This blog has been broken down into the following sub-topics.

- Why should we know how much to score?
- Previous year GATE marks vs GATE rank?
- Marks required to get a top 100 rank in GATE
- Mock test for GATE environmental science?

**It eases your preparation**- as you know your target marks it becomes easy to sort out high-weightage subjects and prepare them thoroughly. Know the weightage of subjects in GATE environmental science here.**It makes you feel relaxed while attending the exam**- if you are aware that you can skip some 10 questions, you can relaxedly skip time-consuming questions and attempt the next easy ones.

Be aware that this is just a guide, and not like "okay I have to score only 75 marks to get AIR 100". Because there are several uncertainties in the GATE exam which are listed below.

- If the exam is easy, you might have to attempt more questions.
- If the exam is difficult, you can afford to skip one or two more questions.

From analysing the previous year GATE environmental science exam results, let's see how much mark is required to get an AIR100, AIR50, and qualifying marks.

- AIR 50 -
**72+****marks** - AIR 100 -
**69+ marks** - Qualifying marks -
**40+ marks**

By seeing the above marks, it is clear that a lesser mark is only required to get an AIR 100 when compared to GATE civil engineering. But, as the average mark was higher last year, qualifying marks was at 40 + for the general category, and therefore more focus is required to qualify the paper. Also, as more PSUs are recruiting from GATE ES paper, it is advisable to get an AIR 50 to grab the opportunity of getting recruited in a PSU.

Aim for 70+ marks to get a top rank

The key to learning is to focus on improving your accuracy. You should be able to believe in what you are solving. This accuracy can be achieved by practising free all India mock tests at APSEd or free mock tests at the IIT Kharagpur website.

Hope you found this post useful. Do refer to the below-attached vlog on the same topic.

https://www.youtube.com/watch?v=jLBl4t8I_rI&t=13s**What exactly is the consolidation of soils**

For loads encountered in geotechnical engineering, it is generally considered that soil grains and pore water are incompressible. Therefore, the volume decrease of soil under stress can be attributed to the expulsion of pore water alone. This is exactly what a consolidation process is i.e., the expulsion of pore water.

**Can consolidation occur in fine-grained soils**

The expulsion of pore water is faster and is generally considered to keep pace with construction in fine-grained soils as it is more permeable. Therefore, the consolidation of fine-grained soils requires no attention. But, in clays expulsion of pore water takes more time as it is less permeable. Therefore, consolidation of clays can happen long after the construction has been completed and requires attention.

**Compaction vs Consolidation of Soil**

Compaction is just the process of mechanical compression resulting in a reduction of pore air and consequent densification of soil. Unlike compaction, consolidation is the expulsion of pore water. Though pore water is expelled, the soil remains fully saturated before, during and even after the consolidation process. Water content alone decreases during the consolidation process. For more details check this out.

**Necessary conditions for consolidation to occur**

- For consolidation to occur, there needs to be a way for pore water to come out i.e., it
**requires drainage**. Drainage could be a layer of sand present above or below or both above and below a fully saturated clay deposit. Based on the presence of the sand layer, there are two drainage systems i.e., single drainage and double drainage, which will be covered shortly. - Consolidation occurs only when there is an
**increase in stress**experienced by the soil. This could happen due to the addition of a new footing which increases the effective stress experienced by the layer. As effective stress varies with depth within the saturated clay layer, average effective stress is taken as the increased effective stress at the middle of the layer for all further computations.

Now that we have understood what consolidation actually is, the difference between compaction and consolidation, and the requirement for consolidation to occur, let us now understand the basic equation to compute consolidation.

Let us consider a fully saturated clay layer that is subjected to increased effective stress. The below image shows the changes it undergoes with time.

The primary settlement, consolidation settlement, and primary consolidation all are necessarily the same term and is represented as Sprimary (ΔH). Expression for the same is derived below.

ΔH = H1 - H2

ΔH/H1 = (ΔH * A)/(H1 * A) = ΔV/V1 = (V1 - V2)/V1, sub V1 and V2

ΔH/H1 = ((Vs + Vv1) - (Vs + Vv2)) / (Vs + Vv1) = (Vv1 - Vv2) / (Vs + Vv1),

dividing each term by Vs,

ΔH/H1 = ((Vv1/Vs) - (Vv2/Vs)) / ((Vs/Vs) + (Vv1/Vs)), as we know void ratio (e) = Vv/Vs,

ΔH/H1 = (e1 - e2) / (1 - e1) = Δe / (1 + e1)

*ΔH = H * (Δe / (1 + e0))*

where,

H - total thickness of clay layer

e0 - initial void ratio = wnat * Gs, wnat - natural water content

Terzaghi published his one-dimensional consolidation theory to find the primary settlement soils and the same is discussed further.

It is defined as a decrease in the void ratio per unit increase in effective stress applied on the soil. Its unit is m^2/kN. According to Dr Terzaghi, the value of av is constant during the consolidation of one type of soil.

**av = (-) Δe/Δ'**

It is defined as the slope of the linear portion of the graph plotted between the void ratio on the y-axis and log(base10)' on the x-axis. It has no units.

**Cc = Δe / log((o' + Δ') / o')**

According to Terzaghi and Peck, the compression index is totally dependent on the liquid limit of soil and the moulding condition of the soil.

*Cc = 0.009 * (Ll - 10%), for natural soil*

*Cc = 0.007 * (Ll - 10%), for re-moulded soil,*

where,

Ll - liquid limit of the clay in %

It is defined as the volumetric strain per unit increase in effective stress. It is the reciprocal of the bulk modulus of soil which is constant for a given soil. Its unit is m^2/kN. It is also called volume compressibility.

mv = ev/Δ' = (Δv/vo) / Δ', ev - volumetric strain,

mv = (Δe/(1+eo)) / Δ' = (Δe/Δ')* 1/(1+eo)

**mv = av / (1+eo)**

Let us now re-write the basic consolidation formula i.e., ** ΔH = H * (Δe / (1 + e0)), **in terms of compression index and coefficient of volume change.

**Primary Consolidation (Sp) in terms of Compression Index (Cc)**

Sp = H * (Δe / (1 + e0))

Cc = Δe / log(o' + Δ'/o')

Δe = Cc * log((o' + Δ') / o'), substituting this Δe in Sp formula,

* Sp = (H * Cc * log((o' + Δ') / o')) / (1 + eo), *same is given in a pictorial form below,

where,

o' - initial effective stress experienced by the soil layer (at the middle of the layer),

Δ' - increase in effective stress experienced by the soil layer (at the middle of the layer),

H - initial thickness of the soil layer,

eo - initial void ratio.

**Primary Consolidation (Sp) in terms of Volume Compressibility (mv)**

*Sp = mv * H * Δ',*

where,

mv - volume compressibility,

H - initial thickness of the layer,

Δ' - increase in effective stress experienced by the soil layer (at the middle of the layer).

An increase in effective stress (Δ') due to footings can be calculated using Bossineq's solution or Westergaurd's solution or 2:1 method unless otherwise specifically mentioned in the question. To know more about stress distribution in soil refer here. Below mentioned are formulas to calculate the same using Bossineq and 2:1 method.

**Bossineq's formula**

*Δ' = (Q/Z^2) * Kb,*

where,

Q - load on the footing (kN),

Z - depth of the layer from the base of the footing, (note that this depth is between the base of the footing and centre of the layer under consideration),

Kb - Bossineq coefficient

**2:1 method**

*Δ' = (q * L * B ) / ((L+Z)*(B+Z)),*

where,

q - load on the footing (kN),

L - length of the footing,

B - breadth of the footing,

Z - depth of the layer from the base of the footing, (note that this depth is between the base of the footing and centre of the layer under consideration)

**In case of newly placed fills**

** Δ' = **fill * tfill

where,

fill - the unit weight of the fill material (in kN/m^3),

tfill - thickness of the fill (in m).

- Soil is homogeneous and isotropic
- The soil is completely saturated
- The soil grains and water are incompressible
- Load is applied in the vertical direction only
- Drainage of pore water is one-dimensional i.e., in the direction of the load
- Primary consolidation is totally governed by the expulsion of pore water i.e., at the initial stage pore water pressure (uw) is 100% and consolidation is 0%. At the final stage, pore water pressure (uw) is 0% and consolidation becomes 100%
- The soil mass remains saturated after primary consolidation
- The coefficient of permeability of the soil is constant
- The coefficient of compressibility (av) and the coefficient of volume change (mv) are constant for a given type of clay
- The time lag of consolidation occurs mainly due to the low permeability of clays
- Darcy law is valid

Terzaghi considered continuity equation, coefficient of compressibility, coefficient of volume change in developing the equation of 1-D theory as given below.

** ∂uw / ∂t = (K / (mv * γw)) * (∂^2uw/∂z^2)**, for better understanding the same formula is given in pictorial form.

where,

uw - pore water pressure,

t - time of primary consolidation,

k - coefficient of permeability of clay,

mv - coefficient of volume change,

z - effective drainage path.

* Cv = k / (mv * γw), *substituting Cv in 1-D equation,

∂uw / ∂t = Cv * (∂^2uw/∂z^2)

The time of consolidation according to Terzaghi is given by,

**Tv = (Cv * t) / d^2,**

where,

Tv - time factor - a dimensionless quantity that is totally dependent on the degree of consolidation

**Tv = (π/4) * (U/100)^2, for U less than or equal to 60%**

**Tv = -0.9332*log(1 - (U/100)) - 0.0851, for U greater than 60%**

where,

U - degree of consolidation = (St / Sf) * 100,

St - settlement at any time,

Sf - total primary settlement of the soil.

