To calculate dead and live load for industrial steel structures and to apply them using TSD

• Brick wall load = 0.15 x 20x 5.2 = 15.6 kN/m^2 (Ground to intermediate floor)
• Brick wall load = 0.15 x 20x6.8 = 20.5 kN/m^2 (Intermediate to roof)
• Brick wall load = 0.15 x 20x14.5 = 14.5 kN/m^2 (Roof to top )
• Roof load = 1.5 kN/m
• Ceiling load = 0.3 kN/m

Procedure:

• Now open Tekla structural designer software
• Open the week 3 file where we have created a steel structure
• Now go to model tab and select slabs on beams option
• Now add slabs to each floor
• Now go to load tab and select load cases and add required loads
• Now select load combinations and add what required
• Select the level we want to apply loads on
• Select the loads required from the list
• Now goto select the area to apply load on slab
• Go to load tab and select UDL to apply on brick walll
• Apply loads on the roof as well

Aim: To Calculate live load in design report based on IS code and apply live load on the model

Procedure:

Based on the IS 875 the live loads are applied, as follows.

• Workshop = 5$\frac{\mathrm{kN}/}{m^2}$
• Staircase =5$\frac{\mathrm{kN}/}{m^2}$
• Corridor = 5$\frac{\mathrm{kN}/}{m^2}$
• Meeting Room = 5$\frac{\mathrm{kN}/}{m^2}$
• Store Room =5$\frac{\mathrm{kN}/}{m^2}$
• Toilet = 2$\frac{k\mathrm{nN/}}{m^2}$
• Canteen = 3 $\frac{k}{m^2}$
• Office Room = 3$\frac{k}{m^2}$
• Conference Room = 5 $\frac{k}{m^2}$
• Equipment Room = 10 $\frac{k}{m^2}$
• Rest Room = 2 $\frac{k}{m^2}$
• Administration Room = 5 $\frac{k}{m^2}$

The Live loads are applied for all the floors, as shown below.

• Now select the level where we want to apply the loads
• Now select the loads as required at the bottom left corner

• Now go to the load tab and select the Area to apply load on the slab
• Now we can see the Area load applied on the slab

Result:

Calculation of live load in design report based on IS code and application of live load on the model is done.

Aim:  Generate a calculation for 5T crane loading based on the following inputs

Procedure:

Given Data:

• Centre to Centre of wheel = 10m
• Weight of crab = 40 KN
• Number of wheels = 4
• Wheelbase = 2m
• Crane Capacity=50 KN
• weight of trolley = 10kN
• hook approach = 1m

Maximum point load on Crane=Crane capacity + weight of the trolley

                            =50+10

                             =60KN

UDL=$\frac{self-weightofcrane}{dis\tan cebetweengrantyrails}$

                            =40/14

                           =2.89KN/m

Moment at A:

$\sum MA$=0

=60x1-RBX14+2.89X14X14/2

RB=24.5KN

$\sum Fy$=0

RA+RB-60-2.89X14=0

RA=60+2.89X14-RB

RA=75.9KN

Now assuming self weight of gantry grider=1.2 KN/m

 assuming self weight of gantry grider=0.3KN/m

Total self weight=1.2+0.3

=1.5KN/m

FACTORED LOAD=1.5X1.5

=2.25KN/m

Now load on each wheel=75.9/2

=37.97KN

Factored load=37.97x1.5

=56.95KN

=57KN

Moment at C:

$\sum Mc$=0

RDx6=2.25x6x6/2+57x4.5+57x2.5

RD=73.25KN

$\sum Fy$=0

RC+RD-57-57-2.25X6

RC+RD=127.5KN

RC+73.25=127.5

RC=127.5-73.25

RC=54.25KN

MAX SHEAR FORCE

$\sum MF$=0

REX6-57-57X4-2.25X6X6/2=0

RE=101.75KN

IMPACT LOAD:

=10% OF WHEEL LOAD

=0.1X101.75

=10.175KN
V=Wl/2

=$\frac{1.5X6}{2}$

=4.5KN

Max shear force=4.5+10.175+101.75

=116.425KN

Lateral load analysis=5% of total weight

=$\left(\frac{5}{100}\right)$x60

=3KN

Result: With this our work is done