ANALYSIS OF SECTION. Behaviour of Beam in Bending

Transcription

1 ANALYSIS OF SECTION Behaviour o Beam in Bening Conier a imply upporte eam ujecte to graually increaing loa. The loa caue the eam to en an eert a ening moment a hown in igure elow. The top urace o the eam i een to horten uner compreion, an the ottom urace lengthen uner tenion. A the concrete cannot reit tenion, teel reinorcement i introuce at the ottom urace to reit tenion.

2 Behavior o Beam in Bening For continuou eam, the loa alo caue the to en ownwar etween the upport an upwar ening over the upport. Thi will prouce tenile zone a hown in igure elow. A the concrete cannot reit leural tenion, teel reinorcement woul e introuce a etail in the igure. Baic Aumption in RC Deign In the eign o reinorce concrete eam the ollowing aumption are mae (See EN 1991: Cl. 6.1 () P.) Plane ection through the eam eore ening remain plane ater ening. The train in one reinorcement, whether in tenion or compreion i the ame a that in the urrouning concrete. The tenile o the concrete i ignore. The tree in the concrete an reinorcement can e erive rom the train y uing tre-train curve or concrete an teel.

3 Baic Aumption in RC Deign Ditriution o Stree an Strain Figure elow how the cro ection o a RC eam ujecte to ening an the reultant train an tre itriution in the concrete. Top urace o cro ection are ujecte to compreive tree while the ottom urace ujecte to tenile tree. The line that introuce in etween the tenile an compreion zone i known a the neutral ai o the memer. Due to the tenile trength o concrete i very low, all the tenile tree at the ottom ire are taken y reinorcement.

4 Ditriution o Stree an Strain ε cc cc η c ηα cc /γ c = λ F cc h z ε t t y F t (1) () (3) For < 50 N/mm : η = 1 (eining the eective trength), ε c = , α cc = 0.85, λ = 0.8, γ c = 1.5, c = / 1.5 = Stre itriution in the concrete Ditriution o Stree an Strain The triangular tre itriution applie when the tre are very nearly proportional to the train, which generally occur at the loaing level encountere uner working loa conition an i, thereore, ue at the erviceaility limit tate. The rectangular-paraolic tre lo repreent the itriution at ailure when the compreive train are within the platic range, an it i aociate with the eign or ultimate limit tate. The equivalent rectangular tre lo i a impliie alternative to the rectangular-paraolic itriution.

5 Ditriution o Stree an Strain The itriution o train acro the eam cro ection i linear. That i, the normal train at any point in a eam ection i proportional to it itance rom the neutral ai. The teel train in tenion ε t can e etermine rom the train iagram a ollow: Thereore ; t ( ) cc t cc t 1 cc Introuction Since, ε cc = or cla C50/60 an For teel with = 500 N/mm an the yiel train i ε t = By utituting ε cc an ε t, Hence, to enure yieling o the tenion teel at limit tate the epth o neutral ai, houl e le than or equal to

6 Type o RC Beam Failure A applie moment on the eam ection increae eyon the linear elatic tage, the concrete train an tree enter the nonlinear tage. The ehavior o the eam in the nonlinear tage epen on the amount o reinorcement provie. The reinorcing teel can utain very high tenile train however, the concrete can accommoate compreive train much lower compare to it. So, the inal collape o a normal eam at ultimate limit tate i caue y the cruhing o concrete in compreion, regarle o whether the tenion teel ha yiel or not. Type o RC Beam Failure Depening on the amount o reinorcing teel provie, leural ailure may occur in three way: Balance : Concrete cruhe an teel yiel imultaneouly at the ultimate limit tate. The compreive train o concrete reache the ultimate train ε cu an the tenile train o teel reache the yiel train ε y imultaneouly. The epth o neutral ai, = Uner-reinorce : Steel reinorcement yiel eore concrete cruhe. The area o tenion teel provie i le than alance ection. The epth o neutral ai, < The ailure i graual, giving ample prior warning o the impening collape. Thi moe i ailure i preerre in eign practice. Over-reinorce : Concrete ail in compreion eore teel yiel. The area o teel provie i more than area provie in alance ection. The epth o neutral ai, > The ailure i uen (without any ign o warning) an rittle. Over-reinorce are not permitte.

