September 14, 2016

Design of Industrial shed

Judul: Design of Industrial shed
Penulis: Pawan Jeph

CEL332 - DESIGN OF STEEL
STRUCTURES
COURSE PROJECT
DESIGN OF INDUSTRIAL SHED
SUBMITTED BY:-
Pawan Jeph (2010CE10421)
Sushant Singh(2010CE10408)
Sachin Arya(2010CE10398)
Lalit mohan meena(2010CE10361)
GROUP – Friday - (6F)
PROBLEM STATEMENT An industrial structure in steel construction Jaipur Roofing in galvanized iron (GI) sheets Live load 0.75 kN/m2 Crane capacity 225kN Safe bearing capacity of soil(SBC) 19 kN/m2 Span length (L) 10 m Bay length (10 no.) 5 m Height (H1) 8 m Height (H2) 5 m CODES USED
IS800:2007 – GENERAL CONSTRUCTION IN STEEL-CODE OF PRACTICE
IS875(PART 1,2,3): 1987 – DEAD LOADS, LIVE LOADS AND WIND LOADS ON STRUCTURES
MATERIAL USED
STEEL GRADE E250 (Fe 410): fy = 250MPa, fu = 410MPa, G = 0.77*105MPa

AREA ,ANGLE , NOS. OF PURLINS :-
Plan area for one truss, A = 10 × 5 = 50 m2
Depth (L/6 to L/5) = 10/6 10/5 = 1.66 m 2 m
So depth taken as 2 m
α = tan-1 (2/5) = 21.80
Inclined length = 2 × cosec(21.8) = 5.385 m
Plan area of inclined surface, Ai = 2 × (5.385 × 5) = 53.85 m2
Number of purlins = 9, Spacing of purlins @ 1.346 m
DESIGN OF TRUSS:-


Fig. Of Truss Members

Load Calculations : -
Dead Load
Weight of sheeting of inclined area =131x5.385x5x2=7.05 kN
Weight of Purlins
Number of purlins = 9
Assuming weight of purlin to be 100N/m
Weight of purlin =100x9x5 =4.5 kN
Self weight of truss on plan area, w = 20+6.6(span length) = 86 N/m2
Total self weight of Truss = 86 × 10 × 5 = 4.3 kN

Weight of Servicing and fixing:
For fixing = 0.025 kN/m2
For service= 0.1 kN/m2
Total for fixing and servicing = 0.125 kN/m2
Weight for fixing and serviceing = 0.125*plan area = 6.25 kN
Total Dead Load = 22.10 kN
Assume 20% extra, Total dead load = 26.52kN
Live Load
Live Load on Sloping Roof = 750 – 20 (α – 10) = 514 N/m2
Therefore, Live Load on truss = ×514 × plan area = 17130 N = 17.13 kN
Wind Load
Basic Wind Speed for Delhi, VB = 47 m/s [IS875 B Appendix A]
VZ = K1 × K2 × K3 × K4xVB
K1 = 1.0 [general buildings and for 50 years life]
K2 = 1.0 [Terrain, height of structure & size factor, category 2]
K3 = 1 [Assuming flat area with angle less than 3 degree]
K4 = 1.15 [for all industrial structures]
So, VZ = 54.05 m/s
Basic Wind Pressure = 0.6 VZ2 = 1.75 kN/m2

θ = 0º

Internal Wind Pressure = Cpi = ± 0.2 p
External Wind Pressure = Cpe
h/w = 8/10 =0.8 So [0.5<h/w<1.5]
Degree Wind Angle θ = 0º Wind Angle θ = 90º
EF GH EG FH
20º -0.7 -0.5 -0.8 -0.6
21.8º -0.61 -0.5 -0.8 -0.636
30º -0.2 -0.5 -0.8 -0.8

CASE (I): θ = 0º, external + internal
Windward: Cpe + Cpi = -0.61 + 0.2 = -0.41
Leeward: Cpe + Cpi = -0.5 + 0.2 = -0.3
CASE (II): θ = 0º, external – internal
Windward: Cpe – Cpi = -0.61 – 0.2 = -0.81
Leeward: Cpe – Cpi = -0.5 – 0.2 = -0.7
CASE (III): θ = 90º, external + internal
Windward: Cpe + Cpi = -0.8 + 0.2 = -0.6
Leeward: Cpe + Cpi = -0.636 + 0.2 = -0.436
CASE (IV): θ = 90º, external – internal
Windward: Cpe – Cpi = -0.8 – 0.2 = -1.0
Leeward: Cpe – Cpi = -0.636 – 0.2 = -0.836
Most Severe Effect
Most Severe Effect = -1.0
Wpe = -1.0 × Design Wind Pressure = -1.0 ×1.75 = 1.75 kN/m2 (away from the rafter)
Wind Load on Principal Rafter = 1.0 x1.75 × 2 × 5.385 × 5 = 94.23 kN
We applied the following unit loads on panel points and analyzed in STAAD (results attached along with report )
Load case 1
1.5(dead laod + live load ) = 67.18 kN
Load 1 (green) = 67.18*1.25/10 = 8.40kN
Load 2 (blue) = 8,40 /2 =4.10 kN

Load case 2:
1.5(dead load +wind load)
Load 1(y) = 10.26 kN Load 1(x) = 6.1 kN
Load 2 (y) = 5.133 kN Load 2 (x) = 3.05 kN ( At corner nodes only )

Load case 3 :
1.2(dead load + live load + wind load)
Load 1(y) = 5.64 kN Load 1 (x) = 4.87 kN
Load 2 (y) = 2.82 kN Load 2 (x) = 2.43 kN (At corner nodes only )

