Design of Concrete Filled Circular tubes

• Materials
• Input data of concrete encased composite columns
• Design Check of concrete encased composite columns

Restrictions on Simplified Design Method
a) columns → doubly symmetrical & uniform cross section
b) steel contribution ratio δ → 0.2 ≤ δ ≤ 0.9
c) non-dimensional slenderness λ ≤ 2
d) 0.3% ≤ As/Ac ≤ 6%
e) minimum & maximum cover are restricted
               minimum cover → 40mm
               maximum cover → cz = 0.3h & cy = 0.4b
f) concrete filled sections → can be fabricated without any reinforcement however concrete encased steel secions → minimum As/Ac = 0.3%
g) ratio of depth to width, 0.2 ≤ h/b ≤ 5

Design Check of Concrete Filled Circular tubes
Type		Data			Check	Reference
Composite column specifications
	Column length	L=		m	Input data
	Effective length y-y	Ley=		m	Input data
	Effective length z-z	Lez=		m	Input data
	Column Type	Fully Encased
Design value of actions
	Design axial force	Nsd=		kN	Input data
	Design bending moment
	about y-y (major) axis	My,top,sd=		kNm	Input data
		My,bot,sd=		kNm	Input data
	about z-z (minor) axis	Mz,top,sd=		kNm	Input data
		Mz,bot,sd=		kNm	Input data
Material properties
Structural steel		S235≤16 S235≤40 S235≤63 S235≤80 S235≤100 S275≤16 S275≤40 S275≤63 S275≤80 S275≤100 S355≤16 S355≤40 S355≤63 S355≤80 S355≤100			Choose the steel grade
	Characteristic yield strength	fy=		N/mm2
	Modulus elastic of steel	Ea=		N/mm2
	Partial safety factor	Ya=
	Design strength	fyd=		N/mm2
Concrete
	Concrete grade	C20/25 C25/30 C30/37 C35/45 C40/50 C45/55 C50/60 C55/67 C60/75			Choose the concrete grade
	Type of concrete	Normal Weight Concrete
	Characteristic value of compressive strength	fck=		N/mm2
	Partial safety factor	Yc=
	Design value of compressive strength	fcd=		N/mm2
	Secant modulus of elasticity	Ecm=		N/mm2
Reinforcement
	Characteristic yield strength	fyk=		N/mm2	Input data
	Partial safety factor	Yc=
	Design yield strength	fsd=		N/mm2
	Design value of modulus of elasticity	Es=		N/mm2
Connectors
	Diameter	d=		mm	Input data
	Overall nominal height	hsc=		mm	Input data
	Ultimate tensile strength	fu=		N/mm2	Input data
Cross section geometry and section propertises of the selected section
Structural Steel					Choose the steel section
	Diameter	d=		mm	Input data
	Thickness	t=		mm	Input data
	Section area	Aa=		cm2
	Second moment of area /yy	Iay=		cm4
	Elastic modulus /yy	Wel,y=		cm3
	Plastic modulus /yy	Wpl,y=		cm3
	Second moment of area /zz	Iaz=		cm4
	Elastic modulus /zz	Wel,z=		cm3
	Plastic modulus /zz	Wpl,z=		cm3
	check for local buckling	d/t=
Concrete
	Area of concrete	Ac=		mm2
	Second moment of area about major axis: y-y (of columns)	Icy=		cm4
	Second moment of area about minor axis: z-z (of columns)	Icz=		cm4
Reinforcement
	The number of longitudinal bars	n=			Input data
	Bar diameter	d=		mm	Input data
	No. of Rebars in 1st row, n1 =				Input data
	No. of Rebars in 2nd row, n2 =				Input data
	Distance from centre of section to 1st row, y1 =			mm
	Distance from centre of section to 2nd row, y2 =			mm
	Spacing of Rebars in vertical direction, s =			mm
	Rebars area in 1st row, A1 (top & bot) =			mm
	Rebars area in 2nd row, A2 (top & bot) =			mm
	Total section area	As=		cm2	0.3%<As/Ac<6% Ok
	Concrete cover			mm	Input data
	Second moment of total area about major axis: y-y (of columns)	Isy=		cm4
	Second moment of total area about minor axis: z-z (of columns)	Isz=		cm4
