Challenge

Project 2

SSouvik Kundu25 Feb 2023

PROJECT-2

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

Steel column

Height of column

Yield strength

250

N/mm²

Elastic modulus

200000

N/mm²

Material factor

γm

1.1

Step 1

Section classification

Properties of selected section

ISHB

400

10466

mm²

400

374.6

250

10.6

12.7

166.1

51.6

cm³

Flange criterion

b/T

9.84

Web criterion

d/t

35.34

1.00

b/T

10.5

d/t

The section is classified as COMPACT

Step 2

Effective Length

T7.5

As both ends pin-jointed

KLx

9.00

KLy

9.00

KLz

9.00

Slenderness ratio

KLx/rx

54.18

KLy/ry

174.42

Non-dimensional effective slenderness ratio

Cl 7.1.2.1

1.964

Step 3

To calculate φ

φ = 0.5[1+α(λ-0.2)+λ²]

h/bf

1.60

1.2

12.7

T 7.1

so, buckling class

0.34

2.73

Step 4

Calculation of χ

χ =

0.216

N/mm²

Step 5

Calculation of fcd

fcd

χfy/γm

49.17

Step 6

Factored Axial Load

Afcd

514.614

Rafter

Span of I beam

Dead Load

kN/m

Imposed Load

kN/m

Length of bearing

100

Yield strength

250

N/mm²

Step 1

Design load calculation

Factored Load

87.00

kN/m

Factored bending moment

97.88

kNm

Step 2

Section modulus - Required

Z reqd

430650.0

mm³

430.65

cm³

Step 3

Section classification

Properties of selected section

ISMB

250

475.5

mm²

250

125

6.9

12.5

Izz

5131.6

cm⁴

Iyy

344.5

cm⁴

459.76

cm³

Flange criterion

B/2T

Web criterion

(D-2T)/t

32.61

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

(Zpxfy)/γm

104.491

kNm

97.88

Hence section ISMB 250 is adequate for flexure

Step 4

Shear resistance of the cross section

Shear capacity

0.6fyAv/γm

1725

mm²

235.23

Factored shear

130.50

V/Vc

0.55

0.6

Step 5

Check for Web Buckling

Slenderness ratio of web

2.5d/t

225

70.33

Design compressive stress

203

N/mm²

315158

315.16

130.50

Hence web is safe against shear buckling

Step 6

Check for web crippling at support

Root radius of ISMB 250

tf + R

25.5

Dispersion length

63.75

Pcrip

245625

245.63

130.50

Hence section adequate for web crippling

Step 7

Check for serviceability - Deflection

Design load

kN/m

Deflection

5.96

Allowable deflection

Hence serviceability is satisfied

Base plate Design


Step 1		Strength of concrete	fcu	=	40	N/mm²
		Yield strength of steel	fy	=	250	N/mm²
		Material factor	γm	=	1.1	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		=	0.4 fcu
				=	16	N/mm²
		Area required		=	93750	mm²

		Let size of plate	Bplate	=	350	mm
			Dplate	=	350	mm
		Area of plate		=	122500	mm²
		Let projection on each side	a = b	=	25	mm
			w	=	12.24	N/mm²

	Cl 7.4.3.1	Thickness of slab base	ts	=	7.7	mm
				<	12.70	mm

	So use base plate of size 350mm x 350mm x 12mm

Pedestal

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

14. 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

23. 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

30. Step :- 3

Anchor bolt design
Area of the plate = 157500mm^2
Stress Max pressure ζ1 = F/A+6*Mz/BL^2
= 10.60N/mm^2
< 16

OK
Min pressure ζ2 = F/A-6*Mz/BL^2
= -9.04N/mm^2
< 16
C = L*ζ1/(ζ1+ζ2)
= 242.89mm
a = L/2-C/3
= 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

65. Step :- 4

Shear check
Factored shear force Vsb = 1.63kN
Vd,sd = 81290.9N
= 21.291kN
Factored Vd,sb = 65.03kN

77. 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
= 72kN

Tn,b = fybAsb(γmb/γmo)
= 123316N
= 123.316kN
Lesser of two =123.32kN So, Td,b = 65kN

94. 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

107. Step :- 7

Anchor bolt length
Bond strength in tension bd = 1.4N/mm^2 Anchor length required = Tb/(3.14*d*ηbd)
= 863.44mm
Let anchor bolt bolt length = 900mm

Z- purlin

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

WL2

l load

WL1)	LL+WL2)
ad
l load

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

WL2

l load

)+(Myy/Zyy)

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

WL2

Actual stress

m of squares

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

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

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


Step 1		Span of gantry girder				=	7	m
		Span of crane girder				=	9	m
		Crane capacity				=	250	kN
		Self weight (trolley,hook,motor)				=	40	kN
		Self weight of crane girder				=	100	kN
		Minimum hook approach				=	1	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				=	290	kN
		Self weight of crane girder				=	100	kN

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

			200 kN
	A	1	14			B

		15 m
		Reaction at A			R_A	=	501	kN
		Load on gantry girder from each wheel				=	251	kN
		Factored wheel load				=	375.833	kN

					R_D	=	281.875	kN
		Maximum moment			ME	=	634.22	kNm
		let self weight of girder				=	2	kN/m
		Dead load (self weight + rails)				=	2.2	kN/m
		Factored DL				=	3.30	kN/m
		M due to DL				=	20.21	kNm
		M due to impact load				=	158.55	kNm
		Factored moment due to all vertical loads
					M	=	812.99	kNm