7 Introuction For a ingly reinorce eam EC limit the epth to the neutral ai, to 0.45 ( 0.45) or concrete cla C50/60 to enure that the eign i or the uner-reinorce cae where ailure i graual, a note aove. For urther unertaning, ee the graph hown elow. Analyi o Section Section 6.1 EN , eal with the analyi an eign o ection or the ultimate limit tate eign conieration o tructural element ujecte to ening. The two common type o reinorce concrete eam ection are: Rectangular ection : Singly an ouly reinorce Flange ection : Singly an ouly reinorce

8 Singly Reinorce Rectangular Beam Beam cro ection, train an tree itriution at ULS o ingly reinorce rectangular eam h = 0.8 F cc Neutral ai z = 0.5 A ε t Notation: h = Overall epth = Eective epth = With o ection = Depth o tre lo A = Area o tenion reinorcement = Neutral ai epth = Characteritic trength o concrete z = Lever arm = Characteritic trength o reinorcement F t Singly Reinorce Rectangular Beam Tenion orce o teel, F t F t = Stre Area A Compreion orce o concrete, F cc F cc = Stre Area ( 0.8) For equilirium, total orce in the ection houl e zero. Fcc F t A A

9 Singly Reinorce Rectangular Beam oment reitance with repect to the teel Let; F cc z Area o tenion reinorcement, Thereore; oment reitance with repect to the concrete F z t 0.87 A 0. 4 K A K ( 0.4) Singly Reinorce Rectangular Beam To enure that the ection eigne i uner-reinorce it i neceary to place a limit on the maimum epth o the neutral ai (). EC ugget: 0.45 Then ultimate moment reitance o ingly reinorce ection or al can e otaine y; al 4 al al [ (0.45 )].[ 0.4(0.45 )] ( ).(0.8 ) al Kal...

10 Singly Reinorce Rectangular Beam Thereore; = K... al = K al... where; K al = I; al or K K al : Singly reinorce rectangular eam (Tenion reinorcement only) > al or K > K al : Douly reinorce rectangular eam (Section require compreion reinorcement) Eample 3.1 The cro ection o rectangular eam i hown in igure elow. Uing tre lo iagram an the ata given, etermine the area an the numer o reinorcement require. Data: Deign moment, ED = 00 kn.m = 5 N/mm = 500 N/mm = 50 mm = 450 mm

11 Solution o Eample 3.1 Calculate the ultimate moment reitance o ection, al al (5)(50)(450 ) 11.36kNm 00kNm Singly reinorce ection Neutral ai epth, (5)(50)( )( ) Solution o Eample 3.1 = mm Ue = 188 mm Cheing; Lever arm, z = ( 0.4) = ( (188)) = mm Area o reinorcement, A A 0.87 z (500)(374.8) 17mm Provie 4H0 ( A prov 157mm )

12 Eample 3. Figure elow how the cro ection o a ingly reinorce eam. Determine the reitance moment or that cro ection with the aitance o a tre lo iagram. Given = 5 N/mm an = 500 N/mm. 50 mm 450 mm H5 Solution o Eample 3. A tre lo iagram i rawn with the important value an notation. = = 450 = 0.8 F cc =0.454 Neutral ai z = 0.4 For equilirium; Fcc F t A = 98 mm A F t =0.87 A A

13 Solution o Eample (500)(98) 151 mm 0.454(5)(50) Cheing; oment reitance o ection; F cc ( )( (151)) 167 knm F t z Douly Reinorce Rectangular Beam When the loa applie increae graually an it will reach a tate that the compreive trength o concrete i not aequate to take aitional compreive tre. Compreion reinorcement i require to take the aitional compreive tre. Thi ection i name a ouly reinorce ection. A h A

14 Douly Reinorce Rectangular Beam Strain an tre lo iagram o ouly reinorce eam A ε c = 0.8 F c F cc h Neutral ai z 1 = z = 0.4 A ε t F t Douly Reinorce Rectangular Beam Internal orce; Fcc F A t an Fc 0.87 A ' Lever arm; z 0. 4 For equilirium o internal orce; F F F z 1 t cc c 0.87 A A ' '

15 Douly Reinorce Rectangular Beam Taking moment aout the centroi o the tenion teel, Fcc. z Fc. z1 ( ).( 0.4) (0.87 A ').( ') For eign purpoe, = 0.45 ( ).[ 0.4(0.45 )] (0.87 A ').( ') (0.87 A ').( ( 0.87 A ').( ') al ') The area o compreion reinorcement, A A ' ( al) 0.87 ( ') or A ' ( K K 0.87 al ) ( ') Douly Reinorce Rectangular Beam The area o Tenion reinorcement, A ultiplie equilirium internal orce equation y z, Limiting = 0.45 an z = 0.4(0.45) = A z z 0.87 A ' z 0.87 A z (0.45 )(0.8 ) 0.87 A ' z 0.87 A z A ' z Kal A A ' or A A ' 0.87 z 0.87 z

