Design of Rolled sections beams by Limit State Method as per IS 800 : 2007 :

Planning for Systematic design of Rolled beams :

Step 1 : Geometrical properties of the cross-section of the Beam.

Step 2 : Classfications of cross- sections : (CLAUSE 3.7)

i.e : The local buckling can be avoided before the limit state is achieved by limiting

the width to thickness ratio of each element of a cross-section subjected to

compression due to axial force, moment or shear as per Table 2.

a) Plastic sections

b) Compact sections

c) Semi-compact sections

d) Slender sections

Step 3 : Reisistance to shear buckling shall be verified as specified in

Clause 8.4.2.1

> 67ε

Step 4 : Check for maximum effective slenderness ratios

(Table 15 : Effective length for simply supported beams)

Step 5 : Section 8 : Design of members subjected to bending

The factored design moment,M at any section, in a beam due to external actions shall

satisfy

M ≤

M = Factored design moment

= Design bending strength of the section

Case 1 : Laterally supported beam :

A beam may be assumed to be adequately laterally supported if restraint member

is capable of resisting a lateral force not less than 2.5 percent of the maximum

force in the compression flange of the member.

Case 2 : Laterally unsupported beam

Case 1 : Laterally supported beam :

when , V <

V = Factored design shear force

= Design shear strength of the section

=

i.e : d / tw

Md

Md

0.6Vd

Vd

Md βbZpfy / ϒmo

To avoid irreversible deformation under serviceability loads,

shall be less than incase of simply supported

beams.

otherwise, when

V >

=

= Design bending strength under hogh shear as defined in 9.2

Step 6 : Holes in the tension zone

The effect of holes in the tension and compression flange, in the design bending strength need

not be considered if it satisfies the criteria given in : Clause 8.2.1.4

Step 7 : Shear lag effects

The simple theory of bending is based on the assumption that plane sections remain plane after

bending.In reality, shear strains cause the section to warp.The higher stresses are produced near

the junction of a web and lower stresses at points away from the web.This phenomenon

is known as shear lag.It results in a non-uniform stress distribution across the width of the flange.

The shear lag effects in flanges may be disregarded provided it validate Clause 8.2.1.5

Case 2 : Laterally unsupported beam (CLAUSE 8.2.2)

Resistance to lateral torsional buckling need not to be checked seprately in the following cases :

a) Bending is about the minor axis of the section - As you are designing the section for moment about the

minor axis only.

b) Section is hollow or solid bars - These sections has high moment of resistance about both the axis.

c)

=

As the value of , is less than 0.4, it means you are overdesiging your section for a

particular force and moment and then, there is no need to check it

for lateral torsional buckling moment.

The design bending strength of laterally unsupported beam as governed by lateral torsional buckling

is given by:

=

Md

1.2 Ze fy / ϒmo

0.6Vd

Md Mdv

Mdv

In case of major axis bending, λLT (as defined in laterally unsupported beam design) is less than 0.4

λLT sqrt (fy / fcr,b)

λLT

Md βbZpfbd

Step 8 : Check for deflection ( Table 6 of IS 800 defines the deflection limits)

Step 9 : Web buckling and Web crippling (As per 10.11 of Design of steel structures by N.Subramanian)

Web crippling strength of the web also called as the web bearing capacity at supports.

Member subjected to combined forces : (Section 9 of IS 800 2007)

Case 1 : Combined shear and bending :

when , V <

V = Factored design shear force

= Design shear strength of the section

=

otherwise, when

V >

=

= Design bending strength under hogh shear as defined in 9.2

a) Plastic or Compact section :

= ≤

b) Semi-compact section :

=

Case 2 : Combined Axial force and Bending moment : (Clause 9.3)

Under combined axial force and bending moment, section strength as governed by

material failure and member strength

The Indian code (IS 800 : 2007) provisions are based on the Eurocode provisions and the code

requires the following two checks to be performed :

a) Local capacity check

b) Overall buckling check

a) Section strength : (Local capacity check)

i) Palstic & compact sections

0.6Vd

Vd

Md βbZpfy / ϒmo

0.6Vd

Md Mdv

Mdv

Mdv Md - β(Md - Mfd) 1.2 Ze fy / ϒmo

Mdv Ze fy / ϒmo

Members subjected to combined axial force (compression & tension) and bending moment, the

following should be satisfied :