Step 3	Maximum moment due to lateral load
		Hzl. Force transferred to rails				=	29	kN
		Horizontal force on each wheel				=	7.25	kN
		Factored horizontal force				=	10.875	kN



		Maximum moment				=	18.35	kNm
		Vertical shear due to
		wheel loads				=	563.75	kN
		impact				=	140.94	kN
		self weight				=	9.90	kN
		Total shear				=	714.59	kN
		Lateral shear due to surge				=	20.68	kN


Step 4	Preliminary section determination
		Minimum economic depth			D	=	L/12
						=	583.333	mm
		Width of compression flange			b	=	L/40 to L/30
						=	175	233
		Required plastic modulus			Zp	=	1.4 x M / f_y
						=	4552721
						=	4552.72	x10³ mm³

	Let us try	ISMB 600 with ISMC 400 on compression flange

Step 5	Properties of selected section

	ISHB	450 @ 0.92 kN/m			ISMC	400 @ 0.49 kN/m
	A	11789	mm²		A	6380	mm²
	h	600	mm		h	400	mm
	b	250	mm		b	100	mm
	tf	20.3	mm		tf	15.3	mm
	tw	11.3	mm		tw	8.8	mm
	Izz	40350	x10⁴mm⁴		Izz	15200	x10⁴mm⁴
	Iyy	3045	x10⁴mm⁴		Iyy	508	x10⁴mm⁴
	R	18.5	mm		Cyy	2.42	mm

Step 6	Determination of section modulus
	Let the distance of N.A from the tension range by y'

					y'	=
					y'	=	407.58	mm
					Izz	=	797.166	x10⁶mm⁴
					Zez	=	1955830	mm³

	For compression flange about y-y axis
					I	=	17843.2	x10⁴mm⁴
					Zey	=	892161	mm³
		Total area of section				=	18169	mm²

	Let plastic NA be at a distance Yp from tension flange.
					Yp	=	375.12	mm
					Zpz	=	5261194	mm³
		For top flange			Zpy	=	1171868	mm³


Step 7	Section classification
		For ISMB			b/t	=	5.88
							<	9.4
					d/t	=	46.23
							<	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
						=	1195.73	kNm
						=	1.2 Zefy / 1.1
						=	533.41	kNm
					Mdz	=	533.41	kNm
		For top flange
					Mdy	=	266.33	kNm
						=	243.317	kNm
					Mdy	=	243.32	kNm

Step 9	Check for combined local capacity
					1.60	>	1


Step 10	Check for buckling resistance
					Md	=	βb Zp fbd
		For plastic section			βb	=	1

					L_LT	=	7000	mm
					E	=	200000	N/mm²
					hf	=	597.5	mm
					Iy	=	18245	x10⁴mm⁴
					A	=	18169	mm²
					ry	=	100.21	mm
					fcr,b	=	503.24	N/mm²

					λ	=	0.705
					ϕ	=	0.91
					X	=	0.99	N/mm²
					fbd	=	166.591

					Mdz	=	876.47	kNm
					Mdy	=	207.33	kNm

				Hence	1.02	>	1
		Hence section is unsafe against torsional buckling

Step 11	Check for shear
					Vz	=	714.59	kN
					Shear capacity	=	Avfyw / √3 1.1
						=	889644	N
						=	889.644	kN
						>	714.59
					Now	=	533.787	kN
							low shear

Step 12	Check for web buckling
					b	=	175	mm
					n1	=	232.6	mm
					d	=	522.4	mm
					t	=	11.3	mm
					λ	=	111.877
	For buckling Class a			Table 9a
					Fcd	=	128.6	N/mm²
					Buckling resistance	=	592316	kN
						>	714.59	kN

Step 13	Check for deflection at working load
	Vertical deflection
		Serviceability vertical wheel load				=	251	kN

		Deflection at mid span
					a	=	2000	mm
					Izz	=	797.17	x10⁶mm⁴
					δ	=	42.82	mm

	Horizontal deflection
					I	=	17843.23	x10⁴mm⁴
					δ	=	4.58	mm
					L/750	=	9.33	mm
	Unsafe in vertical direction
	Safe in lateral direction

Project 2

14. Step :- 1

23. Step :- 2

30. Step :- 3

65. Step :- 4

77. Step :- 5

94. Step :- 6

107. Step :- 7

e) Purlin :- Step :- 1

Step :- 2

Live loads

Wind loads

For maximum upward wind force

For maximum upward wind force

Step :- 3

Summary of loads

Step :- 4

Check for basic properties

Check for stress

Step :- 5

WL2

m of squares

Step :- 6

Ship work like this
in your own course.

Talk to a career counsellor

14. Step :- 1

23. Step :- 2

30. Step :- 3

65. Step :- 4

77. Step :- 5

94. Step :- 6

107. Step :- 7

e) Purlin :- Step :- 1

Step :- 2

Live loads

Wind loads

For maximum upward wind force

For maximum upward wind force

Step :- 3

Summary of loads

Step :- 4

Check for basic properties

Check for stress

Step :- 5

WL2

m of squares

Step :- 6

Ship work like thisin your own course.

Talk to a career counsellor

Ship work like this
in your own course.