Monday, November 28, 2016

Recruitment of site Engineers ( both fresher/experienced)

Recruitment of site(CIVIL) Engineers:-

                                     Baffin engineering projects ltd, a Central Public Sector Enterprise, under the control of Ministry of Heavy Industries & Public Enterprises is engaged in Civil Construction/ Bridge Construction projects all over India and urgently requires the following Executives:
Sl.NoPostNo. of PostAge
Maximum
Experience
(Minimum)
1.Site Engineer25135 yearsFresher
2.Manager
(Material Management)
0145 years12 Years
3.Deputy Manager
(Civil)
02
SC/ST-01, Gen-01
40 years08 Years
4.Asstt. Engineer
(Civil)/ (Mechanical)
04
Civil-03, Mech-01
34 years02 Years

Criteria:
Qualification:
Sl. No. 1: Full time B.E/ B. Tech/ A.M.I.E. in Civil/Mechanical/Electrical/ Electronics/Instrumentation & Control Engineering from a recognized University/Institution.

Sl. No. 2: Graduate in any discipline. Full time Degree/ Diploma in Material Management from a recognized University/ Institution. Graduate in Civil Engineering preferred.

Sl. No. 3: Full time B.E/ B. Tech/ A.M.I.E. in Civil Engineering from a recognized University/ Institution.

Sl. No. 4: Full time Diploma in Civil/ Mechanical Engineering from a recognized University/ Institution.

Experience:
Sl. No. 1: Freshers . (Only 2015-2016 Pass out Candidate can apply)

Sl. No. 2: Experience in PSU/ large Construction Company preferably in bridge/building projects.

Sl. No. 3: Experience in PSU/ large construction Company in the field of steel/ concrete bridges (erection of bridge girders, fabrication and erection of structural steel work for industrial buildings etc.). Experience at Civil construction sites particularly (bridge/building) projects is essential.

Sl. No. 4: Working experience in PSU/ Large Construction Company in the field of Steel/ Concrete Bridge (Structural fabrication) /Civil Work/ Foundation work etc.

Method Of Selection 
  • I.The recruitment shall comprise of two stages – Written Examination followed by Personal Interview.

  • II.The Written examination shall be in the respective subject in the form of an objective type paper.

  • III.Only those candidates who qualify in the Written examination will be called for personal interview.


Scale of Pay: 
Rs.35,000/- To 65,000/- p.m + HRA and medical allowance as per Company norms. 

Important Dates: 
I. Last Date of Application: 30th November 2016.(Wednesday) 
II. Date of Written Examination: 18th December 2016.(Sunday) 
II. Admit card download date starts from: 16th December 2016.(Friday) 

Application Fees: 
I. Application Fees is Rs.1550/- .(Refundable in case of disqualifying examination.) 
Bank Name:State Bank of India
Account No:36190521461
Account name:BAFFIN P N T
IFSC CODE:SBIN0012022
General Instructions: 
1)The Instructions are to be read carefully before proceeding further. The candidate has to select the particular post for which he/she intends to submit application by clicking the tab “Apply Online” at the bottom of this page. 

2)The candidate has to fill in the required details against the various items in the Form,with correct information. 

3)Candidates should upload the scanned (digital) image of their photograph and signature as per the process given below. They should note that only jpg/ jpeg format is acceptable: 

Photograph Image:
Photograph must be a recent passport size colour picture not older than 6 months [in JPG or JPEG format only, size limit 50KB– 100KB]. 

Signature Image: 
The signature must be signed only by the applicant and not by any other person.[Size of file should be between 10kb – 20kb in jpg/png format] 

Deposit Slip : 
The applicant has to submit payment proof as the deposit slip collected from the bank.[Size of file should be between 100kb – 400kb in jpeg/png/pdf format] 

NOTE : 
After applying the online Application form successfully the applicant receive an email on his/her registered email id,which contains the information about the information about written examination.

No TA will be paid to the candidates for appearing in the test and interview. 
http://online.baffineng.in/

                                            APPLY NOW


Friday, November 25, 2016

CIVIL ENGINEERING QUESTION & ANSWERS

CIVIL ENGINEERING QUESTION & ANSWERS:-

                                                                       SET - I
Q1).The process of finishing the joints with mortar of rich mix in brick and stone masonry is called
(1) facing
(2) pointing
(3) plastering
(4) guniting
Answer:  (2) pointing

Q2). The tapering steps which are provided to change the direction of a stair is
(1) Landing
(2) Spiral steps
(3) Winders
(4) Going
Answer:  (3) Winders

Q3). The volume of one bag of cement weighing 50 kg is
(1) 0.05 m3
(2) 0.0245 m3
(3) 0.0345 m3
(4) 0.0445 m3
Answer: (3) 0.0345 m3

Q4). A window projecting outward from the walls of a room is termed as
(1) Dormer window
(2) Louvered window
(3) Bay window
(4) Skylight
Answer:  (3) Bay window

Q5). Suggest a most suitable foundation when the loads are heavy and clay soil is soft with a basement floor
(1) Combined footing
(2) Isolated footing
(3) Raft footing
(4) Wall footing
Answer: (3) Raft footing

Q6) The ultimate load cannot be clearly recognized in case of
(1) Local shear failure
(2) General shear failure
(3) Punching shear failure
(4) None of these
Answer: (3) Punching shear failure

Q7). The required minimum crushing strength of brick for construction purpose is
(1) 2.5 MPa
(2) 3.5 MPa
(3) 5.5 MPa
(4) 10.5 MPa
Answer:(2) 3.5 MPa

