Consider a project with design procedure for a reinforced concrete building (Residetial
building), consisting of four flours and a plan area of (383.0)m2.
The building while the design will be divided in to its structural component.
The chapters of this
project in part (Ι) discuss the
component or the parts of the building ; i.e, ribs, beams, columns, footings
and walls.
Part (ΙΙ) of this project is
about concrete water tank and its design , using working stress method to
design It .
For the design procedure in this
project , The building code requirement for reinforced concrete(ACI 318-99) Used to design the whole building.
For Design &
analysis of this project:
fc' = 25
Mpa
fy =
414 Mpa
fyv = 280 Mpa
The most important characteristic of any structural member
is its actual strength, which must be large enough to resist, with some margin
to spare, all foreseeable loads that may act on it during the life of the
structure, without failure or other distress. It is logical, therefore, to
proportion members, i.e., to select concrete dimensions and reinforcement, so
that member strengths are adequate to resist forces resulting from certain
hypothetical overload stages, significantly above loads expected actually to
occur in service. This design concept is known as strength design.
For reinforced
concrete structures at loads close to and at failure, one or both of the
materials, concrete and steel, are invariably in their nonlinear inelastic
range. That is, concrete in a structural member reaches its maximum strength
and subsequent fracture at stresses and strains far beyond the initial elastic
range in which stresses and strains are fairly proportional. Similarly, steel
close to and at failure of the member is usually stressed beyond its elastic
domain into and even beyond the yield region. Consequently, the nominal
strength of a member must be calculated on the basis of this inelastic behavior
of the materials.
A member
designed by the strength must also perform in a satisfactory way under normal
service loading. For example, beam deflections must be limited to acceptable
value and the number and width of flexural cracks at service loads must be
controlled. Serviceability limit conditions are an important part of the total
design, although attention is focused initially on the strength
concrete is a stone-like material obtained by permitting a carefully
proportioned mixture of cement, sand and gravel or other aggregate, and water
to harden in forms of the shape and dimensions of the desired structure. The
bulk of the material consists of fine and course aggregate. Cement and water
interact chemically to bind the aggregate particles into a solid mass.
Additional water, over and above that needed for this chemical reaction is
necessary to gibe the mixture the workability that enables it to fill the forms
and surround the embedded reinforcing steel prior to hardening, concretes in a
wide range of properties can be obtained by appropriate adjustment of the
proportions of the constituent materials. Special cements (such as high early
strength cement), admixture (such as plasticizers, air-entraining agents,
silica fume, and fly ash), and special curing methods (such as steam-curing)
permit an even wider variety of properties to be obtained .
These properties depend to a very substantial degree on the proportions
of the mix, on the thoroughness with which the various constituents are
intermixed, and on the condition of humidity and temperature in which the mix
is maintained from the moment it is placed in the forms until it is fully
hardened. The process of controlling these conditions is known as curing. To
protect against the unintentional production of substandard concrete, a high
degree of skillful control and supervision is necessary throughout the process,
from the proportioning by weight of the individual components, through mixing
and placing, until the completion of curing .
The factor that make concrete a
universal building material are so pronounced that it has been used, in more
primitive kinds and ways than at present, fro thousands of years, probably
beginning in Egyptian antiquity. The facility with which, while plastic, it can
be deposited and made to fill forms or molds of almost any practical shape is
one of these factors. Its high fire and weather resistance are evident
advantages. Most of the constituent materials, with the exception of cement and
additives, are usually available at low cost locally or at small distances from
the construction site, its compressive strength, like that of natural stones,
is high, which makes it suitable for members primarily subject to compression,
such as columns and arches. On the other hand, again as in natural stones, it
is a relative brittle material whose tensile strength is small members that are
subject to tension either entirely (such as in tie rods) or over part of their
cross sections (such as in beams or other flexural members) .
To offset this
limitation, it was found possible, in the second half of the nineteenth
century, to use steel with its high tensile strength to reinforce concrete,
chiefly in those places where its small tensile strength would limit the carrying
capacity of the member. The reinforcement, usually round steel rods with
appropriate surface deformation to provide interlocking, is placed in the forms
in advance of the concrete. When completely surrounded by the hardened concrete
mass, it forms an integral part of the member. The resulting combination of two
materials, known as reinforced concrete, combines cost, good weather and fire resistance, good compressive strength, and
excellent formability of concrete and the high tensile strength and much
greater ductility and toughness of steel. It is this combination that allows
the almost unlimited range of uses and possibilities of reinforced concrete in
the construction of buildings, bridges, dams, tanks, reservoirs, and a host of
other structures.
