Project 2

1] Roof Slope Calculation:

Roof Slope = 1 in 12

Width, b = 30 m

X / (b/2) = 1/12

X = (30/2) / 12 = 15/12

X = 1.25 m

Total height =9 +1.25 =10.25

2] Modeling in staad pro:

Create structural model using STAAD Pro Connect Edition:

Step 1: Model Creation:

Step 2: Assign Properties:

3D View:

Material : Steel

BETA Angle 90 degree for wind / gable columns

Step 3: Specification:

• Member Tension
• End MY MZ
• Start MY MZ

Step 4: Support:

• Pinned

Step 5: Load Case Definition:

Seismic Definition:

Wind Load Definition:

Step 8:Define parameters:

Step 9:Define commands:

Step 10:Run analysis:

3] Prepare DL, LL, WL and EQL load calculations as per IS 875 standards

Calculation of Design Loads

1)Calculation of Dead load:

UDL on Rafter (Main frame):

Weight of roof sheeting = 5 Kg/m2

Weight of sag rods and braces = 5 Kg/m2

Collateral load = 10 Kg/m2

Total load = 20 Kg/m2 = 0.2 KN/m2

UDL on Rafters:

Bay spacing = 6m

UDL = 0.2 X 6 = 1.2 KN/m

UDL on Rafters from purlin:

Choose suitable purlin section for weight of purlin (IS-811)

Say Z = 270 X 75 X 20 X 2.55 = 8.77 Kg/m

Weight on rafters from purlin = (Number of purlin X Length of purlin)/Length of rafter

= ( 5 X 6 X 8.77)/15

= 17.54 Kg/m = 0.175 KN/m

UDL on main frame = 1.2 + 0.175 = 1.375 KN/m

UDL on gable frame = 1.375/2 = 0.687 KN/m

UDL on Column (Main frame):

Weight of side wall sheeting = 5 Kg/m2

Weight of sag rod and braces = 5 Kg/m2

Load from sag rods & braces = 10 Kg/m2 = 0.1 KN/m2 = 6 X 0.1 = 0.6 KN/m

Weight of grit (270 X 75 X 20 X 2.55) = 8.77 Kg/m

Weight on column = (5 X 6 X 8.77)/6 = 43.85 Kg/m = 0.438 KN/m

Total Load = 0.6 + 0.438 = 1.038 KN/m

UDL on Column (Gable frame):

Total load = 1.038/2 = 0.565 KN/m

Load on wind column:

Load from sheet and sag rods = 0.1 X 6 = 0.6 kN/m

Weight of grits (230 X 75 X 20 X 2) = 6.32 KG/m

Load on wind column = (6X 6.32 X 6)/6 = 37.92 Kg/m = 0.38 KN/m

Total load = 0.6 + 0.38 = 0.98 KN/m

Dead Load on eave strut & Gutter:

Point load:-

Weight of Eave strut = CS 270 X 75 X 20 X 3.15 = 10.7 Kg/m

Weight of eave gutter = 0.25 X 0.25 X 0.001 = [0.25 + 0.25 + 0.25] X 0.001 X 7850

=5.89 Kg/m

Load due to eave strut & gutter = 10.7 + 5.89 = 16.59 Kg/m

Load due to eave strut & gutter (Main frame) = 16.59 X 6 = 99.54 Kg

= 0.995 KN

Load due to eave strut & gutter (Gable frame) =0.497 KN

2) Calculation of Live load:

Live load on roof = 0.75 KN/m2 (IS 875- II)

UDL on Rafter:

Main frame rafter = 0.75 X 6 = 4.5 KN/m

Gable frame rafter = 0.5 X 4.5 = 2.25 KN/m

Eave gutter (250 X 250 X 1mm)

Load due to weight of water in gutter:

Main frame = [0.25 X 0.25 X 9 X 6] = 3.375 KN

Gable frame = [0.5 X 3.375] = 1.687 KN

3) Calculation of wind load:

Basic wind speed = 50 m/s for Chennai.

