Design of T and L Beams in Flexure

Transcription

1 Lecture 04 Design of T nd L Bems in Flexure By: Prof. Dr. Qisr Ali Civil Engineering Deprtment UET Peshwr drqisrli@uetpeshwr.edu.pk Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design Topics Addressed Introduction to T nd L Bems ACI Code provisions for T nd L Bems Design Cses Design of Rectngulr T-bem Design of True T-bem References Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 2 1

2 Introduction to T nd L Bem The T or L Bem gets its nme when the slb nd bem produce the cross sections hving the typicl T nd L shpes in monolithic reinforced concrete construction. Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 3 Introduction to T nd L Bem In csting of reinforced concrete floors/roofs, forms re built for bem sides, the underside of slbs, nd the entire concrete is mostly poured t once, from the bottom of the deepest bem to the top of the slb. Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 4 2

3 Introduction to T nd L Bem Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 5 Introduction to T nd L Bem Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 6 3

4 Introduction to T nd L Bem Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 7 Introduction to T nd L Bem b b b b Compression Tension Tension Compression Section - Section b-b Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 8 4

5 Introduction to T nd L Bem Positive Bending Moment In the nlysis nd design of floor nd roof systems, it is common prctice to ssume tht the monolithiclly plced slb nd supporting bem interct s unit in resisting the positive bending moment. As shown, the slb becomes the compression flnge, while the supporting bem becomes the web or stem. Compression Tension Section - Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 9 Introduction to T nd L Bem Negtive Bending Moment In the cse of negtive bending moment, the slb t the top of the stem (web) will be in tension while the bottom of the stem will be in compression. This usully occurs t interior support of continuous bem. Tension Compression Section b-b Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 10 5

6 ACI Code Provisions for T nd L Bems For T nd L bems supporting monolithic or composite slbs, the effective flnge width shll include the bem weidth plus n effective overhnging flnge width in ccordnce with ACI Tble Slb Effective Flnge Width Flnge h Web or Stem s w Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 11 ACI Code Provisions for T nd L Bems Clcultion of Effective Flnge Width ( ) (ACI ) T - Bem h 2 + s w s w s w 3 + l n /4 Lest of the bove vlues is selected Where is the width of the bem, h is the slb thickness, s w is the cler distnce to the djcent bem nd l n is the cler length of bem. Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 12 6

7 ACI Code Provisions for T nd L Bems Clcultion of Effective Flnge Width ( ) (ACI ) L - Bem 1 + 6h 2 + s w /2 3 + l n /12 b w Slb s w s w Effective Flnge Width Flnge h Web or Stem Lest of the bove vlues is selected Where is the width of the bem, h is the slb thickness, distnce to the djcent bem nd l n is the cler length of bem. s w is the cler Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design Design Cses In designing T-Bem for positive bending moment, there exists two conditions: Condition 1. The depth of the compression block my be less thn or equl to the slb depth i.e. flnge thickness ( h) In such condition the T-Bem is designed s rectngulr bem for positive bending with the width of compression block equl to. h d N.A A s Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 14 7

8 Design Cses Condition 2. The compression block my cover the flnge nd extend into the web ( h) In such condition the T-Bem is designed s true T-bem. ( - )/2 d h N.A A st Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 15 Design of Rectngulr T-bem Flexurl Cpcity When h h d N.A A s Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 16 8

9 Design of Rectngulr T-bem Flexurl Cpcity ( F x = 0) 0.85f c = A s f y d h ε c N.A 0.85f c /2 C l = (d - /2) = A s f y / 0.85f c ( M = 0) A s ε s T M n = T*l = A s f y (d /2) As ΦM n = M u ; ΦA s f y (d /2) = M u Therefore, A s = M u /Φf y (d /2) The other checks remins s tht of the rectngulr bem design. Note: In clculting A smx nd A smin, use, not. Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 17 Design of Rectngulr T-bem Exmple 01 The roof of hll hs 5 thick slith bems hving 30 feet c/c nd 28.5 feet cler length. The bems re hving 9 feet cler spcing nd hve been cst monolithiclly with slb. Overll depth of bem (including slb thickness) being 24 in nd width of bem web being 14 in. Clculte the steel reinforcement re for the simply supported bem ginst totl fctored lod (including self weight of bem) of 3 k/ft. Use f c = 3 ksi nd f y = 60 ksi. Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 18 9