Q8).A load bearing brick with a compressive strength of 10 MPa having the dimensions 19 cm × 9 cm × 9 cm then the maximum loading capacity of this brick is
(1) 850 kg
(2) 1710 kg
(3) 171 kg
(4) 17.1 kg
Answer:(2) 1710 kg

Q9). The most commonly used admixture to accelerate the setting time of concrete is
(1) Gypsum
(2) Calcium Carbonate
(3) Calcium Chloride
(4) Calcium Sulphate
Answer: (3) Calcium Chloride

Q10). The major cementing compound present inOPC is
(1) Tri-calcium silicate
(2) Di-calcium silicate
(3) Tri-calcium aluminate
(4) Tetra-calcium alumino ferrite
Answer: (1) Tri-calcium silicate

                                                       SET - II

Q1). Gypsum is added at the time of grinding of cement clinker for the purpose of
(1) quick setting
(2) delay the setting time
(3) improve the strength
(4) improve workability
Answer: (2) delay the setting time

Q2). The presence of excess alumina in the clay is reflected in the bricks as
(1) Makes the bricks brittle and weak
(2) Makes the bricks crack warp on drying
(3) Changes the colour of the brick from red to yellow
(4) Improves impermeability and durability of the brick
Answer: (1) Makes the bricks brittle and weak


Monday, November 21, 2016

Quiz - Only for Brilliants

Answer the below questions & develop ur civil engineering skills..

Lintel & its Types

What is Lintel?
A lintel is a horizontal member which is placed across the openings like doors, windows etc. It takes the load coming from the structure aboveit and gives support. It is also a type beam, the width of which is equal to the width of wall, and the ends of which are built into the wall.
These are very easy to construct as compared to arches.Bearing of lintel:The bearing provided should be the minimum of following 3 cases.
i) 10 cm.
ii) Height of lintel beam
iii) 1/10thto 1/12th of span of the lintel.

Classification of lintels:Lintels are classified based on the material of construction as:
1. Timber
2. Stone
3. Brick
4. Steel
5. Reinforced Concrete

Different styles of Staircases

Architectural view of staircases

Saturday, November 12, 2016

Shear Force & Bending Moment Diagram of Simply Supported Beam

Shear force and bending moment diagram of simply supported beam can be drawn by first calculating value of shear force and bending moment. Shear force and bending moment values are calculated at supports and  at points where load varies.
SIMPLY SUPPORTED BEAM WITH POINT LOAD EXAMPLE
Draw shear force and bending moment diagram of simply supported beam carrying point load. As shown in figure below.
Shear force, bending moment, simply supported beam, example
Solution
First find reactions of simply supported beam.
Both of the reactions will be equal. Since, beam is symmetrical. i.e.,
R1 = R2 = W/2 = 1000 kg.
Now find value of shear force at point A, B and C.
When simply supported beam is carrying point loads. Then find shear force value in sections. Shear force value will remain same up to point load. Value of shear force at point load changes and remain same until any other point load come into action.
Shear force between ( A – B ) = S.F (A-B) = 1000 kg
Shear force between (B – C) = S.F (B -C) = 1000 – 2000
S.F (B – C) = – 1000 kg.
Shear Force Diagram
shear force diagram, simply supported beam


Bending Moment
In case of simply supported beam, bending moment will be zero at supports. And it will be maximum where shear force is zero.
Bending moment at Point A and C = M(A) = M(C) = 0
Bending moment at point B = M(B) = R1 x Distance of R1 from point B.
Bending moment at point B = M (B) = 1000 x 2 = 2000 kg.m
Bending Moment Diagram
bending moment diagram, solved example, simply supported beam, point load

SIMPLY SUPPORT BEAM WITH UDL & POINT LOAD EXAMPLE
Draw shear force and bending moment diagram of simply supported beam carrying uniform distributed load and point loads. As shown in figure.
simply supported beam with udl and point load, shear force diagram, bending moment diagram

Solution
First find reactions R1 and R2 of simply supported beam.
Reactions will be equal. Since, beam is symmetrical.
R1 = R2 = W/2 = (600 +600 + 200 x4)/2 = 1000kg
Hence, R1 = R2 = 1000 kg.
Shear Force
Shear force between  section A – B = S.F (A – B) = 1000 kg.
Shear force at right side of point B = S.F (B) = 1000 – 600
S. F (B) right = 400 kg.
Now shear force at left side of point C.Because of uniform distributed load, value of shear continuously varies from point B to C.
Shear force at point C (Left) = S.F (L) = 400 – (200×4)
Shear force at point C (Left) = S.F (L) = -400 kg
Shear force between section C – D = S.F (C-D) = -400 – 600
Shear force between section C – D = S.F (C-D) = -1000 kg.
Shear Force Diagram
shear force diagram, simply supported, uniform distributed load, example

From Shear force, one can see;
  • Shear force is maximum at point A and remain same until point load.
  • At point B shear force value decreases, because of point load.
  • From B to C shear force continuously decreases, because of udl.
  • At point C shear force gradually falls, because of point load.
  • From point C to D, shear force remain same, because no other point load is acting in this range.
 B.M WILL BE ZERO AT SUPPORTS. I.E.,
M(A) = M(D) = 0
B.M at points B and C = M(B) = M(C) = 1000 x2 = 200 kg.m
Now, how to find maximum bending moment?
Bending moment will be maximum at point, where shear force is zero.  Hence, bending moment will be maximum at mid point.
M (max) = 1000×4 – 600×2 -200×2(2/2)
M (max) = 2400 kg.m
Bending Moment Diagram