K1 - 1.00

K2 - 1.0

K3 - 1.00

K4 - 1.15

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

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 Column in (+X)-direction:

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 Load on Column in (-X)-direction:

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 Load on Column in (+Z)-direction:

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 Load on Column in (-Z)-direction:

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 in (+X)-direction:

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 Load on Rafters in (-X)-direction:

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 Load on Rafters in (+Z)-direction:

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 Load on Rafters in (-Z)-direction:

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

4) 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 force.

4] Design using MS – Excel

1. Steel column
2. Rafter
3. Base plate
4. Pedestal
5. Z- purlin

1.DESIGN OF STEEL COLUMN:

Flange thickness = T = 12.7 mm.

Overall height of Column ISMB400 = h = 400 mm.

Clear depth between flanges = d = 400 – (12.7 x 2)

= 374.6 mm.

Thickness of web = t = 10.6mm.

Flange width = 2b = bf = 250 mm.

Hence, half Flange Width = b = 125 mm.

Self –weight = w = 0.822 kN/m.

Area of cross-section = A = 10466 mm2.

Radius of gyration about x = rx = 166.1 mm.

Radius of gyration about y = ry = 51.6 mm.

Type of section:

b/T = 125/12.7= 9.8 < 10.5

d/t =374.6/10.6 =35.3 < 42

(Table 3.1 of IS: 800)

Hence, cross-section can be classified as “COMPACT”.

Effective Sectional Area, Ae = 10,466 mm2

(Since there is no hole, (Clause 7.3.2 of IS: 800)

no reduction has been considered)

Effective Length:

As, both ends are pin-jointed effective length, KLx = KLy = 3m

Slenderness ratios:

KLx/rx = 9000/166.1 =54.1

KLy/ry = 9000/51.6 =174.41

Non-dimensional Effective Slenderness ratio, λ">λ:

λ">λ

= (sqrt2.5X(174)^2)/(pi^2X2X10^5`= 3.83

Value of ϕ">ϕ from equation:

Hence, α">α = 0.34 for buckling class ‘b’ will be considered.

Hence, ϕ">ϕ= 0.5 x [1+0.34 x (0.654-0.2)+0.6542] = 0.791

Calculation of x:

= 0.809

Calculation of fcd from the following equation:

= 0.809 x 250/1.1 = 183.86 N/mm2

Factored axial load in kN.

pd = Ag fcd

= 10466 x 183.86/1000 = 1924.28 kN.

2. DESIGN OF RAFTER:

Span of Rafter = 6 m

Dead load = 18KN/m

Imposed load = 40KN/m

support bearing = 100mm

yield strength = 250N/mm2

Design load calculation:

Factores load 1.5(DL+LL) =87 KN/m

Factores Bending moment = Wl28">Wl28">Wl28��28FA+6XMzBL2= 65.25 KN/m

Section modulus Required:

Z reqd = (65.25 x106106 x1.1) /250 =287100 mm3 = 287.1 cm3

Section classification:

ISMB-200

A = 323.3 mm2

D = 200mm

B = 100mm

t = 5.7 mm

T = 10.8 mm

Ixx = 2235.4 cm4

Iyy = 150 cm4

Zp = 375.35 cm3

Moment of resistence of the cross-section:

Md = βbZpfyγm0">βbZpfyγm0��������0LXσ1σ1+σ2

= (1 x 375.35 x 250) / ( 1.1)

Md = 85.306 > 65.25 KN/m

3. DESIGN OF BASE PLATE:

Strength of concrete, Fcu = 40 N/mm2

Yield strength of steel, fy = 250 N/mm2

Material factor, γm">γm= 1.1 KN

Factoresl Load = 1500 KN

Steel column section:

Thickness of flange, T = 12.7 mm

Area required:

Bearing strength of concrete = 0.4fcu = 0.4 x 40 = 16 N/mm2

= (1500x1000) /( 16)

= 93750 mm2

Let size of plate , Bplate = 450 mm

Dplate = 300 mm

Area of plate= 135000 mm2

projection on each side = a=b =25mm

W = (1500x1000) /( 450 x 300)