Cv - coefficient of consolidation,

t - time of primary consolidation,

d - length of drainage path. It is dependent on the number of drainages as given below,

In the case of a single layer of sand (i.e., **single drainage**) the max distance travelled by the pore water is equal to the total thickness of the clay layer i.e., *d = H*

In the case of two layers of sand at the top and bottom (i.e., **double drainage**), the maximum distance travelled by the pore water is equal to half of the total thickness of the layer i.e.,

**d = H/2**

**Note**

- In case of double drainage, the time of consolidation gets reduced to 1/4th of the time required in case of single drainage (i.e., t double drainage = (1/4) * t single drainage)
- The total consolidation will remain the same in the case of both single and double drainage.
- In the case of 100% consolidation, Tv becomes infinity and eventually, the time required for consolidation (t) also becomes infinity. This means that 100% consolidation can never be achieved.

It is the ratio of pre-consolidation pressure experienced in the past to the effective stress action at present on the soil.

*OCR = **' (past) / ' (present)*

Based on OCR, soils (clays) are classified as,

OCR > 1, Over consolidated clays

OCR = 1, Normally consolidated clays

OCR < 1, Under consolidated clays

Two practise problems based on primary consolidation and Terzaghi's 1-D theory are given below. Practise them and get clarified with the concept.

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]]>Based on the position of weld on the members there are four different types of weld that are used in practice.

- Fillet welds
- Groove welds
- Slot welds
- Plug welds

These are the most commonly used weld due to their **economy, ease of fabrication, and site adaptability**. They are approximately **triangular in cross-section** and requires overlapping of pieces that are to be welded. Because of this overlapping, they require less precision while fitting.

Listed below are the specifications of a fillet weld according to IS 800: 2007, and IS 816.

- The size of the fillet weld is represented by S. The size of a fillet weld is the length of the smaller side of the fillet weld (S1/Smin) or it is taken between S1(S min) and S2 (S max).
- According to IS800: 2007 the
**minimum size (S1)**of a fillet weld is**based on****the****thickness of the thicker plate**as given below.

**Note:**If S1 (Smin) > thickness of the thinner plate (t thin), then S1 is taken as t thin.**S2 or S max**is**based**on the**thickness of the thinner plate**;- S2 = t thin - 1.5 (for square edge)
- S2 = (3/4) * t thin (for round edge)

**Size of fillet weld (S)** is therefore either adopted between S1 (S min) and S2 (S max) or simply **taken as S1 (S min) **in most cases.

As the fillet weld is in the form of a triangle in cross-section, the effective size of a fillet weld is required for designing purposes. This effective size is expressed in terms of throat thickness and is measured as the **perpendicular distance drawn on the hypotenuse of the triangular cross-section from the root of the fillet.**

According to IS800: 2007, the throat thickness (tw) of a fillet weld is totally dependent on the fushion angle (the angle at which two members are joined)__,__ as shown below.

**Note:**

- Throat thickness
**(tw) shall not be less than 3mm,**and - Throat thickness
**(tw) shall not be greater than 0.7 * thickness of the thinner plate (t thin)**.

The effective length of a fillet weld is the **total length of fillet weld (l) - (2 * weld size (S))** i.e., *leff = l - (2*S)*

**Note: **According to IS 800: 2007,

- leff must have a minimum value of 4 times the weld size (S), i.e.,
**leff > 4*S always** - In case of intermittent weld, the
**minimum leff**value shall be**greater of 4*S or 40 mm**

**Note:** The length of the weld shown in the diagram is always taken as the effective length (leff) and the extra length i.e., 2 * S, is provided by the welder.

Spacing of intermittent weld according to IS 800: 2007, shall not exceed

- for tension member - 16 * thickness of thinner plate (t thin) or 200mm
- for compression member - 12 * thickness of thinner plate (t thin) or 200 mm

According to IS 800: 2007 the maximum permissible shear stress (f weld) or (fw) in a fillet weld shall be,

*f weld = fu/(√3*γmw),*

where,

fu - smaller of the ultimate street of the weld or of the parent metal (but mostly taken as the ultimate tensile strength of the parent metal),

γmw = 1.25 for shop welding (refer table 5 of IS 800: 2007),

γmw = 1.5 for field welding (refer table 5 of IS 800: 2007).

**Note:** The **direct shear force **applied in the weld is a**lways less than or equal to the tensile strength of the smaller plate (given by ((fy/1.1) * (b*t))).**

**P = fw * leff * tw**

As said in the above note, P must be less than or equal to the tensile strength of the smaller plate. Therefore, according to the Limit State Method (IS 800), the design strength of a fillet weld per mm length of the weld is given by,

**P = (fy/1.1) * (b * t (smaller)) = (fu/√3*γmw) * leff * tw,**

where,

fy - yield strength of the smaller plate,

b - breadth of the smaller plate,

t - thickness of the smaller plate.

A groove weld is also called a butt weld as this type of weld is often used for forming butt joints. Edge preparation is necessary for making groove welds. A single V-butt weld is shown below.

*Clause 10.5.7.1.2 Butt welds of IS 800: 2007* says, "Butt welds shall be treated as parent metal with a thickness equal to the throat thickness, and the stresses shall not exceed those permitted in the parent metal. Therefore, according to the Limit state method, the design strength of a butt weld (Pw) is given by,

**Pw = (fy/γmw) * lw * tw,**

where,

tw - (5/8) * thickness of thinner plate, for single V- Butt weld;

tw - thickness of the thinner plate, for double V-Butt weld;

fy - yield strength of the plate,

γmw = 1.25 for shop welding (refer table 5 of IS 800: 2007),

γmw = 1.5 for field welding (refer table 5 of IS 800: 2007).

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]]>This blog has been broken down into the following sub-topics.

- Why should we know how much to score?
- Previous year GATE marks vs GATE rank?
- Marks required to get a top 100 rank in GATE
- How many questions can we skip and still get a top 100 rank in GATE?

**It eases your preparation**- as you know your target marks it becomes easy to sort out high-weightage subjects and prepare them thoroughly. Know the weightage of subjects in GATE civil engineering here**It makes you feel relaxed while attending the exam**- if you are aware that you can skip some 10 questions, you can relaxedly skip time-consuming questions and attempt the next easy ones

Be aware that this is just a guide, and not like "okay I have to score only 75 marks to get AIR 100". Because there are several uncertainties in the GATE exam and the same are listed below.

- If the exam is easy, you might have to attempt more questions.
- If the exam is difficult, you can afford to skip one or two more questions.

Going by the previous year GATE marks vs rank statistics, we can see that the range of marks to get a top 100 rank in GATE varies from 75+ marks to 80+ marks. And this figure is almost consistent for the past 5 years.

From the above data, it is clear that a mark in the range of 75 to 85 will help you to get the desired AIR 100 rank in GATE 2022.

Aim for 75+ marks to get a top rank

The key to learning is to focus on improving your accuracy. You should be able to believe in what you are solving. This accuracy can be achieved by practising free all India mock tests at APSEd or free mock tests at the IIT Kharagpur website.

Provided that you have improved your accuracy to 95% and you want to score 75+ marks, you can actually skip 13 questions.

Attempt 52 questions at 95% accuracy to get 75+ marks

Refer to the below-attached vlog on the same topic.

https://www.youtube.com/watch?v=FONvJps_g-0There are two predominant types of joints in a bolted connection namely, lap joint and butt joint. There are sub-types within these two types i.e., eccentric connections, pure moment connections etc. but are beyond the scope of this blog. The nature of the joints and sub-types within these joints are explained as follows.

In a lap joint, the main members to be connected are placed over one another to form an overlap between the members, and then the bolting is done on the overlapped portion. Because of the very nature of the connection, an eccentricity is produced.

In this type of joint, a cover plate is used to join two members. Based on the number of cover plates there are two types of butt joint namely, single cover butt joint and double cover butt joint.

To know a little more about bolts, types of bolts, other different types of butt and lap joints click here.

Out of all the above-mentioned joints, the **double cover butt joint** is the most preferred for the following reasons.

- The shear capacity of the bolt is more than in a lap joint
- No eccentricity is present in the connection

In a bolted connection either the connecting plate might fail or the bolt might fail. Therefore, it becomes important to consider the "Limit States" or failure modes of both bolt and the plate. Possible limit states by which a bolted connection might fail are mentioned below.

- Shear Failure of Bolts
- Bearing Failure of Bolts
- Tensile Failure of Bolts

- Shear Failure of Plate
- Bearing Failure of Plate
- Tensile Failure of Plate

As the name suggests, this failure occurs due to shear force at the interface of surfaces in a joint. Depending on the number of shear surfaces there are two types of shear failure that could occur in a bolted connection, namely, the single shear failure and double shear failure.

Here, the bolt is subjected to a single shear force which could cause the failure of the bolt. This type of failure occurs in single cover butt joint and lap joint.

Here, the bolt is subjected to two shear forces at two separate shear planes. This type of failure occurs in the double cover butt joint.

In this failure, the bolt fails in bearing due to contact with the plates. This type of failure occurs in cases where a low-strength bolt is used with a plate of very high grade, which usually doesn't occur in practice.

The tensile strength of the bolt is the amount of pull the bolt can withstand in the perpendicular direction to the plane of loading. If the pull on this perpendicular axis exceeds the tensile strength of the bolt, then the bolt will fail in tension.

Both shear and bearing failure of the plate can be avoided by providing sufficient centre to centre distances between the bolts as mentioned in section 10 of IS 800: 2007.

Due to a reduction in the net area (i.e., due to bolt holes) of the plate along the bolt line, the tensile strength of the plate will be lesser than the actual value at this section. Because of this, the plate might fail under tension. Therefore, it becomes important to calculate the least net area among different bolt lines to find the least tensile strength of the plate and check it for safety for the applied load.

- All bolts in a connection are stressed equally
- Friction between the plates in a connection is neglected
- The shear stress distribution is uniform for the bolt
- The bearing stress distribution is uniform for the bolt
- The bending of the bolt is neglected in case of smaller grip length, in case of larger grip lengths reduction factors must be considered for bending of the bolt

- Pitch distance (p) - centre to centre distance between two adjacent bolt holes in the direction of the applied load
- Gauge distance (g) - centre to centre distance between two adjacent bolt holes in the perpendicular direction of the applied load
- Edge distance (e) - the distance between the edge of the plate to the nearest centerline of a bolt hole in the perpendicular direction to the applied load
- End distance (e') - the distance between the end of the plate to the nearest centerline of the bolt hole in the direction of the applied load

For detailed specifications regarding all the above-mentioned terms refer to IS 800: 2007, *Section 10 Connections.*

Black bolts are the least expensive bolts available and are also called ordinary, unfinished, or common bolts. They are primarily used in light structures under static loading such as small trusses, bracings, etc.

