The post Civil Engineering Basic Knowledge – Every Engineer Must Know appeared first on The Civil Engineering.
]]>As civil engineers, it is our job to plan, construct, and maintain the infrastructure that allows our communities to function. We work on everything, including water treatment facilities, airports, roads, and bridges. Every site engineer should be familiar with a certain set of fundamental concepts, Civil Engineering Basic Knowledge and Practical construction knowledge to succeed in this industry.
Get started with the fundamentals of civil engineering basics and practical. This blog post provides an overview of the essential concepts, principles, and applications that form the backbone of the discipline. Learn about structures, materials, surveying, and about the Basic Knowledge of Civil Engineering in this comprehensive guide.
As a Civil Engineer, there is some fundamental practical knowledge that you should be aware of. Let’s go through each one of them to ensure that you have a clear understanding:
Aspiring builders and architects must have a clear understanding of building construction principles to ensure that their projects meet safety standards and are structurally sound. Here are some important factors to consider:
As a construction professional or site engineer, it is important to have a basic knowledge of the terminologies and standards used in the industry. Here are some key points to remember:
Lapping of bars (Reinforcement) | Diameter of the bar (Reinforcement) is less than 36mm. |
Circular RCC Column | Use Minimum of 6 longitudinal reinforcement bars |
Thickness of the slab should | should be Minimum is 0.125m |
PH value of water | should be More than 6 should be used for building purposes. |
The compressive strength of Bricks should | should be 3.5 N /mm^{2} |
Dead Load of structure | Self-weight of Structure (like Slab, beam-column, etc) |
Moisture content and | Sand that have more than 5% must not be used for Concrete mix. |
DPC (Damp Proof Course) | Thickness should not be less than 2.5cm |
RMC (Ready Mix concrete) | This type of concrete is made at a factory and transported to the site, making it ideal for construction sites with limited space or large amounts of concrete needed. |
Height of floor | The standard height for a floor is 3m or 10ft. |
Cantilever beam | This type of beam have one end fixed support and the other end is free. |
Simply supported beam | A minimum of two supports are needed |
PCC | This type of concrete is used when no tensile forces are acting on the member. |
Weight of first-class clay brick and crushing strength | 3.85 Kg and 10.5 MN/m^{2} |
Impermeability of concrete | It is concrete that resists the entry of water or moisture into it. |
Curing Time of RCC | 28 days |
Minimum Sill Level height | 44 inches |
Thermal expansion co-efficient of concrete and steel | The thermal expansion coefficient of concrete and steel is 12×10−6/°C. |
Number of Bricks necessary for 1m^{3} of Brick masonry | 550 bricks |
Specific gravity of cement The specific gravity of brick Specific gravity of sand |
3.16g/cm^{3}
2g/cm^{3} 2g/cm^{3} |
Unit weight of PCC
Unit weight of RCC Unit Weight of STEEL |
24KN/m^{3}
25 KN/m^{3} 7850Kg/m^{3} |
Volume of 50 kg cement bag | 1.3 cft |
TMT bars | Thermo Mechanically treated bars |
Length of each bar from factory | 12m |
Cement: Cement is produced by crushing limestone and clay gravel. It acts as a binder, facilitating the cohesion of cement, sand, and water. There are several important types of cement.
Aggregate: There are two main types of aggregate:
Concrete: Concrete is formed by combining cement, aggregate, and water according to a specific mix. The water-to-cement ratio plays a crucial role in determining the strength of the concrete, with a higher ratio resulting in weaker concrete.
The strength of concrete is indicated by a term such as M-25, where “M” stands for mix and “25” represents the compression strength at 28 days, measured using concrete cubes of 15 cm on each side.
Concrete has two types of setting:
The setting time of concrete is determined using a Vicat apparatus.
With the aid of admixtures, concrete setting time can be modified based on environmental factors. Retarders and accelerators are terms used to describe mixtures that affect how quickly the setting time passes.
Staircase construction is an important aspect of building design and requires careful consideration to ensure safety and functionality. Here are the minimum design requirements for staircase construction:
Fire Safety: Staircases must be built by all applicable fire safety regulations, including the provision of materials that are fire-resistant and a suitable means of escape.
