10/14/2011. Types of Shear Failure. CASE 1: a v /d 6. a v. CASE 2: 2 a v /d 6. CASE 3: a v /d 2

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1 V V Types o Shear Failure a v CASE 1: a v /d 6 d V a v CASE 2: 2 a v /d 6 d V a v CASE 3: a v /d 2 d V 1

2 Shear Resistance Concrete compression d V cz = Shear orce in the compression zone (20 40%) V a = Interlocking between aggregates (35 50%) V Steel tension V d = Dowel action (35 50%) Shear orce is transmitted through the crack member by a combination o the uncracked concrete in compression zone, V cz, the dowelling action o the lexural reinorcement, V d and aggregate interlocking across tension cracks, V a EC2 Shear Design EC 2 introduces the strut inclination method or shear capacity checks. In this method the shear is resisted by concrete struts acting in compression and shear reinorcement acting in tension. 2

3 z = 0.9d 10/14/2011 EC2: Cl EC2: Cl Assumed truss model or the strut inclination method Concrete strut in compression Vertical shear steel in tension d z cot Longitudinal steel in tension b w V Ed z z cos cd b w z cos V Rd, max V Ed sin V Ed cot V Ed 3

4 (1) Diagonal Compressive Strut V Rd, max = [ cd (b w z cos θ)] sin θ = cd b w z cos θ sin θ In EC2 this equation is modiied by the inclusion o a strength reduction actor or concrete cracked in shear v 1 and the introduction o coeicient taking account o the state o the stress in compression chord α cw thus, V Rd, max = α cw v 1 cd b w z / (cot θ + tan θ) = α cw v 1 ( ck /1.5) b w (0.9d) / (cot θ + tan θ) = α cw v ck b w d / (cot θ + tan θ) (1) Diagonal Compressive Strut It is set by the EC2 to limit the θ value between 22 to 45. The recommended value or α cw and v 1 are given in Cl EC2. For the purpose o this module the ollowing values are used: α cw = 1.0 and v 1 = 0.6 (1 ck /250), hence V Rd,max 0 ck.36 ckbwd (1 / 250) (cot tan ) 4

5 (2) Vertical Shear Reinorcement The shear resistance o the link is given by V Ed = V Rd, s = ywd A sw = ( yk / 1.15) A sw = 0.87 yk A sw (2) Vertical Shear Reinorcement I the links are spaced at a distance s apart, then the shear resistance o the link is increased proportionately and is given by; V Ed = V Rd, s = 0.87 yk A sw (z cot θ / s) = 0.87 yk A sw (0.9d cot θ / s) = 0.78 yk A sw d (cot θ / s) 5

6 (2) Vertical Shear Reinorcement Thus, rearranging them gives; A sw s V Ed 0.78 ykd cot (2) Vertical Shear Reinorcement EC2 (Cl ) speciies a minimum value or A sw /s such that; A sw s b w yk ck EC2 (Cl ) also speciies that the maximum spacing o vertical link, s should not exceed 0.75d. 6

7 Redesign section 10/14/2011 (3) Additional Longitudinal Force The longitudinal tensile orce, F td is caused by the horizontal component required to balance the compressive orce in the inclined concrete strut; Longitudinal orce = (V Ed /sin ) cos = V Ed cot Assumed hal o this orce is carried by the reinorcement in the tension zone o the beam, then the additional tensile orce provided in the tensile zone is given by; F td = 0.5V Ed cot θ Shear Design Procedure in EC2 Start Determine shear orce, V Ed Determine V Rd, max or cot = 1.0 ( = 45) and cot = 2.5 ( = 22) V Rd,max 0 ck.36 ckbwd (1 / 250) (cot tan ) I V Ed V Rd, max cot = 1.0 I V Ed V Rd, max cot = 2.5 I V Rd, max cot θ = 2.5 V Ed V Rd, max cot θ = 1.0 A B 7

8 Shear Design Procedure in EC2 A Calculate shear reinorcement using cot = 2.5 Asw VEd V s 0.78 d cot d yk yk Ed C Shear Design Procedure in EC2 B Calculate 0.5sin 1 VEd 0.18bwdck (1 ck / 250) Calculate shear links A s sw VEd.78 d cot 0 yk C 8

9 Shear Design Procedure in EC2 C Calculate minimum links required by EC2: Cl (5) and s 0.75d D Yes Flanged beam? No Calculate additional longitudinal tensile orce caused by shear End Example 1 Design the required shear reinorcement rom the beam section shown below. Take yk = 500 N/mm 2 and ck = 30 N/mm kn/m 225 mm 8 m 2H16 d = 500 mm 3H25 (1473 mm 2 ) 9

10 Example 1: Solution Maximum shear orce; V Ed = wl/2 = /2 = 400 kn Concrete strut capacity; V Rd, max = 0.36b w d ck (1 ck /250) / (cot + tan ) = (1 30/250) (cot + tan ) or = 22, cot = 2.5 V Rd, max = 371 kn = 45, cot = 1.0 V Rd, max = 535 kn V Rd, max cot = 2.5 (371 kn) V Ed (400 kn) V Rd, max cot = 1.0 (535 kn) Thereore, 22 Example 1: Solution = 0.5sin -1 [V Ed / 0.18b w d ck (1 ck /250)] = 0.5sin (1 30/250) = 24.2 tan = 0.45 ; cot = 2.22 Shear links; A sw /s = V Ed / 0.78 yk dcot = / ( ) =

11 Example 1: Solution Try link: H10 A sw = 157 mm 2 Spacing, s = 157/0.923 Minimum links; A sw /s = 170 mm s max = 0.75d (375 mm) = 0.08 ck 1/2 b w / yk = 0.08 (30) 1/2 225 / 500 = Try link: H10 A sw = 157 mm 2 Spacing, s = 157/0.197 Provide H mm = 797 mm s max = 0.75d (375 mm) Provide H mm Example 1: Solution V min = (A sw /s)(0.78d yk cot ) H H H m 4.33 m 1.83 m