¯
Aggregates:
Natural aggregates are generally
classified as fine and coarse, fine aggregate, or sand, is any material that
will pass a No. 4 sieve, i.e., a sieve with four openings per linear inch.
Material coarser than this is classified as coarse aggregate, or gravel. When
favorable gradation is desired, aggregate are separated by sieving into two or
three size groups of sand and several size groups of coarse aggregate. These
can then be combined according to grading charts to result in a densely packed
aggregate. The maximum size of coarse aggregate in reinforced concrete is
governed by the requirement that it shall easily fit into the forms and between
the reinforcing bars. For the purpose it should not be larger than one-fifth of
the narrowest dimension of the forms of one-third of the depth of slabs, nor
three-quarters of the minimum distance between reinforcing bars. Requirements
for satisfactory aggregates are found in ASTM C33, “Standard Specification for
Concrete Aggregates,” .
¯
Cement:
A cementations material is one
that has the adhesive and cohesive properties necessary to bond inert
aggregates into a solid mass of adequate strength and durability. This
technologically important category of materials includes not only cements
proper but also limes, asphalts, and tars as they are used in road building,
and others. For making structural concrete, so-called hydraulic cements are
used exclusively. Water is needed for the chemical process (hydration) in which
the cement powder sets and hardens into one solid mass. Of the carious
hydraulic cements that have been developed, Portland cement, which was first
patented in England
Portland cement is a finely
powdered, grayish material that consists chiefly of calcium and aluminum
silicates. The common raw materials from which it is made are lime-stones,
which provide CaO, and clays or shale’s, which furnish SiO2 and Al2O3.
These are ground, blended fused to clinkers in a kiln, cooled, and ground to
the required fineness. The material is shipped in bulk or in bags containing 94
lb of cement.
Over the years, five standard types of Portland cement
have been developed, Type I: normal Portland cement, is used for over 90
percent of construction in Jordan
achieved using Type I at 28 days. Type III Portland cement contains the
same basic compounds as Type I, but the relative proportions differ and it is
ground more finely.
When cement is mixed with
water to from a soft paste, it gradually stiffens until it becomes a solid.
This process is known as setting and hardening. The cement is said to have set
when it has gained sufficient rigidity to support an arbitrarily defined
pressure, after which it continues for a long time to harden, i.e., to gain
further strength. The water in the paste dissolves material at the surfaces of
the cement grains and forms a gel which gradually increases in volume and
stiffness. This leads to a rapid stiffening of the paste 2 to 4 hours after
water has been added to the cement. Hydration continues to proceed deeper into
the cement grains, at decreasing speed, with continued stiffening and gardening
of the mass. In ordinary concrete the cement is probably never completely
hydrated. The gel structure of the gardened paste seems to be the chief reason
for the volume changes that are caused in concrete by variations in moisture,
such as the shrinkage of concrete as it dries.
¯
Admixtures:
In addition to main components of
concretes, admixtures are often used to improve concrete performance. There are
admixtures to accelerate or rated setting and hardening, to improve
workability, to increase strength, to improve durability, to decrease
permeability, and to impart other properties. The beneficial effects of
particular admixtures are well established. Chemical admixtures should meet the
requirements of ASTM C494, “Standard Specification for Chemical Admixtures for
Concrete.”
Air-entraining agents are probably the
most commonly used admixtures at the present time. They cause the entrainment
of air in the form of small dispersed bubbles in the concrete. These improve
workability and durability (chiefly resistance to freezing and thawing), and
reduce segregation during placing. They decrease concrete density because of
the increased void ratio and thereby decrease strength; however, this decrease
can be partially offset by a reduction of mixing water without loss of
workability the chief use of air-entrained concretes is in pavements, but they
are also used for structures, particularly for exposed elements.
Accelerating admixtures are used to
reduce setting time and accelerate early strength development. Calcium chloride
is the most widely used accelerator because of its cost effectiveness, but it
should be used with caution in prestressed concrete, or in reinforced concrete
in a moist environment, because of its tendency to promote corrosion of steel.
