Project 1_Analyze and design a steel building to 10T CRANE as per IS standard code using TEKLA STRUCTURAL DESIGNER.

 

Ans  - 

PROCEDURE:

Initially open the Tekla Structural Designer software. Now click on New option to create a new model workspace. Using the Model Ribbon using the grid line and Quick parallel grid line option the vertical and horizontal grids was drawn.     1. Grid -

 

 

 

2. Levels - 

 

in the Model toolbar select Construction levels icon to open the construction levels dialogue box.

The construction level dialogue box appears and name each level like base, GF, 1st floor and Roof and heights was assigned to each level as shown in the cross section pdf.

 

 

 

 

 

 

3. Pedestal Column - 

 

Now to add Pedestal columns in the Home menu click on Manage property sets options and the manage property sets dialogue box appears.

Click on New followed by members and select concrete column. and off the auto design option and chose m30 grade of concrete and size is 500x500.

 

 

 

Rename the column for our own convenience and apply the properties as required on the side.

 

 

 

Now using the column option in the ribbon tab under the model menu columns was placed at locations as required. Columns was placed at base floor and till ground floor

 

 

 

4. Steel Column - 

 

In the similar way concrete column was created steel columns was created and was renamed as Ismb 450

Now using the column option in the ribbon tab under the model menu columns was placed at locations as required. Columns was placed at ground floor and till top roof.

 

 

 

 

 

 

 

 

5. Beams - 

 

 

In the similar way concrete column was created steel beams was created.

Now using the beam option in the ribbon tab under the model menu beams was placed at locations as required.

 

Creting Beam in manage property set - 

 

 

 

 

 

 

 

 

 

 

Applying beam to roof level - 

 

 

 

 

creating fram for bracing and rafter - 

 

 

 Taking 2 paraller grid from 1m. offset and start placing beam for cran load.

 

 

 

 

 After that placed bracing on Roof level.

 

 

 

 3D view - 

 

 

 

 

 STEP .  6 

. Next we want to create the steel brace

. So go the home tab and select manage property set

. Next we pick the new option and select members and pick the steel brace option

. And go the general setting and go to the section option

. Select the equal angles and select the 75  75  5  6  8  10

. And select the X brace option on the model tab

. Go to the properties box select the created brace option

. Next go to the frame 

. And draw the X brace on the floor

. And turn the screen on 3d and draw the X brace on the roof top

. The completed brace image as been shown below

. Next go to the roof panel option

. place the roof panel on the top of the roof

. The roof panel placed image as been shown below

. Next we want to make slab on the beam

. so go to the manage properties sets and make new slab and give a values 

. Finnaly place the slab on beam.

 

 

 

 

 

Creating brace on Frame A and F as well. _

 

 

 

 

 

 

 

7. Creating slab property for structure - 

 

 

 

 

Applying Slab - 

 

 

 

 

 

 

 

 

 

CALCULATION OF DEAD LOADS : 

. Brick wall thickness = 150mm

. Height of the wall = 5m

. unit weight = 20 KN/m^3 

. Brick wall loading (gf) = 0.15 x 20 x 5 = 15 KN/m

. Brick wall loading (ff) = 0.15 x 20 x 7 = 21 KN/m

. Unit weight of concrete = 24 KN/m^3

. Finishes = 50mm x 24 KN/m^3 = 0.05 x 24 = 1.2 KN/m^2

Floor height  

. Assume floor finish thickness = 50mm

. floor finish = 0.05 x 24 = 1.2 KN/m^3

. Roofing load based on purlin size : 1.5 KN/mm^2

. Ceiling load of 0.3 KN per sq m

Calculation of Imposed load : 

. The live load has been considered by IS 875 PART 2 

Calculation of crane load : 

GIVEN : Crane capacity= 100 KN

. Weight of crab = 35 KN

. Weight of crab = 160 KN

. Weight of trolley car = 10 KN

. The approximate minimum approach of the crane hook to gantry girder = 1m

. Span of crane girder (c/c of wheel) = 20 m

. Span of granty girder (c/c of wheel) = 6 m

. Self weight of the crane = 160/2 = 80

. Wheelbase = 3m

SOLUTION:

