- 1 -
Chapter one
Concrete
1.1.1 Introduction
Concrete is one of the most popular materials for buildings because it has high compressive strength , flexibility in its form and it is widely available . The history of concrete is age dates back for over a thousand year. Contemporary cement concrete has been used since the early nineteenth century with the development of portland cement despite the high compressive strength , concrete has limited tensile strength , only about ten percent of its compressive strength and zero strength after chocks develop . In the late nineteenth century , reinforcing materials such as iron or steel rods began to be used as common reinforcing material usually steel bars have over (100) times the tensile strength of concrete but the cost is higher than concrete (1) . Therefore, it is most economical that concrete resists compression and steel provides tensile strength . Also it is essential that concrete and steel deform together and deformed reinforcing bars are being used to increase the capacity to resist bond stresses.
Concrete have many advantages can be summarized as follows:
-It has a relatively high compressive strength.
-It has better to resistance fire than steel.
-It has long service life with low maintenance cost.
-In some type of structure such as dams and footing it is the most economical structural material.

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-It can be cast to take any shape required . Making it widely used in pre east structural components.
When concrete it freshly produced, it is plastic in behavior and is after called plastic concrete . The requirement, for fresh concrete include the time available for placing it into formwork and workability. There are many concrete structural members such as beams ,slabs and columns. In this project the design of reinforced concrete beam (continuous beam) and slab (one way slab) will be done.
.(1.1) Description for general concrete beam.

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Continuous beam is the structure which has more than two supports.
The deformation of the structure under load or stresses depends not only on load but also an rotations and slopes of the element at its ends. The magnitudes of the slope and rotation depend on the rigidity (or flexibility) of the joint that is relative rigidities of the adjoining members, which normally comprise horizontal beam elements and vertical column elements. The nominal strength of a continuous beam ( beams general) is calculated ,based on the best current knowledge member and material behavior .That nominal strength is modified by a strength reduction factor
Φ less than unity, to obtain the design strength . The required strength, should the hypothetical overload stage actually be realized , is found by applying load factors , greater than unity , to the loads actually expected.
These expected service loads include the calculated dead load,the calculated or legally specified live load , and environmental load such as those due to seismic action or temperature . Thus reinforced- concrete members are proportioned so that
M u
≤M d
=
M n
P u
≤P d
=
p n
V u
≤V d
=
V n
Where the subscripts u indicate the required strength at factored loads , the subscripts d denote the design strength and subscripts n the nominal strength .The strength reduction factors
normally differ ,depending

- 4 upon the type of strength to be calculated ,the importance of the member in the structure, and the other considerations .
1.1.3 Slab(one way slab)
The slab provides a horizontal surface and it is usually supported by columns , beams or walls there are two types of slabs one way slab and two way slab.
One way slab : a-one way slab /one way flat slab
These slabs are supported on two opposite sides and all bending moment and deflection are resisted in the short direction .A slab supported on four sides with length to width ratio equal to or greater than two , should be designed as one way. b-One way joist floor system
In this type bending in one direction only and this type of slab . Also called ribbed slab , is supported by reinforced concrete ribs or joists .The ribs are usually tapered and uniformly spaced and supported on girders the at rest on columns .
A one way slab is designed as a single reinforced width beam strip.
Loading for slabs is normally specified in kN/m 2 . One has to distribute the reinforcement over the 1m strip and specify the center –to center spacing of the reinforcing bars . In slab design , a thickness is normally assumed , and the reinforcement is calculated using a trial level arm
(d-a/2).

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One way slabs do not normally require shear reinforcement for typical loads particularly if low tensile reinforcement ratios are used ,it will be found that the shear capacity of the concrete V c
will almost without exception be well above the require shear strength V u
at factored loads.
Transverse reinforcement has to be provided perpendicular to the direction of bending in order to resist shrinkage and temperature stresses.
The reinforcement ratio for a slab can be determined by divided the area of one bar by the area of concrete between two successive bars ,the later area being the product of the depth to the center of the bars and the distance between them center to center .The spacing between bars that is necessary to furnish a given area of steel per one meter width of the slab by dividing the number of bars required to furnish this area into
1m.According to ACI-CODE limitation the maximum spacing for reinforcement in structural slabs must not be greater than 5h for temperature and shrinkage reinforcement and 3h for main reinforcement, but the maximum spacing of this reinforcement shall not exceed 450 mm.