Concrete Mixes |
Slump range in mm |
Columns and Retaining walls | 75-150 mm |
Beams and Slabs | 50-100 mm |
Cement Concrete Pavements | 20-30 mm |
Decks of Bridge | 30-75 mm |
Vibrated Concrete |
12-25 mm |
Huge Mass constructions | 25-50 mm |
Grades of Concrete | Proportion |
M-5 | 1:5:10 |
M-7.5 | 1:4:8 |
M-10 | 1:3:6 |
M-15 | 1:2:4 |
M-20 | 1:1.5:3 |
M-25 | 1:1:2 |
RCC Footing | 50 mm |
Top Raft Foundation | 50 mm |
Bottom/ Sides Raft Foundation | 75 mm |
Strap RCC Beam | 50 mm |
Grade Slab | 20 mm |
RCC Column | 40 mm |
Shear Wall | 25 mm |
RCC Beam | 25 mm |
Slab | 15 mm |
Flat Slab | 20 mm |
Stair-case | 15 mm |
Retaining Wall | 20 – 25 mm |
Water Retaining Structures | 20 – 30 mm |
Concrete | 25 kN/m^{3} |
Brick | 19 kN/m^{3} |
Steel | 7850 Kg/m^{3} |
Water | 1000 Lt/m^{3} |
Cement | 1440 Kg/m^{3} |
In Compression | 38 diameter |
In Tension | 47 and 60 diameter |
Enlist below are various techniques used for the curing of concrete:
Using cube samples that accurately reflect the concrete’s volume is essential when evaluating the compressive strength of the material. The quantity of cube samples needed to ensure accurate measurements varies depending on the amount of concrete being tested. The suggested cube sample sizes for various concrete volumes are listed below:
S. No | Volume of Concrete | Number of samples |
1 | from 1 to 5 m^{3} | 1 sample |
2 | from 6 to 15 m^{3} | 2 samples |
3 | from 16 to 30 m^{3} | 3 samples |
4 | from 30 to 50 m^{3} | 4 samples |
5 | 50 plus or above 50 m^{3} | 5 samples |
The weight of the steel (reinforcement) bar in meters per kilogram is given in the table below:
S. No | Dia of the bar in (mm) | Weight of steel in Kg per meter |
1 | 6 | 0.22 |
2 | 8 | 0.39 |
3 | 10 | 0.61 |
4 | 12 | 0.88 |
5 | 16 | 1.57 |
6 | 20 | 2.46 |
7 | 25 | 3.85 |
8 | 32 | 6.31 |
9 | 40 | 9.86 |
Note:
The formula is used in this table is D^{2}/162.162 in Kg/m. this formula is used when the dia of bar (reinforcement) is in mm and the length of the bar is in meters.
There are different members of formwork such as foundation, column, beam and slabs, etc. after poring of concrete the shuttering should be removed after some time the time of de shuttering is given below:
S. No | Members of structure | Days |
1 | for Sides of foundation, beam, columns and walls | 2 days |
2 | for Sides of slab under 4.5 meter span | 7 days |
3 | for Sides of slab above 4.5 meter span | 14 days |
4 | for Side of beams and arches up to 6 meter span | 14 days |
5 | for Side of beams between 6 meter to 9 meter span | 21 days |
6 | for Side of beams and arches above 9 met | 28 days |
In conclusion, the field of civil engineering is difficult and complex, requiring a wide variety of abilities. The topics we covered in this article are just a few of the basic and practical knowledge of civil engineering that every engineer should be familiar with. Engineers can create an infrastructure that is secure, effective, and sustainable by mastering these ideas and putting our communities’ needs first.
FAQ’s:
Civil engineers are responsible for planning, designing, and supervising the construction of various infrastructure projects. They assess the site’s conditions, carry out feasibility studies, create design plans, compute structural loads, monitor construction activities, guarantee that regulations are followed, and evaluate the structures’ long-term performance and durability.
A combination of technical, analytical, and communication skills are required for civil engineers. Success in this field depends on having strong mathematical and problem-solving skills, mastery of computer-aided design (CAD) software, familiarity with engineering concepts and building materials, an eye for detail, and effective communication abilities. Other valuable traits include the capacity for teamwork and project management.
Concrete, steel, wood, masonry, asphalt, and various composites are typical building materials. Steel is used for structural support in buildings and bridges, whereas concrete is used extensively in construction due to its strength and adaptability. In addition to masonry materials like bricks and stones being used for walls and facades, wood is frequently used in residential construction.
A key component of civil engineering is surveying. Collecting information for planning and building infrastructure projects involves measuring and mapping land and other physical features. Establishing property lines, figuring out elevation and topography, planning construction layouts, and ensuring that structures are placed precisely are all made easier by surveying.
The post Civil Engineering Basic Knowledge – Every Engineer Must Know appeared first on The Civil Engineering.
]]>The post Compound Wall Estimate Guide with Bar Bending Schedule appeared first on The Civil Engineering.
]]>If you are planning a building project, it’s important to get an accurate estimate of how much the compound wall will cost. Using bar bending schedule to plan is a useful tool that not only helps with budgeting, but also enables you to have better control over your project. Its also help to prepare estimating and costing of earthwork excavation and block masonry construction cost of wall.
A compound wall for a building is a wall that surrounds the outside of a building or collection of buildings, like an apartment building or a campus of office buildings. It acts as a boundary between private property from public areas.
Its major function is to define the boundaries of the property and to give residents privacy and security. Depending on the level of security necessary, compound walls can be constructed from a variety of materials, including concrete, brick, stone, or wood. They can also vary in height and thickness.
Let’s now calculate the price of a compound wall for a 40′ × 50′ site with a 10 ft gate as shown below:
We have provided the RCC columns at 10 ft. c/c.
The No. of columns required
= site perimeter ÷ c /c column’s distance
= [ ( 50 ft.× 2 nos.) + ( 40 ft. × 2 nos.)] ÷ 10 ft.
= 180 ft ÷ 10 ft.
= 18 nos. ( as shown below.)
The number of excavation pit required for the rcc column footing = 18 nos.
Let us provide 1.5 ft × 1.5 ft size footings, and the dimension of the pit to accommodate these footings shall be 2ft. × 2ft.× 2.5 ft. as shown in the drawing below.
The total volume of excavation for footing = Volume of Single Footing x Nos of Footings
= (2′ × 2′ × 2.5′) × 18 nos.