The members as follows:
Vertical 15,21,8,19,13 2000 38.1 25.2
Vertical (0 Force) 23,17 500 0 0
Inclined 24,31,26,27,30,29 1952 16.4 24.8
Tie Members 1,2,3,4,5,6,7,10 1250 86.7 73.5
Table: Forces in truss members


Roof Truss Design
Principal Rafter

L = 1.35 m
Design Compressive Load = 80 kN
Design Tensile Load = 97 kN
Effective length = 1 × 1350 = 1350 mm
For = 100,
Buckling Class 'C' [Table 10, IS800]
Therefore, fcd = 107 MPa [Table 7(c), IS800]
P = fcdA
Therefore, required cross section area = 80 × 103/107 = 747.66 mm2
Take 2 ISA 50 ×5 0 × 5 back-to-back [Steel Table]
Total area = 958 mm2 > 747.66 mm2
Minimum Moment of Inertia is @ z-z axis = 2Izz rzz = rmin = 15.2mm

= 1 × 1350/15.2 = 88.81 < 180 → OK

fcd = 122.785 MPa [Table 9(c), IS800]
Load capacity = 95.281 kN > 80 kN → OK
For local buckling,
b/t = 50/4 = 12.5<15ε (ε=1)
d/t = 50/4 = 12.5<15ε → OK
Check for Tension
Yielding [6.2, Pg 32, IS Code 800: 2007]

Tdg = Agfy/γmo = 776 × 250/1.1 = 176.36 kN ( > 97 kN → OK)

Rupture of Net Section [6.3.3 Pg 33, IS Code 800: 2007]
Tdn = 0.9Anfu/γm1
This depends on length of connection,
Using welded connection of size of weld = 4 mm
Weld Strength/mm = 0.707 × S × fuw/√3 × 1/γm1 = 535.54 N/mm
Weld length required for each ISA = 97 × 103/ 2× 535.54 = 90.56 mm ≈ 100mm.

l1/l2 = 2. 546
Use l1 = 70 mm and l2 = 30 mm
Lc = length of connection = l1 = 70 mm



Therefore,
Tdn = 0.9 × Anc × fu/γm1 + β × Ago × fy/ γm0
β = 1.4 – 0.076 (
bs = shear leg width = 50 mm
w = width of outstanding leg = 50 mm
t = thickness = 5 mm
Lc = 70 mm
β = 1.4 – 0.076 (50/5)(250/410)(50/70)= 1.0689 ( > 0.7 → OK)
Also, 1.0689< fu γm0/fyγm1 = (410x1.1)/ (250x1.25) = 1.44 → OK

Tdn = 0.9 × 475 × + 1.0689 × 776 × = 328.735 kN ( > 97 kN → OK)

(c) Block Shear Strength [6.4.1 Pg 33, IS 800:2007]
As at the end, connected leg is only welded along the length of the member, block shear failure will only be along weld line by shear

Total weld length = 70 mm for ISA 50 × 50 × 5
Tdb = Avg × fy /(1.732x γm0)= 131.215 > 97 kN → OK

Or Tdb = 0.9 × Avn x fu /(1.732x γm1) = 170.433 >97→OK

Vertical Member

Design Compressive Strength = 39 kN
Design Tensile Strength = 26 kN
L = 2.00 m
Effective length = 1 × 2.00 = 2000 mm
For = 50, Buckling Class 'C' [Table 10, IS800]
Therefore, fcd = 107 MPa [Table 7(c), IS800]
P = fcdA
Therefore, required cross section area = 39 × 103/107 = 364.48 mm2
Take ISA 70 × 70 × 6 [Steel Table]
Total area = 806 mm2 > 364 mm2
Minimum Moment of Inertia is @ z-z axis = 2Izz rzz = rmin = 13.6

= 1 × 2000/13.6 = 147.05 < 180 → OK

fcd = 61.265 MPa [Table 9(c), IS800]
Load capacity = 49.38 kN > 39 kN → OK
For local buckling,
b/t = 70/6 = 11.67<15ε
d/t = 70/6 = 11.67<15ε → OK

Check for Tension
(a) Yielding [6.2 Pg 32, IS 800:2007]

Tdg = Agfy/γmo = 183.18 kN ( >26 kN → OK)

Tdn = 0.9Anfu/γm1
This depends on length of connection,
Using welded connection of size of weld = 4 mm
Weld Strength/mm = 0.707 × S × fuw/√3 × 1/γm1 = 535.54 N/mm
Weld length required for each ISA = 26 × 103/535.54 = 48.55 mm ≈ 60mm.

Use l1 = 55 mm and l2 = 20 mm
Lc = length of connection = l1 = 55 mm

Anc = (70 – 3) × 6 = 402 mm2

Therefore, Tdn = 0.9 × Anc × fu/γm1 + β × Ago × fy/ γm0
β = 1.4 – 0.076 (
bs = shear leg width = 70 mm
w = width of outstanding leg = 70 mm
t = thickness = 6 mm
Lc = 55 mm
β = 0.711( > 0.7 → OK)
Also, 0.711< fyγm1 = 1.44 → OK

Tdn = 248.912kN ( > 26 kN → OK)

(c) Block Shear Strength [6.4.1 Pg 33, IS 800:2007]
As at the end, connected leg is only welded along the length of the member, block shear failure will only be along weld line by shear