≤ 1

Conservatively,

≤ 1

ii) Semi-compact sections

In the absence of high shear force, semi-compact section design is satisfactory under combined

axial force and bending, if the maximum longitudinal stress under combined axial force and bending,

≤

For cross-section without holes, the above criteria reduces to,

≤ 1

b) Overall member strength : (Overall buckling check)

Members subjected to combined axial force and bending moment shall be checked for overall

buckling failure as given in this section.

i) Bending and axial tension

=

ii) Bending and axial compression

1

1

Important Points :

1) For design purpose, the platform beams shall be designed as a laterally unsupported beam.

As there is no slab over these beams (except roof paltform), there exists an unsupported length

main beam which has to be designed.Laterall bending of main beam will take palce between this

2) When a member is capable of resisting a lateral force not less than 2.5percent of the maximum force

in the compression flange of the member, the member will as a full lateral restraint to the compression

flange of the main beam.

(My / Mndy) 1α + ((Mz / Mndyz)α2

N / Nd + My / Mdy + Mz / Mdz

fx satisfies the following criteria :

fx fy / ϒmo

N / Nd + My / Mdy + Mz / Mdz

Meff (M-Ψ T Zec / A) ≤ Md

P / Pdy + Ky Cmy My / Mdy + KLT Mz / Mdz ≤

P / Pdy + 0.6 Ky Cmy My/Mdy+ KZ Mz / Mdz ≤

Ly , of certain dimensions between two connecting beams which will act as lateral restraint to the

unsupported length Ly.So, the beam will act as a laterally unsupported Beam.

Design of Plate girders as per IS 800 : 2007 (Limit state method) : Beam B1Lateral status of the beam = Unsupported

= 11739 kN-m L/C = 3 = 0 kN-m

= 300 kN = 0 kN

= 44.31 mm = 0.00 mm

= 240 Mpa 0 kN

= 240 MpaE = 2.0E+05 Mpa Section used =

= 7.4E+04 mm² 2-flange = (400mm x 40mm)

= 5.21E+10 1-web = (2100mm x 20mm)

= 4.28E+08

= 4.78E+07 400mm40mm

= 2.14E+06

= 839.43 mm

= 76.06 mm

= 24.3 m 2180mm

= 2.509 m 2100mm =d

c = 1.5 m

c/d = 0.71 20mm

= 1.02

= 1.02

Flange classification = Plastic section 40mm=tfWeb classification = Compact section

Section classification = Compact section 400mm = b

Buckling class about

z-z axis = b Check for minimum web thickness : (As per clause 8.6.1.1(b)

Buckling class about Condition : When only transverse stiffeners are provided

y-y axis = c Case 1: when 3d ≥ c ≥ d ≤

Depth of neutral axis from Case 2: when 0.74d ≤ c < d ≤

top flange about z-z axis = 1090 mm Case 3: when c < 0.74d ≤

Depth of neutral axis about Case 4: c > 3d Web shall be considered as unstiffened

y-y axis = 200 mm Governing case : = c < 0.74d

C.G. of the compression section from the ,d/tw ≤ 270 εw

neutral axis about z-z axis 760.68 mm 105 ≤ 275.4

C.G. of the compression section from the Web satisfies the Serviceability criteria

neutral axis about y-y axis 46.08 mm

Plastic modulus of section about

5.63E+07 Condition : When transverse stiffeners are provided

Plastic modulus of section about Checked to prevent the comp. flange from buckling into the web.

