Project 2

PROJECT 2

QUESTION-1:

Design a warehouse building located in Chennai using STAAD Pro connect edition. The specification must be as follows:

Width – 30 m

Length – 50 m

Eave height – 9 m

Bay spacing – 6 m

Soil type – medium

Safe bearing capacity – 200 kN/m^2

Roof slope – 1 in 12

Assume suitable sections for structural elements. Follow IS 800:2007, IS 1893 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, rafter, base plate, pedestal and Z-purlin
• Design foundations using STAAD foundation.
• Attach necessary sketches and drawing wherever required.
• All stipulations, assumptions and design parameter must adhere to Indian standards.

ROOF SLOPE CALCULATION:

Roof slope = 1 in 12

Width, b = 30 m

X/(b/2) = 1/12

X = (30/2)/12 = 15/12 = 1.25 m

GEOMETRY/ MODEL CREATION:

PROPERTIES:

3D VIEW OBJECT:

Material steel

SPECIFICATION:

• Member tension
• End MY MZ
• Start MY MZ

SUPPORT:

• Pinned

LOAD CASE DEFINATION:

EQ +X

• IS 1893 load -X 1

EQ +Z

• IS 1893 load -Z 1

Dead load

• Self-weight Y -1.15
• Main frame rafter GY -1.55 KN/m
• Gable frame rafter GY -0.775 KN/m
• Main frame column GY -1.2 KN/m
• Gable frame column GY -1.6 KN/m
• Gable/ wind column GY -1.04 KN/m
• Wind frame point load FY -1.14 KN/m
• Gable frame point load FY -0.57 KN/m

Live load

• Main frame rafter GY -4.85 KN/m
• Gable frame rafter GY -2.425 KN/m
• Wind frame point load FY -4.15 KN/m
• Gable frame point load FY -2.06 KN/m

SEISMIC DEFINATION:

Zone parameters:

MEMBER WEIGHT:

Assign to edit list from followed by the dead load

• Self-weight Y -1.15
• Main frame rafter GY -1.55 KN/m
• Gable frame rafter GY -0.775 KN/m
• Main frame column GY -1.2 KN/m
• Gable frame column GY -0.6 KN/m
• Gable/ wind column GY -1.04 KN/m

JOINT WEIGHT:

Assign to edit list from followed by the dead load (point load)

• Wind frame point load FY -1.14 KN/m
• Gable frame point load FY -0.57 KN/m

DESIGN OF STEEL COLUMN:

Solution:

Flange thickness = T = 12.7 mm

Overall height of column ISMB 400 = h = 400 mm

Clear depth between flanges = d = 400-(12.7*2) = 374.6 mm

Thickness of web = t = 10.6 mm

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 mm^2

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 = 10466 mm^2

(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 = 3 m

Slenderness ratios:

KLx/rx = 9000/166.1 = 54.1

KLy/ry = 9000/51.6 = 174.41

Non-dimensional effective slenderness ratio, λ” > λ3.3X7^2/8:

= square root (fy/fcc) = square root (fy(KL/r)^2/π^2E)

λ” > λ = square root (2.5)*(174)^2 / π^2*2*10^5 = 3.83

value of φ” > φ from equation:

φ = 0.5[1 + α(λ-0.2) + λ^2]

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

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

Calculation of x from equation x:

1/[φ+(φ^2-λ^2)^0.5] = 0.809

Calculation of fcd from the following equation:

Fcd = fy/ϒmo / φ+(φ^2-λ^2)^0.5 = Xfy/ϒmo <= fy/ϒmo

= 0.809*250/1.1 = 183.86 N/mm^2

Factored axial load in KN:

Pd = Agfcd = 10466*183.86/1000 = 1924.28 KN

DESIGN OF RAFTER:

Solution:

Span of rafter = 6 m

Dead load = 18 KN/m

Imposed load = 40 KN/m

Support bearing = 100 mm

Yield strength = 250 N/mm^2

Design load calculation:

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

Factored bending moment = Wl28” > 10^6Wl^2/8 = 65.25 KN/m

Section modulus required:

Z reqd = (65.25*1.1)/250 = 287100 mm^3 = 287.1 cm^3

Section classification:

ISMB-200

A = 323.3 mm^2

D = 200 mm

B = 100 mm

t = 5.7 mm

T = 10.8 mm

Lxx = 2235.4 cm^4

Lyy = 150 cm^4

Zp = 375.35 cm^3

Moment resistance of the cross-section:

Md = ϐbZpfy / ϒmo = (1*375.35*250) / 1.1

Md = 85.306 > 65.25 KN/m

DESIGN OF BASE PLATE:

