Two way slab method of analysis :

1- Design of two way slab : a. ACI Direct Design Method. b. Equivalent Frame Method. c. Yield Line Theory. 2- Deflection and control of two way slab. 3- Shear of two way slab. 4- Prestressed of simply supported R.C beams. References : - Design of concrete structures. By : Arthur H. Nilson. Two way slab method of analysis : 1. Method One ( coefficient method ) : see ACI-318 code 1963. 2. Method two (coefficient method) : see ACI-318 code 1963. a. Slab supported by brick wall. b. Slab supported by concrete wall. 3. Method three( coefficient method) : see ACI-318 code 1963, edge supports two way slabs. a. Slab supported by brick wall. b. Slab supported by concrete wall. c. Slab supported by steel beam. 4. ACI Direct Design Method : It is a semi empirical procedure for slab analysis and it is application : a. Slab supported by beams or walls. b. Flat slab system. c. Flat plate slabs system. d. Two way grid slabs system. 1 Dr.Laith

5. Equivalent Frame Method : It is an approximate elastic analysis of slab system and applied to : a. Slab supported by beam or wall. b. Flat slabs. c. Flat plate slab. d. Two way grid slab analysis. 6. Yield Line Theory : It is a method of slab based on elastic considerations and the condition obtained in the structures just prior to failure. The method is usually used to analysis slab of irregular as well as regular slabs for various loading and support condition. fixed Yield line Simply sup. supporte 1. ACI Direct Design Method : It is an approximate semi empirical method for analysis two way slab system it's applied for : 1. Slab supported by beam or wall. 1 1 Sec (1-1) 2 Dr.Laith

2. Flat slabs System. slab 1 1 column capital column drop panel 3. Flat plate slab system. slab column 4. Two way grid slab analysis ( waffle slab or two way ribbed slab). 1 1 Sec.(1-1) Limitations D.D.M : 1. There shall be a minimum of three continuous spans in each direction. 2. Panels shall be rectangular with a ratio of longer span to shorter span center to center of supports with in a panel not greater than 2. 3. Successive span lengths center to center of supports in each direction shall not differ by more than one-third the longer span. 4. Column may be offset a maximum of 10% of span in the direction of offset from either axis between center lines of successive columns. 5. Loads must be due to gravity only and live load must not exceed 2 times the load ( L.L. / D.L. 2 ). 3 Dr.Laith

6. For a panel with beams between supports on all sides, the relative stiffness of beams in two perpendicular direction given by the ratio : Main steps of D.D.M. : 1. Calculate the total factored static moment Mₒ. 2. Distribution of Mₒ into positive and negative moments. 3. Distribution of each positive & negative moments into column & middle strip moments. Where : :width of frame measured from center to center of panels for interior frame and from center of panel to edge of panels to edge of exterior frame. :clear span of l1 measured from face to face of support column, column capital, bracket, wall.. Note : Positive and negative moments : a. For interior span : M+ = 0.35 Mₒ M- = 0.65 Mₒ 4 Dr.Laith

b. For exterior span, go to the table of (13-6-3-3). - Column strip moment : c.s (width)=0.25( or )+ (0.5 + )-c.s(width) - For interior frame :- - For exterior frame :- a- For positive moment (+M) :- From table (13.6.4.1) in ACI code. b- For negative moment exterior (- ) :- From table (13.6.4.2) length of torsional member. for exterior strip = for interior strip = 5 Dr.Laith

[ ] y y 2 x 2 x y 1 y 1 x x If no beam then " " (torsion member ): 1- t 2- T Slab Drop panel Capital column 3-4- 6 Dr.Laith

Note : (13.6.4.3) Negative moment are uniformly distributed cross when the support width is : Column or shear wall Note : (13.6.4) Wall (R.C or Brick) Moment in beams: Moment in beams form : 1. Slab moment. 2. Beam self weight. 3. Partition. 1- Moment in beam from slab weight If 2- If moment in beam from slab weight =0 7 Dr.Laith

3- Moment in beam from slab weight between 0-0.85 of col. Strip moment. IF - Moment of slab from col. Strip mom. = - Moment of middle strip = mom. Slab in middle strip only. Summary procedure of D.D. M. 1. find 2. find 3. find M- and M+ 4. find from = (? ) *(M- or M+) Then find from = (M- or M+) 5. If beam found a- b- 6. a- (if no beam found ). 7. (M- or M+) 8. To find moment per meter :- a- b- H.W :- chose for the below roof if the D.D.M applied or not :- Assume :- - L.L = 2.5 - Imposed D.L =1 - concrete density = 2.4 - thick of slab =150 mm 6 6 7 4 4 6 8 Dr.Laith

