Design of Plate girders as per IS 800 2007-LSM - Unsupported - Latest.xlsx

Design of Rolled sections beams by Limit State Method as per IS 800 : 200 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 i.e : d / tw

Step 4 :

>

67ε

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

Case 1 :

Md

M

≤

M Md

=

Factored design moment

=

Design bending strength of the section

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 ,

0.6Vd

V

<

V Vd

=

Factored design shear force

=

Design shear strength of the section

Md

=

βbZpfy / ϒmo

To avoid irreversible deformation under serviceability loads, 1.2 Ze fy / ϒmo shall be less than incase of simply supported beams. otherwise, when

Step 6 :

V

>

0.6Vd

Md

=

Mdv

Mdv

=

Design bending strength under hogh shear as def

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

As the value of

=

sqrt (fy / fcr,b)

, is less than 0.4, it means you are overdesiging your section fo

particular force and moment and then, there is no need to chec for lateral torsional buckling moment. The design bending strength of laterally unsupported beam as governed by lateral torsional buckling is given by: 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 Vd

=

Factored design shear force

=

Design shear strength of the section

Md

=

>

0.6Vd

otherwise, when V

0.6Vd

V

βbZpfy / ϒmo

Md

=

Mdv

=

Mdv Design bending strength under hogh shear as defined in 9.2

a) Plastic or Compact section : Mdv

=

Md - β(Md - Mfd)

≤

1.2 Ze fy / ϒmo

b) Semi-compact section : Mdv

Case 2 :

=

Ze fy / ϒmo

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 Members subjected to combined axial force (compression & tension) and bending moment, the following should be satisfied : (My / Mndy)α1 + ((Mz / Mndyz)α2 Conservatively, N / Nd + My / Mdy + Mz / Mdz 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,

fx satisfies the following criteria : fx

≤

fy / ϒmo

For cross-section without holes, the above criteria reduces to, N / Nd + My / Mdy + Mz / Mdz 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 Meff

=

(M-Ψ T Zec / A)

≤

Md

ii) Bending and axial compression P / Pdy +

Ky Cmy My / Mdy +

KLT Mz / Mdz ≤

P / Pdy +

0.6 Ky Cmy My/Mdy+

KZ Mz / Mdz ≤

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 Ly , of certain dimensions between two connecting beams which will act as lateral restraint to the main beam which has to be designed.Laterall bending of main beam will take palce between this unsupported length Ly.So, the beam will act as a laterally unsupported Beam. 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.

hod as per IS 800 : 2007 :

ved by limiting

ernal actions shall

n moment strength of the section

pported if restraint member

percent of the maximum

n shear force

rength of the section

Md

case of simply supported

strength under hogh shear as defined in 9.2

sign bending strength need

sections remain plane after

r stresses are produced near

b.This phenomenon

cross the width of the flange.

ate Clause 8.2.1.5

ately in the following cases :

g the section for moment about the

resistance about both the axis.

ed beam design) is less than 0.4

ou are overdesiging your section for a

and then, there is no need to check it

rned by lateral torsional buckling

uctures by N.Subramanian)

pacity at supports.

ction 9 of IS 800 2007)

ar as defined in 9.2

2 Ze fy / ϒmo

as governed by

de provisions and the code

) and bending moment, the

≤

1

≤

1

satisfactory under combined

mbined axial force and bending,

≤

1

hall be checked for overall

1 1

ally unsupported beam.

xists an unsupported length act as lateral restraint to the will take palce between this

2.5percent of the maximum force

ateral restraint to the compression

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

= 11739 kN-m L/C = 3 300 kN

=

44.31 mm

fyf

=

240 Mpa

fyw

= =

240 Mpa 2.0E+05 Mpa

=

7.4E+04 mm²

Izz

=

5.21E+10 mm

Iyy

=

4.28E+08 mm4

Zzz

=

4.78E+07 mm3

Zyy

=

2.14E+06 mm3

rzz

=

839.43 mm

ryy

=

76.06 mm

Lz

=

24.3 m

Ly

=

2.509 m

c

=

1.5 m

c/d εflange

=

0.71

=

1.02

εweb

=

1.02 =

h,act

=

Axial force, Fa

= =

2-flange

=

(400mm x

1-web

= (2100mm x y 400mm

2180mm z tw = 20mm

top flange about z-z axis =

400mm y

b

Check for minimum web thickness : (As per clause

c

Condition : When only transverse stiffene ,d/tw Case 1: when 3d ≥ c ≥ d ,c/tw Case 2: when 0.74d ≤ c < d