Set-retarding admixtures are used
primarily to offset the accelerating effect of high ambient temperature and to
keep the concrete workable during the entire placing period. This helps to
eliminate cracking due to form deflection and also keeps concrete workable long
enough that succeeding lifts can be placed without the development of “cold”
joints.
For most effective reinforcing action, it
is essential that steel and concrete deform together, i.e., that there be a
sufficiently strong bond between the tow materials to ensure that no relative
movements of the steel bars and the surrounding concrete occur, this bond is
provided by the natural roughness of the mill scale of hot-rolled reinforcing
bars, and by the closely spaced rib-shaped surface deformations with which
reinforcing bars are furnished in order to provide a high degree of
interlocking of the two materials.
In the history of reinforced concrete various
steels were used, ranging from soft steels, with yield strength of 30Ksi
(207Mpa), to hard steels of yield strength of 60 Ksi (413 Mpa).
The elongation or strain at failure depends
on the type of steel. The modulus of elasticity of steel, which is the slope of
the initial strength line portion of the diagram, is constant equal to 29000Ksi
(200Gpa) for all type of steel.
Reinforcing steel is manufacturing in three forms:
Ø Plain bars
ØDeformed bars
Ø Plain deformed bars
Reinforcing steel is manufacturing in four yield
levels:
1- 40 Ksi
2- 50 Ksi
3- 60 Ksi
4- 75 Ksi
Steel
is used in two different ways in concrete structures: as reinforcing steel and
as prestressing steel. Reinforcing steel is placed in the forms prior to casing
of the concrete. Stresses in the steel, as in the gardened concrete, are caused
only by the loads on the structure, except for possible parasitic stresses from
shrinkage or similar causes. In contrast, in prestressed concrete structures
large tension forces are applied to the reinforcement prior to letting it act
jointly wit the concrete in resisting external loads, the steel for these two
uses are very different and will be discussed separately. And for prestressing
steel it is used in large structures such as bridges and dams.
Ribs
can be one-way or two-way according to slab dimension, for large and
discontinuous slab spans it is recommended to use the two-way rib system which
depend on transfer loads in tow directions, this means that the load value will
be reduced causing the deflection to be small.
In the
over side the one-way rib, the most popular system, is used when the slab spans
are relatively small and there is continuity between them, the load will
transfer in the direction of the ribs only (in one direction) so the deflection
will be controlled by the continuity of the spans.
Analysis of ribs & beams was by using
the Prokon program,
which determines the shear, max. Moments & deflections through taking all
possible cases of loading (by the Automatic Envelop)
to find the max. & Critical moments & shear forces.
Loads
Ø The Loads that act on the
structures can be divided into two main categories:
Dead loads & Live loads.
Dead Loads: are constant in magnitude 7 fixed in
locations through out the lifetime of the structure. Usually the major part of
the dead load is the weight of the structure itself. This can be calculated
with good accuracy from the design configuration, dimension of structure, &
the density of the materials for buildings, floor fill, and finish floor. &
plastered ceiling are usually including as dead loads.
Live Loads: are consisting chiefly of occupancy loads in
building. They may be either fully or partially in place or not presence at all
& may also change in location. There magnitude & distribution at any
given time are uncertain & even their maximum intensities through the
lifetime of the structure are not known with precision. The minimum live loads
for which the floor & roof of a building should be designed are usually
specified in the building code that governs at the site of the construction.
Ø And we can mention another Category Of loads That is the
Environomental load;
Environmental Loads: Environmental load are those
loads caused by environment such as snow load, wind pressure and section,
earthquake loads, soil pressure on subsurface portion of structure, loads form
possible pounding of rainwater on the surface and forces caused from
temperature differentials.
These loads
like live loads, at any given times are uncertain both in magnitude and
distribution.