= 11.11 N/mm2

Therefore, Thickness of Base Plate, clause 7.4.3.1

ts= 7.3 mm < 12.7 mm

Size of Base plate 450 x 350 x 16 mm

4.DESIGN OF PEDESTAL:

Grade of concrete = 40 N/mm2

Load = 200 KN

Moment = 120 KN

Horizontal shear = 20 KN

Yield strength = 250 N/mm2

Length of base plate = 450 mm

Width of base plate = 350 mm

C/C distance of bolt in group-Z = 300 mm

C/C distance of bolt in group-X = 180 mm

Bearing strength of concrete Fc = 16 N/mm2

Depth of Column = 300 mm

Width of Column = 250 mm

Anchor Bolt Details:

Dia of anchor bolts =24 mm

No:of anchor bolts in each side = 4

Total no. of anchor bolts, n = 8

Gross area of the bolt ;Asb' = 452.16 mm2

Net area of bolt 'Anb' = 352 mm2

Ultimate tensile strength of bolt 'fub' = 400 N/mm2

Fyb (anchor bolts) = 240 N/mm2

Base plate Details:

Ultimate tensile strength of plate 'fu' = 490 N/mm2

Thickness of plate = 16 mm

Yield stress of plate = 330 N/mm2

Anchor bolt design:

Area of the plate = 157500 mm2

Stress,

Stress, Maximum pressure =

= 10.6 N/mm2 < 16 Hence OK

Minimum pressure =FA-6XMzBL2">FA−6XMzBL2��-6�����2

= -9.04 N/mm2 < 16 Hence OK

Centroid = = 242.89 mm

a =L/2-C/3 = 144.04 mm

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

y = (L - C/3-e) = 294.04 mm

Tension in anchor bolt along the length of plate, FT = (Mz - Fa)/Y

= 364.38

Tension per bolt = 91 KN

Shear per bolt = 1.63 KN

Shear Check:

Factored shear force = Vsb = 1.63 KN

Vd,sb = 81290.9 N

= 81 KN

Factored shear force = 65.03 KN

Tensile Check:

Factored tensile force in bolt, Tb = 91.1 KN

Tensile strength of bolt:

Ts,b = Tn,b /γmb">γmb���= 126 KN

Td,b = 98.65 KN

Combined Unity Check:

Vsb/Vdb = 0.025

Tb/Tdb = 0.92

Unity check = 0.85 < 1, Hence OK

Anchor Bolt Length:

Bond strength in tension, τbd">τbd���= 1.4 N/mm2

Anchor length required = Tb(3.14*τbd">τbd���)

= 863.44 mm

Let Anchor Bolt Length = 900 mm

5.DESIGN OF Z-PURLIN:

Span of the purlin = 6 m

Spacing of purlin = 1.5 m

No. of sag rods = 1

Slope of the roof = 4.76 = 5 degree

Dead load:

Weight of sheeting = 6kg/m2

self weight of purlin = 4.22 kg/m2

Additional load = 10% = 0.42 kg/m2

Total Dead load = 0.106 kg/m2

Live load:

Live load on roof = 75 kg/m2

Wind Load:

Basic wind speed = 50 m/s

Terrain caterogy = 2

Building class = B

K1 = 1

K2 = 1

K3 = 1

Design wind speed, Vz = 57 m/s

Design wind pressure, Pz = 1949.9 m/s = 1.94 KN

Length of building, L = 50 m

Breadth of building, W = 30 m

Height of building, H = 10.25 m

Height of eaves, = 9 m

h/w = 0.34

L/w = 1.667

External Pressure Co-eeficient:

Maximum downward =Cpe = -0.4

Maximum upward =Cpe = -0.7

Internal Pressure Co-eeficient:

Maximum positive =Cpi = 0.5

Maximum negative =Cpi = -0.5

For Maximum upward wind force:

Max upward Cpe = - 0.7

Cpi = - 0.5

Cpe + Cpi = -1.2

Pz = 1.95

Wind Pressure for Purlin Design = - 2.339 KN/m2

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 = 0.195 KN/m2

Design Load Calculation:

