QRG_CE.pdf

QUICK REFRESHER GUIDE For

Civil Engineering

By

www.thegateacademy.com

Quick Refresher Guide

Contents

CONTENTS Part Mathematics

Page No. 1 – 45

1.1 Linear Algebra 1.2 Probability & distribution 1.3. Numerical Methods 1.4. Calculus 1.5. Differential Equations 1.6. Complex Variables 1.7: Laplace Transform

1–8 9 – 14 15 – 19 20 – 30 31 – 37 38 – 42 43 – 45

Structural Engineering

46 – 141

2.1 Mechanics 2.2 Structural Analysis 2.3 Concrete Technology 2.4 Steel Structures

46 – 59 60 – 79 80 – 100 101 – 141

Geotechnical Engineering

142 – 171

3.1 Soil Mechanics 3.2 Foundation Engineering

142 – 160 161– 171

#4

Water Resource Engineering

172 – 239

#5

Environmental Engineering

243 – 264

5.1 Water Quality & Standards 5.2 Water Supply and Its Treatment 5.3 Waste Water Treatment 5.4 Air pollution

243 – 245 246 – 253 254 – 256 257 – 264

Transportation Engineering

265 - 274

6.1 Highway Planning 6.2 Highway Materials

265 – 269 270 – 274

#7

Surveying

275 – 287

#8

Reference Books

288 – 289

#1

#2

#3

#6

4.1 Fluid Mechanics 4.2 Hydrology 4.3 Irrigation

172– 218 219– 225 226 – 242

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Mathematics

Part - 1: Mathematics 1.1 Linear Algebra 1.1.1

Matrix Definition: A system of “m n” numbers arranged along m rows and n columns. Conventionally, single capital letter is used to denote a matrix. Thus,

A=[

a

a a

a a

a

a

a a a

a a a a

]

ith row, jth column

1.1.1.1 Types of Matrices 1.1.1.2 Row and Column Matrices  Row Matrix [ 2, 7, 8, 9] 

Column Matrix

[1 ] 1 1

single row ( or row vector) single column (or column vector)

1.1.1.3 Square Matrix     -

Same number of rows and columns. Order of Square matrix no. of rows or columns Principle Diagonal (or Main diagonal or Leading diagonal): The diagonal of a square matrix (from the top left to the bottom right) is called as principal diagonal. Trace of the Matrix: The sum of the diagonal elements of a square matrix. tr (λ A) = λ tr(A) , λ is scalartr ( A+B) = tr (A) + tr (B) tr (AB) = tr (BA)

1.1.1.4 Rectangular Matrix Number of rows

Number of columns

1.1.1.5 Diagonal Matrix A Square matrix in which all the elements except those in leading diagonal are zero. e.g. [

]

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Page 1


Mathematics

1.1.1.6 Unit Matrix (or Identity Matrix) A Diagonal matrix in which all the leading diagonal elements are ‘1’. 1 e.g. I = [ ] 1 1 1.1.1.7 Null Matrix (or Zero Matrix) A matrix is said to be Null Matrix if all the elements are zero. e.g.

0

1

1.1.1.8 Symmetric and Skew Symmetric Matrices:  Symmetric, when a = +a for all i and j. In other words  Skew symmetric, when a = - a In other words = -A

=A

Note: All the diagonal elements of skew symmetric matrix must be zero. Symmetric Skew symmetric a h g h g f] [h b f ] [h g f c g f

Symmetric Matrix

𝐀𝐓 = A

Skew Symmetric Matrix 𝐀𝐓 = - A

1.1.1.9 Triangular Matrix  A matrix is said to be “upper triangular” if all the elements below its principal diagonal are zeros.  A matrix is said to be “lower triangular” if all the elements above its principal diagonal are zeros. a a h g [ ] [ g b ] b f f h c c Upper Triangular Matrix Lower Triangular Matrix 1.1.1.10

Orthogonal Matrix: If A. A = I, then matrix A is said to be Orthogonal matrix.

1.1.1.11

Singular Matrix: If |A| = 0, then A is called a singular matrix.

1.1.1.12

̅) Unitary Matrix: If we define, A = (A Then the matrix is unitary if A . A = I

= transpose of a conjugate of matrix A


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1.1.1.13

Mathematics

Hermitian Matrix: It is a square matrix with complex entries which is equal to its own conjugate transpose. A = A or a = a̅̅̅

1.1.1.14

Note: In Hermitian matrix, diagonal elements

1.1.1.15

Skew Hermitian matrix: It is a square matrix with complex entries which is equal to the negative of conjugate transpose. A = A or a =

a̅̅̅

Note: In Skew-Hermitian matrix , diagonal elements 1.1.1.16

always real

either zero or Pure Imaginary

Idempotent Matrix If A = A, then the matrix A is called idempotent matrix.

1.1.1.17

Multiplication of Matrix by a Scalar:

Every element of the matrix gets multiplied by that scalar. Multiplication of Matrices: Two matrices can be multiplied only when number of columns of the first matrix is equal to the number of rows of the second matrix. Multiplication of (m n) , and (n p) matrices results in matrix of (m p)dimension , =, . 1.1.1.18

Determinant:

An n order determinant is an expression associated with n

n square matrix.

If A = [a ] , Element a with ith row, jth column. For n = 2 ,

a D = det A = |a

a a |=a

a

-a

a

Determinant of “order n”

D = |A| = det A = ||

a a

a

a

a

a

a a

| |

a


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1.1.1.19 

Mathematics

Minors & Co-Factors:

The minor of an element in a determinant is the determinant obtained by deleting the row and the column which intersect that element. Cofactor is the minor with “proper sign”. The sign is given by (-1) (where the element th th belongs to i row, j column).



1.1.1.20 Properties of Determinants: 1. A determinant remains unaltered by changing its rows into columns and columns into rows. 2. If two parallel lines of a determinant are inter-changed, the determinant retains its numerical values but changes its sign. (In a general manner, a row or a column is referred as line). 3. Determinant vanishes if two parallel lines are identical. 4. If each element of a line be multiplied by the same factor, the whole determinant is multiplied by that factor. [Note the difference with matrix]. 5. If each element of a line consists of the m terms, then determinant can be expressed as sum of the m determinants. 6. If each element of a line be added equi-multiple of the corresponding elements of one or more parallel lines, determinant is unaffected. e.g. by the operation, R R + pR +qR , determinant is unaffected. 7. Determinant of an upper triangular/ lower triangular/diagonal/scalar matrix is equal to the product of the leading diagonal elements of the matrix. 8. If A & B are square matrix of the same order, then |AB|=|BA|=|A||B|. 9. If A is non singular matrix, then |A |=| | (as a result of previous). 10. 11. 12. 13.

Determinant of a skew symmetric matrix (i.e. A =-A) of odd order is zero. If A is a unitary matrix or orthogonal matrix (i.e. A = A ) then |A|= ±1. If A is a square matrix of order n, then |k A| = |A|. |I | = 1 ( I is the identity matrix of order n).

1.1.1.21

Inverse of a Matrix



A

 

|A| must be non-zero (i.e. A must be non-singular). Inverse of a matrix, if exists, is always unique. a b d If it is a 2x2 matrix 0 1 , its inverse will be 0 c d c



=

| |

b 1 a

Important Points: 1. IA = AI = A, (Here A is square matrix of the same order as that of I ) 2. 0 A = A 0 = 0, (Here 0 is null matrix) 3. If AB = , then it is not necessarily that A or B is null matrix. Also it doesn’t mean BA = . 4. If the product of two non-zero square matrices A & B is a zero matrix, then A & B are singular matrices. 5. If A is non-singular matrix and A.B=0, then B is null matrix. 6. AB BA (in general) Commutative property does not hold 7. A(BC) = (AB)C Associative property holds 8. A(B+C) = AB AC Distributive property holds THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com

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9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

Mathematics

AC = AD , doesn’t imply C = D ,even when A -. If A, C, D be matrix, and if rank (A)= n & AC=AD, then C=D. (A+B)T = A + B (AB)T = B . A (AB)-1 = B . A AA =A A=I (kA)T = k.A (k is scalar, A is vector) (kA)-1 = . A (k is scalar , A is vector) (A ) = (A ) ̅ ) (Conjugate of a transpose of matrix= Transpose of conjugate of matrix) (̅̅̅̅ A ) = (A If a non-singular matrix A is symmetric, then A is also symmetric. If A is a orthogonal matrix , then A and A are also orthogonal.

21. If A is a square matrix of order n then (i) |adj A|=|A| (ii) |adj (adj A)|=|A|( ) (iii) adj (adj A) =|A| A 1.1.1.22 Elementary Transformation of a Matrix: 1. Interchange of any 2 lines 2. Multiplication of a line by a constant (e.g. k R ) 3. Addition of constant multiplication of any line to the another line (e. g. R + p R ) Note:  Elementary transformations don’t change the ran of the matrix.  However it changes the Eigen value of the matrix. 1.1.1.23

Rank of Matrix

If we select any r rows and r columns from any matrix A,deleting all other rows and columns, then the determinant formed by these r r elements is called minor of A of order r. Definition: A matrix is said to be of rank r when, i) It has at least one non-zero minor of order r. ii) Every minor of order higher than r vanishes. Other definition: The rank is also defined as maximum number of linearly independent row vectors. Special case: Rank of Square matrix Rank = Number of non-zero row in upper triangular matrix using elementary transformation. Note: 1. 2. 3. 4.

r(A.B) min { r(A), r (B)} r(A+B) r(A) + r (B) r(A-B) r(A) - r (B) The rank of a diagonal matrix is simply the number of non-zero elements in principal diagonal. 5. A system of homogeneous equations such that the number of unknown variable exceeds the number of equations, necessarily has non-zero solutions. THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com

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Mathematics

If A is a non-singular matrix, then all the row/column vectors are independent. If A is a singular matrix, then vectors of A are linearly dependent. r(A)=0 iff (if and only if) A is a null matrix. If two matrices A and B have the same size and the same rank then A, B are equivalent matrices. 10. Every non-singular matrix is row matrix and it is equivalent to identity matrix. 6. 7. 8. 9.

1.1.1.24

Solution of linear System of Equations:

For the following system of equations A X = B a a

a

x x

a a

Where, A =

, [a

a

a

]

=

,

B =

[x ]

[

]

A= Coefficient Matrix, C = (A, B) = Augmented Matrix r = rank (A), r = rank (C), n = Number of unknown variables (x , x , - - - x ) Consistency of a System of Equations: For Non-Homogenous Equations (A X = B) i) If r r , the equations are inconsistent i.e. there is no solution. ii) If r = r = n, the equations are consistent and there is a unique solution. iii) If r = r < n, the equations are consistent and there are infinite number of solutions. For Homogenous Equations (A X = 0) i) If r = n, the equations have only a trivial zero solution ( i.e. x = x = - - - x = 0). ii) If r < n, then (n-r) linearly independent solution (i.e. infinite non-trivial solutions). Note: Consistent means:

one or more solution (i.e. unique or infinite solution)

Inconsistent means:

No solution

Cramer’s Rule Let the following two equations be there a

x +a

x = b ---------------------------------------(i)

a

x +a

x = b ---------------------------------------(ii)

a D = |b

a b |


Page 6


b D =| b

a | a

a D =| a

b | b

Mathematics

Solution using Cramer’s rule: x =

and x =

In the above method, it is assumed that 1. No of equations = No of unknowns 2. D 0 In general, for Non-Homogenous Equations D 0 single solution (non trivial) D = 0 infinite solution For Homogenous Equations D 0 trivial solutions ( x = x =………………………x = 0) D = 0 non- trivial solution (or infinite solution) Eigen Values & Eigen Vectors 1.1.1.25

Characteristic Equation and Eigen Values:

Characteristic equation: | A λ I |= 0, The roots of this equation are called the characteristic roots /latent roots / Eigen values of the matrix A. Eigen vectors: [

]X=0

For each Eigen value λ, solving for X gives the corresponding Eigen vector. Note: For a given Eigen value, there can be different Eigen vectors, but for same Eigen vector, there can’t be different Eigen values. Properties of Eigen values 1. The sum of the Eigen values of a matrix is equal to the sum of its principal diagonal. 2. The product of the Eigen values of a matrix is equal to its determinant. 3. The largest Eigen values of a matrix is always greater than or equal to any of the diagonal elements of the matrix. 4. If λ is an Eigen value of orthogonal matrix, then 1/ λ is also its Eigen value. 5. If A is real, then its Eigen value is real or complex conjugate pair. 6. Matrix A and its transpose A has same characteristic root (Eigen values). 7. The Eigen values of triangular matrix are just the diagonal elements of the matrix. 8. Zero is the Eigen value of the matrix if and only if the matrix is singular. 9. Eigen values of a unitary matrix or orthogonal matrix has absolute value ‘1’. 10. Eigen values of Hermitian or symmetric matrix are purely real. 11. Eigen values of skew Hermitian or skew symmetric matrix is zero or pure imaginary. | | 12. is an Eigen value of adj A (because adj A = |A|. A ). THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com

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13. If i) ii) iii) iv) v)

Mathematics

λ is an Eigen value of the matrix then , Eigen value of A is 1/λ Eigen value of A is λ Eigen value of kA are λ (k is scalar) Eigen value of A I are λ + k Eigen value of (A I)2 are ( )

Properties of Eigen Vectors 1) Eigen vector X of matrix A is not unique. Let is Eigen vector, then C is also Eigen vector (C = scalar constant). 2) If λ , λ , λ . . . . . λ are distinct, then , . . . . . are linearly independent . 3) If two or more Eigen values are equal, it may or may not be possible to get linearly independent Eigen vectors corresponding to equal roots. 4) Two Eigen vectors are called orthogonal vectors if T∙ = 0. ( , are column vector) (Note: For a single vector to be orthogonal , A = A or, A. A = A. A =  ) 5) Eigen vectors of a symmetric matrix corresponding to different Eigen values are orthogonal. Cayley Hamilton Theorem: Every square matrix satisfies its own characteristic equation. 1.1.1.26

Vector:

Any quantity having n components is called a vector of order n. Linear Dependence of Vectors  If one vector can be written as linear combination of others, the vector is linearly dependent. Linearly Independent Vectors  If no vectors can be written as a linear combination of others, then they are linearly independent. Suppose the vectors are x x x x  

Its linear combination is λ x + λ x + λ x + λ x = 0 If λ , λ , λ , λ are not “all zero” they are linearly dependent. If all λ are zero they are linearly independent.


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Mathematics

1.2 Probability and Distribution 1.2.1

Probability

Event: Outcome of an experiment is called event. Mutually Exclusive Events (Disjoint Events): Two events are called mutually exclusive, if the occurrence of one excludes the occurrence of others i.e. both can’t occur simultaneously. A

B =φ, P(A

B) =0

Equally Likely Events: If one of the events cannot happen in preference to other, then such events are said to be equally likely. Odds in Favour of an Event = Where m n

no. of ways favourable to A

no. of ways not favourable to A

Odds Against the Event = Probability: P(A)=

=

. .

P(A) P(A’)=1 Important points:  P(A B) Probability of happening of “at least one” event of A & B  P(A B) ) Probability of happening of “both” events of A & B  If the events are certain to happen, then the probability is unity.  If the events are impossible to happen, then the probability is zero. Addition Law of Probability: a. For every events A, B and C not mutually exclusive P(A B C)= P(A)+ P(B)+ P(C)- P(A B)- P(B C)- P(C A)+ P(A B C) b. For the event A, B and C which are mutually exclusive P(A B C)= P(A)+ P(B)+ P(C) Independent Events: Two events are said to be independent, if the occurrence of one does not affect the occurrence of the other. If P(A B)= P(A) P(B)

Independent events A & B

Conditional Probability: If A and B are dependent events, then P. / denotes the probability of occurrence of B when A has already occurred. This is known as conditional probability. P(B/A)=

(

) ( )


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For independent events A & B

Mathematics

P(B/A) = P(B)

Theorem of Combined Probability: If the probability of an event A happening as a result of trial is P(A). Probability of an event B happening as a result of trial after A has happened is P(B/A) then the probability of both the events A and B happening is P(A B)= P(A). P(B/A),

[ P(A) 0]

= P(B). P(A/B),

[ P(B) 0]

This is also known as Multiplication Theorem. For independent events A&B P(B/A) = P(B), P(A/B )= P(A) Hence P(A B) = P(A) P(B) Important Points: If P 1. 2. 3. 4.

& P are probabilities of two independent events then P (1-P ) probability of first event happens and second fails (i.e only first happens) (1-P )(1-P ) probability of both event fails 1-(1-P )(1-P ) probability of at least one event occur PP probability of both event occurs

Baye’s theorem: An event A corresponds to a number of exhaustive events B , B ,.., B . If P(B ) and P(A/B ) are given then, P. /=

( (

). ( ) ). ( )

This is also known as theorem of Inverse Probability. Random variable: Real variable associated with the outcome of a random experiment is called a random variable. 1.2.2

Distribution

Probability Density Function (PDF) or Probability Mass Function: The set of values Xi with their probabilities P constitute a probability distribution or probability density function of the variable X. If f(x) is the PDF, then f(x ) = P( = x ) , PDF has the following properties:  Probability density function is always positive i.e. f(x)  ∫ f(x)dx = 1 (Continuous)  f(x ) = 1 (Discrete)


Page 10


Mathematics

Discrete Cumulative Distribution Function (CDF) or Distribution Function The Cumulative Distribution Function F(x) of the discrete variable x is defined by, F (x) = F(x) = P(X x) =

P(x ) =

f(x )

Continuous Cumulative Distribution function (CDF) or Distribution Function: If F (x) = P(X x) =∫ f(x)dx, then F(x) is defined as the cumulative distribution function or simply the distribution function of the continuous variable. CDF has the following properties: ( ) i) = F (x) =f(x) 0 ii) 1 F (x) 0 iii) If x x then F (x ) F (x ) , i.e. CDF is monotone (non-decreasing function) ) =0 iv) F ( v) F ( ) = 1 vi) P(a x b) =∫ f(x)dx = ∫ f(x)dx - ∫ f(x)dx = F (b) F (a) Expectation [E(x)]: 1. E(X) = x f(x ) (Discrete case) 2. E(X) = ∫ x f(x )dx (Continuous case) Properties of Expectation 1. E(constant) = constant 2. E(CX) = C . E(X) [C is constant] 3. E(AX+BY) = A E(X)+B E(Y) [A& B are constants] 4. E(XY)= E(X) E(Y/X)= E(Y) E(X/Y) E(XY) E(X) E(Y) in general But E(XY) = E(X) E(Y) , if X & Y are independent Variance (Var(X)) Var (X) =E,(x

) ]

Var (X)= (x x

) f(xx )

Var (X)=∫ (xx Var (X) =E(

(Discrete case)

) f(x)dx (Continuous case)

)-,E(x)-

Properties of Variance 1. Var(constant) = 0 2. Var(Cx) = C Var(x) -Variance is non-linear [here C is constant] THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com

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Mathematics

3. Var(Cx D) = C Var(x) -Variance is translational invariant [C & D are constants] 4. Var(x-k) = Var(x) [k is constant] 5. Var(ax+by) = a Var(x) + b Var(y) 2ab cov(x,y) (if not independent) [A & B are constants] = a Var(x) + b Var(y) (if independent) Covariance Cov (x,y)=E(xy)-E(x) E(y) If independent

covariance=0,

E(xy) = E(x) . E(y)

(if covariance = 0, then the events are not necessarily independent) Properties of Covariance 1. Cov(x,y) = Cov(y,x) (i.e. symmetric) 2. Cov(x,x) = Var(x) 3. |Cov(x,y)| Standard Distribution Function (Discrete r.v. case): 1. Binomial Distribution : P(r) = C p q Mean = np, Variance = npq, S.D. =√npq 2. Poisson Distribution: Probability of k success is P (k) = no. of success trials , n no. of trials , P success case probability mean of the distribution For Poisson distribution: Mean = , variance = , and =np Standard Distribution Function (Continuous r.v. case): 1. Normal Distribution (Gaussian Distribution): f(x) =

√

e

(

)

Where and are the mean and standard deviation respectively  P(
P(x1 < x < x2) = ∫

2. Exponential distribution : 3. Uniform distribution: 4. Cauchy distribution :

√

e

(

)

dx = Area under the curve from x1 to x2

f(x) = λ e , x , here λ = , x f(x)= , b f(x) a = , otherwise f(x)= .( )

5. Rayleigh distribution function : f(x) =

e

,

x

Mean:  For a set of n values of a variant X=( x , x , … . . , x ) THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com

Page 12


Mathematics

The arithmetic mean, ̅=   

For a grouped data if x , x , … . . , x are mid values of the class intervals having frequencies f , f ,….., f ,then, ̅= If ̅ is mean for n data; ̅ is mean for n data; then combined mean of n +n data is ̅

̅

̅= If ̅̅̅ , be mean and SD of a sample size n and m , SD of combined sample of size n +n is given by, (n

n )

D = m -m ( n)

=n

+n

be those for a sample of size n then

+n D +n D

(m , = mean, SD of combined sample)

= (n

)

(n D )

Median: When the values in a data sample are arranged in descending order or ascending order of magnitude the median is the middle term if the no. of sample is odd and is the mean of two middle terms if the number is even. Mode: It is defined as the value in the sampled data that occurs most frequently. Important Points:  Mean is best measurement ( all observations taken into consideration).  Mode is worst measurement ( only maximum frequency is taken).  In median, 50 % observation is taken.  Sum of the deviation about “mean” is zero.  Sum of the absolute deviations about “median” is minimum.  Sum of the square of the deviations about “mean” is minimum. Co-efficient of variation = ̅

100

Correlation coefficient = (x,y) =      

( , )

-1 (x, y) 1 (x,y) = (y,x) |(x,y)| = 1 when P(x=0)=1; or P(x=ay)=1 [ for some a] If the correlation coefficient is -ve, then two events are negatively correlated. If the correlation coefficient is zero, then two events are uncorrelated. If the correlation coefficient is +ve, then two events are positively correlated.

Line of Regression: The equation of the line of regression of y on x is y The equation of the line of Regression of x on y is (x

̅̅̅̅

y= x) =

̅̅̅̅

(x

̅̅̅̅

x) (y

y)

is called the regression coefficient of y on x and is denoted by byx.


Page 13


̅̅̅̅

Mathematics

is called the regression coefficient of x on y and is denoted by bxy.

Joint Probability Distribution: If X & Y are two random variables then Joint distribution is defined as, Fxy(x,y) = P(X x ; Y y) Properties of Joint Distribution Function/ Cumulative Distribution Function: 1. F ( , ) = 2. F ( , ) = 1 3. F ( , ) = { F ( , ) = P( y) = 0 x 1 = 0 } ) = F (x) . 1 = F (x) 4. F (x, ) = P( x 5. F ( , y) = F (y) Joint Probability Density Function: Defined as f(x, y) = Property: ∫

∫

F(x, y) f(x, y) dx dy

= 1

Note: X and Y are said to be independent random variable If fxy(x,y) = fx(x) . fy(y)


Page 14


Mathematics

1.3 Numerical Methods 1.3.1

Solution of Algebraic and Transcendental Equation / Root Finding : Consider an equation f(x) = 0

1. Bisection method This method finds the root between points “a” and “b”. If f(x) is continuous between a and b and f (a) and f (b) are of opposite sign then there is a root between a & b (Intermediate Value Theorem). First approximation to the root is x1 =

.

If f(x1) = 0, then x1 is the root of f(x) = 0, otherwise root lies between a and x1 or x1 and b. Similarly x2 and x3 . . . . . are determined.  Simplest iterative method  Bisection method always converge, but often slowly.  This method can’t be used for finding the complex roots.  Rate of convergence is linear 2. Newton Raphson Method (or Successive Substitution Method or Tangent Method) ( ) xn+1 = xn – (

)

This method is commonly used for its simplicity and greater speed. Here f(x) is assumed to have continuous derivative f’(x). This method fails if f’(x) = . It has second order of convergence or quadratic convergence, i.e. the subsequent error at each step is proportional to the square of the error at previous step.  Sensitive to starting value, i.e. The Newton’s method converges provided the initial approximation is chosen sufficiently close to the root.  Rate of convergence is quadratic.    

3. Secant Method x

=x

(

)– (

)

f(x )

 Convergence is not guaranteed.  If converges, convergence super linear (more rapid than linear, almost quadratic like Newton Raphson, around 1.62). 4. Regula Falsi Method or (Method of False Position)  Regula falsi method always converges.  However, it converges slowly.  If converges, order of convergence is between 1 & 2 (closer to 1). THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com

Page 15


Mathematics

 It is superior to Bisection method. Given f(x) = 0 Select x0 and x1 such that f(x0) f(x1) < 0 x =x -

– (

)

(

)

, f(x ) =

(

)–

(

)

(

(i.e. opposite sign)

( ) )

Check if f(x0) f(x2) < 0 or f(x1) f(x2) < 0 Compute x

………

which is an approximation to the root. 1.3.2 1.

Solution of Linear System of Equations Gauss Elimination Method Here equations are converted into “upper triangular matrix” form, then solved by “bac substitution” method. Consider a1x + b1x + c1z = d1 a2x + b2x + c2z = d2 a3x + b3x + c3z = d3 Step 1: To eliminate x from second and third equation (we do this by subtracting suitable multiple of first equation from second and third equation) a1x + b1y + c1z = d1’ (pivotal equation, a1 pivot point.) b ’y + c ’ z = d ’ b ’y + c ’ z = d ’ Step 2: Eliminate y from third equation a1x + b1y + c1z = d1’ b ’y + c2z = d ’ c ’’z = d ”

(pivotal equation, b ’ is pivot point.)

Step 3: The value of x , y and z can be found by back substitution. Note: Number of operations: N =

2.

+n -

Gauss Jordon Method  Used to find inverse of the matrix and solving linear equations.  Here back substitution is avoided by additional computations that reduce the matrix to “diagonal from”, instead to triangular form in Gauss elimination method.  Number of operations is more than Gauss elimination as the effort of back substitution is saved at the cost of additional computation. Step 1: Eliminate x from 2nd and 3rd THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com

Page 16


Mathematics

Step 2: Eliminate y from 1st and 3rd Step 3: Eliminate z from 1st and 2nd 3.

L U Decomposition It is modification of the Gauss eliminiation method. Also Used for finding the inverse of the matrix. [A]n x n = [ L ] n x n [U] n x n a11 a12 a13 1 0 0 a21 b22 c23 L21 1 0 = a31 b32 c33 L31 L32 1

U11 U12 U13 0 U22 U23 0 0 U31

Ax = LUX = b can be written as a)LY=b and b) UX=Y Solve for from a) then solve for from b). This method is nown as Doolittle’s method.  Similar methods are Crout’s method and Choles y methods. 4. Iterative Method (i) Jacobi Iteration Method a1x + b1y + c1z = d1 a2x + b2y + c2z = d2 a3x + b3y + c3z = d3 If a1, b2 , c3 are large compared to other coefficients, then solving these for x, y, z respectively x = k1 – l1y – m1z y = k2 – l2x – m2z z = k3 – l3x – m3y Let us start with initial approximation x0 , y0 , z0 x1= k1 – l1y0 – m1z0 y1= k2 – l2y0 – m2z0 z1= k3 – l3y0 – m3z0 Note: No component of x(k) is used in computation unless y(k) and z(k) are computed. The process is repeated till the difference between two consecutive approximations is negligible. In generalized form: x(k+1) = k1 – l1 y(k) – m1z(k) y(k+1) = k2 – l2 x(k) – m2z(k) z(k+1) = k3 – l3 x(k) – m3y(k) (ii) Gauss-Siedel Iteration Method Modification of the Jacobi’s Iteration Method Start with (x0, y0, z0) = (0, 0, 0) or anything [No specific condition] THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com

Page 17


Mathematics

In first equation, put y = y0 z = z0 which will give x1 In second equation, put x = x1 and z = z0 which will give y1 In third equation, put x = x1 and y = y1 which will give z1 Note: To compute any variable, use the latest available value. In generalized form: x(k+1) = k1 – l1y(k) – m1z(k) y(k+1) = k2 – l2x(k+1) – m2z(k) z(k+1) = k3 – l3x(k+1) – m3y(k+1) 1.3.3

Numerical Integration

Trapezoidal Formula: Step size h = ∫

f(x)dx =

h

*( first term

last term)

(remaining terms)+

Error = Exact - approximate The error in approximating an integral using Trapezoidal rule is bounded by h (b 1

a) max |f ( )| , , -

Simpson’s One Third Rule (Simpson’s Rule):

∫

f(x)dx =

h

*( first term

last term)

(all odd terms)

(all even terms)+

The error in approximating an integral using Simpson’s one third rule is h (b 1

a) max |f ( ) ( )| , , -

Simpson’s Three Eighth Rule: ∫

f(x)dx =

h ( first term {

last term)

(all multiple of terms) } (all remaining terms)

The error in approximating an integral using Simpson’s / rule is (b

a)

max |f ( ) ( )| , , -

1.3.4 Solving Differential Equations (i) Euler method (for first order differential equation ) Given equation is y = f(x, y); y(x0) = y0 THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com

Page 18


Mathematics

Solution is given by, Yn+1 = yn + h f(xn,yn) (ii) Runge Kutta Method Used for finding the y at a particular x without solving the 1st order differential equation = f(x, y) K1 = h f(x0, y0) K2 = h f(x0 + , y0 + ) K3 = h f(x0 + , y0 + ) K4 = h f(x0 +h, y0 + k3) K = (k1 + 2k2 + 2k3 + k4) Y(x0+h) = y0 + k


Page 19


Mathematics

1.4 Calculus 1.4.1

Limit of a Function

Let y = f(x) Then lim f(x)= 0< |x a|< , |f(x)

i.e, “ f(x) |<

as x a” implies for any

(>0), (>0) such that whenever

Some Standard Expansions (1

x) = 1

x

a =x a

x

(

nx x

e =1+x+

+

log(1

x) = x

log(1

x) =

a

Sin x = x

x

n(n

a

1)(n

)

x

.........x

.........a

......... +

x

)

......... ......... .........

Cos x = 1

+

Sinh x = x

......... .........

Cosh x = 1

+

.........

Some Important Limits lim

sinx = x

lim (1 lim(1 lim lim

1 ) = x x) =

a

1 x

e

1 x

= log a =1


Page 20


lim lim

log(1 x x x

x)

Mathematics

=1

a = a

a

lim log|x| = L – Hospital’s Rule  When function is of limit.

or

form, differentiate numerator & denominator and then apply

Existence of Limits and Continuity: 1. f(x) is defined at a, i.e, f(a) exists. 2. If lim f(x) = lim f(x) = L ,then the lim f(x) exists and equal to L. 3. If lim

f(x) = lim

f(x)= f(a) then the function f(x) is said to be continuous.

Properties of Continuity If f and g are two continuous functions at a; then a. (f+g), (f.g), (f-g) are continuous at a b. is continuous at a, provided g(a) 0 c. |f| or |g| is continuous at a Rolle’s theorem If (i) f(x) is continuous in closed interval [a,b] (ii) f’(x) exists for every value of x in open interval (a,b) (iii) f(a) = f(b) Then there exists at least one point c between (a, b) such that

( )=0

Geometrically: There exists at least one point c between (a, b) such that tangent at c is parallel to x axis

C C 2

C1 a

b


Page 21


Mathematics

Lagrange’s Mean Value Theorem If (i) f(x) is continuous in the closed interval [a,b] and (ii) f’(x) exists in the open interval (a,b), then atleast one value c of x exist in (a,b) such that ( )

( )

= f (c).

Geometrically, it means that at point c, tangent is parallel to the chord line.

Cauchy’s Mean Value Theorem If (i) f(x) is continuous in the closed interval [a,a+h] and (ii) f (x) exists in the open interval (a,a+h), then there is at least one number such that

(0< <1)

f(a+h) = f(a) + h f(a+ h) Let f1 and f2 be two functions: i) f1,f2 both are continuous in [a,b] ii) f1, f2 both are differentiable in (a,b) iii) f2’ 0 in (a,b) then, for a ( ) ( )

1.4.2

( ) = ( )

b ( ) ( )

Derivative:

’( ) = lim

(

)

( )

Provided the limit exists ’( ) is called the rate of change of f at x. Algebra of derivative:i. (f g) = f g ii. (f g) = f – g iii. (f. g) = f . g f .g iv. (f/g) =

.

.


Page 22


Mathematics

Homogenous Function Any function f(x, y) which can be expressed in from xn . / is called homogenous function of order n in x and y. (Every term is of nth degree.) f(x,y) = a0xn + a1xn-1y + a2xn-2y2 f(x,y) = xn

………… an yn

. /

Euler’s Theorem on Homogenous Function If u be a homogenous function of order n in x and y then,  x +y = nu  1.4.3

x

+ 2xy

+y

= n(n

1)u

Total Derivative

If u=f(x,y) ,x=φ(t), y=Ψ(t) =

.

u=

+ x+

. y

Monotonicity of a Function f(x) 1. f(x) is increasing function if for , f( ) Necessary and sufficient condition, f’ (x) 2. f(x) is decreasing function if for , , f( ) Necessary and sufficient condition, f (x)

f( ) f( )

Note: If f is a monotonic function on a domain ‘D’ then f is one-one on D. Maxima-Minima a) Global

b) Local

Rule for finding maxima & minima:  If maximum or minimum value of f(x) is to be found, let y = f(x)  Find dy/dx and equate it to zero and from this find the values of x, say x is , , …(called the critical points). 

Find

at x = ,

If

, y has a minimum value

If

,y has a maximum value


Page 23


If

Mathematics

= , proceed further and find at x = .

If

, y has neither maximum nor minimum value at x =

But If

= , proceed further and find

If

, y has minimum value

If

, y has maximum value

If

at x = .

= , proceed further

Note: Greatest / least value exists either at critical point or at the end point of interval. Point of Inflexion If at a point, the following conditions are met, then such point is called point of inflexion

Point of inflexion i) ii)

=

,

=0,

iii)

 Neither minima nor maxima exists

Taylor Series: f(a

h)= f(a)

h f’(a)

f”(a)

.........

