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Project 2

1) Design a Warehouse Building located in Chennai using STAAD Pro Connect Edition. The specification must be as follows: Width 30m Length 50m Eave Height 9m Bay spacing 6m Soil type Medium Safe Bearing Capacity 200 kN/m2 Roof slope 1 in 12 Assume suitable sections for structural elements. Follow IS800:2007, IS1893…

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VARSHA MOHAN WAGH
updated on 12 Oct 2022

1) Design a Warehouse Building located in Chennai using STAAD Pro Connect Edition. The specification must be as follows:

Width	30m
Length	50m
Eave Height	9m
Bay spacing	6m
Soil type	Medium
Safe Bearing Capacity	200 kN/m2
Roof slope	1 in 12

Assume suitable sections for structural elements. Follow IS800:2007, IS1893 and IS 875

1) Prepare a Design Basis Report for the project :-

PDF formate attach below

2) Create structural model using STAAD Pro Connect Edition :-

3) Prepare DL, LL, WL and EQL load calculations as per IS 875 standards :-

1) Design Foundations using STAAD Foundation :-

4) Design using MS – Excel

a) Steel column :-

Height of column h = 9m

Yeild strength fy = 250N/mm^2

Elastic modulus E = 200000N/mm^2

Material factor γm = 1.1

Step :- 1

Section classification

Properties of selected section

ISHB	400
A	10466mm^2
h	400mm
B	250mm
t	10.6mm
T	12.7mm
rx	166.1mm
ry	51.6mm
Zp	cm^3

Flange criterion b/T = 9.84

Web criterion d/t = 35.34

ε = 1

b/T < 10.5

d/t < 42

The section is classified as COMPACT

Step :- 2

Effective length

As both ends pin-jointed

KLx = 9m

KLy = 9m

KLz = 9m

Slenderness ratio

KLx/rx = 54.18

KLy/ry = 174.42

Non-dimensional effective slenderness ratio

λ = 1.964

Step :- 3

To calculate φ

h/bf = 1.60

> 1.2

T = 12.7mm

So, buckling class = b

α = 0.34

φ = 0.5(1+α(λ-0.2)+λ^2)

φ = 2.73

Step :- 4

Claculation of χ

χ = 0.216N/mm^2

Step :- 5

Calculation of fcd

fcd = χfy/γm

= 49.17

Step :- 6

Factored Axial Load

Pd = Afcd

= 514.614kN

b) Rafter :-

Span of I beam = 6m

Dead load = 18kN/m

Imposed load = 40kN/m

Length of bearing = 100mm

Yield strength = 250N/mm^2

Step :- 1

Design load calculation

Factored load = 87kN/m

Factored bending moment = 391.5kNm

Step :- 2

Section modulus required

Zreqd = 1722600mm^3

= 1722.60cm^3

Step :- 3

Section classification

Properties of selected section

ISMB	250
A	475.5mm^2
D	250mm
B	125mm
t	6.9mm
T	12.5mm
Izz	5131.6cm^4
Iyy	344.5cm^4
Zp	459.76cm^3

Flange criterion B/2T = 5

Web criterion (D-2T)t = 32.61

ε = 1

B/2T < 9.4

(D-2T)/t < 83.9

The section is classified as PLASTIC

Step :- 3

Moment of resistance of the cross section

Moment Md = (Zpχfy)/γm

Md = 104.491kNm

< 391.50

Hence section ISMB 250 is adequate for flexure

Step :- 4

Shear resistance of the cross section

Shear capacity Vc = 0.6fyAv/γm

Av = 1725mm^2

Vc = 235.23kN

V = 261kN

V/Vc = 1.11

.> 0.6

Step :- 5

Check for web buckling

Slenderness ratio of web = 2.5d/t

d = 225mm

= 70.33mm

Design compressive stress fc = 203N/mm^2

Pw = 315.16kN

> 261

Hence web is safe against shear buckling

Step :- 6

Check for web crippling at support

Root radius of ISMB250 R = 13mm

tf + R = 25.5mm

Dispersion length n^2 = 63.75mm

Pcrip = 245625N

= 245.63kN

< 261

Hence section adequate for web crippling

Step :- 7

Check for serviceability Deflection

Design load = 58kN/m

Deflection δ = 95.36mm

Allowable deflection = 30mm

Hence serviceability is satisfied

c) Base plate :-

Step :- 1

Strength of concrete fcu = 40 N/mm^2

Yield strength of steel fy = 250 N/mm^2

Material factor γm = .kN

Factored load = 1500 kN

Step :- 2

Steel column section

Thickness of flange T = 12.7 mm

Step :- 3

Area required

Bearing strength of concrete = 0.4fcu

= 16 N/mm^2

Area required = 93750 mm^2

Let size of plate

Bplate = 350 mm

Dplate = 350 mm

Area of plate = 122500 mm^2

Let projection on each side

a = b

= 25 mm

w = 12.24 N/mm^2

Thickness of slab base ts = 7.7 mm

< 12.70 mm

So use base plate of size 350mmx350mmx12mm.

