Shear and Tranverse Design

21
SHEAR REINFORCEMENT Ved = 55.255 kN Vrd= 233.798 kN for 45 degree 161.018 kN for 22 degree If Ved < Vrd,max Asw/s= Ved/(0.78(fyk)(d)(cotϴ) 0.173 spacing,s Asw/(Asw/s) 328 Max spacing= 0.75d= 0.75x 328 246 SHEAR REINFORCEMENT Ved = 86.259 kN Vrd= 225.245 kN for 45 degree 155.127 kN for 22 degree If Ved < Vrd,max Asw/s= Ved/(0.78(fyk)(d)(cotϴ) 0.280 spacing,s Asw/(Asw/s) 202 (0.36(bw)(d)(fck)(1-fck/250))/(cotϴ+tanϴ) cotϴ=2.5, use cotϴ=2.5, and calculate the shea (0.36(bw)(d)(fck)(1-fck/250))/(cotϴ+tanϴ) cotϴ=2.5, use cotϴ=2.5, and calculate the shea

description

rc

Transcript of Shear and Tranverse Design

Page 1: Shear and Tranverse Design

SHEAR REINFORCEMENT

Ved = 55.255 kNVrd=

233.798 kN for 45 degree

161.018 kN for 22 degree

If Ved < Vrd,max

Asw/s= Ved/(0.78(fyk)(d)(cotϴ)0.173

spacing,s= Asw/(Asw/s)328

Max spacing= 0.75d= 0.75x 328 246 use H6-225

SHEAR REINFORCEMENT

Ved = 86.259 kNVrd=

225.245 kN for 45 degree

155.127 kN for 22 degree

If Ved < Vrd,max

Asw/s= Ved/(0.78(fyk)(d)(cotϴ)0.280

spacing,s= Asw/(Asw/s)202

(0.36(bw)(d)(fck)(1-fck/250))/(cotϴ+tanϴ)

cotϴ=2.5, use cotϴ=2.5, and calculate the shear reinforcement as follows:

(0.36(bw)(d)(fck)(1-fck/250))/(cotϴ+tanϴ)

cotϴ=2.5, use cotϴ=2.5, and calculate the shear reinforcement as follows:

Page 2: Shear and Tranverse Design

Max spacing= 0.75d= 0.75x 316 237 mm use H6-200

SHEAR REINFORCEMENT

Ved = 44.422 kNVrd=

233.798 kN for 45 degree

161.018 kN for 22 degree

If Ved < Vrd,max

Asw/s= Ved/(0.78(fyk)(d)(cotϴ)0.139

spacing,s= Asw/(Asw/s)407

Max spacing= 0.75d= 0.75x 328 246 use H6-225

SHEAR REINFORCEMENT

Ved = 86.259 kNVrd=

225.245 kN for 45 degree

155.127 kN for 22 degree

If Ved < Vrd,max

Asw/s= Ved/(0.78(fyk)(d)(cotϴ)0.280

spacing,s= Asw/(Asw/s)202

(0.36(bw)(d)(fck)(1-fck/250))/(cotϴ+tanϴ)

cotϴ=2.5, use cotϴ=2.5, and calculate the shear reinforcement as follows:

(0.36(bw)(d)(fck)(1-fck/250))/(cotϴ+tanϴ)

cotϴ=2.5, use cotϴ=2.5, and calculate the shear reinforcement as follows:

Page 3: Shear and Tranverse Design

Max spacing= 0.75d= 0.75x 316 237 use H6-200

SHEAR REINFORCEMENT

Ved = 55.255 kNVrd=

233.798 kN for 45 degree

161.018 kN for 22 degree

If Ved < Vrd,max

Asw/s= Ved/(0.78(fyk)(d)(cotϴ)0.181

spacing,s= Asw/(Asw/s)313

Max spacing= 0.75d= 0.75x 328 246 use H6-225

(0.36(bw)(d)(fck)(1-fck/250))/(cotϴ+tanϴ)

cotϴ=2.5, use cotϴ=2.5, and calculate the shear reinforcement as follows:

Page 4: Shear and Tranverse Design

Ved= 55.255 Med,max= 1.88bw= 150 z= 312d= 328fck= 30cotϴ= 1 ϴ= 0.785 45 Additional longitudinalcotϴ= 2.5 ϴ= 0.384 22fyk= 500 ∆Ftd =hf 125 69.07L 3600 tan 45 = 1b 1158 tan 22 = 0.404 Med,max/z 6.026fctm = 2.9

As,req= ∆Ftd/0.87(fyk)158.78

try link = H6 = Asw= 56.6 mm

Ved= 86.259 Med,max= 51.136bw= 150 z= 281d= 316fck= 30cotϴ= 1 ϴ= 0.785 45 Additional longitudinalcotϴ= 2.5 ϴ= 0.384 22fyk= 500 ∆Ftd =hf 125 107.82L 3600 tan 45 = 1b 1158 tan 22 = 0.404 Med,max/z 181.979fctm = 2.9

As,req= ∆Ftd/0.87(fyk)247.87

try link = H6 = Asw= 56.6 mm

0.5(Ved)(cotϴ)

0.5(Ved)(cotϴ)

Page 5: Shear and Tranverse Design

Ved= 44.422 Med,max= 21.218bw= 150 z= 312d= 328fck= 30cotϴ= 1 ϴ= 0.785 45 Additional longitudinalcotϴ= 2.5 ϴ= 0.384 22fyk= 500 ∆Ftd =hf 125 55.53L 3600 tan 45 = 1b 1158 tan 22 = 0.404 Med,max/z 68.006fctm = 2.9