Maclaurian Series: f(x) = f( )

x f’( )

f ( )

h

f ( )

Maxima & Minima (Two variables) r= 1.

= 0,

2. (i) if rt (ii) if rt (iii) if rt (iv) if rt

,s= =

, t= solve these equations. Let the solution be (a, b), (c, d)…

s and r maximum at (a, b) s and r minimum at (a, b) s < 0 at (a, b), f(a,b) is not an extreme value i.e, f(a, b) is saddle point. s > 0 at (a, b), It is doubtful, need further investigation.


Page 24


1.4.4

Mathematics

Standard Integral Results

1. ∫ x dx =

, n

1

2. ∫ dx = log x 3. ∫ e dx = e 4. ∫ a dx = (prove it ) 5. 6. 7. 8. 9. 10. 11.

∫ cos x dx = sin x ∫ sin x dx = cos x ∫ sec x dx = tan x ∫ cosec x dx = cot x ∫ sec x tan x dx = sec x ∫ cosec x cot x dx = cosec x dx = sin ∫ √

12. ∫

√

dx =

sec

13. ∫ dx = sec x √ 14. ∫ cosh x dx = sinh x 15. ∫ sinh x dx = cosh x 16. ∫ sech x dx = tanh x 17. ∫ cosech x dx = coth x 18. ∫ sech x tanh x dx = sech x 19. ∫ cosec h x cot h x dx = cosech x 20. ∫ tan x dx = log sec x 21. ∫ cot x dx = log sin x 22. ∫ sec x dx = log( sec x tan x) = log tan( ⁄ 23. ∫ cosec x dx = log(cosec x cot x) = log tan

x⁄ )

24. ∫ √

dx = log(x

√x

a ) = cosh ( )

25. ∫ √

dx = log(x

√x

a ) = sinh ( )

26. ∫ √a

x dx =

27. ∫ √a

x dx = √x

a

log(x

√x

a )

28. ∫ √x

a dx = √x

a

log(x

√x

a )

29. ∫

dx =

tan

30. ∫

dx =

log (

) where x
31. ∫

dx =

log (

) where x > a

32. ∫ sin x dx = 33. 34. 35. 36.

√

sin

sin x

sin x ∫ cos x dx = ∫ tan x dx = tan x x ∫ cot x dx = cot x x ∫ ln x dx = x ln x x

37. ∫ e

sin bx dx =

(a sin bx

b cos bx )

38. ∫ e

cos bx dx =

(a cos bx

b sin bx )


Page 25


39. ∫ e ,f(x)

Mathematics

f (x)-dx = e f(x)

Integration by parts: ∫ u v dx = u. ∫ v dx

∫(

∫ v dx)dx

I L A T E E

Selection of U & V Inverse circular (e.g. tan 1 x)

Exponential

Logarithmic

Algebraic Trigonometric

Note: Take that function as “u” which comes first in “ILATE” 1.4.5 Rules for Definite Integral 1. ∫ f(x)dx =∫ f(x)dx+∫ f(x)dx 2. ∫ f(x)dx =∫ f(a 3. ∫ f(x)dx =∫

/

b

x)dx

f(x)dx+∫

=0 4. ∫ f(x)dx =2 ∫ f(x)dx =0

/

a
x)dx

f(a x)dx ∫ f(x)dx = ∫ if f(a-x)=f(x) if f(a-x)=-f(x) if f(-x) = f(x), even function if f(x) = -f(x), odd function

/

f(x)dx

Improper Integral Those integrals for which limit is infinite or integrand is infinite in a then it is called as improper integral.

x

b in case of ∫ f(x)dx,

1.4.6 Convergence:  ∫ f(x)dx is said to be convergent if the value of the integral is finite.  If (i) f(x) g(x) for all x and (ii) ∫ g(x)dx converges , then ∫ f(x)dx also converges  If (i) f(x) g(x) for all x and (ii) ∫ g(x)dx diverges, then ∫ f(x)dx also diverges ( ) ( )



If lim



diverge. is converges when p ∫



∫ e



The integral ∫



The integral ∫

= c where c 0, then both integrals ∫ f(x)dx and ∫ g(x)dx converge or both

dx and ∫

1 and diverges when p

1

e dx is converges for any constant p

(

)

(

)

is convergent if and only if p

1

is convergent if and only if p

1

and diverges for p


Page 26


1.4.7

Mathematics

Vector Calculus:

Scalar Point Function: If corresponding to each point P of region R there is a corresponding scalar then (P) is said to be a scalar point function for the region R. (P)= (x,y,z) Vector Point Function: If corresponding to each point P of region R, there corresponds a vector defined by F(P) then F is called a vector point function for region R. F(P) = F(x,y,z) = f1(x,y,z) ̂ +f2(x,y,z)ĵ f3(x,y,z) ̂ Vector Differential Operator or Del Operator:

=.

ĵ

̂

/

Directional Derivative: ⃗⃗ is the resolved part of f in direction N ⃗⃗ . The directional derivative of f in a direction N ⃗⃗ = | f|cos f. N ⃗ is a unit vector in a particular direction. Where ⃗N Direction cosine: l

m

n =1

Where, l =cos , m=cos , n=cos , 1.4.8

Gradient:

The vector function f is defined as the gradient of the scalar point function f(x,y,z) and written as grad f. grad f = f = î    1.4.9

ĵ

+̂

f is vector function If f(x,y,z) = 0 is any surface, then f is a vector normal to the surface f and has a magnitude equal to rate of change of f along this normal. Directional derivative of f(x,y,z) is maximum along f and magnitude of this maximum is | f|. Divergence:

The divergence of a continuously differentiable vector point function F is denoted by div. F and is defined by the equation. div. F = . F THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com

Page 27


F=f + ĵ

Ψ̂

div.F= . f = . =  

Mathematics

+

̂

ĵ

Ψ̂)

/ .( f + ĵ

+

. f is scalar . = is Laplacian operator

1.4.10 Curl: The curl of a continuously differentiable vector point function F is denoted by curl F and is defined by the equation. ĵ Curl F =

̂

f =|

| f

φ

Ψ

F is vector function 1.4.11 Solenoidal Vector Function If .A = 0 , then A is called as solenoidal vector function. 1.4.12 Irrotational Vector Function If

A =0, then A is said to be irrotational otherwise rotational.

1.4.13 DEL Applied Twice to Point Functions: 1. div grad f = 2. 3. 4. 5.

f=

+

+

---------- this is Laplace equation

curl grad f = f=0 div curl F = . F =0 curl curl F = ( f) = ( . f) grad div F = ( . f)= ( F) +

F F

1.4.14 Vector Identities: f, g are scalar functions & F, G are vector functions 1. (f g) = f + g 2. . (F G) = . F .G (F G) = 3. F G 4. (fg) = f g + g f 5. . (fG)= f. G f. G 6. (fG) = f G f G 7. (F. G) = F ( G) G ( F) 8. . (F G) = G.( F) F. ( G) THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com

Page 28


(F

9.

G) = F(

G)

G(

Mathematics

F)

Also note: 1. (f/g)= (g f – f g)/g 2. (F.G)’ = F’.G F . G’ 3. (F G)’ = F’ G + F G’ 4. (fg) = g f + 2 f. g + f

g

1.4.15 Vector product 1. Dot product of A B with C is called scalar triplet product and denoted as [ABC] Rule: For evaluating the scalar triplet product (i) Independent of position of dot and cross (ii) Dependent on the cyclic order of the vector [ABC] = A B. C = A. B C = B C. A= B.C A = C A. B = C.A B A B. C = -(B A. C) ⃗ B ⃗ = (extreme adjacent) Outer ⃗) C 2. (A = (Outer. extreme) adjacent (Outer. adjacent) extreme ⃗⃗⃗⃗ ⃗⃗⃗ ⃗ = (C ⃗ .A ⃗ )B ⃗ .B ⃗ ⃗ - (C ⃗ )A  (A B) C ⃗ (B ⃗ ) = (A ⃗ .C ⃗ )B ⃗ .B ⃗ ⃗ C ⃗ - (A ⃗ )C  A ⃗ ⃗ ⃗ ⃗ ⃗ ⃗  (A B ) C A (B C ) 1.4.16 Line Integral, Surface Integral & Volume Integral 

 

Line integral = ∫ F(R)dR If F(R )= f(x,y,z) ĵ (x,y,z) + ̂ Ψ(x,y,z) dR = dx ĵ dy ̂ dz dy Ψ dz ) ∫ F(R )dR = ∫ ( f dx ⃗ ds, Where N is unit outward normal to Surface. Surface integral: ∫ ⃗F . ⃗⃗⃗⃗ ds or ∫ ⃗F . ⃗N Volume integral : ∫ F dv If F(R ) = f(x,y,z)î +

(x,y,z)ĵ

∫ F dv = î∫ ∫ ∫ fdxdydz

Ψ (x,y,z) ̂ and v = x y z , then

ĵ ∫ ∫ ∫ dxdydz + ̂ ∫ ∫ ∫ Ψdxdydz

1.4.17 Green’s Theorem If R be a closed region in the xy plane bounded by a simple closed curve c and if P and Q are continuous functions of x and y having continuous derivative in R, then according to Green’s theorem. ∮ (P dx

Q dy) = ∫ ∫ .

/ dxdy


Page 29


Mathematics

1.4.18 Sto e’s theorem If F be continuously differentiable vector function in R, then ∮ F. dr = ∫

F .N ds

1.4.19 Gauss divergence theorem The normal surface integral of a vector point function F which is continuously differentiable over the boundary of a closed region is equal to the ∫ F .N.ds =∫ div F dv


Page 30


Mathematics

1.5: Differential Equations 1.5.1

Order of Differential Equation: It is the order of the highest derivative appearing in it.

1.5.2

Degree of Differential Equation: It is the degree of the highest derivative occurring in it, after expressing the equation free from radicals and fractions as far as derivatives are concerned.

1.5.3

Differential Equations of First Order First Degree:

Equations of first order and first degree can be expressed in the form f (x, y, y ) = or y = f(x, y). Following are the different ways of solving equations of first order and first degree: 1. Variable separable : f(x)dx + g(y)dy = 0 ∫ f(x)dx

∫ g(y)dy = c is the solution

2. Homogenous Equation: 

=

( , ( ,

) )

To solve a homogeneous equation, substitute y = Vx =V+x



Separate the variable V and x and integrate.

Equations Reducible to Homogenous Equation: The differential equation:

=

This is a non - homogeneous but can be converted to homogeneous equation Case I: If Substitute x = X + h

y=Y+k

(h and K are constants)

Solve for h and k ah b c=0 ah b c =0 = Case II: If

= = =

=

(say)

( (

) )


Page 31


Mathematics

Substitute ax +by = t, so that, (

=

)

a

Solve by variable separable method. 3. Linear Equations: The standard form of a linear equation of first order: + P(x) y = Q(x) , where P and Q are functions of x Second order linear equation:

d y dx

P(x)

dy dx

Q(x)y = R(x)

Commonly nown as “Leibnitz’s linear equations” Integrating factor, I.F. = e∫ ye∫

= ∫ Q. (I. F)dx

C

y(I. F. ) = ∫ Q. (I. F)dx

C

Note: The degree of every linear differential equation is always one but if the degree of the differential equation is one then it need not be linear. Ex:

x . /

y

= 0.

.1 Bernoulli’s Equation: +Py=Qy

where, P & Q are functions of x only.

Divide by y y Substitute, y

Py

=Q

=z (1

n)Pz = Q (1-n)

This is a linear equation and can be solved easily

4. Exact Differential Equations: M (x, y) dx + N (x, y) dy = 0


Page 32


Mathematics

The necessary and sufficient condition for the differential equations M dx +N dy = 0 to be exact is = Solution of exact differential equation: ∫

M dx

∫(terms of N not contaning x ) dy = C

4.1 Equation Reducible the Exact Equation: Integrating Factor: Sometimes an equation which is not exact may become so on multiplication by some function known as Integrating factor (I.F.). Rule 0: Finding by inspection 1. x dy + y dx = d (x y) 2. =d( ) 3.

= d [log (

4.

=-d( )

5.

= d [tan (

)-

6.

=d[

)-

log(

)]

Rule 1: when M dx + N dy = 0 is homogenous in x and y and M x + N y Rule 2: If the equation f (x, y) y dx + f (x, y) x dy = 0 and M x – N y / = f(x), then I.F. = e∫

Rule 3: If the M dx + N dy = 0 and . Rule 4: If the equation M dx + N dy = 0 and

1.5.4

0 then I.F. = 0 then I.F. =

( )

/ = f(y) , then I.F. = e∫

.

( )

Linear Differential Equation with Constant Coefficients: -------

The equation can be written as (D

y=X D

-----

)y = X {Where, D =

}

f(D) y = X ; f(D) = 0 is called Auxiliary Equation. Rules for Finding Complimentary Function: Case I :

If all the roots of A.E. are real and different


Page 33


(D

m ) (D

m ) - - - - - - (D

m )y=0

So, the solution is: y = C e Case II:

Mathematics

C e

-- - - - -+ C e

If two roots are equal i.e. m = m y = (C

C x)e

Similarly, if m = m = m y = (C Case III:

C x +C x ) e

If one pair of roots are imaginary i.e.

m =

i , m =

y = e (C cos x Case IV:

i C sin x)

If two pairs of root are imaginary i.e. repeated imaginary root y=e

1.5.5

i ,

C ) cos x

,(C x

i (C x

C ) sin x ]

Rules for finding Particular Integral P.. =

X=

( )

.X

Case I: When X = P.I. =

( )

P.I. = x

( )

P.I. =

( )

put D = a

[ ( )

0]

put D = a

[ ’( )

0, ( ) = 0]

put D = a

[ ( ) = 0, ’( ) = 0, ’’( )

0]

Case II: When X = sin (ax + b) or cos (ax +b) P.I. = =x =

(

(

)

(

(

) (

)

)

put )

(

=-

, (-

)

0]

put

=-

, ’(-

)

, (-

) = 0]

) put

=-

, ’’(-

)

, ’(-

)


Page 34


= , (-

Mathematics

) = 0]

Case III: When X =

, m being positive integer P.I. =

= [ ( )-

( )

=

(D) ,1

=

(D) [1

( )

-

( )

( )

( )

-

Case IV: When X =

V where V is function of x P. I. =

V

( )

=

(

)

V then evaluate

(

)

V as in Case I, II & III

Case V: When X = x V(x) P.I. =

( )

( )

x V(x) = 0

( )

1

( )

V(x)

Case VI: When X is any other function of x P.I. = Factorize f(D) = (D

( )

X

) (D

and then apply,

) - - - - - - - (D

X=

∫

) and resolve

( )

into partial fractions

on each terms.

Complete Solution: y = C.F. + P.. 1.5.6

Cauchy-Euler Equation: (Homogenous Linear Equation) .

Substitute

------ -

=X

x=e x

= Dy = D (D-1) y = D (D-1)(D-2) y


Page 35


Mathematics

After substituting these differentials, the Cauchy – Euler equation results in a linear equation with constant coefficients.

1.5.7

Legendre’s Linear Equation: (

)

(

ax + b =

)

- - - - -- -

=X

t = ln (ax + b)

(ax + b)

=aDy

(

)

=

D(D-1)y

(

)

=

D(D-1)(D-2)y

After substituting these differentials, the Legendre’s equation results in a linear equation with constant coefficients. 1.5.8

Partial Differential Equation: z = f(x, y) =p,

1.5.9

=q,

= r,

= s,

=

Homogenous Linear Equation with Constant Coefficients: ------ -

= f( x, y)

this is called homogenous because all

terms containing derivative is of same order. (

-------

) = f(x, y)

{ where D =

and D’ =

}

f (D, D’) = f(x,y) Step I: Finding the C.F. 1. Write A.E. Where m = 2. CF = (y + CF = (y + CF =

(y +

----= 0, and the roots are , ---- x) + (y + x) + - - - - - - , are distinct x) + x (y + x) + (y + x) + - - - - - - , x) + x

(y +

x) +

(y +

x) + - - - -

,

, ,

two equal roots. three equal roots.


Page 36


Mathematics

Step II: Finding P.I. P.. =

( ,

)

f (x, y)

1. when F( ax +by ) = , put [ D = a, = b] 2. when F( x, y) = sin (mx +ny), put ( = , 3. when F(x, y) = , P. = ( , ) =[ ( , 4. when F(x, y) is any function of x and y. P. =

( ,

= ))

,

=

f (x, y) , resolve

)

( ,

)

into partial

fractions considering ( , ) as a function of D alone and operate each partial fraction ) on f(x, y) remembering that f(x, y) = ∫ ( , where c, is replaced by (

)

y + mx after integration.


Page 37


Mathematics

1.6: Complex Variables =

is a complex no., where x & y are real numbers called as real and imaginary part of z.

Modulus or absolute value = | | = √

, Argument of

=

=

( )=

. /

1.6.1 Function of a Complex Variable: It is a rule by means of which it is possible to find one or more complex numbers ‘w’ for every value of ‘z’ in a certain domain D, then w = f (z) Where z = x + iy, w = f (z) = u(x, y) + i v(x, y) 1.6.2   

Continuity of f (z):

( ) = ( ). A function = f (z) is said to be continuous at = if Further f (z) is said to be continuous in any region R of the z-plane, if it is continuous at every point of that region. Also if w = f (z) = u(x, y) + i v(x, y) is continuous at = , then u(x, y) and v(x, y) are also continuous at x= & y = .

1.6.3

Theorem on Differentiability:

The necessary and sufficient conditions for the derivative of the function f( ) to exist for all values of in a region R. i)

,

,

,

ii)

=

1.6.4

Analytic Functions (or Regular Function) or Holomorphic Functions

,

, are continuous functions of x and y in R. =

,

Cauchy-Riemann equations (CR equations)

 A single valued function which is defined and differentiable at each point of a domain D is said to be analytic in that domain.  A point at which an analytic function ceases to possess a derivative is called Singular point.  Thus if u and v are real Single – valued functions of x and y such that , , , are continuous throughout a region R , then CR equations =

,

=-

are both “necessary and sufficient” condition for the function f(z) = u

iv to be analytic in R.

 Real and imaginary part i.e. u, v of the function is called conjugate function.  An analytic function posses derivatives of all order and these are themselves analytic.


Page 38


1.6.5

Mathematics

Harmonic Functions:

If f(z) = u + iv be an analytic function in some region of the z – plane then the C –R equations are satisfied. =

,

=

Differentiating with respect to x and y respectively, =

,

=

=0

(Laplace Equation)

Note: (1) For a function to be regular, the first order partial derivations of u and v must be continuous in addition to CR equations. (2) Mean value of any harmonic function over a circle is equal to the value of the function at the centre. 1.6.6

Methods of Constructing Analytic Functions:

1. If the real part of a function is given then, ’( ) = -i Integrate with points at (z, 0) f(z) = ∫ . / dz - i ∫ . / ( , )

( , )

dz + c

Similarly in case v(x, y) is known, then f’ (z) = +i f (z) = ∫ . /

( , )

dz + i ∫ . /

( , )

dz + c

2. If u (x, y) is known, then to find v(x, y) we have dv = dx + dy dv =

dx +

dy

Integrate this equation to find v. f (z) = u(x, y) + i v(x, y) 3. If a real part of the analytic function f(z) is given which is harmonic function u (x, y), then f(z) = 2u . , / - u(0, 0) 1.6.7 Complex Integration  Line integral = ∫ ( ) , C need not be closed path Here, f(z) = integrand , curve C = path of integration  Contour integral = ∮ ( ) , if C is closed path


Page 39


Mathematics

If f(z) = u(x, y) + i v(x, y) and dz = dx + i dy ∫

( )

=∫(

)

∫(

)

Theorem: f(z) is analytic in a simple connected domain then ∫

( )

= f( )

( ), i.e.

integration is independent of the path Dependence on Path: In general “Complex line integration” depends not only on the end points but also on the path (however analytic function in simple connected domain is independent of path.) 1.6.8

Cauchy’s Integral Theorem:

If f(z) is analytic in a simple connected domain D, then for every simple closed path C in D, ∮𝐶 𝑓(𝑧)𝑑𝑧 = 0 Note: In other words, by Cauchy’s theorem if f(z) is analytic on a simple closed path C and everywhere inside C (with no exception, not even a single point) then ∮ ( ) = D C

1.6.8.1 Cauchy’s Integral Formula: If f(z) is analytic within and on a closed curve and if a is any point within C, then

.

( )=

∫

’( ) =

∫

”( ) =

∫

( )

( ) (

) ( )

(

. ( )=

)

. ∫

( ) (

)


Page 40


1.6.9 Morera’s Theorem: If f(z) is continuous in a region and ∫ simple closed C then f(z) is analytic in that region.

Mathematics

( )

= 0 around every

1.6.10 Taylor’s Series: If f(z) is analytic inside a circle C with centre at a then for z inside C f(z) = f(a) f(z) = where

( )

f’(a) (z-a) + ( =

(z-a) + - - - - - - -

) ∫

(

( ) )

Other form, put z = a + h f(a+h) = f(a) + h ’( ) +

”( ) + - - - - - - -

1.6.11 Laurent’s Series: If f(z) is analytic in the ring shaped region R bounded by two concentric circles and of radii and ( ) and with centre at a then for all z in R (

f(z) = where,

=

∫

(

)

(

)

(

)

(

)

( ) )

If f(z) is analytic inside the curve then

= and Laurent series reduces to Taylor’s series.

1.6.12 Zeroes of Analytic Function: The value of z for which f(z) = 0 If f(z) is analytic in the neighbourhood of a point z = a then by Taylor’s theorem.

where if

=

f(z) =

(

)

=

(

)

= =

(

)

(

)

( )

=------

= 0, then f(z) is said to have a zero of order n at z =a.

1.6.13 Singularities of an Analytic Function: A “singular point” of a function as the point at which the function ceases to be analytic. 1. Isolated Singularity: If z =a is a singularity of f(z) such that f(z) is analytic at each point in its neighbourhood (i.e. there exists a circle with centre a which has no other singularity 1, then z =a is called an isolated singularity). 2. Removable Singularity: If all the negative powers of (z-a) in Laurent series are zero then THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com

Page 41


Mathematics

f(z) = ( ) singularity can be removed by defining f(z) at z = a is such a way that it becomes analytic at z =a ( ) exists finitely, then z = a is a removable singularity. Example: f(z) = , then z = 0 is a removable singularity. 3. Essential singularity: If the numbers of negative power of (z-a) in Laurent’s series is infinite, then z =a is called an essential singularity. ( ) does not exist in this case 4. Poles: If all the negative power of (z-a) in Laurent’s series after singularity at z = a is called a pole of order n. A pole of first order is called a “simple pole”.

are missing then. The

1.6.14 Residue Theorem If f(z) is analytic in and on a closed curve C except at a finite number of singular point within C then ∫ f(z)dz = i (sum of the residue at the singular point within C) Calculation of Residues 1. If f(x) has a simple pole at z=a , then Res f(a) = ,( ) ( )( ) ) ( ), ( ) 2. If ( ) = ( ) where ( ) = ( Res ( ) =

( ) ( )

)

𝐶

𝑎

3. If ( ) has a pole of order n at z=a , then ( )=(

𝑎

2

,(

)

𝐶

( )-3

C

𝐶 𝐶

Here n =order of singularity Note: If an analytic function has singularities at a finite number of points, then the sum of residues at these points along with infinity is zero.


Page 42


Mathematics

1.7: Laplace Transform 1.7.1

Introduction

Laplace Transform (LT) is a method to get generalized frequency domain representation of a continuous time signal and is generalization of CTFT (Continuous Time Fourier Transform). Definition of Laplace Transform , ( )- = ( ) = ∫

. ( )

, ( )- = ( )= ∫ 1.7.2

. ( )

: One sided/ unilateral LT, where S = (

J ω)

: Two sided/ bilateral LT.

Properties of Laplace transform

Frequency shift L [e-at f(t) ] = F(s + a) and

[eat f(t) ] = F(s - a)

Time shift L [f(t – to)] =

. F(s)

Differentiation in Time domain [

( ) ] = s F(s) – f(0) where f(0) is initial value of f(t).

If initial conditions are zero (i.e. f(0) = 0),differentiating in time domain is equivalent to multiplying by s in frequency domain. Similarly,

[

( )]=

F(s) –s f(0) -

(0) where

(0) is the value of [

( ) ] at t = 0

Integration in Time domain L 0∫

( ) 1=

( )

and

2∫

( ) 3=

( )

∫

( )

Integration in time domain is equivalent to division by s in frequency domain, if f(t) = 0 for t < 0. Differentiation in Frequency Domain L [ t f(t) ] =

( )

and

*

( )+ = ( 1)

(F(s))

Differentiation in frequency domain is equal to multiplication by t in time domain. Integration in Frequency Domain L0

( )

1 = ∫

( )


Page 43


Mathematics

Integration in frequency domain is equal to division by t in time domain.

1.7.3

Initial Value Theorem

If f(t) and its derivative

( ) are Laplace transformable, then

( )=

( )

This theorem does not apply to the rational function F(s) in which the order of numerator polynomial is equal to or greater than the order of denominator polynomial. 1.7.4

Final Value Theorem

If f(t) and its derivative

(t) are Laplace transformable, then

( )= ( ) For applying final value theorem, it is required that all the poles of s- plane (strictly) i.e. poles on axis also not allowed. 1.7.5

Convolution theorem L, ( ). ( )- = ( )

( )

L, ( )

( )

( )- =

( ).

( ) be in the left half of

1.7.6 Laplace Transform of the Periodic Function If f(t) is periodic function with period T, then ( ( )) = ( 1.7.7

)

.

(s) where

(s) = ∫

( )

Laplace Transform of Standard Functions Table. Laplace Transform of Standard Functions

S. No 1. 2.

Function, f(t) ( ) u(t)

4. 5.

u(t)

6.

t.u(t)

8.

1

1

3.

7.

Laplace transform of f(t), L{f(t) = F(s) 1

. ()

(

)

1⁄

. () f(t).

1⁄ 1⁄

√ ⁄ ()

1

F(s-a)


Page 44

Quick Refresher Guide 9.

sin at. u(t)

10.

cos at. u(t)

11.

sin hat. u(t)

12.

cos hat. u(t)

13. 14. 15.

f (t) f (t)

16.

19. 20.

f(t⁄a) f(at)

21.

f1 (t) f (t)=∫ e

t

22

e

t

23

1

24 25

t

s . F(s) s. f(o ) –f (o ) 1⁄ F(s) s 1 S

F(s)

u)du

1 S

where f 1 (o ) = ∫

f(u)du

. F(s) dn ( 1)n . n (F(s)) ds |a|. F(as) 1 F(s⁄a) |a| F1 (s). F (s) where * is convolution operator e

f1 (u). f (t

f 1 (o ).

(s

sin ωt

ω

as

) ⁄(( s

⁄((s ) ∫s F(s)ds

)

ω )

ω )

√

t t

)

. cos ω t

. f(t)

1

⁄( ⁄(s

)

a ) s.F(s)-f(o )

∫ f(u) du t ∫ f(u)du f(t-a).u(t-a) tn . F(t)

)

s

t

17. 18.

⁄( ⁄(

Mathematics

1

s

√


Page 45



Part – 2: Structural Engineering 2.1 Mechanics Simple Stress and Strain relationship stress and strain in two dimensions, principle stresses, stress transformation, mohr circle  

Normal stress σ = Eε Shear stress ζ = G γ

Where, ε, γ are normal and shear strain respectively E, G are Young’s modulus and shear modulus respectively A – Proportional limit point B – Elastic limit point C – Upper yield point D – Lower yield point E – True breaking stress point F – Nominal breaking stress point E B Stress

A

C F

D

Strain Stress-Strain Relationship For mid steel   

Proof stress is equivalent to 0.1% to 0.2% stress Proof stress = Yield stress Factor of safety =



Let k = bulk modulus of elasticity, μ = Poisson’s ratio a) E = 2G (1+ μ) b) E = 3 K (1- 2μ) Relations Between E,K,G,μ c) E = d) μ =



Theoretically poisson’s ratio μ varies from -1 to +0.5, Practically poisson’s ratio μ varies from 0 to 0.5.

Elongation of bars of different section 1. For uniformly tapered circular bars THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 46



4PL Ed d

=

2. Uniformly tapered rectangular bars

=

(

)

3. Elongation of a uniform section bar due to self weight =

2 Where W = Weight of bar A = Area of cross section 4. Elongation of a bar uniformly tapered due to self weight

=

= 2 6 Where A = Average area of cross section 5. Elongation of a stepped bar

= Unit weight of material of a bar

THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 47


= =

+ [

+


+ +

]

Analysis of a compound bar 1. Condition of equilibrium , ,

P +P =P 2. Condition of compatibility = PL PL = E E 3. Equivalent modulus of elasticity PL PL PL = = E E E P +P =P PL E PL E + =P E L R L , L =L 1 , E + E -=1 E E + E E= + , = 

For composite material consisting of steel and copper rod, For thermal expansion, as > Steel steeve ( E ) Copper will be in compression and steel will be in Cu - rod ( E ) tension. Steel steeve ( Governing equation will be (

E)

)∆t =

+

Temperature Stresses 1. In uniform section fixed from both sides change in length = tL where = coefficient of thermal expansion t = change in temperature



σ =( ) E σ = tE L It support yields by ‘a’ ( a) then σ = E = ( tL L

a)


E L

2. In a tapered circular section fixed from both sides Force ‘P’ induced due to temperature change ‘t’ tL a = ( ) 4 3. Composite sections

Governing equation P P ( = + ) E E E + E =( ) Lt E + E Where t is temperature change 

Strain energy = ∪ = Module of resilience is the strain energy per unit volume = u =

∪

= = Eε Proof resistance is applicable when σ = yield stress = σ up = 

If σ and σ are stress acting on the body as shown below, along x-x σ X Normal stress, σ = σ

σ

θ

X

σ σ σ

θ

ζ

Shear stress ζ =

If, θ = 90 + θ σ = ζ =

+

cos 2θ

Sin 2

cos 2θ

sin 2 θ’

σ σ ’ are complementary normal stress σ Ζ, ζ’ are complementary shear stress σ +σ =σ +σ Ζ + ζ’ = 0 

If shear stress

also acts on body along with σ and σ



Normal stress, σ = . Shear stress, = 

/+.

/ cos2θ -

sin 2θ +


Sin 2θ

= cos 2θ

Principle stresses are direct normal stresses acting on mutually perpendicular planes on which shear stresses are zero. If σ and σ are the principle stresses (major and minor respectively) /±√.

σ ,σ =.

/ +



The plane of maximum shear stress lies at 45 to the plane of principal stress = and on that plane of ,σ =



If the plane on which σ is maximum makes θ angle with vertical the plane on which is maximum, makes (45+ θ) angle with vertical, Tan 2 θ = σ maximum Cot 2 θ =



ζ maximum

In case of pure shear element, the principle stresses act at 45 to the plane of pure shear stress

Stress invariants I = (σ + σ + σ ) = σ + σ + σ = σ + σ + σ i.e. sum of stresses on any plane is constant I = (σ σ + σ σ + σ σ ) = σ σ + σ σ + σ σ Mohr’s Circle

( =(

+ +

) =

)

=

= =

,

Transformation equation for strains



ε = =

2


ε +ε ε ε γ +. / cos 2θ + = sin 2θ 2 2 2 ε +ε 2

ε = .

ε . ε

2

ε 2

γ = sin 2θ 2

/ cos 2θ

/ cos 2θ +

γ = sin 2θ 2

Maximum normal strain =

ε +ε ε ε γ + √. / +. / 2 2 2

=

ε +ε 2

ε √.

ε 2

γ / +. / 2

To find plane of principal strain use tan 2

=

Maximum shear strain To find plane of maximum shear strain tan 2

=

=

( ε √.

ε

ε γ ε 2

) γ / +. / 2

Sign convention for strain

+

+

, +




σ =+ σ =σ

σ

σ

Properties of Mohr’s Circle a) Center of Mohr’s Circle, C = b) Radius of Mohr’s Circle r =

= =

= √.

/ +

xy

c) Major principle stress, σ = C + r d) Minor principle stress, σ = C – r e) In case of pure shear elements σ = + and σ = - and centre of the circle coinsides with the origin Octahedral stresses Stress on a octahedral plane (a plane which is equally inclined it all three principal axis of reference For such plane =

=

Where

,

= =  Let,

,

+

=

1

=

√3 are direction cosines

+

=

3 (

3

) +(

) +(

)

9 √2 √ 3

3

Simple bending theory, flexural and shears tresses, unsymmetrical bending y = distance of any particular section from neutral axis stress due to bending, σ =

y




Where M = Bending moment I = Moments of inertia about neutral axis Also,

=

Where E = Young’s modulus R = Radius of curvature Flexure formula =

=

Some Special Cases 

For the same square beam, and for a given stress, the ratio of moments of resistance,

= √2

y y

b x

X bx

X

b

y b

y Position (ii)

Position (i) 

To achieve same strength under same stresses, the relation regardings weight among rectangular, circular and square section is W
  

The ratio of depth to width of the strongest beam that can be cut from a circular log is 1.414 Moment of resistance m = σZ M Z where Z section modulus As at the top of beam, y = maximum and at the neutral axis y = 0, bending stress at top is maximum and at neutral axis is zero

 Shear stress for Different Section i) For rectangular section with shear force V,  Shear stress =

.

y /, d = depth

 Average shear stress =

, b = width

 Maximum shear stress = ii) For circular section with shear force v, (r  Shear stress = y ), r = radius of circle  Average shear stress =




 Maximum shear stress =  At y = local shear stress is equal to the average shear stress

iii) For triangular section, with shear force v,  Shear stress = (h-y)y, b = base width h = height  Average shear stress =  Maximum shear stress = iv) For tubular section, with shear force V  Average shear stress = (

 Maximum shear stress 4 + + = =( 3 + Where

= (

)

)

)

Same typical shear distribution diagram 2h 3 r

d

3 v 2 bh

b Rectangular section

4 v 3 r

h 3

Circular section

h 8 v 3 bh

b Triangular section

= T-section

F

I-section

 Shear centre, thin walled pressure vessels, uniform torsion, bulking of column, combined and direct bending stress.  Shear centre  Lateral load acting on a beam through shear centre will produce bending without torsion  Shear centre is the centre of flexure  Shear centre always lies in the axis of symmetry THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 54



For channel sections,

h/2

tf

tw

Shear centre distance

c e v

h/2

e= Where, I = moments of inertia about axis of symmetry

b

For semi circular sections, e r e= 0

V Thin walled Pressure vessels Let p = internal pressure, t = thickness d= diameter μ =

= Poisson’s ratio

E = Young’s modulus 

For Cylinders 1. Hoop stress or circumferential stress, F = 2. Longitudinal stress, F = 3. Maximum shear stress, = 4. Hoop strain, E =

.2

5. Longitudinal strain, ε = 6. Volumetric strain ε = 

= / .1 .5

/ /

For Spherical shells 1. Hoop stress (f ) = longitudinal stress (f ) = 2. Maximum shear stress = 0 3. Strain in any direction, ϵ = .1 / 4. Volumetric strain, ϵ =

.1

/

Uniform Torsion Let T = Total torque acting on the section THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 55



J = Polar moment of inertia = Shear stress at any radial distance R = Radius of shaft G = Rigidity modulus = Angular momentum due to strain in length of shaft L = Length of shaft θ = Angular strain in cross-section T = J

=G

θ L



For solid shaft of diameter d, T =



For hollow shaft of outer diameter do and inner diameter di, T =



Power transmitted by shaft, p = KW Where N = Rotation per minute (rpm) T = means Torque (KN.m)

(d

d )

 Strain energy in torsion 1 U= T θ 2  For shaft in series, Dia = d A

B Dia = d

C

l l Torque will be same for both the shafts 1. Twists will be different for both the shafts Let twist are θ for AB and θ for BC =. / T A

d

d

T

B

C l T +T =T ⁄at =

l

l

l

l =

+

+

⁄at



=

+


+

= Shaft of varying diameter 1.