d) Pedestal :-

Grade of concrete = 40N/mm^2

Load = 123kN

Moment = 116kNm

Horizontal shear = 13kN

Yield strength = 250N/mm^2

Length of base plate L = 450mm

Width of the base plate B = 350mm

C/C distance of (Ld) of bolt group in Z = 300mm

C/C distance of (Ld) of bolt group in X = 180mm

Bearing strength of concrete fc = 16N/mm^2

Depth of column D = 300mm

Width of column W = 250mm

Step :- 1

Anchor Bolt Details

Dia of anchor bolt = 24mm

No of anchor bolt on each side = 4

Total number of anchor bolt,n = 8

Gross area of the bolt 'Asb' = 452.16mm^2

Net area of the bolt 'Anb' = 352mm^2

Ultimate tensile strength of the bolt fub = 400N/mm^2

Fyb(Anchor bolt) = 240N/mm^2

Step :- 2

Base plate details

Ultimate tensile strength of a plate,fu = 490N/mm^2

Thickness of plate = 16mm

Yield stress of plate = 330N/mm^2

Step :- 3

Anchor bolt design

Area of the plate = 157500mm^2

Stress Max pressure σ1 = F/A+6*Mz/BL^2

= 10.60N/mm^2

< 16

Min pressure σ2 = F/A-6*Mz/BL^2

= -9.04N/mm^2

< 16

C = L*σ1/(σ1+σ2)

= 242.89mm

a = L/2-C/3

= 144.04mm

e = (L-Ld)/2
= 75mm

y = (L-C/3-e)

= 294.04mm

Tension in anchor bolts along the length of plate

FT = (Mz-Fxa)/y

= 364.38

Tension per bolt = 91.10kN

shear per bolt,V = 1.63kN

Step :- 4

Shear check

Factored shear force Vsb = 1.63kN

Vd,sd = 81290.9N

= 21.291kN

Factored Vd,sb = 65.03kN

Step :- 5

Tensile check

Factored tensile force in bolt Tb = 91.10kN

Tensile strength of bolt Td,b = Tnb/γmb

Tn,b = 0.9fubAn

= 126720N

= 126.72kN

Tn,b = fybAsb(γmb/γmo)

= 123316N

= 123.316kN

Lesser of two =123.32kN

So, Td,b = 98.65kN

Step :- 6

Combined unity check

(Vsb/Vdb)^2+(Tb/Tdb)^2

Vsb/Vdb = 0.025

Tb/Tdb = 0.92

Unity check = 0.85

< 1

Hence OK

Step :- 7

Anchor bolt length

Bond strength in tension bd = 1.4N/mm^2

Anchor length required = Tb/(3.14*d*τbd)

= 863.44mm

Let anchor bolt bolt length = 900mm

e) Purlin :-

Step :- 1

Span of the purlin = 7m

Spacing of the purlin = 1.5m

Number of sag rods = 1

Slope of the roof = 5deg

Step :- 2

Dead loads

Weight of sheeting = 5kg/m^2

Self weight of purlin = 4.22kg/m^2

Additional load = 10%

= 0.42kg/m^2

Total dead load = 0.096kN/m^2

Live loads

Live load on roof = 75kg/m^2

= 0.75kN/m^2

Wind loads

Basic wind spped = 50m/s

Terrain category = 2

Building class = B

k1 = 1

k2 = 1

k3 = 1

Design wind speed Vz = 57m/s

Design wind pressure Pz = 1949.40N/m^2

= 1.95kN/m^2

Length of the building I = 50m

Breath of the building w = 30m

Height of the building h = 10.5m

Height of the building at eaves = 9m

h/w = 0.35

I/w = 1.67

External pressure coefficient

Maximum downward Cpe = -0.4

Maximum upward Cpe = -0.7

Internal pressure coefficient

Maximum positive Cpi = 0.5

Maximum negative Cpi = -0.5

For maximum upward wind force

Maximum upward Cpe = -0.7

Cpi = -0.5

Cpe + Cpi = -1.2

Pz = 1.95

Wind pressure for purlin design = -2.339kN/m^2

For maximum upward wind force

Max upward Cpe = -0.4

Cpi = 0.5

Cpe + Cpi = 0.1

Pz = 1.95

Wind pressure for purlin design = 0.195kN/m^2

Step :- 3

Design load calculation

Spacing of purlin = 1.5m

Slope of roof = 5deg

Total dead load = 0.096kN/m^2

DL normal component = 0.144kN/m

DL tangential component = 0.013kN/m

Total live load = 0.75kN/m^2

LL normal component = 1.121kN/m

LL tangential component = 0.098kN/m

WL1 wind load = -2.339kN/m^2

WL normal component = -3.496kN/m

WL2 wind load = 0.195kN/m^2

WL normal component = 0.291kN/m

Summary of loads

	DL+LL	DL+WL1	DL+LL+WL2
Normal load	1.265	-3.351	1.556
Tangential load	0.111	0.013	0.111