	Reinforcement ratio	As/Ac=			As/Ac<0.3% Not Ok
Plastic resistance of the composite cross section to compression:
		Npl,Rd=		kN
	Steel contribution factor	δ=			<0.2 Not Ok
Checking long term loading:
Efective elastic flexural stiffness:
		Ec,eff=		N/mm2
		φt=			Input data
	assuming permanent load is % of design load	NG,Ed=		kN	Input data
	About the axis (y-y) or z-z:
		Ke=			Input data
		(EI)ey=		kNm2
Elastic buckling load:
	About the major axis (y-y) or z-z:
		Ncry=		kN
Plastic resistance to compression:
		Npl,Rk=		kN
Non-dimesional slenderness ration:
	About the major axis (y-y) or z-z:
					< 2 Ok
Evaluate the resistance of the composite column under axial compression:
	Reduction factor:
	strut curve b for major axis and strut curve c for minor axis
	y-y axis
		ay=			Input data
		fy=				Eqn 16 & 17 CE5509
		χy=
	Where: a is the imperfection parameter which allows for different levels of imperfections in the columns
Confinement effect of circular cross-section
ecceentricity of loading		e =		mm
consider confinement if		e/d =			< 0.1 Ok
					< 0.5 Ok
					consider confinement
		Npl,Rd =		kN
the compression resistance of cross-section is:
		Npl,Rd =		kN
		Nb,Rd=		kN
					< 1 ok
Checking Resistance of composite section to under combined compression & Bending
	Major axis bending (y-y):
	Wps: Plastic section modulus for reinforcement
		Wps=		mm3
	Wpsn: Plastic section modulus for reinforcement within the region of 2hn from the middle line
		Wpsn=		mm3
	Neutral axis position:
		hn=		mm
		ez=		mm	Not ok
	Wpc: Plastic section modulus for concrete:
		Wpc=		mm3
	Wpan: Plastic section modulus of steel within the region of 2hn from the middle line:
		Wpan=		mm3
	Wpcn: Plastic section modulus of concrete within the region of 2hn from the middle line:
		Wpcn=		mm3
	The bending resistance
		Mmax,Rd=		kNm
		Mpl,Rd=		kNm
	The resistance force
		Npm,Rd=		kN
Interaction Diagram:
	Major axis bending (y-y)
Point
A	Bending Moment M (kNm) M=0	MA=		kNm
	Compression force N (kN) N=Npl,Rd	NA=		kN
B	Bending Moment M (kNm) M=Mpl,Rd	MB=		kNm
	Compression force N (kN) N=0	NB=		kN
C	Bending Moment M (kNm) M=Mpl,Rd	MC=		kNm
	Compression force N (kN) N=Npm,Rd	NC=		kN
D	Bending Moment M (kNm) M=Mpm,Rd	MD=		kNm
	Compression force N (kN) N=0.5Npm,Rd	ND=		kN
Checking for combined compression and bending:
		(EI)eff,II,y=		kNm2
		Ke,II=			Input data
		Ko=			Input data
		Ncr,eff,y=		kN
	for end moment	β=			Input data
		k1,y=
hence	for bending moment from menber imperfection	β=			Input data	equivalent moment factor
		k2,y=			Recalculate
		e0,y=		m
		e0,z=		m
		My,Ed=		kNm
		μdy Mpl,y,Rd=		kNm	Not Ok
		αM=			Input data
		Mz,Ed=		kNm
		Mz,Ed/μdzMpl,z,Rd =
		αM=			Input data
					Ok
				imperfection only considered in plane in which failure is expected to occur.
vertical shear
	major axis y-y
	The vertical shear is
		Vy,Ed=		kN
	The design shear resistance is
		Vpl,a,Rd,y=		kN	Not Ok
	Minor axis z-z
		Vz,Ed=		kN
		Vpl,a,Rd,z=		kN	Not Ok
Longitudinal shear
	longitudinal shear force is
		NEd,c=		kN
	longitudinal shear stress				there is no well-established method for calculating longitudinal shear stress, usually based on
				N/mm2	shear connectors should be provided
	perimeter of steel section	pa=		mm
	load introduction length	lv=		mm
	design shear strength			N/mm2
	if
		PRd1=		kN
		PRd2=		kN
		hsc/d=
		α=
		PRd=		kN
	number of headed studs
		n=