= 180 cft.
Let us excavate 4″ (inch) extra on both side of the plinth beam for formwork removal as shown in the drawing.
The volume of excavation for the plinth beam (RCC)
= [ perimeter of site – ( No. of footings × footing excavation width.)] × plinth excavation width × plinth excavation depth
= [180 ft. – (18 nos. × 2 ft.)] × 1.416 ft. × 0.9166 ft.
= 144 ft. × 1.416 ft. × 0.9166 ft.
= 186.90 cft.
The total excavation of earthwork for compound wall
= footing excavation + plinth excavation
= 180 cu ft + 186.90 cu ft.
= 366.90 cft.
Let us make a boulder soling of 9″ (inch) thick ( 0.75 ft.) for the footing as shown below.
The volume of soling for the footing = Volume of Soling per footing x No. of footings
= ( 2 x 2 x 0.75) × 18 Nos
= 3 cft × 18 Nos.
= 54 cft.
Let us prepare 4 inch thick ( 0.33 ft.) PCC bed for the footing & plinth beam.
volume of PCC for the footings:
The volume of PCC for the footings = Volume of PCC per footing x Nos of Footings
Vp1 = ( 2 ft x 2 ft x 0.33 ft) × 18 cft.
= 1.32 cft × 18 Nos.
= 23.76 cft.
As we know that the volume of excavation for the plinth beam is 186.90 cu ft.
From the above drawing, depth of excavation = 0.9166 ft.
PCC thickness = 4″ ( 0.33ft.).
Volume of PCC for the plinth beam
Volume of PCC for the plinth beam = X-Sec Area of PCC under Beam x [Beam Parameter – (Width of footing x Nos of Footings)]
Vp2 = (1.416 x 0.33) x [ 180 ft – ( 1.5 ft x 18 Nos)]
= 0.47 sft × 153 ft
= 71.91 cft.
= Vp1 + Vp2
= 23.76 cft.+ 71.91 cft.
= 95.67 cu ft.
Let us make PCC (Plain Cement Conc.) in the 1:2:4 mix.
Quantity of cement bags required for PCC (Plain Cement Conc.)
= 17.942 bags × ( 95.67 cft ÷ 100 cu ft.)
= 17.16 bags.
The volume of sand required for PCC
= 44 cu ft. × ( 95.67 cft ÷ 100 cft.)
= 42.09 cft.
The volume of aggregates required for PCC
= 88 cu ft. × ( 95.67 cft ÷ 100 cft.)
= 84.19 cft.
Let us make a RCC footing of size 1.5 ft × 1.5 ft. having 10″ ( 0.833 ft. ) thickness as shown in the drawing
Given data :
Footing length = 1.5 ft.
Width = 1.5 ft.
Thickness = 0.833 ft.
Rebar diameter = 10 mm.(0.0328 ft.),
Spacing = 5″ (0.416 ft. ) c/c
Cover = 2″ (0.166 ft ) on all the sides.
As we know that, the number of footings = 14nos.
= [total nos. × length × breadth × thickness]
= [18 nos. × 1.5 ft. × 1.5 ft. × 0.833 ft.]
= 33.74 cu ft. i.e. 0.95 cum.
No. of bars along the x-axis
= [ {( footing length ) – ( 2 × cover )} ÷ spacing ] + 1
= [ {( 1.5 ft.) – ( 2 ×0.166 ft.)} ÷ 0.416 ft.] +1
= [ { 1.168 ft. } ÷ 0.416 ft. ] +1
= 2.807 +1
= 4 nos.
( By rounding off )
No. of bars along the y-axis
= [ {( footing width ) – ( 2 × cover )} ÷ spacing ] + 1
= [ {( 1.5 ft.) – ( 2 ×0.166 ft.)} ÷ 0.416 ft.] +1
= [ { 1.168 ft. } ÷ 0.416 ft. ] +1
= 2.807 +1
= 4 nos.
( By rounding off )
Cutting length of the bar along the x-axis
= [ {bar length in x-axis } + { 2 nos. × ( L – bend length)}] – 2nos. × ( 2 times bar dia. for 90° bend.)
( we have deducted 2 times bar dia i.e. 2d for the 90° bend of the bar. )
= [ { footing length – 2 × cover } + 2nos.×{ footing height – 2 × cover}] – 2× ( 2 × bar dia. )
= [ { 1.5 ft. – 2 × 0.166 ft. } + 2 × { 0.833 ft. – 2 × 0.166 ft. } ] – 2 × ( 2 × 0.0328 ft.)
= [ 1.168 ft. + 1.002 ft. ] – 0.131ft.
= 2.17 ft. – 0.131 ft.
= 2.039 ft. i.e. 0.6214 m.
Cutting length of the bar along the y -axis
= [ {bar length in y-axis } + { 2 nos. × ( L- bend length)}] – 2nos. × ( 2 times bar dia. for 90° bend.)
= [ { footing width – 2 × cover } + 2nos. × { footing height – 2 × cover}] – 2× ( 2 × bar dia. )
= [ { 1.5 ft. – 2 × 0.166 ft. } + 2 × { 0.833 ft. – 2 × 0.166 ft. } ] – 2 × ( 2 × 0.0328 ft.)
= [ 1.168 ft. + 1.002 ft. ] – 0.131ft.