EX(1): for the slab system shown in fig. below find all moment in x direction? assuming: concrete cover = 30 mm. S.L.L =2.728 S.D.L.(including self weight )=4.82 =21 mpa, =420 mpa Col. Dia.= (250 250)mm Edge beam depth =600 mm Int. beam depth =500 mm y 6 x Frame A 6 Frame B 4 4 Sol :- For frame A :- 9 Dr.Laith

950mm 150mm 500mm b=350 250mm Or use less 700mm (1) 150mm 600mm (2) b=450mm 250mm - For Or use less 10 Dr.Laith

- For - For For frame B :- 700mm 150mm 600mm b 250mm - Interior negative mom. - Interior positive mom. 11 Dr.Laith

Table (1) moment at interior frame : Mom. End span Int. span Ext. Pos. Int. Int. Pos. Int. C.S.S M.S.S beam 2.29 2.17 12.67 5.49 24.41 31.13 6.75 29.9 38.23 6.2 27.9 35.5 3.36 15.0 19.04 6.2 27.9 35.5 Table (2) moment at exterior frame : Mom. End span Int. span Ext. Pos. Int. Int. Pos. Int. C.S.S M.S.S beam 1 1.7 6.1 2.8 12.6 16.16 3.5 15.6 19.89 3.26 14.5 18.49 1.75 7.8 9.94 3.26 14.5 18.49 Deflection requirement according to ACI code (9.5.3): - Slab with beam :-the min. thickness of slabs depends on as follow : 1- go to (9.5.3.2) 2-3- clear span on long direction. 12 Dr.Laith

- at discontinuous edge and edge beam should provided with a stiffness ratio not less than 0.8 ; or minimum thickness required of eq. 9.12 and 9.13 should increased by 10%. EX(1): for the slab beam system, check the thickness requirement due to deflection? assuming : a- all col. (300 300)mm b- Edge beam (300 650)mm c- Int. beam (300 550) mm d- =150 mm e- =420 mpa 6m C D 6m A B 4m 4m Sol:- - to find approximate k then : 1. - - 13 Dr.Laith

150 2. Other wise : - For panel A:- 14 Dr.Laith

- for panel B :- 15 Dr.Laith

- for panel C :- - for panel D :- Use large the slab thickness =150 mm is satisfactory T=150 125.7 o.k Shear Design of Slab : - slabs with beams : a- one way shear (11.1-11.5) (Beam Shear ) )b-y) d d x x no need shear reinforcement need shear reinforcement 16 Dr.Laith

b- tow way shear action (punching shear ): To find (11.12.2.1) - - - hole int. col. edge col. corner col. y x EX(1): for typical interior panel of the flat plate slab system shown. determine the shear strength of slab at support with the allowable one? increase the shear strength by reinforcement if it's not adequate? Assuming : 0066 x Col. Size =(300 300) mm 17 Dr.Laith

Sol :- 0066 1- One way shear ( ) 2- two way shear : Use less d/2 18 Dr.Laith

2- Equivalent Frame Method : Step (1): Equivalent slab beam section. Transfer slab with beam in to equivalent plate. Step (2): Equivalent columns. Equivalent frame method steps :- 1- there is no limit or boundary for the E.F.M. 2- the three dimensional frame can be transform to the more than one frame of two dimensional frame. 3- the frame of 3D to be solve it can solve by cutting the section parallel the center line of columns. 4- If we have only dead and live loads we can tales one story and analysis it. 5- If the building exposed to lateral loads (wind,blast,or seismic load ) we must analyzed the overall frame. 6- The equivalent frame consists of : a- Equivalent columns. b- Equivalent slab beams. Procedure of E.F.M. 1. Equivalent slab beams. a- Find Where :- (width of support parallel to ) Go to table (A13-A,B) Find :- 1- (stiffness factor ) 19 Dr.Laith

2- Carry over factor. 3- (factor of F.E.M) slab beam 2. Equivalent columns. where (stiffness factor ) Torsional member Where :- equivalent columns stiffness. stiffness column. torsional stiffness of transverse beam. ( ) for end frame for interior frame = 20 Dr.Laith

