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

PROJECT 1
AIM:
The aim of this project is to,

Analyze and design a steel building to 10T CRANE as per IS standard code in TEKLA STRUCTURAL DESIGNER. Refer the attached plan and elevation. Provide bracings and moment connection for lateral stability.

Consider dead, live, equipment and wind loading. Consider the brick wall loading for 150mm thick and 1.5 KN per sq m for wall and roof cladding.

Assume wind loading basic speed as 39m/s

Report of each member to be generated and extract drawings of structural plans from the software

INTRODUCTION:

Tekla Structural Designer is used for analysis and design of structures. The first step is to model the structure,then to apply dead load,live load, wind load, seismic load over it and to study the response of the structure. The response of the structure can be its deflection, shear force or bending moment. Accordingly suitable measures are taken for increasing the member size, providing UB/UC beams, restraining top flange continuity etc. Finally all the members are checked for failures.

PROCEDURE:

Study the drawings of structural layout

Launch TSD

Save the new files as Project Now go to Home – Settings to check the Units, Design codes, Structure defaults etc

Also go to Home – Materials to check the material grades

Construction levels:

Go to Model – Construction levels

Study the section details drawing to enter heights of GF,FF, RBL and RL

To find the height of roof level assume roof angle as 17 deg

tan 17 = Height of the roof / (Span / 2) = H / (38000/2)

Therefore, H = tan 17 * 19000 = 5808mm = 5.808 m

Grid Lines:

Go to Model – Grid lines

Draw 16 vertical grids of 6m spacing

Number it accordingly

Similarly construct the horizontal grid lines A-G of appropriate spacing as given in the layout

Pedestals:

Check the drawing to see whether any construction lines are needed to model the pedestals

Draw a construction line of 14m from B using Parallel(quick) in the construction line option

To create pedestals, go to Home – Manage Property Sets – New – Members – Concrete column

Create a pedestal of size 500*1000mm , major alignment – centre and releases – fixed top and bottom

Under Model – Concrete – Column , pick PF1 from the Properties window and start placing the columns

Select the levels as Base level to GF level

Modelling Steel Columns:

Check the section layout drawing. Notice that the interior columns are of different heights

So create 2 additional levels next to RB level. As we have set the roof level at 5.81m from RBL , subtract 1m for the roof level = 4.81m

Model the steel column of ISMB 550

Now go to GF level

Click on Steel – Column

From the properties, choose C1(MB 550) and then the levels from GF level till the roof beam level for outer grids and C2(MB 600), levels from GF to Int L1 for grid B and levels from GF to Int L2 for columns on construction line

Modelling Steel Beams:

Create a beam B1 of MB 550 ,alignment – bottom,centre for GF and top,centre for other floors

Model B1(MB 550) beam for all 6m span

Create another beam B2 MB 600 for beams between A and B grids

Model B3 MB 350 for secondary beams connecting primary beams

Now model beams for staircase

Draw a grid line of 3m from the 9th grid to model staircase beam

Now coming to the first floor, check the layout

Notice that there are no intermediate beams between B and G

So we need to model the beams accordingly

Model secondary beams between A and B

Similarly model staircase beams between grids 9 and 10

Similarly proceed to the roof level

Frame modelling:

Inorder to create rafters, frame modelling needs to be done

Go to Model – Frame and pick all grid lines

Creating Rafters:

Model the grid line at the midspan(38000/2 = 19000) for the apex of the rafter

Then go to Structure 3D to Edit- Copy rafter to all frames by picking the reference node

Now provide the ridge beam connecting the apex of all rafters

Also model purlins by using ISMC section

Modelling Bracings:

Similarly model bracings for roof also

Creating slab on beams:

Find ly/lx ratio and classify the slab as either one way or two way and assign accordingly

Create Slab S1(one way) and S2(two way) with the following properties

Creating Wall and Roof panels:

Select frame A and G and frame 1 and 16 to model the wall panels

Bracket for Crane Support:

On checking the section drawing, brackets are placed at 950 + 2050 from the FF level

So we need to create a construction level at 3000mm from the FF level

Now go to Crane Support level and draw a parallel construction line 1m from gridline B.