Total weld length = 75 mm
Tdb = Avg × fy/ √3γm0 =59.047> 26 kN → OK

Or Tdb = 0.9 × Avn × fu/ √3γm1 =76.695 > 26 kN → OK

Inclined Members
L = 1.952 m
Design Compressive Strength = 17 kN Design Tensile Strength = 25 kN
For = 100, Buckling Class 'C' [Table 10, IS800]
Therefore, fcd = 107 MPa [Table 7(c), IS800]
P = fcdA
Therefore, required cross section area = 17 × 103/183 = 158.87 mm2
Take ISA 60 × 60 × 6 [Steel Table]
Total area = 684 mm2 > 158.87 mm2
Minimum Moment of Inertia is @ z-z axis = Izz rzz = rmin = 11.5 [along vv axis]
= 1 × 1952/11.5 = 169.74 < 180 → OK

fcd = 48.1 MPa [Table 9(c), IS800]
Load capacity = 32.9 kN > 16 kN → OK
For local buckling,
b/t = 60/6 = 10 <15ε [=15] d/t = 60/6 = 10<15ε → OK

Check for Tension
Yielding [6.2 Pg 32, IS 800:2007]

Tdg = Agfy/γmo = 155.45 kN ( > 25 kN → OK)

Rupture of Net Section [6.3.3 Pg 33, IS 800:2007]

Tdn = 0.9Anfu/γm1
This depends on length of connection,
Using welded connection of size of weld = 4 mm
Weld Strength/mm = 0.707 × S × fuw/√3 × 1/γm1 = 535.54 N/mm
Weld length required for each ISA = 65 mm

l1/l2 = 41.9/18.1 = 2.315
Use l2 = 45 mm and l1 = 20 mm
Lc = length of connection = l2 = 45 mm

Anc = (60 – 3) × 6 = 342 mm2

Therefore,
Tdn = 0.9 × Anc × fu/γm1 + β × Ago × fy/ γm0
β = 1.4 – 0.076 (
β = 0.782( >0.7→OK)

Also, 0.782 < fuγm0 / fyγm1 = 1.44 → OK

Tdn = 222.523 kN ( > 25 kN → OK)

(c) Block Shear Strength [6.4.1 Pg 33, IS 800:2007]

As at the end, connected leg is only welded along the length of the member, block shear failure will only be along weld line by shear

Total weld length = 65 mm for ISA 60 × 60 × 6
Tdb = Avg × fy/ √3γm0 = 65x6x250/(1.732x1.1) = 51.174 > 25 kN → OK

Or Tdb = 0.9 × Avn × fu/ √3γm1 = 0.9x65x6x410/(1.732x1.25) = 66.469 > 25 kN → OK


(IV) Tie Member
L = 1.25 m
Design Compressive Strength = 87 kN Design Tensile Strength = 74 kN
For = 100, Buckling Class 'C' [Table 10, IS800]
Therefore, fcd = 107 MPa [Table 7(c), IS800]

P = fcdA
Therefore, required cross section area = 87 × 103/107 = 813.08 mm2
Take 2 ISA 50 × 50 × 5 back-to-back [Steel Table]
Total area = 958 mm2 > 813.08 mm2
Minimum Moment of Inertia is @ z-z axis = 2Izz
rzz = rmin = 15.2

= 1 × 1250/15.2 = 82.23 < 180 → OK

fcd = 132.655 MPa [Table 9(c), IS800]
Load capacity = 127.083 kN > 87 kN → OK
For local buckling,
b/t = 50/5 = 10<15ε [=15]
d/t = 50/5 = 10<15ε → OK
Check for Tension
Yielding [6.2 Pg 32, IS 800:2007]
Tdg = Agfy/γmo = 958 × 250/1.1 = 217.727 kN ( > 74 kN → OK)

Rupture of Net Section [6.3.3 Pg 33, IS 800:2007]
Tdn = 0.9Anfu/γm1
This depends on length of connection,
Using welded connection of size of weld = 4 mm
Weld Strength/mm = 0.707 × S × fuw/√3 × 1/γm1 = 535.54 N/mm
Weld length required for each ISA = 74 × 103/ 2× 535.54 = 69.089 mm ~ 75 mm
l1/l2 = 2.546
Use l1 = 65 mm and l2 = 25 mm
Lc = length of connection = l1 = 65 mm

Anc = 2 × (50 – 2.5) × 5 = 475 mm2

Therefore, Tdn = 0.9 × Anc × fu/γm1 + β × Ago × fy/ γm0
β = 1.4 – 0.076 (
bs = shear leg width = 50 mm
w = width of outstanding leg = 50 mm
t = thickness = 5 mm Lc = 65 mm
β = 1.4 – 0.076 (50/5)(250/410)(50/65)= 1.043
(1.043 > 0.7→ OK)
Also, 1.043< fuγmo /fyγm1 = 1.44 → OK

Tdn = 0.9 × 475x410/1.25 + 1.043 × 475x250/1.1 = 252.816 kN ( > 74 kN → OK)

(c) Block Shear Strength [6.4.1 Pg 33, IS 800:2007]

As at the end, connected leg is only welded along the length of the member, block shear failure will only be along weld line by shear

Total weld length = 90 mm for ISA 50 × 50 × 5
Tdb = Avg × fy /(1.732x γm0 ) = 85.29 > 74 kN → OK
Or Tdb = 0.9 × Avn × fu /(1.732x γm0 ) = 110.781 > 74 kN → OK
(V) ZERO Force Vertical Member

L =500 mm
Provide single ISA 30 × 30 × 5 section
l1/l2 = 2.26
let total weld length = 30 mm
Weld length, l1 = 20 mm
l2 = 10 mm

It will be safe in both tension and compression, as it is a zero force member.