= 2.17 ft. – 0.131 ft.
= 2.039 ft. i.e. 0.6214 m.
Note: The cutting length & number of bars in both ( x-axis & y-axis ) directions will be the same, in the case of square footing having a similar bar diameter.
Now, we will prepare BBS (Bar Bending Schedule) of the footing, from calculated data.
sl bar dia. no. length total weight total
no. type mm. in m. length in kg/m weight
1. x- axis 10 4 0.6214 2.4856 0.62 1.54
2. y – axis 10 4 0.6214 2.4856 0.62 1.54
Total weight of the bars = 3.08 kgs
Add 2 % wastage = 0.0616 kgs
A grand total of rebar for a footing = 3.1416 kgs.
Note : Weight of 10mm dia bar /meter is 0.62 kg.
The total weight of the 10mm dia bar for all the footings
= [18 nos. × 3.1416 kgs.]
= 56.55 kgs.
Let us make a 9″×12″ RCC plinth beam as shown in the drawing.
Given data :
Plinth beam size = 9″× 12″
Main bar dia = 12 mm., no. of bars = 4 nos.
Stirrups 8mm @ spacing 150mm c/c , clear cover = 25 mm from all the sides.
= length × breadth × depth
= [ (perimeter of site – no.of columns × column width) × breadth × depth]
= [(180 ft – 18 nos. × 0.75 ft.) × 0.75 ft × 1 ft.]
= [166.5 ft. × 0.75 ft. × 1ft.]
= 124.875 cu ft. i.e. 3.53 cum.
Note: We have deducted the column width from the plinth beam length, as we have included them in the column volume.
The perimeter of site = 180 ft. = 54864 mm.
We will provide Ld at the corner rcc columns of the compound wall, passing the plinth bar through the intermediate columns.
The cutting length of the main bar
= [ (perimeter of site) + (8 nos × Ld ) + (4 nos. × lap length) – (8nos × column width) ]
Note: we assume that one overlapping for each bar & we will provide a lap length of 50d.
Let us provide development length Ld = 40d for the main bar.
= [ (54864 mm.) + (8 nos. × 40 × 12mm) + (4 nos × 50 × 12 mm) – (8 nos.× 230mm)]
= [54864 mm + 3840mm + 2400mm – 1840mm ]
= 59264 mm. i.e. 59.264 m.
Cutting length of the stirrup
= 2 nos. × (a +b ) + hook length – 90° bend – 135° bend
Where a = beam width – 2 × cover, & b = beam depth – 2 × cover
= 2 nos. × [ ( 230 mm – 2 × 25mm.) + ( 300 mm – 2 × 25mm ) ] + (10d ) – (3 nos. × 2d ) – (2 nos. × 3d)
Here, 10d is taken for hook length.
We have deducted 2d for 90° bend – 3nos., & 3d for 135° bend – 2nos. as shown in the above drawing.
= 2 nos. × [ ( 180 mm ) + ( 250 mm ) ] + (10 × 8mm) – ( 3 nos. × 2 × 8mm ) – ( 2 nos. × 3 × 8 mm.)
= 2 nos. × [ 430 mm ] + 80 mm – 48mm – 48mm.
= 860 mm + 80 mm – 96 mm.
= 844 mm i.e. 0.844 m.
Number of stirrups
= ( length of the plinth beam ÷ stirrup spacing ) + 1
where Length of Beam = Parameter – (0.75 x No. of footings) = 180 – (0.75 x 18)
here, length of the plinth beam = 166.5 ft = 50749 mm
= ( 50749 mm. ÷ 150 mm) +1
= 338.33+ 1
= 339.33 nos.
By rounding off, the no. of stirrups required = 340 nos.
Now, let us prepare a BBS (Bar Bending Schedule) table for the plinth beam.
sl. bar dia. no. length total weight total
no. ( mm) (m.) length kg/m weight
1. main bar 12 4 59.264 237.056 0.89 210.98
2. stirrups 8 340 0.844 286.96 0.395 113.35
Total weight of bars = 324.33 kgs.
Add 2% wastage = 6.49 kgs.
Grand total of rebars = 330.82 kgs.
Let us make a rcc column of size 9″ × 9″ as shown in the drawing.
Given data :
Column height above GL = 6 ft.+ 5″(0.416 ft.) = 6.416 ft., below GL = 7″ (0.583ft.)
Size of Column = 9″ × 9″ (228.6mm × 228.6 mm )
Longitudinal bars 12mm (0.03936 ft.) – 4 nos, cover – 40mm.
Lateral ties dia d1 – 6mm @ 6″ (150 mm.) c/c
From part 1, number of columns = 18 nos.
= total nos. × height × length× breadth
= total nos. × (height below GL + height above GL ) × length × breadth
= 18 nos. × ( 6.416 ft. + 0.583 ft. ) × 0.75 ft. × 0.75 ft.
= 18 nos. × 7 ft. × 0.75 ft × 0.75 ft.
= 70.875 cft. i.e. 2.0 cum.
Length of the longitudinal bar
= above GL + GL to footing top + development length ( Ld )
= 6.416 ft. + 0.583 ft. + ( 50d )
( we have taken Ld as 50d, where d = bar diameter.)
= 7 ft. + (50 × 0.03936 ft.)
= 7 ft. + 1.968 ft.