	DL+LL	0.75(DL+WL1)	0.75(DL+LL+WL2)
Normal load	1.265	-2.514	1.167
Tangential load	0.111	0.009	0.083

Maximum normal component = 1.265kN/m

Step :- 4

Purlin section

Section selected = Zx200x6x2.3

Yield stress of material = 2400kg/cm^2

Flange width b = 60mm

Depth of section d = 200mm

Thickness t = 2.30mm

Length of Lip lip_I = 20mm

Internal bending radius = 3mm

Area = 8.07cm^2

Zxx = 47.72cm^3

Zyy = 10.22cm^3

Ixx = 477.18cm^4

Iyy = 61.34cm^4

Weight of purlin = 6.335kg/m

=4.223kg/m^2

Check for basic properties

t = 2.30mm

w = 49.40mm

Fy = 2400kg/cm^2

w/t = 21.48

< 60

Check for stress

Span major axis = 7m

Span minor axis = 2m

BM coefficient for Mxx = 10

BM coefficient for Myy = 10

	DL+LL	DL+WL1	DL+LL+WL2
Normal load	6.198	16.422	7.625
Tangential load	0.060	0.007	0.060
Mxx/Zxx	129.878	344.133	159.789
Myy/Zyy	5.895	0.672	5.895
(Mxx/Zxx)+(Myy/Zyy)	135.773	344.805	165.684

Maximum compressive stress = 344.80N/mm^2

1435/sq.root of f = 24.44

> 21.48

Hence section OK

Step :- 5

Combined Bending And Shear stresses in web

h/t = 84.96

	DL+LL	DL+WL1	DL+LL+WL2
Fbw	506.539	673.697	637.697
Fv	735.223	977.847	977.847

Actual stress

fbw	129.878	344.133	159.789
fv	5.895	0.672	5.895
fbw/Fbw	0.256	0.511	0.237
fv/Fv	0.008	0.001	0.006
Sqrt of sum of squares	0.257	0.511	0.237

Maximum combined stresses = 0.511

< 1

Step :- 6

Check for deflection

Number of spans = 3

Deflection = (3/384)(wl^4/EI)

w = 3.351kN/m

Span = 6m

E = 207400N/mm^2

Ixx = 477.18cm^4

δ = 34.29mm

Allowable deflection δ = Span/180

= 33.33mm

> 34.29

Not OK

2) Design a simply supported gantry girder to carry electric overhead travelling crane

Step :- 1

Span of gantry girder = 7 m

Span of crane girder = 9 m

Crane capacity = 250 kN

Self-weight of trolley, hook, electric motor etc. = 40 kN

Self-weight of crane girder excluding trolley = 100 kN

Minimum hook approach = 1.0 m

Distance between wheels = 3 m

Self-weight of rails = 0.2 kN/m

Step :- 2

Maximum moment due to vertical load

Weight of trolley + weight of load lifted = 290kN

Self weight of crane girder = 100kN

For maximum reaction on girder, the moving load should be as close to the gantry as possible

Reaction at A RA = 501kN

Reaction at B RB = 281.875kN

Load on gantry girder from each wheel = 251kN

Factored wheel load = 375.833kN

Maximum moment ME = 634.21kNm

Letr self weight of girder = 2kN/m

Dead load = 2.2kN/m

Factored DL = 3.3kN/m

M due to DL = 20.21kNm

M due to impact load = 158.55kNm

Factored moment due to all vertical loads

M = 812.99kNm

Step :- 3

Maximum moment due to lateral load

Horizontal force transferred to rails = 29kN

Horizontal force on each wheel = 7.25kN

Factored horizontal force = 10.875kN

Maximum moment = 18.35kNm

Vertical shear due to wheel loads = 563.75kN

Impact = 140.94kN

Self weight = 9.90kN

Total shear = 714.59kN

Lateral shear due to surge = 20.68kN

Step :- 4

Preliminary section determination

, ,, Minimum economic depth D = L/12

= 583.333mm

Width of compression flange b = L/40 to L/30

= 175 to 233

Required plastic modulus Zp = 1.4M/fy

= 4552721

= 4552.72x10^3

Let us try ISMB600 with ISMC400 on compression flange.