Final sectional design and self-weight:
S.No. Member group Quantity Length (m) Section Weight(kg/m) Total weight(kg)
1 Tie members 8 1.25 2 ISA 50 × 50 × 5 3.8 76
2 Principle Rafter members 8 1.346 2 ISA 50 ×5 0 × 5 3.8 81.84
3 Inclined members 6 1.952 ISA 60 × 60 × 6 5.4 63.25
4 Vertical members 6 2.0 ISA 70 × 70 × 6 6.3 75.6
5 Vertical member 2 0.5 ISA 30 × 30 × 5 2.2 2.2
∑ = 298.89kg
Assumed self-weight of truss = 430 kg > 298.89 kg okay
Hence, OK
Purlin Design
Dead Load:
Weight of GI sheet = 131*1.34625 = 176.35 N/m
Assumed distance between purlin = 1.34625 m.
Assumed self-weight = 100 N/m
Total = 276.35N/m
Live Load:
Live load = 750 - 20 (21.8-10) = 514 N/m2
Live load on purlin per metre = 514*1.25= 645 N/m
Wind Load: (calculated earlier)
Wpe = 1750 N/m2 (away from rafter)
Wind load on purlin per metre = 1750*1.34625 = 2360 N/m
Design Summary:
Dead Load = 280 N/m
Wdy = 280 *cos(21.8) = 259.97 N/m
Wdz = 280*sin(21.8) = 103.98 N/m
Live Load = 645 N/m
Wly = 645*cos(21.8) = 598.87 N/m
Wlz = 645*sin(21.8) = 239.53 N/m
Wind Load = 2360 N/m
Wwy = -2360 N/m (outwards)
Wwz = 0
Factored Design Load:
Load combination 1: 1.5 (DL+LL)
Wy1 = 1.5*(260+600) = 1290 N/m
Wz1 = 1.5*(104+240) = 516 N/m
Load combination 2: 1.5 (DL+WL)
Wy2 = 1.5*(260-2360) = -3150 N/m
Wz2 = 1.5*(104 + 0) = 156 N/m
Load combination 3: 1.2 (DL+LL+WL)
Wy3 = 1.2*(260+ 600 - 2360) = -1800 N/m
Wz3 = 1.2*(104 + 240 + 0) = 412.8 N/m
Bending Moment Calculation:
Plastic Analysis: Failure of two spans is considered
Collapse of span Mp = Wu*L2/8 (as per code )
assumption : simply supported purlin - safer case.
Among three combinations, 1.5 (DL + WL) governs

Mz = Wy*L2/8 = 3150*52/8 = 9843.75 Nm
My = Wz*L2/8 = 516*52/8 = 1612.5 Nm
Selection of section:
fy = 250 MPa
Ԑ = √(250/fy) = 1
Try ISJC 150 @ 9.9 kg/m weight = 97.1 N/m < 100 (assumed) Hence O.K.
b/tf = 55/6.9 = 7.97 < 9.4Ԑ (Section B plastic, Table 2, page 18)
d/tw = 136.2/3.6 = 37.83 < 42Ԑ
βb = 1(for plastic section)
Bending about y-y:
for Zyy
A = 1249.32 mm2
Zyy = 18.682 mm3
M(design-y) = βb*Zp*fy/ vm0 (Zp = Ze = Zyy)
= 1*18.682*103*250/1.1 = 4.24 kNm > 1.6125 kNm
Hence O.K.
Bending about z-z:
Mdz = βb*Zzz*fy/ m0
= 1*72.04*103*250/1.1 = 16.3 kN/m > 9.843 kNm
Hence O.K.
Check for biaxial bending:
(Mz/Mdz) + (My/Mdy) = (9.843/16.3) + (1.6125/4.24) = 0.98 < 1
Hence O.K.
Check for Shear:
The above check for Md is applicable when V < 0.6*Vd
To obtain shear at penultimate support, find 'R' at end support
Mpz = 9843 = -R*5 + (3150*5*5/2)
R = 5906.4 N
V at penultimate support = (3150*5) – 9843 = 5.907 kN
0.6*Vd = 0.6*h*tw*fyw/ 1.732*Vm0 = 0.6*150*3.6*250/(1.732*1.1) = 42.513 kN
V < 0.6* Vd
Hence O.K.
Check for deflection:
Limiting value of deflection for GI sheets = L/180 = 5000/180 = 27.75 mm
(Table 4 page 29)
LOAD COMBINATION AT SERVICE LOAD :-
Load combination 1: (DL+LL)
Wy1 = 260+600= 860 N/m
Wz1 = 104+240= 344 N/m
Load combination 2: (DL+WL)
Wy2 = 260-2360= -2100 N/m
Wz2 = 104 + 0= 104 N/m
Load combination 3: (DL+LL*0.8+WL*0.8)
Wy3 = 260+ (600*0.8) – (2360*0.8) = -1148 N/m
Wz3 = 104 + (240*0.8) + 0 = 296 N/m
Taking load combination 2 and considering simply supported span for purlin
Wy2 = - 2100 N/m
δy = 5*Wy*L4/(384*E*I) = 5*2.1*54 *1012 / (384*2*105*471.1*104)
=18.2 mm
(maximum deflection in y direction)
Wz2 = 104 N/m
δz = 11.16 mm

