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
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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:
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:
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:
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)
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)
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)
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
Ved= 55.255bw= 150d= 328fck= 30cotϴ= 1 ϴ= 0.7854 45cotϴ= 2.5 ϴ= 0.38397 22fyk= 500
try link = H6= Asw= 57175
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
bw= 150d= 328fck= 30cotϴ= 1 ϴ= 0.7854 45cotϴ= 2.5 ϴ= 0.38397 22fyk= 500
try link = H6= Asw= 56.6275
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
hf= 125L= 1675bw= 150d= 316fck= 30cotϴ= 1 ϴ= 45cotϴ= 2 ϴ= 26.5fyk= 500b= 1158fctk = 2
Therefore transverse reinforcement required
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)
L= 3600Ved= 55.255bw= 150d= 328fck= 30cotϴ= 1 ϴ= ϴ=cotϴ= 2 ϴ= ϴ=fyk= 500b= 1158hf= 125
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)
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)
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)
L= 3600Ved= 55.255bw= 150d= 328fck= 30cotϴ= 1 ϴ= ϴ= 45cotϴ= 2 ϴ= ϴ= 26.5fyk= 500b= 1158hf= 125fctm= 2.9
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