Example Beam Column Design IS800

Design Doc. No. SVNIT_ASHD_03 Rev. No. 0Project TitleClientName of UnitDrawing Ref.Designed by: G.C. Checked by: G.C. Approved by: Date: 10.03.08Reference

BASIC DATA

The followings are the basic data for the Staggered Vierendeel Trusses for ASHD Building at SVNIT, Surat: Type of Structure : Staggered Vierendeel Truss Span : 23 m Height : 4 m Roof Covering : RCC Deck / Roof SlabsTruss Panels : 5 panels @ 4.6 M c/c.Spacings : 6.75 m c/c.generally and 4.0 m c/c. at Central Stair LocationsSupports : On Trussed Tripod Steel Columns DESIGN BASIS

The structure has been designed as per latest Indian Codes of Practices for the following loadingsDead Load : As per IS 875 ( Part 1 ) - 1987Live Load : As per IS 875 ( Part 2 ) - 1988Wind Load : As per IS 875 ( Part 3 ) - 1989Seismic Load : As per IS: 1893 : 2002 and Other Special Literatures for Wind and Seismic LoadingsThe design is done as per IS 800 - 2007, IS: 4923 - 1997, IS 806 - 1968 with the structural properties of steel tubesobtained from IS 1161 - 1998 and IS: 4923 - 1997

Design of Top Chord Members of Vierendeel Trusses (using Hot Rolled Sections):

Factored Axial Load, N = 5 T 4.6 mFactored BM, Mz, at bottom = 4 T-m Factored BM, Mz, at top = 1.5 T-mFactored BM, My, at bottom = 0.5 T-m Factored BM, My, at top = 0.2 T-m

ISPAT NIKETAN', 1st. FloorSARDAR VALLABHBHAI NATIONAL INSTITUTE OF TECHNOLOGY, SURAT

CALCULATIONS

Span of Vierendeel Beam =

RCC DESIGN OF VIERENDEEL TRUSS MEMBERS FOR ASHD BUILDING, SVNIT, SURAT

52 / 1A Ballygunge Circular RoadKolkata - 700019

Institute For Steel Development & Growth (INSDAG)VERTICAL EXPANSION OF ASHD BUILDING, SVNIT, SURAT

STAGGERED VIERENDEEL TRUSS MEMBERS OF ASHD BUILDINGINS/SVNIT/GA/01, INS/SVNIT/GA/02

Using ISMB 250

12.5 mm ; 250 mm

6.9 mm ; 125 mm

47.55 cm2 ; 5131.6 cm4

334.5 cm4 ; 410.5 cm3

53.5 cm3

; 10.39 cm

2.65 cm ; 37.3 kg/m

466 cm3

; 84 cm3

250 N/mm2 ; 1

1.1 200000 N/mm2

9.4 84

10.5 105

15.7 126

i)

b / tf = 62.5 / 12.5 = 5 < 9.4e = 9.4 x 1 = 9.4 The Section is Plastic

d / tw = 225 / 6.9 = 32.61 < 84e = 84 x 1 = 84 The Section is Plastic

Coefficient for Plastic section =

Coefficient for Semi-compact section =

Hence, The Section is Plastic for direct load and The Section is Plastic for Bending

Coefficient for Semi-compact section =

Classification of Sections:

Moment of Inertia, I yy =

Partial Safety Factor for Materials, γm0 =

Coefficient for Compact section =

For Outstanding element of Compression Flange:

Elastic Modulus, E =

For Web Element of I, H or box section:

Type of section used = Rolled Section

Coefficient for Plastic section =

Coefficient for Compact section =

Sectional Properties:

Overall Depth, D =Thickness of Flange, t f =

Plastic Section Modulus, Zpz =

εεεε =

Radius of Gyration, r y =

Yield stress of Steel, f y =

Mass of the section, m =

Section Modulus, Zzz =

Width of Flange, b =

Moment of Inertia, I zz =

Section Modulus, Zyy =

Plastic Section Modulus, Zpy =

Radius of Gyration, r z =

Thickness of Web, t w =

Sectional Area, Aa =



CALCULATIONS





ii)

Af y / γγγγ m0 = 1080.68 kN βb = 1.0

50 kN

Hence, n = N/Nd = 0.05 ≤ 0.2

Now, Mdz = βb x Zpz x fy / γm0 = 105.91 kN-m

∴∴∴∴ Mndz = 1.11 x Mdz x (1 - n) ≤ Mdz

∴∴∴∴ Mndz = 1.11 x 105.91 x (1 - 0.05) = 111.68 kN-m > 105.91 kN-m

∴∴∴∴ Mndz = Mdz = 105.91 kN-m

Actual Bending Moment, Mz = 40 kN-m

For n ≤ 0.2, Mndy = Mdy = βb x Zpy x fy / γm0 = (1 x 84 x 250) / (1.1 x 1000) = 19.09 kN-m

Actual Bending Moment, My = 5 kN-m α1

Actual direct load, N =

Check for resistance of the section against material failure due to the combined effects of the loading: (Clause- 9.3.1)

Axial Strength of the section, N d =

(b b = 1.0 for calculation of M dz and M dy as per clause 8.2.1.2 )

1 αα

0.4 ≤ 1 Hence O.K.

