThe
shear wall needs minimum reinforcement.
Ø Area of steel As for vertical direction:
Ast = 0.0025*Ag
=
0.0025*200*1000
=
500 mm2.
Use 5Φ12mm/m.
(As=565mm2) .
Ø Area of steel As for the
horizontal direction:
Ast = 0.0015*Ag
=
0.0015*200*1000
=
300 mm2.
Use 5Φ10mm/m.
(As=339mm2) .
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Foundation
Design
|
Introduction:
Foundation is the part of a structure that usually placed below the
surface of the ground that transmits the load to the soil or rock, soil
compressive stresses caused the supported structure to settle .
The foundation is generally considered to
be the entire lowermost supporting part of the structure. Normally, a footing
is the last, or nearly the last, structural element of the foundation through
which the loads pass. A footing has as its function, the requirement of
spreading out the superimposed load so as not to exceed the safe capacity of
the underlying material, usually soil, to which it delivers the load.
Additionally, the design of footings must take into account certain practical
and, at times, legal considerations.
Square footing will be chosen for most
footing system, and one combined footing, and wall footing surrounding under
stone walls and shear wall for staircase.
The design of foundation of structures
generally requires knowledge of such factors as (a)
the load that will be transmitted by the superstructure to the foundation
system, (b) the requirements of the
local building code, (c) the behavior
and stress-related deformability of soils that will support the foundation
system, (d) the geological conditions of
the soil under consideration.
Ä The more common types of
footings may be categorized as follows:
1)
Individual column footings
Are often termed
isolated spread footings and are generally square. However, if space
limitations exist, the footing may be rectangular in shape.
2)
Wall footing
Support walls,
which may be either bearing or nonbearing walls.
3)
Combined footings:
Support two or
more columns and may be either rectangular or trapezoidal in shape. If two
isolated footings are joined by a strap beam, the footing is sometimes called a
cantilever footing.
4)
Mat foundations:
Large continuous
footings that support all columns and walls of a structure. They are commonly
used where undesirable soil conditions prevail.
5)
Pile caps or pile footings:
Serve to transmit column loads o a group of piles, which will, in turn,
transmit the loads to the supporting soil though friction or to underlying rock
in bearing.
Ø Foundations may be also classified based
on where the load is carried by the ground, producing:
¯Shallow foundations:
Termed bases, footings, spread footings, or mats, the depth is generally
D/B <= 1 but may be somewhat more.
¯Deep foundations:
Piles,
drilled piers, or drilled caissons. Lp/B >= 4+ with.
Design Procedure:
¯ Single footings:
Determine
the area of footing:
This is the first step on designing the foundation; this is done by
dividing the ultimate un-factored load carried by the column supported by the
footing on the soil bearing capacity, and as the following equation:
Ÿ
where: is the bearing capacity of the soil.
is the bearing capacity of the soil.
: 1.1 is a factor to include footing own
weight in calculations.
According to the number calculated from this equation will be used to
determine the footing dimension (B L), and the new area is notified as Aactual .
L), and the new area is notified as Aactual .
Determine
the effective depth of footing:
The depth of a
foundation is found according to shear loads carried by the footing, but
footings are subjected to two types of shear forces:
Wide
beam shear:
The critical
section for this type of shear is at distance d from the face of the supported
column in the long side of footing, the rest length of the long side footing
will generated a shear force at this section, which must resist this force
safely without failure.
This force should be less than or equal to the shear strength of the
wide beam shear, which is calculated by the following
equation:
Ÿ 
Where:
= 0.85 (according to
ACI-99 code )
= 0.85 (according to
ACI-99 code )
The ultimate load at this critical section is calculated from the
following equation:
Ÿ 
Where: q is the actual soil pressure under the foundation and equal:
Ÿ 
For safe design:
Ÿ 
So the design equation will be the next:
Ÿ 
Find d to
satisfy wide beam shear.
Punching
shear:
The critical section for this type of shear is at distance d/2 from the
face of the supported column around it (at long side and short one of footing).
This area of the footing will generate a shear force at this section, which
must resist this force safely without failure.
This force should be less than or equal to the shear strength of the
punching shear, which is calculated by the following equation:
Ÿ 
The ultimate shearing force at this critical
section is calculated from the following equation:
Ÿ Vu = Pu – qu
(d+W1)*(d+W2)
qu is the same calculated from equation 9.4.Find d to satisfy
punching shear.
Ÿ Choose the largest d calculated from the both
critical sections.