16 Douly Reinorce Rectangular Beam Stre in compreion reinorcement. The erivation o eign ormula or ouly reinorce ection aume that the compreion reinorcement reache the eign trength o 0.87 at ultimate limit tate. From the train iagram a hown in igure elow. h A ε c c ( ') ') ( c A ε t ' c Douly Reinorce Rectangular Beam For the eign trength 0.87 to e reache, ε c = 0.87 / E c ' 0.87 E (500) Thereore, i / < 0.38 the compreion reinorcement can e aume reach the eign trength o I / > 0.38, a reuce tre houl e ue. c E. c 3 c 0010 (0.0035)(1 '/ ) 700(1 '/ )

17 Eample 3.3 The cro ection o rectangular eam i hown in igure elow. Uing the ata given, etermine the area an the numer o reinorcement require. Data: Deign moment, ED = 450 kn.m = 5 N/mm = 500 N/mm = 50 mm = 50 mm = 500 mm Solution o Eample 3.3 Ultimate moment reitant o ection, al al (5)(50)(500 )(10 6 ) 60.94kNm 450kNm Compreion reinorcement i require Area o compreion reinorcement, A A ' ( al ) / 0.87 ( ) mm 6 ( ') / 0.87(500)(500 50)

18 Solution o Eample 3.3 Cheing / ratio ' / (500) 5mm 50/ Compreion teel achieve it eign trength at 0.87 Area o tenion teel, A A 0.87 al z 49mm A ' (0.8500) Provie H5 (A Prov. = 98 mm ) Compreion reinorcement 5H5 (A Prov. = 454 mm ) Tenion reinorcement Eample 3.4 Calculate moment reitance o the ouly reinorce ection hown in igure elow. Given = 30 N/mm an = 500 N/m. = 50 mm = 50 mm 3H0 = 500 mm 5H5

19 Solution o Eample 3.4 A tre lo iagram i rawn with the important value an notation = 50 mm = 50 mm 3H0 F c = 0.87 A 0.8 F cc = = 500 mm Z1 Z Neutral Ai 5H5 F t = 0.87 A Solution o Eample 3.4 Reinorcement ue 3H0, A = 943 mm & 5H5, A = 455 mm Neutral ai epth, 0.87 ( A A ') 0.87(500)( ) (30)(50) 193mm Cheing the tre o teel / 193/ ' / 50/ Steel achieve it eign trength 0.87 y a aume

20 Solution o Eample 3.4 oment reitance o ection, Fc. z 1 Fcc. z 0.87 A '( ') ( 0.4) 0.87(500)(943)(500 50) 0.454(30)(50)(193)( (193)) 10 46kNm 6 Flange Beam Flange eam occur when eam are cat integrally with an upport a continuou loor la. Part o the la ajacent to the eam i counte a acting in compreion to orm T- an L-eam a hown in igure elow. e w h e h Where; e = eective lange with w = reath o the we o the eam. h = thine o the lange. T-Beam w L-Beam

21 Flange Beam The eective with o lange, e i given in Sec o EC. e houl e ae on the itance l o etween point o zero moment a hown in igure elow. The eective lange with, e or T-eam or L-eam may e erive a: Where e e, i e, i 0.i 0.1l o 0. l Flange Beam w o an e, i i

22 Eample 3.5 Bae on igure elow, etermine the eective lange with, e o eam B/ FS1 (150 thk.) FS (150 thk.) FS3 (150 thk.) A B C D Solution o Eample 3.5 l o (itance etween point o zero moment) mm 4500 mm l o = = 550 mm l o = 0.15 ( ) = 115 mm l o = = 385 mm Eective lange with, e e e, i w

23 Solution o Eample 3.5 e e,1 e, A B C w = 00 mm = 500/ = 150 = 4000/ = 000 Span 1- = = 350 mm e1 = 0.(150) + 0.1(550) = 505 mm < 0.l o = 510 mm < 1 = 150 mm e = 0.(000) + 0.1(550) = 655 mm > 0.l o = 510 mm < = 000 mm e = ( ) + 00 = 115 mm < 350 mm Solution o Eample 3.5 e e,1 e, A B C w = 00 mm = 500/ = 150 = 4000/ = 000 Span -3 = = 350 mm e1 = 0.(150) + 0.1(385) = 63.5 mm < 0.l o = 765 mm < 1 = 150 mm e = 0.(000) + 0.1(385) = 78.5 mm > 0.l o = 765 mm < = 000 mm e = ( ) + 00 = mm < 350 mm

24 Solution o Eample 3.5 Span 1- Span -3 e = 115mm e = 1598 mm h = 150 mm h = 150 mm w = 00 mm w = 00 mm Flange Beam The eign proceure o lange eam epen on where the neutral ai lie. The neutral ai may lie in the lange or in the we. In other wor, there are three cae that houl e coniere. Neutral ai lie in lange ( < ) Neutral ai lie in we ( > ut < al ) Neutral ai lie in we ( > al ) e e