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1.2 Steel
1.2.1 Introduction
Steel is an alloy ,which is a metal made from various elements .The two main elements that comprise steel are iron and carbon .Iron is by far the main ingredient in terms of percentage, usually making up roughly
98%(2) of all the components found in select .Carbon typically will comprise less than 0.5%of steel but is very important because it effects the steel strength and hardness .In mild-carbon steels an increasing carbon content will produce stronger and harder steels .At the same time ,this increase in carbon content will result in a reduction of ductility. This loss of ductility is sometimes referred to as brittles ,which is not an advantageous quality for any structure material .
Other alloying elements ,such as silicon ,nickel ,manganese and copper and also be added to steel in varying amounts to enhance certain properties such as strength ,hardness and corrosion resistance .There are many shape of steel sections such as W,C,or MC,L ,S ,WT,atc and WT shape. W section is the most common structural shape used for beam and columns .In this project we used W shape in designing the beam and we selectW
24*192 section.It checking for deflection and strength (ΦM n
).
W
24*192 means that the rolled shape is a wide- flange section approximately 24 inches in depth and 192 weight pounds per foot of length .The depth and weight are particularly important pieces of information to designer because they are constantly concerned with clearance height ,dead weight and cost per pound of the structure .The cross sectional properties of specified rolled shapes they will be referred to the table in the AISC manual.
There are two methods for design any steel structures ASD and LRFD method. We use the method of LRFD to design beam in this project.

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Fig.(1.2.1)Description for general beam.
Fig .(1.2.2) Description for general steel beam.
1.2.2 LFRD philosophy
Over last 25 years, the structures design community has been very inserted in developing a design procedure that uses a probability- based rationale .With the adoption (1986) of the AISC(load Resistance factor design specification) .The steel design industry has given its endorsement

- 8 for such a method .Although (LRFD)has not received the quick and over whelming acceptance in the design community some would like ,few doubt that it will achieve majority usage by designers in the next 15 years.
1.2.3 Beam design philosophy using LRFD
A beam is a structural member that is subjected primarily to transverse loads and negligible axial loads. Transfers loads cause internal shear forces and bending moments. These internal shear forces and bending moments cause longitudinal axial stresses and shear stresses in the cross section .The most basic requirement of beam design using the LRFD method is the design moment capacity( Ф Mn) must be greater than or equal to the required (factored ) flexural strength M u.
The required flexural strength may be referred to by some as the
"ultimate moment″ .This most basic requirement can be simply as follows:
Φ M n
≥ M u where Φ=0.9 for bending
Bending strength is affected by unbraced length of the compression flange,the compactness of the member and the axis of bending ,since rolled shapes are primarily bent about the strong axis and are compact ,we will restrict our discussion of the LRFD requirements to those that focus on the unbraced length of compression flange .Also a serviceable of the beam usually means that the deformation primarily the vertical slug or deflection must be limited. The maximum deflection of the designed beam is checked of the service -level loads the deflection due to service -level loads must be less than the specified values. Serviceability shall be evaluated using appropriate load combination for the serviceability limit

- 9 states identified and the deflection should be checked appropriate limits for deflection can be found from the governing building code for the region .
1.3 Methods of analysis
There are several methods of analysis of statically indeterminate structures have been developed over the years .Some of them are more exact than others ,and some include approximations and simplifications that facilitate relatively quick solution when computer utilization is not readily possible or justified. The classical methods are based primarily on physical understanding of the structural deformational behavior and are particularly important in interpreting the response of the system to the type and sense of applied load .
*Slope of deflection method
The slope deflection method is a displacement method involves writing two equation for each span of a continuous structure ,one at each end , expressing the end moments as the sum of the following four contributions:
-The restraining moment resulting from an assumed fixed-end condition of the loaded span.
-The moment due to the rotation of the tangent to the elastic curve at one end of the span.
-The moment due to the rotation of the tangent to the elastic curve at the other end of the span.
-The moment resulting from the translation of one end of the span with respect to the other.