Ÿ The total depth of foundation is then equal to d +
concrete cover.
Determine
the reinforcement:
This is the last step in designing the footing; this step is done by
taking a unit strip from the face of column to the border of footing and
determines the moment generated by it.
This
strip should be taken from the other direction.
The moment will equal to:
This moment will be used in equation 3.3 which is:
to calculate the term 
.
.
Then
using 
table, find
the value of .
Find 
(steel ratio) by substitution the value of which is:
(steel ratio) by substitution the value of which is:
As for
beams and ribs this value of should be tested for 
and
.
Finally
find the steel area by using the following equation:
O
I will use allowable bearing capacity
of soil 250 KN/m2 in my
project and 5 cm cover.
And I will
use 1.5m to the face of footing.
Design of Second footing (F1):
Column = 0.2 *
0.4
Pu = 731.3 KN
P service = 731.3
/ 1.55 = 471.8 KN
qall.
= 250 KN/ m2
Cover = 5 cm
gSoil=18
KN/m3
qe = 250– 18 * 1.5 – 24 * 0.3
= 215.8 KN/ m2
Area = 471.8 / 215.8 = 2.2 m2
Use Length = 1.5
m & Width = 1.5 m with High = 0.3m
d = H – cover = 0.3
– 0.05 = 0.25 m
qu = Pu / A = 731.3 / 2.25 = 325 KN/ m2
Check for punching shear:
W1 = 400 mm & W2 =200 mm
bo =
2*(0.4+0.25) +2*(0.2+0.25) = 2.2 m
But, Vu = Pu - qu
(C1+d)*(C2+d)
Vu = 731.3– 325*(0.4+0.25)*(0.2+0.25)
= 636.2 KN.
And, 
From Vu = Ø Vc
Then d = Vu /
Ø(1/3)*(fc)0.5*bo
Ü d= 204.13
mm. < d= 250 mm O.K.
Check for Wide beam shear:
ÄFirst
Direction :
L/ =(1.5/2)-(0.4/2) – d = 0.3 m
Vu = Ø
Vc
But Vu = qu*L2* L/ ; where L2: the other
direction.
Vu = 325*1.5*0.3
= 146.3 KN.
Then d = Vu/ Ø (1/6)*(fc) 0.5*b d= 137.7mm
Ü d= 137.7
mm. < d= 250 mm O.K. .
ÄSecond
Direction :
L/ =(1.5/2)-(0.2/2) – d = 0.4 m
Vu = Ø
Vc
But Vu = qu*L2* L/ ; where L2: the other
direction.
Vu = 325*1.5*0.4
= 195 KN.
Then d = Vu/ Ø (1/6)*(fc) 0.5*b d
= 183.5 mm
Ü d= 183.5
mm. < d= 250 mm O.K.
Ø Use footing depth(h)= 30cm
= 0.3 m
Find reinforcement in
footing:
ÄFirst
Direction :
L = 1.5 m & W1 = 0.4m
with width = 1 m
L1/
=(1.5/2)-(0.4/2)= 0.55 m
Mu = (qu) *( L1/
)2 / 2 = 325 * ( 0.55)2
/ 2 = 49.2 KN . m
Mu = f * b * d 2 * fy * r ( 1 – 0.59 * r * (fy / fc) )
49.2 * 106 = 0.9 * 1000 * (250)2 * 414 * r ( 1 – 0.59 * r
* (414/ 25) )
r = 0.001332 < rmin = 0.00338 So use
rmin
Ü As = r
* b * d = 0.00338 * 1000 *250
= 845.4 mm2
(Use 6 F
14 / m.r = 924 mm2)
ÄSecond
Direction :
L = 1.5 m & W1 = 0.2 m
with width = 1 m
L1/ =
0.65 m
Mu = (qu) *( L1/
)2 / 2 = 325 * ( 0.65)2
/ 2 = 68.7 KN . m
Mu = f * b * d 2 * fy * r ( 1 – 0.59 * r * (fy / fc) )
68.7* 106 = 0.9 * 1000 * (250)2 * 414 * r ( 1 – 0.59 * r
* (414/ 25) )
Ü r = 0.0029 < rmin
= 0.00338 use rmin
As = r * b * d = 0.00338 * 1000 * 250
= 845.4 mm2
(Use 6 F
14 / m.r = 924 mm2)
Design of Second footing (F2):
Column = 0.3 *
0.4
Pu = 1007.7 KN
P service = 1007.7
/ 1.55 = 650.2 KN
qall.