EX(1): Data :- 1- Three story building 2- All beam =0.3 0.6 m 3- All columns =0.3 0.3 m 4- Slab thickness =0.2 m 5- Story height =3m 6- D.L=10 (total) 7- L.L=5 8- =25 mpa, =400 mpa Required :- 5 5 5 7 5 Find all negative moment at interior of x-direction. Sol:- Equivalent slab beam section : For ext. frame (AB) From table (A13.A) A B C D a b c d 0.6m 0.2m 0.3m 21 Dr.Laith

For int. frame (BC):- Eq. columns : Column (aa): 0.6 0.7 0.2 0.3 ( ) ( ) 22 Dr.Laith

Column (bb):- 1.1m 1 6.2 3 2 2 1 0.6 0.3m ) For ) ) For ) ( ) A 279970 B 198018 C 277776 D 77522.3 77957.5 77757.5 77522.7 a b c d Joint Mem. D.F F.E.M D.M C.O.F D.M C.O.F D.M C.O.F D.M a aa 6 6 6-21.05 6 5 6-1.2 0-19 Aa 0.2 0-42.1 0 10 0-2.4 0 0.568-39 A AB 0.8 210.5-168.4-50.27 40 12-9.6-2.84 2.272 39 BA 0.5-210.5-100.5-84.2 24 20-5.68-4.8 1.375-401 B Bb 0.14 0-28.15 0 6.72 0-1.6 0-0.385-27 bb 6 6 6-14.1 0 3.36 0-0.8 0-13 BC 0.36 411.6-72.4 36.2 17.28-8.64-4.1 2.05-0.99 408 C CB 0.36-411.6 72.4-36.2-17.28 8.64 4.1-2.05 0.99 23 Dr.Laith

3- Yield Line Theory of Slab Analysis : Hint analysis:- It's known that concrete does not respond elastically to loads of more than about one half the ultimate loads therefore there is a certain in consistency in design of reinforced concrete cross-section by ultimate strength method in to account in elastic consideration while design moment have been found by elastic analysis.this procedure (elastic analysis coupled with in elastic design of section ) is not consistent but it's present acceptable due to safety and connectivity. For astatically indeterminate structure (beam,slab,frame ) failure will not occur when the ultimate moment capacity at just one critical crosssection is reached. After formation of plastic hinges at more highly stressed sections,substantial changes may occur in the ratio of moments at critical section as loads are further increased before collapse of the structure take place. 1- elastic : 2- plastic : The third hinge forms when :- Plastic analysis P.H 24 Dr.Laith

P.H P.H P.H Increment percentage = Plastic hinge and collapse mechanism :- Kd (under reinforcement section) 0.85 T T 25 Dr.Laith

due to increase in lever arm some increase in of reinforcement. is due to strain hardening After the calculation of resisting moment " " is reached. continued plastic rotation is assumed to occur with no change in the applied moment the beam behaviors as if there where a hinge at that points,this hinge differs from the free fraction hinge. If such a plastic hinge forms in a determinate structure will collapse,the resulting system is referred to as a mechanism. EX(1): Find the max. load that can be applied at simply supported beam shown: Sol:- External work = internal work P EX(2): find the max. load that can be applied at un determinate beam shown: Sol:- W H.W: find the max. load that can be applied at un determinate beam shown: 26 Dr.Laith

Sol:- - for right part : 10m 8m δ - for left part : - for two part : Yield Line Analysis of Slabs : The yield line theory is a method for the limit analysis of R.C slabs. the theory first developed by K.W Johansen (Denmark) at 1943. It permits the analysis of irrgular as well as regular slabs of avariety of support and loading condition :- Simply supported hinge Fixed hinge Free edge 27 Dr.Laith b

L =1 L-xL xl b EX(1):- Sol:- ( ) ( ) ( ) 28 Dr.Laith

( ) H.W:- A continuous one way slab is uniformly loaded and has an ultimate moment capacity of 24 kn.m/m at "A" 30 kn.m/m,at "B" and 18 kn.m/m at "C" using the principle of virtual work find the collapse load " "? Hint :- A B 6m x 6-x Sol:- 29 Dr.Laith