Step :- 5

Properties of selected section

ISMB	600@1.22kN/m
A	15600mm^2
h	600mm
b	210mm
tf	20.3mm
tw	12mm
Izz	91800x10^4mm^4
Iyy	2650x10^4mm^4
R	20mm

ISMC	400@0.49kN/m
A	6380mm^2
h	400mm
b	100mm
tf	15.3mm
tw	8.8mm
Izz	15200x10^4mm^4
Iyy	508x10^4mm^4
Cyy	2.42mm

Step :- 6

Let the distance of N>A> from, t,he tensio,n range by y'

y' = 388.93mm

Izz = 1348.13x10^6mm^4

Zez = 3466240mm^3

For compression flange about y-y axis

I = 16766.7x10^4mm^4

Zey = 838333mm^3

Total area of section = 21980mm^2

Let plastic N>A> be at a distance

Yp = 580.88mm

Zpz = 4410317mm^3

For top flange Zpy = 1078488mm^3

Step :- 7

Section classfication

For ISMB b/t = 4.90

< 9.4

d/t = 43.28

< 84

For ISMC b/t = 5.96

< 9.4

Hence section is plastic

Step :- 8

Check for local moment capacity

Local moment capacity for bending in vertical plane

Mdz = fyZp/1.1

= 1002.34kNm

Mdz = 1.2Zefy/1.1

= 945.34kNm

For top flange

Mdy = 245.11kNm

Mdy = 228.64kNm

In the above which ever the minimum value is taken

Mdz = 945.34kNm

Mdy = 228.64kNm

Step :- 9

Check for combined local capacity

(812.99/945.34)+(18.35/228.64)<1

0.947<1

Step :- 10

Check for buckling resistance

Md = βbZpfbd

βb = 1

Lt = 7000mm

E = 200000N/mm^2

hf = 597.5mm

Iy = 17850x10^4mm^4

A = 21980mm^2

ry = 90.12mm

fcr,b = 417.43N/mm^2

λ = 0.744

φ = 0.95

= 0.97N/mm^2

fbd = 166.591

Mdz = 734.72kNm

Mdy = 202.84kNm

Hence 1.20>1

Hence section is unsafe against torsional buckling

Step :- 11

Check for shear

Vz = 714.59kN

Shear capacity = Avfyw/(3^0.5x1.1)

= 944755N

= 944.755kN

> 714.59

Now = 566.853kN

Low shear

Step :- 12

Check for web buckling

b = 175mm

n1 = 232.6mm

d = 519.4mm

t = 12mm

= 104.746

Fcd = 128.6N/mm^2

Buckling resistance = 629008kN

> 714.59kN

Step :- 13

Check for deflection at working load

Vertical deflection

Serviceability vertical wheel load = 251kN

Deflection at mid span a = 2000mm

Izz = 1348.13x10^5mm^4

δ = 25.32mm

Horizontal deflection

I = 16766.65x10^4mm^4

δ = 4.88mm

L/750 = 9.33mm

Unsafe in vertical direction.

Safe in lateral direction.

Question

1. Design a Warehouse Building located in Chennai using STAAD Pro Connect Edition. The specification must be as follows:

Width	30m
Length	50m
Eave Height	9m
Bay spacing	6m
Soil type	Medium
Safe Bearing Capacity	200 kN/m2
Roof slope	1 in 12

Assume suitable sections for structural elements. Follow IS800:2007, IS1893 and IS 875

Prepare a Design Basis Report for the project
Create structural model using STAAD Pro Connect Edition
Prepare DL, LL, WL and EQL load calculations as per IS 875 standards
Design using MS – Excel
1. Steel column
2. Rafter
3. Base plate
4. Pedestal
5. Z- purlin

Design Foundations using STAAD Foundation.
Attach necessary sketches and drawings wherever required.
All stipulations, assumptions and design parameter must adhere to Indian Standards.

2. Design a simply supported gantry girder to carry electric overhead travelling crane

Given:

Span of gantry girder = 7 m

Span of crane girder = 9 m

Crane capacity = 250 kN

Self-weight of trolley, hook, electric motor etc. = 40 kN

Self-weight of crane girder excluding trolley = 100 kN

Minimum hook approach = 1.0 m

Distance between wheels = 3 m

Self-weight of rails = 0.2 kN/m

Your Answers

1. Design a Warehouse Building located in Chennai using STAAD Pro Connect Edition. The specification must be as follows:

Width	30m
Length	50m
Eave Height	9m
Bay spacing	6m
Soil type	Medium
Safe Bearing Capacity	200 kN/m2
Roof slope	1 in 12

Assume suitable sections for structural elements. Follow IS800:2007, IS1893 and IS 875

Prepare a Design Basis Report for the project
Create structural model using STAAD Pro Connect Edition
Prepare DL, LL, WL and EQL load calculations as per IS 875 standards
Design using MS – Excel
1. Steel column
2. Rafter
3. Base plate
4. Pedestal
5. Z- purlin

Design Foundations using STAAD Foundation.
Attach necessary sketches and drawings wherever required.
All stipulations, assumptions and design parameter must adhere to Indian Standards.

Step 1: Model is prepared in STAAD Pro.

Step 2: Section properties and Materials are applied to all the members.

Step 3: Releases and specifications are applied.

Step 4: Support conditions are applied.