= 8.968 ft. i.e. 2.733 m.
Length of the lateral ties
= perimeter of lateral ties + total hook length – no. of bends
= 2 sides × ( a – 2 × cover ) + 2 sides × ( b – 2 × cover ) +( 2nos × hook length) – (3 nos. × bend )
( Here, we have taken hook length = 10d1 for 135°∠ & bend = 2d1 for 90°∟)
={ [ 2 × (228.6mm – 2 × 40mm.) ] + [ 2 × ( 228.6 mm – 2 × 40 mm.) ] } + { 2 × 10 × 6mm } – {3 × 2 × 6mm }
={ [ 2 × 148.6 mm ] + [2 × 148.6 mm ]} + 120 mm – 36 mm.
= {297.2 mm + 297.2 mm} + 84 mm
= 678.4 mm i.e. 0.678 m.
Total number of lateral ties ( stirrups )
={ [ length of the longitudinal bar – Ld ] ÷ stirrup spacing } + 1
Note: Ld is deducted from the length, as no stirrups are provided over that length.
= {[ 2733 mm – (50 × 12 mm )] ÷ 150 mm.} + 1
= {[ 2733mm – 600mm ] ÷ 150 mm.} + 1
= {2133 mm ÷ 150 mm.} + 1
= 14.22 + 1
= 15.22 nos.
Rounding off, the number of stirrups required = 15 nos.
Now, let us prepare BBS (Bar Bending Schedule) for a column.
sl. bar dia. no. length total weight total
no. ( mm) (m.) length kg/m weight
1. longitudinal 12 4 2.733 10.932 0.89 9.729
2. lateral 6 15 0.678 10.17 0.22 2.237
Total weight of bars = 11.966 kgs.
Add 5% wastage = 0.5983 kgs.
Grand total of rebars = 12.564 kgs.
The total weight of bars for compound wall columns
= 18 nos × 12.564 kgs = 226.152 kgs.
Let us build this compound wall of 6″ (inch) thickness having 6 ft. height as shown in the drawing.
The total length of the compound wall
= site perimeter – gate length
= (50ft. × 2 nos.) + (40 ft. × 2 nos.) – 10 ft.
= 100 ft. + 80 ft. -10 ft.
= 170 ft.
The total length of the block masonry wall
= compound wall length – (no. of columns × width of a single column.)
= 170 ft. – (18 nos.× 0.75 ft.)
= 170 ft. – 10.5 ft.
= 156.5 ft.
Height of the block masonry wall
= compound wall height – coping thickness
= 6 ft – 0.33 ft.
= 5.67 ft.
= length × height × thickness
= 156.5 ft. × 5.67 ft. × 0.5 ft.
= 433.68 cft.
Number of concrete blocks required
= 210 nos. × ( 433.68 cu ft. ÷ 100 cu ft.)
= 911 nos.
The number of cement bags required
= 1.038 bags × ( 433.68 cu ft. ÷ 100 cu ft.)
=4.501 bags.
The volume of sand required
= 7.634 cu ft. × ( 433.68 cu ft. ÷ 100 cu ft.)
= 33.12 cu ft.
Note: The above-given quantities are taken from the article “Calculating the quantity of materials in a 100 cubic ft. block wall.”
Let us make coping over block masonry work having 4″ (0.33 ft.) thickness in M15 grade.
= coping length × width × thickness
= block masonry length × masonry width × coping thickness
= 156.5 ft. × 0.5 ft. × 0.33 ft.
= 25.82 cu ft. i.e. 0.731 cu m.
Let us make this coping in M15 grade concrete.
The number of cement bags required for the coping work
=17.942 bags × (25.82 ÷ 100 cu ft. )
= 4.632 bags.
The total volume of sand required for coping
= 44 cu ft × (25.82 ÷ 100 cu ft. )
= 11.361 cu ft.
The total volume of coarse aggregates required
= 88 cu ft. × (25.82 ÷ 100 cu ft. )
= 22.722 cu ft.
Note: The above quantities for the calculation purpose is taken from “ Calculating the quantity of materials in 100 cu ft. & 1 cum. of M15 (1:2:4 ) grade concrete“.
Volume of Backfilling for plinth beam
picture
= Volume of excavation for plinth beam – beam PCC volume – plinth beam volume up to GL.
= [(186.90 cu ft.) – (52.33 cu ft.) – (124.875 cu ft. × 0.583 ft ÷ 1 ft.)
( By volume ratio with plinth beam 👆)
= 186.90 cu ft. – 52.33 cu ft – 72.802 cu ft.
= 61.768 cu ft.
Backfilling for footing:
=[ volume of excavation for footing – {footing vol. – soling vol. – PCC vol. – column vol. up to GL.}]
= [180 cu ft.- {33.74 cu ft. + 54 cu ft. + 23.76 cu ft.+ ( 18 nos × 0.583 × 0.75 × 0.75) } ]
= [180 cu ft – 117.40 cu ft.]
= 62.60 cu ft.
The total volume of backfilling
= backfilling for plinth beam + backfilling for footing
= 61.768 cu ft. + 62.60 cu ft.
= 124.368 cu ft. i.e. 3.52 cum.
First, we will sum up the total quantity of materials from all parts of this compound wall.
The volume of RCC for compound wall
= [for footings+ for plinth beam + for columns]
= [0.95 cum + 3.53 cum. + 2.0 cum.]
