Page 1
Design Doc. No. Project Title Client Name of Unit Drawing Ref. Designed by:
SVNIT_ASHD_03 Rev. No. 0 RCC DESIGN OF VIERENDEEL TRUSS MEMBERS FOR ASHD BUILDING, SVNIT, SURAT VERTICAL EXPANSION OF ASHD BUILDING, SVNIT, SURAT Institute For Steel Development & Growth (INSDAG) SARDAR VALLABHBHAI NATIONAL INSTITUTE OF TECHNOLOGY, SURAT ISPAT NIKETAN', 1st. Floor 52 / 1A 1A Ballygunge Circular Road STAGGERED VIERENDEEL TRUSS MEMBERS OF ASHD BUILDING Kolkata - 700019 INS/SVNIT/GA/01, INS/SVNIT/GA/01, INS/SVNIT/GA/02 G.C. Checked by: G.C. Approved by: Date: 10.03.08 CALCULATIONS
Reference BASIC DATA
he followings are the basic data for the Staggered Vierendeel Vierendeel Trusses for ASHD Building at SVNIT, Surat: ype of Structure : Staggered Vierendeel Truss Span : 23 m Height :4m Roof Covering : RCC Deck / Roof Slabs russ Panels : 5 panels @ 4.6 M c/c. Spacings : 6.75 m c/c.generally and 4.0 m c/c. at Central Stair Locations Supports : On Trussed Tripod Steel Columns DESIGN BASIS
he structure has been designed as per latest Indian Codes of Practices for the following loadings Dead Load : As per IS 875 ( Part 1 ) - 1987 Live Load : As per IS 875 ( Part 2 ) - 1988 Wind Load : As per IS 875 ( Part 3 ) - 1989 Seismic Load : As per IS: 1893 : 2002 and Other Special Literatures for Wind and Seismic Loadings he design is done as per IS 800 - 2007, IS: 4923 - 1997, IS 806 - 1968 with the structural properties of steel tubes obtained from IS 1161 - 1998 and IS: 4923 - 1997 Design of Top Chord Members of Vierendeel Trusses (using Hot Rolled Sections):
Factored Axial Load, N = Factored BM, M z at bottom = Factored BM, M y at bottom =
5 4 -m 0.5 -m
Span of Vierendeel Beam = Factored BM, M z at top = Factored BM, M y at top =
4.6 m 1.5 -m 0.2 -m
Sectional Properties: Using
ISMB 250
Thickness of Flange, t f =
12.5 12.5 mm
;
Overall Depth, D
=
250 mm
Thickness of Web, t w =
6.9 6.9 mm
;
Width of Flange, b
=
125 mm
=
47.55 cm2
;
Moment of Inertia, I zz =
5131.6 cm
Moment of Inertia, I yy =
334.5 cm4
;
Section Modulus, Z zz =
410.5 cm
Section Modulus, Z yy =
53.5 cm
;
Radius of Gyration, r z =
10.39 cm
Radius of Gyration, r y =
2.65 2.65 cm
;
Mass of the section, m
Plastic Section Modulus, Z pz =
466 cm
;
Plastic Section Modulus, Z py =
Yield stress of Steel, f y =
250 N/mm2
;
Sectional Area, A a
Partial Safety Factor for Materials, γ m0 m0 = Type of section used =
i)
=
37.3 37.3 kg/m kg/m 84 cm ε=
1 200000 N/mm
Elastic Modulus, E =
1.1 Rolled Section
For Outstanding element of Compression Flange:
For Web Element of I, H or box section:
Coefficient for Plastic section =
Coefficient for Plastic section =
84
9.4
Coefficient for Compact section =
10.5
Coefficient for Compact section =
105
Coefficient for Semi-compact section =
15.7
Coefficient for Semi-compact section =
126
Classification of Sections:
b / tf =
62.5 / 12.5 =
d / tw =
225 / 6.9 =
5
<
9.4e =
9.4 x 1 =
9.4
The Section is Plastic
32.61
<
84e =
84 x 1 =
84
The Section is Plastic
Hence, The Section is Plastic for direct load and The Section is Plastic for Bending
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Design Doc. No. Project Title Client Name of Unit Drawing Ref. Designed by:
SVNIT_ASHD_03 Rev. No. 0 RCC DESIGN OF VIERENDEEL TRUSS MEMBERS FOR ASHD BUILDING, SVNIT, SURAT VERTICAL EXPANSION OF ASHD BUILDING, SVNIT, SURAT Institute For Steel Development & Growth (INSDAG) SARDAR VALLABHBHAI NATIONAL INSTITUTE OF TECHNOLOGY, SURAT ISPAT NIKETAN', 1st. Floor 52 / 1A 1A Ballygunge Circular Road STAGGERED VIERENDEEL TRUSS MEMBERS OF ASHD BUILDING Kolkata - 700019 INS/SVNIT/GA/01, INS/SVNIT/GA/01, INS/SVNIT/GA/02 G.C. Checked by: G.C. Approved by: Date: 10.03.08
Reference
CALCULATIONS Check for resistance of the section against material failure due to the combined effects of the loading: (Clause- 9.3.1)
ii)
Axial Strength of the section, N d =
Af y / γ γ m0 =
Actual direct load, N =
1080.68 kN
βb =
1.0
105.91
kN-m
50 kN
Hence, n = N/N d =
0.05
≤
Now, Mdz =
βb x Zpz x fy / γ m0 m0 =
∴ Mndz =
1.11 x Mdz x (1 - n) ≤ Mdz
∴ Mndz =
1.11 x 105.91 x (1 - 0.05) =
∴ Mndz = Mdz =
0.2 105.91 105.91 kN-m
>
111.68 111.68 kN-m
105.91 105.91 kN-m
Actual Bending Moment, M z =
40
kN-m
βb x Zpy x fy / γ m0 m0 =
For n ≤ 0.2, M ndy = Mdy =
(1 x 84 x 250) / (1.1 x 1000) =
19.09
kN-m
(b b = 1.0 for calculation of M dz and M dy as per clause 8.2.1.2 ) Actual Bending Moment, M y =
M
1
y ∴ M ndy
5
α1
kN-m
α 2
M z + M ndz
0.4
=
{α1= 5n but ≥1; ∴ α1 = 5 x 0.05 =
0.25
≤
=
1
Hence O.K.