= 250 KN/ m2
Cover = 5 cm
gSoil=18
KN/m3
qe = 250 – 18 * 2 – 24 * 0.35
= 205.6 KN/ m2
Area = 650.2 / 205.6 = 3.162 m2
Use Length = 2.0
m & Width = 1.7 m with High = 0.35 m
d = H – cover = 0.35
– 0.05 = 0.30 m
qu = Pu / A = 1007.7 / 3.4 = 296.4 KN/ m2
Check for punching shear:
W1 = 400 mm & W2 =300 mm
bo =
2*(0.4+0.3) +2*(0.3+0.3) = 2.6 m
But, Vu = Pu - qu
(C1+d)*(C2+d)
Vu = 1007.7– 296.4*(0.4+0.3)*(0.3+0.3)
= 883.2 KN.
And,
From Vu = Ø Vc
Then d = Vu /
Ø(1/3)*(fc)0.5*bo
Ü d= 239.8 mm. < d= 300 mm O.K.
Check for Wide beam shear:
ÄFirst
Direction :
L/ =(2/2)-(0.4/2) – d = 0.5 m
Vu = Ø Vc
But Vu = qu*L2* L/ ; where L2: the other
direction.
Vu = 296.4*1.7*0.5
= 252 KN.
Then d = Vu/ Ø (1/6)*(fc)
0.5*b d = 209.3 mm
Ü d= 209.3 mm. < d= 300 mm O.K. .
ÄSecond
Direction :
L/ =(1.7/2)-(0.3/2) – d = 0.4 m
Vu = Ø Vc
But Vu = qu*L2* L/ ; where L2: the other
direction.
Vu = 296.4*2*0.4 =
237.1 KN
Then d = Vu/ Ø (1/6)*(fc)
0.5*b d = 167.4 mm
Ü d= 167.4 mm. < d= 300 mm O.K.
Ø Use footing depth(h)= 35cm
= 0.35 m
Find reinforcement in footing:
ÄFirst
Direction :
L = 2.0 m & W1 = 0.4m
with width = 1 m
L1/
=(2/2)-(0.4/2)= 0.8 m
Mu = (qu) *( L1/
)2 / 2 = 296.4 * ( 0.8 )2
/ 2 = 94.8 KN . m
Mu = f * b * d 2 * fy * r ( 1 – 0.59 * r * (fy / fc) )
94.8 * 106 = 0.9 * 1000 * (300)2 *
414 * r ( 1 – 0.59 * r
* (414/ 25) )
r = 0.001332 < rmin = 0.00338 use rmin
Ü As = r
* b * d = 0.00338 * 1000 *300
= 1014.5 mm2
(Use 6 F
16 / m.r = 1206 mm2)
ÄSecond
Direction :
L = 1.7 m & W1 = 0.3 m
with width = 1 m
L1/ =
0.7 m
Mu = (qu) *( L1/
)2 / 2 = 296.4 * ( 0.7 )2
/ 2 = 72.6 KN . m
Mu = f * b * d 2 * fy * r ( 1 – 0.59 * r * (fy / fc) )
72.6* 106 = 0.9 * 1000 * (300)2 *
414 * r ( 1 – 0.59 * r
* (414/ 25) )
r = 0.0029 < rmin
= 0.00338 use rmin
Ü As = r
* b * d = 0.00338 * 1000 * 300
= 1014.5 mm2
(Use 6 F
16 / m.r = 1206 mm2)
Design of Third footing (F3):
Column = 0.3 *
0.4
Pu = 1162.6 KN
P service = 1162.6
/ 1.55 = 750.1 KN
qall.
= 250 KN/ m2
Cover = 5 cm
gSoil=18
KN/m3
qe = 250 – 18 *2 – 24 * 0.4
= 204.4 KN/ m2
Area = 750.1/ 204.4= 3.67 m2
Use Length = 2 m & Width = 2 m with High = 0.4 m
d = H – cover = 0.4
– 0.05 = 0.35 m
qu = Pu / A = 1162.6 / 4 = 290.7 KN/ m2
Check for punching shear:
W1 = 400mm & W2 =300 mm
bo = 2*(0.4+0.35)
+2*(0.3+0.35) = 2.8 m
But, Vu = Pu - qu (C1+d)*(C2+d)
Vu = 1162.6 –
290.7*(0.4+0.45)*(0.3+0.45) = 977.3 KN.
And,
From Vu = Ø Vc
Then d = Vu / Ø(1/3)*(fc)0.5*bo
Ü d= 246.4 mm. < d= 350 mm O.K.