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*Moment Distribution Method
This method is a numerical application of the displacement method in which the desired moments, shear, or stresses are obtained by a method of successive approximation suitable for longhand computation .The method lends itself to simple physical interpretation. Hence it can be used for quick approximate as well as exact solutions, depending on the number of successive cycles chosen. It is essentially an iterative solution of the slope deflection equations and has been used extensively since its development by Hardy cross in the United States in 1939.
*Displacement Method and the Stiffness Matrix
The displacement method is analogous to the force method except that the nodal displacement are considered as the unknowns instead of the redundant forces or moments .Essentially the slope deflection method it can be considered the direct link to computer method of structural analysis. Since the joint displacement represent the freedom to move or rotate, the term ″degrees freedom″ represents the joint displacement as a measure of the kinematic degrees of indeterminacy.
*ACI Moment Coefficients
The ACI 318 Building Code contains approximate coefficients for calculating shear and moments, which can be used when (and only when) the following conditions are satisfied :
-There are two or more spans.
-Spans are approximately equal, with the larger of two adjacent spans not greater than the shorter by more than 20 percent.

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-Loads are uniformly distributed .
-Unfactored live load, L, does not exceed three times unfactored dead load , D.
-Members are prismatic .
1.4 AIM OF STUDY
Concrete and steel analysis and design of aeration tanks beam by using
ACI-CODE 2002 and LRFD in saad. pro program and made comparison between the results for cost and strength and choosing the best section in steel and reinforced concrete and depends on the results which appears from this comparison to be realized aim from this project

Chapter two
Hand calculation
2.1 Concrete design
*Dimensions of beam
-Height of beam=0.85m
-Length of span =15m
-Width of beam=0.3m
* Material properties
-f′ c=
24 N/mm 2
-E c=
4700* √𝑓′𝑐
-f y
=400 N/ mm 2
-density =24k N/m 2
* Loadings
-Dead load =1.3kN/m
-Live load=2.2kN/m
Aeration load(concentrated mid span load)
-Max. aerator load =11.5kN
-Grating +hand rail=1 kN/m
-self weight of single beam=24*0.85*0.3=6.12 kN/m
-self weight of slab=0.15*24*1=3.6 kN/m
Analysis the structure
1- Find the reactions by using moment distribution method(F.E.M).
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2-Draw shear and moment diagram and find maximum shear and maximum moment .
W u
=1.2D.L+1.6L.L
P u
=1.6L.L
W u
=1.2(1.3+1+(6.12*2)+3.6)+1.6(2.2)=25.48 kN/m
For concentrated load
P
L.L
=11.5*1.5=17.25 kN( 1.5facter for impact load)
P u
=1.6*17.25=28 kN
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Fig (2.1)shear and moment diagram

Design for flexure
M max ve
=831.06 k N.m
(max. negative moment) from F i g .(2.1)
M max+ve
=474 KN .m
(max. positive moment) from Fig.(2.1)
V =263.7KN (max. shear) from Fig(2.1) d=h-50-12-(Φ/2) d=850-50-12-(25/2) d=775 mm
μ =( f y
)/(0.85×f c
μ =(400)/(0.85*24)=19.61
R u
=(M u
)/(
Φ *b*d²)
R u
=(831.06*10^6/2)/(0.9*300*775²)=2.56 N/mm² (pre single beam)
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1/μ*(1√1 −
2∗𝑅∗𝜇
)
𝐹𝑦 𝜌 =
1
19.61
∗ (1 − √1 − (
2∗2.56∗19.61
400
))
ρ
=6.86*10^ 3
A req
.at internal support(per single beam)
Α s req
. =6.86*10^ 3*300*755=1594.95 mm 2
No.of bar =(1594.95)/(491)=3.25
use 4Φ25 mm top bars per each beam (to be in the safe side)
M
+ve
=474 kN.m
R u
=(237*10^6)/(0.9*300*775²)=1.46 N/mm²(per single beam)
ρ
=1/19.61*(1 √1 − (
2∗19.61∗1.46
) =3.79*10^
400
3
A s
=3.79*10^ 3*300*775=881.175 mm²

