Week 5 Challenge
1) d/tw=(1200-2x50)x10=111>67ξ
check for shear buckling
minimum thickness requirement of web
d/tw<200ξ
1100/10=111<200ξ(ok)
shear capacity check
Vd=Avxfyw/√3xγm0
=1100x10x250/1.732x1.1
=1443418N
1443KN
Not safe in shear
Check for shear buckling as per IS800:2007, P-59
Vcr=Avxζb
ζb=fyw/√3xλw²
λw²=√(fyw/√3xζcr)
ζcr=Kvxπ²xE/12(1-μ²)/(d/tw)²
Kv=5.35 when transverse stiffeners are provided only at support.
E=2X105N/mm², μ=0.3
ζcr=5.35xπ²x2x105/12(1-0.3²)/(1100/10)²=79.84
λw=√(fyw/√3xζcr)=√(250/1.732x79.84)=1.3445
ζb=250/1.732x1.3445²=79.849
Vcr=1100x10x79.849=878339N=878.339<1000KN
Result=Thus it is concluded that web is not adequate to carry a factored shear force of 1000KN
2)d/tw=(1500-2x30)x12=120>67ξ
Shear buckling check required
Minimum web thickness requirement
d/tw=120<200ξ(ξ=1)ok
Shear capacity check
Vd=Avxfyw/√3xγm0=1440x12x250/√3x1.1=2267478.47N=2267.47KN
0.6Vd=1360.48KN<V=2000KN
(Not safe in shear)
Check for shear buckling as per IS800:2007, P-59
Vcr=Avxζb
Shear stress corresponding to web buckling
ζb=fyw/√3xλw²
λw²=√(fyw/√3xζcr)
Kv=5.35 when transverse stiffeners are provided only at support.
E=2X105N/mm², μ=0.3
ζcr=Kvxπ²xE/12(1-μ²)/(d/tw)²
ζcr=5.35xπ²x2x105/12(1-0.3²)/(1440/12)²=67.09
λw=√(fyw/√3xζcr)=√(250/1.732x67.09)=1.466
ζb=250/1.732x1.466²=67.16
Vcr=1440x12x67.16=1160524.8N=1160.52<2000KN
Check for the requirement of stiffener
If d/tw>85 then vertical stiffener required.
d/tw>200 then longitudinal stiffener required.
d/tw>250 then additional vertical stiffener required.
d/tw=(1500-2x30)/12=120>85ξ
only vertical stiffener need to be required.
Check for bearing capacity IS800 P-67
Fw=(b1+n2)xtwxfyw/γm0
b1=2x(2.5xtf)=2x(2.5x30)=150mm
Fw=(0.0+150)x12x250/1.1=409090.9=409KN<2000KN
Web is not capable to resist the factored shear force so stiffeners are required.
Considering the thickness tq of stiffener as 12mm(IS800 P-65)
Outstand L0=14twxξ=14x12=168mm
Outstand of the face of the web on either side of stiffener=20tqξ=2x20x12x1=480mm
core area=2x168x12+480x12=9792mm²
Ic or e=480x12³/12+12{12x168³/12+168x12x90²}=42.211x106mm4
ymin=√(I/A)=√(42.211X106/9792=65.65
Slendernessratio KL/y=0.7X1440/65.65=15.35
Fcd from table 9c of IS800 by interpolation Fcd=225.395N/mm²
Fc=AcorexFcd=9792x225.395=2207067.84N=2207.067KN>Vumax=2000KN
hence safe.
Spacing of intermediate stiffener(c)
3d>=c<=d
let c/d=1.4
c=1.4x1440=2016 say 2000mm c/c
3x1440>=2000<=1440
4320>=2000<=1440(ok)
c/d=2000/1440=1.38<√2
Is=1.5d³xtw³/C²=1.5x1440³x12³/2000²=19.402x106mm4
Iprovided=12x1440³/12=29.859
Result=Thus it is calculated that intermediate stiffeners are required for the given PG to resist shear force of 200KN and c/c spacing of stiffener is 2000mm.
3)d/tw=(1500-2x30)/12=120>67ξ
Shear buckling check required
Minimum web thickness requirement
d/tw=120<200ξ (ξ=1) ok
Shear capacity check
Vd=Avxfyw/√3xγm0=1440x12x250/√3x1.1=2267478.47N=2267.47KN
0.6Vd=1360.48KN<V=2000KN
(Not safe in shear)
Check for shear buckling as per IS800:2007, P-59
Vcr=Avxζb
Shear stress corresponding to web buckling
ζb=fyw/√3xλw²
λw²=√(fyw/√3xζcr)
ζcr=Kvxπ²xE/12(1-μ²)/(d/tw)²
Kv=5.35 when transverse stiffeners are provided only at support.
E=2X105N/mm², μ=0.3
ζcr=5.35xπ²x2x105/12(1-0.3²)/(1460/12)²=65.27