Resultant deflection = √( δy2 + δz2) = 21.35 mm < 27.75 mm
Hence O.K.
LOADING OF CRANE
Weight of crane girder/truss = 180 KN
Crane capacity= 225KN
Weight of crab + motor = 50 KN
Span of crane girder/truss (Lc) = 10m
Min. hook approach (L1) = 1.2m
Center-to-center distance between gantry columns (L) = 5m
Weight of rail= .25KN/m
Distance Between wheels (c)= 2m
Vertical Loading
Maximum static wheel load due to weight of crane = 225/4 = 56.25kN
Assuming L1 (minimum hook approach to centerline of crane beam)=1.2m
Lc (span of crane) = 10m
Maximum static wheel load due to crane load = W1 = [Wt(Lc – L1)]/(2*Lc)
=[(225+50)*(10-1.2)]/(10*2) = 121kN
Total load due to weight of crane and crane load = 50+121 = 171kN
To allow for impact etc. this load is increased by 25% (clause 6.1, IS875 Part B)
Design load = 1.25 * 171 = 213.75kN
Factored Wheel load on each wheel = Wc = 213.75* 1.5 = 320.625kN
Lateral (horizontal) Surge load
Lateral load (per wheel) = 10% of (hook + crab load)/4 = 0.1 * (225+50)/4 = 6.875kN
Factored lateral load = 1.5 * 6.875 = 10.3kN
Longitudinal (horizontal) brake load
Horizontal force along the rails = 5% of wheel load = 0.05 * 213.75 = 10.68kN
Factored load Pg = 1.5 * 10.68 = 16.03kN
Lateral shear force due to sway load:
Vy = 10.3 x (2 – 2/5) = 16.48 kN
MAXIMUM BENDING MOMENT OF CRANE:
Vertical Maximum bending moment:
L (centre to centre distance between columns= span on gantry girder) = 5m
Without considering self weight
M1 = Wc*L/4 = 320.625 * 5/4 = 400.78 KNm
M2 = 2*Wc*(L/2 – C/4)2/L = 2*320.625*(5/2 – 2/4)2/5= 513KNm
Hence, M = 513KNm
Self- weight of rail = .25KN/m
Assuming self-weight of gantry girder is 1.6kN/m
Total dead load = (1.600+ self weight of rail) = (1.6+.250) = 1.85kN/m
Factored dead load = 1.5 * 1.9 = 2.78kN/m
Bending Moment due to dead load = WL2/8 = 2.78*52/8 = 8.68 KNm
Horizontal maximum bending moment
Moment due to surge = 2 x 10.3 x (5/2 – 2/4)2 /5
My = 16.48 kN-m
Bending moment due to drag (rail height = 0.15 m, depth of girder = 0.6 m):
Reaction due to drag force= Pg*e/L
= 16.03 x (0.3+0.15)/5
= 1.44 kN
Mz = 1.44 (5/2 – 2/4)
= 2.88 kNm
Total design bending moment (Mz) = 513 + 8.68 + 2.88
= 524.56 KN-m = 525KNm
SHEAR FORCE:
Vertical shear force:
Shear force due to wheel load = WL (2-c/L) = 320.625 x (2 – 2/5) = 513 kN
Shear force due to Dead Load = WL/2 = 2.78 x 5/2 = 6.95 kN
Maximum ultimate shear force Vz = 513+6.95 =519.95KN
Reactions due to drag force = 1.28 KN
Maximum ultimate reaction = 519.95+1.28
=521.23 = 522KN
Design of Gantry Girder:
Preliminary Section Selection:
Since L/12 = 5000/12 = 416.66mm
We choose depth = 600 mm
Approximate width of beam =L/30 = 5000/30 = 166.6m
Design moment for laterally unsupported beam, Md = βb.Zp.fbd (clause 8.2.2, pg 54, IS800)
Required Design moment (Md) = 525 KNm
(From Book Pg 1042) Zp (required) = 1.4M/fy= (1.4*525 x 106)/250 = 2.94 x 106 mm3
Select ISMB 600 @ 1.22 KN/m with a ISLC 300 @.325KN/m on top
Bf of ISMB600= 210 mm
Properties of the sections:
ISMB 600 @ 1.22 kN/m ISLC 300 @ 0.325 kN/m
A= 15621 mm2 A= 4211 mm2
tf=20.8 mm tf=11.6 mm
tw=12 mm t-w=6.7 mm
B= 210 mm B= 100 mm
Izz = 91813 * 104mm4 Izz = 6048 * 104mm4
Iyy = 2651 * 104mm4 Iyy = 346 * 104mm4
R= 20 mm Cy= 25.5 mm