1.) Maximum wheel Ioad

1. Maximum concentrated load on crane = 35 + 100 = 135 KN

2. Self weight of crane will act as uniformly distributed load of intensity = 35 /10 = 3.5 KN/m

Taking Moment about B

 (RA X 20) - (135 X 19) - (8 X 20 X 10) = 0

 RA = 208.25 KN 

Taking Moment about

RB = 295 - 208.25 = 86.75

RB = 86.75 KN   

RA + RB = 295 KN

The reaction of the crane gider is distributed equally on two wheel at the end of the crane girder.

Maximum wheel load on each wheel of crane (RA/2) = 208.25/2 = 104.12 KN

2.) Maximum Bending Moment: 

Assume self-weight of gantry girder as 1.5 KN/m

Assume self-weight of rail as 0.3 KN/m

Total dead load = 0.3 + 1.5 = 1.8 KN/m

At D :

  (RC X 6) = 104.12  X (2.25+3) + 104.12  X 2.25

RC = 130.15 KN 

At C : 

RD = 78.09 KN  

RC + RD = 208.24 KN

Bending Moment under a wheel load due to live load

RD X 2 = 78.09 X 2 = 156.18 KNm

Bending moment due to impact = 0.10 X 156.182 ( 10 % due to M.o.T) = 15.61 KN m

Total bending moment due to live load and impact load = 171.79 + 8.1 = 179.89 KNm

Bending moment due to dead load WI^2/8 = 1.8 X (6 x 6) /8 = 8.1 KNm

Maximum bending moment = 179.89 + 8.1 = 187.99 KN m

3.) Maximum shear force: 

At D : (RC X 6) - (104.125 x 6) - (104.125 x 3)  = -312.375 RC = 156.18 KN 

 Hence the Maximum shear force due to wheel load is = 156.18 KN

Lateral Forces : Lateral force transverse to rails = 5% of the weight of crab and weight lifted = 0.05 x (1.125) = 2.25 KN

Lateral forces each wheel F1 = 2.25/2 = 1.125 KN

Maximum horizontal reaction due to lateralforce by proportion at C

= Lateral force x reaction at c due to vertical load / Maximum wheel load due to vertical load

= 1.125 x 156.18 / 104.12

= 1.68 KN

Horizontal reaction due to lateral force by proportion at D 2.25 - 1.68 = 0.57 KN

Bending moment due to lateral load = (1.125 / 104.12) X 130.15= 1.406 KNm 

Calculation of WIND load : 

Given Data : Basic wind speed = 39m/s Terrain category 2 .

. Total length of the building (l)  = 90m

. Total width of the building  (w) = 38m

. Total height of the building (h) = 16.5m

. Class of the structure = Class A

. Life of the structure = 50 Years

. l/w  = (90/38)    = 2.36

. h/w = (16.5/38) = 0.43

STEP :1 EXTERNAL PRESSURE CO-EFFICIENT (Cpe)

As per the IS 875 Part 3, Table 5

Building height ratio

Building height ratio = 1/2

Therefore, the building plan ratio = 1

Hence the plan we choose is given below

STEP :2

Finding the Factors (k1, k2, k3, k4)

k1: From Table 1 for the basic wind speed for 39 m/s,

Risk Coeffiecent, K1 = 1.0

K2 : From Table 2, 10m height = 1 for terrain category 2

15m height = 1.05

by using the interpolation method,

k2 = 1.02

K3: Topography factor, k3 = 1 (from clause 6.3.3)

K4: Importance factor K4 = 1.15 (from Clause 6.3.4)

STEP :3 Vz=Vb x k1 x k2 x k3 x k4

= 39x1x1.02x1x1.15

= 45.747 m/s

STEP :4

Pz = 0.6 Vz^2

= 0.6 x 45.747^2

= 1255.6728 n/sq.m

= 1.2556 kn/sq.m

STEP :5 Wind direction (up to roof level)