Ø Minimum thickness of ribbed slab
Using the table 9-5.a in the ACI code that gives the minimum thickness
of beam or one way slabs:
|
Minimum thickness, h
|
||||
|
|
Simply supported
|
One end continuous
|
Both end continuous
|
Cantilevers
|
|
2.1.1
Solid
one-
2.1.2
Way
Slab
|
L/20
|
L/24
|
L/28
|
L/10
|
|
2.1.3
Beams
or Ribbed
2.1.4
One-way
Slab
|
L/16
|
L/18.5
|
L/21
|
L/8
|
ÄOne end continues:
L= 4.5 H= L/18.5
H=4.5/18.5=0.243243 m
ÄSimply supported:
L= 4.93 H= L/16
H= 4.93/16= 0.308125 m
So that we use thickness for slab = 0.31m = 31 cm
¯ Using
24cm Block+7cm slab h=310mm
Þ No need to check
deflection
Load Calculations:
¯ Dead
load Calculations:
study area = (1) *(1.04) =
1.04 m2
Weights:
Plaster = 1.04 * 1 * (20/10) *21 = 43.68
Kg/ 1.04 m2
Top mat = 0.07 * 1.04* 1 * 2500 = 182 Kg/ 1.04 m2
Sand = 0.1 * 1.04
*1 * 1800 = 187.2 Kg/ 1.04 m2
Mortar = 0.03 * 1.04 *1 * 2000 = 62.4 Kg/ 1.04 m2
Tile = 1.04*50 = 52 Kg/ 1.04
m2
Ribs = 0.12 * 1 *0.24 *2500*2 = 144 Kg/ 1.04 m2
Weight of Hollow Block = 0.15 KN weight of (5) block = 0.75 KN
For partitions:
Add 150 Kg/ m2
Load /m =940*0.52=488.8kg/m of rib
DL=488.8*9.81/1000=4.8KN/m
¯ live
load Calculations:
LL= 200 Kn/ m2 ….
(For Residual Rooms)
LL=200*0.52*9.81/1000=1.04
KN/m
Design of Ribs:
Note:
h = 310mm d = 285 mm bf = 520m
bw = 120mm .
Ø= 0.9 f’c = 25 MPa fy = 414 MPa
œ Flexural Design Procedure:
• The moments are determined by using Moment
distribution and (Prokon program), which is “Prokon
structural analyses”.
• The ultimate moment can be calculated by
where ; Ø= 0.9 .
cod 9.3.2.1 &
•
Calculate the value w ( 1 – 0.59 w ) = 
• From ( w ) table and the value
of w ( 1 – 0.59 w ); find w .
Find (steel ratio) by
substituting the value of w in:
• Compare the value of calculated in the
previous step with and
Where ;
but
not less than ;
• Find the steel area by using this equation:
•
Choose an appropriate steel diameter to satisfy the steel area required.
= 0.00337*120* 295 = 119 mm2
• For negative
moment at end supports, use the following value for negative moment where the
support is a beam :
œ Shear Design Procedure:
Shear force in rib
is relatively small especially when compared with shear force in beams, which
considered a large force and must be designed with care, so the rib shear
reinforcement is usually be minimum (1 stirrup 
8mm at 20cm). And the procedure is explained next:
8mm at 20cm). And the procedure is explained next:
•
the
ultimate shear force (Vu) should be less or equal the shear strength
capacity (Vn), in formula:
Vn), in formula:
where
Vn= Vs +Vc
• According to ACI-99 code, Vc is given by the following
equation :
• According to ACI-99 code, Vc,rib
shall be permitted to be
10% more than Vc ,beam as
follow:
Vc,rib =
1.1 Vc
• Calculate the
spacing by using the formula :
• the
limits of the spacing (s) according to ACI- code is given as the smallest value of the
following:
(use the smallest value)
O if 
use minimum steel reinforcement for shear:1 8mm at 20cm.
use minimum steel reinforcement for shear:1 8mm at 20cm.