Check for Wide beam shear:
ÄFirst
Direction :
L/ =(2/2)-(0.4/2) – d = 0.45 m
Vu = Ø
Vc
But Vu = qu*L2* L/ ; where L2: the other
direction.
Vu = 290.7*2*0.45
= 261.6 KN.
Then d = Vu/ Ø (1/6)*(fc) 0.5*b Ü d = 184.7 mm
ÄSecond
Direction :
L/ =(2/2)-(0.3/2) – d = 0.5 m
Vu = Ø Vc
But Vu = qu*L2* L/ ; where L2: the other
direction.
Vu = 290.7*2*0.5 = 290.7
KN.
Then d = Vu/ Ø (1/6)*(fc)
0.5*b Ü d = 205.2 mm
Ø Use footing depth(h)= 40cm
= 0.4 m
Find reinforcement in
footing:
ÄFirst
Direction :
L = 2m & W1 = 0.4 m
with width = 1 m
L1/
=(2/2)-(0.4/2)= 0.8 m
Mu = (qu) *( L1/
)2 / 2 = 290.7 * ( 0.8 )2
/ 2 = 93.02 KN . m
Mu = f * b * d 2 * fy * r ( 1 – 0.59 * r * (fy / fc) )
93.02 * 106 = 0.9 * 1000 * (350)2 * 414 * r ( 1 – 0.59 * r
* (414 / 25) )
r = 0.0020802< rmin = 0.00337816 use rmin
Ü As = r * b * d =
0.00338 * 1000 *350
= 1183.6 mm2
(Use 5 F18
/ m.r =1272 mm2)
ÄSecond
Direction :
L = 2 m & W1 = 0.3 m
with width = 1 m
L1/ =(2/2)-(0.3/2)=
0.85 m
Mu = (qu) *( L1/ )2 / 2
= 290.7 * ( 0.85)2 / 2 = 105
KN . m
Mu = f * b * d 2 * fy * r ( 1 – 0.59 * r * (fy / fc) )
105* 106 = 0.9 * 1000 * (350)2 * 414 * r ( 1 – 0.59 * r
* (414 / 25) )
r = 0.002354
Ü use rmin
As = r * b * d = 0.00338 * 1000 *350
= 1183.6 mm2
(Use 5F
18 / m.r = 1183.6 mm2)
Design of fourth footing (F4):
Column = 0.3*
0.5
Pu = 1553.8 KN
P service =1553.8
/ 1.55 = 1002.45 KN
qall.
= 250 KN/ m2
Cover = 5 cm
gSoil=18
KN/m3
qe = 250 – 18 * 2 – 24 * 0.45
= 203.2 KN/ m2
Area = 1002.45 / 203.2 = 4.93 m2
Use Length = 2.5 m & Width = 2 m with High = 0.45 m
d = H – cover = 0.45
– 0.05 = 0.4 m
qu = Pu / A = 1553.8 / 5 = 310.8 KN/ m2
Check for punching shear:
W1 = 500mm & W2 =300 mm
bo =
2*(0.5+0.40) +2*(0.3+0.40) = 3.2 m
But, Vu = Pu - qu
(C1+d)*(C2+d)
Vu = 1553.8 –
310.8*(0.5+0.40)*(0.3+0.40) = 1358 KN.
And,
From Vu = Ø Vc
Then d = Vu /
Ø(1/3)*(fc)0.5*bo
Ü d= 300 mm. < d= 400 mm O.K.
Check for Wide beam shear:
ÄFirst
Direction :
L/ =(2.5/2)-(0.5/2) – d = 0.6 m
Vu = Ø
Vc
But Vu = qu*L2* L/ ; where L2: the other
direction.
Vu = 310.8*2*0.6
= 373 KN.
Then d = Vu/ Ø (1/6)*(fc) 0.5*b Ü d = 263.3 mm
ÄSecond
Direction :
L/ =(2/2)-(0.3/2) – d = 0.45 m
Vu = Ø Vc
But Vu = qu*L2* L/ ; where L2: the other
direction.