Depth of neutral axis from 1090 mm

Case 3:

=

200 mm

neutral axis about z-z axis

760.68 mm

C.G. of the compression section from the neutral axis about y-y axis

46.08 mm

Plastic modulus of section about 3 z-z axis, (Zpzz) = 5.63E+07 mm Plastic modulus of section about 3 y-y axis, (Zpyy) = 3.41E+06 mm 24.3 m 2.51 m

KLz

=

KLy

=

KLz /rzz

=

28.95

KLy / ryy

=

33

c > 3d

Governing case :

C.G. of the compression section from the

,d/tw

when c < 0.74d

Case 4: when

Depth of neutral axis about y-y axis

=

Section used =

Compact section Compact section

Buckling class about y-y axis

=

Plastic section

Section classification = Buckling class about z-z axis

Vh

4

Flange classification = Web classification

=

Member no. = 8

v,act

E At

Unsupported My

Web shall be considere

,d/tw

= ≤

c < 0.74d 270 εw

105

≤

275.4

Web satisfies the Serviceability crite Check for compressiom flange buckling requireme clause 8.6.1.2(b) :

Condition : When transverse stiffeners Checked to prevent the comp. flange from buckling into when

c ≥ 1.5d

,d/tw

when Governing case :

c < 1.5d

,d/tw

Case 1: Case 2:

,d/tw

= ≤

c < 1.5d 345 εf

105

≤

351.9

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

=

105

67εw

=

68.34

Shear buckling analysis is required

Check for resistance to shear buckling : (As per clause 8.4.2) d / tw Kv

= = 67εw Sqrt(Kv/5.35)=

105.00 14.61 112.93

web with stiffeners

Shear Buckling design methods : Simple post-critical method : a) Poissons ratio, µ

 cr ,e 

λw

=

k v 12 1  

2

2

0.3

Factored maximum shear force =

E



d  



2

= 239.54 Mpa Shear carrying capacity =

t w 

=sqrt(240/(sqrt(3)x239.54)) = τb =

Nominal shear strength, Vn =Vcr = Shear area abt. major axis , Vdz =

Check for shear (at support)

therefore, Vn / ϒmo

Avz

=

0.76 138.56 Mpa Av x τb 42000sqmm

Vn = Vcr =

5819.52 kN

=

5290.47 kN

Section is safe in shear Section is in low shear Shear area abt. minor axis , Vdy =

Therefore, Vn / ϒmo

Avy

=

32000sqmm

Vn = Vcr =

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 shear Check for design capacity of the section : Simply supported beam Md where, βb = 1 = βb . Zp . fbd fbd λLT

= =

Mcr

=

fcr,b Mcr

= =

λLT

=

χLT fy / ϒmo χLT , sqrt( βb Z pfy / Mcr) ≤

=

2 2 0.5 1 , ФLT = 0.5[1 + αLT + ( 1 / {ФLT+ [ Ф LT -λ LT] } ≤ sqrt(1.2 Zefy/Mcr) = otherwise, λLT

(2EIy hf)/(2L2LT) [1+ (1/20){(LLT/ry)/(hf/tf)}2]0.5 (1.1 E)/(LLT/ry) 2

2

1.01E+11N-mm 0.37

, LLT

[1+ (1/20){(LLT/ry)/(hf/tf)} ]

, αLT , fcr,b

2 0.5

≤

0.37

or

λLT

=

3012 mm

= 0.49 = 1.40E+03N-mm = 0.41

Therefore, λLT

=

0.37

Member may be designed as laterally suppo

For laterally supported beam : Mdzz = ≤ 12281.52 kN-m Mdzz = 12281.52 kN-m Mdzz > Mz Mz/Mdzz

Safe in bending = 11739/12282 0.956

Mdyy Mdyy

12524.8kNm

=

0.956

<

1

= =

744.0kNm 560.38 kN-m

≤

Mdyy > My

Section is safe

Check for Deflection : Unfactored moment abt z-z axis, M = v,act = L / 325 max,per.= = z

7826.00kNm 44.31 mm 74.77 mm

Safe in Deflection

Symbols: Mz

=

Factored moment about major axis z-z.

My

=

Factored moment about minor axis y-y.

Vz

=

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

Vh

=

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

δv,act.

=

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

δh,act.

=

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

fy

=

Yield strength of steel.

E

=

Youngs modulus of steel.

