How to generate reinforcement drawing in Autodesk Advance Concrete 2016

Md Humayun Kabir has made this nice tutorial for construction professionals. In this tutorial, Kabir shows some useful tips for creating reinforcement drawing in Advance Concrete 2016 developed by Autodesk. Advance Concrete 2016 is used to model reinforced concrete structures instantly. It supports Windows 7 64-bit and Windows 8 64-bit.

This construction program contains the following features :-

• Modeling structural elements like Column, Beam, Slab, Isolated footing, Continuous footing, Pile as well as Stairs, Roofs.

• Contain material & section libraries with several concrete, masonry and full bricks materials as well as parametric sections.

• Create view automatically that includes plan views, elevation views, cross section views, isometric views, reinforcement views etc.

• Create reinforcement drawing efficiently for manual reinforcement like stirrups, straight bars, polygonal bars, meshes etc.

• Create documents like organizing drawings on layout, inserting frame & title block, generating BOM for concrete volume & formwork area.

• BIM integration with Graitec compatible software and Revit, IFC 2×3 export/import.

• Structural analysis for finite element.

Download a free trial Autodesk Advance Concrete 2016

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Published By

Arka Roy

www.constructioncost.co

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Detailed Process for effective slab reinforcement

This construction video will highlight the detailed process for complicated slab reinforcement. All steel bars are connected following the design details. Here the concrete ratio is 1:2:4. This video is very useful for Civil Engineering students as well as various structural engineering professionals.

The slabs containing ratio of longer length to its shorter length (Ly/Lx) larger than 2 is known as one way slab or else as two way slab. In one way slab main reinforcement is equivalent to shorter direction and the reinforcement equivalent to longer direction is defined as distribution steel. In two way slab main reinforcement is rendered alongside both direction.

Detailing Requirements of RCC Slab as per IS456: 2000

a) Nominal Cover:

For Mild exposure – 20 mm

For Moderate exposure – 30 mm

However, if the diameter of bar is less than 12 mm, or cover may be decreased to 5 mm. Therefore for core reinforcement up to 12 mm diameter bar and for mild exposure, the nominal cover will be 15 mm.

b) Minimum reinforcement: The reinforcement in either direction in slab must not be lower than

• 0.15% concerning the total cross sectional area for Fe-250 steel

• 0.12% concerning the total cross-sectional area for Fe-415 & Fe-500 steel.

c) Spacing of bars: The utmost spacing of bars must not surpass

• Main Steel – 3d or 300 mm whichever is lesser.

• Distribution steel –5d or 450 mm whichever is lesser Where, ‘d’ denotes the operative depth of slab. Note: The least clear spacing of bars must not be lower than 75 mm (Preferably 100 mm) even if code does not suggest any value.

d) Maximum diameter of bar: The maximum diameter of bar in slab, must not surpass D/8, where D denotes the total thickness of slab.

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Published By

Arka Roy

www.quantity-takeoff.com

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Detailed Process for effective slab reinforcement

This construction video will highlight the detailed process for complicated slab reinforcement. All steel bars are connected following the design details. Here the concrete ratio is 1:2:4. This video is very useful for Civil Engineering students as well as various structural engineering professionals.

The slabs containing ratio of longer length to its shorter length (Ly/Lx) larger than 2 is known as one way slab or else as two way slab. In one way slab main reinforcement is equivalent to shorter direction and the reinforcement equivalent to longer direction is defined as distribution steel. In two way slab main reinforcement is rendered alongside both direction.

Detailing Requirements of RCC Slab as per IS456: 2000

a) Nominal Cover:

For Mild exposure – 20 mm

For Moderate exposure – 30 mm

However, if the diameter of bar is less than 12 mm, or cover may be decreased to 5 mm. Therefore for core reinforcement up to 12 mm diameter bar and for mild exposure, the nominal cover will be 15 mm.

b) Minimum reinforcement: The reinforcement in either direction in slab must not be lower than

• 0.15% concerning the total cross sectional area for Fe-250 steel

• 0.12% concerning the total cross-sectional area for Fe-415 & Fe-500 steel.

c) Spacing of bars: The utmost spacing of bars must not surpass

• Main Steel – 3d or 300 mm whichever is lesser.