3D Rendering is as follows

Step 5: Loading are calculated and Design Parameters are applied.

Calculation of Design Loads on PEB Structure – Project 2

Dead Load

Dead Load on Rafter

Weight of Roof Sheet = 5 Kg/m2
Weight of Purlin and sag rods = 5 Kg/m2
Additional weight = 5 Kg/m2

Total weight of rafter = 15 Kg/m2

Dead load on rafter = 15 x 10 x 8 = 1200 N/m = 1.2 kN/m

Dead Load on Purlin = 10 x 10 x 8 = 800 N/m = 0.8 kN/m
Mezz Slab self-weight (150 mm thk slab) = 0.15 x 25 kN/m3 = 3.75 kN/m2
Floor Finishes and additional load on Mezz Floor = (150 + 500) kg/m2 = 0.65 kN/m2
Weight of 230mm thick brick wall (4-meter height) on Mezz floor = 20 kN/m3 x 0.23 x 4 = 18.4 kN/m

Live Load

Live Load on Roof and canopy = 57 kg/m2 x 10 x 8 = 4.56 kN/m
Live Load on Mezz Floor = 400 kg/m2 = 4 kN/m2

Collateral Load

Collateral Load on Roof

Load of Electrical fittings = 0.1 kN/m2
Load of Sprinklers = 0.03 kN/m2
Load of rockwool Insulation and bubble wrap insulation = 0.15 kN/m2
Load of Solar Panels = 0.15 kN/m2
Load of Ceiling = 0.1 kN/m2

Total Collateral load on Roof = 0.53 x 8 = 4.24 kN/m

Collateral Load on Mezz Floor (due to false ceiling & electrical fittings)

= 0.2 x 8 = 1.6 kN/m

Collateral Load on column due to Bracket Beam

Load due to service line supported on the beam spanning 8 m width= 150 kg/m x 8 x 10 = 12 kN

Moment acting on column due to service line = 12 x 0.75 = 9 kNm

Collateral Load on Jack Beam due to service line= 2.5 kN/m
Collateral Load on Mezz False Ceiling = 3.2 kN/m

Wind Load

Basic wind speed = 50 m/s for chennai

K1 - 1.00

K2 - 1.05 (Terrain Catogory-1. Building height at eaves = 10.50 m)

K3 - 1.00

K4 - 1.00 (For other structures)

Coefficient of cyclonic wind may be taken as 1,

Kd -0.9

Ka (Rafter) - 0.8 or as per tributary area.

Ka (Column) - 0.8245 or as per tributary area.

Ka (Purlin) - 0.9721 or as per tributary area.

Ka (Side Runner) - 0.978 or as per tributary area

Kc -0.9

Design wind pressure = pd = kd x ka x kc x 0.6 x (Vb x k1 x k2 x k3 x k4)²= 0.652 kN/m2

But should not be less than 0.7 Pz = 0.7 x 1.006 = 0.704 kN/m2

Cpe - As per table 5 & 6 of IS: 875 (Part 3) – 2015

h/w = 10/60 = 0.6

L/w = 170/60 = 2.83

Cpe On wall:

Direction	A	B	C	D
0 deg.	0.7	-0.25	-0.6	-0.6
90 deg.	-0.5	-0.5	0.7	-0.1

Cpe On Roof:

Roof angle = 2.86

Direction	EF	GH	EG	FH
0 deg.	-0.86	-0.4
90 deg.			-0.8	-0.4

Cpi = 0.2

Wind Load on Columns = (Cpe – Cpi) x A x Pd

Wind about X Direction + Cpi = (0.7 – 0.2) x 8 x 0.704 = 2.816 kN/m on Wall A

Wind about X Direction + Cpi = (-0.25 – 0.2) x 8 x 0.704 = -2.534 kN/m on Wall B

Wind about X Direction + Cpi = (-0.6 – 0.2) x 6.75 x 0.704 = -3.802 kN/m on Wall C & D

Wind about X Direction - Cpi = (0.7 + 0.2) x 8 x 0.704 = 5.069 kN/m on Wall A

Wind about X Direction - Cpi = (-0.25 + 0.2) x 8 x 0.704 = -0.282 kN/m on Wall B

Wind about X Direction - Cpi = (-0.6 + 0.2) x 6.75 x 0.704 = -1.9 kN/m on Wall C & D

Wind about Z Direction + Cpi = (-0.5 – 0.2) x 8 x 0.704 = -3.942 kN/m on Wall A

Wind about Z Direction + Cpi = (-0.5 – 0.2) x 8 x 0.704 = -3.942 kN/m on Wall B

Wind about Z Direction + Cpi = (0.7 – 0.2) x 6.75 x 0.704 = 2.376 kN/m on Wall C

Wind about Z Direction + Cpi = (-0.1 – 0.2) x 6.75 x 0.704 = -1.425 kN/m on Wall D