The grade of a bolt indicates the properties of the bolt such as nominal ultimate tensile strength__,__ and yield strength. It varies from 4.6 to 10.9. Generally, a bolt of grade 4.6 is used in construction. A grade of 4.6 means, 4 indicates 1/100 the of the nominal ultimate tensile strength of the bolt in N/mm^2, and 0.6 indicates that the yield strength of the bolt is 60% of the ultimate tensile strength of the bolt. Thus a bolt with grade 4.6 has an ultimate tensile strength of 400 N/mm^2 and a yield strength of 240 N/mm^2. The same is indicated in the picture below.

As per IS 800: 2007, the ** Design Bearing Strength of the Bolt (Vdpb)** is given by,

**Vdpb = Vnpb/γmb,**

where,

γmb - partial safety factor;

**Vnpb = Nominal Bearing Strength of the Bolt = 2.5 * Kb * d * t * fu;**

d - nominal diameter of the bolt;

t - the sum of the thickness of the connected plates experiencing bearing stress in the same direction;

fu - ultimate tensile strength of the plate (i.e., for Fe410, fu=410N/mm^2);

Kb - smaller of the following values:

*e/(3 * do),**(p/(3 * do)) - 0.25,*and,*fub/fu,**1*where,

e,p - end and pitch distances of the fastener along bearing direction;

**e = 1.5 * do; p = 2.5 * d;**do - diameter of the hole (refer

*Table 19 (clause 10.2.1) of IS 800: 2007*, a simplified version is provided below) ;fub - ultimate tensile strength of the bolt;

fu - ultimate tensile strength of the plate.

As per IS 800: 2007, the* Design Shear Strength of the Bolt (Vdsb)* is given by,

**Vdsb = Vnsb/γmb,**

where,

γmb - partial safety factor;

*Vnsb = Nominal Shear Capacity of the Bolt = (fub/√3) * ((nn * Anb) + (ns * Asb));*

fub - ultimate tensile strength of a bolt;

nn - number of shear planes with threads intercepting the shear plane;

ns - number of shear planes without thread intercepting the shear plane (i.e., shank area intercepts the shear plane), usually taken as zero (0);

Anb - net shear area of the bolt at threads, taken as the area corresponding to root diameter at the thread;

* Anb = 0.78 * ((π*d^2)/4); d - diameter of the bolt;* and

Asb - nominal plain shank area of the bolt.

- If the joint is too long - when the distance between the first and last rows of bolts in the joint, measured in the direction of load transfer exceeds 15 times the diameter of the bolt (d);
- If the grip length is large - when the grip length (equal to the thickness of the connected plates) exceeds 5 times the diameter of the bolt (d);
- If a packing plate of more than 6 mm in thickness is provided

Refer to *section 10.3.3 shear Capacity of Bolt *of IS 800: 2007 to get the formulas for reduction factors of each of the above-mentioned cases.

Bolt value is the **least of Design Bearing Strength (Vdpb) and Design Shear Strength (Vdsb) of the bolt.** Let us understand the same using an example problem.

**Question: **Calculate the bolt value of a 20mm diameter bolt of grade 4.6 to join main plates of 12mm thick under the following cases. Assume Fe410 for all plates.

- Lap joint
- Single cover butt joint with cover plate = 10mm thickness
- Double cover butt joint with cover plate = 8mm thickness

**Solution:**

Given data,

Diameter od the bolt (d) = 20mm

for Fe410 grade plates, fu = 410 N/mm^2

for bolt grade 4.6, fub = 400 N/mm^2

Common data,

Diameter of the bolt hole (from table 19 of IS 800: 2007) = 22mm

γmb = 1.25 (from Table 5 of IS 800: 2007)

It is assumed that the threaded portion alone is crossing the shear plane (i.e., ns = 0).

Anb = 0.78 * *(π*d^2)/4)*

*Anb = 0.78 * *((π*20^2)/4) = 245 mm^2

e = 1.5 * do = 1.5 * 22 = 33 mm

p = 2.5 * d = 2.5 * 20 = 50 mm

**Case 1: Lap joint (nn =1)**

**1. Design bearing strength of the bolt**

**Vdpb = Vnpb/γmb,**

**Vnpb = 2.5 * Kb * d * t * fu,**

**Kb least of:**

*e/(3 * do) = 33/(3*22) = 0.5,**(p/(3 * do)) - 0.25 = (50/(3*22)) - 0.25 = 0.5,*and*fub/fu, = 400/410 = 0.97,**1*

Therefore Kb = 0.5.

t = 12 mm (thickness of the main plate)

*Vnpb = 2.5 * 0.5 * 20 * 12 * 410 = 123 kN*

*Vdpb = 123/1.25 = 98.4 kN*

**2. Design shear strength of the bolt**

**Vdsb = Vnsb/γmb,**

*Vnsb = (fub/√3) * ((nn * Anb) + (ns * Asb)),*

*Vnsb = (400/√3) * (1 * 245) = 56.58 kN*

*Vdsb = 56.58/1.25 = 45.26 kN*

Therefore, the bolt value or design strength of the bolt in a lap joint is = 45.26 kN (least).

**Case 2: Single cover butt joint (nn =1)**

**1. Design bearing strength of the bolt**

**Vdpb = Vnpb/γmb,**

**Vnpb = 2.5 * Kb * d * t * fu,**

**Kb least of:**

*e/(3 * do) = 33/(3*22) = 0.5,**(p/(3 * do)) - 0.25 = (50/(3*22)) - 0.25 = 0.5,*and*fub/fu, = 400/410 = 0.97,**1*

Therefore Kb = 0.5.

t = 10 mm (thickness of the cover plate)

Note:

- if the thickness of the cover plate < thickness of the main plate, then t = the thickness of cover plate;
- if the thickness of the cover plate > thickness of the main plate, then t = the thickness of the main plate;
- if two cover plates are used, t = summation of the thicknesses of both the cover plates provided that the sum is less than the thickness of the main plate.

*Vnpb = 2.5 * 0.5 * 20 * 10 * 410 = 102.5 kN*

*Vdpb = 123/1.25 = 82 kN*

**2. Design shear strength of the bolt**

**Vdsb = Vnsb/γmb,**

*Vnsb = (fub/√3) * ((nn * Anb) + (ns * Asb)),*

*Vnsb = (400/√3) * (1 * 245) = 56.58 kN*

*Vdsb = 56.58/1.25 = 45.26 kN*

Therefore, the bolt value or design strength of the bolt in a single cover butt joint is = 45.26 kN (least).

**Case 3: Lap joint (nn =2)**

**1. Design bearing strength of the bolt**

**Vdpb = Vnpb/γmb,**

**Vnpb = 2.5 * Kb * d * t * fu,**

**Kb least of:**

*e/(3 * do) = 33/(3*22) = 0.5,**(p/(3 * do)) - 0.25 = (50/(3*22)) - 0.25 = 0.5,*and*fub/fu, = 400/410 = 0.97,**1*

Therefore Kb = 0.5.

As the sum of the thickness of the two cover plates is 16mm (8+8), the t value is taken as the thickness of the main plate (a lesser value must be taken).

t = 12 mm (thickness of the main plate)

*Vnpb = 2.5 * 0.5 * 20 * 12 * 410 = 123 kN*

*Vdpb = 123/1.25 = 98.4 kN*

**2. Design shear strength of the bolt**

**Vdsb = Vnsb/γmb,**

*Vnsb = (fub/√3) * ((nn * Anb) + (ns * Asb)),*

*Vnsb = (400/√3) * (2 * 245) = 113.16 kN*

*Vdsb = 56.58/1.25 = 90.52 kN*

Therefore, the bolt value or design strength of the bolt in a double cover butt joint is = 90.52 kN (least).

As per IS 800: 2007, *Section 6 Design of Tension Members*, the design strength in tension of a plate, Tdn as governed by rupture of net cross-sectional area (An) is given by,

*Tdn = (0.9 * fu * An)/γm1,*

where,

*An - the net effective area of the plate = (b - n*do)*t, for a chain of holes in a row;*

*An = (b - n*do + *Σ[(psi^2)/(4*gi)])*t*, for staggered arrangement;*

fu- ultimate tensile strength of the plate;

γm1 = 1.25, refer table 5 of IS800: 2007.

The strength of a bolted connection is the **least of bolt value (design strength of the bolt) and the design tensile strength of the plate.**

Refer to the below attached video lectures on Introduction to bolted connection and codal provisions for bolted connection.

https://www.youtube.com/watch?v=8IsnrpN2cF8&list=PLSNhedsleX1335_jU9sx8sW7_LcMW4EZc&index=7https://www.youtube.com/watch?v=uANirTSfaD8&list=PLSNhedsleX1335_jU9sx8sW7_LcMW4EZc&index=8Want to contribute to the student community through APSEd? Join our APSEd community by filling out the form below.

]]>Bhabha Atomic Research Centre is India's premier nuclear research facility, headquartered in Trombay, Mumbai, Maharashtra. The organization is currently recruiting scientific officers in the department of atomic energy which is a Group-A post of Government of India, through the Bhabha Atomic Research Training school. The selection process and eligibility criteria are discussed in brief.

The official announcement is here, check it out.

There are two programs on offer at Bhabha Atomic Research Training school. Both training programs also offer stipends for students which could go up to 55,000 INR/ month.

- One-year Orientation Course for Engineering Graduates and Science Postgraduates for the academic year 2022-2023 (OCES-2022)
- Two-year DAE Graduate Fellowship Scheme for Engineering Graduates and Physics Postgraduates for the academic session beginning in 2022 (DGFS-2022)

The selection process is a two-step process i.e, screening of candidates and interview process.