1.) Wind direction along y- direction (Face C & D): 

	Cpe +	Cpi -
Height of the building	16.5m	16.5m
External pressure co-efficient Cpe	-0.6	-0.6
Internal pressure co-efficient Cpe	0.7	-0.7
Net pressure co-efficient Cp = Cpe - Cpi	1.3	0.1
Design wind pressure, Pz	1.2	1.2
Wind load on Wall (Cp x Cz) Kn/m^2	-1.625	0.125

2.) Wind direction along X- direction (Face A): 

	Cpe +	Cpi -
Height of the building	16.5m	16.5m
External pressure co-efficient Cpe	0.7	0.7
Internal pressure co-efficient Cpe	0.7	-0.7
Net pressure co-efficient Cp = Cpe - Cpi	0	1.4
Design wind pressure, Pz	1.2	1.2
Wind load on Wall (Cp x Cz) Kn/m^2	0	1.75

3.) Wind direction along X- direction (Face B): 

	Cpe +	Cpi -
Height of the building	16.5m	16.5m
External pressure co-efficient Cpe	-0.25	-0.25
Internal pressure co-efficient Cpe	0.7	-0.7
Net pressure co-efficient Cp = Cpe - Cpi	-0.95	0.45
Design wind pressure, Pz	1.2	1.2
Wind load on Wall (Cp x Cz) Kn/m^2	-1.187	0.562

STEP :6

Roof Calculation: W.k.t

h/w = 0.6

In order external coefficient from the table 6

Slope = Sin theta = opp/hyp

= 2.813/8.939

theta = 18.34

by using interpolation method

(18.34-10)/ (20-10) = (x+1.2)/(-1.4+1.2)

therefore, x =-1.36

Similarly
(18.34-10)/(20-10) = (x+0.4)/(-0.4+1.4)

X = -1.24

Hence, EF = -1.36

FG = -1.24

STEP :7

1.) Wind direction along Y- direction (Face EF): 

	Cpe +	Cpi -
Height of the building	16.5m	16.5m
External pressure co-efficient Cpe	-0.4	-0.4
Internal pressure co-efficient Cpe	0.7	-0.7
Net pressure co-efficient Cp = Cpe - Cpi	-1.1	0.3
Design wind pressure, Pz	1.2	1.2
Wind load on Wall (Cp x Cz) Kn/m^2	-1.375	0.375

 

2.) Wind direction along y- direction (Face GH): 

	Cpe +	Cpi -
Height of the building	16.5m	16.5m
External pressure co-efficient Cpe	-0.4	-0.4
Internal pressure co-efficient Cpe	0.7	0.7
Net pressure co-efficient Cp = Cpe - Cpi	-1.1	-0.3
Design wind pressure, Pz	1.2	1.2
Wind load on Wall (Cp x Cz) Kn/m^2	-1.375	0.325

 

APPLY THE LOAD ON TSD : 

. Go to the load tab 

. pick the load cases option

.  The load cases dialouge box as been opened automatically

. In there create a new loads like dead, imposed, wind, crane, servises, seismic..,

. Next go to the load combination

. And go to the generate option on the load combination dialouge box

. And select the 1st option and select next

. Again give the next option as 2 time

. and finally pick the finish option

. The load combination as generate sucessufully 

. Next we want to apply the dead load 

. So go to the below the screen on show process option

. And select the dead load option

. Next go to the home tab and select the manage properties set

. And select the new option and select the slab items and rename it as (one way slab and two way slab)

. Next go to the slab on beam option and select the slab item on general box

. As per the IS rules apply the one way slab and two way slab on the floors

. Next go to the area load 

. apply the loads on room as per same IS rules for all the floors

. After complete the load applyed on all the floors

. Next apply the load on roof panel

. The load applyed image as been shown below.