3.41E+06 Case 1: when ≤

= 24.3 m Case 2: when ≤

= 2.51 m Governing case : = c < 1.5d

= 28.95 ,d/tw ≤ 345 εf

= 33 105 ≤ 351.9

Mz My

Vz Member no. = 8 Vh

dv,act dh,act

fyf Axial force, Fa =

fyw

For E250 STEELAt

Izz mm4

Iyy mm4

Zzz mm3

Zyy mm3

rzz

ryy

Lz

Ly

tw =

εflange

εweb

,d/tw 200 εw

,c/tw 200 εw

,d/tw 270 εw

when

Check for compressiom flange buckling requirement : (As per clause 8.6.1.2(b) :

z-z axis, (Zpzz) = mm3

y-y axis, (Zpyy) = mm3c ≥ 1.5d ,d/tw 345 εf

2

KLz c < 1.5d ,d/tw 345 εf

KLy

KLz /rzz

KLy / ryy

z z

y

y

Member is within Slenderless limit Compression flange buckling requirement is satisfiedCheck for shear buckling before yielding : (As per clause 8.2.1.1)

= 105

= 68.34 Shear buckling analysis is required

Check for resistance to shear buckling : (As per clause 8.4.2)= 105.00= 14.61 web with stiffeners

67εw Sqrt(Kv/5.35)= 112.93

Shear Buckling design methods :a) Simple post-critical method : Check for shear (at support)

= 0.3 Factored maximum shear force

= 239.54 Mpa= 1554 kN

Shear carrying capacity

= 5290 kN

Hence safe

=sqrt(240/(sqrt(3)x239.54)) = 0.76

138.56 Mpa

42000sqmm

therefore, 5819.52 kN= 5290.47 kN

Section is safe in shear

Section is in low shear

32000sqmm

Therefore, 4433.92 kN= 4030.84 kN

Section is safe in shear

Section is in low shear

Classification of the section based on shear capacity :

About major axis z-z = Section is in low shear

About minor axis y-y = Section is in low shearCheck for design capacity of the section :

Simply supported beam = βb . Zp . fbd 1 1.1

1 /otherwise,

3012 mm 2140mm

0.491.01E+11N-mm 1.40E+03N-mm

0.37 ≤ 0.37 or 0.41

d/tw

67εw

d / tw

Kv

Poissons ratio, µ

λw

τb =

Nominal shear strength, Vn =Vcr = Av x τb

Shear area abt. major axis , Avz =

Vn = Vcr =Vdz = Vn / ϒmo

Shear area abt. minor axis , Avy =

Vn = Vcr =Vdy = Vn / ϒmo

Md where, βb = , ϒmo =

fbd = χLT fy / ϒmo , χLT = {ФLT+ [ Ф2LT -λ2

LT]0.5} ≤ 1 , ФLT = 0.5[1 + αLT + ( λLT - 0.2) + λLT 2]λLT = sqrt( βb Z pfy / Mcr) ≤ sqrt(1.2 Zefy/Mcr) λLT = sqrt( fy/fcr,b)

Mcr = (p2 EIy hf)/(2L2LT) [1+ (1/20){(LLT/ry)/(hf/tf)}2]0.5 , LLT = ,hf =

fcr,b = (1.1p2E)/(LLT/ry)2 [1+ (1/20){(LLT/ry)/(hf/tf)}2]0.5 , αLT =Mcr = , fcr,b = λLT = λLT =

2

2

2

,

112

w

vecr

td

Ek

p

0.37 Member may be designed as laterally supported beamTherefore, λLT =

For laterally supported beam :

12281.52 kN-m ≤ 12524.8kNm 744.0kNm ≤ 560.4kNm12281.52 kN-m 560.38 kN-m

Safe in bending Safe in bending

= 11739/12282 = 0.956

0.956 < 1

Section is safe

Check for Deflection :Unfactored moment abt z-z axis, M = 7826.00kNm

= 44.31 mm

= 74.77 mm

Safe in Deflection

Symbols:

= Factored moment about major axis z-z.

= Factored moment about minor axis y-y.

= Maximum shear stress in the transverse direction paralel to y-y axis.

= Maximum shear stress in the lateral direction paralel to z-z axis.

= Actual deflection in the transverse direction parallel to y-y axis.

= Actual deflection in the lateral direction parallel to z-z axis.

= Yield strength of steel.

E = Youngs modulus of steel.

= Total cross-section area.

= Moment of inertia of the section about z-z axis.

= Moment of inertia of the section about y-y axis.

= Elastic modulus of the section about z-z axis.

= Elastic modulus of the section about y-y axis.

= Radius of gyration about z-z axis.