θ

=

(

)

0

1

2. Indeterminate system of shafts

( ) ( )

=

Column stability Short column – fails by crushing Long column – fails by bucking Intermediate columns – fails by combinations of crushing & buckling Buckling Of Columns A.

Euler’s Theory  Assumptions 1. Flexural rigidity EI is uniform 2. After unloading axis the column should be perfectly straight 3. Material is isotropic and homogeneous 4. The line of thrush will coincide exactly with the unrestrained axis of the column 

Buckling load, P =



=


. /

L Where = called stenderness ratio r Where, L = effective length of the column End conditions Effective length (of original length L ) L =L Both end hanged L = Both end fixed L = 2L One end fixed, Other end free L = √ One end fixed Other end hanged B. Variation of bucking load/Euler load with slenderness ratio If is less than a certain limit , . / We cannot use Euler’s theorem

L ( ) r

 Rankine – Gordon formula (for intermediate columns) 1 1 1 = + P P P Where P = Rankine buckling load P = Direct compressive load P = Euler’s load = Yield Stress i. e. , = . EI P = L . .,

=

1+ L where. = called slenderness ratio r THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 58


 Combined bending and direct stress Let normal stress σ is due to bending M and of diameter d,  Then, σ =


is shear stress due to tourism T for shaft

= = If the major and minor principal stress are σ = σ

respectively, σ

σ

σ

,

=

Maximum shearing stress = 

σ =

= ±√. / +

(M + √M + T ) and =



±√. / +

d= √

√M + T (M + √M + 2)

Where, σ = allowable working stress intension Again d = √ Where,

√M + T

= allowable working stress in shear




2.2 Structural Analysis (Analysis of statically determinate trusses, arches, beams, cables and frames)

Three hanged arch Standard cases Case1:

three hinged parabolic arch of span L rise ‘h’ carries a udl of w over the whole span W unit per run h H

H L

(a) The horizontal reaction at each support is H = (b) The net bending moment and shear force at any section on the parabolic three hinged arch is zero M = Beam moment – H moment = 0 Case2:

three hinged semicircular arch of radius ‘R’ carries a udl of w even the whole span. W unit per run C

X y

H

R θ 0

WR

WR

(a) The horizontal thrust at each end; H = (b) The maximum bending moment for the arch is Mmax = (Hogging) which occurs at θ = 30 from the horizontal and the distance of point of maximum bending moment from the crown is, R cos 30 =

√




Case 3: A three hinged arch consisting of two quadrant parts AC and CB of radii R and R . The arch carries a concentrated load of W on the crown W C

A

H

R

0 B

0

H

V V V =V =H= Case 4: A symmetrical three- hinged parabolic arch of span land rise carries a point load w, which may be placed anywhere on the span W K h

y H A

B

x V

H

V

L

H= The absolute maximum bending moment occurs at a distance of

on either side of the crown

√

Case 5: A three hinged parabolic arch of span has its abutments at depth crown the arch carries a udl of w per length over the whole span

h H A

and

below the

h L B

A

L L

H

V



The horizontal thrust at each support is give by H =

(√

√


)

Case 6: A three hinged parabolic arch of span l has its abutments A and B at depth h and h below the crown C. The arch carries a concentrated load W at the crown. The horizontal thrust at each support is given by H=

(√

√

)

Hinged Arches Two hinged arch is an indeterminate structure. V and V can be determined by taking moment about either end. The horizontal thrust at each support may be determined from the condition that the horizontal displacement of the either hinge with respect to other is zero W

W C y

h

x

H

B

A V =

∫

W

L

H

V

.

∫

Where, M is beam moment Standard Cases Case 1: A two hinged semicircular arch of radius R carries a concentrated load W at the crown W C

A

B H

H V =

V =

The horizontal thrust on each support is given by H =




Case 2: A two hinged semicircular arch of radius R carries a load W at a section, the radius vector corresponding to which makes an angle with the horizontal W C D

A

H

H V

V

The horizontal thrust at each support is given by H =

Sin

Case 3: A two hinged semicircular arch of radius R carries a udl W per unit length over the whole span W unit per run C R H

A

B

θ

H

WR

WR

The horizontal thrust at each support is given by H =

Case 4: A two hinged semicircular arch of radius R carries a distributed load uniformly varying from zero at the left end to w per unit rup at the right end. W x R B

A H

0 V

H

V

The horizontal thrust at each support for this case is H = .




Case 5: A two hinged parabolic arch carries a udl of W per unit run on entire span of the span of the arch is L and its rise is h W unit per run C H

h

y

A

B H L

V =

V =

The horizontal thrust at each is given by H = Case 6 : When half of the parabolic arch is loaded by udl. Then the horizontal reaction at support is given by H= Case 7 : When two hinged parabolic arch carries uniformly varying distributed load, for zero to w the horizontal thrust is given by H = Case 8: A to hinged parabolic arch of span L and rise h carries a concentrated load W at the crown W C

y H

A

h

x

V =

B

H

V =

The horizontal thrust at

Static & Kinematic Indeterminacy I. Static Indeterminacy 1. Statically Determinate Structures Can be analyzed with the help of equation static equilibrium alone For a plane frame f = f = M = 0 For a space from f = f = f = M = M = M = 0




2. Statically Indeterminate Structures Cannot be analyzed using equations of equilibrium alone Number of forces> Number of equilibrium equations  Degree static indeterminacy = Number of unknowns – number of static equilibrium equations A. For frames D =D +D where D = Internal indeterminacy = m (2j 3) for pin joined plane frame (i.e. truss) = m (3j 6) for pin joined space frame (i.e. truss) = 3C for rigid jointed plane frame = 6C for rigid jointed space frame D

= External indeterminacy = r 3 for a plane frame = r 6 for a space frame

Where m = Number of members in the structure j = Number of joints in the structure c = Number of curs required to obtain an open configuration B. Simplified formula = ( + ) 2 for pin jointed plane frame = ( + ) 3 for pin jointed space frame = (3C + r) 3 for rigid jointed plane frame = (6C + r) 6 for rigid jointed space frame II.

Kinematic Indeterminacy (or Degrees of freedom) The number of independent components of joint displacements (x,y,z & )

Kinematically Indeterminate Structures If the displacement components of the joints cannot be determined by compatibility equations alone i.e. additional equations based on equilibrium are required = 2j

r for pin joined plane frame

= 3j

r for pin joined space frame

= 3j

r for rigid jointed plane frame

= 6j

r for rigid jointed space frame

If axial strains are also considered in members then



=3

(

+ ) for rigid jointed plane frames

=6

(

+ ) for rigid jointed space frames


Method of Analysis 1. For statically determinate structures a. Double integration method b. Maxwell’s reciprocal theorem c. Moment area method 2. For statically indeterminate structures i. Force methods a. Conjugate beam method b. Theorem of three moments c. Column analogy method ii. Displacement methods a. Slope deflection method b. Moment distribution method c. Kani’s method iii. Energy methods a. Castiglione’s theorem b. Principal of virtual work 1. Beth’s theorem 2. Unit load method 3. Maxwell’s reciprocal theorem Truss 1. 2.

Truss members are subjected to only axial forces i.e. no bending moment and shear force. Joints are assumed to be pin joints.

3. 4.

Strain energy, u= , where P is member force. In a truss for the given load system, If force in individual member be P1 and to find the deflection at any node of the truss in any direction, apply a unit load in that direction at that node now, suppose the forces in member due to this unit load is So, deflection at node in that direction, Ui Pi hi =∑ n = no of member i Ei

Beam & Frame 1. Beam carries transverse load ie. bending moments and shear force but no axial force. 2. Frame carries both axial force and transverse load. 3. For beam &Frame, ∆= ∫ a) When M is original bending moment due to external loading and m be moment induced due to unit load/virtual load, Then deflection = THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 66



Cables 1.

2.

Cable theorem – Product of horizontal components of cable tension and vertical distance between the cable chord and cable at any point is equal to bending moment at that point on a horizontal simply supported beam subjected to the same external loading as that on the cable. Horizontal thrush (H)= ( )

Some Simple Results Configuration

Slope

A

( 6

=

P a

b

B

l

Deflection )

(2 6

=

)

(

=

=√

) 9√3

,

3

At center, if a>b, =

(3l

4b

P A

H2

C H2

B

l

= Udl=w

=

A

= =

16

24

max =

48 =

=

5wl 384 EI

l

M A

B

=

6

max = ml / 9 √3 EI At x = L / √3

l

=

t centre, = 3




P

θ =

max =

θ =

max =

θ =

max =

θ =

max =

θ =

max =

A l P a

A

B

l

=

(3l - a)

Udl = ω A

l

ω A

l

M A

l

Method for Determinate Structures 1. Double Integration Method Involves successive integration of expressions for BM. Under pure bending d 1 M = = = dx R EI Where = Curvature y = Deflection at any point x R = Radius of curvature THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 68



EI = Flexural rigidity Slope dy 1 = ∫ Mdx dx EI 1 Deflection, y = ∫ ∫ Mdx EI 2. Macaulay’s Method It is same as double integration method but the integration of expression for B.M is used in such a way that same constants of integration are valid for all portions of the beam even through the law of bending moment difference form portion to portion. Rules for applying Macaulay’s Method 1. 2. 3. 4. 5.

The origin (where x = 0) is to be taken on the extreme end The section P (at which BM is found out) is to be taken between the last load and the other extreme end (where x = l) While integrating, the terms within the brackets are to be kept intact i.e. the brackets are not to be cleared. For example the expression ∫(x a ) dx = (x a) + C If for any value of x, the quantity within the brackets is –ve, the term is to be neglected. That is, if the expression (x-a) becomes –ve, after substituting the value of x, the term containing the factor (x a) is to be omitted If a beam carries U.D.L for a portion of its whole length, the U.D.L is to be extended upto the extreme end of the beam and we are to superimpose a U.D.L equal and opposite to that which has been added while extending the given U.D.L

3. Deflection by moment area theorem(or Mohr’s method) Original shape B

A

A

B Only deflected shape

Deflected shape

x

x

x

According to 1st moment area theorem, the change in the slope of elastic curve between any two points for an initially straight continues beam, is equal to the area under the diagram between these two points. i.e. θ

= ∫

( )

dx

According to 2nd moment area theorem the tangential deviation of pt. B measured from the tangent to the elastic curve at point A is equal to the moment of the area under the diagram A and B about point B. i.e. t

= ∫ (x

x )

dx




4. Deflection by Conjugate Beam Method Shear force at any section of conjugate beam Original Beam

Conjugate Beam

i)

i)

ii)

ii)

iii)

iii)

1.

Overviews of Force Method → Degrees of indeterminacy should be calculated and all the reactions should be located →The redundant reactants whose number is same as degree of indeterminacy should be picked up. →The reaction other than redundant reaction should be expressed in terms of redundant reactions. → Compatibility conditions should be applied to calculate redundant reaction. →The redundant reaction should be calculated back from the redundant reaction knowing equilibrium conditions →The deflections are calculated from the forces and reactions. a. Shear force of any section of the conjugate beam gives the rotation t that section of the original beam b. Bending moments of any section of the conjugate beam gives the deflection at that section of the original beam

2. Analysis by Three Moment Theorem Generally for continues indeterminate beam it is applied B

C

A

L

L

,

diagram for L and

are area of bending moment L span respectively

x , x are the CG distance

x

x




According to this theorem, M

+ 2 MB .

+

/+M

+ 6E .

.

= -6 . +

/+

+

/

/

Where I and I are moment of inertia for span L and L respectively. ,

,

are support settlement at support A,B,C respectively

Method for a Indeterminate Structures I.

Force methods Conjugate beam method 1. Slope at any point in the elastic curve of active beam is equal to the shear force at the point in its conjugate beam. 2. Deflection at any point in the active beam is equal to the bending moment at the corresponding point in the conjugate beam 3. diagram of the original beam will be the loading for the conjugate beam to determine shear force and bending moment as require in point 1 and point 2 mentioned above.

II.

Energy Methods For bending strain energy = ω = ∫ Deflection, =

= ∫m

dx

dx

If m be the bending moment at the point due to external loading and m be the bending moment due to unit load applied at that point Deflection = ∫ For which moment m applied at that point Slope, = ∫ 1. Maxwell’s Reciprocal Theorem In any beam or truss the deflection at any point D due to a load W at any other point C is the same as the deflection at C due to the same load W applied at D. W

W C ∆C

D ∆d

D

C c

∆

So, C = ∆d




2. Castilian’s Theorem a. Castiglione’s First Theorem The partial derivative of the strain energy of a linear elastic structure expressed in terms of displacements with respect to any displacements with any displacement is equal to the force at coordinate: ∪ . ., = The partial derivative of strain energy has been used because the change of strain energy due to an increment in displacement at coordinate ‘ ’ has only been considered keeping all other displacement unchanged b. Castiglione’s Second Theorem The partial derivative of the strain energy of a linearly elastic structure expressed in terms of forces with respect to any force applied at coordinate is equal to the displacement at coordinate j. ∪ . ., = Where, Ss ds ∪= ∫ for axial forces E Mm ds ∪= ∫ for moment E

First Theorem of Castigliano In any beam or truss subjected to any load system, the deflection at any point r is given by the partial differential co-efficient of the total strain energy stored with respect to a force Pr, acting at the point r in the direction in which the deflection is desired. P P P

P

So, y =

r yr

A here, ω = Corresponding strain energy stored III. Analysis by displacement methods (Slope deflection and moment distribution methods)  Degrees of freedom are the number of independent displacement components of a structure required to be known to completely find out the deformed shape of the structure.  Degrees of freedom for flowing frame for different conditions




a) All members are axially rigid and the beam is also flex rally rigid: DOF = 1 b) All members are only axially rigid : DOF = 3 c) Members are neither axially nor flexuarally rigid : DOF = 6 Overview of displacement method     

ll the unknown displacement components i.e. DOF’s are located The relationship between force/moments and displacement rotations are obtained using compatibility conditions. Joints equilibrium conditions are used and then expressed in terms of displacement or rotations with the help of the compatibility conditions. Now the simultaneous equations are expressed in terms of displacements and solved to obtain the unknown displacements. Unknown reactions are obtained from unknown displacement

1. Analysis by slope-Deflection Method In any continuous beam, considering one span as below C

B

A

D

Considerin g

B

A

Only AB L

(Settlement ) Again

∆ (Settlement of support B)

∆

M

M

M

M

L

M L∆

Where M M

are fixed end moments

M =

(2θ + θ )

M =

(2θ + θ )

M = M = -6 EI

M

∆

So, M = M

+

(2θ

θ ) - 6 EI

M =M

+

(2θ + θ ) - 6 EI

∆

∆






Some common fixed end calculation Udl = ω A

B L P

A

B b

a L

M

=

,M

=

M = ,M = 2. Analysis by Moment Distribution Method  Moment required to create unit rotation in a beam with near end hanged and far end hanged or fixed is called flexural stiffness factor  Carry over factor is the ratio of moment created at the far end to the moment created at the near end End Condition

Far end Fixed 4 EI / L

Stiffness Factor

Far end hanged 3EI L 0

1 2

Carry over factor A A

B

B

0 E , L ,I

M

M

E , L ,I

I , E ,L

C

M

M 0

M

C

M

Common rotation at 0 = θ M =.

/ θ

M =.

/θ

M =.

/ θ

Total M = M + M + M = .

+

+

/θ

Or, θ =



M =M


Distribution factor

M =M

M =M

Influence Lines for Determinate and Indeterminate Structure Influence line diagram (ILD) for a specific effect (e.g Reaction R) Shear force (v), Moment (m), Torsion (t) at a specific point may be defined as a curve, the coordinates of which show the variation of that effect caused by a unit moving along the span Muller Breslor’s Principle →The influence line for any stress function of a structure such as shear force, bending moment or any reactive force or movement is given by the deflected curve of a structure obtained by unit distortion or deformation in the direction of stress function of the structure Different Cases For a Single Concentrated Load Moving From One End to Another 1.0 B

A L

    

Reaction at one end will be maximum when the load is at that one end and reaction at other end will be zero simultaneously Maximum shear force (may be (+)ve, may be (-)ve at a section across when the load is at the section itself Absolute maximum shear force anywhere in beam equal to the reaction itself Maximum bending moments at a section occurs when the load itself is at the section Absolute maximum BM occurs when the load is placed at the centre of the span



Structural Engineering 1.0

1.0 B

A

A

B

C

L a Absolute max. SF 1.0

1.0 A

+ ILD for R + ILD for R

a

L

L

1+ -

C a L

a(1- ) B +

1.0 A

B

ILD for M 1.0

Absolute max. SF L/2

ILD for V Reaction

L L/2

Shear Force

B

A Absolute max. BM Bending Moments

For UDL Moving From One End to Another Type A: For UDL greater than span of girder: (d > L)

   

Maximum positive SF occurs when the tail of the load is at the section itself Maximum negative SF occurs when the head of the load is at the section itself Maximum BM occurs at any section when the entire span is fully loaded With the entire span fully loaded, absolute maximum BM occurs at the centre of the span

Type B: For UDL less than span of girder (d < L)   

Maximum positive SF occurs when the tail of the load is at the section itself Maximum negative SF occurs when the head of the load at the section itself Maximum BM occurs at any section only when the load is placed with the equal ratio as of span of the two length divided by the section itself






Absolute maximum BM occurs at the midpoint of the span when the load is placed equally divided on the span For Several Concentrated Load Moving From One End To Another A

B

C L a

b

x W (Resultant)

Now, to find critical load we have to place loads over the span w.r.t. c let, W , W , W are on the hand side of C and W , W , W are on the right hand side of C Let, > Again putting W on the right hand side of C, If < W is the critical load Now, the load W is nearly to resultant W. So, maximum BM will occur under W and the position like following figure. W

A

B

⁄

⁄

L/2

L/2

i.e. load position of W is at a distance . 

/for maximum BM

The absolute maximum bending moment generally occurs under the heavier loads, which is very near to the CG of the load system

Matrix Method of Structural Analysis (Basic Concepts)  Coordinate Systems 1. Global Coordinates: If the location and direction of all the displacement vectors and the forces are written w.r.t a single coordinate system, the coordinates are called global coordinates. 2. Local Coordinates: If the location & direction of all the displacement vectors and the forces are written by shifting the origin to the starting node of each element, the coordinates obtained are called local coordinates. Global and local coordinates are interchangeable




Flexibility & Stiffness The flexibility of a structure is defined as the displacement caused by a unit force. ∆

i. e. f = | Stiffness is the force required for unit displacement. K = ∆ Cases 1. Axial Displacement 2 1

P1

(4)

3 ∆1 = ∴ axial flexibility

11

=

∆

=

Axial stiffness K11 = ∆ =

=

2. Transverse displacement P2 (2) D2

A ∆2 = Transverse flexibility

11

=

and Transverse stiffness K22 =

If the further end (end ) is hinged then ∆2 =

∆

∆2

So, transverse flexibility,

22

=

and transverse stiffness K22 = 3. Bending or flexural displacement ∆3 = ∴ Flexural flexibility

33

=




and flexural stiffness K33 = If the end is hinged Then ∆3 = ∴ Flexural flexibility

33

=

Flexural stiffness K33 = 4. Torsional Displacement or Twist The angle of twist ‘∆4’ due to torque P4 is given by ∆4 = Where, G = Shear modulus of elasticity K = Torsion constant. ∴ torsional flexibility,

44

=

Torsional stiffness K44 = GK L Some basic Relations i. e upto K

=

If NF = Total number of known member forces NP = No. equation can be formed or no. of global degree of freedom Global 1. *deformation vector+ = ,Deformation matrix- x { } displacement vector Or *e+

= ,B-

x *X+

2. *element force vector+ = ,Stiffness matrix- x *Deformation vector+ Or *F+ = ,Sx *e+ , Local stiffness matrix 3. Global Stiffness matrix, [k] = [A] [S] , - , Transformation matrix Step for solution a. b. c. d.

Calculate [k] = [A] [S] , Calculate *X+ = ,k- {f} Calculate {e} = , - {X} Calculate {f} = [S] {e}




2.3 Concrete Technology Properties of Concrete & Basics of Mix Design         

Water cement and aggregates are basic ingredient of concrete C S give long term stability in concrete& generate minimum heat C S gives short term strength C gives maximum amount of heat of hydration Max C means rapid hardening cement During manufacturing gypsum is added to cement to prevent fast setting due to excess heat of hydration caused by ettringite (calcium sulfur ferrite) Plasticizer work as water reducers to obtain higher workability without using excess of water Retenders slower the processes of hydration e.g. Gypsum Accelerators are used to increase the rate of early strength development in concrete

Properties of concrete are of two types 1) Fresh 2) Hardened Fresh properties of concrete are a) Setting time i) Initial and (30 to 60 minutes) ii) Final which can be found out by penetrometer test (5 to 6 hours) & should not be exceed 10 hours b) Workability: Degree of fluidity or mobility. Measurement of work ability is done by following methods i) Slump test: Low slump – slump height 25 – 75 mm Medium slum – slump height 50 – 100 mm High slump – slump height 100 – 150 mm ii) Compacting factor test : work ability C. F =

compacting factor (C.F)

weight of partially compacted concrete weight of finally compacted concrete

iii) Flow test iv) Vee-bee compaction meter test c) Segregation: Separation of constituent materials of concrete d) Bleeding: One type of segregation in which laitance (formation of cement paste at the surface) is occurred e) Bulking of aggregates: Maximum increase in volume is 40% and maxm bulking occurs at a moisture content means 5% THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 80



f) Consistency of concrete: If P is the standard consistency of cement the amount of water used in conducting initial setting time test on cement is 0.85 P, water of ( + 3.0) percent of combined weight of cement and sand is to be added to cement and sand for strength test g) Alkali aggregate reaction:  Reaction between the alkalis of cement and the active silica or carbonates aggregates  Alkali silica gel is produced  Can be avoided by using non-reactive aggregates and low alkali cement 1. Hardening Properties of Concrete  Strength 1. Compressive strength – lower water cement ratio higher compressive strength 2. Tensile strength a) Direct tensile strength b) Flexural strength (σ) P

l M= σ= , Z = section modulus c) Splitting tensile strength (σ ) Load P, cylinder diameter D cylinder length = L σ = = ⁄ 3. Modulus of elasticity (E ): E = 5000 √fck fck = characteristic strength of concrete  Durability: Indicates lifetime of concrete structure Other Properties of Concrete a) Shrinkage in concrete→ change in volume from fresh state to hardening state of concrete b) Creep in concrete : Increment of strain in concrete due to sustained load Concrete mix design Fck = characteristic strength of concrete Fck = target strength of concrete t = acceptance criteria s = standard deviation  

Relation is fck = fck + t ×s Normal mix is only applied m 20 or below grade concrete Grade of concrete nominal mix (cement: stone: sand) M 20 1 : 1.5 : 3 M 15 1:2:4 M 10 1:3:6



Volume of 1 bag cement = 34.5 litres THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 81



Important Method of Concrete Proportioning 1. Maximum density method Fuller’s maximum density d P = 100√ D 2. Fineness modulus & water cement ratio method proportion of fine aggregate to coarse aggregate + fine aggregate is given by P P R= P P Where , are fineness modulus of coarse and fine aggregate respectively Q and P is desired fineness modulus Impact of Water Content on Concrete Properties 1. High water cement ratio increases workability but reduces strength 2. Lower water cement ratio also reduces strength and workability 3. Optimum water cement ratio for required strength of concrete should be decided from graphs and expressions Practical Water Cement Ratio M 15 → 0.50 M 20 → 0.50 M30 → 0.45 Concrete Design Basic working stress and limits state design concept analysis of ultimate load capacity and design of members subjected to flexure shear compression and torsion by limit state method. Working stress method of design: - {

f

= characteristic strength of concrete } f = characteristic yield stress of steel



Assumption 1. Based on the behavior of structure at working load 2. Stress – strain relationship is linear under working load 3. Factor of safety for concrete in compression & bending and for steel in tension & bending are 3.0 and 1.8 respectively 4. Real factor of safety = 0.67 fck/0.33 fck = 2.0 for concrete



Consideration 1. Modular ratio (m) =

where σ

= permissible compressive stress in concrete in bending = 0.33 fck

2. Permissible stress in tension in steel

=

.

= 0.55




3. From definition m = Where E = Young modulus of steel E = Young modulus of concrete = 5000 √f 

Some Constants 1. Neutral axis constant, k = 2. Lever arm constant, j = (1 - k⁄3) 3. Moment of resistance constant



=.

k

j

σ

/

Position of Neutral Axis Case I: (section is known) √(

)

X= Where = Area of steel reinforcement m = Equivalent concrete area Case II: (when stresses developed in the section are known) x = kid 

Different Type of Sections x

σ

σ

σ

σ

=σ

x x

σ m Balanced section

σ σ = m m Under reinforced section

σ σ m m over reinforced section

xC = critical depth of neutral axis (for balanced section) x &x = actual depth of N.A When balanced section is not possible under reinforced section is preferred as the steel reaches the yield stress first






Some basic concepts

d

Nominal reinforcement D b

Stirrup

Main reinforcement

Enlarging Q Q Nominal cover

Nominal cover

b = width of section D = over all depth of section = dia of stirrup = dia of main reinforcement Section

Nominal cover (mm) 15 25 40 50

Slab Beam Column Footing

Clear cover = Nominal cover + Effective cover = clear cover + So, effective depth, d = D – effective cover 

Moment of resistance for balanced section b

σ C = ×σ

x N.

b

Z = lever arm D

d

T=σ

Alternatively MR = T × Z σ ×Z=σ jd (for under reinforced sec) Or = For balance section Qbd = σ

jd



Or

× 100% = ×

k


100

Chose shear stress ( C) according to p check whether max shear stress ( corresponds to p is greater than or less than C

Cmax)

a) If ( C)max> Design shear force VS = V - T bd Provide stirrup with spacing (SV) of minimum of these three σ = permissible stress in stirrup for ahear = σ for fc 250 fc 415 i) S = ( ) = rea of stirrup .

ii) S = . iii) S = 0.75 d b) If ( c)max< c (Sectors are needed to be redesigned) Spacing of main reinforcement should not be greater then 300 mm or width of beam whichever is less.

Deflection Cheek a) Using Basic Value Only Basic value = 7 (for cantilever) (for span upto 10 m) = 20 (for simply supported) = 26 (for continuous) Effective depth (d) = For spans above 10 m, the above value shall be multiplied by

in meters except

for cantilever b) Using Modification Factor Also d=( ) where k = modification factor for % of main reinforcement (tension) k = modification factor for % of compression reinforcement (for double reinforcement beam) k = modification for T beam or L.Beam Development length (Ld) cheak

σ = Bond stress L pullout force = frictional force σ = ×L σ Or L = 4 For end anchorage + L >L bd




Where m1 max moment of resistance at particular section V = max shear force at that section L = Effective depth (d) or 12 which ever is greater. The value of

may be

increased by 30% when ends of the reinforcement by compressive reaction effective length for simply supported beam or slab reaction It is smaller of (L + d) or (L + a) Where l = clean distance between the support a = support width (if the support width is unequal take the lesson one) for simple beam generally b is taken as and D is taken as . to / 

Base rule for Design Formulation of Slab:

l (y denotes longer direction)

If > 2, one way slab l (x denotes shorter direction) If ≤, 2 way slab

Effective Span (Same as beam) Deflection check (Same as beam) Minimum reinforcements – 0.15% mid steel -

0.12% for high strength deformed bar

Maximum diameter ≯ th of total thickness of slab no shear reinforcement should be used spacing of bar i) Main reinforcement - ≯ 3d or 300 mm whichever is ii) Secondary reinforcement - ≯ 5d or 450 mm which over is smaller






Design formulation for double reinforced beam

Asc

b

d

d’

Ast 2

Ast1 Balanced section

Compression steel and additional tensile steel

Total moment of resistance, m = m + m m = movement of resistance of balanced section m = moment of resistance for compression steel Ast1 = , Ast2 = ( ) Total = Ast1 + Ast2 Asc = x x  



.

(where, x = x = kd)

For same cross sectional area, moment of resistance of T-section is greater then that of rectangular section Some important codal provisions (IS: 456, 2000) Tensile stress = Where F = Total tension on the member minus pretension in steel = Cross sectional area of concrete Design formulation column Eccentricity. e = + ≯ 20 mm If or ≤ 12, short column or > 12, longer slender column Where, L, D, b are length, diameter and breath of column respectively For short column, Permissible loads, P = σ + σ Asc Where, σ = Permissible stress in concrete for direct compression (e.g. m →4.0n mm , m → 5.0 n mm ) σ = Permissible compressive dress for column bars Ac = Cross-sectional area of concrete excluding reinforcement of steel Asc = Cross sectional area of longitudinal bars For long column permissible load, P = Cr. (σ + σ Where, Cr = Reduction factor < 1 = 1.25 – B = least lateral dimension of column

) = Cr. P

For Short Column with Reinforcement Permissible load, P =1.05 (σ Ac + σ Asc) ⋆ I.S code requirements for design of column THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 87



a. % of longitudinal reinforcement, Asc≮ 0.8 % ≱6 For practical purpose Asc≱ 4% b. Minimum number of longitude bars is 4 for square and rectangular column and 6 for circular column c. Minimum diameter of bar to be used is 12 mm d. Spacing of longitude bars measured along periphery shall not exceed 300 mm. e. Diameter of lateral ties (transverse reinforcement) ≮ th of dia of longest longitudinal bar or minimum 6 mm f. Pitch (vertical spacing) for lateral ties are be minimum of these three i) Least lateral dimension of column ii) 16 times smallest dia. of longitudinal bars iii) 300 mm 

Design Formulation for Footing a. Minimum depth of footing at the end of sloped footing should not be less than 150 mm b. Permissible shears tress (for punching shear) at critical section is K Where K = (0.5 + Bc) but ≯ 1 B = ratio of short side to long side of column = 0.16 √f (Working stress method) = 0.25 √f (Limit state method) c. Shear force i) One way shear Critical section for shear shall be assumed at c vertical section located from the face of the column at a distance equal to depth of footing (d)

Column

d

d d d ii) Two way shear (punching shear) The critical section shall be of a distance of d/2 from the periphery of the column d 2

d 2




d. Bending moments: Critical section shall be at the face of column

Critical section Limit State Method of Design  Assumptions 1. Stress and strain relationship in concrete of parabolic are parabolic & linear respectively . 2. Measure compressive stress . = 0.446 fck 3. Maximum strain concrete in compression is 0.0035 in bending irrespective of strength of concrete 4. Maximum strain in tensile reinforcement at failure should not be less than. + 0.002 / .

Considerations: 1. Two types limit states. 2. Limit state of collapse → a) Flexure b) Shear c) Torsion d) Compression ii) Limit states serviceability a) Cracking b) Deflection c) Vibration d) Fire resistance 2. Partial factor of safety for limit state of collapse for concrete 1.5 and for steel 1.15 3. Load combination design load = DL + L.L 

Or D.L + W.L Or = D.L + 0.8 L.L + 0.8 W.L Whichever is maximum 

Moments of resistance of a rectangular section (single reinforced) .

. =

.

. =

= .

.

+

.

=

.

= .



x =


3x 4x = 0.430 x = = 0.51 7 7

Also, according to IS: 456, 2000, For balanced section, Fy X /d 250 0.53 415 0.48 500 0.46 Design compressive force in concrete (c) = 0.36 f bx Design tensile force in steel = 0.87 f For a balanced section C=T 0.36 f b x = 0.87 f 0.87 f x = 0.36 f b From strain diagram

,

+ x , lim ε = d ε +ε ε x , lim = ε +ε

.d =

0.0035 0.0035 + .0.002 +

.