A cantilever beam of MB 550 is to be modelled as the bracket supporting the crane

Now model this beam from grid B to the construction line.

Notice that it throws an error on validating

Double click on the beam to change its properties

Mark releases at end 2 as Cantilever end

Copy and paste the brackets to other grid lines

Mirror the brackets to the construction line drawn parallel to the grid

Application of loads:

Dead Load application:

Under load- loadcases, add wind and seismic loads

Under Combination, select Generate to generate the load combinations

Lets now proceed with the Dead load calculations by opening a MS Excel file and calculating Floor finishes load, Brickwall load( from GF to FF and FF to RBL) and roof loading

Lets now go to GF level to apply these loads

Under loadcases below, select Dead load

Now we are going to apply the floor finish load of 1.2 kN/m2 to each slab

Select Area load under Panel loads

Change the load intensity to 1.2 kN/m2 and select each panel one by one

Similarly apply the floor finishes load of 1.2 kN/m2 to the other floors also

Validate the model for any errors

Check the applied dead loads in 3D view

Now the brickwall loading is applied for each floor by referring to the sketches

For the GF level, brickwall loading of 15 kN/m is applied

The loading is first applied along the perimeter

Under Member loads, select Full UDL for the brick walls on the member

Also under Panel loads – Line loads to model the interior brick walls according to the sketch

Repeat the same procedure for FF level too

Now apply the roof loading of 1.5 kN/m2

Live Loads Application:

Referring to IS 875 Part 2, the following table is framed

The various rooms in our structure are listed one by one

The corresponding live loads are obtained from the IS 875 Part 2 code book

Now select Imposed load from the load cases below

Go to Panel loads – Area loads/ Patch to apply loadings on floors referring to the sketch

Apply the roof loading of 0.75 kN/m2 on the RB level

Crane Load:

Now lets generate a calculation for 10T crane loading based on following inputs

Crane capacity = 10T = 10000kg = 100 kN

Centre to Centre of wheel = 14m

Weight of crab = 40 KN

Total Number of wheels = 4

Wheel base = 2m

No of wheels on each side = 2

Weight of the crane = 60 kN

Span of the gantry girder(max c/c distance between 2 columns) = 6 m

Maximum Wheel load:

There are 2 loads acting on the crane girder.

UDL due to Self weight of the crane

Concentrated load due to weight of the crab / trolley and crane capacity(weight lifted by the hook)

Lets find the loads one by one.

UDL due to Self weight of the crane = S.W of the crane / c/c between the wheels

=60/14 = 4.3 kN/m

Concentrated load due to weight of the crab / trolley and crane capacity = 40 + 100 = 140 kN

Minimum hook approach = 1m

Applying the equilibrium equations,

Ra + Rb = 140 + (4.3* 14) = 200 kN

∑M b = 0

(14 * Ra) – ( 140 * 13) – ( 4.3 * 14 * 7) = 0

Ra = 160 kN

Rb = 200 – 160 = 40 kN

Max reaction is taken always

Ra = 160 kN = Static wheel load

This load is shared by 2 wheels , Load transferred by 1 wheel = 160/2 = 80 kN

Adding 25% for Impact = 100 kN

Factored load = 100 * 1.5 = 150 kN

Maximum Bending Moment:

Assume self weight of the gantry girder = 1.6 kN/m

Self Weight of the rails = 0.4 kN/m

Total UDL = 1.6 + 0.4 = 2 kN/m

For maximum BM, the wheel loads should be kept so that the C.G of 2 wheel loads and the wheel loads recline equidistant from the centre of the span

i.e Max BM occurs under wheel load ‘a’ when centre of the span is at equal distance from the wheel load and the C.G of the wheel loads