12 Example 1: Solution Additional longitudinal reinorcement; Additional tensile orce, F td = 0.5V Ed cot = = 445 kn Additional tension reinorcement, A s = F td / 0.87 yk = / = 1022 mm 2 Provide 3H25 ( 1473 mm 2 ) Flanged Beam Shear between the web and langed o a langed section 12

13 Flanged Beam The longitudinal shear stress, v Ed at the web-lange interace is determine according to; where: F d v Ed M ( b b x ( d h / 2) b w ( h ) / 2 F d. x) M = the change in moment over the distance x x = hal the distance between the sections with zero moment and that where maximum moment occurs. Where point loads occur, x should not exceed the distance between the loads. Flanged Beam The concrete strut capacity o the lange is given by; v Rd = v cd sin cos = 0.6 (1 ck /250) ( ck /1.5) sin cos = 0.4 ck (1 ck /250) (cot + tan ) The permitted range o the values o cot are recommended as ollows: 1.0 cot 2.0 or compression langes ( ) 1.0 cot 1.25 or tension langes ( ) 13

14 Flanged Beam: Design Procedure D Calculate longitudinal design shear stress, v Ed v Ed Fd ( h. x) F Yes v Ed 0.27 ctk 0.4 ctd = 0.4( ctk /1.5) = 0.27 ctk No Check the stresses in the incline strut. Compare v Ed with v Rd, max F Yes No Use min v Ed v Rd, max E Flanged Beam: Design Procedure E Calculate and cot 0.5sin ck ved (1 ck / 250) F Calculate transverse shear reinorcement A s s vedh 0.87 cot yk Calculate minimum transverse steel area where b = 1000 mm Calculate additional longitudinal tensile orce caused by shear End 14

15 d = 530 mm d = 45 mm 10/14/2011 Example 2 Design the required shear reinorcement rom the beam section shown below. Take yk = 500 N/mm 2 and ck = 25 N/mm kn/m 9 m 600 mm 2H mm 3H mm Example 2: Solution Maximum shear orce; V Ed = wl/2 = /2 = 405 kn Concrete strut capacity; V Rd, max = 0.36b w d ck (1 ck /250) / (cot + tan ) = (1 25/250) (cot + tan ) or = 22, cot = 2.5 V Rd, max = 373 kn = 45, cot = 1.0 V Rd, max = 537 kn V Rd, max cot = 2.5 (373 kn) V Ed (405 kn) V Rd, max cot = 1.0 (537 kn) Thereore, 22 15

16 Example 2: Solution = 0.5sin -1 [V Ed / 0.18b w d ck (1 ck /250)] = 0.5sin (1 25/250) = 24.5 tan = 0.46 ; cot = 2.19 Shear links; A sw /s = V Ed / 0.78 yk dcot = / ( ) = Example 2: Solution Try link: H10 A sw = 157 mm 2 Spacing, s = 157/0.893 Minimum links; A sw /s = 176 mm s max = 0.75d (398 mm) = 0.08 ck 1/2 b w / yk = 0.08 (25) 1/2 250 / 500 = Try link: H10 A sw = 157 mm 2 Spacing, s = 157/0.200 Provide H mm = 786 mm s max = 0.75d (398 mm) Provide H mm 16

17 Example 2: Solution V min = (A sw /s)(0.78d yk cot ) H H H m 4.76 m 2.12 m Example 2: Solution Transverse steel in the lange; x = 0.5(L/2) = 9000/4 = 2250 mm Change o moment over distance x rom zero moment: M = (wl/2)(l/4) (wl/4)(l/8) = knm Change in longitudinal orce; M ( b bw ) / 2 Fd x ( d h / 2) b = ( ) (530 55) (2 600) = 420 kn 17

18 Example 2: Solution Longitudinal shear stress; v Ed = F td / (h x) = / ( ) = 1.70 N/mm 2 Since v Ed (1.70 N/mm 2 ) 0.27 ctk = = 0.49 N/mm 2 Transverse steel reinorcement is required Example 2: Solution Concrete strut capacity in the lange; v Rd, max = 0.4 ck (1 ck /250) / (cot + tan ) = (1 25/250) (cot + tan ) or = 27, cot = 2.0 v Rd, min = 3.59 N/mm 2 = 45, cot = 1.0 v Rd, max = 4.50 N/mm 2 v Ed (1.70 N/mm 2 ) v Rd, min cot = 2.0 (3.59 N/mm 2 ) and v Ed (1.70 N/mm 2 ) V Rd, max cot = 1.0 (4.50 N/mm 2 ) Thereore, use = 27 ; tan = 0.50 ; cot =

19 Example 2: Solution Transverse shear reinorcement; A s / s = v Ed h / 0.87 yk cot = / ( ) = 2.0 Try H10: A s = 79 mm 2 Spacing, s = 79/0.21 = 367 mm Minimum transverse steel area; A s, min = 0.26 ( ctm / yk ) bh = 0.26 (2.60/500) bh = bh bh = = 147 mm 2 Provide H (A s = 262 mm 2 /m) Example 2: Solution Additional longitudinal reinorcement; Additional tensile orce, F td = 0.5V Ed cot = = 444 kn Additional tension reinorcement, A s = F td / 0.87 yk = / = 1021 mm 2 Provide 3H25 ( 1473 mm 2 ) 19

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- Rectangular Beam Design -

Semester 1 2016/2017 - Rectangular Beam Design - Department of Structures and Material Engineering Faculty of Civil and Environmental Engineering University Tun Hussein Onn Malaysia Introduction The purposes