Ribs Calculations:
|
Design Rib # 1 ( R1)
|
Length of the rib ( L) = 4.55 m
, Cover = 2.5 cm = 25 mm , H = 31 Cm = 310 mm
D = H – cover = 310 – 25 = 285 mm
Shear & Moment Diagram :
ρ
min = 1.4 / fy = 1.4 / 414 = 0.0033816
ρ
min = (fc )1/2 / 4 fy = (25 )1/2 / 4 * 414 = 0.0030193
ρ
balance = 0.85 * β (fc
/ fy) [ 600 / 600 + fy]
=
0.85 * 0.85 (25 / 414) * [600 / 600 +414]
=
0.02581611
ρ max =
0.75 ρ balance
= 0.75 *
0.02581611
=
0.019362
Mu / Ф * fc * b * d2
= w ( 1 – 0.59 w )
Span (1) :
@ M +ve
= 22.0 KN . m
Mu / Ф * fc *
b * d2 = w ( 1 – 0.59 w )
22.0 * 106 / [ 0.9 * 25 * 520 * 2852
] = 0.0231497 = w ( 1 – 0.59 w )
w = 0.0234749
ρ
= w (fc / fy)
= 0.0234749 ( 25 / 414) = 0.00141757
ρ < ρ min so that we use minimum steel
As = ρ * b * d
= 0.00141757 * 120 * 285 = 115.6 mm2
( use 2 Ф 12 =
226 mm2 )
Design
of Shear
Span (1)
= 0.85
* (25)0.5 * 120 * 285 / 6
= 24.225 KN
Vu = 16.3 KN
Ü Vu
< fVc
So that we
use minimum reinforcement
Assume 2-leg Ø 8 mm (Av
=101 mm2)
Min reinforcement must be used (Smax)
S= 3 * Av*fyv / b
=3
*101 *280 / 120 = 707 mm
S=0.75d=0.75*285=213.75 mm
S=500mm
Ü Use
1 f 8 / 200 mm
|
Design Rib # 2 ( R2 ]
|
Length of the rib ( L) = 4.55—4.55--4.55—4.55—4.55--4.55 m , Cover = 2.5 Cm = 25 mm , H = 31 Cm = 310 mm
D = H – cover = 310 – 25 = 285 mm
Shear & Moment Diagram :
Span (1) :
@ M +ve
= 15.2 KN . m
Mu / Ф * fc *
b * d2 = w ( 1 – 0.59 w )
15.2 * 106 / [ 0.9 * 25 * 520 * 2852
] = 0.0159944 = w ( 1 – 0.59 w )
w = 0.016148
ρ = w (fc / fy) = 0.016148 (
25 / 414) = 0.000975
ρ < ρ min so that we use minimum steel
As = ρ * b * d
= 0.003381 * 120 * 285 = 115.6 mm2
( use 2 Ф 12 =
226 mm2 )
Span (2) :
@ M +ve
= 9.47 KN . m
Mu / Ф * fc *
b * d2 = w ( 1 – 0.59 w )
9.47
* 106 / [0.9 * 25 * 520 * 2852] = 0.009964 = w (1 – 0.59 w )
w = 0.010024219
ρ = w (fc / fy) = 0.010024219
( 25 / 414) = 0.00060533
ρ < ρ min so that we use minimum steel
As = ρ * b * d
= 0.003381* 120 * 285 = 115.6 mm2
( use 2 Ф 12 =
226 mm2 )
Span (3) :
@ M +ve
= 10.7 KN . m
Mu / Ф * fc *
b * d2 = w ( 1 – 0.59 w )
10.7
* 106 / [0.9 * 25 * 520 * 2852] = w (1 – 0.59 w )
w= 0.0113350
ρ = w (fc / fy) = 0.0113350 (
25 / 414) = 0.0006102
ρ < ρ min so that we use minimum steel
As = ρ * b * d
= 0.003381 * 120 * 285 = 115.6 mm2
( use 2 Ф 12 =
226 mm2 )
Span (4) :
@ M +ve
= 10.7 KN . m
Mu / Ф * fc *
b * d2 = w ( 1 – 0.59 w )
10.7
* 106 / [0.9 * 25 * 520 * 2852] = w (1 – 0.59 w )
w= 0.0113350
ρ = w (fc / fy) = 0.0113350 (
25 / 414) = 0.0006102
ρ < ρ min so that we use minimum steel
As = ρ * b * d
= 0.003381 * 120 * 285 = 115.6 mm2
( use 2 Ф 12 =