Total dead load intensity = 1.2 (ISMB) + 0.325 (ISLC) + 0.3 (Rails)= 1.825 kN/m =~ 1.9 kN/m
As we have assumed approx. Same value of the dead load, there is no need to change.
Elastic Properties of combined section:
Total area, A= 15621+4211 = 19832 mm2
Distance of NA from extreme fibre of tension flange= [15621*300 +4211*(600+6.7-25.5)]/19832 = 359.71 mm
Izz= 91813* 104 +15621*59.712 +346* 104 +4211* 221.52
= 1.18*109 mm4
Calculation of plastic modulus:
The plastic neutral axis divides the area into two equal areas, i.e. 9916 mm2.
15621/2 + (x)*12 = 15621/2 - (x)*12 + 4211
x = 175.45 mm = distance from centre of I-section to plastic neutral axis
Zpz = 20.3*210*(475-20.3/2) + (475-20.3)*12*(475-20.3)/2 + 4211*(106.2) + 210*20.3*(131.7-6.7-10.15) + (559.4/2 – 175)2*12/2
= (3222.2 + 1002.6)*103 mm3 = 4224.8*103 mm3
For the top flange only
Zpy = 20.3*2102/4 + (300-2*11.6)2*6.7/4 + 2*100*11.6*(150-11.6/2) = 558.8*103 mm3
Moment capacity of section:
Mcr= (C1 π2EIy /KL)*{[(K2Iw /Kw2 Iy) + GIt (KL)2/π2EIy+ (c2 yg - c3 yj)2]0.5– (c2 yg - c3 yj)}
(Clause E-1.2)
K = 0.8, C1= 1.03, C2= 0.422, C3= 1.22
Kw=Warping restraint factor = 1.0
Yg= 306.7mm
Ift= bd3/12 = 20.8 * 2103/12 = 1.60 * 107 mm4
Ifc= 1.60 * 107 + 6.048 * 107 = 7.64 * 107 mm4
βf = Ifc /( Ifc + Ift) = 7.64/(7.64+1.64) = 0.83> 0.5
Consider a plate of area equal to ISLC300 of thickness 't'
Then t= 4211/210 = 20 mm
t1= 20+20.8 =40.8 mm
hy= 620-40.8/2-20.8/2 = 589.2mm
Yj= 0.8*(2βf– 1)* hy /2 = 0.8*(2*0.83-1)*583.2/2 = 153.96 mm
Iyy= 6048*104+2651*104= 8699*104mm4
Llt= 0.8*5000 = 4000mm
Iw= (1-βf)* βf*Iy*(hy)2= (1-0.83)*0.83*8699*104*589.22= 4.26*1012mm6
It= Torsional Constant = ∑bt3/3 = [300*27.53+ 558.4*123+ 200*20.83]/3 =30.01*105mm4
G = 0.77*105MPa
E=2 * 105 Mpa
Plugging all the values: Mcr =3118.3 KNm
Laterally unsupported beam, using (clause 8.2.2, Pg.54, IS800)
Mdz= βb*Zp*fbd
λLT= √( βb Zp fy/ Mcr) ≤√(1.2*Ze*fy/ Mcr)
λLT = √(1*42.24*105*250/3118.3*106) = 0.85
αlt= 0.21 for rolled section
ΦLT= 0.5*[1+ αlt (λLT - 0.2)+ λLT 2] = 0.5*[1+0.21(0.85-0.2)+0.852] = 0.93
xLT= 1/[ ΦLT + (ΦLT 2- λLT 2)0.5] = .76 ≤ 1.0
fbd = 0.947*250/1.1 = 172.8 MPa
Mdz= βb *Zp* fbd = 1.0*42.24*105*172.8 = 729.9 kN.m > Md
Moment capacity of transverse section:
Plastic moment capacity about y-y axis = Mdy= 1*250*5.59*105/1.1
=127.04 kNm
Check for biaxial bending:
(Mz/ Mdz)+(My/ Mdy) = (525/729.9) +(16.48/127.04)=0.85 ≤1 (OK) (clause 9.3.1.1, pg 70 IS800)
Check for shear capacity
Design shear force for vertical load Vz= 445kN
Shear capacity = Av*fyw*/(√3*1.1) = 12*558.4*250/(√3*1.1)
= 879.25 kN > 445kN (OK)
Design shear force= 445 kN < 0.6*879.25 = 527.55 kN
Hence, it is safe in vertical shear and no reduction in moment capacity required.
Check for deflection
Deflection at mid span Δ= wL3(3a/4L – a3/L3)/(6EI)
where a =(L-C)/2 = (5000-3500)/2 = 750
1. Vertical: Combined Izz= 1.18 *109 mm4
Δ = [171*103*50003/(6*2*105*1.18*109)] * [3*750/4*5000 – 7503/50003] =1.64mm < 5000/750 = 6.66 mm Safe
2. Lateral: only the compound top flange will be assumed to resist applied surge load as in bending check.
I = (Iz)ch+ IF= 6048*104 + 20.8*2103/12 = 7653.24104 mm4
Δ = [6.875*103*50003/(6*2*105*7653.24*104)] * [3*750/4*5000 – 7503/50003] =1.02 mm< 6.66mm Safe
Check for Web Buckling: At points of concentrated loads (wheel load/reaction) the web of the girder must be checked for local buckling. The dispersion length under the wheel (assuming diameter of wheel = 150mm, angle of dispersion = 45˚)
b1=150mm, h1=600/2 + 2*6.7 = 313.4mm
Web slenderness λ= 2.5*d/t = 2.5*(600- 2*(20.3+20))/12 = 108.2
Stress reduction factor (table8, IS800) = 0.426
fcd=0.426*250/1.1 = 96.8MPa
Buckling resistance =(b1+n1)*t*fcd= (150+313.4)*12*96.8*10-3
=538.3 kN
Maximum wheel load = 171kN< 538.3kN Safe
Web bearing: Load dispersion at support with 1:2.5 dispersion
Minimum stiff bearing b1= Rx/ (tfyw/1.1)-n2,
n2= (20.3+20)*2.5 = 100.75mm
Rx= 445 kN (support reaction)
b1= 445*103/ (12*250/1.1) – 100.75 = 62.41 mm
Web bearing at the support requires a minimum stiff bearing of 62.41 mm
Weld design
The required shear capacity of the weld is given by, q= VAy/Iz
y=606.7-359.71-6.7= 240.29
A= 4211 mm2, V= 445 kN
Iz= 1.18*109 mm4
q= 445*103*4211*240.29/(1.18*109)
= 381.6 N/mm