10 Design of Rectngulr T-bem Exmple Solution: Spn length (l c/c ) = 30 ; cler length (l n ) = 28.5 W u = 3 k/ft d = = 21.5, = 14 ; h = 5 5 Effective flnge width ( ) is minimum of, + 16h = = A s 19 + s w = = l n /4 + = /4 = Therefore, = 94 Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 19 Design of Rectngulr T-bem Exmple Solution: Check if the bem behviour is T or rectngulr. M u = w u l 2 /8 = 3 x 30 2 x12 / 8 = 4050 in-kips Tril # 01 Let = h = 5 A s = M u /Φf y (d /2) = 4050/{ (21.5 5/2)} = 3.94 in 2 Tril # 02 = A s f y /(0.85f c ) = / ( ) = 0.98 h = 5 Therefore, design s Rectngulr bem. Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 20 10

11 Design of Rectngulr T-bem Exmple Solution: A s = M u /Φf y (d /2) = 4050/{ ( /2)} = 3.56 in 2 Tril # 03 = A s f y /(0.85f c ) = / ( ) = 0.89 A s = M u /Φf y (d /2) = 4050/{ ( /2)} = 3.56 in 2 Therefore A s = 3.56 in 2 Try #8 brs, No of Brs = 3.56 / 0.79 = 4.50, sy 5 brs Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 21 Design of Rectngulr T-bem Exmple Solution: Check for mximum nd minimum reinforcement llowed by ACI: A smin = 3 ( f / f y ) d (200/f y ) d 3 ( f /f y ) d = 3 ( 3000 /60000) d = x 14 x21.5 = 0.82 in 2 (200/f y ) d = (200/60000) x = 1.0 in 2 A smin = 1.0 in 2 A smx = 0.27 (f c / f y ) d = 0.27 x (3/60) x = 4.06 in 2 A smin (1.0) < A s (3.95) < A smx (4.06) O.K! Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 22 11

12 Design of Rectngulr T-bem Exmple Solution: Check design cpcity your self. Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 23 Design of Rectngulr T-bem Exmple Solution: = (3+2),#8 brs =14 Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 24 12

13 Design of Rectngulr T-bem Exmple 2: Design the Bem B1 yourself for the following moments f c = 4 ksi, f y = 60 ksi, bem width = 15, Slb thickness = 6 Overll bem depth (including slb thickness) = 24, K K 24 cler length 24 cler length K B1-Moment Digrm Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 25 Design of True T-bem Flexurl Cpcity When > h ( - )/2 ( - )/2 d h d = + d N.A A st A sf A s bw ΦM n ΦM n1 ΦM n2 ΦM n = ΦM n1 + ΦM n2 Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 26 13

14 Design of True T-bem Flexurl Cpcity ΦM n1 Clcultion: From stress digrm d ( - )/2 h h/2 C 1 N.A l 1 = d - h/2 T 1 = C 1 A sf C 1 = 0.85 f c ( - )h T 1 T 1 = A sf f y ΦM n1 A sf f y = 0.85f c ( - )h Everything in the eqution is known except A sf Therefore, A sf = 0.85f c ( - )h / f y ΦM n1 = T 1 x l 1 = ΦA sf f y (d h/2) Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 27 Design of True T-bem Flexurl Cpcity ΦM n2 Clcultion: From stress digrm T 2 = C 2 d ( - )/2 h /2 C 2 l 2 = d - /2 N.A A s C 2 = 0.85 f c T 2 T 2 = A s f y ΦM n2 A s f y = 0.85f c = A s f y / (0.85 f c ) ΦM n2 = T 2 x l 2 = Φ A s f y (d /2) Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 28 14