Vu = 310.8*2.5*0.45
= 349.7 KN
Then d = Vu/ Ø (1/6)*(fc)
0.5*b Ü d = 197.5 mm
Ø Use footing depth(h)= 45cm
= 0.45 m
Find reinforcement in
footing:
ÄFirst
Direction :
L = 2.5 m & W1 = 0.5 m
with width = 1 m
L1/ = 1.0
m
Mu = (qu) *( L1/
)2 / 2 = 310.8 * ( 1 )2
/ 2 = 155.4 KN . m
Mu = f * b * d 2 * fy * r ( 1 – 0.59 * r * (fy / fc) )
155.4 * 106 = 0.9 * 1000 * (400)2 * 414 * r ( 1 – 0.59 * r
* (414 / 25) )
r < rmin
= 0.00338 use rmin
As = r
* b * d = 0.00338 * 1000 * 400
= 1352.7 mm2
(Use 6 F
18 / m.r = 1527 mm2)
ÄSecond
Direction :
L = 2 m & W1 = 0.3 m
with width = 1 m
L1/ = 0.85
m
Mu = (qu) *( L1/
)2 / 2 = 310.8 * ( 0.85 )2 / 2 = 112.23 KN . m
Mu = f * b * d 2 * fy * r ( 1 – 0.59 * r * (fy / fc) )
112.23* 106 = 0.9 * 1000 * (400)2 * 414 * r ( 1 – 0.59 * r
* (414 / 25))
r
< rmin
= 0.00338 use rmin
Ü As
= r
* b * d = 0.00338 * 1000 * 400
= 1352.7 mm2
(Use 6 F
18 / m.r = 1527 mm2)
Design of fifth footing (F5):
Column =
0.3*0.5
Pu = 1843 KN
P service =
1843 / 1.55 = 1189.3 KN
qall.
= 250 KN/ m2
Cover = 5 cm
gSoil=18
KN/m3
qe = 250 – 18 * 1.5– 24 * 0.5
= 211KN/ m2
Area =1189.3 / 211 = 5.64m2
Use Length = 3 m & Width = 2 m with High = 0.5 m
d = H – cover = 0.5
– 0.05 = 0.45 m
qu = Pu / A = 1843 / 6 =
307.2 KN/ m2
Check for punching shear:
W1 = 500 mm & W2 =300 mm
bo =
2*(0.5+0.45) +2*(0.3+0.45) = 3.4m.
But, Vu
= Pu - qu (C1+d)*(C2+d)
Vu = 1843 –
307.2*(0.5+0.45)*(0.3+0.45) = 1624.12 KN.
And,
From Vu = Ø Vc
Then d = Vu /
Ø(1/3)*(fc)0.5*bo
Ü
d= 337.2 mm. < d= 450 mm O.K.
Check for Wide beam shear:
ÄFirst
Direction :
L/ =(3/2)-(0.5/2) – d = 0.8 m
Vu = Ø
Vc
But Vu = qu*L2* L/ ; where L2: the other
direction.
Vu = 307.2*2*0.8
= 491.52 KN.
Then d = Vu/ Ø (1/6)*(fc) 0.5*b Ü d = 347 mm
.
ÄSecond
Direction :
L/ =(2/2)-(0.3/2)
– d = 0.4 m
Vu = Ø
Vc
But Vu = qu*L2* L/ ; where L2: the other
direction.
Vu = 307.2*3*0.4
= 368.65KN.
Then d = Vu/ Ø (1/6)*(fc) 0.5*b Ü d = 173.5mm
Ø Use footing depth(h)= 50cm
= 0.5 m
Find reinforcement in
footing:
ÄFirst
Direction :
L = 3 m & W1 = 0.5 m
with width = 1 m
L1/
=(3/2)-(0.5/2)= 1.25 m
Mu = (qu) *( L1/
)2 / 2 = 307.2 * ( 1.25)2
/ 2 = 240 KN . m
Mu = f * b * d 2 * fy * r ( 1 – 0.59 * r * (fy / fc) )
240 * 106 = 0.9 * 1000 * (450)2 *
414 * r ( 1 – 0.59 * r
* (414 / 25) )
r = 0.003286 < rmin = 0.00338 So use rmin
Ü As = r
* b * d = 0.00338 * 1000 *450
= 1521.7 mm2
(Use 7 F18
/ m.r =1780.4 mm)
ÄSecond
Direction :
L = 2 m & W1 = 0.3 m
with width = 1 m
L1/ =
0.85 m
Mu = (qu) *( L1/
)2 / 2 = 307.2 * ( 0.85)2
/ 2 = 111 KN . m
Mu = f * b * d 2 * fy * r ( 1 – 0.59 * r * (fy / fc) )
111* 106 = 0.9 * 1000 * (450)2 * 414 * r ( 1 – 0.59 * r
* (414/ 25) )
Ü r = 0.001493 < rmin = 0.00338
So use rmin
As = r
* b * d = 0.00338 * 1000 *450
= 1521.7 mm2
(Use 7 F
18 / m.r = 1780.4 mm2)
 |
Load from slab
= 19.31/ 0.52 = 37.14 KN/m
Load from wall
= 20.25 KN/m
P service form
slab = 20.64/1.55= 23.96 KN/m
Ü Uniform service load (W service) = 23.96
* 4 + 20.25 * 4
= 176.84 KN/m
qall.