**Screening** is based on two alternative methods i.e.,

- a candidate can choose to either appear for an online examination conducted by BARC in April 2022, or
- a candidate can choose to be screened on the basis of GATE (2021 or 2022) score, except for candidates with a degree in "nuclear engineering"

**Interview** for shortlisted candidates could occur in **June or July 2022. **The** final selection** is based on the medical fitness of the candidate and will be displayed around **September 2022 **on BARC's official website.

Following engineering disciplines are eligible. It is necessary to have 60% aggregate marks in any of the following disciplines. Candidates opting for GATE screening must have a GATE exam in the same discipline as the qualifying discipline. Candidates awaiting GATE 2022 results shall also apply.

- Mechanical engineering
- Chemical engineering
- Metallurgical
- Electrical engineering
- Electronics communication and engineering
- Computer science
- Instrumentation
- Civil engineering
- Nuclear engineering

Similarly, qualifying degrees that are eligible under different categories like fast reactor technology, quality assurance & control, physics disciplines, chemistry disciplines, radiological safety, geology disciplines etc. is given in detail in the official notification.

**Age limit**

- General category - 26
- OBC - 29
- SC/ST - 31

Only online application is available. A non-refundable fee of 500 INR is applicable. Do check out the online application form here.

- Commencement of Online Application Process for OCES/DGFS-2022 - 17th Jan 2022
- Last date for Registration for Online Application - 11th Feb 2022
- Last date for Submission of Online Application - 12th Feb 2022
- Online Examination Slot Booking (the window will be open for 2 weeks/ 10 days) - 4th to 18th March 2022
- Online Examination - 7th to 13th April 2022
- Last date candidates to upload the GATE Score - 13th April 2022
- Display of List of candidates short-listed for Interview on Online Application Portal - 28th April 2022
- Availability based option on Online Application Portal to select Interview Slot for qualified candidates - 30th April to 4th May 2022
- Selection Interviews - 14th June to 31st July 2022
- Display of List of Candidates finally selected for OCES-2022 on Online Application Portal - 18th August 2022
- Last Date for Selected OCES-2022 Candidates desirous of DGFS to give details of M.Tech I M.Chem. Engg. admission in a DGFS institute - 20th August 2022
- Declaration of List of Applicants Selected for DGFS-2022 on Online Application Portal - 2nd week of September 2022

*Disclaimer: We have sourced the information from the notification. Kindly check the official website, notification*,* and re-check the information before applying. *

Subscribe to APSEd Blog by providing your mail Id below and get to know the latest ,__j__,ob notifications and much more.

- Bernoulli's equation is also called the
**energy equation**. - It is based on the ,
__law of conservation of energy____,__for more details on the law of conservation on fluid mechanics click ,__here__. - According to Bernoulli's equation, the total energy of a flowing fluid at any section consists of three parts as mentioned below.
- Pressure or Static head
**(P/γ)** - Kinetic or Velocity head
*(v^2/2*g)* - Datum or Potential head (
*Z)*

- Pressure or Static head

By combining these three parts, we get Bernoulli's equation as,

* (P/γ) + (v^2/2*g) + Z = constant, *a pictorial representation is given for better understanding,

In the above equation,

* (P/γ) + Z, *is called the

** (v^2/2*g), **is called the

The piezometric head is represented by __hydraulic gradient line (HGL)__ Both piezometric head and kinematic head combined represents the total energy line (TEL). It could be easily understood that HGL is always below the TEL. Also, TEL always decreases along the flow direction except at the pump section.

Based on Bernoulli's equation there are different flow measuring devices as mentioned below.

- Venturimeter
- Orifice meter
- Pitot tube
- Nozzle meter
- Rotameter
- Elbow meter

Out of the above-mentioned devices, Venturimeter is the most popular flow measuring device. Pitot tube and orifice meter are also used to some extent. Therefore, it's important to know the concept, design, and formulas, of these devices and the same is discussed further.

Venturimeter is a device to measure the rate of flow of fluid in pipes. Venturimeter consists of three portions i.e., convergent, throat, and divergent portion. Design aspects of these three portions are discussed further.

**1. Convergent portion**

- Diameter at convergent and divergent portions are taken as the diameter of the pipe to which the venturimeter is connected.
**Therefore, Dinlet = Dpipe = Doutlet** - The
**angle of convergent**is maintained as (21°± 1°), mostly**20°** - The length of the convergent portion is taken as 2.7*(D - d), where, D - diameter of the pipe, and d - diameter of the throat

**2. Throat portion**

- The
**diameter of the throat(d)**is an important factor for the safety of both the pipe and the Venturimeter. If d is too small, velocity head increases and pressure head decreases to a great extent. This negative pressure creates bubbles that explode on the downstream side (where pressure increases) and causes__notching__of the material. To avoid this d is taken as**half of the diameter of the inlet pipe**, i.e., 1/2*D - The
**length of the throat**portion (Lthroat) is taken as the**diameter of the throat**portion (d)

**3. Divergent portion**

- The divergent portion is provided to protect the Venturimeter from cavitation i.e., pressure is gradually increased
- The
**angle of the****divergent portion**is taken between 5° to 15°, mostly**6°**

Other general considerations include,

- To measure the pressure difference between inlet and outlet a Utube manometer is adopted
- Head loss in convergent portion is neglected (therefore Venturimeter gives theoretical discharge only)
- To get the actual discharge, the
**co-efficient of discharge (Cd)**is multiplied with theoretical discharge. Cd value varies between**0.95 to 0.99** - The
**size of a Venturimeter**is specified as the**diameter of the main pipe and the diameter of the throat of the venturimeter**. For example, venturimeter with size 150 * 75 mm represents main pipe diameter (D) as 150 mm and throat diameter (d) as 75mm

Refer to the below attached pictorial representation of the Venturimeter.

**Formula Derivation for Venturimeter**

Considering the reference line of U-Tube manometer,

Py = Pz

*P1 + γ1*x = P2 + γ1*(x-h) + γ2*h,*

P1 - P2 = -γ1*x + γ1*x - γ1*h + γ2*h,

P1 - P2 = -γ1*h + γ2*h,

P1 - P2 = h*(γ2 - γ1),

P1 - P2 = (h*γ1) * [(γ2/γ1) - 1],

(P1 - P2)/γ1 = h*[((S2*γw)/(S1*γw)) - 1],

(P1 - P2)/γ1 = h*[(S2/S1) - 1]

The difference in piezometric head between inlet and outlet (ΔH),

ΔH = [(P1/γ)+Z1] - [(P2/γ)+Z2], (here, Z1 = Z2)

ΔH = h*[(S2/S1) - 1]

* **ΔH = **h*[(Sm/S) - 1], if Sm > S*

* **ΔH = **h*[1 - (Sm/S)], if Sm < S*

where,

h - the manometric reading as shown in the diagram

Sm - specific gravity of the mercury (i.e., 13.6)

S - specific gravity of the flowing fluid

Continuing derivation, apply Bernoulli's equation between inlet and throat i.e., between section 1 & section 2

(P1/γ) + (v1^2/2*g) + Z1 = (P2/γ) + (v2^2/2*g) + Z2 + hL (head loss - neglected)

[(P1/γ) + Z1] - [(P2/γ) + Z2] = [(V2^2) - (V1^2)]/(2*g)

**ΔH = [(V2^2) - (V1^2)] / (2*g) --- 1**

Applying continuity equation, Q1 = Q2

A1*V1 = A2*V2 = Q

** V1 = Q/A1, V2 = Q/A2,** substituimg V1 and V2 in equation 1,

ΔH = [(Q/A2)^2 - (Q/A1)^2] / (2*g)

2*g*ΔH = Q^2[(1/A1^2) - (1/A2^2)]

2*g*ΔH*(A1^2)*(A2^2) = Q^2 * ((A1^2)-(A2^2))

*Qth = (A1*A2*√(2*g*ΔH)) / √((A1^2) - (A2^2)), *

Here discharge is theoretical (Qth) as the head loss was neglected. Actual discharge is calculated by multiplying the coefficient of discharge (Cd) with theoretical discharge.

*Qactual = Cd * Qth,*

The same formulas are represented in a pictorial form below.

where,

A1 - area of the pipe at the inlet (i.e., diameter = D)

A2 - area of the pipe at the throat (i.e., diameter = d)

ΔH - the difference in the piezometric head (as explained earlier)

Cd - coefficient of discharge 0.95 to 0.99

g - acceleration due to gravity (9.8 m/sec^2)

An orifice meter is a similar device used to measure the discharge of the fluid. It consists of a flat plate containing a circular orifice which is provided concentrically with the pipe across the flow. This device works on the same principle as that of the venturimeter.

Orifice meter is not most commonly used because of the below-mentioned disadvantages which are not present in a venturimeter.

- Low accuracy than a venturimeter
- Subjected to wear
- Dirt and sediments are collected due to the obstructive design
- It has low efficiency than a venturimeter

Though the orifice meter has some disadvantages it can be still easily made as it is simple in design and is cheap.

A pitot tube is a simple device used for measuring the velocity of a flowing fluid at any point. It appears as an L-shaped bent pipe in its simplest form and is used to measure the velocity of flow which are open to the atmosphere.

Applying Bernoulli's equation,

(P1/γ) + (v1^2/2*g) + Z1 = (P2/γ) + (v2^2/2*g) + Z2 + hL + ΔH,

here,

(P1/γ) = Z1 = (P2/γ) = v2^2 = Z2 = hL(headloss neglected) = 0, therefore we get,

(v1^2/2*g) = ΔH

** Vth = √(2*g*ΔH), **here velocity is theoretical (Vth) as the head loss was neglected. To get the actual velocity, the coefficient of velocity (Cv) must be multiplied with theoretical velocity.