Wind about Z Direction - Cpi = (-0.5 + 0.2) x 8 x 0.704 = -1.689 kN/m on Wall A

Wind about Z Direction - Cpi = (-0.5 + 0.2) x 8 x 0.704 = -1.689 kN/m on Wall B

Wind about Z Direction - Cpi = (0.7 + 0.2) x 6.75 x 0.704 = 4.276 kN/m on Wall C

Wind about Z Direction - Cpi = (-0.1 + 0.2) x 6.75 x 0.704 = 0.475 kN/m on Wall D

Wind Load on Rafters = (Cpe – Cpi) x A x Pd

Wind about X Direction + Cpi = (-0.86 – 0.2) x 8 x 0.704 = -5.970 kN/m on Side A Rafter

Wind about X Direction + Cpi = (-0.4 – 0.2) x 8 x 0.704 = -3.378 kN/m on Side B Rafter

Wind about X Direction - Cpi = (-0.86 + 0.2) x 8 x 0.704 = -3.716 kN/m on Side A Rafter

Wind about X Direction - Cpi = (-0.4 + 0.2) x 8 x 0.704 = -1.125 kN/m on Side B Rafter

Wind about Z Direction + Cpi = (-0.8 - 0.2) x 6.75 x 0.704 = -4.752 kN/m Side C Rafter

Wind about Z Direction + Cpi = (-0.4 - 0.2) x 6.75 x 0.704 = -2.851 kN/m Side D Rafter

Wind about Z Direction - Cpi = (-0.8 + 0.2) x 6.75 x 0.704 = - 2.851kN/m Side C Rafter

Wind about Z Direction - Cpi = (-0.4 + 0.2) x 6.75 x 0.704 = -0.950 kN/m Side D Rafter

Wind Load on Canopy

Cp - As per table 8 of IS: 875 (Part 3) – 2015

Roof Angle 2.86 deg.

Cp -ve = 1.08

Cp +ve = 0.35

Wind Load on Canopy (Upward) = 1.08 x 8 x 0.704 = 6.082 kN/m

Wind Load on Canopy (Downward) = 0.35 x 8 x 0.704 = 1.971 kN/m

Seismic Load

As per IS: 1893 - 2016

Seismic zone - III - Z = 0.16

Rf - 4.0

I - 1.0

SS - 3.0

25% of Live Load on roof considered for calculation of seismic forces.

100% of collateral load on roof considered for calculation of seismic forces.

50% of (Live load + Collateral load) on mezzanine floor considered for calculation of seismic forces.

100% of Dead load considered for calculation of seismic forces.

Design Parameters:

IS 800 LSD
FYLD = 345000 for Portal frame and Mezz Beams
FYLD = 250000 for Bracings
BEAM = -1
STP = 2 for welded sections
MAIN = 250 for tie beam
Ly = Lx = 1.5m for Rafter
Ly = Lx = 3.25m for Columns
Ly = Lx = 4.4m for columns supporting canopy
Lz = 11m for Internal columns
Ly = Lx = 0.5m for Mezz floor beams
Lz = 6.75m for Mezz Floor beams
Ly = 1.7m for Canopy Beam
RATIO = 1
STEEL TAKE OFF

Results of the analysis is as follows which shows are members are passing and no member property has failed.

Foundation Design:

Check for Max/min Bearing Pressure

a) wt of soil above the footing

= (area of footing-Area of pedestel)xdepth of soil x unit weight of soil

= ((13.35x3.35) - (0·3×0.45)) x 1.2 × 18 =239.45 KN

b) self wt of footing.

= (vol of footing + vol of pedestall)x unit wt of concrete. =

((3.35×3·35x0·7)+(0.3×0.4x1))× 25

199.77 KN

Total load on footing

= a+b+ Imposed load

=239.45+ 199.47 + 1500

1939.26 kN

Section modulus of footing

Z=bd^2/6

3.35×(3.35)^2/6

Maximam bearing pressure

P/A+M/Z

1939.26/(3.35+3.35)+150/6.26

= 196.7 kN/m^2

Min bearing Pressure

P/A+M/Z

M=148.8 kN/m^2

which is less than safe bearing capacity ie 200kN/m^2

As per Staad pro calculation.