/

When x = x , lim section is balanced x x lim section is under reinforced x x lim section is over reinforced Lever arm = distance between C.G of compressive force (C) & tensile force (T) i.e., Z = d 0.416x = d .1 0.416 / ∴ Moment of resistance, (MOR) 1. For balanced section, (x = x , lim)(both concrete & steel fail together) x , lim x lim MOR = C Z = T Z = {0.36f b ( ) d} {d (1 0.416 )} d d x , lim x lim M = 0.36 f (1 0.416 ) bd d d 2. For under reinforced section ( MOR = T

Z = 0.87f

d .1

) steel fails first desired x 0.416 / 4



M = 0.87 f

d *1


f + f bd

3. For over reinforced section (concrete fails first not desired) x , lim MOR = C Z = 0.36 f b x , lim [d 0.416 ] bd d MOR = C Z = 0.36 f b x , lim,d 0.416x , limx , lim x , lim M = 0.36 f (1 0.416 ) bd d d According to IS 456 : 200 f x , lim d 250 0.53 415 0.48 500 0.46 Doubly reinforced rectangular section Moment resisted by compression reinforcement = Moment applied (M) – limiting moment of resistance of singly reinforced section Case 1: is less than the limiting value , ( = 0.36 0.416 ) + ( 0.446 = 0.87 ( 0.416 ) Case 2: M , lim

)

,

= = 0.36f x , lim b(d

Case 3:

)(

0.416x , lim) + (f

0.446f )

(d

d)

,

This corresponds to over reinforced section 

Flanged Beam

(FLANGED WIDTH)

D




Flange width (bf) For T beams: = min (

6

+

+6

+

+

)

For Isolated T beams = min (

+4

+

, )

For L-beams = min ( + 12

+ 3

,

+

)

For isolated L- Beams = min (

0.5 +4

+

,

)

bf= effective width of flange bw = breadth of rib or web b = actual width of flange lo = distance b/w points of zero moments in beam (may be taken as 0.7 time effective span for continuous beams) x1, x2= half the clear distance b/w/ two adjacent beams df = thickness of flange

Moment of Resistance The depth of NA is first calculated assuming it falls in flange ie (xu
0.87 0.36

If the xu come out to be less than Df ,then it is treated as a rectangular section of width bf. Otherwise xu falls outside the flange and compression of web has to be taken into account (N inside the flange) Case 1:  Neutral axis lies within the flange, The moment of resistance can be found out from expressions applicable for rectangular sections by replacing b by bf.  = 0.87

,

= 0.36

)*

(1 ,

.1

0.416

+ ,

/

*

+

When xu>xu, max, this become over-reinforced section & should be redesigned however moment of resistance is limited to Mu, lim.



Case 2:

1. 2.


(N outside the flange)

Two situation can arise in this case (N falls outside the rectangular portion of stress block). ≤ (N falls inside the rectangular portion of stress block)

D

b -b

Figure 4.5 The compressive force Cuv from the web contribution is given by = 0.36 If reinforcement Asw is the component of tensile reinforcement required to balance the compressive force Cuv and remaining reinforcement Asp if required to balance the compressive force Cuf in the flange, we have Ast = Asw +Asfthe tensile force Tuv balancing the compressive force Cuv will be 0.87 fyAsw 0.36 fckxubw= 0.87 fyAsw. A rectangular stress block of depth yf , can be assumed instead of actual rectangular-cum-parabolic stress block for computation of compression force Cuf. The assumed stress blocks have same peak stress value 0.446fck. y = 0.15 x + 0.65 D but less thanD Compressive fore in flange (Cuf)= 0.446fck (bf –bw) yf Tensile force Tuf balancing the Cuf= 0.87 fy Asf Total Cu = 0.36 fckxubw + 0.446 fck (bf-bw) yf Total Tensile force Tu = Tuf = 0.87 fyAst equating Tu and Cu 0.36 f x bw + 0.446 f (b b ) y = 0.87 f As THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 93



If yf>Df Then yf is taken as Df Therefore, 0.36 fckxubw +0.446 fck (bf-bw) Df = 0.87 fyAst Under Reinforced section : (xu = xu, max) 1. ≤ = 0.36 = 0.87 = 0.446 ( ) = 0.87 =

(

0.416

) = 0.36

01

0.416

1

= C (d – 0.416x ) = 0.446f (b - b )D (d – 0.5D ) Where Muw & Muf are the moment of resistance of web and flange respectively. Similarly w and f in subscripts denotes the web & flange respectively Mu = Muw +Muf 2. y = (0.15 x + 0.65 D )orD whichever is lesser = 0.36 = 0.87 = 0.446 ( ) = 0.87 Therefore, ( = 0.416 ) = 0.36 .1 0.416 = Hence

(

= 0.36 

0.5 ) = 0.446 .1

0.416

(

/

0.5

) ( + 0.446

(

/ ) ) (

0.5 )

Vertical stirrups 5 mm to 15 mm dia steel bars are provided as stirrups by bending them around tensile reinforcement. The stirrups may be two legged, one legged, four legged or multi-legged the strength of shear reinforcement as vertical stirrups is given by 0.87 = ,( ) =

0.87

=

0.87 (

)

Minimum & maximum shear reinforcement  As per Is 456 : 2000, minimum shear reinforcement shall be provided in form of stirrups such that: 0.4 0.87 Where, = Characterstic strenght of stirrups and shall not be taken greater than 415 N mm Hence spacing based on minimum shear reinforcement 0.87 2.175 ≤ ≤ 0.4  The maximum spacing of shear reinforcement measured along the axis of member shall not exceed 0.75 d for vertical stirrups and d for inclined stirrups at 450 where d is the effective depth of section under consideration. In no case shall the spacing exceeds 300mm. THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 94



Limit State of Collapse Bond The external load on a RC element is first borne by concrete and then transferred to steel reinforcement. This transfer of force from concrete to steel is effected only when there is no slip, sliding or relative movement b/w any one of them when any one of these is strained. The force which prevents the slip is called bond. Development length of bars (Ld) This is the length of bars required on either side of any section to develop the required stress in steel. =

4

Where , = Nominal diameter of bar = Design bond stress for plain bars in tension.

Limit state of collapse in Torsion Torsion is dominant in peripheral beams in each floor of any multi-storeyeed building in which slabs are casted monolithic with beam giving the L-beam configuration. In the ring beam also provided at bottom of an elevated water tank, tensional forces comes. Torsional reinforcement is calculated with each of the shear reinforcement & longitudinal reinforcement resulting in fictitious shear and fictitious bending moment respectively which are function of torsion with shear force and torsion with bending moment. Shear & Torsion Equivalent shear as a result of shear & torsion is calculated by following expression. =

+ 1.6

Where, = Equivalent shear = hear = Torsional moment = Breadth of beam =

= Equivalent nominal shear stress




Reinforcement in Members Subjected to Torsion If veexceeds c the reinforcement is required otherwise not. It is provided as longitudinal reinforcement and transverse reinforcement. 1. Longitudinal Reinforcement It should be provided to resist an equivalent bending moment Me1 = Mu + Mt. Where, = Bending moment at cross section 1+ = [ ] 1.7 = Torisonal moment = Overall depth of beam If Mt exceeds Mu, then torsional reinforcement shall be provided on compression flexural compression face, such that equivalent BM Me2 can also be resisted = Where, is taken as acting in opposite sense to 2. Transverse Reinforcement =

(0.87

)

+

2.5 , (0.87

)

However, total reinforcement shall not be less than ( )( ) 0.87  Transverse torsional reinforcement shall be closed rectangular stirrups and placed perpendicular to the axis of member. The spacing shall not exceed the least of x, , 300 mm. where x1&y1 are respectively short and long dimension of stirrups.  Longitudinal torsional reinforcement shall be placed as close as possible to corners of cross section for all cases, there should be at least one longitudinal bar in each corner of ties. One way slab Analysis is done by assuming it to be a beam of 1 m width and reinforcement is calculated and distributed accordingly =

0.5

*1

√1

4.16 + 1000 (1000 )

Spacing of main reinforcement 1000 S= → rea of cross section of 1 bar Transverse reinforcement Min → 0.15 if mild steel is used → 0.12 if HYSD bars used Two way slabs, along x. M = M =

w lx w ly



Where , IS 456:2000 

are constant depending upon


ratio and different edge condition as given

Design Formulation for column Axial load on member, P = 0.4 fck + 0.67 fy → = 0.45 + 0.67 Where, Ac = Area of concrete Asc = Area of longitudinal reinforcement Members subjected to combined axial load and biaxial bending may be designed satisfying following equation 0

1

+[

] ≤ 1.0

Where, M , M = moments about x and y axes due to design loads M , M = maximum uniaxial moment capacity for an axial load P , bending about x and y respectively. = And P = 0.45 fck Ac + 0.75 fy Asc The additional moments for slender compression members M calculated by M

=

.

and May shall be

/

M = . / Where, Lex = effective length in respect of the major axis. Ley = effective length in respect of the minor axis D = depth of the cross-section at right angles to the major axis B = width of the member Some codal provision (IS, 456: 2000)  To prevent sulphate attack, the water soluble sulphate content of the concrete Max, should not exceed 4 percent by mass of the cement in the mix  Concrete in sea water shall be at least m20 grade for pcc and m30 grade for Rcc  The maximum permissible free fall of concrete is 1.5m  Under transient wind load the lateral sway at top should not exceed H/500, where H is the total height of building  Where a member is built into a masonry wall which develops only partial restraint, the member shall be designed to resist a negative moment at the face of the support of where w is the total design load and L is the effective span 

Effective width of flange



Slenderness limit for a simply supported or continuous beam is 60db or whichever is less. For a cantilever beam is 25b or



whichever less

The final deflection due to all loads including the affects of temperature, creep and shrinkage a. And measure from the as cast level of the support of floors, roofs and all other horizontal member, should be limited to span/250




b. Occurring after erection of partitions and the application of finishes should be limited to or 20 mm which is less 

To calculate the development length of bars, a. The depth stress values (as IS: 456,2000) should be increased 60% for deformed bar b. For bars in compression the values of bond stress be increased by 25% bars in tension  The anchorage value of a. Bend shall be taken as 4times the diameter of the bar for each 45 bend subject to maximum of 16times the diameter of the bar. b. Standards u-type hook shall be equal to 16 times the diameter of the bar.  Bars in a bundle shall terminate at different points spaced apart by not less than 40 times the bar diameter except for bundle stopping at a supports  Lap splices shall not be used for bars larger than 36 mm.  Lap length including an change value of books for bars a. In flexural tension is L or 30Q, whichever is greater b. For direct tension is 2L or 30Q, whichever is greater c. Straight length of the lap shall not be less than 15 Q or 200 mm d. In compression L or 24Q, whichever is greater  Minimum horizontal distance between individual bars is the greater of the following: 1. Diameter of bar for equal bar diameters 2. Diameter of length bar for unequal bar diameter 3. 5mm more than the nominal maximum size of coarse aggregate  Tension Reinforcement: . a. Minimum: =

 

b. Maximum: Ast = 0.04 bD Where, D = overall depth Compression Reinforcement Maximum Asc = 0.04 bD Side face reinforcement

D)750 mm Total 0.1% Applicable for depth of web in beam exceeding 750mm total area of side face of web area reinforcement = 0.1% of web area to be distributed equally on two faces at a spacing not exceeding 300 mm or web thickness whichever is less  

One or more expansion joints are provided for structures exceeding 45m in length For deep beam, < 2.0 for simply supported beam < 2.5 for continue beam



Lever arm for deep beam, Z = 0.2 (l+2D) for 1 ≤ ≤ 2 (simply supported beam) Z = 0.2 (l+1.5D) for 1 ≤ ≤ 2.5 (for continues beam)



  


Minimum thickness of walls shall be 100 mm Permissible bearing stress = 0.25 fck (W.S.M) = 0.45 fck (L.S.M) A member subjected to lesser load than 0.2 fck Ac may be considered as flexural member for the purpose of crack control

3. Basic Elements of Pre-stressed concrete, analysis of beam section at transfer and service loads  Concrete strength requirements Pre-tensioning member M 40 Post-tensioning member M 30  

Ultimate tensile strength of steel is about 1500 N/mm for wires of 5 to 7 mm diameter Cover Requirements i) Pre-tensioning member – 20 mm ii) Post-tensioning member 30 mm

Pre stressing system a. Pre-tensions →hoyer system b. Post-tensioning 1. Fressmet system 2. Magnel Blaton system 3. Gifford udall system 4. Lee-Mccall system → nalysis of Beam Case 1: Concentric Tendon P/A

+

M/Z +

+

+ Cross section

P/A Direct stress Due to prestress

M⁄Z

-

Stress due to external bending moments

Final Stress

Case 2: Eccentric Tendon Pe⁄Z P/A e P

P

e Cross section

M/Z

-

+

+

+

+ P/A

-

-

M⁄Z Direct stress Due to prestressStress due to Stress due to eccentricity of external bonding prestress moment Pe⁄Z

+ Final Stress




Stresses  At Transfer a. At top fiber = -

+

b. At bottom fiber = +

-

 At working loads/ service loads a. At top fiber = η. /+. b. At bottom fiber = η.

+

/+.

/ +

/

Where M and M dead load and live load moments respectively Z andZ are section modulus at top and bottom respectively η = loss factor, (Unless mentioned η=1 (take))  Two critical situations 1. At transfer

Maximum Pre-stressing force Minimum service force

2. At service

Maximum Service load Minimum Pre-stressing force.

 Loss of Pre-stress Loss in pre-tensioning is more than loss of post tensioning  Loss due to elastic shorting = mfc Where fc = Average stress in concrete at the level of steel  Loss due to shrinkage = EcsE Where E = 300 x 10 for pre tensioned =

(

)

for post tensioned

And t = Age of concrete at the time of transfer in day 

Loss due to creep = m Fc Where, = Creep co-efficient

 Use of Pre-stress Pre-tensioning is more economical for large number of small scale production e.g. railway sheepers where as, post-tensioning is more suitable for very long span structure line box girden type bridge deck




2.4 Steel Structures Introduction Types of Structural Steel 1. Mild steel (standard steel) 2. High tensile strength steel Properties of Structural Steel 1. Physical properties Density = 7850 Modulus of elasticity E = 2 10 N mm Poisson’s radio μ = 0.3 Modulus of rigidity G = 0.769 10 N mm Coefficient of thermal expansion = 12 10

C

2. Mechanical properties Yield stress Ultimate stress Determined by tensile tests Maximum strain allowed

Stress-strain curve for meld steel 3. Analysis and Design of Tension Members Tension Member Tie A member carrying direct tension is called a Tie. Angle section are mostly used as tie




Various design philosophies 1. Working stress method (WSM) IS 800:1984  Oldest analytical method  It considers yield point as failure points which as not true  Permissible stress (f ) = 

Gives over safe and economical sections so, no need to check for service ability

2. Ultimate load design (ULD) or load factor method (LFM)  Formation of several plastic hinges  Allowed redistribution of forces is accounted  Load factors are introduced instead of safety factor for material behavior Design load Load factor = Working load Load combination

Allowed stress (WSM)

1.7

DL+LL DL + WL DL + WL + WL/EL

Minimum load factor

1.7 1.33

1.3

3. Limit state design (LSD) IS 800:2007  Limit states are the states beyond which the structure no longer satisfies the specified performance requirements  Categories limit states of strength  Categories limit states of service ability  LSD is a probabilistic approach Characteristic strength ( ) The strength (ultimate or yield) shown by 95% of samples tested. Design strength = Where partial safety factor For designed governed by Yielding Buckling Fracture . .,

=

Or =

1.10 1.10 1.25

1.10 1.25




Bolted Connections Classification of Bolts 1. Bearing Type Unfinished (black) bolts Finished (Turned) bolts 2. Friction type : HSFG Bolts For unfinished bolts → Hole dia = nominal bolt dia +mm For finished bolts hole dia. = normal bolt dia. + 1.50 mm In friction grip bolts, bolts are tightened to a proof load and hence the plates are pulled together to develop a huge friction force so, no need to check for shearing. Its surface is kept unfinished like black bolts Classification of bolted Connections 1. Based on arrangement of bolts and plates a. Lap joint bolting  Single bolting  Chain bolting  Staggered bolting Bolts are subjected to single shear b. Butt joint  Single bolted butt joint  Chain bolted butt joint  Staggered bolted butt joint Terminology

1. Pitch of bolts → 2. Gauge distance → THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 103



3. Edge distance → 4. End distance → 5. Staggered distance or staggered pitch → IS 800 :2007 Recommendations 1. Clause 10.2.2 min p → ≮ 2.5 d nominal dia of bolt 2. Clause 10.2.3 max p → ≯ lesser of a. 16 t or 200 mm → tension member b. Lesser of 12 t or 200 mm → compression c. For staggered joints p may be increased by 50% of values mentioned in above clauses (2.9 or 2.6) if gauge distances is less than 15 mm d. For butt joint → p ≯4.5 d for a distance of 1.5 times the width of plate from the butting surface

.

→

( ) →≯ .

Possible Failure in Bearing Bolts 1. 2. 3. 4. 5.

Shear of bolts Bearing of bolts Bearing of plate Cracking of plate(tension) Block shear failure

1. Shear Strength of Bearing Bolts Design shear strength of bolts = = Partial safety factor for the material of the bolt 1.25 → Nominal shear strength of bolts , = + √

= Ultimate tensile strength of the bolt = Number of shear planes intercepting the bolt at its thread (root) not section area = Number of shear planes intercepting the bolt at shank not section area = Not section area, at thread = Area of shank of the bolt THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 104



Reduction Factors for Shear Capacity Of Bolts a. Reduction factor for long joint ( If joint length 15d then = 1.075

)

0.005

=Subjected to 0.75 ≤

≤ 1.0

b. Reduction factor is grip length is large If 5 , then is applied = sum of thickness of plates jointed . 8 = 3 + c. Reduction factor is packing plates are used 6 = 10.0125 =

(

√3

)

+

2. Bearing capacity of bolts (

)

= Where = Design bearing strength of bolts = Nominal bearing strength of bolts = Partial safety factor for the material of bolt = 1.25 Where,

= 2.5 +

is smaller of

=

0.25

1.0

= diameter of bolt and hole = end & pitch distances , = ultimate tensile stress of the bolt & plate = summation of thicknesses of connected members Efficiency of a joint = Strength of solid plate is generally lesser in yielding than in rupture i.e., strength of solid plate = strength in yielding Eccentric connections 1. When load is lying in the plane of bolts force in any bolt due to eccentricity . . =




Where = Distance of ith bolt from the C.G of bolts = Force due to moment in the ith bolt. acts perpendicular to the radial distance of that bolt and makes an angle vertical shear force p Direct shear force f = n Net force in ith bolt = √F + F

+ 2F F

with the direct

cos θ

The farthest bolt is subjected to maximum force 2. When load is lying perpendicular to the plane of bolts Bolts are subjected tension plus shear

Tensile force in any bolt Where, M =

=

p. e

1+[

]

T V then check ( ) + ( ) ≤ 1.0 T V




Shear Capacity of HSFG Bolts = = Coefficient of friction (called slip factor) = Number of effective interface offering frictional resistance to the slip = 1.0 for fasteners in clearness holes = 0.85 for fasteners in oversized and short slotted holes and for long slotted holes loaded perpendicular to the slot = 0.70 for fasteners in long slotted holes loaded parallel to the slot F = Minimum bolt tension at installation =

f

= Net area of bolt at thread = 0.78 d f

= Proof stress = 0.70f

Design step resistance ∪ γ

=

= 1.10 for service load design =1.25 if the step resistance is designed at ultimate load

For commonly used HSFG bolts (grade 88), yield stress f f = 800 MPa

= 640 MPa(0.8

800) and

Prying Forces (Q) To be applied to HSFG bolts connections subjected to tension due to flexibility of connected plates G = l 2l (T +

f bt ) 27l l +




Where, = 2 for non-pretension bolts = 1 for pretensional bolts η = 1.5 b = effective width of flange per pair of bolts

Welded Connections Types of Welded Joints 1. Butt weld 2. Fillet weld 3. Slot weld and plug weld

C D = toe of weld C, D = leg size of weld B = throat thickness of weld = root of weld CB = 45 → standard filled weld or 30 , 60 Fillet weld 1. Size of fillet weld (s) – smaller of two leg:- (root to toe) (a) Minimum size: Thickness of thicker part S 1) Upto 10mm 3mm 2) 10mm – 20mm 5mm 3) 20mm – 30mm 6mm 4) 30mm – 50mm 8mm in first run and 10mm minimum (b) Maximum size: 1.5mm less than the thinner member THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 108



2. Maximum throat thickness → 0.7 times the thickness of the thinner plate. 3. Effective length = overall length – 2 weld size →Only effective length is shown on the drawing and the addition length (2 x weld size) is provided by the welder → Effective length ≮ 4x weld size. 4. Overlap lap

≮5 Where t = thickness of thinner plate 5. Intermittent fillet weld → Length ≮ 4 x weld size or 40mm whichever is more → Length clear spacing of intermittent weld shall be 12t for compression joints and 16t for tensile joints, where t is the thickness of thinner plate Design Stress in Welds 1) Butt Welds: Butt welds are treated as parent metal with a thickness equal to throat thickness and the stress shall not exceed than permitted in the metal. 2) Fillet weld, sbt or plug welds. (It is based on its throat area) F = √

Where, f = smaller of the ultimate stress of the weld or of the parent metal. γ = 1.25 for shap welds. = 1.5 for field welds  Reduction in design stress for long joints If the length of welded joint l > 150t, where t is throat thickness, design capacity is reduced by the factor B

= 1.2 –

.

≤ 1.0

y

 Eccentric Connections 1. Plane of moment is the same as the plane of welds. → direct shear

P

e γ θ

d

X q =(

)

→maximum shear stress due to moment

b



q =

(


)

Where γ I

= distance of the extreme weld from the C.G of the ground = polar moment of inertia = I

+I

∴ Total shear stress = √q + q + 2q q Cosθ 2. Moment at Right Angle to The Plane at Weld In this case, the weld is subjected to combined axial and shear stress. As per IS 800 : 2007 recommends that the equivalent stress shall satisfyf = √f + 3q ≤ √

Where, f = axial stress, direct or due to bending q = shear stress due to shear force or tension P e h

Direct shear stress q = Bending stress f =

=

. (

)

t

=

∴ Equivalent stress f = √Fa + 3q ≤

√

Design of Tension Members Tension member are called as tie members. Angle sections are mostly used tension members Permissible Stress Lowest of the following a) Design strength due to yielding of gross section Tdg b) Rupture of critical cross – section Tdn c) The block shear Tdh 1. Strength Due to Yielding of Gross – Section T = Where, fy = yield stress of the material Ag = gross area of cross – section r = 1.1 – partial safety factor for yielding 2. Design Strength Due to Rupture of Critical Section T

=

.



Where, An is the net effective area at critical section i) For plates a) Aligned bolts f = (b - nd )t


f

d n bolts in a line

t = thickness of the plate

g

b) Staggered bolts

g

= (b - nd )t + Additional term giving more strength to the joint ii) For angles T

.

=

+

g S

( contribution of outstanding leg is less so a factor ) t

t

w

w

wi b =w + wi - t of the connected leg Where, b ==wnet area = gross area of the out standing leg = 1.4 – 0.076 . / . / ≤

0.7

Where, L = length of the end connection that is the distance between the outer most bolt in the end joint measured along the load direction or length of weld along the load direction. 3. Design strength due to block shear When failure occurs along a path such that one face fails in shear and the perpendicular face fails in tension this failure is called black shear failure. f

1

2

4

3

f

Shear failure along 1 – 2 and 3 - 4 Tension failure along 2 – 3

IS 800 : 2007 recommendations for Block shear Strength T Or T Where,

=

= . √

& &

–

√

.

+ = minimum gross and net shear area (1 – 2 & 3 – 4) = minimum gross and the area in tension (2 – 3)




Lug Angle When the length of splice required for any tension member comes act be fairly large, an extra and short length of an angle section is used. This angle is called lug angle Recommendation of IS 800: 2007 1. The connection of the lug angle should start in advance of the member and terminate at the end of the member. 2. Minimum two bolts or equivalent weld should used for attaching lug angle to the gusset plate 3. If the main member is an angle a) The whole area of the member shall be taken as the effective rather then net effective section (i.e with reduction for outstanding leg area) The whole area of the member is the gross area minus deduction for bolt holes. b) The strength of lug angles and fastener connecting lug angle to the gusset plate should be at least 20% more than the force in outstanding leg c) The strength of the fastener connecting lug angle and main member shall be at least 40% more than the force carried by the outstanding leg. 4. If the main member is channel a) The strength of fasteners connecting lug angle to the gusset should be at least 10% more than the force in outstanding leg b) The strength of fasteners connecting lug angle to main member shall be at least 20% more than the force in outstanding leg Net Area in Tension  Net area of any section = Gross Area – Area for the holes  For angles the gross width shall be the sum of the width of the legs less the thickness  The hole diameter is 1.5 mm in excess if nominal diameter of rivet is ≤ 24 mm  The hole diameter is 2 mm in excess if nominal diameter of rivet 24 mm  The area of a leg of an angle = Thickness of angle

.Length of leg

Thickness of leg/

 The area of a web or a tee = Thickness of web (Depth Thickness of flange)  In general, the net effective width (b ) can be calculated by the following equation: b

=b

id + j . /

(1.1) where, b = width of the section

g g g g

P d p

b

P

p = staggered pitch g = gauge i = no. of rivet holes j = no. of zigzags or inclined lines




Net Effective section for Angles and Tees Case I Single Angles in Tension, connected by one leg only Net effective area = a + k b

(1.2) where, a = net sectional area of the connected leg b = area of the outstanding leg k =

Case II Pair of Angles back to back (or a single Tee) in Tension, connected by only one leg of each angle (or by the flange of a tee) to the same side of the gusset Net effective area = a + k b

(1.3) where, a = net area of the connected legs (or flange of the tee) b = area of the outstanding leg (or web of the tee) k =

Case III Double Angles or Tees in Tension, placed back to back and connected to both side of the gusset plate or to both sides of a part of a rolled section Net effective area = Gross area

Deduction for holes

(1.4)

3. Plastic Analysis of Beams and Frames Notations f : Yield stress of material σ : Ultimate stress σ : Stress corresponding to working load σ : Maximum permissible stress in axial compression using actual length as effective length : Effective cross section of the member P : Yield strength of axially loaded section




Assumptions  Plane section normal to the axis of bending remains plane after bending  The stress-strain relationship is idealized to consist of two straight lines as shown  There is no axial load on beam  Shear strains are neglected  The deformations are assumed to be small so that slope of beam is given by its tangent at any point Shape Factor  Under plastic condition ,neutral axis divides the section into two equal areas  Plastic moment is equal to yield stress multiplied by the sum of moments of areas in tension and compression zones about the neutral axis  The ratio of plastic moment to yield moment is called the Shape Factor (f). It is also the ratio of plastic modulus to elastic modulus Hence, f = =  The Shape Factor depends only on the geometry of the section Load Factor  The Load Factor ( ) is the ratio of ultimate collapse load and working load Hence, =

=

=

=f

Factor of Safety

Compression Member Design of Compression Member  A structural member which is subjected to compressive force along its axis is called a compression member  If the net end moments are zero, the compression member is required to resist load acting concentric to the original longitude axis of the member and is termed axially loaded column or simply column.  If the net end moment are not zero the member will be subject to an axial load and bending moment along its length. Such member are called beam columns, otherwise an axially loaded column Possible failure modes 1. Local buckling: Failure occurs by buckling of one or more individual plate elements eg: flange or web with no overall deflection in the direction normal to applied load. The failure mode may be presented by selecting suitable width to thickness ratios of components plates.




2. Squashing: When the length is relatively small (stocky column) and its component plate elements are prevented from local buckling then the column will be able to attain its full strength or squash load (yield stress X area of cross section). 3. Overall flexural buckling: This mode of failure normally controls the design of most compression member. In this mode failure of the member. In this mode, failure of the member occurs by excessive deflection in the plane of the weaker principal axis. 4. Torsional and flexural – torsional bucking :Torsional bucking failure occurs by twisting about the shear center in the longitudinal axis. A combination of flexure and twisting called flexural – torsional buckling is also possible Torsional buckling is a possible mode of failure for point symmetric section. Flexural torsional buckling must be checked for open section that is singly symmetric and for section that have no symmetry Behavior of compression member 1. Short compression member: For very short compression member the failure stress will equal the yield stress and no buckling will occur for short column ≤ 88.85 2. Long compression member: For these compression member the Euler formula predicts the strength of long compression member very well where the axial buckling stress remain below proportional limit such compression member will buckle elastically 3. Intermediate length compression member: For intermediate length member some fiber would have yielded and some fiber will still be elastic. This compression member will fail both by yielding and buckling and their behavior is said to be inelastic. Effective length of compression member =

(

⁄ )

Where k = 1.0 for column with both ends pinned k = 0.5 for columns with both ends fixed k = 0.707 for columns with one end fixed and the other end pinned k = 2.0 for columns with one end fixed and the other end free k ≤ 1.0 for columns partially restrained at each end and k 2.0 for columns with one end restrained and the other end rotation partially restrained Maximum slenderness ratio of compression member Types of member 1. Carrying loads resulting from dead load and superimposed loads 2. Carrying loads resulting from wind and seismic loads only provided the deformation of such a member does not adversely affect the stress is any part of structure 3. Normally acting as a tie in a roof truss or a bracing system but subject to

⁄ 180 250 350




possible reversal of stress resulting from the action or seismic force 4. Lacking bars in columns 5. Elements (components) in built up sections

145 50

Built up compression members  For large loads and for efficient use of material built up columns (also called as combined columns or open web columns) are often used. They are generally made up of two or more individual section such as angle channel or I section and properly connected along their length by lacing or battening so that they act together as a single unit Lacings 1) The radius of gyration of the combined column about the axis perpendicular to the plane of lacing should be greater than the radius of gyration about the axis parallel to plane of racing 2) Lacing should be uniform throughout the length of column. 3) Single and double laced system should not be provided on the opposite sides of the same member similarly lacings and battens should not be provided on opposite sides of same member 4) Simple laced system on opposite sides of the main component shall be in the same direction viewed from either side so that one is shadow of the other. 5) The lacing shall be designed to resist a total transverse shear V t at any point in the member equal to 2.5% of the axial force in the member and this shear be divided among the lacing system in parallel planes 6) The lacing in additional should be designed to resist any shear due to bending moment or rate load or member 7) The slenderness ratio of lacing should not exceed 145 8) The effective length shall be taken as the length between inner and bolts/rivets of the bar for simple lacing and 0.7 times the length for double lacing effectively connected at intersection. For welded Bars, the effective length taken as 0.7 times the distance between the inner ends of the welds connecting the single bars to the member. 9) The minimum width of the lacing bar shall no be less than approximately three time the diameter of the connecting bell/rivet the thickness shall not be than 1⁄40th of the effective length for single lacing and 1⁄60th for double lacing 10) The spacing of lacing bars shall be such that maximum slenderness ratio of the component of the main member between two consecutive lacing connection is not greater than 50 or 0.7 times the most unfavorable slenderness ration of the combined column. 11) When welded lacing bars overlap the main member the amount of lap should be not less than four times thickness of the bar and the welding is to be provided along each side of the bar for the full length of lap Where lacing bars are fitted between main member they should be connected by filet welds on each side or by full penetration butt weld. 12) Where lacing bars are not lapped to form the connection to the components of member they shall be so connected that there is no appreciable interruption in triangulated system. 13) Plates shall be provided at the ends of laced compression member and shall be designed as batters. 14) Flats angle channels or tube may be used as lacings.




15) Lacing bars whether in double or single shear shall be inclined at an angle of 40 0 to 700 to the axis of the built up member. 16) The effective slenderness ration . / of the laced column shall be taken as 1.05 times . /, where . / is the maximum actual slenderness ratio of the column to account for shear defamation effects.

Battens Rules for Design Of Battens 1. 2.

3.

4. 5.