= 6.48 cum .i.e. 228.8 cu ft.
Let us make this RCC in M20-grade concrete.
The no. of cement bags required for RCC work
= [8.06 bags × ( 6.48 cum ÷ 1 cum )]
= 52.23 bags
The volume of sand required for RCC work
= [ 0.42 cum × ( 6.48 cum ÷ 1 cum )]
= 2.722 cum. i.e. 96.1 cu ft.
The volume of coarse aggregates required for RCC work
= [0.84 × ( 6.48 cum ÷ 1 cum )]
= 5.44 cum. i.e. 192.2 cu ft.
Note: The values i.e. directly added above is taken from “Calculating the quantity of materials in different grades of concrete“.
Now, we will sum up the total quantities of cement, sand, & aggregates required for the compound wall construction.
Note: The values i.e. mentioned below are taken from all above 8 parts of the series.
The total number of cement bags for compound wall.
= [for PCC + for masonry + for RCC + for coping]
= [17.16 + 4.501 + 52.23 + 4.632]
= 78.523 bags
The total volume of sand required in the compound works
= [for PCC + for masonry + for RCC + for coping ]
= [42.09 + 33.12 + 96.1 + 11.361]
= 182.671 cu ft.
The total volume of coarse aggregates required in the compound work.
= [for PCC + for RCC + for coping]
= [84.19 + 192.2 + 22.722]
= 299.11 cu ft.
The total weight of rebars required for the compound work.
= for footing + for plinth beam + for columns
= [56.55 kgs + 330.82 kgs. + 226.152 kgs.]
= 613.522 kgs.
In conclusion, preparing a compound wall estimate with a bar bending schedule is an essential part of the construction process. It ensures that the right amount of material and resources are utilized, making the construction process more efficient and cost-effective. A bar bending schedule also helps in ensuring the strength and stability of the wall, making it a crucial aspect of any construction project.
The post Compound Wall Estimate Guide with Bar Bending Schedule appeared first on The Civil Engineering.
]]>The post Flight of Stairs | How Many Flight of Stairs per Floor | Design Criteria appeared first on The Civil Engineering.
]]>Need help figuring out how many flights of stairs you should use per floor? This guide provides easy-to-follow steps and explanations for calculating the correct number of steps in a flight of stairs, we will also discuss the meaning, definition, benefits, design criteria and how tall it is!
The term “stair” refers to a flight or series of steps that connect one floor to another. It is designed to provide simple and quick access to multiple floors.
A stair’s steps can be constructed as a flight of open, horizontal treads with room in between (like a ladder or foot-over bridge) or as closed steps with a vertical face between the treads, known as the riser. A staircase is an enclosure or area of a building with stairs.
What is Flight of Stairs Meaning:
The Flight of Stairs refers to a series of steps or a staircase that leads from one level (floor) of a building to another.
It can also refer to a set of stairs that connects multiple levels in a building, such as in a multi-story structure. The term can also be used to describe a group of stairs in an outdoor setting, such as a set of stairs that lead to a lookout point or the top of a hill.
They can be made of a variety of materials, including wood, concrete, and stone, and can be found in both residential and commercial buildings. They are used to provide access to different levels of a building and are typically located inside, but can also be found outside.
A Stairway or set of steps connecting one floor or landing to the next.
Between the landings, the flight is made up of a continuous staircase of stairs. If there are too many stairs (or steps) in one flight without landings between them, it can be tedious to climb and confusing to walk down, and the likelihood that a fall will result in serious injury is increased.
The number of steps required in a flight of stairs depends on several factors including the height of the floor and the desired height of each step.
The standard height of a stair step is 7 inches (17.78 cm) and the standard height of a floor is 8 feet (2.44 meters). This means that a flight of stairs that connects two floors with a height difference of 8 feet should have a minimum of 11 steps (8 feet / 7 inches per step = 11.428 steps).
Most of the flights of stairs average out at 12 steps or 13 steps. The precise number of steps, however, will depend on the needs of the building or structure, as well as factors like the width of the stairwell, the presence of landings, and the demand for accessibility.
The right number of steps for a particular flight of stairs should be determined by consulting an architect or building code specialist.
To find the number of flights of stairs per floor in a building, you can do the following:
It’s important to note that the method above is only an estimate and the actual number of stairs may vary.
Alternatively, you can use the building blueprints or architectural drawings to count the number of flights of stairs per floor.
There are several design criteria that must be considered when designing a staircase, including:
The height of the floor is generally known. The procedure for determining the number of treads and risers is as follow:
The positions of first and last risers are determined with regard to the positions of doors, windows and internal circulation area.
A convenient height of the riser is assumed.
Number of risers = Total floor height/Height of riser
Number of treads in a flight = number of risers – 1
Provision of headroom is must. Should preferably be not less than 2m.
It is not desirable to provide a flight with more than 12 steps or at the most 15 steps and not less than 3 steps.
Suitable landings should be provided for user’s comforts. The placement of a landing after a certain number of steps is regulated by statutory instruments and serves to guarantee improved safety.
Should be avoided as it is a discomfort in circulation.
Should be avoided. If at all required should be designed properly.
When a flight consists of more than three steps should be provided with a handrail. If the width of the stair is more, should be provided at both ends. In public buildings for wider steps should be provided in the center also. The height of the handrail should not be less than 80 cm.