{α1= 5n but ≥1; ∴ α1 = 5 x 0.05 = 0.25 = 1 and α2 = 2 (As per Table 9.1)}

Alternatively,

0.685863561 ≤ 1 Hence O.K.

iii)

KL y = KL z = 0.85 x 460 = 391 cm

KL y / r y = 147.55

KL z / r z = 37.63

Therefore, Non-dimensional slenderness ratios, λy 1.66

and, λz 0.42

Since, h/b f = 2 > 1.2

214.66 N/mm2

Hence, P dz = 1020.71 kN

Since, t f = 12.5 ≤ 40

65.91 N/mm2

Hence, P dy = 313.4 kN

313.4 kN and ny = 0.16nz = 0.05

Therefore, P d = P dy =

(b) Determination of M dz (Clause 9.3.2.2 & 8.2.2.1)

For major axis bending buckling curve ‘a’ is applicable (Refer – Table 10 of IS 800 : 2007)

For minor axis bending buckling curve ‘b’ is applicable (Refer – Table 10 of IS 800 : 2007)

From Table 9(a) for KLz / rz = 37.63, fcdz =

From Table 9(b) for KLy / ry = 147.55, fcdy =

Check for resistance of the section against material failure due to the combined effects of the loading: (Clause- 9.3.1)

(a) Determination of P dz , P dy and P d (Clause 7.1.2)

=

+

∴

21 αα

ndz

z

ndy

y

M

M

M

M

=++dy

y

dz

z

d M

M

M

M

N

N

==cc

y

f

f

=

E

r

KLf

y

yy

2

2

π

=

E

rKL

fz

zy

2

2

π==

cc

y

f

f



CALCULATIONS





Where, L T = KL

ππππ = 3.14 αααα LT = 0.21 (Refer Clause 8.2.2)

Therefore, Mcr = (3.14² x 200000 x 3345000 x 237.5 / 2 x (3910)² x √(1 + 1/20(147.55 / (237.5/12.5))²)

= 102666447.7 N-mm

Now, √(1 x 466000 x 250 / 102666447.65) = 1.07 λLT

and, 0.5[1 + 0.21(1.07 - 0.2) + 1.07²] = 1.16

Hence, 1 / [1.16 + {1.16² - 1.07²}½] = 0.62 < 1

Now, 0.62 x 250 / 1.1 = 140.91 N/mm2

Hence, M dz = β β β β b. Z pz. f bd = 1 x 466000 x 140.91 = 65664060 N-mm

= 65.66 kN-m

M dy = β β β β b. Z py . f y / γγγγ m = 1 x 84000 x 250 / 1.1 = 19090909.09 N-mm

(c) Determination of M dy (Clause 8.2.1.2)

5.02

2

2

/

/

20

11

)(2

+=

ff

yfy

crth

rKL

KL

hEIM

π

== crypbLT MfZ /..βλ

( )[ ]=+−+=2

2.015.0 LTLTLTLT λλαφ

{ }[ ] =−+

=5.022

1

TLLTLT

LT

λφφχ

== myLTbd ff γχ /.

= 19.09 kN-m

ψ ψ ψ ψ = 0.38

C mz = 0.752 = C mLT

ψ ψ ψ ψ = 0.4

C my = 0.76

K y = 1.23 > 1.128

Hence, K y = 1.128

K z = 1.01 ≤ 1.04

Hence, K z = 1.01

K LT = 1 - (0.1λLT ny /(CmLT - 0.25) = 0.97 < 0.97

Now,

50/313.4 + 1.128 x ((0.76x5)/19.09) + 0.97x (40/65.66) = 0.97 ≤ 1.00

50/1020.71 + 0.6x1.128x((0.76x5)/19.09) + 1.01x((0.752 x 40)/65.66) = 0.76 ≤ 1.00

(d) Determination of C mz and C mLT (Clause 9.3.2.2)

1 + (λz - 0.2)nz =

(e) Determination of C my (Clause 9.3.2.2)

0.6 + 0.4 x 0.4 =

(f) Determination of K y , K z , and K LT (Clause 9.3.2.2)

1 + (λy - 0.2)ny =

0.6 + 0.4 x 0.38 =

=++dz

zLT

dy

ymy

y

dy M

MK

M

MCK

P

P

=++dz

zmz

z

dy

ymy

y

dz M

MCK

M

MCK

P

P6.0

Example Beam Column Design IS800

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