Check for ultimate Bearing pressure

Max ultimate bearing pressure

P/A+M/Z

= (239.45 +199.77 +1.5x1500)/ 3.35×.35 +(1.5×50/6.26)

=275.5 kN/m2

Min ultimate Bearing Pressure

P/A+M/Z

=239.6-35.9=203.7kN/m^2

which is less than safe bearing capacity for ultimate Loads - 400 KN/m2

As per Staad pro calculations

2. Design a simply supported gantry girder to carry electric overhead travelling crane

Given:

Span of gantry girder = 7 m

Span of crane girder = 9 m

Crane capacity = 250 kN

Self-weight of trolley, hook, electric motor etc. = 40 kN

Self-weight of crane girder excluding trolley = 100 kN

Minimum hook approach = 1.0 m

Distance between wheels = 3 m

Self-weight of rails = 0.2 kN/m

W= 250 +40= 290 KN

Self-Weight=100 KN

w= 100/9=11.11 kN/m

(

\sum "> \sum

$F y "> F_{y}$

=0). RA+ RB = 3900 KN

(

\sum M a = 0 "> \sum M_{a} = 0

) 290x1 + 100x 9/2 = $R b * 9 "> R_{b} * 9$

Rb =(290+450)/9=82.2kN

Ra: 390-82.2=308kN

Load on gantry girder from each wheel = 308/2=154kN

(Σ \cdot F y = 0) R c + R d "> (Σ \cdot F_{y} = 0) R_{c} + R_{d}

= 293.5 + 293.5

= 587 kN

\sum M c = 0 "> \sum M_{c} = 0

293.5(3.5 - 1.75 - 0.875) + 293.5(3.5 + 0.875) =7 $R d "> R_{d}$

R d "> R_{d}

=220.12 kN

R c "> R_{c}

= 587 - 220.12= 366.8kN

M = 366.8x (0<=x<=0.875)

M= 366.8x - 293.5(x - 0.875)

73.3x-256.8 (0.875 <= x <= 4.375)

M = 366.8x - 293.5(x - 0.875)- 293.5(x - 4.375)

= 220.2x+1540.86 (4.375 <= x <= 7)

M(at x=0)=366.8x0=0

M(at x = 0.875) = 366.8x 0.875 = 320.425KNm

M (at x = 4.375)=-220.12x4.375 + 1540.86=577.8 KN

M (at x = 7) = -220.12X7 +1540.86 = 0

Mmax = 577.8kNm

appx equal to 578kNm

Moment due to additional impact load. of electric overhead travelling.

= 25% in case of electric overhead travelling

= 25% of 578 = 144.5kNm

Assume self weight of girder=2kN/m

Total weight of girder + rails= (2+0.2) KN/m = 2.2 KN/m.

Factored dead load =2.2x 1.5=3.3 KN/m

Maximum moment = WL^2/8

=3.3x7^2/8

=20.21KN

M = (wLx)/2 - (wx^ 2)/2

For Mmax (dM)/(dx) = 0

Wl/2 - wx = 0

x = L/2

Mmax (at x=L/2)= wL/2 * L/2 - w/2 ( L/ 2 )^2

= (wL^ 2)/8

Total moment = 578+144.5+20.21

=742.71kNm-Mz

Lateral forces: [Horizontal forces transverse to rails ]

10% of weight of (crab+ lifted weight) for EOT Cranes = 10% of (40+250)

=29kN

This is transferred to all 4 wheels:

Load on each wheel =29/4 = 7.25kN

Factored 1oad = 7.25 * 1.5 = 10.875kN

My =(10.875/293.5)X578=21.4kNm

R c + R d "> R_{c} + R_{d}

= 21.75

R d "> R_{d}

=(10.875 x 0.875)+(10.875x4.375)/7

8.16KN

R d "> R_{d}

=21.4-8.15

=13.59kN

M e = R c "> M_{e} = R_{c}

x4.375-10.875x3.5

=21.39kNm

Shear forces: -

\sum F y "> \sum F_{y}

R c + R d "> R_{c} + R_{d}

=587 kN

\sum M c "> \sum M_{c}

293.5x3.5=7

R d "> R_{d}

R d "> R_{d}

=146.75kN

R c "> R_{c}

=440.25kN

V c = R c "> V_{c} = R_{c}

=440.25kN

Impact load 25 % extra

25% of 440.25=110kN

Shear force due to self weight = WL/2

= 4.375X7/ 2 = 15.31 KN

:: Total shear force = 440.25+110.06+15.31= 565.62kN

Lateral forces:

Lateral shear = (10.875/293.5)x440.25

=16.31kN

Longitudinal force :

5% of Static wheel load

5% of 293.5

14.67 kN

Preliminary section

D=L/12=7000/12=583.33

b=L/40 - L/30

=7000/40 - 7000/30

=175 to 233.33mm

Z p r e q d = M f y "> Z_{p} r e q d = \frac{M}{f_{y}}

x1.4 (k=1.3-1.6)

(740.21 x 10^6/250) x1.4

=4145.176x10^3mm^3

We will try ISMB 550 with ISMC 250

b t f "> \frac{b}{t_{f}}

= (250/2 - 14.1)/(7.1 + 19.3) = 4.2 [< 9.4 $ε "> ε$

= 9.4 ]

d t w "> \frac{d}{t_{w}}

= (550 - 2 x19.3)/(11.2) =45.7[<84 $ε "> ε$

=84]

therefore, Plastic section [

β b "> β_{b}

= 1]

yy - axis remains the same

zz-axis (NA):

