Week 4 Challenge

For application of Loads in the Model we have create Steel Braces,Horizontal Bracings,X Bracings,Wall panels,Roof Panels,Slab

Bracings

Top Bracings

Wall Panels on Grid 1

Wall Panels on Grid G

Wall Panels in 3D

Slab Items > Slab on Roof

Slab S1 is created

Slab on Ground floor

Simillarly slab is applied on First floor and Roof

1. Calculate dead load in design report based on IS code and apply dead load on the model

• Finishes of 50mm
• Slab as per design
• Brickwall 150mm thickness
• Roofing load based on purlin size
• Ceiling loading 0f 0.3KN per sq m

 

Answer:

Unit weight of concrete = 24KN/m3

Dead Load:

1.Finishes Loading = 0.05*24 = 1.2 KN/m2 (Area Loading)

2.Slab load will be Designed by Software

3.Brickwall Loading 

Thickness of Brickwall = 150 mm

Unit weight = 20KN/m3

Floor Height = From Ground Floor to First Floor = 5.2m

Brickwall loading = 0.15*20*5.2 = 15.6 KN/m

Floor Height = From First floor to Roof = 6.8m

Brickwall Loading = 0.15*20*6.8 = 20.4 KN/m

4.Roof Loading based on Purlin size = 1.5 KN/m2

5.Ceiling loading 0f 0.3 KN/m2

 Load combinations are Generated

 Dead load 1.2 KN/m2 apllied on Ground Floor , First Floor , Roof

2. Calculate live load in design report based on IS code and apply live load on the model

• Assume the loading based on IS 875
• Roof loading
• Consider equipment loading as 5KN per sq m

Answer:

Ground Floor Rooms:

Stiar case = 5KN/m2

Workshop = 5 KN/m2 ( Considering Light Machinery)

Corridor = 5KN/m2

Meeting Room =5 KN/m2

Store room = 5 KN/m2

Toilet = 2 KN/m2

Canteen =3 KN/m2

Office = 3KN/m2

Conference = 5KN/m2

Equipment = 10KN/m2

First Floor Rooms:

Staircase = 5KN/m2

Restroom = 2 KN/m2

Meeting = 5KN/m2

Administration room = 5KN/m2

Office =3 KN/m2

Toilet = 2KN/m2

Live load applied = 10KN/m2 on the slab for Equipment Room

 

 Live load applied = 3KN/m2 on the slab for Office Room

 

Live load applied = 5KN/m2 on the slab for Stair case

Simillarly the same loadings applied on slab for First floor and Roof also

Final load applied view in 3D

3. Generate a calculation for 5T crane loading based on following inputs

• Centre to Centre of wheel = 10m
• Weight of crab = 40 KN
• Number of wheels = 4
• Wheel base = 2m

Answer:

Given that

Crane Capacity                                          = 50KN

Self weight of crane girder excluding trolley = 40KN

Weight of trolly Car or Crab                           = 10KN

Aprroximate minimum approach of the crane hook to gantry girder = 1m

c/c distance between gantry rails                  = 14m

c/c distance between column                        = 5.5m

Self weight of rail section                             = 300 N/m = 0.3 KN/m2

Yield stress of steel                                     = 250 N/m2 = 0.25 KN/m2

Wheel base                                                = 3m

To find Maximum Wheel load of the Crane

Maximum Point load on the crane = 50 + 10 = 60 KN 

(Crane capacity + weight of Crab)

Self weight of crane = 40KN 

To convert this in to UDL = (Self weight of crane/c/c distance between gantry rails)

                                    = 40KN/14m = 2.88KN/m

 

Consider moment about B

RA*14-60*13-2.88*14*7 =0

RA = 75.85 KN

Maximum wheel load on each wheel = 75.85/2 =  37.92 = 38 KN

For calculation of Maximum Bending moment 

1.Considering the Dead loads

Assume self weight of gantry girder as 1.5 KN/m

Given Self weight of rail section = 0.3 KN/m

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

For absolute maximum bending moment

1.The load and

2.The resultant        are equidistant from mid-span

also we can do as shown in diagram

 

BM due to deadload M1 = wl2/8 = 1.8*5.5^2/8 = 6.81 KN-m

2.Considering the Live loads

Taking moments about  C   

RD*5.5 - 75.85*3.4375 - 75.85*0.6875 = 0

RD = 56.9 kN

Taking moments about  D  

RC*5.5 - 75.85*4.8125 - 75.85*2.0625 = 0

RC = 94.81 KN

Maximum Bending Moment will be below wheel load at  M2 = RD x 2.0625
                                                                = 56.9 x 2.0625
                                                                 =  117.35 kN-m

Impact load = 10% of maximtim static wheel load

Maximum BM due to impact, M3 = 10% of M2

                                               = 0.1*117.35 = 11.735 kN-m

M  = M1 + M2  + M3 = 6.81 + 117.35 + 11.735  = 135.90 KN-m

For calculation of Maximum Shear Force Due to Vertical Loads

1.For Dead load

Shear force due to self weight V = wl/2 = 1.8*5.5/2 = 4.95 KN

For maximum shear force due to wheel loads, one of the wheel loads is at the support.

2.For Live loads

Taking moments about F

 
RE*5.5 - 75.85*5.5- 75.85*2.75 =0

RE = 113.775 kN

3.For Impact loads

Impact load = 10% of maximum wheel load

Maximum shear force due to impact =0.1*113.775 = 11.38 KN

Total Maximum shear Force = 1+2+3 = 4.95+113.775+11.38 = 130 KN

Lateral Load Analysis

Lateral load = 5% of (the weight of the trolley + weight lifted by the crane)

                  = 0.05*(10+50)

                  = 3 KN

For one wheel, the lateral load = 3/4 = 0.75 kN

since we have four wheels)

The maximum bending moments and shear force can be calculated by proportion method :

Maximum Horizontal reaction due to lateral force by proportion at C

= Lateral force*lateral force due to reaction at C due to vertical load/Max wheel load due to vertical load

=0.75*56.9/75.85

=0.56 KN

Maximum Horizontal reaction due to lateral force by proportion at D

= Lateral force*lateral force due to reaction at D due to vertical load/Max wheel load due to vertical load

=0.75*94.81/75.85

= 0.94 KN

Total Lateral load = 0.56+0.94 = 1.5 KN

Bending moment due to lateral load = 0.75*117.35/75.85 = 1.16 KN-m