Lesson 1 In Design of Reinforced Concrete By Ayman Gamal
1. Load oads , and Loa Load d Ca Cases Load types • • • • •
حما Dead Loads الدائمة م Live Loads التشغي م احم الر م احم Wind Loads ر ال م احم Earth-Quack Loads زل Lateral Loads ( Pressure , Earth-Pressure ……….)
نبيةالج م حما •
Dynamic Loads ( Machines ,…………………….. ,…………………….. ) م حما
يكيةد الد •
Other Loads
Load Cases Eccs 203 Ultimate Method 2.
According to Egyptian Code
1-
له نى ح شأ وى المائمة فقط ع م احم و لة وفى ح U = 1.4*DL يً دوث نسبال ن In case of Dead Load Only 2U = 1.4*Dl +1.6*LL U = 1.5*( DL + LL)
when L.L < = 0.75 DL When L.L > 0.75 DL
3U = 1.4*DL 1.4*DL + 1.6*( 1.6*( LL + E ) Where E : Lateral Load Load 1
And Not less Than (2 لهال ع ق ا ج ) و U = 1.4*Dl +1.6*LL when L.L < = 0.75 DL U = 1.5*( DL + LL) When L.L > 0.75 DL 4U = 0.8*( 1.4*Dl + 1.6*LL +1.6*W) where W : Wind Load U = 0.8*( 1.4*Dl + 1.6*LL +1.6*S) where S : Earth-Quack Load And wind load , Earth-Quack Not Combined in One load Case
م لةح القصى ف حمب اس و يتواحد مي لةح ف م ت زو ال الر حم الدائمة حمى اع زال حم نجم خر ر و ر التشغي حم الدائمة و حمى ا عالر د ىعنصر ع ك و ذالتشغي حم و. We use the Max of U = 0.8*( 1.4*Dl + 1.6*LL +1.6*W) U = 0.8*( 1.4*Dl + 1.6*LL +1.6*S )
3. Strength Reduction Factor
يد التسنة و حدرة لوض المقخ We Use Strength Reduction Factor because the variation of Material properties such as steel and concrete , Differences in Calculations and Numbering .
و سفى ال و نتيجه التقر قصاو ال ل سالم ع اة الموق اخت سب دست و فى الم روال γc and γs and
Concrete Reduction Factor γc = 1.5 Steel Reduction Factor γs = 1.15
And In case of Eccentric Compression we Obtain γc , γs From the Equations e / t ) ≥ 1.5 3 e / t s = 1.15 (7 / 6 − ) ≥ 1.15 γ 3 c = 1.5(7 / 6 − γ
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4. Ultimate Design Concepts
Strain in concrete 0.003 in Flexure that makes apart of section under tension force 0.002 in Compression force acts on the c.g of the section 3
فى الشك م فىءك يمست نهرغط فى الال يم ت و , ي نة فى الشدرة الوق ما ت ىع تمد الم مي لى و تر لك 3 ا0.003 نه عرفى ال نا سم و ةوس نةرغط فى الال ائم ك ت الق ا م و , القيمة ذ ع نا زعد ددالشد فى ال لق, ةخرا عطفى ا ها ا ا - ةد لمرح ص د لدال ك غط وة لوق صىنة الى ارال ص و فيه : ضغا ها هص ى نه ورال يان سب ( ىء يان ) ص ايان الق يان فيك ال ينا ذا ال ع ي و م دالب ج و ينا ار انذا اخ و رح لىالت فى الجداو اال القيع يد نسبة التس دع ينذا ا دث و - ص ظةال غط و فى نة لوق صىنه الى ار ال و ت لهال ذ فى : ازت ها ددنه و الرفى ال ينا ك و ة الرح د الىدال - الى د و ددال ا فى حيالقص توق ب نه لرال ك لهال ذ و فى: شدا ها حتى أخذ فتر و شر دث حيقد له ك ينا ذا ال و فى , حد ال نيةرال عالق صمي فى ن اف ذا و الق غطال يى انع االتصمي ت حيازالمت ين الشد – اي انلتيل ر لىو الت