This shear is taken by the welds. Hence use a minimum weld of 4 mm (442 N/mm per weld) connection the channel to the top flange of the I-beam.
For lateral shear force,
Fy=16.48 kN
Shear capacity Vny= Avfyw/(3*1.10)
= 250/(3*1.10)*(210*4)*10-3
= 110.2 kN > 16.48 kN
Hence it is safe for resisting lateral shear.
Wind load calculation for columns
Basic wind speed = Vb= 47m/s (for Jaipur)
V2=k1*K2*k3*k4*Vb
K1=1
K2=1
K3=1
k4=1.15
V2= 1*1*1*1.15*47= 54.05 m/s
Basic wind pressure= 0.6*v22 = 0.6*54.052 =1.75 KN/m2
Cpe= external pressure coefficient, Cpi= internal pressure coefficient,
A= surface area of structural element or cladding unit, and
For Cpe: (Table 5, Page 16)
h/w = 8/10 (>1/2 , <3/2)
l/w = 10/5 =2.0
Roof Angle Wind Angle
ϴ=00 ϴ=900
A B C D A B C D
21.80 -0.7 -0.3 -0.7 -0.7 -0.5 -0.5 0.7 -0.1
Cpi= -0.2 or +0.2
Cpi Cpe-Cpi
ϴ=00 ϴ=900
A B C D A B C D
0.2 -0.5 -0.1 0.5 0.5 -0.5 -0.1 -0.5 -0.5
-0.2 -0.9 -0.5 -0.9 -0.9 -0.7 -0.7 0.5 -0.3
Critical Wind load per unit area: (critical case)
ϴ=00:
A= 0.9*1.75=1.575 KN/m2
B = 0.5*1.75=0.875KN/ m2
C= 0.9*1.75=1.575KN/ m2
D=0.9*1.75=1.575 KN/ m2
ϴ=900
A = 0.7*1.75=1.225 KN/ m2
B = 0.7*1.75=1.225 KN/ m2
C=0.9*1.75=1.575 KN/ m2 (inside)
D=0.3*1.75=0.525 KN/ m2
Area supported by wall for one column:
Length of column* Bay Length = 8*5 m2 = 40 m2
Critical Wind load:
ϴ=00
A= 63KN
B = 35 KN
C= 63 KN
D= 63 KN
ϴ=900
A = 49
B = 49 KN
C= -63 KN
D= 21 KN
Design Load =63 ≈ 65 KN
Column Design:
Maximum reaction force from truss = 33.06 kN = 34kN
Assuming Cladding Weight = 15KN
Spacing provided between two ISHB columns = 0.5m
Maximum Bending Moment = -34*0.25(From reaction of truss) + 16.48 (From Surge Load of crane) + 530*0.25 (From vertical crane load) -15*0.25 (From cladding) +65*8/2 (From wind) = 461.73 KN-m
Using ISHB300 @ 63kg/m (Pg 138 IS code) – (section is determined by KL/ryy)
ryy = 52.9mm , D = 300mm, tf =10.6mm, tw =9.4mm, bf=250mm, Area= 8025mm2 R=11.0 Izz=12950.2cm4 Iyy=2246.7 cm4
Net Axial Force = 34KN (Due to truss) +530KN (Due to crane) + 63*5/100KN (Due to column of height H2) + 63*8/100KN (Due to column of height H1) +15KN (Due to cladding) = 587.19KN = 590KN
Moment of inertia of combined section = 2*{2246.7*104+8025*2502} = 1048.06*106 mm4
Za = Iyy/(B/2+s/2) = 2.79*10^6 mm3
ra = √(Iyy/2*Area) = 255.53 mm
Checking local buckling: (Table 2, pg 18)
For fy=250 ϵ =(250/fy)1/2 = 1
b/tf = 125/10.6 = 11.79 > 10.5 (Therefore, section is semi-compact)
Checking local capacity of section:
Nd= Agfy/γmo = 2*8025*250/(1.1*1000) = 3647 KN
n= N/Nd = 590/3647 = 0.161
Mda = Ze*fy/γmo (Semi-compact section) = 634.09 KN-m
Reduced plastic moment for rolled section = Mnda = 1.1*Mda{1-n} = 585.2 KN-m< Mda
Interaction equation for cross-section: (Section 9.1, Page 70):(Mz/Mnda)α2 < 1.0 (α2 = 2 (Table 17))
= (415.5/585.2)2 =0.504 < 1.0
Hence O.K
Compressive strength:
KLa/ra = 1.5*5000/255.53 = 29.35
KL/rzz = 0.85*5000/127 = 33.46
Local slenderness between lacings
(Assuming 45o lacing, spacing = 750mm) =750/ryy = 14.57 < 50
14.57 < 0.7*46.85 =32.79 (o.k) (Clause 7.6.6)
L/r of compound column = 1.05*46.85 = 49.19
h/bf = 300/250 = 1.2 tf <100mm
for built-up section, buckling class=cHence, fcd = 187.5 N/mm2 (Table 9(c))
Check for overall buckling:Assuming bending about z-z axis produces axial forces in the two I-sections of the compound column.
Axial force = moment/centroidal distance between the two I sections=461.73/0.75 = 615.64 kN
Maximum compression force in one I-section = 590/2 +615.64 = 910.64 kN
Compression resistance of section = Agfcd/γm0 = 8025 * 187.5
= 1504.6kN > 910.64 kN
Hence O.K
Design of Lacing:
Inclination=45 (between 40-70) (Cl. 7.6.4, Pg.50)
Assuming gauge distance = 50mm
Hence, spacing of lacing bar = 250+50+50mm = 350mm
Slenderness ratio of compound member between consecutive connections = (350+350)/ryy = 700/52.9 = 13.29 (< 50, <0.7*105)(Cl. 7.6.5.1, Pg.50)
Width of lacing bar = 3*(nominal diameter of bolt) = 3*20mm = 60mm (Cl. 7.6.2, Pg.50)
Thickness of lacing = leff/40 = 350*√2/40 = 12.37mm ≈ 15mm
Hence, size of lacing is 350mm X 60mm X 15mm
Checking slenderness of lacing:
r = √(I/y) = t/√12 = 4.33mm
leff/r = 350*√2/4.33 = 114.31 < 145 (safe)
Design shear force = 2.5% of maximum compression in I-section = 2.5*(910.64)/100 = 21.56 KN
Hence, compressive force = 21.56/sin(45) ≈ 30 KN