25 Flange Beam Neutral ai lie in lange ( < ) Thi conition occur when the epth o tre lo (0.8 ) le then the thine o lange, h a hown in igure elow. e h 0.8 F cc Z = 0.4 A F t w Flange Beam oment reitance o ection, F cc z cu e 4 For thi cae, maimum epth o tre lo, 0.8 are equal to h.567 h h / 0 Where, = Ultimate moment reitance o lange. Thereore, i the neutral ai lie in lange an the eign can e treate a rectangular ingly reinorce eam. A 0.87 z or A 0.87 ( 0.4)

26 Eample 3.6 The T-eam with imenion a hown in igure elow i ujecte to eign moment, = 50 knm. I = 30 N/mm an = 500 N/mm have een ue, etermine the area an numer o reinorcement require. e = 1450 mm h = 100 mm = 30 mm A w = 50 mm Solution o Eample 3.6 oment reitance o lange,.567 h h / 0 e / kNm 50kNm Since <, Neutral ai lie in lange Compreion reinorcement i not require (30)(1450)( )(30 0.4) mm Ue = 41.74mm

27 Flange Beam Cheing Lever arm,z z (41.74) mm Area o tenion reinorcement, A A mm Provie 4H5 (A prov = 1964 mm ) z 0.87(500)(303.3) Neutral ai lie in we ( < < al ) Flange Beam I the applie moment i greater than the neutral ai lie in the we a hown in igure elow. e h F cc F cc1 z z 1 A F t w

28 Flange Beam From the tre lo, internal orce; Lever arm, z Fcc 1 (0.567 )( w.0.8) F ) h cc (0.567 )( e F t z oment reitance, F z F. cc1 1 cc. z A w w z 0. 5h (0.567 )( ) h ( 0.5h ) 0 w e w Ultimate moment reitance o ection, al (When = 0.45) al Thereore; Flange Beam (0.45 ) w ( )( e w ) h ( 0.5h Divie oth ie y e, then; h al w e al e e al e 1 w e 1 ) h

29 Flange Beam I applie moment < al, then compreion reinorcement are not require. Area o tenion reinorcement can e calculate a ollow y taking moment at F cc. A 0.87 A ( 0.5h ) [( 0.5h ) ( 0.4)] Uing; = 0.45 A F t. z Fcc 1.( z z1) w ( w ( [ h 0.5h [ h ) ) ] h ] w Eample 3.7 The T-eam with imenion a hown in igure elow i ujecte to eign moment, = 670 knm. I = 30 N/mm an = 500 N/mm have een ue, etermine the area an numer o reinorcement require. e = 1450 mm h = 100 mm = 30 mm w = 50 mm

30 Solution o Eample 3.7 oment reitance o lange,.567 h h / 0 e / kNm 670kNm Since >, Neutral ai lie in we al e (30) 6 al 0.153(30)(1450)(30 ) 10 68kNm al 68kNm 670kNm Compreion reinorcement i not require Solution o Eample 3.7 Area o tenion reinorcement, A A mm w ( [ h h ) ] 0.1(30)(50)(30)[0.36(30) 100] 0.87(500)(30 50) Provie 8H3 (A prov = 6433 mm )

31 Flange Beam Neutral ai lie in we ( > al ) I the applie moment i greater than al the neutral ai lie in the we an the compreion reinorcement houl e provie. The tre lo are hown in igure elow. e A 1 h F c F cc 0.8 F cc1 z 3 z z 1 A F t w Flange Beam From the tre lo, internal orce; Fcc 1 (0.567 )( w.0.8) F ) h cc (0.567 )( e Fc 0.87 A ' F A t Lever arm, z w w z z 0. 5h z 3 ' oment reitance, F z F. z F. cc1 1 cc c. z 3

32 Flange Beam (0.567 )( ) h ( 0.5h ) 0 cu w cu e w 0.87 A '( ') When = 0.45, then al 0.87 A '( ') Area o compreion reinorcement, A A ' al 0.87 ( ') Flange Beam For equilirium o orce F F F t cc1 cc F c 0.87 A w (0.45 ) ( e w ) 0.87 A ' Area o tenion reinorcement, A A 0. w h ( e w ) A '

33 Deign Formula Deign Proceure or Rectangular Section Suppoe the eign ening moment i, eam ection i, concrete trength i an teel trength i, to etermine the area o reinorcement, procee a ollow. Deign Formula

34 Deign Formula Deign Formula Deign Proceure or Flange Section Suppoe the eign ening moment i, eam ection i, concrete trength i an teel trength i, to etermine the area o reinorcement, procee a ollow.

35 Deign Formula Deign Formula