The stairway’s flight of stairs has the following benefits or advantages:
One flight of stairs can conserve space in a house and also add to its architectural appeal because long, straight lines create nice sightlines in a room.
The ceiling height in a house is typically between 8 and 10 feet, with 8 feet being the most typical.
Houses with these high ceilings almost always have at least one flight of stairs, unless you have a split-level home.
In a typical home, there isn’t enough room for a landing and it isn’t necessary, therefore flights of stairs are obviously more common.
Landings are always only seen in buildings with more space between floors, even if long staircases may need them owing to space restrictions.
One straight run or stairway might be appropriate, but it might not be practical in your home.
A staircase can become a focal point by combining a consistent aesthetic with remarkable finish elements.
A landing is not necessary to break up a flight of steps that has been completed adequately.
The use of a landing may be necessary, in which case there would be two distinct “flights” of stairs as opposed to one.
If you choose floating stairs or adequately finished hardwood treads, a flight of steps can become a focal point in your home.
Calculating the number of steps required for a flight of stairs can be done in various ways, but the most straightforward method is to consider the overall height of the floor, including the width of the joists supporting the floor above and the thickness of the subfloor.
For example, in a house with 8-foot ceilings, a flight of stairs with a step rise of 7 ¾ inches would require 14 steps.
If the ceilings are 9 feet, then 16 steps would be needed with a step rise of 7 ¾ inches.
And if the ceiling is 10 feet, then a flight of stairs would require 17 steps with a step rise of 7 ¾ inches.
For example, your house has 8feet ceilings, with 10inches wide joists supporting the floor above.
To calculate the total number of steps for a flight of stairs with 8 feet ceilings and 10 inches wide joists supporting the floor above, you can use the following formula:
Total number of steps = (ceiling height (in inches) + joist width (in inches)) / height of each step (in inches)
If the height of each step is 7 inches and the joist width is 10 inches, the total number of steps for an 8-foot ceiling would be:
(8 feet x 12 inches/foot + 10 inches) / 7 inches = 113 inches / 7 inches = 16.14 steps
It’s important to note that this is an estimate, and the actual number of steps may vary depending on the specific design and construction of the stairs. Therefore, it’s best to consult with an architect or building code expert to determine the appropriate number of steps for a specific flight of stairs.
For this case, you will have to round up the number to 17 steps.
The length of a flight of stairs depends on several factors including the number of steps, the height of each step, and the width of the stairs.
Standard stair tread width is 10 inches (25.4 cm), and step height is 7 inches (17.78 cm) and the no. of steps are 12.
You can use the following formula to determine the length of a flight of stairs:
Formula: Waist Slab Length or Length of Flight of Stairs = c2 = (a2+b2)
For Example:
The Total Rise of Stair = A = Size of Riser x (Nos. of Steps)
The Total Rise of Stair = A = 7 x 12 = 84 in
The Total Run of Stair = B = Size of Tread x (Nos. of Steps)
The Total Run of Stair = B = 10 x 12 = 120 in
Waist Slab Length or Length of Flight of Stairs = c2 = (a2+b2)
Waist Slab Length = c2 = (842 + 1202)
Waist Slab Length = c = √(21,456)
Waist Slab Length = c = 146.48 in = 372.06 cm
It’s important to remember that the formula above is only a rough estimate, and the precise design and construction of the steps may affect how long a flight of stairs actually is.
The length of the flight of stairs will also vary depending on the stairway’s width; the broader the stairway, the longer the flight will be.
The height of a flight of stairs, also known as the total rise, is the vertical distance between the floor level of the starting and ending point of the stairs. It is determined by the number of steps and the height of each step.
A standard step height is 7 inches (17.78 cm) and the standard height of a floor is 8 feet (2.44 meters).
To calculate the height of a flight of stairs, you can use the following formula:
Formula: Height of flight of stairs = (number of steps x height of step)
For example, a flight of stairs with 14 steps and a step height of 7 inches would have a height of: 14 x 7 inches = 98 inches (2.489 meters)
It’s important to note that the formula above is an estimate and the actual height of a flight of stairs may vary depending on the specific design and construction of the stairs.
It’s also important to mention that, if the flight of stairs is not straight, but with some turns or landings the height will change, it’s important to measure the height of each section separately and then sum them up to get the total height of the flight of stairs.
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]]>The volume calculation is one of the most important steps in constructing a staircase. Without knowing the right stair formula you may not be able to calculate the right amount of concrete needed for your staircase. Also discuss about the stairs quantity takeoff in excel calculator, concrete for steps, dog legged staircase, cost estimation of staircase with landing and much more.
Before, reading about “How to Calculate Quantity of Concrete Volume for Staircase” you should read first about the Introduction of Dog Legged Staircase.
A staircases is a part of a building that helps us move from one floor to another. It’s an important part of many buildings. The sizes of staircases vary, depending on the kind of building they’re meant to fit into.
Definition of Staircase: A flight or series of flights of steps and a supporting structure connecting separate levels.
A dog legged Staircase is a stair configuration that has a quarter-landing before turning at a right angle and continuing upwards. The flights do not have to be equal, and frequently are not.
A stair’s slab that slopes upward from the floor slab to the landing slab is referred to as the waist slab.