The number of battens shall be such that the member is divided into not less than three bays Battens shall be designed to resist simultaneously longitudinal shear Vb = Vt ⁄ and moment M = ⁄2 Where = transverse shear force = distance between centre to centre of battens longitudinally = number of parallel planes of battens = minimum transverse distance between the centroids of the bolts / rivet group / welding connecting the batten to the main member When plates are used for battens the effective depth between the end bolts / rivets or welds shall not be less than twice the width of one member in the plane of battens nor less than three quarters of the perpendicular distance between centroids of the main member for intermediate battens and not less than the perpendicular distance between the centroids of main member for end battens. The thickness of battens plates shall not be less than 1⁄50th of the distance between the inner most connecting transverse belts / rivets or welds. The effective slenderness ratio of b attended column shall be taken as 1.10 time ( ⁄ )o, where ( ⁄ )o is the maximum actual slenderness ratio of the column to a count for shear deformation effects

Plastic Analysis Classification of structural analysis The structural analysis can be classified into two groups (i) Elastic analysis (ii) Plastic analysis and Limit analysis The simple plastic theory takes the advantage of the ductility of steel and a key parameter is redistribution of moments (i) When bending moment at critical section is less that yield moment then beam behaves elastically (M



Stress Distribution

Cross section

M M

Fully elastic M
(ii) When bending moments (M | at critical section reaches the value M then yielding gradually spreads in to the web fibers (M = M ) σ E M M

σ

Initial yielding M
(iii) When bending moment M reaches the maximum moment carrying capacity (M ) of the structure, then practically all the fibers at the section of maximum moment have fielded and the section is fully plastic (M =M ). at this stage deformation increase rapidly without no increase in load and beam is wear collapse σ

E

M

Elastic zone Neligible M

σ

M =M

→ The load corresponding to stage (iii) is known as limit load or collapse load or ultimate load. It is sometimes represented as P, where is the load factor and P is working load. The corresponding moment is known as ultimate moment or plastic moment M →During the formation of the plastic moment, a small elastic zone is left is the mid depth of the section. Even through the stress is less than the yield value σ in the elastic zone at the centre the distribution is idealized into rectangular stress blocks for purpose of derivations of equations. →The load factor can be expressed as THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 118


l=


=

→The saving of material achieved by designing according to plastic theory instead of elastic theory is given by % Savings in materials = 01

1 x 100

Plastic Sections Modulus σ y

Plastic N.A

y

σ Z = Z =

.y + y y + y

Shape Factor Shape factor ( ) =

=

Where Z is elastic section modulus. → Load factor ( ) =

=

= F.O.S x Shape factor = F. S x Characteristics of plastic hinges (i) Plastic hinge is formed at sections, where the moments are maximum (ii) Plastic hinges is a zone of yielding due to flexure (iii) Its length depends on geometry and loading but in most of the analytical work it is assumed that all plastic rotations occur at a point. (iv) Plastic hinge is also called rusty hinge or friction hinge (v) Where two members meet, a hinge will be form in the member whose capacity is less. Where three or more members meet, plastic hinge will be formed in all members. (vi) Plastic hinges are reached first at sections subjected to greater curvature (bending moment) (vii) Formation of plastic hinges allows a subsequent redistribution of moment until M is reached at each critical section (viii) At those section where plastic hinges are formed, the member acts as if it were hinged except with a constant restraining moment M (ix) The total number of plastic hinges required at collapse mechanism is equal to number of static indeterminacy plus one. THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 119



Length of Plastic Hinge ( ) →It is the length of the beam over which the moment is greater than the yield moment (M ) (a) For simply supported beam carrying concentrated load, length of plastic hinge is given by L = (For rectangular section) = (For I section) L = L 01 1 (b) For simply support beam carrying udl, length of plastic hinge is given by L = L √1 Where is shape factor and L is total length of beam →The length of plastic hinge depends on loading and geometry Fundamental Conditions of Plastic Analysis 1. Condition of statically equilibrium 2. The plastic moment Mp must not exceed anywhere in structure 3. At ultimate load, there must be just enough plastic hinges that a mechanism is formed Plastic Hinges Minimum no. of plastic hinges p required to reduce a structure of degree of indeterminacy equal to ‘i’ into a mechanism is ‘i+1’ Theorem of Plastic Analysis There are two fundamental theorems in plastic analysis to find out load factor ( ): 1. Statically Method or Safe Theorem (Lower Bound Theorem) 2. Mechanism Method or Kinematic or Unsafe Theorem (Upper Bound Theorem) Independent Mechanism No. of Independent Mechanism = No. of Plastic Hinges – Degree of Redundancy Possible Mechanism 1. Beam Mechanism 2. Panel Mechanism 3. Gable Mechanism 4. Combined Mechanism

Plastic Design THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 120



Load Factor for different load as per IS 800:1984 Load

Load Factor ( )

Dead Load

1.7

Live Load

1.7

Dead Load + Wind or Earthquake

1.7

Dead Load + Imposed Load + Wind or Earthquake

1.3

Beams Moment capacity of beam, M = z

f

Tension Member Load capacity of tension member, P = 0.85

f

Strut Load capacity of strut carrying axial load only, P = 1.7

σ

Beam-Column  Maximum moment capacity acting in beam-column when P ≤ 0.15P M

=M

 Maximum moment capacity acting in beam-column when P M

= 1.18

(1

)

0.15P

M

Shear  Maximum shear capacity of beam or beam-column, V = 0.55

f

Analysis and Design of Compression Member Notations f n E l r f σ

: : : : : : : : :

Yield stress of steel Imperfection index, taken as 1.4 Modulus of elasticity of steel, taken as 2 10 Slenderness ratio Effective length of compression member Appropriate radius of gyration of member Effective cross section area of member Elastic critical stress in compression Maximum permissible stress in axial compression

Definition  Structural member carrying axial compression without bending is called a compression member  Structural member carrying compressive load in truss are called struts  The vertical compression members in buildings are called columns, posts or stanchions  Compression member of a crane is called boom  The main compression member in roof truss are called rafters THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 121



Buckling  The sudden bending of column under unstable equilibrium is called buckling of column  At a particular load, the column buckles under slightest disturbances, the particular load is the buckling or critical load of column Allowable Stress for Steel in Compression σ

= 0.6

; where, f [

=

and =

]

Axial Load of Compression Member 

The maximum load (P) in axial compression is given by, P = σ



The main compression member in roof truss are called rafters

Effective Area of Strut 

The effective area in case of struts is equal to the gross area of section



For empty holes, full deduction of hole area is done



For holes filled with black bolts, 25% reduction of hole area is made

Effective Length of Strut The following notation can be used for better understanding: Fixed End

:

Effectively held in position and restrained against rotation

Hinged End

:

Effectively held in position but not restrained against rotation

Free End

:

Not held in position and not restrained against rotation

Degree of Restraint

Effective Length

1. 2. 3. 4.

0.65L 0.80L 1.00L 1.20L

Both end fixed One end fixed & other end hinged Both end hinged One end fixed and other end restrained against rotation but not held in position 5. One end fixed and other end partially restrained against rotation but not held in position 6. One end hinged and other end restrained against rotation but not held in position

1.50L 2.00L




Maximum Slenderness Ratio Component with Loading 1. 2. 3. 4.

Maximum

Compression member carrying dead and superimposed load Compression member carrying load due to wind and seismic 250 Tension member under reversal of stress due to wind & seismic 350 Tension member 400

180

Tacking Rivets  Spacing of tacking rivets should not exceed 600mm  For compression member connected back to back, the slenderness ratio of each member should not greater than 40 nor 0.6 times the most unfavorable slenderness ratio of the whole strut Build-up Compression Member Lacing General Specification 1. The compression member comprising two main components laced, should have a radius of gyration about the axis perpendicular to the plane of the lacing not less than the radius of gyration at right angles to that axis 2. In laced system, cross members perpendicular to the longitudinal axis of the strut should not be used 3. The single laced systems on opposite sides of the main components should preferably be in the same direction so that one be the shadow of the other 4. Laced members should be provided with tie plates at ends of lacing system and at points where the lacing system are interrupted Design Specification 1. The angle between the lacing bars and the axis of the member is kept between 40° and 70° 2. Slenderness ratio . / for the lacing bar 145, where, l =

length between inner end rivets for single lacing

= 0.7 times length between inner end rivets for double lacing = 0.7 times length between inner ends of weld for welded lacing 3. For flat bar of width b and thickness t, ≯ 40 for single lacing ≯ 60 for double lacing



4. For lacing system,


≯ 50 or 0.7 times most unfavorable slenderness ratio of the whole strut

where, L = Distance between the centre of consecutive connection of laced bars r = Radius of gyration of compression member 5. The lacing members should be designed to resist a transverse shear, V = 2.5 of axial force in the member 6. Minimum width of lacing bar is approximately 3 times the nominal rivet diameter 7. The overlap in welded connection lap joint ≮ 4t of bar or member whichever is less Batten General Specification 1. The compression member composed of two main components battened, should have a radius of gyration about the axis perpendicular to the plane of the batten not less than the radius of gyration at right angles to that axis 2. The number of batten should be such that the number is divided into at least three parts longitudinally 3. The effective length of battened column should be increased by 10%

Design Specification 1. For battened compression member  If ≤ 0.8 , then . / ≤ 50 or 0.7 

If

0.8

, then . /

≤ 40 or 0.6 . /

whichever is less whichever is less

Where, L

= effective length of battened compression member

r

= radius of gyration of buildup section about an axis perpendicular to plane of

r

= radius of gyration of buildup section about an axis parallel to plane of battens

c

= spacing of battens

r

= minimum radius of gyration of component

battens

2. If d is the effective width of batten and a is the centroid distance of member then  d . / a for intermediate battens 

d

a for end battens

3. For flat bar of width b and thickness t, ≯ 40 for single lacing ≯ 60 for double lacing




4. Thickness of battens, t where l is the distance between innermost connecting lines of rivets or welds 5. Batten should be designed to carry bending moment and shear arising from transverse shear, V = 2.5 of axial force in the member 6. Moment in batten plate, M = and longitudinal shear, V = where, N is the no. of parallel planes of batten and S = Minimum transverse distance between cg of rivet group or weld 7. The overlap in welded connection lap joint ≮ 4t of bar or member whichever is less

Beams Introduction →The load transfer by a beam is primarily by bending and shear. ny structural member could be considered as a beam if the loads cause bending of the member. Classification of Beam 1. Floor Beams: A major beam of a floor system usually supporting joints in building, a transverse beamin bridge floors. 2. Girder: Girders are the same as floor beams. Also a major beam in any structure. Floor beams are often referred to as girders. 3. Girt: A horizontal member fastened to and spanning between peripheral columns of an industrial buildings used to support wall cladding such as corrugated metal sheeting. 4. Joist: A beam supporting floor construction but not a major beam 5. Lintels: Beam members used to carry all loads over wall openings for doors, windows etc. 6. Purlin: A roof beam, usually supported by root trusses. 7. Laffer: A root beam, usually supporting purlins 8. Spandrels: Exterior beams at the floor level of buildings, which carry a part of the floor load and the exterior wall. 9. Stringers: Members used in bridges parallel to the traffic to carry the deck slab they will be connected by transverse floor beams. Beam Types Type Of Beam Angles

Optimum span range (m) 3-6

Relled I – section Castellated beams Plate girders

1-30 6-60 10-100

Box girders

15-200

Open web joints Trusses

4-40 10-100

Application For lightly loaded beams such as purlin and sheeting rail Most frequently used as a beam Long spans and light loads Long spans with heavy loads such as bridge grinders Long spans and heavy loads Such as bridge girders Fabricated for larges span, using angles or tubes as chords and round bars for web diagonals Long spans and moderate loads such as industrial roofing




Main Failure Modes of Hot Rolled Beams 1) Category I – Excessive bending triggering collapse 2) Category II - Lateral torsional buckling of long beams which are not suitably braced in the lateral direction 3) Category III – Failure by local buckling of a flange in compression or web due to shear or web under compression due to concentrated loads. 4) Category IV – Local failure by (a) shear yield of web. (b) Local crushing of web and (c) Buckling of thin flanges Effective Length The concept of effective length of the compression flange incorporates the various types of support conditions. Conditions at supports Effective length 1. Compression flanges at the end unrestrained L* against lateral bending (Free toratate is plan) 2. Compression Flange partially restrained against 0.85L* lateral bending (partially free to rotate in plane at the bearing) 3. Compression flange restrained fully against lateral 0.7L* bending (rotation fully restrained in plan) Design for Shear The factored design shear force V in a beam due to external actions should satisfy V ≤V Where V , the design strength is given by V =

V⁄ γ

The nominal shear strength of a cross – section V may be governed by plastic shear resistance or the strength of the web governed by shear buckling. The nominal plastic shear resistance under pure shear is given by V = V where = V =

√

is the shear area and f is the yield strength of web. The shear area may be calculated as given in table 1 for different cross sections. Note: Fastener holes need not be accounted for in the plastic design shear strength calculation provided that Y n . /. ⁄Y

/

⁄0.9




If does not satisfy the above condition, the effective shear area may be taken as that satisfying the above limit. Block shear failure criteria may be verified at the end connections. Table 1 – Shear areas Section

Shear area

b t

htw

h

t b tw h

htw

tf

d

h

tw

a) Built-up

htw, dtw

tf

b

b

2btf

tf b) Built- up sections or rolled (minor axis bending) b

h

Ah/(b+h)




c) Rectangular hollow section h b

Ab/(bth)

d) Rectangular hollow section (loaded parallel to width) 2

e) Circular hollow section

A f) Plates and solids bar Web Buckling and Web Crippling A heavy load or reaction concentrated on a short length produces a region of high compressive stresses in the vertical elements of the web either under the load or at the support. Away from the loading point, stresses are spread widely. But just under a load or above a reaction point they may cause web failures such as web buckling as shown in figure and web crippling or web crushing

Web Buckling

Web Crippling

(a) Local Buckling of the web (b) Design Procedure For Channel / I-section purlins Various steps involved in the design are as follows 1. The span of the purlin is taken as the centre to create distance of adjacent trusses. 2. The gravity loads P, due to sheeting and line load, and the load H due to wind are computed. The components of these loads in the direction perpendicular and parallel to the sheeting are determined. These loads are multiplied with partial safety factor Y for loads (See table 4 of code) for the various loads combination




3. The maximum bending moment (M and M ) and shear forces (F and F ) using the factored loads are determined. 4. The required value of plastic section modulus of the section may be determined by using the following equations. γ γ Z = Mz ⁄F + 2.5 . / [M ⁄F ] Where, γ = Partial safety factor for material = 1.1 d = depth of the trial section b =breadth of the trial section M andM are the factored bending moments about z and axis f = yield stress of steel 5. Check for the section classification (table 2 of code) 6. Check for the shear capacity of the section for both the Z and y axis (for purlins shear capacity will always be high and may not govern the design) f VdZ = ⁄ (√3γ ) f Vdy = ⁄ (√3γ

)

And

= htω Avy = 2b t Where h is the height, tω is the thickness of web b - breadth of flange and t is the thickness of the flange of I-channel section, respectively. 7. Compute the design capacity of the section is both the axis. f F ⁄γ ≤ 1.2 Z ⁄γ Md =Z Md = Z

F

⁄γ

≤Z

f

⁄γ

8. Check for local capacity by using the interaction equations .

/+(

)≤ 1.0

Simplified Method For Design Of Angle Purlins As per IS 800-2007, angle purlins should be designed, For biaxial bending. However IS 8001984 and present British code permit design of angle purlins by assuming that load normal to sheeting is resisted by purlin and load parallel to sheeting is resisted by sheeting provided the following conditions are fulfilled Design Of Grillage Beams Let the length of grillage beam be L and length of plate in this direction be ‘a’ maximum moment occurs at centre of beam carries a total load p. ( ) M= x x - x x = Maximum shear occurs at distance ‘a’ from the centre of the beam and its value is ( ) F= . /= The grillage beams should be designed for the above moment and shear. It should be checked for web crippling.




Bending Strength of a Laterally Supported Beam If d / f ≤ 67ε. IS 800- 2007 considers two cases one with design shear less than 0.6V and other with design shear strength more then 0.6V where is design shear strength, when d/ > 67ϵ, buckling of web is likely to take place. (a) If V ≤ 0.6 V The design bending strength M shall be taken as M = Z f x ≤ 1.2Z f x for simply supported beam shear ≤ 1.5Z Where

for cantilever beam

= 1.0 plastic and compact sections = for semi- compact sections.

Z , Z = plastic and elastic section modulus of the cross – section, respectively. (b) If V > 0.6V In such cases, M = Mdn Where M = design bending strength height shear this reduced value is recommended to account for the effect of higher shear on the bending strength of section Md is to be calculated as given below (clause 9.2.2 in IS 800: 2007) a) Plastic or compact section M = M | (M M )≤ 1.2 Z f x Where =.

1/

M = plastic design moment of the wholes section V = Factored applied shear force V = Design shear strength M = Plastic design strength of the area of cross section excluding the shear area, considering partial safety factor γ b) Semi compact section. M = Shear Strength of a Laterally Unsupported Beam The design shear strength of a section is given by (clause 8.4 of IS 800:2007) V =

√

x

Where = Shear area and can be taken as shown in table shear area The shear are may be calculated as given below Notations f M V n E I

: : : : : :

y D

: :

Yield stress of steel (MPa) Bending moment at any section Shear force at any section Imperfection index, taken as 1.4 Modulus of elasticity of steel, taken as 2 10 Moment of inertia of section about centroidal axis perpendicular to plane of bending Distance of fibre from neutral axis (N.A.) of bending Overall depth of beam



T r d t b l

: : : : : : :

ω f c ,c k ,k

: : : :


Mean thickness of beam Radius of gyration of section about its axis of minimum length Depth of web Thickness of web Width of any section where stress is to be calculated Effective laterally unsupported length of member Ratio of minimum section to maximum section of total area of both flanges Ratio of I of compression flange to I of whole section Elastic critical stress in bending The lesser and greater distances from section N.A. to the extreme fibres Coefficient depends on and ω [refer table of IS:800-1984]

Bending Stress 

Bending stress in compression or tension is given by, σ

(

)

=

Allowable Bending Stress   

Permissible bending stress in tension, σ = 0.66f Permissible bending stress in compression for laterally restrained beam, σ = 0.66f Permissible bending stress in compression for laterally unrestrained beam, σ

= 0.6 [

where, 

f

= k (X + k

]

Y) ; Y =

. ( )

The calculated f is to be increased by 20% if

MPa and X = Y√1 + ≯ 2 and

(

)

≯ 85

Effective Length of Compression Flange 

For simply supported beam of span L with each end restrained against torsion, the effective length Lateral Bending

Effective Length

Unrestrained

0.85L

Partially restrained

0.75L

Fully restrained

0.50L

Note : The above values to be increased by 20% if the beam is not restrained against torsion 

For cantilever beam of projected length L, effective length is given by




Support End

Free End

Effective Length

Build-in

Unrestrained

0.85L

Build-in

Restrained against torsion

0.75L

Build-in

Restrained against torsion

0.50L

and lateral deflection Continuous and unrestrained

Free at end

3.0L

Free at end

2.0L

Free at end

1.0L

against torsion

Continuous and partially restrained against torsion

Continuous and fully restrained against torsion Shear Stress 

Shear stress is given by, =

where, y = Moment of area about N.A. of the section of the part of the section beyond the fibre where shear stress is to be calculated 

Average shear stress is given by,



Allowable shear stress is given by  when ≯ 85, = 0.45f & 

when

=

, where, d = depth of beam = 0.4f

85, stiffeners are provided and

depends on dimension of panel

Deflection  Allowable deflection for simply supported beam should not exceed  Allowable deflection for cantilever beam should not exceed




Web Crippling  For intermediate concentrated load

(

 For end support

(

where,

√ ) √ )

≯ 0.75

≯ 0.75

= Concentrated load on beam = End reaction at support = Bearing length = Depth of root of the fillet from the top of the flange

Web Buckling 

≯

where, = +Length of stiff portion of bearing+Thickness of compression flange for simply supported beam = +Length of stiff portion of bearing+Thickness of compression flange for continuous beam over bearing = Permissible axial compression stress for slenderness ratio of where,

√

is given by the clear depth of web between root fillets

Buildup Beam Symmetrically Buildup Beam Area of each cover plate is given by,

=

, where

= Section modulus of top section

= Section modulus of primary I beam = Depth of beam Un-symmetrically Buildup Beam Area of each cover plate is given by,

=

. (

)

Connecting rivets for Buildup Beam Pitch of the rivets,

=

(

, where

)

= No. of row of rivets = Width of plate = Thickness of plate = Depth of primary beam = Rivet value




Unsymmetrical Bending of Purlins 

If

is the angle of major principal axis with horizontal axis(

then, 

=

(

)

)

Condition of simplified analysis of purlins  Slope of the roof ≯ 30 to horizontal  Depth of angle section≮  Width of other leg ≮  Bending moment in purlins = where,

=Span of purlin =UDl on purlin including wind and minimum SIDL of 0.75

⁄

Gantry Girder Deflection of gantry girder is given by Loading type

Deflection

Manually operated crane Electric operated crane of upto 50 Electric operated crane of over 50 Other moving load *

= Span of crane runaway girder

3. Plate Girder Notations

V

: : : : : : : : : : : :

Yield stress of steel (MPa) Bending moment at any section (kNm) Shear force at any section Modulus of elasticity of steel, taken as 2 10 (MPa) Transverse force to be taken by stiffeners (kN) Number of row of rivets Pitch of rivets Rivet value Total load applied on girder (kN) Overall depth of beam (mm) Outstand of stiffener (mm) Maximum thickness of compression flange



: : : : : : :


Area of flange plates Gross area of web Depth of web Thickness of web Span length of plate girder Permissible bending stress in tension Permissible bending stress in compression

Weight and Economic Depth of Plate Girder  If web area is assumed to share some part of applied bending moment, the economic depth of the section is given by,

= 1.1√

 If web area is not assumed to share any part of applied bending moment, the economic depth of the section is given by,

=√

 Actual depth is usually taken 10

less than the depth given by above equations

 In short girder, where a uniform flange section without curtailment is adopted, the economic depth of the section is given by,

= 1.21√

 The self weight of girder to start with, may be assumed as,

=

Design of Web Average shear stress in the web, where,

= 0.4

,

=

≯

for unstiffened web and

= as per cl. 6.4.2 of

800

1984 depending on

ratio for stiffened web

Web Stiffeners 800

1984 recommends the provision of web stiffeners as follows Condition

1.

≤

2.

≤

3.

≤

√ √ √

Provision of Web Stiffeners √

,

85

No Stiffener

200

Vertical Stiffener

250

Vertical Stiffener + Horizontal Stiffener t

from

Vertical Stiffener + Horizontal Stiffener at

from

compression flange 4.

≤

√

400

compression flange + Horizontal stiffener at N.A.



where,


is given by

1. For vertically stiffened web without horizontal stiffeners = The clear distance between flange angles and where there are no flange angles, the clear distance between flanges ignoring fillets 2. For vertically stiffened web with horizontal stiffeners = The clear distance between the tension flange and horizontal stiffeners 3. Where tongue plates having a thickness of not less than twice the thickness of web plates used = The depth of girder between the flange – sum of depths of tongue plates or 8 sum of thicknesses of tongue plates, whichever is less and is given by =2

clear distance from compression flange to the N.A.

Note: In no case the grater clear dimension of web panel should exceed 270 clear dimension exceed 180

nor the lesser

Vertical Stiffeners Spacing of Stiffeners  

The spacing of stiffeners should be kept between 0.33 to 1.5 where is defined as earlier when no horizontal stiffener is provided The spacing of stiffeners should be kept between 0.33 to 1.5 where is the clear distance between horizontal stiffener and tension flange ignoring fillets when horizontal stiffener is provided

Moment of Inertia of Stiffeners .



The moment of inertia of stiffener angle, ≮ ; where, = Minimum required thickness of web and = Maximum permitted clear distance between vertical stiffener for thickness



Increase in moment of inertia due to moment



Increase in moment of inertia due to lateral load or shear

= =

.

Horizontal Stiffeners Moment of Inertia of Stiffeners  

The moment of inertia of first horizontal stiffener at of distance between compression flange and neutral axis, ≮ 4 ; The moment of inertia of second horizontal stiffener at . ., ≮

where, = Minimum required thickness of web and = Actual distance between vertical stiffener THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 136



Connection of Web Stiffeners 

Stiffeners are designed to withstand a shearing force

≮

Load Bearing Stiffeners 

The moment of inertia of stiffeners, ≮

; where,

= End Reaction

Design of Flange Tension Flange Area of tension flange,

=

Compression Flange Area of compression flange,

=

Permissible Bending Stress 

The maximum compressive stress, calculated on gross flange area,



The maximum tensile stress, calculated on net flange area,

,

=

,

=

≯

≯

Curtailment of Flange Plates The length to be curtailed of the plate, where

,

,

,

= √

;

= no. of plates to be curtailed counting 1, 2, 3 ... from outer plate

Riveted Connection Connection between flange plates and flange angles The rivets in between are designed to resist horizontal shear force at flange plate level, the pitch of rivet is given by,

=

.

/

Connection between flange angle and web plates The pitch of rivet is given by,

= √(

.

/

)




4. Column Bases Slab Base 

The thickness of a rectangular slab base as per IS:800-1984, is given by =√

.

/

where, = The slab thickness (mm) = The pressure or loading on the underside if the base

(MPa) = The grater projection of the plate beyond the column (mm) = The lesser projection of the plate beyond the column (mm) = The permissible bending stress in slab bases =

185 (MPa)



The thickness of a square slab base plate under a solid round column as per IS:800-1984, = 10√

(

)

where, = The thickness of plate (mm) = The total axial load (kN) = The length of the side of cap or base (mm) = The diameter of the reduced end of the column

(mm) 185 (MPa) 

= The permissible bending stress in slab bases =

The cap or base plate should not be less 1.5(

+ 75) mm in length or diameter

Gusseted Base The thickness of base plate in case of gusseted base is less than slab base as the bearing area of column on base plate increased by gusset plates




Grillage Foundation: (As per IS: 800-1962)  

Concrete cover for grillage base is 100 A minimum clear spacing of 75 is kept between the flanges of beams for proper placing of concrete Permissible stresses in grillage beams encased in concrete is increased by 33 (50 when the effect of wind, seismic or erection is considered) Consider a tier of grillage beam having a length and carrying a load of over a central length  The maximum bending moment at the middle of the beam, =

 

(

)

 The maximum shear force at the edge of the load, 

=

(

)

The bearing pressure on the web at the root of fillet should not exceed the safe bearing pressure, i.e.

(

≤ 1.33

)

√

189

; where

= no. of beams in tier

6. Beam-Column and Beam-Column Connections Notations: : : : : : : : : : : : : : : : :

r

, ,

x, y l, b, t M, P x

Yield stress of steel Modulus of elasticity of steel, taken as 2 10 Slenderness ratio Effective length of compression member Appropriate radius of gyration of member Effective cross section area of member Elastic critical stress in compression Calculated average axial compressive stress Permissible stress in axial compression Calculated bending compressive stress in extreme fiber Permissible bending compressive stress Represents x-x and y-y axis Length, width and thickness of base plate Represents moment and axial force on base plate Distance of extreme compression fibre of base plate from N.A. Permissible stress of concrete

Beam-Column 

Member subjected to axial compression and bending needs to satisfy the following equation: ,

+

where, f

, ,

0

.

=

1

+

,

[

, .

]

≤ 1.0 ;

and =





,



,

If the ratio +

,


is less than 0.15 then following equation needs to satisfy: ,

+

≤ 1.0

The value of coefficient C is given by Condition a) Member in frame where side away is not prevented b) Side sway prevented, members not subjected to transverse load between their supports in the plane of bending c) Side sway prevented, members loaded transversely i. Ends of member restrained against rotation ii. Ends not restrained against rotation

Value of 0.85 0.6 0.4 ≮ 0.4 0.85 1.00

Eccentrically Loaded Base Plate 

Criteria for base plate to not take any tension by anchor bolts ≤



The thickness of base plate is found by the following equation M=



Length and width of base plate can be found by the following equations i. P = ii.



185

=

In case of tension in base plate, the value of tension and size of anchor bolts can be found by the following equations i. P + T = ii.

M = T .n

/+P.

/

5. Connections- Simple and Eccentric Bolt & Rivet Connection  

 Some Definitions Lap joints: The plates are to be connected overlapping each other. Butt joints: A cover plate is provided to connect the plates.  Gross diameter of rivet  =Nominal diameter of rivet + 1.5 mm ( Upto nominal diameter of 24 mm)  =Nominal diameter of rivet + 2 mm (Nominal diameter of greater than 24 mm)  Minimum pitch distance = 2.5 times the diameter of rivet hole.  Maximum pitch distance for  Any two adjacent rivets = 32t or 300mm whichever is less  Rivets lying in parallel to the force in the member: i) In tension = 16t or 200 mm whichever is less ii) In compression = 12t or 200 mm whichever is less

Maximum edge distance = 12tε THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 140



Where, t = thickness of thinner outside plate. ε=(

) 0.5

 Strength of joints (Rivet value) is minimum of three  Strength of plate against tearing σtf x (p-d) x t  Strength of rivet against single shear vf x (Π 4) x d2  Strength of rivet against double shear 2 x vf x (Π 4) x d2  Strength of rivet against bearing σpf x d x t Where, d = gross diameter of rivet p = pitch of rivet Stresses have usual notations.  Union’s empirical formula d = 6√t  Weld Connection  Minimum size of fillet weld should not be less than 3 mm or nor more than the thickness of the thinner part.  Maximum size of fillet weld  Equals to the thickness of the plate when the plate thickness is less than 6 mm.  Equals to at least 1.5 mm less than the edge thickness when the plate thickness is more than 6 mm  Should not exceed ¾ th of the thickness of the section at the toe where weld is applied to the rounded toe of the rolled section.  Effective length of fillet weld= overall length – 2 x size of weld  Effective throat thickness = K x Size of weld and should not be less than 3 mm. Where, k = cos(α 2) , α = angle between fusion faces.  For side fillet,  Length of weld should not be less than width of plate  Width of plate should not be greater than 16 times effective throat thickness  Lap length should not be less than 5 times lower thickness of plate.




Part – 3: Geotechnical Engineering 3.1 Soil Mechanics

Classification On basis of transport agency   

Alluvial soil: soil by cumming water Lacustrine soil: lakes Marine soil: sea water

Special Classification      

Loess→ Windblown silt. Bentonite→ Volcanic ash Muck→ Organic matter Talus→ Action of gravity Maul→ Marine Origin Black soil contains montmorellonite responsible for swelling.

Properties of Soil

1. Water content = 2. Void ratio, e = 3. Porosity, Ƞ =

x 100% x 100%

x 100%

4. Degree of saturation, S = (

)

5. Air content, aC =



6. % air voids =


x 100

7. Relative compaction or density under = ID = 8. Specific gravity, Gm = 9.

=

or e =

Ƞ Ƞ

10. e = When w: Water content (

11. 12.

sat

13.

d

=

)

(

)

=

(

)

14. ` =

(

)

15. d = 16.

=(

)=

where Bulk unit weight Saturated unit weight Dry unit weight  Specific gravity of solids by Pycnometer G=  If perosine is used against water G=*

+xK

Where, w

Weight of dry sample



w

Weight of pycnometer + soil +water

w

Weight of pycnometer + water


Method for in-situ weight determination a. Core cutter method b. Sand replacement method c. Water displacement method 1. Sieve analysis (for coarse grained soils) Coarse sieve analysis = 63, 20, 10 & 4.75 mm Fine sieve analysis = 2mm, 1mm, 600, 425, 300, 212, 150 &275 m

2. Sedimentation analysis For fine grained soils < 75 m Strokes law is used (

V

)

(

)

Where V

Terminal velocity (m/s) Viscosity [in absolute of power] Particle size in mm

If

is the effective reading of hydrometer, then √ (

Where,

)

√

(h

t )

Reading of hydrometer V

Volume and of bulb of hydrometer

A

Cross sectional area of jar




Fraction retained on 4-75mm sieve is called gravel fraction which is subjected to wave stieve analyst. 

finer

x

0.002mm Clay Siet

0.075mm

80

Fine Medium Coarse Fine Coarse Sand

Gravel

300 Cobble Boulder

US 200 sieve = 0.075mm US 4 sieve = 4.75mm Consistency of soil

1. Liquid limit (w ) The minimum water content at which the soil is still in liquid state 2. Plastic limit (w ) It is the minimum water content after which, the soil starts crumbling 3. Shrinkage limit (w ) Maximum water content after which reduction in water content well not cause shrinkage of the soil mass Shrinkage limit w Where, w

(

(w

)

)

Water content of orginal sample of volume v

Plasticity index

w

w

Liquidity index (IL) = Consistency index( ) = Flow index ( ) = Where, w w

(

)

water contents corresponding n

n blows respectively




Toughness index (It) = Sensitivity =

SR = Where, V1 = Volume of soil mass at water content, W1% V2 = Volume of soil mass at water content W2% Vd = Volume of dry solids W1& W2 = Water Content. Activity = Relative density = A line, IP = 0.73 (WL – 20%) f soil’s

P

is above A low, soil is more clayly

f soil’s

P

is belowAline, it is ML, OL or MH or ML.

Well graded gravels, Cu> 4; CC → to where Cu =

., called as GW

CC = Gravel with fines, GM & GC for gravel with silt and gravel with clay respectively. Sand: SW →C → to Cu> 6.

Soil structure and clay minerals For same void ratio, W.C. of clay > W.C. of sand. Clay minerals Kaolinite, mont morillonite, illite Mont morillonite has large specific area and dwelling. Illite has potassium in adjacent places. THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 146



Soil Compaction =(

(

)

)

= = standard proctor test: Weight of hammer = 2.6 Kg. modified proctor test: Weight of hammer = 4.9 Kg ratio of energy in modified to standard proctor test is 4.5 embankments are compacted on higher side of OMC. dam embankments are also on lower side of O.M.C. but dam cores are compacted on higher side to ensure high permeability Factors Effecting Compaction. 1. Type of soil,

is maximum for GW and minimum for CH.

2. Water content 3. Method of compaction 4. Amount of compactive method. Capillary and Permeability 1. Seepage Velocity v

ki → where K is permeability constant

i= hydraulic gradient  V x A = KiA  Discharge q = K i A  Pi = x z x i

z

z iz

 Quick condition ic =

=




Where R is permeability of soil, i is hydraulic gradient and A is area (cross-section), z is vertical height of sample.  R is given by, R =

x

xCx

Determination of coefficient of permeability 1. Empirical methods Where

m is the grain size (in an) occurring with great frequency

2. Allen Hazen formula  Allen Hazen gave R = C Where D10 = is in cm C = 100. 3. Laboratory Methods a. Constant head permeability test

Overhead Tank

Supply

h

Overflow

h

Air Value L L

Soil Specimen

Soil Specimen

Bottom Tank Measuring Jar

Figure: Constant Head Test q=KiA  q = Q/t = Ki A K= x x




b. Falling Head Permeability Test Funnel ;

Time t1

dh Stand Pipe

Time t2 h1

h h2

L

Figure Falling Head Test If a

cross sectional area of stand p

L

length of the sample

h h t

p

initial and final head

time clasped to fall the head h to h

R=

.

log 10

.

c. Capillary permeability test.

(h0)2 Q (h0)1

StopperA

B

Dry Soil

Spring

P

Stopper

Screen

Air Vent hc

Figure Capillarity Permeability Test THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 149


(

)=

h


h

 Average permeability parallel to the bedding planes .

Kx =

 Average permeability perpendicular to bedding planes Kx =

.

Deflection of flow lines at interface of dissimilar soil

Governing equation is

Seepage through the body of a dam

L C

0.3L D

Directrix P(x,y) H y F A

G H

Q s

E

D

Figure :Casagrande’s Method of determining phreatic line in a dam with horizontal drainage filter



Discharge q


s

Where Permeability s

√

Where,

Head of water Distance of arbitarary orgin of phreatic line as shown in figure

Well hydraulics 1. Unconfined aquifer :→ also called as non – artisan aquifer ( upuit’sThoery) Ground Level r2 r1 Initial water table

s2

s1

s

Cone of depreciation H

h1

h2

(x, y) Cone of depreciation

h

y x

q=KiA q=Kx q

x 2π x y

π (

h )

.

(

log ( )

log

h ) ( )

In artisan aquifer case/confined aquifer(Thiem equation ) q=Rx

x 2π x b

Where, b= is depth of confined strata q

2πbk(

h)

log ( )

2. 2 bk( log

h)

( )




 Field determination of K and T: Pumping out test: K=

(

for unconfined aquifer (Using Dupuit equation)

)

and q

r log ( ) in confined equation (using Thiem equation) h) r

2πb(

 Recuperation test in open well q = KH also q =

.

log10( ) .

