Project 2

1. 

Solution:

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

GEOMETRY / MODEL CREATION:

PROPERTIES:

3D VEIW OBJECT:

Material: STEEL

BETA Angle 90 degree for wind / gable columns

SPECIFICATION:

• Member Tension
• End MY MZ
• Start MY MZ

SUPPORT:

• Pinned

LOAD CASE DEFINATION:

1. EQ+X

• IS 1893 Load-X 1

2. EQ+Z

• IS 1893 Load-Z 1

3. 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
• Wind frame point load FY - 1.14 KN/m
• Gable frame point load FY -0.57 KN/m

4. 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

LOAD CASE DEFINATION:

5. Wind load (0+CPi)

6. Wind load (0- CPi)

7. Wind load (90+CPi)

8. Wind load (90- CPi)

DESIGN OF STEEL COLUMN:

Solution:

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>

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">λ3.3X728:

λ">λ =

    =2.5X(174)2π2X2X105</math>">√2.5X(174)2π2X2X1052.5�(174)2�2�2�105= 3.83

Value of ϕ">ϕϕts=√2.5w(a2−0.3b2)γm0fy>tf from equation: 

Hence, α">αα = 0.34 for buckling class ‘b’ will be considered.
Hence, ϕ">ϕϕ = 0.5 x [1+0.34 x (0.654-0.2)+0.6542] = 0.791

Calculation of x from equation 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.

DESIGN OF RAFTER:

Solution: 

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">106Wl28</math>">Wl28��28= 65.25 KN/m

Section modulus Required:

Z reqd = (65.25 x 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</math>">βbZpfyγm0��������0

      = (1 x 375.35 x 250) / ( 1.1)

Md = 85.306 > 65.25 KN/m

DESIGN OF BASE PLATE:

Strength of concrete, Fcu = 40 N/mm2

Yield strength of steel, fy = 250 N/mm2

Material factor, γm">γmγmFA−6XMzBL2= 1.1 KN

Factoresl oad = 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

LXσ1σ1+σ2

ts= 7.3 mm < 12>

Size of Base plate 450 x 350 x 16 mm

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,   Maximum pressure =FA+6XMzBL2</math>">FA+6XMzBL2��+6�����2

                                         = 10.6 N/mm2 < 16>

             Minimum pressure =

                                         = -9.04 N/mm2 < 16>

 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                      = 65.03 KN

Tensile Check

Factored tensile force in bolt, Tb = 91.1 KN

Tensile strength of bolt        Ts,b = Tn,b /γmb</math>">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</math>">τbd���= 1.4 N/mm2

Anchor length required = Tb(3.14*τbd</math>">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 = 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 = 5degree

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

Foundations using STAAD Foundation.

2. 

Solution:

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=290X89+1002=307.77KN</math>">290X89+1002=307.77KN290�89+1002=307.77��

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 &

The wheel are equidistant from the ccentre of girder, as shown below.

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

                                7

Max. moment, Me = 148.40 x 2.25 = 333.90 KN-m

Moment due to impact = 0.25 x 333.90 = 83.47 KN-m

Assume self weight of girder = 2KN-m

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

Factored DL = 2.2 x 1.5 = 3.3 KN-m

Moment due to DL = = 20.21 KN-m

Factored moment due to all vertical loads,

Mz = 333.90 + 83.47 + 20.21 = 437.58 KN-m

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 byproportioning we get.

My = (10.875/230.83) x 333.90 = 15.73 KN-m.

Shear force:

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

Vertical shear due to wheel loads = 230.83 + (230.83 x 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 = 1.4X437.58X106250</math>">1.4X437.58X1062501.4�437.58�106250 = 2450 x 103</math>">103103mm3

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