• Distribution steel –5d or 450 mm whichever is lesser Where, ‘d’ denotes the operative depth of slab. Note: The least clear spacing of bars must not be lower than 75 mm (Preferably 100 mm) even if code does not suggest any value.

d) Maximum diameter of bar: The maximum diameter of bar in slab, must not surpass D/8, where D denotes the total thickness of slab.

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Published By

Arka Roy

www.quantity-takeoff.com

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Strength of Doubly Reinforced Concrete Beam (compression steel not yielding)

Some useful construction tips to calculate the nominal flexural strength Mn concerning the reinforced concreterectangular section. It is illustrated in figure 9-3(a).

The provided section is reinforced twice with steel in tension and compression zone of the section. The guidelines of ACI-318 is followed for calculating the nominal flexural strength Mn. The highest value of applicable strain at the utmost concrete fiber is supposed to be 0.003.

For fc’ grerater than 4000 psi the value of ß1 is calculated as given below;

ß1 = 0.85 – 0.05 {( fc’ -4000)/1000} = 0.8

Assume that the compression steel has yielded when the strength is reached (strain in concrete is 0.003).

Given that tension steel consists of 4 bars of #10 (dia 1.27 in.).

Area of one bar of #10 = 1.27 in2.

As = 4-#10 bars. = 4 (1.27) = 5.08 in2.

Area of compression steel , As’ = 2-#7 = 2(0.6) = 1.2 in2. The internal forces acting on the section shown in figure 9-3(c) are calculated as given below;

Cc = 0.85 fc’ ba = 0.85 (5) (14) a = 59.5 a

Cs = (fs’ – 0.85 fc’) As’ = (60 – 0.85*5) 1.2 = 66.9 kips

T = As fy = (5.08) (60) = 304.8 kips

Applying static equilibrium, we get Cc + Cs = T;

59.5 a + 66.9 = 304.8

therefore depth of stress block, a = (304.8- 66.9)/59.5 = 3.998 in Depth of neutral axis x = a / ß1 = 3.998/0.8 = 4.997 in

 

Read More: civilengineer.webinfolist.com/

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Published By

Arka Roy

www.quantity-takeoff.com

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How to estimate a Civil Engineering Project

This construction video will provide step-by-step guidance for creating estimates for various Civil Engineering Projects.

The estimate includes detailed as well as abstract estimation.

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Published By

Arka Roy

www.constructioncost.co

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Brief overview of Pin-jointed Truss and different Analysis methods

Truss belongs to a structure that contains straight members to build up one or more triangular units. Generally Pin-joints are used to tie the members of truss at the end. The joints concerning a truss are defined as nodes. External forces and reactions can only function at the nodes and produce forces in the members to be axial forces (tensile or compressive).

If all the members and nodes are positioned inside a two dimensional plane, it is called plane truss. If a truss consists of members and nodes and expands into three dimensions, it is called space truss.

Trusses are inherent parts of various structures like bridges, roof supports, transmission towers, space stations etc.

Truss are defined as:-

a) Statically establish; all the unidentified forces (support reaction and member forces) can be specified with the use of equations of static equilibrium. (provided m+r = 2j).

b) Indeterminate; equations of static equilibrium can’t solely find out unknown forces, provided m+r> 2j.

c) Unstable; not appropriate to bear load; provided m+r < 2j; Where ‘m’ denotes member’s no in a truss; ‘r’ denotes number of reaction components; And ‘j’ denotes number of joints in a truss;

Analysis methods:

Two methods are available for determining the forces in the members concerning a truss;

(i) Methods of Joints : This method focuses on the equilibrium of the all joints concerning the truss. There exists two equations of static equilibrium ?Fx & ?Fy.So one has to start with the joint containing lower than 2 unknown forces. Go through problem 3-1 (civilengineer.webinfolist.com)

(ii) This method is useful while finding out the forces in a few members. Under this method, a fictitious is passed over the members in which forced should be set and to obtain unidentified forces, select the equilibrium of the left hand side or the right hand side of the truss.Go through problem 3-2 (civilengineer.webinfolist.com)

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Published By

Arka Roy

www.quantity-takeoff.com

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Some useful construction tips for laying brick columns & soldier course

The following construction video is presented to provide some useful tips for constructing a brick column by applying some effective brick laying patterns & soldier course.