Ax( )H=

log10( ) m3/m4

Quantity is called as specific yield.

Seepage Analysis  iC = where iC is critical gradient for upflow condition (quick condition)  Velocity pointed is ∅ such that ∅

= u and

 Kx

∅

+ Ky

∅

= v.

∅

= 0.

 Stream function φ is such that =u 

=v =0

 Flow across flow net is q=KxHx Where,

x Total hydraulic head causing flow

N

Total number of flow channels in the flow net




N

Total number of potential drops in complete flow net

b

Width and length of the field

ield -> The portion enclosed between two successive equipotential lines and successive flow lines

Stress Distribution in Soil Boussinesy’s Solution for Point Load ⁄

σz =

[

( )

]

where, σz is vertical stress at depth z Q is load z is vertical depth r is radial depth.  Maximum vertical stress occurs for

.

Where, r z Line load

σ

2q [ πz

] ( )

Where, q

Load intensity,

x

Horizontal offset of the point from line load

Strip load σ

q π

Where

sin Angle made by the width of the strip at the point




Circular Load ⁄

σ

q[

] (

Where, a

)

Radius of the circular load

 Boussinesy’s theory for rectangular loading σz =

)(

[(

where m = ,

√

)

tan

[

√

]]

n=

Consolidation of Soil  Compression index, CC =  CC = 0.007 (WL – 10%) for remoulded clay CC = 0.009 (WL – 10%) for field clay  Coefficient of compressively, a  m

.

Where m is coefficient of volume change.  m  Coefficient of consolidation, C = Where, K is permeability m is coefficient of volume change

is density of water.

Consultation Settlement Any time to during consolidation m

σ dz

Ultimate or final settlement m ( σ) Degree of consolidation (U )




Time factor C t d

T

Where, t d

time

drainage path

For single d

[Thickness of drainage layer]

For double drainage d  Tv = (

2

)if U < 60%

 Tv = 1.7813 – 0.9332 log10 (100 – U%) if U% > 60%  Methods to determine CV are 1. Square root of time method 2. Logarithm of time fitting method.  Calculation of void ratio and co-efficient of volume change 

Height of solids method



Change in void ratio method.

 Change in void ratio ratio e

e (

e)

 Height of solid = Md: Mass of dried speciment : Density of water A: Area of speciment. Shear Strength Defined f(σ) as a linear function  Mohr envelops: Shear capacity of a soil is a function of normal stress acting on it S

f(σ)

 Columb envelope S

C

σ tan ∅

 The effective stress principle f

C

σ tan ∅



 Also σ

+


cos 2∝

sin 2∝ for plane to be most critical, ∝ = 450 +

∅

∝ is the angle of failure plane with the horizontal Whereσ3 σ1= Normal stresses σ1 Normal stresses on plane inclined at ∝ to σ3 σ1.  C and ∅ are not fundamental properties of soil. Measurement of shear strength 1. 2. 3. 4.

Direct shear test Triaxial test Unconfined compression test Vane shear test

Depending upon drainage conditions a. Undrained test or quick test b. Consolidation undrained test c. Drained test  In undrained stress, no drainage of water is allowed.  In drained test, drainage is permitted throughout the test during application of normal and shear stress.  In consolidated – undrained test, drainage is permitted under initially applied normal stress only and full softening is allowed to take place. 1. Direct shear test, failure plain is pre-determined and less control on drainage is possible. 2. Triaxial test  σd = σ1 –σ3 is called as deviator stress in triaxial compression test.  Pore pressure during triaxial compression test is measured.  σ  or (σ

∅

σ tan ( σ )

)

cos

2C Tan ( (σ

∅

)

σ ) sin

Above equation is modified in the form THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 156


2

(σ

σ )

d

2

(σ


σ ) tan

 Modified failure envelope is plotted between (σ1`–σ3`) and (σ1

σ3`).

 n modified envelope sin ∅` = tan φ C` =

∅

The calculation of deviator stress must be done on the basis of the changed area of cross section at failure or during any stage of the test. The area A at failure or at any impotent is given by Additional axial load A

evitor stress σ  A2 =

for computation of stress at any point

 ∝ is angle of failure plane with horizontal 3. Unconfined compression test In unconfined compression test, σ2

σ3= 0 and

∅

σ1 = 2 Cu tan (45 + )  This test is used for saturated clays for which ∅

.

4. Vane shear test  Vane shear test is used to determine undrained shear strength of cohesive soil. (πd

T

πd

) (

d 2

2∫

(2πrdr ) r

d )

2

If only the bottom end partakes in the shearing T

πd

(

2

d ) 2

Skempton’s Pore Pressure Perimeter If C

Compressibility of soil skeleton Change in pore pressure



σ

σ

σ

(

C

Increase in the principle stresses

2u)

u=

* ( σ

u = B* σ U

=B


σ

σ σ

A( σ

σ )+ σ + σ

u σ A=  Shear strength of cohesive soils: a. Undrained test on saturated cohesive soil: →  Diameter of total stress and effective stress diameter remains same.  For saturated soil B = 1  A varies with OCR. Shear Strength of Cohesive Soils 1. Undrained Test on Saturated Cohesive Soil The undrained test is carried out on undistributed sample of clay, silt and peat to determine the strength of the natural ground. It is also carried out on remoulded samples of clay to measure it sensitivity.

Mohr stress circles for undrained test on saturated cohesive soil 2. Undrained Test on Partly Saturated Cohesive In the case of earth embankments, which are compacted at optimum water content, the soil remains partly saturated and it is necessary to conduct undrained test to determine the shear parameters of the soil. This test is also sometimes applied to undisturbed samples of soil taken from existing rolled fills or trail sections.




Mohr stress circles for undrained test on partly saturated Soil 3. Consolidated Undrained Test on Saturated Cohesive Soil The consolidated undrained test is carried out on undisturbed sample of clay, silt and pear, on remoulded sample of clay and silt, and on redeposited samples of cohensionless soils such as sand and gravel. The consolidated-undrained tests are performed by two methods:(i) The moulded specimens are first consolidated under the same cell pressure and then sheared under undrained conditions with different cell pressure by increasing the axial stress. (ii) The remoulded specimens are sheared under a cell pressure equal to the consolidation pressure.

Consolidated Undrained tests on saturated clay 4. Consolidation Underained on Partly Saturated Cohesive Soil Consolidated undrained test are required to determine the shear parameters of undisturbed samples or of compacted sample of earth fill where the soil is partly saturated. The test may also be conducted to examine the effect on C and ∅ of flooding foundation strata and earth fill materials, by applying back pressure to the pore space to ensure full saturation. THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 159



Failure envelopes for drained test




Part – 3: Geotechnical Engineering 3.2 Foundation Engineering

Surface Investigation 

Area ratio for sampler =



Inner clearance =

x 100%



Outer clearance =

x 100%

x 100%

Where, D2 is outer diameter of cutting edge. D1: Inner diameter of cutting edge D3: Inner diameter of sample tube D4: Outer diameter of sample tube. 

Area ratio should be as low as possible



Sounding tests are used to measure penetrative resistance.



For SPT, first 150mm settlement is taken as redundant



If N>15 in SPT, corrected Ne = N + (N – 15)



In SPT, dilation correction is Ne = N *

+

Where σ is effective overall under pressure. 

In cone penetration method, resistance profile for first 8cm of penetration is recorded.



In electrical reactivity method, resistivity = 2ΩD R Where R = Resistance D = Spacing between electrodes.

Earth Pressure 

In active state, retaining wall moves away from the soil wedge. THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 161


Active pressure =


vertical pressure

Where, =

=

tan (

)

sin sin

=

= Co-efficient of earth pressure



In passive state, retaining wall moves towards the soil wedge and resistance acts downward.



K P=

, = tan (

)

KP: Coefficient of earth pressure (passive) for C = 0 soil. When the wall is at rest 

K0=

=

,

K0: Coefficient of earth pressure at rest. 

In no submerged condition pressure active, Pa = x x H2



In submerged condition, Pa =

x

x H2 + x

x H2

Back fill with uniform surcharge Total active pressure =

2

Where q is the surcharge 

In code of surcharge, total active pressure = ∫



Back fill with sloping surface *

Ka = Cos β

√

z

q

+

√ *

KP = Cos β

√

+

√





If back fill is submerged, P = √



In case of cohesive backfill


where P1 = x Ka x r x H2

Pa = rH2 Cot2∝2- 2C Cot ∝ H PP = rH2 Tan2∝2 + 2C tan ∝ Where ∝ = (

)



In active case, cohesive soil should be able to withstand the depth of



Depth of tension crack =



φ angle =

-

tan ∝.

tan ∝.

- δ where δ is angle of indication of Pa and normal of wall.

Stability of Slopes Types of Slopes Infinite Slope A representing the boundary surface of a semi infinite soil mass and having soil properties uniform at every depth Finite slope: A slope of limited extended Factor of safety Where

=

= Shear strength =Shear stress acting on the soil



For cohesive soil & non cohesive soil generation formula for factor of safety is FC =

+

If water table / seepage is not present. 

For C = 0 (i.e., for cohesion loss soils) FC =



If seepage is parallel to the slope FC =





If submergence in a certain portion of slope



If seepage is parallel to the slope C cos i sin i

=

.


tan tan i

Again for cohesion less soils C= 0 = 

.

tan tan i

If the slope is submerged =

C

cos i tan cos i sin i

For cohesion less soils =

tan tan

i. e. , same as in dry state

If su mergence in a certain potential Where

=

+ tan

* tan

Z = Total height of slope h = Height of slope above i = Angle of slope = Angle of internal friction C = Cohesion of soil



Critical height of any slope can be found by putting for as applicable

=

and

=

in then the formula

E.g.: critical height of a submerged slope of cohesive soils =  

tan

Stability number =

tan

cos

= Sn

= FC Where

is called the mobilized cohesion



 

= Sn Depth factor , Df = Where,




FC =

H = Height of slope D = Depth of soil between slope and hard strata. ̅

for

= soil.

Where , x̅ = Distance of centroid of slip circle from centre of rotation r = Radius of slip circle. 

For C -

soil.

FC = Where N is sum of normal components of weight is with respect to tangent of slip circle T is sum of tangent components of weight along slip circles. 

Culman method assumes wedge / planar failure.



Swedish method assumes circular failure.

Types of Foundation 

Raft footing Allowable differential settlement = 65 to 100mm on clay. On sand : 40 to 65mm.

Definitions 1. Bearing Capacity The load or pressure developed under the foundation without introducing any damaging movement in foundation and in the supported structure, is called bearing capacity of solid. 2. Gross Pressure Intensity It is the total pressure at the base due to weight of the super structure 3. Net Pressure Intensity It is defined as the different in intensities of the gross pressure after the construction of the structure and the original over burden pressure If D is the Depth of footing q =q-σ=q–rD




4. Ultimate Bearing Capacity It is the minimum gross pressure intensity at the base of foundation at which the soil fails in shear. 5. Net Ultimate Bearing Capacity (qnf) It is the minimum net pressure intensity causing shear failure of soil q =q +σ 6. Net Safe bearing Capacity The net safe bearing capacity is the net ultimate bearing capacity divided by a factor of safety F q = 7. Safe bearing Capacity The maximum pressure which the soil can carry safely without rest of shear failure is called safe bearing capacity q = q + rD = + rD Types of failure of foundations



General shear failure occurs in stiff soils.



Local shear failure soil with single compressibility and sands.



Footings at very shallow depth in loose sand are susceptible to punching failure.



Angle of zone III in Terzaghi analysis is 450 – with horizontal.



For a strip footing qf = CNC + σ ̅ Nq + 0.5 r BNr For bearing capacity factor =

,

and

depends only

a 2 cos (

)



where a = e =( =

tan 2



,

( cos

)

= passive earth pressure coefficient dependent on are also given in standard tables

For qs = ( Whereσ ̅=



)

) cot

Where ,

( .


̅ σ

̅ σ

. r

r )+ σ ̅

.

Terzaghi’seqn for local shear failure. 2 (C = C) and tan

2 = tan

qu = CNC + σ ̅Nq` + r BNr`. Guidelines for Local Shear Failure Condition for Shear Failure (i) Stress strain test (l- soil) General shear failure→ low strain < % Local shear failure → strain of to 2 % (ii) Angle of shear resistance Φ> , general shear failure Φ <2 , local shear failure (iii)

Penetration test ≥ : general shear failure ≤ : local shear failure

(iv)Plate load test Shape of the load ‘settlement curve decides whether it is general shear failure or local shear failure (v) Density Index I > 70, General shear failure I < 20, Local shear failure






or friction cohesive soil → qf = 1.3CNC + qnq + 0.3 NrBr for circular footing



For square footing, qf = 1.3 CNC + qNq + 0.4 rBNr



For rectangular footing, qf = 1.0 CNC*(



For local failure (or local shear).

.

)+ + qNq + 0.5 rBNr*

.2 +

Cn = C Tan

m

= Tan

m.



Nq = Tan2(



NC = (Nq–



Effect of water table on bearing capacity is taken by, RW1& RW2



qf = CNC + qNq x RW1 + 0.5 Nr x B x r2 x RW2.



RW1 = 0.5 *

+

RW2 = 0.5 *

+

Where

)r Tan . Cot .

ZW1 = Depth of water table from surface or ground level. ZW2 = Depth of water table from base of footing.



Max RW1& Max RW2 = 1



Effect of size of plate in settlement on granular…. soil: Where

=*

*

. .

+

δ : Settlement of footing δ : Settlement of plots.



For clayey soil ∫ ∫



=* +

Total settlement of footing is S = Si + SC + SS Where, Instantaneous settlement, Si = *

+ x IwqB




μ = Poison ratio Es = Modulus of elasticity Iw = 0.88 – 1.70, 0.88 for rigid circular footing and 1.70 for rigid rectangular footing. B = Least dimension of footing. 

SC = H x CC log10(

)…

CC : 0.009 (WL – 10) L0 : Initial void ratio C : Coefficient / correction factor depending upon geometry of footing and history of loading on clay. Es : *

+

CC = 0.007 (WL – 10) for remoulded sample. Pile Foundation 

Pile driving is done by drop hammer, single acting hammer, double acting hammer and vibratory hammer.



Dynamic formulae  Engineering news formulae 1. Drop hammer, Qa = Where S : final set per blow. C : 2.5 cm for drop hammer. 2. Single acting steam hammer Qa =

.

3. Double acting steam hammer Qa =

.

a = area of piston p = pressure of steam.






Hiley’s formulae Qf = nb =

if W > iP

nb =

*

+ if W < iP.



In case of submerged loose soil equitation may take place due to dynamic load.



Use of dynamic formulae for clay is meaningless.



Static formulae Qmp = Rf + RP. = As rf + Aprp. rf: Average spin function rp: Point function As: Surface area Ap: Point area.



For cohesive Clay rf = αC̅ (or mC̅ rp = 9Cp Qmp = mC̅Ap + 9CpAp.



For non cohesive soil rf =

Tan

rz

q

rp = 0.3 rBNrq [for circular] rp = rBNrq [for rectangular] 

In clayey soil, group efficiency of function piles may be less than sum of individual efficiencies of pile.



In end bearing piles, group efficiency = n x individual efficiency Where, n = number of piles.






Converse labre formula. ηg = 1 -

*

+

= Tan–1( ) where, m: number of piles in a row n: number of rows. d: diameter of pile s: spacing (C/C) between 2 piles. 

Settlement of pile group in clay δ=

logi(

)

where, H = height of clay strata e = initial void ratio 

One – third height of pile will also contribute in load transfer.



Angle of load dispersion may be assumed to be 300. Qu =Ap x NC x Cb + Aa + NC x Ca

Ca x As

∝Ca As.



Ap : Cross section area of pile stem at soil.



NC: Bearing capacity factor = 1



CP: Cohesion at tos of pile



Aa =



Ca` = Average cohesion of soil around underream drills.



As` = Surface area of cylinder curricumscriling under-reams.



Ca = Average cohesion of soil along pile stems.



Cu =

u

, where Du is dia. of underream, D is diameter of stem.

for clayey soil.



Water Resource Engg

Part – 4: Water Resource Engineering 4.1 Fluid Mechanics Continuum Flow Two factors which are important is determining the validity of continuum model. (i) The distance between molecules (ii) Elapsed time between collision Knudsen Number: ( ) = = Molecular mean free path L = Characteristics length (i) Continuum ( ) – No slip condition (ii) Slip flow ( < n ) (iii) Transition flow ( < n ) (iv) Free molecular flow (Kn > 10) Common temperature scale F C C Compressible and Incompressible Fluids Compressible Fluid – Variable density Incompressible Fluid – Constant density Variation of viscosity with temperature Liquids

Gases

Viscosity

Temperature



Water Resource Engg

Rheelogical Diagram Thixotropic Ideal Plastic

Shear yield stress stress

Non Newtonian (Pseduo plastic) Newtonian Dilatant (Velocity gradient)

Ideal Fluid

Based on property of viscosity, fluids may be classified (i) Ideal Fluid→ μ (ii) Real fluid → Shear stress are induced when fluid is in motion, which possesses viscosity (iii) Newtonain fluid→ Which follows ζ ∝ (Newton law) e.g.; Air, water (iv)Non-Newtonian fluid→ Which does not follows Newton law e.g.; Paints, Ink. Gel, Emulsion Flow Patterns 1. Stream line A line so that tangent at any point represents the velocity vector at a given instant 2. Path line : Actual path traversed by a given fluid particle 3. Streak line : Locus of particles passed through a prescribed point 4. Time line : Set of fluid particles that form a line at a given instant 5. Dilatants fluids: Dynamic viscosity increases with rate of shear e.g. Butter, Printing ink 6. Pseudo plastic : Dynamic viscosity decreases as shear rate increases e.g. clay, Milk, Cement, Colloidal solution 7. Plastic fluid : Begins to flow after reaching a yield value of shear stress (

)

8. Thixotrope fluids : Dynamic viscosity decreases with time when shearing forces are applied e.g. Jelly 9. Rhepectic: increases with time 10. Viscoelastic: Behaves like Newtonian fluid, but of shear stress changes suddenly it behaves as a plastic Surface Tension → Cohesion – Enables liquid to resist tensile stress. Adhesion – Enables liquid to adhere to another body. → Liquid Fluid interfaces Liquid gas interfaces: Free surface Liquid-liquid (immiscible) interface Capillary THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 173


Water Resource Engg

1) Cohesion < adhesion – liquids wets solids, rises at point of contact 2) Cohesion > adhesion – liquid surface depress at point of contact → Capillary rise in water and capillary fall in mercury → Capillary rise (h) ς

Surface tension

θ

Wetting angle

γ

Specific wt. of liquid

r = Radius of tube Pascal’s Law All points in a connected body of a constant density fluid at rest are under the same pressure if they are at same depth below liquid surface Absolute and gauge Pressure

Gauge Pressure Atmospheric Pressure Pressure

Vacuum = negative gauge pressure

Absolute pressure Atmospheric pressure

Absolute pressure Absolute zero

Gauge pressure – Positive – if above atmospheric pressure Negative if below atmospheric pressure p

p

p

 The atmospheric pressure head in 760 mn if Hg or 10.33 of water  Atmospheric pressure at sea level at 15 C is 1013N cm Measurement of Pressure 1) 2) 3) 4) 5)

Barometer Bourdon gauge Pressure transducer Piezometer column Simple nanometer



Water Resource Engg

For on a Horizontal Plane Area F γhA Where force on the plane area = volume of prism area Location of forces on a horizontal (Centre of Pressure) y

F

PA

γhA

xp

x

Y p = ∫ ∫ x d A, X = ∫ ∫ y d A Forces on a Inclined Surface F

γAh

h=y+

Sin θ

y = Centre of gravity Ig = Moment of interia about C.G Force on vertical plane

doffed line ̅

Force,

̅



Water Resource Engg

Forces on curved surface Water Surface

0

dfy

A

df θ θ

dfx B F = √Fx

Fy

Tan θ f

force on vertical projection of the current area = weight of liquid supported by the curved surface upto free surface of liquid

→ The direction of Fy will be taken in upward direction Archimedes Principle: An immersed body is buoyed up by a force equal to the weight of fluid it displaces. → Buoyant Force > Gravity force – object float. → The apparent wt of an object in a liquid is gravitational force (weight) minus the buoyant force (w - FB) → Ship sinks until weight of water displaced by underwater volume is equal to weight of ship If a body floats at surface of separation between two immiscible liquids Then F Where

γ v ,

γ v volumes of parts of the body immersed in the liquids of density

G B



Water Resource Engg

Metacentre M θ

Angle of heel

G Z

B

B

Of angle too large, M moves off centre lines Metacentric height (G-M) 1) Determine size of righting / upsetting arm (For < s < 7) 2) Large GM > large righting arm (stiff) 3) Small GM > Small righting arm (tender) Relationship between G and M 1) G under M : Ship is stable 2) G = M ship neutral 3) G over M : Ship unstable Metacentre

stability curves

Plot Gz (Righting arm) v angle of heel Ship’s G does not change as angle changes Ship’s B always at centre of underwater portion of hull Ship’s underwater portion of hull changes as heel angles changes GZ changes as angle charges ( )

1. 2. 3. 4.

(

)

Time Period of Oscillation T

π√

K = Radius of gyrations THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 177


Water Resource Engg

GM = Metacentric height Liquid Jar subjected to uniform linear horizontal acceleration a) Vertical Pressure distribution P γh b) Horizontal pressure distribution in direction of accelerated motions Tan θ P

γ h and P

γh

θ h

ax

γhda pda

h

h P da

P da

Liquid jar subjected to uniform vertical acceleration

Upward acceleration

γhda

h Pda

P

γh(

)

→ If the tank is accelerated downwards P γh( ) If a = g then P =0 pressure will be atmospheric at all points Acceleration of a fluid mars along a slope (a) Acceleration up the slope Tan θ θ = Inclination of liquid surface with horizontal (b) Acceleration down the slope Tan θ = Pressure variation along a stream line (i.e tangential direction) +

=C

V = tangential velocity



Water Resource Engg

P

r

ds

P + dp Pressure variation in a radial direction =P Here V is a function of r.

Forced Vortex

Fluid is made to move in a curved path under the action of an external force

Rise in the level at the ends

falls in level at centre

P + dP rP dr

Free Vortex

V r = C i.e. product of velocity and radial distance constant Ex → Vortex motion of water in a shallow wash basin discharge through a central hole at bottom Steady and unsteady flow Steady flow:( ) = 0, ( ) Unsteady flow

, ,

, ,

= 0, ( )

, ,

=0



( )

≠ , ( )

, ,

, ,

≠ , ( )

Water Resource Engg

≠0

, ,

Uniform and Non-uniform flow Uniform flow = ( ) = 0 Non uniform flow= ( )

≠

Mach Number = Ma Ma = V/C It is a good indicator of whether or not compressibility affects are important Ma < 0.3 : Incompressible Ma < 1 : Subsonic Ma = 1 : Sonic Ma > 1 : Super sonic Ma > 5 : Hyper sonic Rotational and Irrotational flow Rotational flow Type of flow in which fluid particles while flowing along streamlines rotate about their own axis Irrotational flow The fluid particles while flowing along streamlines do not rotate about their own that type of flow Acceleration field ⃗

a⃗⃗⃗⃗ =

+u

⃗

+v

⃗

+w

⃗

In vector form a⃗ =

⃗

=

⃗

⃗ ⃗D)V ⃗ + (V ⃗

Local acceleration It is zero for steady flow Advective acceleration – u

⃗⃗

+V

⃗⃗

+w

⃗

It accounts for the effect on the fluid particle moving to a new location Total derivative operator is called material derivation Vorticity and Rotationality ⃗ Vorticity vector is defined as the curl of velocity vector ζ = ⃗ x V Vorticity is equal to twice the angular velocity of a fluid particles = ⃗⃗⃗⃗⃗ Circulation: The line integral of the velocity around a closed contour is called circulation Circulation = ∮ v cos ds Continuity equation: The equation based on the principle of conservation of mars is called continuity equations → For compressible flow A V = A V THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 180


Water Resource Engg

→ For Incompressible flow, ∫ = ∫ A V =A V ie Velocity Potential If ϕ is some function of x and y in two dimensional flow, such that U= ,V=Φ is called velocity potential function Here u and v are the velocity components is x and y directions Slope of equipotential line Stream Function: →u and V = → ⃗V = ⃗ x ⃗⃗⃗ → The different between the stream F values at two points gives the volumetric flow rate or (Volumetric flow) | | Flow net in Isotropic soil Q = KH ( ) q = rate of flow or seepage per unit width (L T) k = Hydraulic conductivity (L T) N = No. of flow lines N = No. of potential drops. Flow through pipes < < < > Turnbulent Major losses I. Loss of energy (or heads) due to friction 1. Darcy weisbach formula

Where f f

R

R

(for laminar)

(for turbulent)

2. Chezy’s formula v

C√Rs where s

h L

II. Minor energy losses 1. Loss of head due to sudden enlargement



(

Water Resource Engg

)

Where,

are velocities of section 1 and 2

2. Due to sudden contraction [

]

(Lused if not given ) Where, is the velocity at vena contracta of smaller pipe 3. Loss of head at the entrance of a pipe v h g 4. At the exit

Where v is the velocity in the smaller pipe 5. Due to an obstruction in pipe v A h [ ] g c (A a) A a projected areas of pipe and obstruction perpendicular to the direction of flow 6. Due to bend in pipe v h coefficient of bend g 7. For pipes connected in series *

+

8. For pipes connected in parallel h h h f L V f L V f L V ie gd gd gd Equivalent pipe A pipe if suitable diameter so, that it has same head loss and length is equal to the sum of different pipes connected. Bernolli’s Theorem This theorem states that ‘In a steady continues flow of a friction less incompressible fluid, the sum of the potential head, the pressure head and the kinetic head in the same at all points. Application of Bernoulli’s theorem (Flow measurement) Measurement of flow rate – ventuirmeter orifice meter Measurement of velocity – Pitot static tube THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 182


Water Resource Engg

Euler’s equation of motion ( ) ( )

i.e. Total change in energy per unit mass is zero The Venturi meter Q=C √ gh √

C = Discharge coefficient

(for calculation purpose)

f(R )

A V P

A V P

Velocity at throat is assumed to be constant Converging cone angle Because of the cone and the graded reduction in the area, there is no vena-contract. There will be no change in the in the result whether the venture meter is horizontal vertical or inclined Where, h – It is the difference in level between two piezometer at 1 and 2 in venture meter. The momentum equation “ The net force acting on a fluid mars is equal to change in momentum of flow per unit time in that direction” The orifice meter The is kept 0.5 times the diameter of pipe generally though it may vary from 0.4 to 0.8 times √∫ [ Then Here

(

) (

) ]

for calculations

∝ In this method, pressure loss is minimum



Water Resource Engg

The Pitot tube

or, V √ gy (theoretically) V C √ gy (practically) Pitot tubes are called prandti tubes Its readings for gases are extremely small Force exerted by a Flowing Fluid on a Pipe Bend

V sin θ V θ

(1)

V cos θ

P A

F θ F

P A

F = Q (V V cos θ) + P A - P A Cos θ F = Q ( V Sinϕ) - P A Sin θ F = √F

Fy , tan θ



Water Resource Engg

Flow over a Rectangular Sharp Crested Weir or Notch Water Surface Nappe

H

V

Crestor Z sill

Q = C √ g LH If velocity of approach A = C √ g L [(H ha) ha ] If there are end contractions = C √ g (L n H) [(H ha)

ha

]

Bazin’s Formula m√ g LH m

(

H

)

Flow over a triangular weir Vnotch weir or triangular (V-notch) Water surface

H

θ

Q = C √ g tanθ H If velocity of approach is to be taken into account Q = C √ g tan [(H ha) ha ] Cipolletti weir (or Notch) Slopping sides → Horizontal to 4 vertical Q/2 = Q = 1.842 *(H

)

(

)

+

Time Required Empting A Reservoir with Rectangular Weir THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 185


t=

[

√

√

√

Water Resource Engg

]

Criteria to Classify Weirs 1. Thin plate weir H B< > B 2. Narrow crested weir H B H 3. Broad crested weir H B H Submerged Weir

,

√ (

[, ) *√

(

)

(

)

]

+

Submergence Ratio

Proportional Weir or Sutroweir A weir designed in such a shape for which To design it – Boundary Layer Flow →In the boundary layer

tan

∝H

√

exist so extents shear stress.

But outside boundary layer

= 0, velocity is constant no. shear stress



Water Resource Engg Boundary layer

F

Tubulent

Laminar boundary layer E Laminar sublayer A B Leading edge

Re = 5 x 5x

C

Transistor zone

for laminar in plate for turbulent

Laminar sublayer ( ) ζ μ( ) =u Displacement thickness ( *) * ∫ ( ) dy Momentum thickness (θ) θ=∫ ( ) dy Energy thickness ( ) =∫

*

+dy

Drag force on a flat plate due to boundary layer → = V on karmar momentum integral equations →Fo

∫ ζ b dx

→ Local coefficient of drag C * = → Average coefficient of drag, C = Effect of Pressure Gradient on Boundary layer separation

C B

S

Pressure distribution

A

<0

B

D

>0

P min

C

S

< 0 due to increase in velocity – The entire boundary layer move forward. THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 187


Water Resource Engg

> 0 decrease of velocity Location of separation point: 1. If ( ) is negative, flow has separated 2. If ( )

= 0, flow is on verge of separation

3. If ( )

is positive, the flow will not separate or flow will remains attached with

surface Flow of viscous fluid through circular pipe ζ x 2πrdx R (P+

r

x) πr

Pπ

P. πr – ζ x πr dx – (P Ζ

-

n) πr = 0

.

D y

Velocity distribution

Shear stress distribution (a) Velocity distribution μ [R r ] Umax = Uavg =

.R =

(

)P

= 2.0 (b) Drop of pressure for a given length (L) of pipe = h = 32 THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 188


Water Resource Engg

P - P = drop of pressure Flow of viscous fluid between two parallel plates (ζ

y) n x

A

D (P +

y

Px y

B

n) y x 1

x ζ x

C

Velocity distribution U=-

[ty - y ]

Umax = -

t

Uavg = -

t

Umax Uavg Drop of pressure head for a given length h =

Hydraulics and Hydraulic Machinery Flow in Open Channels Laminar Flow and Turbulent flow Re →

to

Laminar

Re > 2000

Turbulent.

Re →

Transitional

to

Subcritical, Critical & Super Critical Flow Fr = 1, Critical flow.



Water Resource Engg

Fr > 1, Super critical flow. Fr < 1, Sub critical flow. Section Factor (Z) Z = A √D = ( ) D = Hydraulic depth. → To measure velocity distribution in a channel (a) Pitot tube. (b) Current meter.

- Cut Type - Screw or Propeller type.

Velocity Distribution Curve Along A Vertical Line Of Channel Section y V0.2 d = 0.6 y0

V0.6 =Vavg V0.8

..

Uniform Flow in Channels  A uniform flow will be developed if the resistance is balanced by gravity forces.  τ0

γRS0

R = Hydraulic Radius. So = Slope of channel Bottom.  Unsteady – Uniform Flow is practically non – existent.  Turbulent uniform flow is most encountered.

Uniform Flow Formulae Chezy Formula τ0 = ρv2 The Gangwillet – Kutter Formula THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 190


C=

( *

Water Resource Engg

) *

+

√

+

 C=  Stickler formula n= Most economical Section or Most Efficient Section 1. Rectangular Channel Section

y

B

2. Trapezoidal channel section

y

y Q zy

zy

→ Top width

x (one of the slopping sides of channel)

→R →Z

√

or θ

0.

3. Triangular channel section

y θ Q

→θ

Z

0

→R

y ⁄

√



Water Resource Engg

4. Most economical or most efficient circular channel section

r

r

r

θ y

A= P

r (θ

Sin θ)

rθ

Bhy Chezy’s θ

C√

AC √RS

S

discharge is maximum when ( ) is maximum. (

( )= →θ

)

(

)

=0

80 (approx.)

→y

D

→R

D

By Manning’s formula → θ will be maximum when ( ) is maximum. →θ

0

200

→y

8D

Conditions of Maximum When Velocity of Flow Through a Circular Channel Section By Chezy  Mean velocity of flow maximum when (

) is maximum


Quick Refresher Guide ( )

=

 θ

(

)

(

Water Resource Engg

)

(approximately).

0

 y = 0.81 D  R = 0.30D. Open channel section for constant velocity at all depths of flow  For an ordinary channel section with an increase in depth of flow, the velocity is considerably increased.  y = R loge x + √(x

R )

 y=0

y = R loge (x

x=R

C √x

R )

R log R

Q = AV = CA √RS = K√S Where, K = CA√R K is known as Conveyance. Q = AV = K√S 

√

Specific energy and specific energy curve Specific energy, E = y + E = EP + EK

y=E

S < Sc Y > Yc

Upper Limb Flow depth y(n) y

yc

Lower Limb

v/pg

S> Sc, y < yc E



Water Resource Engg

 yC (Critical depth):- Specific energy is minimum.  yC = ( )

For Rectangular section.

 VC = √ y VC – critical velocity.  Emin =

y

 Critical flow condition A T

g

Critical Flow in Channel Sections of Other Shapes (i)

Triangular channel section yC = ( y

(ii)

) E

Parabolic Channel Section y

E

yC = (

(iii)

T

k

⁄

)

Trapezoidal channel section ( )=

( (

) )(

)

 Some of the devices commonly used for measuring the discharge through channel based on critical flow theory are (i) Venture flume. (ii) Standing wave flume. (iii) Parshall flume.