= Radius of gyration about y-y axis.

= Span of the beam about z-z axis.

= Unsupported length of the compression flange.

= Effective length about z-z axis.

= Effective length about y-y axis.

d = Depth of web.

= Thickness of web.

= Partial safety factor.

= Plastic shear resistance under pure shear.

= Shear area about z-z axis.

h = Overall depth of the section.

= Thickness of flange.

= Width of flange.

= Shear area about y-y axis.

Mdzz = Mdyy =Mdzz = Mdyy =

Mdzz > Mz Mdyy > My

Mz/Mdzz

dv,act

dmax,per.= Lz / 325

Mz

My

Vz

Vh

δv,act.

δh,act.

fy

At

Izz

Iyy

Zzz

Zyy

rzz

ryy

Lz

Ly

KLz

KLy

tw

ϒmo

Vn = Vp

Avz

tf

bf

Avy

= Design moment of the whole section disregarding high shear force effect.

= Design moment capacity of the section disregarding high shear force effect about z-z.

= Design moment capacity of the section disregarding high shear force effect about y-y.

= Design moment capacity of the section under high shear.

= Design moment capacity of the section under high shear about z-z axis.

= Design moment capacity of the section under high shear about y-y axis.

= Plastic design strength of the area of the cross-section excluding the shear area.

= Design compreesive stress of axially loaded compression members.

= Effective sectional area

χ = Stress reduction factor for different buckling class,slenderness ratio and yield stress.

α = Imperfection factor

KL/r = Effective slenderness ratio

= Gross area of cross-section

= Design moment capacity of the section about z-z.

= Design moment capacity of the section about y-y.

= Design reduced flexural strength under combined axial force and the respective

uniaxial moment acting alone.

= Constants

= Equivalent uniform momemt factor.

= Design strength under axial compression as governed bu buckling about minor(y) &

major(z) xais respectively.

= ratio of actual applied axial force to the design axial strength for buckling about the y

& z axis respectively.

= Equivalent uniform momemt factor lateral torsional buckling.

Md

Mdzz

Mdyy

Mdv

Mdvzz

Mdvyy

Mfd

fcd

Ae

Ag

Mdz

Mdy

Mndy , Mndz

α1 , α2

Cmy , Cmz

Pdy , Pdz

ny , nz

CmLT

Design of Plate girders as per IS 800 : 2007 (Limit state method) : Beam B3Lateral status of the beam = Unsupported

= 381 kN-m L/C = 3 = 0 kN-m

= 77 kN Member no. = 54 = 0 kN

= 13.89 mm = 0.00 mm

= 250 Mpa 0 kN

= 250 MpaE = 2.0E+05 Mpa Section used =

= 1.5E+04 mm² 2-flange = (300mm x 16mm)

= 6.57E+08 1-web = (465mm x 12mm)

= 7.21E+07

= 2.65E+06 300mm16mm

= 4.80E+05

= 208.11 mm

= 68.9 mm

= 8.48 m 497mm

= 2.7 m 465mm =d

c = 1.35 m

c/d = 2.9 12mm

= 1

= 1

Flange classification = Compact section 16mm=tfWeb classification = Plastic section

Section classification = Plastic section 300mm = b

Buckling class about

z-z axis = b Check for minimum web thickness : (As per clause 8.6.1.1(b)

Buckling class about Condition : When only transverse stiffeners are provided

y-y axis = c Case 1: when 3d ≥ c ≥ d ≤

Depth of neutral axis from Case 2: when 0.74d ≤ c < d ≤

top flange about z-z axis = 248.5 mm Case 3: when c < 0.74d ≤

Depth of neutral axis about Case 4: c > 3d Web shall be considered as unstiffened

y-y axis = 150 mm Governing case : = 3d ≥ c ≥ d

C.G. of the compression section from the ,d/tw ≤ 200 εw

neutral axis about z-z axis 194.83 mm 38.75 ≤ 200

C.G. of the compression section from the Web satisfies the Serviceability criteria

neutral axis about y-y axis 48.53 mm

Plastic modulus of section about

2.96E+06 Condition : When transverse stiffeners are provided

Plastic modulus of section about Checked to prevent the comp. flange from buckling into the web.