5. Reinforcement of Section subjected to Flexure Ultimate Design Method
ة الشد فقط يس ا ( وال ) ءنة لرالم عالق صمي
Sections With Tension Steel Only
Section subjected to flexure if applied Compression P is less than 0.1 Fcu*Ac P ≤ 0.1 Fcu * Ac
ع ةالقيمة الم ع ق الم مال اء إنل ر تبر الق M =
As * fy a (d − ) s 2 γ
Where 4
As * fy γ s 0.67 * ( fcu / γ c) * b
a= and a/d Not Less than 0.1
برا الق ك 0.1 سبة عال ذ لة زفى ح
زال and yct (Moment Arm) Not More Than 0.95d > 0.1 Yct ≤ 0.95 d a / d
Notes
لهفى ح ه القمت صى ع ا ا له الق رالم الك دله عال ذ دست ازالمت ينا ت ازالمت ينلة افى ح ة القمت صى عا برا ى القر عالمؤ ال اف غط يدث ان نه ختىرال د غط لقة ال د فىحد و This Case is Used if the Moment Apllied to Section in < Mmax Where M = R max* fcu * b * d 2 / γ c
ازمتا هة ا فى طة ايتحم اقصى ع
Rmax Steel Grade 240/350 280/450 360/520 400/600 450/520
Cmax/d 0.5 0.48 0.44 0.42 0.40
μmax -4
8.56*10 fcu 7.00*10-4 fcu 5.00*10-4 fcu 4.31*10-4 fcu 3.65*10-4 fcu
Rmax 0.214 0.208 0.194 0.187 0.180
Blanced Section Reinforcement
ط ازمتا هى اع تصما ازالمت ينله افى ح ه القمت صى ع وس رالمؤ ال ك If M applied = Mmax for the section Then M = R max* fcu * b * d 2 / γ c
and Rmax is Given in tables and then As = µ max* b * d where μmax is Given in the Tabel in this Case Falure Accures in Concrete and steel Together
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Compression and Tension Reinforcement ازمتا ها ح قصىا ا ع طى امؤثر عا ا ة زي فى ةخرا قد ي ضغة انط ديد فى و يت.
Used when Moment applied to section is > Mmax ( Blanced Section ) Then we Use Compression steel fy M = ( R max* fcu * b * d 2 / γ c) + ( ) As ' ( d − d ' ) s γ
Note : Compression steel In all cases of Design Must be not less than 0.1 As Because it is Useful in Long Term Deflection
ى المدء عنفى ا عدمسل ل د الشد وحد 0.1 غط عحية الفى ن يالتس ق ا ج ال Stirrups Spacing Mustn’t be less than 15φ for Buckling of Bars
L and T section Reinforcement
M = 0.67 ( fcu / γ c) B * tf * ( d −
tf ) 2
Or M
=(
As * fy tf )( d − ) s 2 γ
Where Tf is flanged thickness B is Effective width of Flange ءحن رما عطة ا ة فىسا حا ة فى ا عم طا تس ةس اق
Note μmin = As/b.d = 1.1/fy where fy (n/mm2) 0.25% Normal Mild Steel 0.15% High Grade Steel Lesson 1 End Best Wishes Eng / Ayman Gamal
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Lesson 2 In Design of Reinforced Concrete By Ayman Gamal
Columns Design in Ultimate Method 1st Design Equations for Columns subjected to Axial compression If the section subjected to P (Axial compression force ) and M (Moment single or biaxial ) and the eccentricity is less than 0.05 t ,0.05 b ( Dimension Under Consideration) Where e = Mu / Pu We use these Equation for Designing the Column .