1 University of Kerbala Predication of yielding patterns for two way slab : 1. yield line are generally straight. 2. Axes of rotation pass over any column. 3. Yield line pass through the intersection of the axes of rotation of adjacent slab segment. 4. Axes of rotation generally lies along line of supported. 5. Any symmetry in the slab is maintained in the yield line patterns. Notation :- Column Simply support Fixed support +ve yield line -ve yield line + Point load..axes of rotation EX(2): for the uniformly loaded isotropically reinforced slab shown below determine the collapse load " " assuming the plastic moment permeter? Sol:- ( ) L A A A L A 30 1 Dr.Laith

Or (using the volume of Pyramid as shown ): ( ) H.W:- Sol:- a=2.88 m 10m a 10m ( ) 10m 1 ( ) Or ( ) 31 Dr.Laith

1 University of Kerbala EX(3): Determine the collapse load for the uniformly loaded slab shown below,assuming the plastic moment of resistance / meter width =? Sol:- r W= EX(4): Determine the collapse load for the uniformly loaded slab shown below,assuming the plastic moment of resistance / meter width =? and and = = L or or mode(2) mode(1) L Sol:-mode(1) )L-xL) 32 Dr.Laith

xl 1 ( ) ( ) Or (using the volume of Pyramid as shown ): ( ( ) ) ( ) 33 Dr.Laith

Sol:- mode(2): xl L-2xL xl L ( ) 34 Dr.Laith

1 University of Kerbala EX(5): Determine the collapse load for the uniformly loaded slab shown below,assuming the plastic moment of resistance / meter width =? L 2L 1 1 Sol:- EX(6):-For the uniformly loaded isotropic ally reinforced slab shown below ; determine the collapse load assume the plastic moment of resistance per meter width of slab for positive and negative section is? 35 Dr.Laith

Sol :- ( ) 5m a 5m 4-x 6m x a= sin36 *(4-x) a=(2.4-0.6 x) EX(7): Determine the collapse load for the uniformly loaded slab shown below,assuming the plastic moment of resistance / meter width =? Sol:- ( ( ) ( )) a=l*sin60 2L (2L-b) 2L 36 b a 60 Dr.Laith 4L

b=l*cos60 EX(8): Determine the collapse load for the uniformly loaded slab shown below,assuming the plastic moment of resistance / meter width =? 1 0.5L 60 a 2L 2L 4L 1 a= 1 2 Sol:- (( ) ( )) Or (using the volume of Pyramid as shown ): ( ) ( ) 37 Dr.Laith

Isotropically reinforced slabs :- Slabs reinforced identically orthogonal direction or the tropically reinforced slabs.slabs with different reinforced ratio in orthogonal direction. For isotropically reinforced slabs the moment at any direction α may be shown to be equal to either of the moment in the two orthogonal direction. Prove: y a c d b x Since 1 moment / unit width of slab 38 Dr.Laith

4- Prestress Concrete : Prestressed concrete is a particular reinforced concrete. prestressing involves the application of an initial compressive load on a structure to reduce or eliminate the internal tensile forces and there by control or eliminate cracking. the initial compressive load is imposed and sustained by highly tensioned steel reinforcement reacting on the concrete. A prestressed reinforced section. section is considerably stiffer than the usually cracked Prestressing may also impose internal forces wich are of opposite sign to the external load any may therefore significantly reduce or even eliminate deflection. Advantage of prestressing : 1- Smaller sections will be required for design (the entire section remain effective for stress ). 2- Large span will be possible due to the weight reduction. 3- Deflection under working load will be reduced (Cambering). 4- Prestress reduces diagonal tension stresses at working load. this load to use modified "I" and "T" section.. Type of Prestressed Concrete : 1- Pretensioned Concrete : the Prestressing tendons are initially tensioned between fixed abutments and anchored with the from work in place, the concrete is cost around the highly stressed steel tendons and cured. When the concrete has reached it's required strength the wires are other wise released from abutment, As the highly stressed steel attempts to contract the concrete is compressed. Prestress is imparted viabond between the steel and concrete. to decrease the construction cycle time, Steam curing may be employed. to facilitate rapid 39 Dr.Laith