Water Resource Engg

Measuring Flumes  At critical state of flow the relationship between the depth of flow and discharge is definite and is independent of the channel roughness and other uncontrollable factors. Momentum Principle and Its Application  Whenever there is an unknown energy loss between any two sections under consideration, the result is a change in the linear momentum of flow. Z̅A

 F=

F – Specific momentum, specific force, force function or momentum function.

y2

y1 & y2 are sequent depths or conjugate depths

y0 y1

  Energy equation contains a term for internal loss, where as the momentum equation contains a term for external resistance. Hydraulic Jumps When a stroting flow (Fr > 1) meets a quiescent flow (Fr < 1).  Sequent depth Rectangular sections. [

√

8F ]

If Fr2 is very small. Then Fr

F

Fr

This expression is valid only when Fr2 is very small and Fr < Energy Loss E = E1 – E2



E=

(

Water Resource Engg

)

*(

+

) [

(

y y (y

)]

y )

Hydraulic jumps.

Gradually Valid Flow Energy Equation for Gradually Varied Flow

Energy eqn for Gradually varied flow ∝1

S

y1

z1

Sw

hf

∝2 y2

S0 L Patum

Z2

Kinetic Energy correction factor  Typically

∝

 With obstructions ∝ > 

x=

Standard Step Method if x > , section ( ) is downstream of ( ) if x < , Section ( ) is upstream of ( ) Types of Channel Slopes  A mild slope on which uniform flow is subcritical  A steep slope on which uniform flow is critical. yn > yC → mild slope THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 196


Water Resource Engg

yn < yC → Steep slope yn = yC → critical slope Classification of Slopes Become mild slope

S 0 < SC

steep slope

S 0 > SC

critical slope

S 0 = SC

horizontal slope S0 = 0 adverse slope S0 < 0 

Fr



S



S



S

= S0

(

) (

)

(

)

(

)

If manning’s equation is used

If Chezy’s equation is used

⁄

( )

⁄ )

( (

⁄ )

 S0 > 0 Sustainabler S0

Unsustainable

Zone 1 M1 NDL

y0 yc

= > 0. y will increase in flow direction. THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 197


as y → y0 y→∞

S f → S0  Fr →

Water Resource Engg

→

Sf →

→ S0 Zone 2 yn > y > yc NDL

CDL

y0 yc

N

Zone 3 yn > yc > y. NDL

y0

CDL yc

Bresse’s Method  Applicable only to wide rectangular channels. Impact of Jet Force exerted by the jet on a stationary flat vertical Plate:V

V Plate

V mass of fluid before striking plate, m Fx

ρ

ρ aV

ρav2



Water Resource Engg

Force exerted by a jet on stationary inclined flat plate V Sin θ

V

Fx 90 – θ Fn

θ

Then mass of water per striking the plate

ρ aV

Force of jet in the direction normal to plate

Fn

ρ (V Sin θ – )

ρ

V Sin θ

Force of jet in the direction parallel to plate = 0. Fx

ρav2 Sin2θ

Fy

ρav2 Sin2θ Cos θ

Force Exerted by a Jet on Stationary Curved Plate V Sin θ

Vθ N Cos θ

V

Fixed curved plate

Leaving velocity in the direction of jet flow = V Cos (180 – θ) Fx

ρ aV [V

Fy

ρ aV [ – V Sin θ]

For θ of jet Fx

( V Cos θ]

ρav (

–V Cos θ

Cos θ)

ρav2 Sinθ

, i e when vane becomes semicircle, then outlet tips will be parallel to the direction

ρav2 (

Cos θ)

ρav2.



Water Resource Engg

Jet Strikes The Curved Plate at One End Tangentially when Plate is Symmetrical V

θ V Cos θ

V Sin θ

Fy Fx V Sin θ V θ V Sin θ

Fx

ρav [v cos θ) ]

ρav2 Cos θ

Fy

ρav [v sin θ – v sin θ]

Jet Strikes the Curved Plate at One End Tangentially When the Plate is Unsymmetrical V

∝1 ∝2

F

V

vix

V Cos ∝1, viy

V Sin ∝1 ,

vox = – v Cos ∝2, voy = – v Cos ∝2, Fx

ρav [v cos ∝

Fy

ρav [– v Sin ∝

( v cos ∝ )] ( v Sin )]

ρav [Cos ∝

Cos ∝ ]

ρav [Sin ∝

Sin ∝ ]



Water Resource Engg

Force Exerted by a Jet on a Hinged Plate ς x θ

x

h

A

A1 B Fn

Two forces are acting on the plate. 1. Force due to jet of water, normal to plate Fn

ρ av2 Sin θ`

θ`

-θ

2. Weight of plate, W. Moment of force Fn about hinge = Fn x 0B. ρav2 Sin (90 – 0) x 0B. ρav2 Cos θ x Moment of weight W, about hinge ρav2 x

W x A Sin θ

e av x A = eav2 x W x Sin θ

w x Sin θ

Sin θ Force Exerted by a Jet On Flat Plate Moving with Uniform Velocity (v – u)

u

V

(v – u)

Mass of water striking the plate per sec

ρa (V – u)

[Initial – final] Fx

ρa (V – u) [(V – u) – 0]



Water Resource Engg

ρa (V – u)2  As the plate is moving, so the work will be done by jet on plate. For the stationary plate, the work done is zero.  Work done = Fx. u ρau (V – u)2 Forces on the Inclined Plate Moving in The Direction of Jet (v – u)

V

u θ

Fx 90 – θ Fn

(v – u)

Mass of water striking per second Fn

ρa(V – u).

ρa (V – u) [(V – u) Sin θ – 0] ρa (V – u)2 Sin θ

Fx = Fn Sin (90 – θ) ρa(V – u)2 Sin2θ Fy = Fn Sin (90 – θ) ρa (V – u) Sin θ Cos θ Work done = Fx.u. Force On the Curved Plate Moving In Direction of Jet (V – u)

(V – u) Sin θ

θ

(V – u) Cos θ V

u



Fx

Water Resource Engg

ρa (V – u) [(V – u) – (–v u) Cos θ] ρa (V – u)2 [

Cos θ]

Work done = Fx.u. ρau(V – u)2 [

Cos θ]

K0E0 per second of jet = ρaV3 Efficiency of jet n = =

(

) (

=

(v

u) (

)

Cos θ)

For efficiency to be maximum = 0  V = u and V = 34 (i)

When u = V, W.D. Per second Hence n = 0.

(ii)

When u = V⁄ , n max = =

( ) (

Cos θ)

Cos2 θ⁄

For semi – circular vane, θ  n max =

= 0.597 = 60%.

Curved Vanes on Wheel Mass of jet striking vane

ρaV

w

V Force on the jet ρaV (V – u) [

u ρaV [(V – u) – (–(V – u) Cos θ)] Cos θ]

Work done by the jet on plate per second = Fx u = eaV (V – u) [ K0 E0 per second of jet =

Cos θ] u

ρav3

Efficiency of jet,



n=

(

=

(V

)(

Water Resource Engg

)

u)(

Cos θ)

For efficiency to be maximum  V = 2u.

=0 Hence, nmax = = ( n

( (

) )

Cos θ)

Cos θ) (

(i)

(

For θ

Cos θ) 0

i.e. for semicircular vane

nmax = 0.50 = 50. (ii)

For θ

, nmax = x 2 = 1 or 100%

Force Exerted By a Jet Of Water On An Un – Symmetrical Moving Curved Vane

vw

u2 V

ϕ

v2 β

V

β

ϕ θ

v1 u1

∝ f

∝

V

V

Note (i ) Angles are measured from negative direction of motion for turbines and positive direction of motion for pumps.



Water Resource Engg

(ii) Angles measuring anti-clock wise direction are taken as positive and these in clockwise are negative. As the jet strikes tangentially, the loss of energy due to impact of jet will be zero. Effective velocity of jet at (1) = V1 – u = Vr1 If the vane is smooth u1 = u2 = u & Vr1 = Vr2 (i)

For β <

0

– Vr2 Cos θ

–(u + V

Fx = ea V [(V

u)

= ea V [V (ii)

For β

(iii)

{ (u

(u

V

ρa V [(V

u)

ρ aV [(V

ρa V [ In general, Fx Work done

)}]

0

as V ( u)]

For β = 900, – V Cos ϕ = (u Fx

V

V ]

V Cos ϕ Fx

will be positive

u){ (u

will be positive. ρaV V V ) as V will be negative.

V )}]

V ]

ρa V [

V ]

ρau V [V

V ]

Change in K.E. of jet issued = * Pa V (V ) = Efficiency, η

( (

) )

(V u =

Pa (V ) +

V u ) (V

V )

Rapidly Varied Flow  Rapidly varied flow has vary pronounced curvature of steam lines.  Separation zone can be one or here.  Classification of star…… hydraulic jump THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 205


Water Resource Engg

1) Pre jump F1 < 1.0 2) Undular jump 1.0 < F1 < 1.7 Sequent depth ratio is very small, 3) Weak jump 1.7 < F1 < 2.5,

=0

is about 5%

4) Oscillating jump 2.5 < F1 < 4.5 = 4.5% 5) Steady jump 4.5 < F1 < 9.0 Ranges from 45% to 70% 6) Strong or choppy jump F1 > 9.0 > 70%  The pressures at the toe of the jump at d at the end of the jump fellow hydraulic pressure distribution. Velocity profile in the jump in Rectangular Channel

Reverse flow xx

x x

Forward flow

xx x

yr

x y 1

x maximum velocity

un

Boundary Layer Region 8

Efficiency The ratio of the specific energy after the jump to that before the jump is defined as the efficiency of jump. =

(

) (

)



Water Resource Engg

Turbines → Fluid Machines - Hydraulic Energy into mechanical energy. → Power generated by turbine, P

γ H Nm S

Water horse power to the turbine W.H.R =

metric H.P

= 13.33QH mHP. = 13.33 QH x 0.736 = 9.8QH [KW]

→ Types of Efficiencies 1. Leakage or volumetric efficiency 2. Hydraulic efficiency. 3. Mechanical efficiency 4. Overall efficiency 1. Leakage or Volumetric efficiency ηv =

2. Hydraulic Efficiency ηm = ηm = 3. Mechanical Efficiency Ηm =

=

4. Overall Efficiency ηO

ηm ηn ηv

Classification of Turbines



Name Pelton Wheel Francis Turbine

Type

Type of Energy

Impulse

Kinetic

Reaction Turbine

Kinetic + pressure.

Head

Discharge

Water Resource Engg

Direction of flow

Specific Speed

High head > Low 250 to 1000 m

Tangential to Runner

Low < 35 Simple 35 – 60 multiple jet

Medium 60m 150m

Radial flow

Medium 60 to 300

Medium to

10W < 30 m

Mixed flow High

Axial flow

High 300 to 1000

Pelton Wheel Breaking Jet: To stop the runner in a short time, a small nozzle is provided which directs the jet of water on the back of bucket with which the rotation of the runner is reserved. This jet is called as Breaking jet. Working Properties of Pelton wheel (Turbines) 1. Ideal velocity or theoretical velocity of jet is called sprouting velocity = √ gH Actual velocity of jet V = CV √ gH CV = 0.97 to 0.99 2. The maximum velocity of wheel u = 0.5V in actual practice u = 0.46V = 0.46 x 0.98 √ gH = 0.45 √ gH = Ku √ gH Here Ku →

to 0.47.

3. Least diameter of jet, d=[

√

]

4. Mean diameter of pelton wheel is called pitch dia



Water Resource Engg

u= (

D=

)

√

5. Jet Ratio = = m→

to

generally m = 12 6. No. of Buckets = ( ) + 15 or 5.4( ) 7. Angle of deflected jet, β

0

to 1700.

 Cut in the bucket is provided so that the bucket exactly opposite the jet gets full jet striking on it and is not intercepted by the lower portion of bucket that follows. Velocity triangle of Pelton wheel

ϕ

V1

β

Vr Inlet Velocity triangle V 1 = VW u

V

(V

u)



Water Resource Engg

Outer Velocity Trainable V

V2

ϕ u2 = u β Fx

0

V

β

V Β

ϕ u2 = u

V V negative

ρ (V

V2 0

V

V V

ϕ u2 = u β

V2

V 0,

V +ve

V Cos β)

Work Done = Fx u ρ u (V

u)(

u)(

Cos ϕ)

ρ (V u=

Cos ϕ)

V

Hence, Pmax = (

Cos ϕ)

Hydraulic efficiency, ηh =

(

)(

) ⁄

=2

)(

(

Cos ϕ)

= Mechanical efficiency, ηm = =

(

)(

)

Overall efficiency η

ηh x ηm

for Pelton wheel ηm →

– 99%

ηh → 8 – 90%



Water Resource Engg

Synchronous Speed The turbine should also run at constant speed N at all so adding conditions and turbines are designed for this speed. f= f = 50 N = P = No. of pairs of poles for generator. Some other types of Impulse Turbine (i) Double overhung pelton wheel turbine Two pelton wheels are provided on a single shaft for high speed or greater power. (ii) Multiple jet pelton turbine The maximum number of jet so far used in some larger units is six. The power of such wheel will be (n x P) where n is the number of jets and P is power due to single jet. Reaction Turbines (i)

Radially outward flow reaction turbine

(ii) Radially inward flow reaction turbine. (iii) Mixed flow reaction turbine: Similar to inward flow reaction turbine. The direction of water is turned from radial at exit to axial at outlet. (iv) Axial flow reach turbine: Water flows parallel to axis of rotation. The lower end of the shaft is made larger to form the boss or the hub. → When the vanes are composite with boss the turbine is called Propeller turbine When the vanes are adjustable the turbine is called a Kaplan Turbine. Design of Francis Turbine Runner (i)

Assume suitable values of ηH (8 to

(ii)

Obtain Q from P =

%)

ηO (80 to 90%).

( W)

Or (HP)



Water Resource Engg

(iii) Ratio of Runner width to runner diameter us disproved by ‘n’ which varies from 0.45

to

(iv) Flow Ratio, √

→ (v)

or V

√ gh

to

The area at inlet runner. V =

πD1b1K1 assume, K1 = 0.95, K1 = Space occupied by the thickness of

(

)

(vi) u1 = ϕ√ gh ϕ = 0.60 to 0.9, ϕ = Speed Ratio. u1 = (vii) Assume hydraulic efficiency for Radial flow, ηH = (viii) Guide blade angle at inlet - ∝, the vane angle at inlet, tan ∝

Tan θ

(ix) For flow to be radial at exit of vanes, β tan ϕ =

=

Run Away Speed When external load on the turbine drops to zero and if governing system also fails, then turbine runner will race up and will attain. This limiting speed of runner is called Runaway speed. Pelton Turbine = 1.8 to 1.9 times normal speed. Francis Turbine = 2 to 2.2 times normal speed. Kaplas Turbine = 2.5 to 3.0 times normal speed.



Water Resource Engg

Unit & quantities 1.

Unit speed, Nu =

2.

Unit discharge, Qu =

3.

Unit power, Pu =

√

√

⁄

Specific speed of a Turbine (NS) The specific speed of a turbine is the speed at which the turbine will run when developing unit power under a unit head. NS =

√ ⁄

Performance of Turbine (i) Performance under unit head Qu =

√

√

√

√

Pu = Nu =

(ii) Performance of turbine of same type Unit turbine which is a turbine having the runner diameter of 1m and operating wider a head of 1m. uu =



√

 Nu = N

√

√

Similarly, Qu = and Pu =

√

⁄

(iii) Performance of turbine of different types Here an imaginary turbine called specific turbine is defined. This is a turbine which is identical in shape, geometrical proportion, blade angle and gale opening etc. as actual turbine but reduced to a size that … Develop HP under unit head NS =

√ ⁄

Model Testing of Turbines P = f1 (P, , μ, g, H, D, N) Since velocity is very high, Re will be very large so that effect of viscosity can be ignored.



Water Resource Engg

P = f2 (ρ, , g, H, D, N) = Flow or discharge number. = Head number. = Power Number. Eliminated from head number and power number. ⁄

(

) ⁄

N = Non – dimensional specific speed or shape number.

⁄

(

)

(

)

(

)

(

)

( (

) √ ⁄ (

)

or (

) (

( )

)

√

(

√ ⁄ (

)

(

⁄

)

√

⁄

)

(

⁄

⁄

)

)

Cavitation in Turbines Thomas’ cavitation Factor, ς ς Ha = Atmospheric pressure head. H = Vapour pressure head. H = Section pressure head.

Rotodynamic Pumps → Pumps : Convert mechanical energy into hydraulic energy. → Rotodynamic pumps: The rotodynamic pumps increase the energy of fluid due to kinetic energy, pressure energy & centrifugal action.

The rotodynamic pumps can be THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 214


(i)

Centrifugal (or radial) pumps.

(ii)

Propeller (or axial) pumps.

Water Resource Engg

(iii) Mixed flow pumps. → positive displacement pumps (i)

Reciprocating type.

(ii)

Rotary type.

Suction: The pressure at inlet to the pump is generally suction or negative pressure and its value is limited to avoid cavitations. To reduce axial thrust, the suction is often branched into two parts and liquid is allowed to center from both sides of impeller, pump is then called double suction pump. (i) Static head (h) h=hs + hd

V2/2g hs hd H

h hf

(ii) Friction head (hf) hf = hfd + hfs hfd = friction head in delivery hfs = friction head in section. (iii)Total Head(H) H = h + hf + (iv) Manometric head (Hm) It is the head against which a pump in to work. Hm = Hmd – Hms



Hs + Hms +

=

0 + Hmd +

= hd +

Hmd = hd + Hm = hd + = hd + h d

hd

(1) h (c

+

hd hd hs

=h+h

Water Resource Engg

D) (2)

hs

h

h H

Note: While H is total energy, Hm is pressure energy only.

Efficiencies (i) Manometric efficiency 𝛈mo 𝛈mo=

√

x 100

(ii) Mechanical efficiency 𝛈m It is the ratio of available head at impulse to the energy given to impeller by the prime moves. Its values ranges from 95 to 98% (iii)Volumetric efficiency 𝛈v 𝛈v = (

) x 100

Where q is the leakage discharge. 𝛈v is about 97 – 98% (iv) Overall efficiency 𝛈O 𝛈O =

√

[In a dimensional form]

Pumps in series (multistage) and in parallel (i) If the head to be developed is higher Here two or more impeller is series. Such pumps are called multistage pumps.



Water Resource Engg

In such a case the total head developed H is the sum of heads developed by individual impellers Hi H=∑

H

Where n is the number of impellers (ii) If discharge regiment is large When two or more pumps is parallel if Qi is the discharge is individual pump the total discharge will be Q=∑ Reciprocating Pumps Main Parts of a Reciprocating Pump

Discharge through a Reciprocating Pump Discharge of the pump per second, Q = Discharge in one revolution x No. of revolution per second. =  Weight of water delivered per second. W

ρxgx

=

 Work done by Reciprocating Pump



Water Resource Engg

Work done per second = weight of water lifted per second x total height through which water is lifted. = W x (hs + hd) Where (hs + hd) = Total height through which water is lifted.74 Weight, W W= Work done per second =

(hs + hd)

∴ Power required to drive the pump in KW. P= (

P

)

(

)

,

W

 Work done by simple – acting reciprocating pump Q=  Work done by double – acting reciprocating pump Q= Slip of reciprocating pump Slip = Qth – Qact Percentage slip = =(

)x

x = (1 – Cd) x 100

Where Cd = coefficient of discharge (

C )



Water Resource Engg

4.2 Hydrology Hydrology deals with the occurrence distribution and movement of the water in earth, including that in atmosphere and below the surface of earth The hydrological cycle



Run – off : The portion of HYDROLOGY precipitation which by a variety of paths above and below the surface of earth reaches the stream channel.



The quantities of water going through various individual paths of the hydrological cycle can be described by continuity equation known as water – budget equation on hydrological equation.



Water budget equation for a catchment/storage mass inflow mass outflow = change in mass storage P–R–G–E–T

S → Change in storage.



The area of land drawing into a stream or a water course at a given location is known as catchment area / drainage area / drainage basin / watershed.



Infiltration does not occur in water budget as infiltration within a low to runoff purpose is a fine to the ground water system.



Runoff co-efficient =



The average duration of a particle of water to pass through, a phase of the hydrological cycle is known as residence time of that phase.





Water Resource Engg

Precipitation – denotes all forms of water that router that earth from the atmosphere. Eg., rainfall, hail, frost and dew Forms of Precipitation 1. Rain – 0.5 mm < drops size < 6 min Light rain – intensity 0 – 2.5 mm/hour Moderate rain - intensity 2.5 – 7.5 mm/hour Heavy rain - intensity > 7.5 mm/hour When the rain fall exceeds 2.5 mm, the day is called a raining day 2. Snow – consists of ice – crystal average density – 0.1 g/cm3 3. Drizzle → fine sprinkles of drop size < 0.5 mm and intensity < 1mm/h 4. Glaze → when rain or drizzle comes in contact with cold grounded at around . The water drops freeze form ice-coating which is called glaze 5. Sleet → frozen rain drops formed when rain falls through air at subfreezing temperature 6. Hail → lumps of ice with size > 8mm formed due to vertical movement of air current at sub freezing temperature

Index of Wetness Rainfall in a given year at a given place Average annual rainfall of that place 60% index of wetness means rain deficiency of 40%   

30 to 40% deficiency large deficiency 45 to 60% deficiency serious deficiency >60% deficiency disastrous deficiency

Aridity index An index used to define deficiency AI

PET AET PET

PET: Potential evapotranspiration AET: Actual evapotranspiration (Amount of water consumed by the plant under existing conditions) 

Front – Interface between two distinct air masses



Cyclone – Large low phrase reign with ………… wind when winds anticlockwise in northern hemisphere pressure increases outwards. Anti Cyclone – Wind clockwise in northern. Southwest monsoon (July – September)

Summer (pre – monsoon) (Mar – May)



Water Resource Engg

Post monsoon (Oct – Mar). Winter season (Dec – Feb) Non recording rain gauges: Symon’s gauge follows IS 4986 – 1968 Recording Gauge :Tipping Bucket type – gives data on intensity of rainfall Weighing Bucket type – mass curve of rainfall. Natural Sypson type – (flow type) (IS: 5235 – 1969) Yelemeteroy Rain Gauges. N=( ) Optimal number of stations ε = Allowable degree of emos in estimate of mean rainfall Cv = Coefficient of variation of rainfall values at existing m stations. Cv =

ς ̅ P=

̅

√

∑ (P m

̅) P

(∑ P )

Normal Rainfall: Is the average value of rainfall at a particular date, month or year over a specified 30 year period. If normal annual precipitation at various stations are within arithmetic average periods. Pn = [P1 + P2 + P3 ……

% of normal annual at station x →

Pn]

Else, normal ratio method Pn =



Water Resource Engg

Evapotranspiration Formula for Evapotranspiration 1. Lake Evaporation : c pan evaporation pan coefficient 0.8 for ISI pan 0.7 for class A pan 2. Empirical formulae Meyer’s formulae e )*

v

E

k (e

E e e v k

Evaporation in mm/day Saturated vapour pressure of air in mm of fly Actual vapour pressure of air in mm of fly Monthly mean wind velocity in km/h at about g about the ground Coeffcient

+

0.36 (for large deep water) 0.50 (for small & shallow waters) ( ) Where v

is the velocity of air at a height h

Formulae for Evapotranspiration penmann’s equation AH A

PET

E γ γ

Where PET

Potential evaoptranspiration in mm/day

A

Slope of saturation vapour pressure

v

Temperature curve at the mean air temperature in mm of mercury per Net radiation in mm of evaporable water per day Psychrometric constant = 0.49 mm of mercury C.

Infiltration Variation of infiltration capacity

with time



Water Resource Engg

Cumulative infiltration capacity ∫ f (t)dt

f

Horton’s equation of infiltration f

f

f

(f

f )

for

t

t

Infiltration capacity at any time t from the start of rainfall

f ,f

Infiltration capacity at t

and t

t

f is also called constant rate or ultimate infiltration capacity

(

⁄ )

Hydrograph



Water Resource Engg

Elements 1. 2. 3. 4.

Rising limb AB The crest segment BC Falling limb CD Lag time (T ): The time interval from the centre of mass of rainfall to the centre of mass of hydrograph 5. T → Time base Factors Affecting Runoff 1. Characteristics of precipitation A. Type of precipitation. E.g. rain snow etc.. B. Rain intensity C. Duration of rainfall D. Rainfall distribution E. Soil moisture deficiency F. Direction of prevailing storm 2. Characteristics of catchment A. Shape → Fan shaped catchments gives high peak and narrow hydrographs. B. Size → Peak discharge ∝ where A is the area of catchment and n is a constant . The time base for larger catchments will be larger. C. Slope steeper slope, large D. Drainage density → Ratio of total channel length to the total drainage area Larger drainage density results in pronounced peak discharge E. Land use → Vegetable cover reduces the peak flow Snyder’s Method of Synthetic Hydrograph (

1. Where,

) Basin log in hours

Basin length measure along the water course from the basin divide to the gauging station in km. Distance along the main water course from the gauging station to a point opposite to the water centroid in km A regional constant representing watershed slope and storage effects. Modified Equation (Linstey’s Equation) ( Where,

√

)

and n are basin constants

Basin slope



Standard duration

Water Resource Engg

hours of effective rain fall

Peak discharge of a unit hydrograph of standard duration 8c A t Where A

Catchment area in km

c

A regional constant

If a non-standard rainfall duration 

Modified basin lag



Modified peak discharge 8



Time base (



8

is adopted instead of

, then

)

Width of unit hydrograph at 50% peak discharge 8 Width of 75% of Where



Water Resource Engg

4.3 Irrigation Irrigation may be defined as the science of artificial application of water to the land, in accordance with the crop requirements. Types of Irrigation 1. Surface irrigation 2. Sub-surface irrigation 1. Surface irrigation a) Flow irrigation → In which the water is supplied from a higher level to a lower level by the action of gravity alone. b) Lift irrigation→ If water is lifted up by external power input such as pumps and then supplied 2. Sub-surface irrigation a) Natural sub-irrigation→ Due to leakage from channels etc. b) Artificial sub-irrigation → When a system of open jointed drains is artificially laid below the soil, the sub-surface irrigation is called artificial sub-irrigation 1. Free or Ordering Flooding → Also called wild flooding → Suitable for steep land → Sub sidy ditches are generally spaced at 20 to 50 meters. → Initial cost of land preparation is low and labour requirement are very usually high.

2. Border Flooding: In this method, the land is divided into a number of strips, separated by low levees called borders. → The land areas confined in each strip of the order of 10 to 20 meters in width and 100 to 400 meters in length. → log ( ) Q = Discharge through supply ditch y = Depth of water flowing over border strip f = Rate of infiltration of soil A = Area of land strip to be irrigated. t = Time required to cover the given area



Water Resource Engg

Supply Channel

Area dA covered With water in time Df

Area A Covered with water in time t

Border Strip

3. Check Flooding → Vertical interval of about 5 to 10 cm → The confined plot areas varies from 0.2 to 0.8 hectare → This method is suitable for more permeable soils as well as for less permeable soils. Coenings Ditch

Levees along the coenings

Connecting Leaves 4. Basin flooding: This method is a special type of check flooding and is adopted specially for orchard tress.

Suesidiary Oitches

Enter A water from bank hole or by a hose syphon

Main Ditch

TREES

Basing

5. Furrow Irrigation method → Furrows vary from 8 to 30 cm deep and may be as much as 400 meters long. → Deep furrows are widely used for row crops. → Small shallow furrows called corrugations



Forrow Or Field

50WN Crop

Water Resource Engg

50WN Crop

6. Sprinkler Irrigation method → Water is applied to the soil is form of a spray through a network of pipes and pumps. → Used for all types of soil and for widely different topographies & slopes. → This is very costly method. The conditions favoring the adoption of this method 1. 2. 3. 4. 5.

Topography is irregular Gradient is steeper. Soil is Excessively permeable Seasonal water requirement is low When the water is available with difficulty

Types 1. Permanent System → piper are permanent buried 2. Semi – permanent system → not permanent 3. Portable system. → In semi-permanent system, the main lines are buried in ground, while laterals are portable. → In portable System: The mains as well as laterals are portable. Advantages of sprinkler irrigation (i) Optimum quantity of water is used in this method. (ii) Land leveling is not required. (iii) Avoid surface runoff. (iv)Upto 80% efficiency can be achieved. (v) Fertilizers can be uniformly applied, because they are mixed with irrigation water itself.

Limitations (i) (ii)

High winds may distort sprinkler pattern. In areas of high temperature and high wind velocity, considerable evaporation lasses of water may takes place.



Water Resource Engg

(iii) They are not suited to crops requiring frequent and larger depth of irrigation, such as paddy. (iv) Initial cost is very high. (v) It requires larger electrical power. (vi) Constant supply of water is requires. (vii) Heavy soil with poor intake cannot be irrigated efficiently. Drip irrigation method: also called Trickle Irrigation  

Latest field irrigation technique. In this method water is slowly and directly applied to the root zone of plants thereby minimizing the losses by evaporation & percolation. This system involves lying of a system of head, mains, sub mains, laterals and drop nozzles. Water oozes out of there small drip nozzles uniformly and at a very small rate, directly into the plant roots area. The head consists of a pump to lift water, so as to produce the desired pressure of about 2.5 atmospheres. This method is however, being used for small nurseries, orchard, or gardens.

     



Crop Period or Base Period The time between the first watering to its last watering Duty and Delta of a crop a) Duty: Area of land irrigation for full growth of a given crop by supply of a unit discharge of water continuously during the entire base period of that crop. b) Delta: The total depth of water (in cm) required by a crop to came to mutant Relation between duty and delta = cm Where → delta (cm) D→ Duty (ha/cumec) B →Base period (days)

Factors on which duty depends 1. Type of crop : Duty will be less for a crop requiring more water and vice versa 2. Climate and season: duty includes the water lost in evaporation and percolation, so duty will be more in rainy season than in summer 3. Type of soil: Less for sandy soil, due to percolation 4. Efficiency of cultivation method: less for Less efficient cultivation method   

Crop season’s and Indian agriculture 1. Rabi: 1st October to 31st march (wheat, barley, gram, linseed, mustard etc…) 2. Kharif: 1st April to 30th September (Rice, Bajra, Jowar, Maize, Cotton, etc..) Paleo irrigation: First watering to crops before Kar watering : First watering



Water Resource Engg

Irrigation Efficiencies 1. Efficiency of water conveyance (η ) =

x 100

Where, W = water delivered to the farm W water delivered from the river or reserve’s 2. Efficiency of water application (η ) = x 100 Where , W = water stored in the root zone during the irrigation 3. Efficiency of water storage (η ) η = x 100 Where, W →as above W = water needed in the root zone, prior to irrigation 4. Efficiency of water use (η ) η = x Where, W =water used bentically or consumptively W = water delivered 5. Water distribution efficiency (η ) η =( ) x 100 D mean depth of water stored during irrigation D = average of the absolute values of deviations from the mean Consumptive Irrigation Requirement (CIR) Amount of irrigation water required in order to meet the evapotranspiration needs of the crop during its full growth ∴ CIR Consumptive use – effective rainfall Net irrigation Requirements (NIR) Amount of water required to meet the evapotranspiration need of the crop as well as other needs, such as leaching So, NIR = CIR + water lost as percolation in satisfying other needs, such as leaching



Soil- Moisture- Irrigation Relationship

Soil zone or root zone Soil Moisture W.T

Intermediate zone Capillary zone Ground water Impervious strata

Moisture content (m.c)



Field capacity moisture content Optimum m.c Available m.c or capillary water

Non available m.c or hygroscopic water

Waiting point m.c

Time

1. Field capacity: The water which can’t be easily drained by gravity So, field capacity is the water content of a soil after tree gravity drainage has take place Field capacity consists of two parts. a) Capillary water: Water attached to the soil by surface tension against gravity and can be extracted by plant by the action of capillarity



Water Resource Engg

b) Hygroscopic water: The water attached to the soil by chemical bonds and is not available to plants Depth of water stored in the root zone will fill the soil upto field capacity = xdxF Where,

    

F = Field capacity m.c D = Depth of root zone r = Unit wt of water r = Dry unit wt of soil Permanent wilting point: The stage at which plant can longer extract sufficient water for its growth and wilts up Readily available moisture: That portion of available moisture which is most easily extracted by the plants and is approximately 75 to 80% of the available moisture Crop ratio = Gross command area (GCA) Total area lying between drainage boundaries which can be commended by a canal system Culturable command area (CCA) The area on which corps can be grown satisfactorily CCA = GCA – unculturable area

Sediment transport and design of irrigation channels Bed Formation (Practical aspect)  The channel bed may be distorted into various shapes by moving water, depending upon the discharge or the velocity of water.  There will be different shape as the velocity is increasing.  When the velocity is gradually increased, then first of all a stage is reached, when the sediment load comes just at point of motion. This stage is known as threshold stage of motion. 1. After further increase in velocity after threshold stage of motion →saw tooth ripples → saw tooth ripples.

2. After further increase in velocity after saw tooth ripples → Dunes with ripples.

→ Ripples do not occur if the size of the bed particle is coarser than 0.6 m. 3. After Dunes with Ripples → Dunes. → Flow condition is subcritical is Dunes & Dunes with Ripple.



Water Resource Engg

4. Dunes → Flat surface.

5. Further increase in velocity, results in formation of sand caves in association with surface waves

6. As the velocity further increased so Froude no. F>1 flow is supercritical and the surface waves become so steep that they break intermittently and move stream and move up stream, although the sediment particles keep on moving down stream only.

→ Direction of movement of bed forms in this regime is opposite to that of dunes. 

Average unit attractive force on the bed of a channel Z = r Rs On the sides average Z = 0.75 r Rs Where, R = ( ) S = Channel bed slope

 Design of stable channels 1. ennedy’s Theory The sediments in the channel are supported by eddies. If the velocity is sufficient to generate these eddies, silting will be avoided, This critical velocity V = C y Where C and C are constants depending upon silt charge C = 0.55, C = 0.64 ∴ V = 0.55 y THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 232


Water Resource Engg

Lateral a factor depending upon the type of soil was introduced, called critical velocity ratio (CVR) Hence, V my m → CVR Design Procedures 1) Calculate critical velocity V 2) Determine dimensions of the channel section 3) Calculate actual velocity by using any of the following formula i)

utter’s Formula V

*

ii) Minning’s Formula V iii) Chezy’s Formula V 2.