*Vactual = Cv * Vth,*

*Vactual = Cv * √(2*g*ΔH),*

where,

Cv - the coefficient of velocity

g - acceleration due to gravity (9.8 m/sec^2)

ΔH - obtained from pitot tube

A flow nozzle is a similar device to an orifice meter and venturimeter. It has a smooth converging portion as that of a venturimeter but with no diverging portion. Flow nozzle is also not a commonly used device.

Check out the complete video lecture on open channel flow below.

https://www.youtube.com/watch?v=smX-Tabzk5A&list=PLSNhedsleX11ykZJbtlIVDH8kZXqB_tO-&index=1Hope you found this post useful. Want us to cover a topic of your interest? Let us know in the comments below and get subscribed to our blog for all the latest updates in the engineering world!

]]>In an ideal environment, populations grow at an exponential rate. The growth curve of these populations is smooth and becomes increases steeply over time. However, exponential growth is not possible because of factors such as limitations in food, competition for other resources, disease etc. Populations eventually reach the carrying capacity or saturation capacity of the environment, causing the growth rate to slow nearly to zero. This produces an **S-shaped curve** of population growth known as the logistic curve.

As discussed, this method uses the logistic curve of population growth. Therefore, it uses the equation of the logistic curve to directly predict the population. Equation of the logistic curve is given as,

** P = Ps/(1+(m*ln^-1(n*t))), **pictorial representation of the same is shown below,

where,

*m = (Ps - Po)/Po,*

Ps - saturation population given by, ** Ps = (2*Po*P1*P2 - P1^2(Po+P2))/(Po*P2 - P1^2)**,

pictorial representation of the same is shown below,

** n = (1/t1)*ln((Po*(Ps-P1))/(P1*(Ps-Po))**, pictorial representation of the same is shown below,

Po - population at to years,

P1 - population at t1 years,

P2 - population at t2 years,

t1 - number of years between to and t1,

t2 - number of years between to and t2,

t - number of years between 'to' and required year,

t2 = 2*t1 (in general).

**Question:** Using the data given below find the population for the year 2021.

**Solution:**

Step 1: Given data,

Po = 80,000 at to = 0 years,

P1 = 250,000 at t1 = 10 years,

P2 = 480,000 at t2 = 20 years,

Step 2: Find saturation population Ps using the formula ** Ps = (2*Po*P1*P2 - P1^2(Po+P2))/(Po*P2 - P1^2)**,

**Ps = 655,602**

Step 3: Find the value of m using the formula *m = (Ps - Po)/Po,*

m = (655602 - 80000)/80000

**m = 7.195**

Step 4: Find the value of using the formula *n = (1/t1)*ln((Po*(Ps-P1))/(P1*(Ps-Po)),*

**n = -0.1489**

Step 5: Find the population for the required year using the formula *P = Ps/(1+(m*ln^-1(n*t))),*

t = 30 (number of years between to and t)

P = 655602/(1 + (7.195*ln^-1(-0.1489*30)))

**P = 6,05,436 is the population for the year 2021.**

** **

For more insights and to know more about other population forecasting methods refer to the video lecture below.

https://www.youtube.com/watch?v=BGPParFfTEoWant to be a part of the APSEd community? Fill out the form below and we will get back to you!

]]>The arithmetical Increase Method is mainly adopted for **old and developed **towns**, **where the rate of population growth is nearly constant. Therefore, it is assumed that the rate of growth of the population is constant. It is similar to simple interest calculations. The population predicted by this method is the lowest of all.

* dP/dt = K* (say), where, dP/dt represents rate of growth of population.

Integrating the above equation over P1 to P2 over a time period of t1 to t2,

*∫dP = K∫dt*

*[P2 - P1] = k * [t2 - t1]*

**P2 = P1 + K * Δt**

**P2 = P1 + x̄ * n**

* P2 = P1 + n*x̄

**Pn = Po + nx̄,**

where,

Po - last known population

Pn - population (predicted) after 'n' number of decades,

n - number of decades between Po and Pn and,

x̄ - the rate of population growth.

The following data (common data) will be used in the example problems for all other methods to be discussed.

**Question: **With the help of the common data find the population for the year 2020 using the arithmetic increase method.

**Solution:**

Step 1: Find the increase in population each decade.

Step 2: Find the average rate of increase of population (x̄)

x̄ = (3000+6000+8000+5000)/4

x̄ = 22000/4

x̄ = 5500

Step 3: Find the number of decades (n) between the last known year and the required year

n = 5 (5 decades elapsed between 1970 and 2020)

Step 4: Apply the formula **Pn = Po + nx̄,**

P[2020] = P[1970] + (5 * 5500)

P[2020] = 47000 + 27500

** P[2020] = 74,500**. Therefore, population at 2020 will be 74,500.

This method is adopted for **young and developing towns**, where the rate of growth of population is proportional to the population at present (i.e., dP/dt ∝ P). Therefore, it is assumed that the percentage increase in population is constant. It is similar to compound interest calculations. The population predicted by this method is the highest of all.

Let's say, for the 0th-year population is **P**

For 1st year/decade, according to this method, the population would become,

* P + (r/100)P*, where r is the growth rate.

For 2nd year/decade, according to this method, population would become,

*[P + (r/100)P] + (r/100)[P + (r/100)P]*

*= P[1+(r/100)]^2*

Generalizing the above equation, we get,

**Pn = Po[1 + (r/100)]^n**

** Pn = Po[1 + (r/100)]^n**,

where,

Po - last known population,

Pn - population (predicted) after 'n' number of decades,

n - number of decades between Po and Pn and,

**r - growth rate =****(increase in population/initial population) * 100 (%).**

r could be found as arithmetic mean (i.e., (r1 + r2 + r3 ... rn)/n) or as a geometric mean (i.e., nth root of (r1 * r2 * r3 ... rn)), for the given data. According to Indian standards r should be calculates using geometric mean method.

**Question: **With the help of the common data find the population for the year 2020 using the Geometrical increase method.

**Solution:**

Step 1: Find the increase in population each decade.

Step 2: Find the growth rate.

Step 3: Find the average growth rate (r) using geometrical mean.

r = ∜(12 * 21.4 * 23.5 * 11.9)

r = 16.37 %

Step 4: Find the number of decades (n) between the last known year and the required year

n = 5 (5 decades elapsed between 1970 and 2020)

Step 5: Apply the formula **Pn = Po[1 + (r/100)]^n**

P[2020] = P[1970][1 + (16.37/100)]^5

P[2020] = 47000[1.1637]^5

** P[2020] = 1,00,300. **Therefore, population at 2020 will be 1,00,300.

This method is adopted for **average-sized**** ****towns**** under normal conditions**, where the rate of population growth is not constant i.e., either increasing or decreasing. It is a combination of the arithmetic increase method and geometrical increase method. Population predicted by this method lies between the arithmetical increase method and the geometrical increase method.

** Pn = (Po + nx̄) + ((n(n+1))/2)* ȳ**,

where,

Po - last known population,

Pn - population (predicted) after 'n' number of decades,

n - number of decades between Po and Pn,

x̄ - mean or average of increase in population and,

ȳ - algebraic mean of incremental increase (an increase of increase) of population.

**Question: **With the help of the common data find the population for the year 2020 using the Incremental Increase method.

**Solution:**

Step 1: Find the increase in population in each decade.

Step 2: Find the incremental increase i.e., increase of increase.

Step 3: Find x̄ and ȳ as average of Increase in population and Incremental increase values respectively.

x̄ = (3000+6000+8000+5000)/4

x̄ = 5500

ȳ = (3000+2000-3000)/3

ȳ = 2000/3

Step 4: Find the number of decades (n) between the last known year and the required year

n = 5 (5 decades elapsed between 1970 and 2020)

Step 5: Apply the formula *Pn = (Po + n**x̄) + ((n(n+1))/2)* *** ȳ**,

P[2020] = (P[1970] + nx̄) + ((n(n+1))/2)* ȳ

P[2020] = 47000 + (5 * 5500) + (((5 * 6)/2) * (2000/3))

** P[2020] = 84,500. **Therefore, population at 2020 will be 84,500.

This method is adopted for a town which is reaching **saturation population**, where the rate of population growth is decreasing. In this method, an average decrease in growth rate (S) is considered.

*Pn = P(n-1) + ((r(n-1) - S)/100) * P(n-1),*

where,

Pn - population at required decade,

P(n-1) - population at previous decade (predicted or available),

r(n-1) - growth rate at previous decade and,

S - average decrease in growth rate.

Due to the very nature of the formula, which requires population data at the previous decade i.e., P(n-1), this method requires the calculation of population at each successive decade (from the last known decade) instead of directly calculating population at the required decade.

**Question: **With the help of the common data find the population for the year 2020 using the decreasing rate of growth method.

**Solution:**

Step 1: Find the increase in population.

Step 2: Find the growth rate (r) as in the geometrical increase method.

Step 3: Find the decrease in the growth rate.

Step 4: Find the average of decrease in growth rate(s).

S = (-9.4-2.1+11.6)/3

S = 0.1/3

S = 0.03%

Step 5: Apply the formula ** Pn = P(n-1) + ((r(n-1) - S)/100) * P(n-1), **and find the population at successive decade till the population at required data is arrived.

P[1980] = P[1970] + ((r[1970] - S)/100) * P[1970]

P[1980] = 47000 + ((11.9 - 0.03)/100) * 47000

P[1980] = 52579

P[1990] = P[1980] + ((r[1980] - S)/100) * P[1980]

P[1990] = 52579 + ((11.87 - 0.03)/100) * 52579, here r[1980] is directly found as 11.9 - 0.03 i.e., r[1970] - S, which equals to 11.87.

P[1990] = 58,804

Similarly, P[2020] could be found.

In this method, the population vs time graph is plotted and is extended accordingly to find the future population. It is to be done by an experienced person and is almost always prone to error.

In this method, the population of a town is predicted by comparing it with a similar town.

This method is used for a completely planned city that is not meant to be developed in a haphazard manner.