 

creatig load propety in model  - 

 

 

 

 

Applied dead load as 1.2 kn - 

 

 

 

 

 

 

 Step : 9 

. Next we want to apply the imposed load or live load 

. AS like the same process go to below the screen "show process" set the imposed load

. Next go to the area load on top of the screen 

. For the all individual rooms have individula live load are there

. It as been the IS rules

. So we want to check the all the rooms live load and apply as per the IS rules

. the applyed  imposed load as given in below the image 

. The same process applyed all the floors

 

 

 

 

 

 

 

 

 

 

 

Step :3 

. Open tekla software and 

. Here we want to set the crane load

. So go to the crane load option on below the screen

. Next go to the load tab 

. pick the point load and go to the left side of general box

. In there set the load type as nodal 

. And give the value we derived 

. Next go to the crane lifing beam and apply the load

 Step :4 

. Deleted all the previous wind load as been created in above

. Next go to the wind wizard option on under the load panel ,after wind wizard box as been appear 

. select the worst case data and unselect the (data for each direction and clade frame) --> Next 

. Give the wanted value on basic data box and click next

. Apply the fetch distance 10 and click next 

. And click next on topography box and results box

. And add the wind load on (load cases box) 

. Next go to the seismic load

. select the code spectra on (site specific spectra box) select next

. And given the basic information and select next

. Pick the next botton on structural irregularites

. select the rc steel concrete on fundamental period

. And select the steel building with OMRF on (Seismic force resisting system box) and pick next

. Again give the next botton and click finish

 

 

 Wind Load:

• Given, basic wind speed , Vb = 39 m/s ; Terrain Category = 2
• Length of the building, l = 90 m
• Breadth of the building, w = 38 m
• Height of the building, h = 18.31 m

Step 1: Finding Design Wind Speed:

• Referring to IS 875 P3, Vz = Vb k1 k2 k3 k4 according to Cl 6.3
  • Where Vz – Design wind speed
  • K1 – probability factor
  • K2 – terrain roughness and height factor
  • K3 – topography factor
  • K4 – importance factor for cyclonic region
• From table 1, k1 = 1
• From table 2, for height of the building 18.31 m and terrain category 2 , k2 is obtained by interpolation.
  • For 15 m , k2 = 1.05
  • For 20 m , k2 = 1.07
  • Therefore, for 18.31 m , k2 = 1.063 = 1.06
• From Cl 6.3.3 , k3 = 1 since, theta < 3>
• From Cl 6.3.4, k4 = 1.15 for Industrial structures
• Substituting all the above values, Vz = 39 * 1 * 1.06 * 1 * 1.15 = 47.54 m/s

Step 2: Finding Design Wind Pressure:

• From Cl 7.2, wind pressure, pz = 0.6 Vz^2
  • pz = 0.6 * 47.54 ^ 2 = 1356.1 N/m2
  • pz = 1.356 kN/m2

Step 3: Design Wind Load:

Wind Load on Wall:

• From Cl 7.3.1, F = (Cpe – Cpi) A pz = Cp * pz
• Cpe is found from Table 5 of IS 875 P3 for h/w and l/w ratios
  • h/w = 18.31 / 38 = 0.48
  • l/w = 90/38 = 2.37

• Lets consider medium openings in which Cpi = + 0.5 / -0.5 according to Cl 7.3.2.2 of IS 875 part 3

Wind angle 0:

A – windward side , B – leeward side

• Lets consider the loading Wind + Y + Cpi , Wind + Y – Cpi for all four sides A, B, C and D

Wind angle 90:

C – windward side , D – leeward side

• Lets consider the loading Wind + X + Cpi , Wind + X – Cpi for all four sides A, B, C and D

Wind Load on Roof:

• For calculating wind load, h/w ratio and roof angle is required to find Cpe
• In our case, roof angle = 17 deg

For a roof angle of 17 deg, we need to interpolate between 10 and 20

Wind angle 0, EF = -0.64, GH = -0.4

Wind angle 90, EG = -0.75 , FH = -0.6

• Go to Load – Load cases to add the wind load cases in X and Y directions for + Cpi and – Cpi