The series of steps from floor to the landing.
The transitional level between flights.
The step is made up of the tread and the riser.
The flat area you step on is known as the tread.
The vertical (up and down) portion of a stairway between each tread is known as a riser.
We need to calculate concrete volume for staircases for each component separately, then add them all up to find the total volume of concrete for dog legged staircases.
From the drawing:
Tread = 10″ = 0.833 ft
Riser = 6″ = 0.5 ft
Height of the Flight = 12 ft
Length of the Landing = 8′ – 6″ = 8.5 ft
Width of the Landing = 3 ft
Thickness of the Landing = 6″ = 0.5 ft
Length of Flight or Waist Slab = ?
Width or Length of Steps = 4 ft
No. of Riser = Height of the Flight / Riser = 12 / 0.5 = 12 Risers
No. of Treads = No. of Risers – 1 = 12 – 1 = 11 Treads
The Volume of Concrete for one step = Area of one Step x Length of Step
As you know that the shape of stair step is right angled triangle so we know the formula for the area of right angled triangle is:
Area of One Step = 1/2 x riser x tread
The Volume of Concrete for one step = 1/2 x riser x tread x Length of Step
Volume of Concrete for one step = 1/2 x 0.5 x 0.833 x 4
The Volume of Concrete for one step = 0.833 cft
Therefore, the total volume of concrete required for steps on first flight = Volume x No. of Steps
= 0.833 x 11 = 9.163 cft
The above calculation is only for one flight. We know that the second flight are having the same measurements:
so, Total Concrete = First Flight Concrete Volume x 2 = 9.163 x 2 = 18.326 cft
As per the given Plan,
Length of the Landing = 8′ – 6″ = 8.5 ft
Width of the Landing = 3 ft
Thickness of the Landing = 6″ = 0.5 ft
Volume of Landing = Length x Width x Thickness = 8.5 x 3 x 0.5 = 12.75 cft
As we know that, It is the right angled triangle in order to find the inclined length we use the Pythagoras theorem;
Inclined Length = √ (Horizontal Length)² + (Height)²
Horizontal Length = Tread Size X No. of Treads = 0.833 x 11 = 9.163 ft
As we know that Height = 12 ft, so,
Inclined Length = √ (Horizontal Length)² + (Height)²
– Inclined Length = 15.09 ft
Concrete Volume of Waist Slab = Inclined Length x width of Slab x Thickness of Slab
Concrete Volume of Waist Slab = Inclined Length x width of Slab = 15.09 x 4 x 0.5 = 30.18 cft
As we know that there are two flights with two waist slabs so,
Total Volume of Waist Slab = Volume of Waist Slab x 2 = 30.18 x 2 = 60.36 cft
Staircases Concrete Volume = Steps Volume + Landing Space Volume + Waist Slabs Volume
Staircases Concrete Volume = 18.326 + 12.75 + 60.36 = 91.436 cft
Staircase Concrete Volume = 91.436 cft
Now we will find the material analysis of dog legged staircases concrete;
Wet Volume of Staircases Concrete = 91.436 cft
Dry Volume of Staircase Concrete = 91.436 x 1.54 = 140.81 cft
Ratio of Concrete = 1 : 1.5 : 3 (c : s : a)
Sum of Ratios = 1 +1.5 + 3 = 5.5
so, Cement Content in Concrete = Dry Volume of Concrete / Sum of Ratios x Ratio of Cement
Cement Content in Concrete = 140.81 / 5.5 x 1 = 25.60 cft
As we know that, 1 Cement Bag (50 kg) = 1.25 cft
No. of Cement Bags = 25.60 / 1.25 = 20.48 = 21 Bags (Say)
Sand Content = Dry Volume of Concrete / Sum of Ratios x Ratio of Sand
Sand Content = 140.81 / 5.5 x 1.5 = 38.40 cft Sand
Stone Chips or Aggregates = Dry Volume of Concrete / Sum of Ratios x Ratio of Aggregates
Stone Chips or Aggregates = 140.81 / 5.5 x 3 = 76.80 cft Aggregate
Now lets calculate the water content of staircase concrete. Suppose, water-cement ratio for staircase concrete is specified 0.45.
That means, water/cement = 0.45, or W/C = 0.45
for 1 bag cement, water is, = 0.45 x 1.25 (as we know, 1 bag cement equal to 1.25 cft),
Water = 0.5625 cft.
As we know 1 cubic feet water is equal to 28.31685 litre,
So we can write, water = 0.5625 x 28.31685 = 15.92 litre, say, 16 litre.
So One bag of Cement needs 16 liter of water for 0.45 W/C ratio.
Required Water Content for Staircase Concrete = Req Quantity of Cement Bags x 16 litres
Total Required Water Content for Staircase Concrete = 21 x 16 = 336 litres
Cement = 21 Bags
Sand = 38.40 cft Sand
Aggregates = 76.80 cft Aggregate
Water = 336 litres
Cement = 21 Bags x 1070 RS/Bag = 22,470 RS
Sand = 38.40 cft x 65 RS/cft = 2,496 RS
Aggregates = 76.80 cft x 50 RS/cft = 3,840 RS
Water = 336 litres x 2.50 RS/litre = 840 RS
Total Cost = 22,470 + 2,496 + 3,840 + 840 = 29,646 RS
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