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

Q1

Aim:- to calculate the dead load and define load cases in the model

Step 1:- calculation of dead load

floor finishes with a thickness of 50mm

floor finishes will be applied as area loading on the slab panels

considering unit weight of the concrete as 24 Kn/m^3

total load is 0.05 x 24 = 1.2 Kn/m^2

self weight of the slab will be taken automatically by the model

brick wall loading

brick wall loads are applied as line loads on the beams

thickness of the wall is given as 150mm

consider the unit weight of the brick as 20kn

floor height as 5.2m

total line load is 5.2 x 20 x 0.15 =15.6

ceiling load of 0.3 Kn/m^2 is given

step 1:- to define the load cases

• click on load from the toolbar
• select load cases
• then add the type of loads in the dialoguebox

• define load combinations as per IS standards
• then select the type of load to be assigned
• define the intensity of loading in the properties dialogue box
• then select the method of assigning the loads as area loads
• then click on the slab panels to assign the loads

to assign brickwall load

Q2

Aim:- to calculate the live load and assign them to the model

• live loads are assumed from IS 875 part 2 based on the type of occupancy

rest room = 2 kn/m^2

toilets = 2 kn/m^2

office room = 3 kn/m^

store room = 5 kn/m^2

common room = 5 kn/m^2

lobby= 5 kn/m^2

staircase = 5 kn/m^2

conference room= 5 kn/m^2

equipment room = 5 kn/m^2

To assign live load to the model

• select the load as imposed from the load cases
• then select the type of load as area load
• specify the load value in the properties dialogue box based on occupancy
• click on the panel to assign the load

Result:- the imposed load of 5kn/m^2 is applied on slab panels

Q3

Aim:- to calculate the crane load

Given data

• center to center of wheel is 10m
• weight of the crab is 40kn
• number of wheels 4
• length of wheel base is 2m

max concentrated load on the crane is 50 + 10= 60

self wt of the crane will act as uniformly distributed load over the entire span = 40/10=4 Kn/m

taking moments about B

Ra x 10-(60 x9)-(4x5x10)=0

10Ra = 740

Ra=74 kn

Rbx 10 -(60 x 1)-(4 x 10x5)=0

Rb=26 kn

Ra+Rb= 100 kn

the reaction from the crane girder is distributed equally on two wheels at the end of the crane girder

maximum wheel load on each of the crane wheel = 74/2 =34Kn

Maximum bending moment

Assume self wt of the girder as 1.5 KN/m

total dead load = 300+1500 =1800

Rc x 5.5-(37 x 4.25)-(37 x 2.25)=0

Rc= 240.5/5.5

Rc=43.72

Rd x 5.5- (37 x 1.25)-(37 x 3.25)=0

Rd= 166.5/5.5

Rd=30.27

bending moment under the wheel load due to live load is =Rd x 2=30.27 x 2 ==60.54

bending moment due to impact load is 10% of the live load

bending moment due to dead load is 1.8 x 5.5 x5.5/2= 27.22

total bending moment is 60.54+6.54+ 27.22=93.814

Maximum shear force

for max shear force

Rc x5.5 -(37x5.5)-(37x2.5)=0

Rc=9.25 +20.35/5.5

Rc=5.381

the max shear force due to wheel load is 5.381KN

Lateral forces

lateral forces traverse to rails = 5% of the weight of crab and weight lifted

=0.05(40 +10)

=2.5

lateral forces on each side wheel =2.5/2=1.25

maximum horizontal reaction due to the lateral force

=1.25 x (43.72/37)=1.44

max bending moment due to lateral load by proportion =(1.25/37)x5.31=0.179

RESULT:- Load calculations for the 5T crane is done