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 .
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
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 Zoro-1982@hotmail.com
6
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
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
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 Zoro-1982@hotmail.com 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
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
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