Strength of concrete, Fcu = 40 N/mm^2

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

Material factor, ϒm” > ϒmF/A – 6*Mz/BL^2 = 1.1 KN

Factored load = 1500 KN

Steel column section:

Thickness of flange, T = 12.7 mm

Area required:

Bearing strength of concrete = 0.4fcu = 0.4*40 = 16 N/mm^2

= (1500*1000)/(16) = 93750 mm^2

Let size of plate, Bplate = 450 mm

Dplate = 300 mm

Area of plate = 135000 mm^2

Projection on each side = a = b = 25 mm

W = (1500*1000)/(450*300) = 11.11 N/mm^2

Therefore, thickness of base plate, clause 7.4.3.1

ts = square root (2.5)*w(a^2 – 0.3b^2)*ϒmofy > tf’ LXϭ1 / ϭ1+ϭ2

ts = 7.3 < 12.7 mm

size of base plate 450x350x16 mm

DESIGN OF PEDESTAL:

Grade of concrete = 40 N/mm^2

Load = 200 KN

Moment = 120 KN

Horizontal shear = 20 KN

Yield strength = 250 N/mm^2

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/mm^2

depth of column = 300 mm

width of column = 250 mm

anchor bolt details:

dia of anchor bolts = 24 mm

1. of anchor bolts in each side = 4

total no. of anchor bolts, n = 8

gross area of the bolt ‘Asb’ = 452.16 mm^2

net area of bolt ‘Anb’ = 352 mm^2

ultimate tensile strength of bolt ‘fub’ = 400 N/mm^2

Fyd (anchor bolts) = 240 N/mm^2

Base plate details:

Ultimate tensile strength of plate ‘fu’ = 490 N/mm^2

Thickness of plate = 16 mm

Yield stress of plate = 330 N/mm^2

Anchor bolt design:

Area of the plate = 15700 mm^2

Stress,

maximum pressure = F/A + 6*Mz/BL^2 = 10.6 N/mm^2 < 16 hence ok

Minimum pressure = -9.04 N/mm^2 < 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 = 81290.9 N = 81 KN

Factored = 65.03 KN

Tensile check:

Factored tensile force in bolt, Tb = 91.1 KN

Tensile strength of bolt Ts = Tn/ϒmb = 126 KN

Td = 98.65 KN

Combined unity check:

Vsb/Vdb = 0.025

Td/Tdb = 0.92

Unity check = 0.85 < 1 hence ok

Anchor bolt length:

Bond strength in tension, τbd = 1.4 N/mm^2

Anchor length required = Tb(3.14*τbd) = 863.44 mm

Let anchor bolt length = 900 mm

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 = 6 kg/m^2

self-weight of purlin = 4.22 kg/m^2

additional load = 10% = 0.42 kg/m^2

total dead load = 0.106 kg/m^2

live load:

live load on roof = 75 kg/m^2

wind load:

basic wind speed = 50 m/s

terrain category = 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-efficient:

Maximum downward = CPe = -0.4

Maximum upward = CPe = -0.7

Internal pressure co-efficient:

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/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 = 0.195 KN/m^2

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

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 cm^2

Weight of purlin = 6.335 kg/m

Foundation using STAAD foundation:

QUESTION-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

SOULTION:

Maximum moment due to vertical load

Weight of trolley + 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 = 290*8 / 9 + 100/2 = 307.77 kN

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

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

Factored load = 158.88*1.5 = 230.83 KN

Maximum moments due to moving loads occur under a wheel when the c.g. of wheel load and the wheel are equidistant from the center of girder.

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

Maximum moment, Me = 148.40*2.25 = 333.90 KNm

Moment due to impact = 0.25*333.90 = 83.47 KNm

Assume self-weight of girder = 2 KNm

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

Factored DL = 2.2*1.5 = 3.3 KNm

Moment due to DL = 20.21 KNm

Factored moment due to all vertical loads,

Mz = 333.90+83.47+20.21 = 437.58 KNm

MAXIMUM MOMENT DUE TO LATERAL FORCE:

Horizontal force transferred to rails = 10% of weight of trolley plus load lifted = (10/100)*(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*7.25 = 10.875 KN

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

My = (10.875/230.83)*333.90 = 15.73 KNm

SHEAR FORCE:

For maximum shear force on the girder, the trailing wheel should be just on the girder.

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

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, L/40 = 7000/40 = 175 mm to L/30 = 7000/30 = 233.33 mm

Required Zp = 1.4 x M/fy = 1.4*437.58*10^6 / 250 = 2450 x 10^3 mm^3

Let us try a ISMB 550 with ISMC 250 on compression flange.