= 250 KN/ m2
gsoil=18
KN/m3
qeff = 250 – 18 *1.5 – 24 * 0.3
= 215.8 KN/ m2
Area = W/ qeff =
176.84 / 215.8 = 0.8195 m2
Use Area = 1 m2
Use Length = 1 m & Width = 1m
qu = Pu / A = 208.1/ 1 = 208.1 KN/ m2
Check for Wide beam shear:
ÄFirst
Direction :
L/ = (1/2)-(0.3/2)– d = 0.1 m.
Vu = Ø
VC
But Vu = qu*L2* L/ =
208.1*0.1*1 = 20.81 KN.
Then d = Vu/ Ø (1/6)*(fc) 0.5*b, then d = 29.37
< 250 mm.
Use h = 300 mm.
Find reinforcement in footing:
¯Short direction (Main steel)
L/ = (L
– t)/2 = 0.35 m
Mu = (qu) *( L1/
)2 / 2 = 208.1 * ( 0.35)2
/ 2 = 12.75 KN . m
Mu = f * b * d 2 * fy * r ( 1 – 0.59 * r * (fy / fc) )
12.75 * 106 = 0.9 * 1000 * (300)2 * 414 * r ( 1 – 0.59 * r
* (414 / 25) )
Ü
r < rmin
= 0.00338 use rmin
As = r
* b * d = 0.00338 * 1000 * 300
= 1014 mm2
(Use 7 F
14 / m.r = 1075 mm2)
¯Long direction:
use shrinkage and temperature steel according
to ACI code 7.12
As=0.0018Ag
As=0.0018*1000*300=540 mm2/m2
(Use 7 F
12 / m.r = 792 mm2)
TOTAL LOAD =
80.64+36.6 +40.4 +10.6
= 168.24 KN/M.R
Ps = 168.24/1.55
=108.54 KN/M.R
qall.
= 250 KN/ m2
gsoil=18
KN/m3
qeff = 250 – 18 * 1.5 – 24 * 0.3
= 215.8 KN/ m2
Area = W/ qeff =
108.54 / 215.8 = 0.51 m2
Use Area = 0.8 m2
Use Length = 1 m & Width = 0.8m
qu = Pu / A = 168.24 /0.8 = 210.3 KN/ m2
Check for Wide beam shear:
ÄFirst
Direction :
L/ = (0.8/2)-(0.2/2) – d = 0.05 m.
Vu = Ø
VC
But Vu = qu*L2* L/ = 210.3*0.05*1
= 10.52 KN.
Then d = Vu/ Ø (1/6)*(fc) 0.5*b, then d = 14.85
< 250 mm.
Use h
= 300 mm.
Find reinforcement in footing:
¯Short direction (Main steel)
L/ = (L
– t)/2 = 0.3 m
Mu = (qu) *( L1/
)2 / 2 = 210.3 * ( 0.3)2
/ 2 = 9.46 KN . m
Mu = f * b * d 2 * fy * r ( 1 – 0.59 * r * (fy / fc) )
9.46 * 106 = 0.9 * 1000 * (300)2 * 414
* r
(1 – 0.59 * r
* (414 / 25) )
r < rmin
= 0.00338 use rmin
Ü As = r * b * d =
0.00338 * 1000 * 300
= 1014 mm2
(Use 7 F
14 / m.r = 1075 mm2)
¯Long direction:
use shrinkage and temperature steel according
to ACI code 7.12
As=0.0018Ag
As=0.0018*1000*400=720 mm2/m2
(Use 7 F
12 / m.r = 792 mm2)
The tie beams are not designed because the load and the type of
the load are not known.
-use tie beams with 30*60 cm
-use 4Φ16mm……(in
the top.)
-use 4Φ16mm…..
(in the bottom.)
-use 2Φ16mm
…..(in the middle.)
Ä Seat:
Using
seat with a dimension (70*70*70) cm.
O In This project there
is no seats because the Maximum distance between two footings is less than 5
meters.
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