concrete strength gain and the concrete is often stressed with 24 hrs. of casting. a-tendons stressed between abutments b-concrete cast and cured c-tendons released and prestress rein. 2- Post-tensioned Concrete : The concrete is cast around hollow ducts which are fixed. to any desired profile the steel tendons are usually in place, un stressed in the ducts during the concrete pour, or alternatively may be threaded through the ducts at same later time when the concrete has reached it's required strength, the tendons are tensioned. Tendons may be stressed from one ends with the other end with the other end. the tendons are then anchored at each stressing ends.after the tendons have been anchored and no further stressing the tendons are often filled with grout under pressure. In this way the tendons are bonded to the concrete and are more efficient in controlling cracking and providing ultimate strength. 40 Dr.Laith

a-concrete cast and cured b-tendons stressed and prestress transfer c-tendons anchored and subsequently grouted - Prestress Losses :- 1- immediate losses : a- bend slip. b- fraction. c- elastic shortening. d- shrinkage. 2- long term losses: a- creep in concrete. b- relaxation in steel. - Prestress Forces : 41 Dr.Laith

Method of Analysis : 1- working stress method. This analysis is done in two stage of loading : a- initial loading stage where the stress are calculated after complete of immediate losses. b- service loading stage,where the stress are calculated after all losses. a-initial loading stage :- h e N.A (-) compression ; (+) tension Stress at mid span of beam : Stress at end span of beam : 42 Dr.Laith

b-service load stage : h e N.A Stress at mid span of beam : Stress at end of span : Additional dead load Live load Sustained load Total load The permissible stresses according ACI -318 code (18-4): 1- initial loading stage : - compression stress 0.6 (at mid and end) 43 Dr.Laith

Where: =initial strength at yang ages. - tensile stress at middle 0.25 - tensile stress at end 0.5 2- Service loading stage : - compressive stress due to sustained load 0.45 - compressive stress due to total load 0.6 - tensile stress 0.5 EX(1): For the beam shown check the stresses according to ACI code? assuming :- - immediate losses =10% - total losses =20% - add. D.L =20 - L.L =10 with 10% sustained. - - jacking prestress = 0.8 5 900mm 20m 300mm 400mm Sol:- 1- initial loading stage : 44 Dr.Laith

- at end section : - at mid span : 2- Service loading stage : - at end sec. (for sustained and total ) 45 Dr.Laith

- at mid span (for sustained loading ): - at mid span (for total loading ): 46 Dr.Laith

- rupture cracking : Where: 2- Ultimate Strength Method : for member with bonded tendons: (pretension prestress member ) [ ( )] If compression steel taken in to account then :- a- d b- where: [ ] ω ρ for member with un bonded tendons : 47 Dr.Laith

(post tension prestress members ) a- if Use min. b- if Use min. Strength of prestress section : Then nominal flexural strength of prestress beam : 1- for rectangular section : a- b a c N.A T Or 48 Dr.Laith

b- d a 0.85 c N.A 2- If section "T": a- if stress block at with in flange depth (Rectangular section) : a c N.A Or to calculate a b- stress block is over the flange depth (T-sect.) 49 Dr.Laith

0.85 0.85 a c N.A For partially prestressed Prestress reinforcement limits according ACI -318 code. 1- the reinforcement ratio must be. a- b- ( ) c- Rupture checking of prestress member Or 50 Dr.Laith

EX(1): A simply supported prestress beam with un bonded tendons for the cross-section shown below find the ultimate moment strength then check the reinforcement limit for the section? also check the rupture case? Assume :- Sol:- Since un bonded tendons 51 Dr.Laith

For the reinforcement limit 1- [ ] 2-3- 52 Dr.Laith

EX(2): A pretensioned S.S double tee-beam have a span (7m), =0.85, tendons steel area (600mm 2 ), b f 250mm f pu =1860 mpa. Req.: check if the factored moment capacity are stratifies with ACI-code requirement (assuming e=149 mm, c t =105 mm, c b =305 mm ). Sol:- C=0.85 f c b f a psf ps a psf ps 0 5f c b f f ps f pu [ p ( p f pu f c 0 d d p )] p [0 5 0 05 (fc 2 )] 0 35 ps 600 bd p 2500( 4 05) 0 000 44 0 5 f ps 60 [ 0 35 (0 000 44 6 )] 4 mpa 30 600 4 a 6 4 0 5 30 2500 a M n f ps ps (d p 2 ) 25 n m M u 0 M n 232 n m M cr M cr f r t 3 2 232 53 Dr.Laith

54 Dr.Laith