226 mm2 )
Span (5) :
@ M +ve
= 9.47 KN . m
Mu / Ф * fc *
b * d2 = w ( 1 – 0.59 w )
9.47
* 106 / [0.9 * 25 * 520 * 2852] = 0.009964 = w (1 – 0.59 w )
w =0.010024219
ρ = w (fc / fy) = 0.010024219
( 25 / 414) = 0.00060533
ρ < ρ min
so that we use minimum
steel
As = ρ * b * d
= 0.003381* 120 * 285 = 115.6 mm2
( use 2 Ф 12 =
226 mm2 )
Span (6) :
@ M +ve
= 15.2 KN . m
Mu / Ф * fc *
b * d2 = w ( 1 – 0.59 w )
15.2 * 106 / [ 0.9 * 25 * 520 * 2852
] = 0.0159944 = w ( 1 – 0.59 w )
w = 0.016148
ρ = w (fc / fy) = 0.016148 (
25 / 414) = 0.000975
ρ < ρ min so that we use minimum steel
As = ρ * b * d
= 0.003381 * 120 * 285 = 115.6
mm2
( use 2 Ф 12 =
226 mm2 )
Ø Negative moments :
Support
(2)
@ M -ve
= 18.6 KN . m
Mu / Ф * fc *
b * d2 = w ( 1 – 0.59 w )
18.6 * 106 / [0.9 * 25 * 120 * 2852]
= 0.08481 = w (1 – 0.59 w )
w = 0.08954
ρ = w (fc / fy) = 0.08954 ( 25 / 414) = 0.0054071
As = 0.0054071 * 120 * 285 = 184.92 mm2
( use 2 Ф 12 = 226 mm2 )
Support (3)
@ M -ve
= 13.3 KN . m
Mu / Ф * fc *
b * d2 = w ( 1 – 0.59 w )
13.3*
106 / [0.9 * 25 * 120 * 2852] = 0.060645 = w (1 – 0.59 w )
w = 0.06298
ρ = w (fc / fy) = 0.06298 ( 25
/ 414) = 0.003803
As
= 0.003803* 120 * 285 = 130.07 mm2
( use 2 Ф 12 = 226 mm2 )
Support (4)
@ M -ve
= 14.9 KN . m
Mu / Ф * fc *
b * d2 = w ( 1 – 0.59 w )
14.9
* 106 / [0.9 * 25 * 120 * 2852] = 0.0679411 = w (1 – 0.59 w )
w = 0.0709075
ρ = w (fc / fy) = 0.0709075 (
25 / 414) = 0.0042818
As
= 0.0042818* 120 * 285 =
146.439 mm2
( use 2 Ф 12 = 226 mm2 )
Support (5)
@ M -ve =
13.3 KN . m
Mu / Ф * fc *
b * d2 = w ( 1 – 0.59 w )
13.3*
106 / [0.9 * 25 * 120 * 2852] = 0.060645 = w (1 – 0.59 w )
w = 0.06298
ρ = w (fc / fy) = 0.06298 ( 25
/ 414) = 0.003803
As
= 0.003803* 120 * 285 = 130.07
mm2
( use 2 Ф 12 = 226 mm2 )
Support (6)
@ M -ve
= 18.6 KN . m
Mu / Ф * fc *
b * d2 = w ( 1 – 0.59 w )
18.6
* 106 / [0.9 * 25 * 120 * 2852] = 0.08481 = w (1 – 0.59 w )
w = 0.08954
ρ = w (fc / fy) = 0.08954 ( 25
/ 414) = 0.0054071
As
= 0.0054071 * 120 * 285 =
184.92 mm2
( use 2 Ф 12 = 226 mm2 )
Design of Shear
For all Supports
= 0.85
* (25)0.5 * 120 * 285 / 6
= 24.225 KN
Vu = 13.4 KN
Ü Vu
< fVc
So that we
use minimum reinforcement
Assume 2-leg Ø 8 mm (Av
=101 mm2)
Min reinforcement must be used (Smax)
S= 3 * Av*fyv / b
=3
*101 *280 / 120 = 707 mm
S=0.75d=0.75*285=213.75 mm
S=500mm
Ü Use
1 f 8 / 200 mm
Design
Rib # 3 ( R3 ]
Length of the rib ( L) =
4.55—4.55—2.75 m , Cover = 2.5 Cm = 25
mm , H = 31Cm = 310 mm
D = H – cover = 310 – 25 = 285 mm
Shear & Moment Diagram :
Span (1) :
@ M +ve
= 15.0 KN . m
Mu / Ф * fc *
b * d2 = w ( 1 – 0.59 w )