جة فى ارال ن و مال ىع راؤ ة و ع غط لة وفى ح ال 0.05 غرا ال جة فى ارال , رال 0.05 غرا رال جفى ا ال 0.05 ى ة المسمر ال ا ه حي المدالم داتا ت عتبا عتبا دالب 0.05 او 20 دت ة المسمرال ey = My/P<0.05t ex = Mx/p <0.05b Where b (width of Column) in the direction of X Where t (Length of Column) in the direction of Y
For tied Columns Pu
= 0.35 fcu
* Ac
+ 0.67 fy
* Asc
where Ac : Area Of Concrete Section. Asc : Area Of Longitude Steel Bars. This Chart Using Reinforcement Ratio 1% Where Min Reinforcement Ratio is Max of : 0.8% of Ac Calculated ; 0.6% Of Ac Actual
ااعتم , مسما اق وية و عح رما عمدا صم ى جم حا ه مسما ير ى اع For Spiral Columns Pu is the min of :
دائريةا العمد
يةا الم غرا صى حما Pu Pu Pu
= 0.35 fcu
* Ak + 0.67 fy * Asc +1.38 fyp * vsp =1.14 (0 .35 fcu * Ac + 0.67 fy * Asc ) = 0.40 fcu * Ac + 0.76 fy * Asc
و ي فةلمس د ونيهح ك عمدا ذا ال ن حظة ا Where 8
Ak: Area Of the Concrete Core of Section ( Surrounded by the stirrups)
نةالك اخ نىرال حة القس Asc: Longitude steel bars Fy : Yield stress of longitude steel bars Fyp: Yield stress of Spiral stirrups Vsp: Ratio of steel Volume for One Turn of Spiral Stirrups Where Vsp
≥
Asp Π
* Dk p
احدال فى الدو يد التسنسبة حد
where Asp : Area of spiral stirrup section Dk: Area of the Concrete Core inside Stirrups P: Pitch ( Spacing between turns ) 30-80mm Ratio of stirrups Reinforcement µ sp
=
0.36 (
fcu fyp
)(
Ac Ak
1)
−
;
µ sp
=
Vsp Ak
و بحمدك لها نحس انت ا ا اشهد ا يا غفرك و ااس The End Of Lesson 2 Best Wishes Eng / Ayman Gamal
[email protected] Y
Column Design In Detailed View . 9
عد ا ئية التىنشر اى ال عمدا قغر لد االب خمسة ا غط عال جفى ا د او Ratio “ H/b ” More Than 5 General Considerations Dimensioning
t X
•
غرد االب ع مبر لا د الق د يست بةالمر عالق ل شم و قل ( غرد االب ) رال خمسة ا ع جا فى اد ال ي نيةرائط الال لم بق صم نية وائط خرء حعا ذ اعتبر و ا
b
Or We consider it (R.c Wall) and Designed As walls Minimum Eccentricity Ex 0.05b Ey 0.05t Or 20mm t ≤ 5b
•
Braced Unbraced Column لم مى الع رالمؤ وال عتبخذ فى اا رة الذلالس م لبق عمدا صم و نبي قيد ير او قيد ن اء ةر لالدني حدو د ع Column is considered braced if :
ىالمب ا ك ستمر نيةائط خرح ع عب دعي ر او اقيدا ا ىتبر المب يةا فى الشروست ى ومبفقى لفى المسقط ا تم ز زعة و The Horizontal load is Carried by Another R.c Members such as Shear Walls And Shear Walls Must Satisfy The following 2 Conditions . 1- Satisfy these Equations α = Hb
α = Hb
N
∑ EI
< 0.6
N < 0.2 + 0.1n ∑ EI
For Buildings consists of 4 Stories Or More For Buildings consists of Less Than 4 Stories
Where n : No Of Stories N : Total Load Acting On all Vertical Members Hb : Total Height of the building Above the Foundation ف ا
شأمل ا 10
EI : Sum of Flexure Rigidity of the Shear Walls in the direction ∑
under consideration
عتبا جنية فى ارائط الء لنا ءس جم 2- Shear Walls Must be connected to the Foundations in a way that allow the transmitting of the Horizontal Loads , and Moments completely to the foundation
جةال وفقى و الا مال نق سم قةر ةتص نيةرائط الال ك ا إلى ا لك . نبي قيد مال نحد ك كذا و را فى ا د ر و ا ف د عمدا ك أ ظة يمكح منش ى اع شروا ط يت
Long – Short Columns 1st let us define these parameters i = 0.3b Rectangular Sections .الذا ر القص نص i = 0.25 D Circular sections λ Slenderness Ratio & λ = He/i فة القن He : Buckling Length of column
ىو الس ال رل مال بي لةى حتمد ع ومل ال نبا سفى ال حق ةدات مل نبا قة حسر ةفى البدا Buckling Length Of Column “He” There are 2 methods for Calculating He 1- Equation Method 2- Simple Method ( Using Tabels ) 1st Equation Method for braced Column He is the Smaller of : نبي المقيد عمدسبة لل He = Ho [0.7 + 0.05 (α 1 + α 2)] ≤ Ho He = Ho [0.85 + 0.05α min] ≤ Ho
for Unbraced Column He is the smaller of : ير المقيد عمدسبة لل
نبي He He
= Ho [1.0 + 0.15 (α 1 + α 2)] ≥ Ho
= Ho [ 2.0 + 0.3α min] ≥ Ho
Where α1 is the Lower End Condition ىالس رل مال بي لةح
مال 11
α2
is The Upper End Condition ال رل مال بي لةح
مال min is the min of α1 , α2 α ∑ EcIc
α
And
=
Lc EbIb
∑
Lb
الكمرا ءس جم / مال ءس جم يةو الس ةة الد الع
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