For more insights please refer to the video lecture below on population forecasting methods.

https://www.youtube.com/watch?v=BGPParFfTEoHope you found this post useful and informative. If you want us to cover a topic of your interest please let us know in the form below. Do subscribe to our blogs to stay updated as well.

]]>lpcd stands for **litre/per capita/per day, **which is the amount of water required in litres per person per day.

According to __IS1172: 1993__, the water demand for Indian towns/cities shall be based on two categories, namely,

- Full flushing system / High-income group
- Low-income group

Water required by both these groups for different purposes is shown below.

It depends on the type of industries present in an area/town and usually varies from **50 to 450 lpcd.**

It is the water required for institutional and commercial establishments like schools, colleges, malls, etc. and it usually varies from **10 to 20 lpcd**.

It is the water required for public usage such as gardening, fountains, etc. and it completely depends on the area where the water supply system is being developed.

It is the amount of water required for fire fighting purposes if in case a fire breaks out in an area. This water is required to be available at a pressure of about **100 to 150 kN/m^2 or 10 to 15m **** head of water**. Fire demand is not calculated for smaller towns where the population is less than 50,000. For larger cities, fire demand is calculated using various formulas which are discussed further.

*Q = 3182 * √P,*

where,

Q - water required in litres/minute

P - population in 1000s (i.e., if population is 1,00,000 then P = 100)

*Q = 1136 * [(P/10)+10],*

where,

Q - water required in litres/minute

P - population in 1000s

*Q = 4637 * √P * [1 - (0.01*√P)],*

where,

Q - water required in litres/minute

P - population in 1000s (valid for population of less than 2,00,000)

*Q = 5663 * √P,*

where,

Q - water required in litres/minute

P - population in 1000s

*Q = 1800 litres/minute,*

For every 50,000 population up to 3,00,000 population,

Above 3,00,000 population, extra water shall be 1,800 lit/min for every 1,00,000 population.

It is the water which could be lost during transport and is generally taken as **15%** of water demand calculated.

Summing it all, per capita demand for an Indian city is about **335 lpcd**. Further, according to IS 1172: 1993, the total quantity of water supplied shall be,

- For High-Income group - 200 (domestic) + 135 =
**335 lpcd** - For Low-Income group - 135 (domestic) + 135 =
**270 lpcd**

Per capita demand or annual average daily demand is calculated as ** (Total yearly demand in litres/(365 * Design Population)). **It is measured in lpcd.

Annual average demand or per capita demand represents only the average amount of water required. But the water demand varies (as shown below) and it becomes important to consider these variations in designing the system.

- Seasonal variation - high demand in summer, less demand in winter
- Daily variation
- Hourly variation - high demand in morning and evening hours

Goodrich formula is used to calculate the per cent of the annual draft that is required as a peak factor in meeting the variations in water demand. It's given by,

*p = 180 * (t)^-0.1,*

where,

p - per cent in the annual draft

t - time in days

Daily peak factor, p ( according to Goodrich) = 180 * (1)^-0.1 = 180%

Maximum daily demand = Daily peak factor * Average daily demand

Maximum daily demand = 1.8 * Average daily demand

Maximum hourly demand = 1.5 * Average hourly demand on the max day

Maximum hourly demand = 1.5 * (Maximum daily demand day/24)

Maximum hourly demand = 1.5 * 1.8 * (Average Daily Demand/24)

Maximum hourly demand = 2.7 * Average hourly demand (based on annual basis)

or in simple terms,

Maximum hourly demand = 1.5 * Maximum daily demand

Weekly peak factor, p ( according to Goodrich) = 180 * (7)^-0.1 = 148%

Maximum weekly demand = Weekly peak factor * Average weekly demand

Maximum weekly demand = 1.48 * Average weekly demand

Monthly peak factor, p ( according to Goodrich) = 180 * (30)^-0.1 = 128%

Maximum monthly demand = Monthly peak factor * Average monthly demand

Maximum monthly demand = 1.28 * Average monthly demand

The design capacity of various water supply systems according to IS 1172: 1993 is shown below,

For more insights please do refer to the attached video lecture (below) on fire demand and coincidental draft. Also, check out the full __environmental engineering course__ at APSEd.

What topics do you want us to cover? Do fill out the form below and let us know!

]]>We understand the importance of very similar tests for GATE 2022, and keeping this in eye, our experts have designed all the tests. The tests are not only content-rich but also very affordable comparing all other test series in the market.

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If you are looking for a course, you can get GATE Geomatics Course here.

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You get the extra edge with APSEd GATE Geomatics Test Series 2022. Tackle new questions of GATE level. The test series has validity till the end of the GATE 2022 exam. You can attempt the test anytime before GATE 2022 through our website or APSEd mobile application.

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]]>

At every horizontal section of a road, the radius of the horizontal curve (R) becomes low as a result of which centrifugal force increases and acts outwards (i.e., away from the center) in the horizontal direction on the outer wheel.

The upper image represents the same with,

l - length of the vehicle,

b - the breadth of the vehicle,

R - radius of the horizontal curve,

P - the centrifugal force which is given by

*P = (mv^2)/R*

where,

m - the mass of the vehicle

v - velocity of the vehicle

Due to high speed and low radius at the horizontal curve, large centrifugal force develops which could lead to "overturning of the vehicle" or "skidding of the vehicle".

The moment caused by centrifugal force about the outer wheel, Overturning moment (Mo),

** Mo = P * h**, h - height of C.G.

The moment caused by self-weight of the vehicle, Restoring moment (Mr),

**Mr = (W * b)/2**

For safe condition, i.e., no overturning, Mr > Mo, by which we get,

**(W * b)/2 > P * h**

**b/2h > P / W**

By substituting value for P = (mv^2)/R and W = mg, we get,

**b/2h > (v^2)/(gR)**

In this condition, no overturning occurs.

Ff - lateral frictional resistance, which is given by,

Ff = f * (Ra + Rb), where,

f - coefficient of lateral friction,

Ra + Rb = W (i.e., mg)

**Ff = f * mg**

For safe condition, i.e., no skidding, Ff > P, by which we get,

**Ff > P**

**f * mg > (mv^2)/R**

**f > (v^2)/(gR)**

In this condition, no lateral skidding occurs.

To have the above-mentioned safe conditions of no overturning and no skidding, the vehicular dimensions must be appropriate and the coefficient of lateral friction must also have an appropriate higher value. Unfortunately, both of which are not in control of a highway designer. Therefore, to have a safe passage through a horizontal curve superelevation is being introduced.

As said earlier, superelevation is the transverse slope along the width of the road, provided to develop centripetal force to counteract the centrifugal force. It is achieved by raising the outer edge with respect to the inner edge in a transverse direction for the total length of the curved section. The below figure could be referred for derivation.

For equilibrium condition in the transverse direction,

W*sinθ - P*cosθ + Ff1 + Ff2 = 0

P*cosθ - W*sinθ = R1 + R2

P*cosθ - W*sinθ = fN1 + fN2

P*cosθ - W*sinθ = f*(N1+N2)

P*cosθ - W*sinθ = f*(P*sinθ + W*cosθ)

P*cosθ - f*P*sinθ = W*sinθ + f*W*cosθ

P/W = (sinθ + f*cosθ)/(cosθ - f*sinθ), dividing by cosθ on R.H.S.,

**(v^2)/gR = (tanθ + f)/(1 - f*tanθ)**

Substituting tanθ = e,

*v^2/(gR) = (e + f)/(1 - e*f)*

As per I.R.C., f=0.15 and e is limited to 0.07 (for plain and rolling zone). Therefore e*f becomes 1.

*v^2/(gR) = e + f*

e = rate of superelevation

f = design value of coefficient of lateral friction = 0.15

v = design speed of the vehicle, m/sec

R = radius of the horizontal curve, m

g = acceleration due to gravity = 9.8 m/sec^2

Substituting g = 9.8 and converting v to kmph, we get,

v = design speed of the vehicle, kmph

R = radius of the horizontal curve, m

With these formulas, e value could be found out. If f is taken as 0 (zero), then equilibrium superelevation could be found out. Suitable radius (R) for the constrained values of e and f could also be found out.

IRC provides limiting superelevation value for different zones, as follows,

If in case for plain zone both e and f exceeds 0.07 and 0.15 respectively, then design speed(v) must be reduced or radius of the horizontal curve(R) must be increased.

Vehicles do not have the same speed on a horizontal curve, therefore in such a case, only mixed traffic flow condition is present. For superelevation calculation in mixed traffic flow conditions, the speed shall be taken as **75% of design speed** i.e., 0.75v, and the lateral friction 'f' shall be neglected for safe conditions. Superelevation formula now becomes,

**e = (0.75 * v^2)/(127 * R)**

**e = (v^2)/(225 * R)**

where,

v - design speed in kmph,

R - radius of the horizontal curve in m,

For further clarifications and insights do check out the video lecture on horizontal curve design attached below.

https://www.youtube.com/watch?v=RNONYN2XiPgWant to stay updated with the latest technical content and job notification blogs? Subscribe to our APSEd blog by providing your mail id below.

]]>Go through the blog post to get all the details regarding the recruitment.

The official announcement is here, check it out.

https://www.youtube.com/watch?v=Xp14Hgo4kGMCandidates will be placed at Projects / Power Stations/ Offices including Joint Ventures & Subsidiary Companies of NHPC located in various parts of the country or abroad.

**Trainee Engineer (Civil):**Total Vacancy: 29**Trainee Engineer (Mechanical):**Total Vacancy: 20**Trainee Engineer (Electrical):**Total Vacancy: 04**Trainee Office**r**:**Total Vacancy: 14

The criteria include educational qualification, essential work experience, GATE 2021 Score Card and age limit.

Only GATE- 2021 score (Marks out of 100) is valid for the recruitment of Trainee Engineer (Civil/ Mechanical/ Electrical). GATE score of 2020 or prior is not valid. Only CA/CMA score is valid for the recruitment of Trainee Officer (Finance). Only CS score is valid for the recruitment of Trainee Officer (Company Secretary).