Brickwork is masonry constructed with a bricklayer. Generally, rows of bricks are termed as courses which are placed on top of one another horizontally with equally sized bricks to construct a brick wall.

The word course is more or less applied along with unit masonry like brick, cut stone, or concrete masonry units (“concrete block”).

Video How To Lay Brick Columns and Soldier Courses

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Published By

Arka Roy

www.constructioncost.co

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Make precious design calculation with RC-spreadsheets package version 4B.2

RC-spreadsheets package version 4 are designed by adhering to both Eurocode 2 and BS 8110-1:1997. The most updated version alias 4B.2 contains enhancements regarding to punching shear, column design and pilecap design as well as resolving bug.

The advanced engineer will be greatly benefitted by obtaining the transparent and precise design calculations.

The post-graduates and newbie engineers can easily gather knowledge on concrete design and familiar with analyzing ‘what if’ scenarios.

The individual user can solve their queries by pursuing through the cells to recognize the logic applied.

 download 

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Published By

Arka Roy

www.quantity-takeoff.com

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A position is vacant for Senior Engineer in Houston, Texas

Bechtel, the globally recognized engineering, project management, and construction company, is seeking a Senior Project Engineer on permanent basis. The location of the job will be Houston, Texas, United State.

Job Role:

• Plans, schedules, manages, and organizes complete level of engineering work for a global business unit generally in one discipline in a project or staff group. Accomplish work for customary engineering practice but may involve complicated features like settling conflicting design requirements, inadequacy of conventional materials and/or difficult coordination requirements.

• Accomplishes work by applying Bechtel Standard Application Programs (BSAPs) and proper 2-D and 3-D computer aided design and computer aided engineering (CAD and CAE) tools.

• Plans, coordinates or produces equipment or work specifications, bid evaluations and award recommendations for equipment.

• Organizes engineering efforts in areas entrusted with specialty and other engineering groups or disciplines, together with the client, suppliers, and contractors and between other groups.

• When delegated, undertakes a lead technical role over other engineers or project subgroups for executing particular tasks.

• Provides support in on-the-job-training of assigned personnel as well as input for their performance evaluations.

• Drafts letters independently to vendors and clients.

• Evaluates bid analyses and makes recommendation.

• Composes or supports in preparing conceptual studies, designs, reports or proposals.

• Accomplishes or helps in solving problem analysis and original design.

Go through the following link to apply for the job jobs.bechtel.com

• Prepares as well as offers support in generating cost estimates, quantity take-offs and staffing requirements for proposals, forecasts and change orders.

• Evaluates and verifies work of subordinate engineers.

• Supports the formation of a work environment that cultivates ingenuousness, faith, communication, teamwork, empowerment, innovation, and satisfaction.

 

Image Courtesy: suproindia.com

 

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Published By

Arka Roy

www.constructioncost.co

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How to use ETABS 2015 for designing a Reinforced Concrete Building efficiently

This is an exclusive construction video that demonstrate how to use ETABS 2015, the incorporated software package toward the structural analysis and design of buildings, to create the design of a reinforced concrete building having ramp and wall.

This most updated version of construction software contains superior 3D object based modeling and visualization tools, ardently rapid linear and nonlinear analytical functionalities, advanced and expansive design capabilities for unlimited materials, and insightful graphic displays, reports, and schematic drawings which facilitate the users to instantly and precisely resolve and understand analysis and design results.

We Provide Videos Tutorial for Civil Engineering Relating to Structural Detailing, Structural Modeling, Analysis and Design Software such as: CSI ETABS, CSI SAP2000, CSI SAFE, CSI Column, Autodesk Robot Structural Analysis, Autodesk Revit Structure, Autodesk AutoCAD Structural Detailing, AFES Foundation Design Software, Mathcad for Civil Engineering, spColumn, RCM ACI Builder and ACCOMS GEAR and GRASP.

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Published By

Arka Roy

www.constructioncost.co

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