Maximum force in lowest diagonal = (48.5+6)/cos(45) = 77.07 KN (Compression)
Hence, total force = (77.07+30)/2 = 53.53KN
For l/r = 86.3, fcd = 128 (Table 9(c) , IS 800)
Hence, design compressive strength = 128*60*15/1000 = 115.2 KN > 53.53 KN
Tensile force = 21.56/sin(45) ≈ 30 KN
Tensile strength = 0.9*(60-20)*15*410/(1.25*1000) = 177.12 KN >53.53 KN
Bolt strength = 2.5*0.49*15*20*410/(1.25*1000) = 120 KN > 53.53 KN
Design of tie plates:
Effective depth of tie plates > 2bf (2*250) = 500mm or c/c between two I-sections = 500 mm
Assuming edge distance = 30mm (>1.7do)
Required overall depth = 500 + 2*30mm =560mm
Length of tie plate = 500 + 250mm =750mm
Hence, required thickness = 1/50 *(750-30-30) ≈ 15mm
Hence, size of tie plate is 750mm X 300mm X 15mm
DESIGN OF BASE PLATE
Total design load on Base Plate = 34KN (Due to truss) +530KN (Due to crane) + 63*5/100KN (Due to column of height H2) + 63*8/100KN (Due to column of height H1) +15KN (Due to cladding) =587.19 ~ 590 KN
Assuming M20 concrete
Bearing capacity of concrete = 0.45*20 N/mm2 = 9.0 N/mm2
Eccentricity, e = M/P
= 461.73/590
= 0.78 m
If we design for entire compression in the base plate, required length of base plate will be = 6e = 4.7 m (Too large)
As this value is too large, we will design it for some tension as well. As the width of the built-up member is 750 mm and taking an edge distance of 200 mm on each side, the total base plate length will be 1.15m
Required Breadth (B) = 2p/(L*0.45fck)
= 2*590*1000/(1150*9)
= 114 ≈ 120mm
As this length will not accommodate the depth of I-section and anchor bolts, we will have to take much larger breath
Take B=600mm
Area (A) = 1150*600 = 690000mm2
Z = bd2/6 = 600*11502/6 = 132.25*106mm3
Hence, required area = 590*1000/9 mm2 = 65555.5 mm2 <690000 mm2 Safe
End Pressures
pmax = P/A + M/Z = 590*1000/690000 + 461.73*106/(132.25*106)
= 4.35 Mpa
pmin = P/A – M/Z = -2.64 Mpa
Base Pressure at section along Depth of I-section
= [(1150 – 200)/1150)*(4.35+2.64) – 2.64 = 3.13N/mm2
Neglecting reduction in moment due to biaxial bending
Moment along given axis
= Wa2/2 (rectangular part) + Moment of triangular part
= 3.31*2002/2 + 0.5*250*(2/3)*200*(4.35-3.31)
= 83.53*103Nmm
Moment Capacity = 1.2fyZe/γm0
Ze = t2/6
83.53*103 = 1.2*250*t2/(6*1.1)
Hence t = 42 .86 mm ≈ 50mm
Dimensions of base plate = 1150mm x 600mm x 50mm
Weld connecting base plate to column
Use a 6 mm fillet weld all round the column section to hold the base plate in place.
Total length available for weld = 2*[ 250+250 -9.4 +300 -10.6 ]
= 1560 mm
Hence effective length for welding = 1560 – 2 ( 4 +2 ) * 2 *6 = 1416 mm
Capacity of the weld : 0.7 * 6 * 189/1000 = 0.7398 KN/m
Hence, length of weld required: 590 / 0.7398 = 797 . 51 mm < 1416 mm
Hence a 6 mm weld is adequate.
Design of Foundation
Safe bearing capacity of soil : 190 KN/m2
Pmax at the base plate : 4.35 N/mm2
Load acting on foundation : 4.35 * 1250 * 500 = 2718.75 KN ~ 2750 kN
Now if foundation's dimension are : L * B
Stress on soil from foundation: 2750 / ( L * B)
Now , stress on soil from foundation should be < safe bearing capacity of soil
i. e 2750/ ( L * B) < 190
Hence, we get L = B =3.8 ~ 4 m
We provide grillage foundation, because of the large dimension of footing obtained, to make the foundation design economical.
In 1st tier, 7 ISMB400 beams at a c/c spacing of 200 mm are provided.
In 2nd tier, 4 ISMB400 beams of 10m length at a c/c spacing of 200 mm are provided per meter length.

Design of wind girds:
We are providing 4 wind girds each 5m long, spaced 2m apart on the overall 10m length of the column. Assuming uniform wind load on each of the gird, hence we design it as a laterally simply supported beam with a total load (uniformly distributed) = 65/5
=13 kN
i.e 13/4 = 3.25 KN/m load on the length of 5m
Maximum bending moment acting on a gird, M= wL2/8 = 10.156 kNm
M= (fy/1.1)*Ze
Ze (required) = 44.68 *103 mm3
Provide ISA 150*75*10 angle @165.8N/m , with Ze =51.6*103 mm3 > 44.68 *103
Check for local buckling: d/t=150/10= 15 < 15.7 ε ( OK )
Max. shear capacity needed = W*l/2 = 3.25 * 5/2 = 8.125 KN
Shear capacity= (Av*fy)/(1.73 *1.1)= 2156 * 250 / 1.1 = 282.9 kN > 8.125 kN (o.k)
Hence safe for Shear and flexure.


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