Ra + Rb = (150 *2) + (2 * 6) = 312 kN

∑M at B = 0

(6 * Ra) – (150 * 1.5) – (150 * 3.5) – (2 * 6 * 6/2 ) = 0

Ra = 131 kN

Rb = 181 kN

Max BM at a = (131 * 2.5) – (2 * 2.5^2/2) = 321.25 kNm

Maximum Shear force in gantry girder:

For the max SF in gantry girder, one of the wheel loads have to be placed on the support

∑M at B = 0

(6 * Ra) – ( 150 * 6) –(150 * 4) – (2 * 6^2 / 2) = 0

Ra = 256 kN

Lateral force:

Surge load along y = 10 % of (crane capacity + crab load) = 0.1 * (100 + 40 ) = 14 kN

This load is shared by 4 wheels = 14 / 4 = 3.5 kN

Finding the BM for this load,

150 -à 321.25 kNm

3.5 -à ( 3.5 * 321.25) / 150 = 7.5 kNm

Brake load along x = 5 % of static wheel load

= 0.05 * 160 = 8 kN

Now lets apply the crane load in the form of nodal loads on the brackets

(Brake, Surge, Max SF) are chosen as (x,y,z) loads repectively

In our case, it is (8 , 14 , 256)

First create a Crane load case from Load – Loadcase option

Under the construction level Crane Supp and load case as crane, go to Load – Member loads – Point loads

Under load type choose nodal load and enter (x,y,z) load values as (8, 14, 256)kN respectively

Pick the nodes in the brackets

Thus the crane loads are applied as nodal loads on the brackets

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 6.3Where 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^2pz = 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 ratiosh/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

Wind + Y + Cpi:

On Walls:

On Roofs:

Wind + Y – Cpi:

On Walls:

On Roofs:

Wind + X + Cpi:

On Walls:

On Roof:

Wind + X – Cpi:

On Walls:

On Roofs:

Seismic Load:

Under Load – Seismic load – Seismic Wizard

Now the load combinations are generated

Validate the model and then run analysis

ANALYSIS:

Analyze – 1st order linear

Select load combinations

Under Drift checks, check for deflection

Notice that all maximum deflections happen in the brackets where cranes are resting

So we need to increase the member size of the bracket and column supporting brackets

Also note that we have very high deflection in the columns and rafters. So increasing their member size is also important.

The BMD of a frame is as follows,

The SFD is as follows

The AFD is as follows,

So now lets increase the member sizes and reanalyze the model

First of all lets increase the size of the rafter and help it pass the design criteria

Lets pick frame by frame and select all the members to understand how many members have passed/failed accordingly redesign the failed members

Select all the members of the frame and check the selection

Select FRM G

Check from frame 1 to 16

Check the support reactions

Slab Design:

Go to Review at the bottom in the GF

In the scene content hide, steel columns,beams,bracing, concrete column

Select slab/ foundation reinforcement

Pad footing:

Go to Base level

Under Foundations – Select Pad Base Column

Click on any support to place the footing

Select the footing to enter the details of top bars.Just enter type as Loose bars,bar diameter will be automatically designed

Right click on the created footing and select Design member

The footing size and the diameter of the bars used will be designed automatically

Result:

• Thus a thorough analysis and design of a industrial structure has been carried out by following the below steps:
• Modelling grid lines
• Modelling pedestals for the base floor
• Modelling steel columns from the ground floor
• Modelling steel beams(primary and secondary)
• Modelling brackets for crane support
• Modelling rafters and purlins
• Modelling bracings
• Modelling wall and roof panels
• Applying dead loads (floor finishes load to floors, brickwall loading, roof level loading)
• Applying live loads to each floors
• Applying crane loads
• Applying wind loads in both X and Y direction
• Generating seismic loads
• Generating load combinations
• Running the analysis
• Checking the deflection
• Identifying the failed members and suitably adjusting the sizes of them
• Slab reinforcement design
• Footing design