7.37E+05 Case 1: when ≤

= 8.48 m Case 2: when ≤

= 2.7 m Governing case : = c ≥ 1.5d

= 40.75 ,d/tw ≤ 345 εf2

= 39.19 38.75 ≤ 345

Mz My

Vz Vh

dv,act dh,act

fyf Axial force, Fa =

fyw

For E250 STEELAt

Izz mm4

Iyy mm4

Zzz mm3

Zyy mm3

rzz

ryy

Lz

Ly

tw =

εflange

εweb

,d/tw 200 εw

,c/tw 200 εw

,d/tw 270 εw

when

Check for compressiom flange buckling requirement : (As per clause 8.6.1.2(b) :

z-z axis, (Zpzz) = mm3

y-y axis, (Zpyy) = mm3c ≥ 1.5d ,d/tw 345 εf

2

KLz c < 1.5d ,d/tw 345 εf

KLy

KLz /rzz

KLy / ryy

z z

y

y

Member is within Slenderless limit Compression flange buckling requirement is satisfiedCheck for shear buckling before yielding : (As per clause 8.2.1.1)

= 38.75

= 67 Shear buckling analysis is not required

77.00kN = 0.00kNVn / ϒmo Vn / ϒmo

= 1.1 = 1.1

= 805.40kN = 1385.64kN= 732.18kN = 1259.67kN

Section is safe in shear Section is safe in shearSection is in low shear Section is in low shear

Classification of the section based on shear capacity :

About major axis z-z = Section is in low shear

About minor axis y-y = Section is in low shearCheck for design capacity of the section :

Simply supported beam = βb . Zp . fbd 1 1.1

1 /otherwise,

3240 mm 481mm

0.493.45E+09N-mm 1.04E+03N-mm

0.46 ≤ 0.48 or 0.490.46 Member shall be designed as laterally unsupported beam

For laterally unsupported beam :

0.67 0.87 197.73 Mpa

584.79 kN-m > 381.00kNm 131.03 kN-m > 0kNm

Safe in bending Safe in Bending

= 381/585 = 0.652 Check for shear (at support)0.652 < 1

Section is safe Factored maximum shear force

= 201 kN

Check for Deflection : Shear carrying capacityUnfactored moment abt z-z axis, M = 254.00kNm = 732 kN

= 13.89 mm Hence safe

= 26.09 mm

Safe in Deflection

d/tw

67εw

Factored design shear force,Vz = Factored design shear force,Vy

Design strength,Vdz (abt. major axis)= Design strength, Vdy (about minor axis) =ϒmo ϒmo

Vn = Vp = Avzfyw/sqrt(3) = (d.tw.fyw)/sqrt(3) Vn = Vp = Avy fyw/sqrt(3) = (2b.tf.fyw)/sqrt(3) Vp Vp

Vdz Vdy

Md where, βb = , ϒmo =

fbd = χLT fy / ϒmo , χLT = {ФLT+ [ Ф2LT -λ2

LT]0.5} ≤ 1 , ФLT = 0.5[1 + αLT + ( λLT - 0.2) + λLT 2]λLT = sqrt( βb Z pfy / Mcr) ≤ sqrt(1.2 Zefy/Mcr) λLT = sqrt( fy/fcr,b)

Mcr = (p2 EIy hf)/(2L2LT) [1+ (1/20){(LLT/ry)/(hf/tf)}2]0.5 , LLT = ,hf =

fcr,b = (1.1p2E)/(LLT/ry)2 [1+ (1/20){(LLT/ry)/(hf/tf)}2]0.5 , αLT =Mcr = , fcr,b = λLT = λLT =

Therefore, λLT =

ФLT = , χLT = , fbd =

Mdzz = Mdyy =Mdzz > Mz Mdyy > My

Mz/Mdzz

dv,act

dmax,per.= Lz / 325

unit wt L nos300 16 125.6 8.38 2 631.5168 998.5859465 12 94.2 8.38 1 367.0691

200 20 157 8.38 2 526.264 1007.269457 16 125.6 8.38 1 481.0053

497

0.651

8.6833560.835