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No. of ba r s = A s
/A sb no.of bars =881.175/491 =1.795
Use 2Φ25 mm bottom bars per single beam
For discontinuous exterior edges, use the same reinforcement used @mid span (2Φ25 mm top bars) in order to be on the safe side .
Design for shear
V max shear force at int.support =263.7 k N
V ud
=V u
(W u
*d)
V ud
=263.7
0.775*(25.48)
V ud
=243.95 k N
Φ V c
=0.75* √24 *300*775*10^ 3/(6)
Φ
V c
=142.38 k N
Φ
V c
< V ud
Reinforcement for shear
Φ V s
=V ud
Φ V c
=243.95-142.38=101.57 kN
4*
Φ
V c
=569.52>
Φ
V
S the section is suitable.
2* Φ V c
=284.76> Φ V s the maximum distance equal to:-
S max
=d/2
S max
=775/2=387 mm
UseΦ12@300 mm

one way slab
Transverse slab (one way simple slab span =1.3+0.3=1.6 m)
Slab dead load =0.15*24=3.6 KN/m
Grating +hand rail=1kN/m
Total dead load =4.6kN/m
Live load =1.2 kN/m
W u
=1.2*D.L*1.6L.L
W u
=1.2*(4.6)+1.6*(1.2)=7.44 kN/m
M max
= 𝑤𝑙²
8
+ 𝑝𝑙
4
M max
=
7.44∗(1.6) ²
8
+
28∗1.6
4
=13.581 kN.m
Per meter width (at aeration region )
D=t-Φ/2-20=124 mm
ρ max
=0.85 β*( f′c
)*( 𝑓𝑦
0.003
0.003+0.004
)
ρ max
=0.0186
𝜌 t
=0.016
M u max
= Φ * 𝜌 max
*b*d²* f y
*(1 0.59* 𝜌 max 𝑓𝑦
* 𝑓´𝑐
)=84.13 KN.M
M umax
>M u the thickness provide flexural requirement
Shear requirement
V ud
=V u
– d* W u
=19.95-0.124*7.44= 19.03 KN
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Φ V c
=0.75* √24 *1000*124*10^ 3/(6) =75.9 kN
Since V ud
< Φ V c
The thickness provided shear requirement .
Use nominal stirrups Φ12@200 c/c
Reinforcement in short direction.
μ = 𝑓 𝑦/(0.85 𝑓 𝑐
′
R u
13.58∗10^6
=
0.9∗1000∗124 ²
ρ= 1/ 𝜇 *(1√1 −
2∗𝑅∗𝜇
) 𝑓𝑦 𝜌 =2.5*10 3 < ρ t ∴ ρ=0.9 as assum
A s
= 𝜌 *d*b
=2.5*10 3 *1000*124 =310 mm²
ρ mim
= 0.0018
A s min
=0.0018*1000* 150 = 270 mm²
A s max
=0.0186*1000*124=2306.4 mm²
A smin
<A s
<A smax o.k
Use A s
=310 mm²
Assume diameter of bars=12mm
A sb
=113
S max
=mi m (3*t , 450)
=mim (450,450)
S max
=450 > S=364.5 o.k
Use Φ12 @ 300 mm
Reinforcement in long direction .
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- 18 -
A s
= A smin
=223.2 mm²
By using 10 mm (A sb
=79 mm²)
S =
1000∗79
223.2
=353.9 mm²
S max
=min(5*t , 450 )
= min (750 ,450 )
S max
= 450 > S o.k
Use 3Φ10 @300 mm
Check for deflection in single beam
I g
= 𝑏ℎ ³
12
I g
=
300∗(850^3)
12
E c
=4700* √24 =23025 N/mm
n=
200000
23025
=8.69
M cr
= 𝑓𝑟∗𝐼𝑔 𝑦𝑡𝑐
=(0.7*15.35*10 9 * √24 )/(10 6 *425 ) =123.45 k N.
m
Find neutral axis bx²/2+(2*n 1)*A ś
*(X d′ ) +n*A s
(d x)=0
150*x²+(2*(8.69)-1)*1964*(x-75)= 8.69*(982)*(775-x)= 0
X=144.6 mm use x =145 mm
I cr
=bx 3 /(12)+(2*n 1)*A s
*(x d′ )²+n*A ś
*(d x)²
=300*(145) 3 /(12)+(2*(8369) 1)*1964*(145 75)² +8.69*(982)*(775 -
145)² =3.62*10 9 mm 4
Service load (per single beam )
M max +ve md
=
335.5
2
=167.65 k N.
m