15 Design of True T-bem A s Clcultion We know tht ΦM n = M u ΦM n1 + ΦM n2 = M u ΦM n1 is lredy known to us, Therefore ΦM n2 = M u ΦM n1 And s, ΦM n2 = T 2 x l 2 = Φ A s f y (d /2) d ( - )/2 A s h /2 C 2 l 2 = d - /2 T 2 N.A ΦM n2 Also ΦM n2 = M u ΦM n1 Therefore, A s = (M u ΦM n1 )/ Φf y (d /2); nd = A s f y / (0.85 f c ) Clculte A s by tril nd success method. Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 29 Design of True T-bem Ductility Requirements T = C 1 + C 2 [ F x = 0 ] A st f y =0.85f c ( )h f c A st f y = A sf f y +0.85f c For ductility ε s = ε t = (ACI tble ) For = β 1 c = β d, A st will become A stmx, Therefore, A stmx f y = 0.85f c β d + A sf f y A stmx f y = 0.85f c β d + A sf A stmx = β 1 (f c /f y )d + A sf For f c 4000 psi, β 1 = 0.85, A stmx = 0.27 (f c / f y ) d + A sf OR A stmx = A smx (singly) + A sf So, for T-bem to behve in ductile mnner A st, provided A stmx Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 30 15

16 Design of True T-bem Exmple 03 Design simply supported T bem to resist fctored positive moment equl to 6500 in-kip. The bem is 12 wide nd is hving 20 effective depth including slb thickness of 3 inches. The centre to centre nd cler lengths of the bem re 25.5 nd 24 respectively. The cler spcing between the djcent bems is 3 ft. Mteril strengths re f c = 3 ksi nd f y = 40 ksi. Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 31 Design of True T-bem Exmple Solution: Spn length (l c/c ) = 25.5 ; cler length (l n ) = 24 d = 20 ; = 12 ; h = 3 Effective flnge width ( ) is minimum of, + 16h = = 60 + s w = = 48 + l n /4 = /4 = 84 Therefore, = 48 Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 32 16

17 Design of True T-bem Exmple Solution: Check if the bem behviour is T or rectngulr. Let = h = 3 A s = M u /Φf y (d /2) = 6500/{ (20 3/2)} = 9.76 in 2 = A s f y /(0.85f c ) = / ( ) = 3.18 > h Therefore, design s T bem. Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 33 Design of True T-bem Exmple Solution: Design: We first clculte A sf A sf = 0.85f c ( ) h/f y = (48 12) 3/40 = in 2 The nominl moment resistnce (ФM n1 ), provided by A sf is, ФM n1 = ФA sf f y {d h/2} = {20 3/2} = in-kip Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 34 17

18 Design of True T-bem Exmple Solution: Design: The nominl moment resistnce (ФM n2 ), provided by remining steel A s is, ФM n2 = M u ФM n1 = = in-kip Let = 0.2d = = 4 A s = ФM n2 / {Фf y (d /2)} = / { (20 4/2)}= 2.95 in 2 = A s f y /(0.85f c ) = / ( ) = 3.85 This vlue is close to the ssumed vlue of. Therefore, A st = A sf + A s = = 9.84 in 2 (13 #8 Brs) Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 35 Design of True T-bem Exmple Solution: Ductility requirements, (A st = A s + A sf ) A stmx A stmx = A smx (singly) + A sf = = in 2 ; A smx (singly) =0.27(f c /f y )bd A st = A s + A sf = = in 2 < O.K. Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 36 18

19 Design of True T-bem Exmple Solution: Ensure tht A st > A smin A st = in 2 A smin = 3 ( f /f y ) d (200/f y ) d 3 (f c )/f y d = 3 ( (3000)/40000) x 12 x 20 = 0.98 in 2 200/f y d = (200/40000) x 12 x 20 = 1.2 in 2 A smin = 1.2 in 2 A st (10.27 in 2 ) > A smin (1.2 in 2 ) O.K. Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 37 Design of True T-bem Exmple Solution: Check design cpcity your self. Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 38 19

20 Design of True T-bem Exmple Solution: =48 3 d=20 (5+5+3),#8 brs =12 Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 39 References Design of Concrete Structures 14th Ed. by Nilson, Drwin nd Doln. Building Code Requirements for Structurl Concrete (ACI ) Prof. Dr. Qisr Ali CE 320 Reinforced Concrete Design 40 20