15.0
* 106 / [ 0.9 * 25 * 520 * 2852 ] = 0.01578= w ( 1 – 0.59 w )
w = 0.015933
ρ = w (fc / fy) = 0.015933 (
25 / 414) = 0.000962
ρ < ρ min so that we use minimum steel
As = ρ * b * d
= 0.003381 * 120 * 285 = 115.6 mm2
( use 2 Ф 12 = 226 mm2 )
Span (2) :
@ M +ve
= 9.52 KN . m
Mu / Ф * fc *
b * d2 = w ( 1 – 0.59 w )
w=0.010077
ρ < ρ min so that we use minimum steel
As = ρ * b * d
= 0.003381 * 120 * 285 = 115.6 mm2
( use 2 Ф 12 = 226 mm2 )
Span (3) :
@ M +ve
= 5.49 KN . m
Mu / Ф * fc *
b * d2 = w ( 1 – 0.59 w )
ρ < ρ min so that we use minimum steel
As = ρ * b * d
= 0.003381 * 120 * 285 = 115.6 mm2
( use 2 Ф 12 = 226 mm2 )
ØNegative moments :
Support (2)
@ M -ve
= 19.3 KN . m
Mu / Ф * fc *
b * d2 = w ( 1 – 0.59 w )
19.3 * 106 / [0.9 * 25 * 120 * 2852]
= 0.088004286
= w (1 – 0.59 w )
w = 0.09312042
ρ = w (fc / fy) = 0.09312042 ( 25 / 414) = 0.00638
As = ρ * b * d
= 0.00638 * 120 * 285 = 218.196 mm2
( use 2 Ф 12 = 226 mm2 )
Support (3)
@ M -ve
= 10.6 KN . m
Mu / Ф * fc *
b * d2 = w ( 1 – 0.59 w )
10.6
* 106 / [0.9 * 25 * 120 * 2852] = 0.04833 = w (1 – 0.59 w )
w = 0.049797
ρ = w (fc / fy) = 0.049797( 25
/ 414) = 0.003007
ρ < ρ min so that we use minimum steel
As
= ρ * b * d
As = 0.003381 * 120 * 285 = 115.63 mm2
( use 2 Ф 12 = 226 mm2 )
Design of Shear
Support (2)
= 0.85
* (25)0.5 * 120 * 285 / 6
= 24.225 KN
Vu = 13.4 KN
Ü Vu < fVc
So that we
use minimum reinforcement
Assume 2-leg Ø 8 mm (Av
=101 mm2)
Min reinforcement must be used (Smax)
S= 3 * Av*fyv / b
=3
*101 *280 / 120 = 707 mm
S=0.75d=0.75*285=213.75 mm
S=500mm
Ü Use
1 f 8 / 200 mm
Support (3)
= 0.85
* (25)0.5 * 120 * 285 / 6
= 24.225 KN
Vu = 13.4 KN
Ü Vu < fVc
So that we
use minimum reinforcement
Assume 2-leg Ø 8 mm (Av
=101 mm2)
Min reinforcement must be used (Smax)
S= 3 * Av*fyv / b
=3
*101 *280 / 120 = 707 mm
S=0.75d=0.75*285=213.75 mm
S=500mm
Ü Use 1 f 8 / 200
mm
Design
Rib # 4 ( R4 ]
Length of the rib ( L) =
4.55—4.55—3.05 m , Cover = 2.5 Cm = 25
mm , H = 31 Cm = 310 mm
D = H – cover = 310 – 25 = 285 mm
Shear & Moment Diagram :
Span (1) :
@ M +ve
= 15.1 KN . m
Mu / Ф * fc *
b * d2 = w ( 1 – 0.59 w )
15.1
* 106 / [ 0.9 * 25 * 520 * 2852 ] = 0.015889 = w ( 1 – 0.59 w )
w = 0.016040
ρ = w (fc / fy) = 0.016040 (
25 / 414) = 0.0009686
ρ < ρ min so that we use minimum steel
As = ρ * b * d
= 0.003381 * 120 * 285 = 115.6mm2
( use 2 Ф 12 = 226 mm2 )
Span (2) :
@M
+ve = 9.44 KN.m
Mu / Ф * fc *
b * d2 = w ( 1 – 0.59 w )
ρ < ρ min so that we use minimum steel
As = ρ * b * d
= 0.003381 * 120 * 285 = 115.6 mm2
( use 2 Ф 12 = 226 mm2 )
Span (3) :