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Check the torque moment
T=15.3kN.m
T=15.3*1.5=22.95kN.m(impact)
T u(live load)
=22.95*1.6=36.72 kN.m
For every couple =
36.72
2
=18.36 kN.m
∴ 𝑠ℎ𝑒𝑎𝑟 𝑓𝑜𝑟𝑐𝑒 =
18.36
1.06
= 17.32 𝑘𝑁 (bearing force for every couple)
Bearing force according ACI=0.7*0.85*f′c* (A
2
/A
1
) 0.5
*A
1
Bearing force =0.7*0.85*24*1*((25*150)/(1000))*1=53.55 kN ∴ 𝑏𝑎𝑒𝑟𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒 𝑜𝑓 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 > 𝑠ℎ𝑒𝑎𝑟 𝑓𝑜𝑟𝑐𝑒
k
- 20 -

- 21 -
2.2 STEEL DESIGN
D.L=0.874 KIP/Ft
L.L= 1.4787 KIP/FT
Surface Load =100I b/ft→0.1 kip/ft
Assume dead load =150 b/ft→0.15 kip/ft (per 2 Beams)→ 0.3 kip/ft
4
Self w u
=(1.2*D.L)+(1.6*L.L) wu=1.2*(0.874+0.1+0.3)+1.6*(1.4787) w u
=3.895 kip/ft ≈ 4 kip/ft
Aerator Load =1150 kg → 2.54 kip
Impact factor=1.5*2.54 =3.81 kip
P u
=1.6*3.81= 6.096 kip≈ 6 kip

Fig(2.2)
Use C b
=1 for more safety
M u max
=662 kip.ftt (From fig (2.2)per single beam) f y
=50 Ksi Φ =0.9
M umax
/C b
= 662 /1= 662 Kip ft
From AISC manual use W ₂₄ *
192
Φ b
M p
=2100 kip.ft
Φ b
M p
>M u max o.k
Check the deflection
∆ max
=
L /360
∆ max
= (50.36*12)/(360)= 1.67in
≃ 1.7in
∆ =))
P* L 3 )/(48*EI))+((5*L 4 W)/(384*EI))
- 22 -

∆=
(50.36∗12)
3
∗(
1.9
12
)
48∗29000∗6260
+
5∗(50.36∗12)^4∗(
1.2
12
)
384∗29000∗6260
∆ = 0.957 in < ∆
MAX
Check local buckling
For flange λ 𝑏𝑓
=
2𝑡𝑓
λ = 4.4
λ p
=0.38*(29000/50)=9.5 > λ o.k ∴ 𝑐𝑜𝑚𝑝𝑎𝑐𝑡 𝑓𝑙𝑎𝑛𝑔𝑒
For web
λ ℎ
= 𝑡𝑤
λ = 26.6 λ
P
=3.76*(Es/fy)^0.5
λ
P
=90.5>26.6 ∴ 𝑜. 𝑘 𝑐𝑜𝑚𝑝𝑎𝑐𝑡 𝑠𝑒𝑐𝑡𝑖𝑜𝑛
Check shear of beam
Vu= 185.6 Kip (from fig (2.2)) ℎ if 𝑡𝑤
≤418/ √𝑓𝑦 26.6≤59.1
∴ Φ V n
=0.6*f y*d* t
w
(where Φ=0.9)
Φ Vn=0.6*50*24*0.81=583.2 Kip
Vu<Φ Vn 𝑜. 𝑘 𝑡ℎ𝑒 𝑠ℎ𝑒𝑎𝑟 𝑦𝑖𝑒𝑙𝑑𝑖𝑛𝑔 𝑐𝑎𝑛 𝑡𝑎𝑘𝑒 𝑝𝑙𝑎𝑐𝑒
- 23 -

- 24 -
Chapter three
Staad pro results
3.1 Introduction
Staad-pro program used for the purpose of analysis and design of beams and slab in the present project the analysis and design of structure designed by research engineer international company which proved for many special programs in this field and it is the first program which obtained of I so 9001 in structure engineering field . The program provide active tools the static and dynamic analysis with the ability of the linear analysis and advanced analysis for non linearity of structures .Staad. Pro program depend on finite element analysis that consider of numerical methods whish can applied in wide space of engineering science another property of this program ability to make progress reports of projects which consists – of analysis results and diagram and charts .