At

=

Total cross-section area.

Izz

=

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

Iyy

=

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

Zzz

=

Elastic modulus of the section about z-z axis.

Zyy

=

Elastic modulus of the section about y-y axis.

rzz

=

Radius of gyration about z-z axis.

ryy

=

Radius of gyration about y-y axis.

Lz

=

Span of the beam about z-z axis.

Ly

=

Unsupported length of the compression flange.

KLz

=

Effective length about z-z axis.

KLy

=

Effective length about y-y axis.

d

=

Depth of web.

tw

=

Thickness of web.

ϒmo

=

Partial safety factor.

Vn = Vp

=

Plastic shear resistance under pure shear.

Avz

=

Shear area about z-z axis.

h

=

Overall depth of the section.

tf

=

Thickness of flange.

bf

=

Width of flange.

Avy

=

Shear area about y-y axis.

Md

=

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

Mdzz

=

Design moment capacity of the section disregarding high shear force effect about

Mdyy

=

Design moment capacity of the section disregarding high shear force effect about

Mdv

=

Design moment capacity of the section under high shear.

Mdvzz

=

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

Mdvyy

=

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

Mfd

=

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

fcd

=

Design compreesive stress of axially loaded compression members.

Ae

=

Effective sectional area

χ

=

Stress reduction factor for different buckling class,slenderness ratio and yield stre

α

=

Imperfection factor

KL/r

=

Effective slenderness ratio

Ag

=

Gross area of cross-section

Mdz

=

Design moment capacity of the section about z-z.

Mdy

=

Design moment capacity of the section about y-y.

Mndy , Mndz

=

Design reduced flexural strength under combined axial force and the respective uniaxial moment acting alone.

α1 , α2

=

Constants

Cmy , Cmz

=

Equivalent uniform momemt factor.

Pdy , Pdz

=

Design strength under axial compression as governed bu buckling about minor(y) major(z) xais respectively.

ny , nz

=

ratio of actual applied axial force to the design axial strength for buckling about th & z axis respectively.

CmLT

=

Equivalent uniform momemt factor lateral torsional buckling.

) : Beam B1 0 kN-m 0 kN 0.00 mm 0 kN

ed = 40mm) 20mm)

For E250 STEEL

40mm

2100mm =d z

40mm=tf =b

s per clause 8.6.1.1(b)

se stiffeners are provided 200 εw ≤ 200 εw ≤ ≤

270 εw

be considered as unstiffened

bility criteria requirement : (As per

stiffeners are provided uckling into the web. ≤

345 εf2

≤

345 εf

uirement is satisfied

1554 kN 5290 kN Hence safe

, ϒmo

=

1.1

[1 + αLT + ( λLT - 0.2) + λLT 2] sqrt( fy/fcr,b) ,hf

3N-mm

=

2140mm

ally supported beam

560.4kNm Safe in bending

ffect.

effect about z-z. effect about y-y.

shear area.

and yield stress.

respective

out minor(y) &

kling about the y

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

= 381 kN-m L/C = 3 77 kN

=

13.89 mm

fyf

=

250 Mpa

fyw

= =

250 Mpa 2.0E+05 Mpa

=

1.5E+04 mm²

Izz

=

6.57E+08 mm

Iyy

=

7.21E+07 mm4

Zzz

=

2.65E+06 mm3

Zyy

=

4.80E+05 mm3

rzz

=

208.11 mm

ryy

=

68.9 mm

Lz

=

8.48 m

Ly

=

2.7 m

c

=

1.35 m

c/d εflange

=

2.9

=

1

εweb

=

1 =

h,act

=

Axial force, Fa

= =

Section used = 2-flange

=

(300mm x

1-web

=

(465mm y x 300mm

497mm z tw = 12mm

top flange about z-z axis =

300mm y

b

Check for minimum web thickness : (As per clause

c

Condition : When only transverse stiffene ,d/tw Case 1: when 3d ≥ c ≥ d ,c/tw Case 2: when 0.74d ≤ c < d

Depth of neutral axis from 248.5 mm

Case 3:

=

150 mm

neutral axis about z-z axis

194.83 mm

C.G. of the compression section from the neutral axis about y-y axis

48.53 mm

Plastic modulus of section about 3 z-z axis, (Zpzz) = 2.96E+06 mm Plastic modulus of section about 3 y-y axis, (Zpyy) = 7.37E+05 mm 8.48 m 2.7 m