- Start date for receiving online applications 22/12/2021 (11:00 Hrs)
- Last date for receiving of online applications 17/01/2022 (17:00 Hrs)

Step-1: Log on to www.nhpcindia.com & click on “Career”.

Step-2: Read all instructions given on the website.

Step-3: Fill the Online application form with relevant details and submit.

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This year, GATE 2022 saw some major changes: ,two new papers got introduced for GATE 2022 and that ,the paper combinations of attempting multiple papers changed a bit. In GATE 2022, one can appear for 2 papers. With the paper wise schedule announced, you can plan your travel and on-exam day plans! Check-

- GATE 2022 Syllabus for Geomatics Engineering and Paper Combinations
- GATE 2022 Syllabus for Naval Architecture & Marine Engineering and Paper Combinations
- GATE 2022 Syllabus for Civil Engineering and Paper Combinations
- GATE ES 2022 Syllabus and Paper Combinations

,The two new papers are, ,,Geomatics Engineering and ,Naval Architecture & Marine Engineering.

The schedule of important GATE 2021 papers are as follows:

- GATE Civil Engineering: 12th Feb 2022, 09:00 AM to 12:00 PM & 12th Feb 2022, 14:30 PM to 17:30 PM
- GATE Mechanical Engineering: 13th Feb 2022, 09:00 AM to 12:00 PM & 13th Feb 2022, 14:30 PM to 17:30 PM
- GATE Electrical Engineering: 05th Feb 2022, 14:30 PM to 17:30 PM
- GATE Electronics and Communication Engineering: 06th Feb 2022, 09:00 AM to 12:00 PM
- GATE Computer Science: 05th Feb 2022, 09:00 AM to 12:00 PM
- GATE Environmental Science and Engineering: 06th Feb 2022, 09:00 AM to 12:00 PM
- GATE ,Geomatics Engineering: 13th Feb 2022, 14:30 PM to 17:30 PM
- GATE ,Naval Architecture & Marine Engineering: 06th Feb 2022, 09:00 AM to 12:00 PM

The detailed schedule is here.

Got 2 papers back to back? Check schedule and plan accordingly.

Have you not checked it yet? You can get to practice exactly how the screen will look in the exam hall.

Like every time, the exam will be a Computer Based Test(CBT). The duration of the entire test is 3 hours.

You can practice these GATE mock tests here

No worry if you have not, we got your back. You just click on the link__,__ and you will be redirected to the GATE CE mock test. You need not enter any username and password, just sign in.

Yes, you are John Smith! Click on next then start the test.

*Disclaimer: We have sourced the information from the GATE official website. Kindly check the official website for any further changes and updates. *

Time management is key to meeting customer/client expectations as well as to meeting business goals and projections. There are various Contract Obligations and milestones that are subjected to the Timely Completion of a task/Activities/Project Segment.

To get a summarized and exact real-time view of a project various techniques are used. Pictorial, Graphical representation is one of the very abundantly used techniques to access the key activities/milestones of a project. Bar Chart, Gantt Chart and Milestone Charts are used for the same purpose.

The milestone chart is better than the bar chart as bar charts will not show the interrelation between the activities. But both the charts show the major activities only. If we keep on adding the minor activities, it will prolong too long to understand. To overcome these difficulties, PERT (Program Evaluation & Review Technique) & CPM (Critical Path Methods) are used.

This technique is used when the time of activity is not certain. So, it's a probabilistic approach.

Let's understand this with an example. Simple construction activity takes for sure the 3 months of duration. If we are constructing the same project in a hilly terrain region we cannot assure the duration, due to the site difficulties which are unpredictable. In those cases, PERT is used. PERT is usually used in research stations.

There are 3 time estimates:

**Optimistic completion of time (To)**: The time taken to complete the project, If everything goes as per plan (1% of chance for this time).**Pessimistic completion of time (Tp):**The time taken to complete the project, If nothing goes as per plan (1% of chance for this time).**Most likely time (Tm):**This is the Practical time, that a project would take to get completed.

In terms of distribution for a certain activity, if we plot these times over the frequency,

The past activities data is plotted over the curve, To, Tp occur very rarely. Most likely time (Tm) occurred with more frequency.

Activities mostly follow the beta distribution. It may not be symmetric all the time, they would be either left/right-skewed too. Based on these 3 estimates the expected time (Te) is determined.

The expected Time, Te = (To + 4TM + Tp)/6

The Standard Deviation = (Tp-To)/6, More is the Standard Deviation, more is the Uncertainty.

CPM is used in the construction/Industrial project where the activity times are certain. Laying the foundation takes X duration of time for sure is a certainty, so we may analyse this under CPM. This technique is deterministic in nature and Cost Optimization is given importance as the time is already certain.

**Early Start (ES):** The earliest time when an activity can start

**Early Finish (EF):** The earliest time when an activity can finish

**Late Start (LS):** The latest time when an activity can start

**Late Finish (LF):** The latest time when an activity can finish

Let's understand this with an example.

Here TE represents the earliest expected time and TL represents the latest allowable Occurance Time.

The Earliest given activity of duration 2 can be started is 8. ES=8. The Earliest given activity that can be finished is 8+2 = 10. EF = 10

EF=ES + tij, Here the ES is the TE of the preceding event

The latest start LS=LF-tij, Here the LF is the TL of succeeding events.

It's all about the basics related to PERT and CPM. Now, we will see some the majorly used terms in Project Management while scheduling a project.

Timespan by which starting of an activity can be delayed from the early start, without delaying the completion of the project.

For the given activity C, ES=6; EF=11; LS=14; LF = 18; activity duration=5. For the given activity C, the latest activity can start is 6, and it cannot exceed 19 as the LF is 19 and the activity itself takes a duration of 5. Therefore the Total float of C is.

(TF)c = LF - ES - tij

TF=LF-EF or LS-ES

Along the critical path Total float is 0.

Amount of time an activity can be delayed without delaying the ES of any following activity.

FF = (ES) of succeeding activity -(ES) of current activity-Duration of current activity

=(ES) of succeeding activity- (EF) of current activity

(FF)C = 18-6-5 = 7

=18-11=7

Okay, this is all for the blog. Solve the question below to understand the concept even more perfectly.

Let me know by commenting, how did you find the blog. What do you want next?

]]>BOD is thus an indirect measure of the sum of all biodegradable organic substances in the water. This also indicates how much dissolved oxygen (milligram per litre) is needed in a given time for the biological degradation of the organic wastewater constituents.

Some oxygen is always dissolved in water and it is known as dissolved oxygen (DO). DO is necessary for the survival of aquatic life and a minimum of 4 PPM of dissolved oxygen is required for the survival of aquatic life. In the case of wastewater, organic matter present requires the DO to decompose them.

Microorganisms that decompose the organic matter in the presence of oxygen are known as aerobic bacteria. BOD is defined as the oxygen required by aerobic microorganisms to oxidise the biodegradable organic matter.

https://www.youtube.com/watch?v=eDTAW0ZvQI8It can be used for finding the quantity of oxygen required to stabilise the biodegradable organic matter. It helps us in deciding the size of the treatment units of the wastewater treatment plant. It gives us an idea of the efficiency of the process i.e. how much BOD is being removed. Other than that we can find the strength of sewage.

We can further classify organic matters into two groups:

- Carbonaceous matter: First Stage BOD
- Nitrogenous matter: Second Stage BOD

Total BOD is the summation of the above two.

The standard method used in the laboratory to find the BOD of a sample is the 'Dilution Method'. It is a time taking process, in 5 days 60% -70% of the organic matter is decomposed. In 20 days 95%-99% of the organic matter is oxidised. The water sample is diluted with aerated water and the initial DO is found, then it is incubated for 5 days at 20 Degrees Celsius. After these 5 days, we again find the DO which is now known as the final DO.

BOD₅ at 20 Degrees Celsius = (Initial DO - Final DO) x Dilution Factor

The Dilution Factor is defined as the ratio of Volume of diluted sample to Volume of the undiluted sample.

Dilution Factor, DF = Diluted Sample Volume/Undiluted Sample Volume

To derive the BOD equation, Oxygen Equivalent of Organic Matter Present is plotted against Time. 'Oxygen Equivalent of Organic Matter Present' is nothing but the BOD remaining.

The rate of change of BOD directly varies with BOD remaining at that point in time (Lt)

**So, dLt/dt ∝ Lt**

Integrating both the sides from interval Lo to Lt in time interval 0 to t,

Lt = Lo x e^(-kt)

BOD Consumed, Yt=Lo-Lt

At any time, Yt = Lo - Lo x e^(-kt)

Yt = Lo[1-e^(-kt)]

If we intend to find out the solution in base 10 instead of base e, then

Lt = Lo x 10^(-k't), Where k' = k/2.303

Here the K' is also known as the deoxygenation Constant, which depends upon the temperature. The formulas given here are in at a standard temperature of 20 Degrees Celsius. To convert to some other Temperature,

K' (T Degree Cel.) = K' (20 Degree Cel.) x [1.047] ^(T-20)

To find the ultimate BOD, that happens when t tends to infinity.

So, Ultimate BOD = Lo

To understand these formulas more properly, I have a solved example for you. Do check it out.

https://www.youtube.com/watch?v=LbNFV0iHe_4BOD is a very important topic asked in GATE. Solve this previous year question to get an idea of the kind of question that gets asked.

Thank you for going through the blog. See you soon in the next one. Till then, take care!

]]>Index properties like grain shape, distribution, consistency, and plasticity help us classify soils. Sieve analysis can be used to determine grain size distribution.

- Coarse grained soil classification by Grain Size Distribution
- Fine grained soil classification by Consistency and Plasticity Index

- Unified soil classification system(USC)
- American Association of State Highways and Transport office system(AASHTO)
- Indian Standard Soil classification system(ISSCS)

Find all details about USC and AASHTO here

The three major groups in ISSCS are coarse grained, fine grained, peat/organic.

