Midland Asphalt Materials Inc is seeking a project estimator in Bloomsburg, PA location

Midland Asphalt Materials Inc, a renowned company that deals with road resurfacing technology and pavement maintenance, is inviting application for the post of Project Estimator for the office located at Bloomsburg, PA.

The candidate has to develop cost estimation for projects (from initial to completion stage), modify the construction projects, and analytics to help in managing project development process efficiently.

Duties & Responsibilities:

• Scrutinize documents for creating time, cost, materials, and labor estimates.
• Arranges and organizes logistics, acquisition, and allotment of materials.
• Evaluates cost effectiveness of projects or services, finds the true costs toward the progression of the project.
• Assesses constructability/production analysis of bid documents together with specifications, site conditions, contracts and building codes.
• Make estimates for the management team toward planning, organizing and scheduling work, choosing vendors or subcontractors.
• Get in touch with clients, vendors, personnel in other departments or construction operations to converse and prepare estimates and find solution for issues.
• Execute components of project plan that is essential prior to construction work crews visit the jobsites.
• Formulate estimates out of different types of bid documents and their major components together with components like bonding, insurance, indemnification, and damage clauses.
• Participate with pre-bid and job site meetings; Carry outs and/or joins pre-construction, progress and other project/activities and any staff meetings
• Locate and report real time cost data tracking real cost costs, schedules corresponding to bid proposals with the progression of the project.
• Make sure that any relevant federal/state prevailing wage rates guidelines are considered, supervised, enforced and retained per project bid requirements.
• Accomplishes other duties as necessary or assigned.

In order to submit your application, send CV and salary requirements tomidland.resumes@midlandasphalt.com 


Eligibilities:


• Posses BS Degree in Engineering or Construction Management preferred.
• Atleast 1-2 years’ experience of project cost estimating or working in a similar capacity.
• Must abide by the company Safety and Loss Prevention Program.


Other perks include medical, vision, dental and life & disability insurance.





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Published By
Rajib Dey
www.constructioncost.co
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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 2x3 export/import.

• Structural analysis for finite element.

Download a free trial Autodesk Advance Concrete 2016

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Published By

Rajib Dey

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

Rajib Dey

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

Rajib Dey

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

Rajib Dey

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
Rajib Dey
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

Rajib Dey

www.quantity-takeoff.com

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