Helix Bridge Modeling in Robot Structural Analysis Using AutoCAD
Title | Helix Bridge Modeling in Robot Structural Analysis Using AutoCAD |
Duration | 60 Mins |
Language | English |
Format | MP4 |
Size | 363 MB |
Title | Helix Bridge Modeling in Robot Structural Analysis Using AutoCAD |
Duration | 60 Mins |
Language | English |
Format | MP4 |
Size | 363 MB |
Retaining walls are built in order to hold back earth which would otherwise move downwards. A surcharge load results from forces that are applied along the surface of the backfill behind the wall. These forces apply additional lateral forces along the back of the wall. This spreadsheet calculates the resulting pressure field acting on the retaining wall due to a point load at a given position behind the wall.
Calculation Reference
Surcharge Loads
A surcharge load is an additional retaining wall load superimposed onto the earth pressure force to yield the total lateral force.
SURCHARGE LOADS:
STRIPS LOADS
TRIANGULAR LOADS
LINE LOADS
UNIFORM LOADS
RAMP LOADS
The TOTAL REACTIONS are also calculated.
Calculation Reference
Geotechnics
Description: Tourist Port – Salvador – Plants
Category: Autocad Drawing / Projects / Tourism – recreation
File extension: DWG
Type: Premium
Project Name: 3d hotel – precast concrete
Description: 3d Hotel
Category: Autocad Drawing / Projects / Tourism – recreation
File extension: DWG
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Calculate whether the wall will slide forward? Calculate the bearing pressure beneath the base? Considering friction acting on the vertical faces, would stability improve?
Geotechnics
Reproduction of a sample calculation from a published calculation reference
Geotechnics
A slip has developed in a long natural slope, determine the FoS along the slip plane using the critical state parameter, and the residual strength parameter. Analysis the stability of the slope using the limit state method.
Craig’s Soil Mechanics
Reproduction of a sample calculation from a published calculation reference.
Geotechnics
Concrete is one of the most commonly used building materials.
Concrete is a composite materialmade from several readily available constituents(aggregates, sand, cement, water).
Concrete is a versatile material that can easily be mixedto meet a variety of special needs and formed to virtually any shape.
Density of Cement = 1440 kg/m3
Sand Density = 1450-1500 kg/m3
Density of Aggregate = 1450-1550 kg/m3
How many KG in 1 bag of cement = 50 kg
Cement quantity in litres in 1 bag of cement = 34.7 litres
1 Bag of cement in cubic metres = 0.0347 cubic meter
How many CFT (Cubic Feet) = 1.226 CFT
Numbers of Bags in 1 cubic metre cement = 28.8 Bags
Specific gravity of cement = 3.15
Grade of cement = 33, 43, 53
Where 33, 43, 53 compressive strength of cement in N/mm2
M-20 = 1 : 1.5 : 3 = 5.5, (Cement : Sand : Aggregate)
Some of Mix is – 5.5
Where, M = Mix
20 = Characteristic Compressive strength
Consider volume of concrete = 1m3
Dry Volume of Concrete = 1 x 1.54 = 1.54 m3 (For Dry Volume Multiply By 1.54)
Cement= (1/5.5) x 1.54 = 0.28 m3 ∴ 1 is a part of cement, 5.5 is sum of ratio
Density of Cement is 1440/m3
= 0.28 x 1440 = 403.2 kg
We know each bag of cement is 50 kg
For Numbers of Bags = 403.2/50 = 8 Bags
We Know in one bag of cement = 1.226 CFT
For Calculate in CFT (Cubic Feet) = 8 x 1.225 = 9.8 Cubic Feet
Consider volume of concrete = 1m3
Dry Volume of Concrete = 1 x 1.54 = 1.54 m3
Sand= (1.5/5.5) x 1.54 = 0.42 m3 ∴ 1.5 is a part of Sand, 5.5 is sum of ratio
Density of Sand is 1450/m3
For KG = 0.42 x 1450 = 609 kg
As we know that 1m3 = 35.31 CFT
For Calculation in Cubic Feet = 0.42 x 35.31 = 14.83 Cubic Feet
Consider volume of concrete = 1m3
Dry Volume of Concrete = 1 x 1.54 = 1.54 m3
Aggregate = (3/5.5) x 1.54 = 0.84 m3 ∴ 3 is a part of cement, 5.5 is sum of ratio
Density of Aggregate is 1500/m3
Calculation for KG = 0.84 x 1500 = 1260 kg
As we know that 1 m3 = 35.31 CFT
Calculation for CFT = 0.84 x 35.31 = 29.66 Cubic Feet
Area of brick wall for plaster = 3m x 3m =9m2
Plaster Thickness = 12mm (Outer-20mm, Inner 12mm)
Volume of mortar = 9m2 0.012m = 0.108m3
Ratio for Plaster Taken is = 1 : 6
Sum of ratio is = 7
Calculation for Cement Volume
Dry Volume of Mortar = 0.108 1.35 = 0.1458 m3
Cement= (1/7) = 0.0208 m3
Density of Cement is 1440/m3
= 0 1440 = 29.99 kg
We know each bag of cement is 50 kg
= (29.99/50) = 0.599 Bags
Calculation for Sand Volume
Sand = (6/7) x 0.1458 = 0.124m3
Density of Sand is 1450/m3
= 0 1450 = 181.2 kg
Now we find how many CFT (Cubic feet) Required
As we know that 1m3 = 35.31 CFT
= 0.124*35.31
= 4.37 CFT (Cubic Feet)
Reference : tutorialstipscivil.com