Z = Σ (A z) Z A "> Z = \frac{Σ (A z)}{Z A}

(190 * 19.3 * 193/2) + [(550 - 2 * 19.3) * 19.2(193 + (59.2 * 93)/2)]+[190 * 193(55 - 193 + 193/2)]+[250*7*1*(550 + 7.4/2 )]+2x[(80-7.1)* 14.1x(550+7.1-(80-7.1)/2)/

2(190 * 19.3) + [555 - 2 * 19.3)x11.2]+(250*7.1)+ 2[14.1(80 - 7.1)]

= 333.33

I y y "> I_{y y}

total=1833.8x10^ 4 +219.1x10^ 4

= 2052.9 x10 ^ 4 mm^4

I z z "> I_{z z}

total= $\sum [A \cdot (d) 2] "> \sum [A \cdot {(d)}^{2}]$

= [190x19.3x(333.33 - (19.3)/2) ^ 2]+[ 112(333.33 - 19.1(333.3 - 19.3) ^ 2)/2]+[11.2 x 197.37 x(193.34/2) ^ 2 ]+[190 x19.3 x(197.37 + 19.3/2) ^ 2+[250x 7.1 x (197.84 + 19.3 + 7.1/2) ^ 2]+2x[14.1 x(80 - 7.1) x(143.74 + (80 - 7.1)/2) ^ 2]

= 80243.5 x10 ^ 4mm^4

therefore,

Z p z "> Z_{p z}

total=Z(A.D)=[190x19.3x(333.3*3-19.3/2 )]+[11.2x(333.3 -19.3)x(333 .33- 19.3)/2)] + [ll.2 x193.37 x 193.37/2] + [190x19.3x(197.37x193/2)] +[250 x7.1(197.37 +19.3+ 7.1/ 2 )]+[2x14.1(89 - 7.5)x(143.73+( 80-7.1/ 2)

=3477.55 x10^ 3 mm^3

Z p y "> Z p y

total= $Σ (A y ¯) "> Σ (A ¯ y)$

= [550 - 2x19.3) * 11.2/2 * (11.2)/4 ]*2+[190/2 x19.3x 190/4 ]*4+[ (250/2 - 14.1) * 7.1*( 250/2 -14.1)/ 2+[14.1 x 80(125 - (14 .1)/2)] * 2

=717.819x 10^3 mm^3

M d z "> M_{d z}

- eq given in annexure E of IS 800 2007

M d y = β b \cdot Z p y \cdot f y Y m o "> M_{d y} = β b \cdot Z_{p y} \cdot \frac{f_{y}}{Y_{m o}}

= 1(717.819 * 10 ^ 3) * 250)/(1.11)

= 163.14KNm [>

M y "> M_{y}

= 15.18KNm]

M y M d y + M z M d z \leq 1 "> \frac{M_{y}}{M_{d y}} + \frac{M_{z}}{M_{d z}} \leq 1

M y "> M_{y}

= 15.18

M d y "> M_{d y}

=163.14

M d z "> M_{d z}

= 527.35

V d = A v f y w 3 \cdot Y m 0 "> V_{d} = A_{v} \frac{f_{y w}}{\sqrt{3 \cdot Y_{m 0}}}

= 818.7KN

Web-buckling and web crippling at support and at wheel load

Assume b = 150mm

n 1 "> n_{1}

= D/2 = (550 + 7.1)/2

278.55mm

n 2 = 2.5 (t f + R 1) "> n_{2} = 2.5 (t_{f} + R_{1})

=2.5x (19.3+R_(1)=93.25

λ = 2.5 d t w "> λ = \frac{2.5 d}{t_{w}}

(2.5 * 475.4)/11.2

Table 9(C)

[fy =250 MPa]

f c d "> f_{c d}

=107- (107-94.6)/10 x6.11 =99.4 MPa

f w b "> f_{w b}

(at support)= $b + n 1 t w \cdot f c d "> b + n_{1} t_{w} \cdot f_{c d}$

=(150 + 278.55)× 11.2 × 99.4=477 kN

f w b "> f_{w b}

(at load)= $b + n 1 t w \cdot f c d "> b + n_{1} t_{w} \cdot f_{c d}$

=787.2kN

f w c "> f_{w c}

(at supports)= $b + n 2 t w \cdot f y w Y m o "> b + n_{2} t_{w} \cdot \frac{f_{y w}}{Y_{m o}}$

=619.1kN

$f w c "> f_{w c}$ (at load)= $b + n 2 t w \cdot f y w Y m o "> b + n_{2} t_{w} \cdot \frac{f_{y w}}{Y_{m o}}$

=856 kN

Deflection check

δ c = W L 3 6 E I \cdot [3 a 4 L - a 3 L 3] "> δ_{c} = \frac{W L^{3}}{6 E I} \cdot [\frac{3 a}{4 L} - \frac{a^{3}}{L^{3}}]

Fatigue check