Spacing of purlin = 1.5 m

Slope of roof = 5 degree

Total dead load = 0.096

DL Normal component = 0.144 KN/m

DL Tangential component = 0.013 KN/m

Total Live load = 0.75

LL Normal component = 1.121 KN/m

LL Tangential component = 0.098 KN/m

Total Wind load = -2.339

WL Normal component = - 3.496 KN/m

WL2 WL load = 0.195 KN/m

WL Normal component = 0.291 KN/m

Maximum Normal component = DL+LL = 1.265 KN/m

Purlin section:

Selected section Z200 x 6 x 2.3

Area = 8.07 cm2

Weight of purlin = 6.335 kg/m.

1] Design Foundations using STAAD Foundation.

Click on Staad foundation:

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.

Solution:

Step:-1

Maximum moment due to vertical load

Weight of trollley + Crane capacity = 40 + 250 = 290 kN

Self-weight of crane girder excluding trolley = 100 kN

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

Ra=

This load is transferred to gantry girder, through two wheels, the wheel base being 3m.

So load on gantry girder from each wheel = 307.77/2 = 158.88 KN

Factored load = 158.88 x 1.5 =230.83 KN

Maximum moments due to moving loads occur under a wheel when the c.g of wheel load.

RD= 230.83 x (3-1.5-0.75) + 230.83 x (3+0.75)/7 = 148.40 KN

Max. moment, Me = 148.40 x 2.25 = 333.90 KNm.

Moment due to impact = 0.25 x 333.90 = 83.47 KNm.

Assume self weight of girder = 2KNm.

Dead load due to self weight + rails = 2+0.2= 2.2 KNm.

Factored Dead Load = 2.2 x 1.5 = 3.3 KNm.

Moment due to Dead Load = 3.3X728">3.3X7283.3�728= 20.21 KNm.

Factored moment due to all vertical loads,

Mz = 333.90 + 83.47 + 20.21 = 437.58 KNm.

Step :- 2

Maximum moment due to Lateral force:

Horizontal force transferred to rails = 10% of weight of trolley plus load lifted = (10/100) x (250+40) = 29 KN

This is distributed over 4 wheels.

So, horizontal force on each wheel = 29/4 = 7.25 KN

Factored horizontal force on each wheel = 1.5 x 7.25 = 10.875 KN

For maximum moment in gantry girder the position of loads is same as earlier except that it is horizontal, hence by proportioning we get.

My = (10.875/230.83) x 333.90 = 15.73 KNm.

Step :- 3

Shear force:

Vertical shear due to wheel loads = 230.83 + (230.83 x 3) / 7 = 329.75 KN

Step :- 4

Priliminary Section:

Minimum economic depth, L/12 = 7000/12 = 583.33 mm

Width of the compression flange may be taken as (1/40) to (1/30) of span;

So, flange width can be taken as, L/40 = 7000/40 = 175 mm to L/30 = 7000/30 = 233.33 mm.

Required, Zp = 1.4 x M/fy = 1.4X437.58X106250">1.4X437.58X1062501.4�437.58�106250 = 2450 x 103">103103mm3

Let us try a ISMB 600 with ISMC 400 on compression flange.

Step :- 5

ISMB	600 @1.22 kN/m.
A	15600 sq.mm.
h	600 mm.
b	210 mm.
tf	20.3 mm.
tw	12 mm.
Izz	91800 X 104">104104mm4">mm4��4
Iyy	2650 X 104">104104mm4">mm4��4
R	20 mm.

ISMC	400 @0.49 kN/m.
A	6380 sq.mm.
h	400 mm.
b	100 mm.
tf	15.3 mm.
tw	8.8 mm.
Izz	15200 X 104">104104mm4">mm4��4
Iyy	508 X 104">104104mm4">mm4��4
Cyy	2.42 mm.

Step :-6

Let the distance of N>A> from the tension range by y:

388.93 mm.

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 whichever 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 but Safe in lateral direction.