Sieve analysis is used for grain size distribution. Different sizes of sieves are used to classify the soil.

Sieve analysis by mechanical shaking (10-15min).

- % retained on a sieve = (Weight of soil retain ×100)/(Total weight taken)
- Cumulative % retained = sum of % retained on all the sieves of larger size and that of a particular size.
- % finer than the sieve under consideration=100-cumulative %retained

If the percentage retained on 75micron is >50%, then it comes under this class.

If the percentage retained on 4.75 mm is >50%.

If percentage fines pass 75 microns <5% and Cu > 4 and 1 < Cc < 3 then GW (Well graded Gravel). If any of this condition fails then it comes under GP (Poorly graded Gravel).

If percentage fines range between 5-12 % and Cu >4 and 1

- If % fines>12% then if PI<4% then GM
- If % fines>12% then if PI>7% then GC
- If % fines>12% then if 4< PI<7% then GM-GC

If % retained on 4.75 mm is <50%, then the category is sand.

If %fines passing 75 microns <5% and Cu>6 and 1

If % fines 5-12% and Cu>4 and 1

- If %fines>12% then if PI<4% then SM
- If %fines>12% then if PI>7% then SC
- If %fines>12% then if 4< PI<7% then SM-SC

Example of soil classification

It is fine-grained when % retained in 75-micron sieve is less than 50%. Further classification depends on the liquid limit, plasticity index and A-line.

- If WL <35% and Ip>7% then CL; If 35%
7% then CI; If WL>50% and Ip>7% then CH. - If WL <35% and Ip<4% then ML/OL; If 35%
50% and Ip<4% then MH/OH. - If WL <35% and 4
50% and 4

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]]>Go through the blog post to get all the details regarding the recruitment.

The official announcement is here, check it out.

https://www.youtube.com/watch?v=svel2O6Hj2MOut of a total of 70 vacancies, 65 are current vacancies and rest 4 are backlog vacancy. Let’s see the number of vacancies for each post/level.

**Dy. Project Manager (Electrical):**Total Vacancy: 10, UR: 06, SC: 01, OBC: 02, EWS: 01**Management Trainee (Civil):**Total Vacancy: 40, UR: 17, SC: 06, ST: 03, OBC: 04, EWS: 01**Management Trainee (Electrical):**Total Vacancy: 15, UR: 07, SC: 02, ST: 01, OBC: 04, EWS: 01**Project Manager (Civil)- Backlog Vacancy:**Total Vacancy: 01**Sr. Stenographer- Backlog Vacancy:**Total Vacancy: 01**Office Assistant- Backlog Vacancy:**Total Vacancy: 03, OBC: 03

The criteria include educational qualification, essential work experience, GATE 2021 Score Card and age limit. We will see the eligibility post by post.

Essential Qualification: Full-Time Degree in Electrical Engineering or equivalent from

Govt. recognized University/ Institute with minimum 60% aggregate marks.

Work Experience: 3 Years

GATE Scorecard: Not Required

Upper age as of 08.01.2022: 33 Years

Essential Qualification: Full-time Degree in Civil Engineering or equivalent from Government recognized University/Institute with minimum 60% aggregate marks.

Work Experience: Not Required

GATE Scorecard: GATE 2021 CE Score Card

Upper age as of 08.01.2022: 29 Years

Essential Qualification: Full-time Degree in Electrical Engineering or equivalent from Government recognized University/Institute with minimum 60% aggregate marks.

Work Experience: Not Required

GATE Scorecard: GATE 2021 EE Score Card

Upper age as of 08.01.2022: 29 Years

Essential Qualification: Full-time Degree in Civil Engineering or equivalent from Government recognized University/Institute with minimum 60% aggregate marks.

Work Experience: 06 Years

GATE Scorecard: Not Required

Upper age as of 08.01.2022: 47, 50, 52 Years (GEN/EWS, OBC, SC/ST respectively)

Essential Qualification: Graduate in any stream. Stenography/Typing speed in English 110/50 wpm OR Stenography/Typing speed in Hindi 100/40 wpm respectively.

Work Experience: 02 Years

GATE Scorecard: Not Required

Upper age as of 08.01.2022: 28 Years

Essential Qualification: Graduate in any stream. Stenography/Typing Speed in English 70/35 wpm OR Stenography/Typing speed in Hindi 70/30 wpm respectively.

Work Experience: Not Required

GATE Scorecard: Not Required

Upper age as of 08.01.2022: 25 Years

- Dy. Project Manager (Electrical): 50000-160000 /-
- Management Trainee (Civil): 40000-160000 /-
- Management Trainee (Electrical): 40000-160000 /-
- Project Manager (Civil): 60000-180000 /-
- Sr. Stenographer: 24640 /-
- Office Assistant: 18430 /-

- Opening Date for Online Registration: 09.12.2021 from 1000 hrs
- Closing Date of Online Registration: 08.01.2022 up to 1700 hrs

The Candidates are required to apply online. The relevant link for the online application will be made available from 10:00 hrs on 09.12.2021 under the head “CAREER within Human Resources” on the NBCC website.

Final Online submission of application will be open till 17:00 hrs on 08.01.2022. No other means/mode of application shall be accepted.

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Go through the blog post to get all the details regarding the recruitment.

Total of 45 posts and the engagement of successful candidates will be on contract basis (and not on regular basis) initially for a period for three years.

Download RLDA recruitment advertisement here

Assistant Project Engineer (Civil)/ Contract: 45 posts

Applicant should have valid GATE score (Subject code: CE)

Full time BE/ B. Tech. (Civil Engineering) with not less than 60%

Candidates who have completed M.Tech/ME in Civil Engineering discipline are also eligible to apply

Duly filled in application form is to sent by 23.12.2021

Tentative Stations: Tirupati & Nellore in Andhra Pradesh, Guwahati/ Kamakhya in Assam, Gaya & Muzaffarpur in Bihar, Chandigarh in Chandigarh (UT), Panipat in Haryana, Surat, Udhna, & Somnath in Gujarat, Jammu Tawi in J & K (UT), Ranchi in Jharkhand, Bangalore Cantt in Karnataka, Ernakulam, & Kollam in Kerala, Gwalior in Madhya Pradesh, Mumbai (3) & Nagpur in Maharashtra, Bhubaneshwar & Puri in Odisha, Puducherry in Puducherry (UT), Ludhiana & Amritsar in Punjab, Udaipur, Jaipur (2) & Kota (2) in Rajasthan, Kaniyakumari, Rameshwaram & Madurai in Tamilnadu, Hyderabad in Telangana, Kathgodam in Uttarakhand, Lucknow, Kanpur, Prayagraj & Ghaziabad in Uttar Pradesh, New Jalpaiguri & Asansol in West Bengal & 04 other stations to be decided later on.

Selection Process: The only parameter for selection is valid GATE percentile for Civil Engineering (Subject Code: CE) of eligible applicants. In case, the valid GATE percentile of two candidates is same, then age-wise senior candidate will get preference in selection as well as choice of posting.

Practice best & relevant Qs for GATE

* and re-check the information before applying. *

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The deflection and slope of any beam (not particularly a simply supported one) primary depend on the load case it is subjected upon. If the load case varies, its deflection, slope, shear force and bending moment get changed.

This article will help you find the deflection and slope developed at any point of simply supported and cantilever beams, subjected to temperature change.

The simply supported beam is one of the most modest structures. The configuration of a simply supported beam is so simple having one hinge support at an end and roller support at the other end. With this setup the beam can only rotate horizontally, any vertical moment is restained.

In a case of uniform temperature change, the linear change in dimension is,

Change = L α (ΔT)

Where, L: Original span of the beam

α: Coefficient of Thermal Expansion

ΔT: Temperature Change

So, if the temperature change in a beam is uniform, we can directly apply the above formula to find the deflection due to temperature change. But if the temperature changes in non-uniform, the temperature of the beam across the depth will be different. Let us see that case as well.

Consider, T0: Initial temperature of the beam

T1: Final temperature of the top of the beam

T2: Final temperature of the bottom of the beam

Assume, T2 > T1.

Take a small part **dx **for analysis.

Now, let us find the change in length at the top and bottom of the beam, considering both are two separate cases.

- For the bottom part, the change in length = α dx (T2-T0). So, the new bottom length, Lb = dx + α dx (T2-T0)
- For the bottom part, the change in length = α dx (T1-T0). So, the new bottom length, Lt = dx + α dx (T1-T0)

Due to different expansion in two sides of the beam, the beam will rotate and form an angle dΘ.

So, we can write:

dΘ x h = Lb - Lt = α dx (ΔT)

⇒ (dΘ/dx) = α ΔT/h (Equation 1)

According to the Double Integration formula of beam deflection,

d²y/dx² = (dΘ/dx) = M/EI (Equation 2)

Now, comparing equations 1 and 2,

**d****²****y/dx****²**** = α ΔT/h **(Equation 3)

This is the final formula, we use to get the slope and deflection of the beam due to nonuniform temperature variation in the beam.

Hope you are very clear with the concept. So what to wait then? Let's solve a question on a similar concept. Don't scroll the blog further, solve by yourself, then match your explanation/answer with mine!

Find the deflection at the centre of the beam, due to temperature variation at the bottom and top of the beam as shown.

According to the equation 3, d²y/dx² = α ΔT/h

Integrating once with respect to x, dy/dx = α ΔTx/h + C1

Integrating once again, y = α ΔTx²/2h + xC1 + C2

Applying the boundary/end conditions

(i) At x=0, y=0, so C2=0

(ii) At x=L, y=0, C1=(-α ΔTL/2h)

Now putting the values of C1 and C2, in the deflection function,

y = α ΔTx²/2h - xα ΔTL/2h

As the question asks to find out the deflection at the centre of the beam, Put x = L/2

**So, y = (-α ΔTL²/8h) [DOWNWARD]**

All Clear now? Solve one more question.

Thank you for going through the blog. Hoe you had an awesome reading. Subscribe to A,PSEd Blog and get to know the latest.

]]>