@ M +ve
= 6.90 KN . m
Mu / Ф * fc *
b * d2 = w ( 1 – 0.59 w )
6.9
* 106 / [ 0.9 * 25 * 520 * 2852 ] = 0.0072606 = w ( 1 – 0.59 w )
w = 0.0072919
ρ = w (fc / fy) = 0.0072919 (
25 / 414) = 0.000440332
ρ < ρ min so that we use minimum steel
As = ρ * b * d
= 0.003381 * 120 * 285 = 115mm2
( use 2 Ф 12 = 226 mm2 )
ØNegative moments :
Support (2)
@ M -ve
= 19.1 KN . m
Mu / Ф * fc *
b * d2 = w ( 1 – 0.59 w )
19.1 * 106 / [0.9 * 25 * 120 * 2852]
= 0.0870923 = w (1 – 0.59 w )
w = 0.092096574
ρ = w (fc / fy) = 0.092096574 ( 25 / 414) = 0.005561387
As = ρ * b * d
= 0.005561387* 120 * 285 = 190.2 mm2
( use 2 Ф 12 = 226 mm2 )
Support (3)
@ M -ve
= 11.3 KN . m
Mu / Ф * fc *
b * d2 = w ( 1 – 0.59 w )
11.3
* 106 / [0.9 * 25 * 120 * 2852] = 0.0515258 = w (1 – 0.59 w )
w = 0.0531953
ρ = w (fc / fy) = 0.0531953 (
25 / 414) = 0.0032122
ρ < ρ min so that we use minimum steel
As = ρ * b * d
= 0.0033818* 120 * 285 = 115.6 mm2
( use 2 Ф 12 = 226 mm2 )
Design of Shear
Support (2)
= 0.85
* (25)0.5 * 120 * 285 / 6
= 24.225 KN
Vu = 14.2 KN
Vu
< fVc
So that we
use minimum reinforcement
Assume 2-leg Ø 8 mm (Av
=101 mm2)
Min reinforcement must be used (Smax)
S= 3 * Av*fyv / b
=3
*101 *280 / 120 = 707 mm
S=0.75d=0.75*285=213.75 mm
S=500mm
Use 1 f 8 / 200 mm
Support (3)
= 0.85
* (25)0.5 * 120 * 285 / 6
= 24.225 KN
Vu = 14.2 KN
Vu
< fVc
So that we
use minimum reinforcement
Assume 2-leg Ø 8 mm (Av
=101 mm2)
Min reinforcement must be used (Smax)
S= 3 * Av*fyv / b
=3
*101 *280 / 120 = 707 mm
S=0.75d=0.75*285=213.75 mm
S=500mm
Use 1 f 8 / 200 mm
Design
Rib # 5 ( R5 ]
Length of the rib ( L) =
4.55—3.05 m , Cover = 2.5 Cm = 25 mm , H
= 31 Cm = 310 mm
D = H – cover = 310 – 25 = 285 mm
Shear & Moment Diagram :
Span (1) :
@ M +ve
= 15.1 KN . m
Mu / Ф * fc *
b * d2 = w ( 1 – 0.59 w )
15.1 * 106
/ [ 0.9 * 25 * 520 * 2852 ] =
0.015884 = w ( 1 – 0.59 w )
w = 0.0160409
ρ = w (fc / fy) = 0.0160409 ( 25 / 414) = 0.000968
ρ < ρ min so that we use minimum steel
As = ρ * b * d
= 0.003381 * 120 * 285 = 115.6
mm2
( use 2 Ф 12 = 226 mm2 )
Span (2) :
@ M +ve
= 5.17 KN . m
Mu / Ф * fc *
b * d2 = w ( 1 – 0.59 w )
ρ < ρ min so that we use minimum steel
As = ρ * b * d
= 0.003381 * 120 * 285 = 115.6 mm2
( use 2 Ф 12 = 226 mm2
)
ØNegative moments :
Support (2)
@ M -ve
= 16.9 KN . m
Mu / Ф * fc *
b * d2 = w ( 1 – 0.59 w )
16.9 * 106 / [0.9 * 25 * 120 * 2852]
= 0.0770607 = w (1 – 0.59 w )
w = 0.0809245
ρ = w (fc / fy) = 0.0809245( 25 / 414) = 0.004886
= 0.004886 * 120 *
285 = 167.126 mm2
( use 2 Ф 12 = 226 mm2 )
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