As,req= ∆Ftd/0.87(fyk)127.65

try link = H6 = Asw= 56.6 mm

Ved= 86.259 Med,max= 52.067bw= 150 z= 279d= 316fck= 30cotϴ= 1 ϴ= 0.785 45 Additional longitudinalcotϴ= 2.5 ϴ= 0.384 22fyk= 500 ∆Ftd =hf 125 107.82L 3600 tan 45 = 1b 1158 tan 22 = 0.404 Med,max/z 186.620fctm = 2.9

As,req= ∆Ftd/0.87(fyk)247.87

try link = H6 = Asw= 56.6 mm

0.5(Ved)(cotϴ)

0.5(Ved)(cotϴ)

Page 6: Shear and Tranverse Design

Ved= 57.852 Med,max= 1.88bw= 150 z= 310d= 328fck= 30cotϴ= 1 ϴ= 0.785 45 Additional longitudinalcotϴ= 2.5 ϴ= 0.384 22fyk= 500 ∆Ftd =hf 125 72.32L 3600 tan 45 = 1b 1158 tan 22 = 0.404 Med,max/z 6.065fctm = 2.9

As,req= ∆Ftd/0.87(fyk)166.24

try link = H6 = Asw= 56.6 mm

0.5(Ved)(cotϴ)

Page 7: Shear and Tranverse Design

Asw/s= Ved/(0.78(fyk)(d)(cotϴ)0.17

spacing,s= Asw/(Asw/s)330

Max spacing= 0.75d= 0.75x 328246

If Vrd,max cotϴ=2.5<Ved<Vrd,max cotϴ=1.0

Page 8: Shear and Tranverse Design

Ved= 55.255bw= 150d= 328fck= 30cotϴ= 1 ϴ= 0.7854 45cotϴ= 2.5 ϴ= 0.38397 22fyk= 500

try link = H6= Asw= 57175

Page 9: Shear and Tranverse Design

Minimum link=Asw/s= 0.08(fck)^(1/2)(bw)/(fyk)

0.131

spacing,s= Asw/(Asw/s)430.5708 <0.75d= 246

use = 225 mm

Shear resisitance of minimum linksVmin= (Asw/s)(.78(fyk)(d)(cotϴ)

80.447

Page 10: Shear and Tranverse Design

bw= 150d= 328fck= 30cotϴ= 1 ϴ= 0.7854 45cotϴ= 2.5 ϴ= 0.38397 22fyk= 500

try link = H6= Asw= 56.6275

Page 11: Shear and Tranverse Design

Transverse steel in flange

∆x= (1675/2)837.5

∆M= (46.53-28.922x0.5x0.8375)34.419

the change in logitudinal force,

∆Fd= (∆M/(d-0.5hf)((b-bw)/2b))59.09

longitudinal shear stress

Ved= ∆Fd/(hf(∆x))0.56

Ved > 0.27(fctk) = 0.54 N/mm2 Therefore transverse reinforcement required

Page 12: Shear and Tranverse Design

hf= 125L= 1675bw= 150d= 316fck= 30cotϴ= 1 ϴ= 45cotϴ= 2 ϴ= 26.5fyk= 500b= 1158fctk = 2

Therefore transverse reinforcement required

Page 13: Shear and Tranverse Design

if Ved<= 0.27fctk

no required transverse (step 4)

Asf/sf == 0.08

try link : H10= Asf= 79spacinf, sf= Asf/(Asf/sf)

974

Ved(hf)/0.87(fyk)(cotϴf)

Page 14: Shear and Tranverse Design

L= 3600Ved= 55.255bw= 150d= 328fck= 30cotϴ= 1 ϴ= ϴ=cotϴ= 2 ϴ= ϴ=fyk= 500b= 1158hf= 125

Page 15: Shear and Tranverse Design

if Ved > 0.27fctk required transverse

45Vedmax=

5.28

26.5Vedmax= 4.22

Ved< 4.22Ved< 5.28

therefore use angle = 26.5

Asf/sf == 0.08

try link : H6= Asf= 28.3spacinf, sf=Asf/(Asf/sf)

349

minimum transverse

As,min= 0.26(b)(hf)(fctm)/fyk use b= 1000= 0.26(fctm/fyk)(b)(hf)= 0.0015 bhf > 0.0013bhf Therefore use == 189

0.40(fck)(1-(fck/250))/(cotϴf+tanϴf)

Ved(hf)/0.87(fyk)(cotϴf)

Page 16: Shear and Tranverse Design

L= 3600Ved= 55.255bw= 150d= 328fck= 30cotϴ= 1 ϴ= 45 tanϴ= 1cotϴ= 2 ϴ= 26.5 tanϴ= 0.5fyk= 500b= 1158hf= 125fctm= 2.9

H6-150(189mm2)

Page 17: Shear and Tranverse Design

Med,max= 207z= 428.9

Additional longitudinal

∆Ftd =55.26

Med,max/z = 482.63

As,req= ∆Ftd/0.87(fyk)127.02

0.5(Ved)(cotϴ)

Page 18: Shear and Tranverse Design

L= 3600Ved= 55.255bw= 150d= 328fck= 30cotϴ= 1 ϴ= ϴ= 45cotϴ= 2 ϴ= ϴ= 26.5fyk= 500b= 1158hf= 125fctm= 2.9