KLz

=

KLy

=

KLz /rzz

=

40.75

KLy / ryy

=

39.19

c > 3d

Governing case :

C.G. of the compression section from the

,d/tw

when c < 0.74d

Case 4: when

Depth of neutral axis about y-y axis

=

Plastic section Plastic section

Buckling class about y-y axis

=

Compact section

Section classification = Buckling class about z-z axis

Vh

4

Flange classification = Web classification

=

Member no. = 54

v,act

E At

Unsupported My

,d/tw

= ≤

38.75

≤

Web shall be considere

3d ≥ c ≥ d 200 εw 200

Web satisfies the Serviceability crite Check for compressiom flange buckling requireme clause 8.6.1.2(b) :

Condition : When transverse stiffeners Checked to prevent the comp. flange from buckling into when

c ≥ 1.5d

,d/tw

when Governing case :

c < 1.5d

,d/tw

Case 1: Case 2:

,d/tw

= ≤

c ≥ 1.5d 345 εf2

38.75

≤

345

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

=

38.75

67εw

=

67

Shear buckling analysis is not required

Factored design shear force,Vz = 77.00kN Design strength,Vdz (abt. major axis)= Vn / ϒmo ϒmo = 1.1 Vn = Vp = Avzfyw/sqrt(3) = (d.tw.fyw)/sqrt(3) Vp = 805.40kN Vdz = 732.18kN

Factored design shear force,Vy Design strength, Vdy (about minor axis) ϒmo = 1.1 Vn = Vp = Avy fyw/sqrt(3) = Vp = Vdy =

=

Section is safe in shear Section 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 shear Check for design capacity of the section : Simply supported beam Md where, βb = 1 = βb . Zp . fbd fbd λLT

= =

Mcr

=

fcr,b Mcr

= =

χLT fy / ϒmo χLT , sqrt( βb Z pfy / Mcr) ≤

=

(2EIy hf)/(2L2LT) [1+ (1/20){(LLT/ry)/(hf/tf)}2]0.5 (1.1 E)/(LLT/ry) 2

2

3.45E+09N-mm λLT = 0.46 Therefore, λLT =

, LLT

[1+ (1/20){(LLT/ry)/(hf/tf)} ]

, αLT , fcr,b

2 0.5

≤ 0.46

For laterally unsupported beam : ФLT = , χLT 0.67 Mdzz = 584.79 kN-m >

0.48

or

λLT

=

= 381/585 0.652

3240 mm

= 0.49 = 1.04E+03N-mm = 0.49

Member shall be designed as laterally unsupp

= 0.87

, fbd

=

381.00kNm Mdyy = Mdyy > My Safe in bending

Mdzz > Mz Mz/Mdzz

2 2 0.5 1 , ФLT = 0.5[1 + αLT + ( 1 / {ФLT+ [ Ф LT -λ LT] } ≤ sqrt(1.2 Zefy/Mcr) = otherwise, λLT

=

0.652

<

1

Section is safe

197.73 131.03 kN-m

Check for shear (at support) Factored maximum shear force =

Check for Deflection : Unfactored moment abt z-z axis, M = v,act = Lz / 325 max,per.= =

Shear carrying capacity 254.00kNm 13.89 mm 26.09 mm

Safe in Deflection

=

) : Beam B3 0 kN-m 0 kN 0.00 mm 0 kN

ed = 16mm) 12mm)

For E250 STEEL

16mm

465mm =d z

16mm=tf =b

s per clause 8.6.1.1(b)

se stiffeners are provided 200 εw ≤ 200 εw ≤ ≤

270 εw

be considered as unstiffened

≥ d

bility criteria requirement : (As per

stiffeners are provided uckling into the web. ≤

345 εf2

≤

345 εf

uirement is satisfied

=

xis)

0.00kN Vn / ϒmo

=

(2b.tf.fyw)/sqrt(3) 1385.64kN 1259.67kN Section is safe in shear

, ϒmo

=

1.1

[1 + αLT + ( λLT - 0.2) + λLT 2] sqrt( fy/fcr,b) ,hf

=

481mm

3N-mm

ally unsupported beam

Mpa >

0kNm

Safe in Bending

201 kN 732 kN Hence safe

300 465

16 12

200 457

20 16

497

unit wt L 125.6 94.2 157 125.6

nos 8.38 8.38

2 631.5168 1 367.0691

8.38 8.38

2 526.264 1 481.0053

998.5859

0.651 8.683356

1007.269

0.835