1
and α2 =
1
Hence O.K.
2
(As per Table 9.1)}
Alternatively, Alternatively, N N d
iii)
+
M z M dz
+
M y M dy
0.685863561
=
≤
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) KL y = KL z =
0.85 x 460 =
391 cm
KL y / r y =
147.55
KL z / r z =
37.63 λy =
herefore, Non-dimensional Non-dimensional slenderness ratios,
λz
and,
=
f y f cc f y f cc
KL
f y
=
y
2
r y 2
π
=
f y
KL z
E
=
1.66
2
r z π
2
E
=
0.42
For major axis bending buckling curve ‘a’ is applicable ( Refer – Table 10 of IS 800 : 2007 ) Since, h/b f =
2 > 1.2 214.66 From Table 9(a) for KLz / rz = 37.63, fcdz = Hence, P dz = 1020.71 kN
N/mm
2
For minor axis bending buckling curve ‘b’ is applicable ( Refer – Table 10 of IS 800 : 2007 ) Since, t f =
12.5
≤
40 65.91
From Table 9(b) for KLy / ry = 147.55, fcdy = Hence, P dy = 313.4 kN Therefore, P d = P dy =
313.4
(b) Determination of M dz (Clause 9.3.2.2 & 8.2.2.1)
kN
N/mm
and
2
ny nz
= =
0.16 0.05
Page 3
Design Doc. No. Project Title Client Name of Unit Drawing Ref. Designed by:
SVNIT_ASHD_03 Rev. No. 0 RCC DESIGN OF VIERENDEEL TRUSS MEMBERS FOR ASHD BUILDING, SVNIT, SURAT VERTICAL EXPANSION OF ASHD BUILDING, SVNIT, SURAT Institute For Steel Development & Growth (INSDAG) SARDAR VALLABHBHAI NATIONAL INSTITUTE OF TECHNOLOGY, SURAT ISPAT NIKETAN', 1st. Floor 52 / 1A 1A Ballygunge Circular Road STAGGERED VIERENDEEL TRUSS MEMBERS OF ASHD BUILDING Kolkata - 700019 INS/SVNIT/GA/01, INS/SVNIT/GA/01, INS/SVNIT/GA/02 G.C. Checked by: G.C. Approved by: Date: 10.03.08 CALCULATIONS
Reference
KL / r y EI y h f 1 + 1 2 20 2( KL) h f / t f
π
M cr =
2
2
0.5
Where, L T = KL
π π = 3.14
0.21
α α LT =
(Refer Clause 8.2.2)
herefore, M cr = (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, λ LT = β b . Z p . f y / M cr = (1 x 466000 x 250 / 102666447.65) = and, φ LT = 0 .5 1 + α LT (λ LT − 0 .2 ) + λ LT
Hence, χ LT =
1
[φ
+ {φ LT − λ L T 2
LT
2
Now, f bd = χ LT . f y / γ m =
Hence,
dz
=
b.
pz.
}
0.5
]
2
LT
= 0.5[1 + 0.21(1.07 - 0.2) + 1.07²] =
=
1.16
1 / [1.16 + {1.16² - 1.07²}½] =
0.62 x 250 / 1.1 =
140.91
1 x 466000 x 140.91 =
bd =
1.07
0.62 N/mm
<
1
2
65664060 65664060 N-mm =
65.66
kN-m
(c) Determination of M dy (Clause 8.2.1.2) dy
=
py . y
b.
γ γ m m =
1 x 84000 x 250 / 1.1 =
19090909.09
=
19.09
N-mm kN-m
(d) Determination of C mz and C mLT (Clause 9.3.2.2)
0.38
ψ =
0.6 + 0.4 x 0.38 =
C mz =
0.752
=
C mLT
(e) Determination of C my (Clause 9.3.2.2)
0.4
ψ =
0.6 + 0.4 x 0.4 =
C my =
0.76
(f) Determination of K y , K z , and K LT (Clause 9.3.2.2) K y = 1 + (λy - 0.2)ny 0.2)ny =
1.23
>
1.128
1.01
≤
1.04
1.128
Hence, K y =
K z = 1 + (λz - 0.2)nz 0.2)nz =
1.01
Hence, K z =
K LT = 1 - (0.1λLT ny /(CmLT - 0.25) =
0.97
<
0.97
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 .0 0
Now, C my M y M z P + K y + K LT = Pdy M dy M dz
P Pdz
+ 0.6
K y
C my M y M dy
+
K z
C mz M z M dz
=