(

)

(

R

)

+ √Rs

√

S

C √RS

Lacey’s Theory He introduced the concept of initial regime and final regime (true regime) and said that a channel which is under initial regime is not a channel in true regime. He is also argued that eddies are produced from sides also and not only from bottom

Design Produced For Lacey’s Theory 1. Calculate velocity from V = (

)

(m/s)

Where, f = Silt factor = 1.76 √D Dmm = Average particle size in mm G = Discharge in cumecs 2. Find hydraulic mean depth (R) R= ( ) Lacey’s regime scour depth R = 1.35 ( ) Where q = discharge per unit width 3. Compute are of the channel section A= 4. Compute wetted perimeter: P = 4.75 √ 5. Determine slope :- S = Diversion head works Constructed across a canal to divert the water, e.g. weir and barrages Layout of a diversion head works



Guide bank

Marginal bund joined to high contours

River flow

Divide roes Off-taking canal

Water Resource Engg

Under sluice portion Normal weir portion i.e weir proper

Weir divided into bays with piers Fish ladder Types of weir i) Masonry weir with vertical drop ii) Rock-fill weirs sloping aprons iii) Concrete weir sloping glacis (modern weir) 

River Training Works These are required near the weir site in order to ensure a smooth and axial flow of water and thus to prevent the river from outflanking The river training works required on a canal head works i) Guide banks ii) Marginal bunds iii) Spurs or groyne Marginal bunds are provided on the upper side of the works in order to protect the area from submergence due to rise in FSL



Canal Head Regulator Provided at the head of the off taking canal to control the flow and hence is provided with gates Functions of Head Regulator i) Regulates the supply of water entering the canal ii) Control the entry of silt in the canal iii) Prevents the river floods from entering the canal.



Silt control Devices a) Silt excluders – constructed on the bed of the river, upstream of the canal head regular. It removes the silt before entering the off talking canal. b) Silt extractor silt ejector:



Water Resource Engg

Extracts the silt, already entered into the off taking canal and is d/s of the head regulator Theories of seepage and design of weirs and barrages  Failure of hydraulic structure, founded on pervious foundation a) By piping (undermining) b) By uplift : Lifting or cracking of the concrete or masonry floor due to seepages pressure BLIGH’S Creep theory Creep length = d + B + d 1) Safety against piping L = CH L = Creep length C Bligh’s creep coefficient H = Total head of water retained by the weir →For no piping I < 2) Safety against uplift pressure of pucca floor Thickness at any section T= Where, h = ordinate of H.G.L above the surface of G = sp gravity of concrete talking a factor of safety = ⇒ t 

Khosla’s Theory → The seepages takes place dong streamlines defined by → Critical exit gradient G = Where



=

√ √ ∝

+

=0

x∝

The three simple profiles which are used for determination of pressure and exit gradient are i) A straight horizontal floor of negligible thickness with a sheet pile line on upper side or down side end

E

C C

b

E

b

d

d D ϕ = 100- ϕ ϕ

= 100 - ϕ

D ϕ = cos (

)

ϕ = cos

(

)

√

Where ∝ ii) Φ A straight horizontal floor with negligible with a sheet pile line at some intermediate point THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 235


ϕ = cos

(

ϕ = cos

( )

ϕ = cos

( √

Where,

) ) √

∝

Where, ∝ =

Water Resource Engg

∝

√

√

∝

E

b

∝

C d

and ∝ =

D

iii) A straight horizontal floor depressed below the bed but without any vertical cutoffs ϕ = (ϕ ϕ )+ ∝

ϕ = 100 - ϕ

d D 

D

1

b

Empirical formulas by Bligh for design of weir 1. For weir proper portion a) For weirs having crest shutters L = 2.21C √ b) For weirs having crest shutters L 2.21C √ c) For weirs having crest shutters L + L = 18C √ d) For no crest shutters L

L = 18c √

e) L = Crest shutter if any

u/s talus

L

d

d/s talus

B L

2. For under sluice portion a) For under sluices having crest shutters L = 3.87 C √

L

d L

and L + L = 27C √

b) For under sluices having no crest shutters



Water Resource Engg

L = 3.87 C √ L + L = 27 C √ Design Of Weir Wall Top width → B = √

Bottom width (B) of weir wall should not be less than B= √

Spillways Energy Dissipaters and Spillway Gates → A spillway is a structure constructed at dam site, for effectively disposing the surplus water from u/s to d/s → A spillway can be located either within the body of the dam or at one end of it or away from it Various Types of Spillway a) Based on permanent features 1. Straight drop spillway or over fall spillway 2. Over flow spillway or ogee spillway 3. Chute spillway or trough spillway or open channel 4. Side channel spillway 5. Shaft spillway 6. Siphon spillway b) Based on utility 1. Main spillway 2. Emergency 1. Straight drop or over fall spillway Characteristics → Constructed on small bends or on arch dam → Ventilation of nappe is required → Fatigue of surface occurs due to cavitations → Serious erosion is caused d s if no apron is provided Tree over fall under gate

Underside of the nappe to be ventilated

2. Agee spillway or overflow: → Least suitable for earthen dams → General equation for the profile of spillway for vertical u s face x = H y where H = dessignal head including velocity head x and y are co-ordinates with origin at crest ‘c’ Discharge Q = C L H Where, L = Effective length of spillway crest THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 237


Water Resource Engg

C = Co-efficient of discharge H = Total head over the crest (including the velocity head)

Designed head

Under nappe

Lower nappe

Sharp crested weir

3. Chute Spillway or the trough spillway → Can be adopted on gravity and earthen dams e g Bhakra dam Normal pord level

Tile drains surrounded gravels



Water Resource Engg

4. Side channel spillway → Suitable for earth or rock fill dams in narrow canyons and for other places where directed flew is not possible

Reservior

Spillway

dam Chate channel

d/s river

5. Shaft Spillway → Can be used where, inadequate space is there for other spillways Control lein

Vertical pipe Called shaft

Horizontal tunnel

Water joining the river d/s



Water Resource Engg

6. Syphon Spillway Instead of allowing water to spill over the crest of a dam or weir, the surplus water is discharge by a syphon spillway consisting of one or more siphon units Air vent

Normal pool level

Tail water level

Syphon inlet

Two types of Syphon Spillway 1) Saddle siphon spillway 2) Volute syphon spillway Discharge through saddle spillway Q = C A √ gh [C

g usually]

Where A = Area of cross section at crown H = Operating head Canal fall When available natural ground slope steeper than the designed bed slope of the channel canal falls are to be provided at certain intervals. Requirement for design of a fall 1. Velocity of approach should be minimum 2. Should be able to adapt the variation of water level in the canal 3. Should be safe against erosion, piping and uplift Types of Falls 1. Ogee fall → Not used now, due to drawdown E.g. Ganga canal (olden days)



Water Resource Engg

Draw down Water surface U/s bed

Rubble masonry

2. Rapid Falls Long slopes – concreted or stone pitched so uneconomical

3. Trapezoidal notch falls →Draw down eliminated → Discharge can also be measured

Cross section

4. Well type or siphon well drop u/s W.L →Better suited, when drops required are high but discharge is low U/S W.L

Soil

d/s W.L

Slump well



Water Resource Engg

5. Sarda Type Fall or vertical Drop type → Economic →Simple construct → No clear hydraulic jump →Can’t be flumed U/S H.F.L

D/S H.F.L Floor of water cushion

Stone pitching

D/S bed

Stone pitching

Cistern

6. Straight Glacis Fall U/s H.F.L d/s H.F.L

2

1

1

5

→ A Straight glacis is provided → Can be flumed or unfulmed → Hydraulic jump occurs on glacis METHODS OF IRRIGATION

ii)Sub-surface irrigation

i)Surface irrigation

1.Flooding

i)Wild Flooding

a)Free flooding

2.Furrow method

iii) Sprinkler irrigation

3.Contour farming

ii)Controlled flooding

b)Contour laterals

c)Border d)Checks or strip Flooding

e) Basin floodin g

f)Zig-Zag method



Environmental Engg

Part – 5: Environmental Engineering 5.1 Water Quality & Standards   

Maximum daily consumption = 1.8 x Average daily demand. Maximum hourly consumption = 2.7 x Annual average hourly demand. Unit processes for water treatment, Aeration → Screening → Sedimentation → Disinfection ← Softening ← Filtration

Sources of Municipal Water Supply 1. Surface Source -

Ponds & lakes Streams and rivers Storage reservoir

2. Subsurface Source Characteristics of water 1. Physical characteristics i. Turbidity: Due to suspended matters like clay, silt etc. measured on silica scale Measure by: Turbidity meters A. Jackson turbid meter : Used to measure high turbidity.Longer the light path lower is the turbidity 10.8 cm 200 JTU 21.5 cm 100 JTU B. Nephelometer: Used for turbidity of range o to 1 ppm Units → NTU: Nephelometer turbidity unit FTU for mazin turbidity unit I.S value for drinking water is 10-25 NTU ii. Colour: Due to decaying vegetation or some inorganic colored soil. Algae etc. Measured by: Platinum cobalt method Permissible limit: 5 to 20 ppm 5 to 25 cobalt unit iii. Taste and odour: Due to dissolved organic matter or inorganic salts, dissolved gasses Measured by: osmoscope Units: Taste FTN (Flavor threshold number) Odour: TON (threshold odour number) Permissible limit : 1 to 3 FTN iv. Temperature: At high temperature C and some other volatile gases are expelled loading to decrease in palatability (taste) desirable limit 5 to 12 v. Specific conductivity: Due to dissolved ions Measured by :Dionic water tester



Environmental Engg

Chemical Characteristics 1. PH Due to bicarbonates of ca and Mg and carbonates ( ), hydroxides (OH) of ca, Mg, K, Na. Measured by: potentiometers, colorimetric method Units: Measured on scale Desirable limit: 6.5 to 8.5 2. Hardness Due to bicarbonates & carbonates (temporary hardness), Non-carbonates (permanent hardness) Measured by: EDTA test (Ethyl diamine tetra – acetic acid) Units: of C C 

Hardness Temporary due to carbonate and bi – carbonates of calcium and magnesium Permanent due to presence of sulphates, chlorides and nitrates.



Total hardness = Ca2+ (mg/L) x



Hardness limits





+ Mg2 (mg/L) x

-

If hardness≤ 75 ppm: Soft water

-

If hardness b/w 75 – 200 ppm: Moderate water.

-

If hardness > 200 ppm: Hard water.

Total hardness (TH) = carbonate hardness (CH) + non carbonate hardness (NCH) -

If TH> alkalinity then CH = Alkalinity

-

If TH < alkalinity then CH = TH

Alkalinity measured in mg/l of CaCO3 HC

com ining weight of CaC com ining weight HC

com ining weight of CaC com ining weight C



Chloride: Content detected by AgNO3 solution with KMnO4 as indicator.



Permissible limit 250 mg/L Nitrogen content Different forms 1. Free ammonia: Indicates the presence of un-decomposed organic matter limit < 0.15 ppm THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 244


Environmental Engg

2. Albuminoid nitrogen: Indicates that decomposition of organic matter has started. Limit.< 0.3 mg/L 3. Nitrites: Indicates presence of partly decomposition organic matter. Permissible: Nil 4. Nitrates: Indicates fully oxidized organic matter, permissible limit < 45 ppm Total kjeldahl nitrogen (TKN) = free ammonia + organic nitrogen Bacterial and Microscopic Characteristics 1. Aerobic Bacteria → Required oxygen survival 2. Anaerobic Bacteria → Do not required dissolve 3. Facultative Bacteria → Can survive with or without  Through some species of bacteria it may be helpful in cleaning of water but other pathogenic bacteria are harmful  The presence of pathogenic bacteria can be tested by counting presence of coil forms Measurement of Coliform 

Filter the water through 30 pore size and cultivate the coloness and count the number.



MPN: Most Probable Number.



Environmental Engg

5.2 Water Supply and Its Treatment

Type of Water Demand 1. 2. 3. 4. 5.

Domestic water demand (55 to 60% of total water demand) Industrial water demand (50 lpcdlitre per capita/day) Institutional and commercial water demand (20 lpcd) Demand for public uses (10 lcpd) Fire demand (11 lpcd) When population exceeds 50,000 Then water required √ Where V is in kilo litre and P is in thousand Kuichling & formula: Q = 3182√ where P is population in thousand.

Variation in Demands 1. Maximum daily demand average daily demand 2. Maximum hourly demand average hourly demand 3. Maximum hourly demand in summer average hourly demand We can obtain these values using Goodrich’s formulas Where P = percent of annual average demand for the time t in days Population Fore Casting 1. Geometric Increase method Pn = PO[

]

Where r = r1 x r2 x r3 … xrn. 2. Average (arithmetic) method P = PO + nx̅. 3. Incremental increase method n n nx̅ y̅ A combination of arithmetic increase method and geometric method the average of increase in population is found from arithmetic method and is added to the average of net incremental increase 4. Decreasing method ̅ D [ ][

̅ D

]… [

̅ nD

]

Where ̅ is the rate of decrease in population growth rate . THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 246




Environmental Engg

Basic units for surface water treatment

Raw Water

Screening PST

Disinfection

Filtration

SST

Coagulation Flocculation

1. Screen: Velocity through screen < 0.8 to 1m/sec 2. Sedimentation: The velocity of flowing water is reduced nearly to zero and the sediments in water are allowed to settle by gravity 

Settling velocity, Vs =

(G – 1)

Important Formulae Pertaining to Sedimentation 

Surface overflow rate =



Settling velocity Vs = Where t =



% of particle that can be removed in sedimentation tank =

x 100%

Where, V` = Velocity of settling from slope’s law V = H/t 3. Coagulation and Flocculation is generally done by addition of Al3+ and Fe 4. Filtration: Help in removing color, adour, turbidity and pathogenic bacteria a. The slow sand gravity filter b. The rapid sand gravity filter



Environmental Engg

Comparison of Slow Sand and Rapid Sand Filters Items

Slow sand filter

Rapid gravity filters

1) Pre treatment

Effluent either from plain sedimentation tank or raw water without any treatment are generally fed into them and coagulation is not at all required

Coagulation flocculation and sedimentation is a must

2) Base material The gravel base supports the sand. It varies from 3 to 65 mm in size and 30 to 75 cm in depth 3) Filter sand

4) Size of each unit

5) Rate of filtration

The effective size of filter sand ranges between 0.2 to 0.4 mm and uniformity coefficient between 1.8 to 2.5 or 3.0 The grain size distribution is generally uniform throughout the depth of filter media except that top 10 to 15 cm may be laid of finer variety

The gravel base support the sand and also distributes the wash water uniformly on the surface of sand. It varies from 3 to 40 mm in size and its depth is slightly more than i.e. about 60 to 90 cm

The effective size of the filter sand rages between 0.35 to 0.55 and uniformity coefficient between 1.2 to 1.8 The sand in layers with smallest grain size at top and corset grain size at the bottom

6) Efficiency

7) Post treatment

8) Method of cleaning

Large such as (30m × 60m) the area varying from 100 to 200 sq. m or more.

Small, such as 100 to 200 liters per hour per sq.m of filter are (50 to 60 ML/Ha/day) Very efficient in removing bacteria (98 to 99%) but less efficient in removing color turbidity removal is upto 50 ppm

Small, such as 5m × 8m. The are varying from 10 to 80 sq.m

Large such as 3000 to 6000 liters per hour per sq.m of filter are (1500 to 3000 ML/Ha/day) Less efficient (80 to 90%) very efficient in color removal



Almost pure water is obtained. It may be disinfected slightly

Scrapping and removing the top 1.5 to 3 cm thick layer and washing down by 0.2 to 0.6% of total filtered water is required.

Environmental Engg

Disinfection is must

Backwashing with or without compressed air large amount (1 to 5%) of total filtered water is needed for washing in every 1 to 3 days.

5. Disinfection: To remove harmful bacteria left after filtration methods a. Boiling b. Treatment with excess lime kills bacteria but increases alkalinity.Hence recarbonation is required c. Treatment with ozone: → (nascentoxygen) kills bacteria d. Chlorination: Dose forms hypochlorous acid which kills bacteria 

Break point chlorination: The point beyond which all the chlorine added will appear as free chlorine is called break point chlorination DPD test is used for determination of breakpoint.

Residual chlorine

Super chlorination: Addition of 5 – 15mg/l of chlorine

B

A Applied chlorine

6. Aeration: Water is brought in contact with atmospheric air to promote exchange of gases between water and air. C

H Sand other volatile substances importing taste and odour to water are easily expelled



Environmental Engg

7. Water softening i. Removal of temporary hardness a. Boiling b. By adding lime ii. Removal of permanent hardness a. Lime soda process b. Zeolite process c. Demineralization process a. Lime soda process - Lime Ca H and soda A C are added - We get CaC and H o seperated - Economical and better for excessively hard water - But large quantity of slued is produced and requires recarbonation b. Zeolite process - Zeolite is a complex compound of Al, silica and soda - Achieving almost zero hardness is possible - But cannot be adopted for highly turbid water c. Demineralization - Similar to zeolite process but the metallic ions in this method are replaced by hydrogen’s ions rather than y sodium ions in zeolite process - Water obtained is free from minerals and quality is almost same as distilled water Layout of Water Distribution Network Layout of Distribution Network  The distribution pipe are generally laid the road pavement, and such their layout generally follow the layout of the roads  We can also develop the pavement duct along the road  There are in general four different type of pipe network are There are four types of system for distribution of water 1) Dead end system 2) Grid iron system 3) Ring system 4) Radial system. Dead end system  It is suitable for old town and cities having no definite pattern of road Advantage 1) Relatively cheap 2) Determination of discharge and pressure easier to due to less number of values THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 250


Environmental Engg

Disadvantage 1) Due to many dead end stagnation of water

Grid iron system  It is suitable for cities with rectangular layout where the water mains and branch are laid in rectangle Advantage 1) Water is kept in good circulation due to the absence of dead ends. 2) In the cases of breakdown in some section water is available from some other directions. Disadvantage 3) Extra circulation of sizes of pipes is not possible due to provision of valves on all branches.



Environmental Engg

Ring system  The supply mainly is laid the peripheral road and sub mains branch out from the mains  Thus this system also follows the grid iron system with the flow patter similar in character to that of dead end system  So determination of the pipe size is so easy Advantages  Water can be supplied to any point from at least two direction Main pipe M M

M

S

S

Sub mains

S M

M

S

S M

M

M



Environmental Engg

Radial System  The area is divided into different zones  The water is pumped into the distribution reservoir kept in the idle if each zone the supply pipes are radially laid ending towards the periphery Advantage  It gives quick service  Circulation of pipe size is easy



Environmental Engg

5.3 Waste Water Treatment Aerobic Decomposition: 

In presence of facultative bacteria and aerobic bacteria Anaerobic Decomposition 1. Nitrogenous organic matter →

: N2↑

H2↑

2. Carbonaous organic matter →

Co2 + Heat

Heat

Note: Organic acids including alcohol are converted to CH4, Co2, etc. 

Total solids: 1000 Kg of sewage contains 0.45 Kg of total solids.



Properly oxidized sewage has pH around 7.3



Humans excrete about 6gm of chloride / person / day.



High chloride in waste water indicate presence of sewage water.



Nitrogen Content  Free ammonia indicates first stage  Nitrate indicates last stage.



Fats, oil and grease are soluble in either.



Sulphides, sulphates and H2S: Sulphide compounds are oxidized to So42 – which is unolyctionble when aerobic decomposition takes place but results in formation of H2S, CH4, Co2 when decomposed aerobically.



Dissolved oxygen  Ensure atleast 4 ppm of D.O. in it for survival of fish and other spaces.



B.O.D. or biological oxygen demand gives the amount of active organic matter whereas C.O.D (Chemical Oxygen Demand) gives total oxygen demand.



B.O.D5: B.O.D at end of 5 days at 200C. B.O.D5 = D.O. consumed at end of five days x dilution factors. Dilution factor =



First stage demand is carbonaceous demand and last stage demand is nitrogenous demand.





Environmental Engg

= – RLt Lt : oxygen equivalent of carbonaceous oxidiable organic matter present in sewage after t days.  Loge = –Kt  Log10 = – 0.439Kt  Log10 = – KDt Where, Lt = Organic matter left after t days.



B.O.D after t days is yt = L – Lt = L – L



BOD5 is 6 % approximately of total B



Total organic carbon =

D at t

∞

= 2.66 TOC (for glucose)

Ratio is important factor for wastewater 

BOD5 = 20 mg/l – permissible limit. F



Total oxygen demand is amount of oxygen required to burn organic substances and some minor inorganic substances.



Primary, secondary and tertiary treatment of wastewater.



Preliminary treatment reduces BOD by 15.50%



Primary treatment: Removes large suspended organic solids.



Environmental Engg

Biological treatment

Primary treatment 

Raw waste Screening & Grit water removal  1. Floating 2. Sedimentation

 1. Activated sludge 2. Tricking filter 3. Anaerobic lagoons

Primary treatment (biological)

Sedimentat ion

Tertiary treatment Final effluent

Disposal

Sludes digestion



Environmental Engg

5.4 Air Pollution Components 1. Emission sources → IC engines, burning of coal and oil 2. Transmitting media→Atmosphere 3. Receptor → People, plants, animals Receptor/ medium remediation not possible/ difficult so control at source is best suited Categories of air pollutions 1. Primary pollutants: Emitted directly from the sources 2. Secondary pollutants: Formed in the atmosphere by physical/chemical/biological interactions among primary pollutants and normal atmospheric constituents

General classification of the gaseous air pollutants Class

Primary

Secondary

Sulphur containing Compounds

SO2, H2S, Etc

SO3, H2SO4, MSO4

Nitrogen containing Compounds

NO, NH3, Etc

NO2, MNO2

Oxides of carbon

CO, CO2, etc

None

Halogen None

HCL, HF

One

CO2 is normally not considered as pollutant, but increased concentration worldwide is the basis for the concern regarding its eventual effects.

Another Classifications 1. Gaseous pollution 2. Particulate pollution a. Dust: Formed by mechanical is integration of material crushing grinding etc. - Fine dust < 100m dia - Coarse dust >100m dia b. Smoke: By condensation of supersaturated vapor composed of materials of low vapor pressure in relatively high concentration c. Fumes: Particles formed by condensation, sublimation size < 1 d. Mist: Suspension of the liquid droplets formed by condensation of vapor size < 10 m



Environmental Engg

Particulate Matter and Its Effect •

Particles less than 0.1 m undergo random motion (Brownian motion resulting from collision with individual molecules)

•

Particles between 0.1 to 1 m have settling velocities in still air.

•

Particles larger than 1 m have significant but small settling velocities.

•

Particles above 20 m have large settling velocities and are removed from air by gravity and other inertial forces.

•

Consequently these particles are airborne for less time.

•

Particulates serve as condensation nuclei and influence the formation of clouds, rain and snow.

•

Visibility: Resulting from absorption and scattering of light by airborne liquid and solid materials.

•

Carbon dioxide, water vapor and ozone change the absorption and transmission characteristics of the atmosphere.

•

Reduction in the visibility not only is unpleasing to an individual, but also may have strong psychological effects.

•

In addition some safety hazards arise due to reduction in visibility at airports, on highways, etc.

•

Effect on Materials: Particles get deposited on painted surface, clothing, curtains, etc.

•

particulate matter can cause damage by intrinsic corrosiveness or by the action of corrosive chemicals absorbed/adsorbed by inert particles into the atmosphere.

•

hygroscopic particles commonly found in atmosphere can corrode metal surfaces

•

Effect on the Vegetation and Animal Health: Very little is known of the effects of particulate matter in general on vegetation.

•

However, some specific substances cause damage to the plants. For example, the fluoride containing particles damage the plants.

•

Polluted vegetation may cause diseases in animals, for example, animal fluorosis through vegetation, arsenic poisoning in animals through vegetation.

•

Effects on human Health: Particulate matter enter through respiratory system, deposit in the respiratory tract and damage respiratory organs.

•

Particulate matter may exert a toxic effect in one or more of the following three ways.



Environmental Engg

Effects of gaseous air pollutants •

Carbon Monoxide: Colorless and odorless gas.

•

It is very stable and has a lifetime of 2-4 months in the atmosphere.

•

CO is removed from atmosphere by soil fungi and CO is oxidized to CO2, though at a very slow rate.

•

No significant material damage or effect on the plants or vegetation.

•

The combination of O2 with hemoglobin leads to oxyhemoglobin, O2Hb.

•

Hemoglobin has a affinity for CO that is approximately 210 times of its affinity for O2.

•

100 ppm CO – Many people experience dizziness, headache.

•

Cigarette smoke contains about 400-500 ppm CO, % of COHb in blood increases as the number of cigarettes smoked per day.

Zone of atmosphere

Mixing depths and inversion •

The vertical extent to which the mixing takes place varies diurnally, from season to season, and is also affected by topographical features.

•

The greater the vertical extent, the larger the volume of atmosphere available for dispersion of the pollutants.



Environmental Engg

•

The depth of the mixing layer is known as the Mean Mixing Depth. Atmospheric conditions influence the way thermal plumes behave.

•

Behaviour of plumes under different conditions

•

Case 1- Looping: This occurs when n > 1.4 and atmospheric conditions are unstable

•

Case 2 – Coning: This occurs when n is between 1 and 1.4 and atmospheric conditions are unstable

•

Case 3 – Fanning: This occurs when n < 1 and inversion takes place

•

Case 4 – Lofting: This occurs when the inversion layer lies below the stack height

•

Case 5 – Fumigation: This occurs when the inversion layer lies above the stack height

•

Case 6 – Trapping: This occurs when double inversion layers occur

•

Case 7: Neutral: This occurs when n = 1 and atmospheric conditions are stable

DALR

Turbulent air

WARM SEASONS WITH CLEAR SKIES

ELR(n>1.4)

Z

UNSTABLE

LOOPING

T°C NIGHT TIME WITH CLOUDY & WET CLIMATE ELR(n:1-1.4) STABLE

Z

Moderate wind Axis of plume

DALR

CONING

T°C Plan

Z

ELR(n<1) Light wind INVERSION

NIGHT TIME AND EARLY MORNING FANNING

Elevation

T°C EVENING TIME IN CLEAR SKIES Moderate wind RADIATION INVERSION BELOW STACK

Z

LOFTING

T°C

INVERSION ABOVE STACK

LATE MORNING TO EARLY AFTERNOON IN CLEAR SKIES MONTLY OF SUMMER Light to moderate wind

Z FUMIGATION

T°C


Quick Refresher Guide DALR

Light to moderate wind ELR (subsidence inversion)

Z

Environmental Engg

ANY TIME OF THE DAY IN ANY SEASON

TRAPPING

T°C

DALR

Light wind

ELR (n-1.4) Z

CLEAR SKIES IN NEUTRAL ATMOSHPERE

Neutral atmosphere

SEUTRAL PLUME

T°C

Looping DALR

Lofting

ELR

Fanning Lofting Fumigation

Trapping Fanning

Control of air pollution 1. 2. 3. 4.

Setting chambers: Works by gravity Inertial separators: Battles and lauvres are used to intercept the effluent’s path Cyclones : Cyclonic action to separate dust particles Filters:

Mechanical Shaker-Baghouse

Reserve-Air Baghouse

Reserve Jet-Baghouse

Filter cleaning methods 1. Online cleaning I.

Automatically timed filter cleaning which allows for continuous, uninterrupted dust collector operation for heavy dust operations.



Environmental Engg

2. Offline cleaning I. II.

Filter cleaning accomplished during dust collector shut down. Allows for maximum effectiveness in dislodging and disposing of dust.

3. On-demand cleaning I.

Filter cleaning initiated automatically when the filter is fully loaded, as determined by a specified drop in pressure across the media surface.

4. Reverse-pulse/Reverse-jet cleaning I.

It delivers blasts of compressed air from the clean side of the filter to dislodge the accumulated dust cake.

5. Impact/Rapper cleaning I. II.

In this high-velocity compressed air forced through a flexible tube results in a random rapping of the filter to dislodge the dust cake. Especially effective when the dust is extremely fine or sticky.

5. Electrostatic precipitators 6. Scrubbers or wet collectors Solid waste Solid waste in dry refuse and includes garbage ashes, rubbish, dust etc. Density varies from 300 g⁄cm to 600 g⁄cm a. Garbage: Putrescible organic wastes e.g.: vegetable, pills, kitchen generated waste b. Rubbish: Non putrescible wastes, e.g. rags, papers, broken glass & furniture, card boards etc. Transformation methods used for the management of solid waste Transformation Processes used for the management of solid waste

Transformation Process

Transformation means or method

Transformation or Principal conversion Products

Physical Component Separation

Manual and/or mechanical separation

Individual components found in commingled municipal waste



Environmental Engg

Volume reduction

Application of energy in the form of force or pressure

The original waste reduced in volume.

Size reduction

Application of energy in the form of Shredding, grinding, or milling

The original waste components altered in form and reduced in size.

Thermal oxidation

Carbon dioxide (C02), sulphur dioxide (S02), other oxidation products

Chemical Combustion

A gas stream containing a verity of gases, tar and or oil and a char Pyrolysis

Destructive distillation

Gasification

Starved air combustion

A low-Btugas, Char containing carbon, and pyrolytic oil

Aerobic biological conversion

Compost (humus-like material used as a soil conditioner)

Anaerobic digestion (lowor high-solids)

Anaerobic biological conversion

Methane (CH4), carbon dioxide (C02), trace gases, digested humus or sludge

Anaerobic composting (Occurs in landfil)

Anaerobic biological conversion

Methane (CHJ, carbon dioxide (CH2), digested waste

Biological Aerobic compositing

Various methods of refuse disposal 1. Sanitary land filling The refuse is dumped into the low lying area and it is covered by good earth layer. The waste gets stablised with 2 to 4 months by aerobic as well as anaerobic processes Leachate: Liquid collected at bottom of a landfill Land fill gas: 90% of the gas produced is CH and C 2. Incineration: Burning the wastes method C tar charcoal etc may gets produced 3. Barging it out into the sea: At a considerable distance away from the coast 4. Pulverisation: Refuse is pulverized in grinding machine so as to reduce its volume THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 263


Environmental Engg

5. Compositing : Can be done under aerobic or anaerobic conditions or both a. Indore method: Uses manual turning of piled up mass (refuse + night soil) under aerobic conditions. T stabilizes the material in shorter time and needs less space no odorous gases produced, so environmental friendly b. Bangalore method: Anaerobic, required longer time, larger space, produces adorous gases,. But does, not involve any turning or handling of the mass hence more clean than indoor method. Evaluation of hazardous contaminant present dc dt

c

or ln

C C

t

Where C C

Concentration at time zero and ‘T’ Rate constant

t

6



Transportation Engg

Part – 6: Transportation Engineering 6.1 Highway Planning



Stopping sight distance is the minimum distance available to stop a vehicle travelling at design speed safely without collision with any other obstruction. IRC has recommended the height of eye level of driver as 1.2m and the height of object as 0.15m above the road surface for measuring the stopping sight distance.



Stopping distance of a vehicle is sum of lag distance and braking distance. Lag distance = 0.278 vt. Braking distance =

Design Speed: The design speed of roads depends upon class of road and terrain. Terrain have been classified as plain, rolling, mountainous and steep depending on the cross slope of the country as given. Cross Slope (in %)

Terrain classification

0 – 10

Plain

10 – 25

Rolling

25 – 60

Mountainous

Greater than 60

Steep

Horizontal Curves: To avoid overturning and lateral skidding on a horizontal curve, the centrifugal ratio ( ) should be less than and f. Super Elevation: The transverse inclination the pavement surface is known as super elevation and is denoted by ‘e’. e+f= IRC recommends that maximum limit of super elevation in plain and rolling terrain as 7% and 10% in hill roads.  Extra widening as curves is given by e=

+

. √



Transportation Engg

Where, n = number of traffic lanes. l = length of wheel base of longest vehicle. v = design speed (in kmph). R = radius of curve (in meters). Length of Transition curve The length of transition curve should be the maximum of three values derived based on following three conditions. (i)

Rate of change of centrifugal acceleration to be developed gradually.

(ii)

Rate of introduction of the designed super elevation to be at reasonable rate.

(iii)

Minimum length as per IRC formula.

 Length of transition curve as per first condition is given by LS =

.

Where C =

m/Sec2 [0.5 < C < 0.8]

 Length of transition curve as per second condition is given by. LS =

(W + We) for pavement rotated about centre line.

= eN (W + We) for pavement rotated about inner edge.  Length of Transition curve as per IRC empirical formula is given by LS = =

.

for plain and rolling terrains For mountainous and steep terrains.

 Shift of transition curve, S =  Grade compensation on horizontal curve is given by grade compensation (in %) = subject to maximum value of 75/R.  As per IRC the grade companion is not necessary for gradients flatter than 4%.



Transportation Engg

Length of Summit Curve For stopping side distance (SSD) can be calculated by using following formulae. (i)

L=

(√

√

)

when L > SSD.

Where, N = deviation angle S = Stopping sight distance H = Height of eye level of driver above roadway surface. h = Height of object above pavement surface. As per IRC standard. H = 1.2 h = 0.15 L= (ii)

.

(√

.

√

L = 2S –

(√

L = 2S –

.

√

) )

=

.

when L < SSD

Length of Summit curve: For safe overtaking sight distance (OSD) can be calculated by using following formulae (i)

L=

When L > S

As per IRC standard, H = 1.2 L=

(ii)

When L > S

.

2S = 2S =

When L < S .

(Taking H = 1.2m).



Transportation Engg

Length of Valley Curve It is designed based on (i)

Allowable rate of change of centrifugal acceleration of 0.6 m/sec3 and

(ii)

Head light sight distance, and the higher of the two value is taken.

 Length of transition curve, LS based on first condition is LS= 0.19 (NV3)1/2  Total length of valley curve = L = 2LS = 0.38 (NV3)1/2  Length of valley curve based on head light sight distance is calculated as: L=

.

for L > SSD.

.

L = 2S –

.

.

for L < SSD.

Traffic Capacity Studies  Traffic volume is the number of vehicles that are passing a given spot is a specified direction and specified unit of time.  Traffic density is the number of vehicles occupy a unit length of lane at a given instant. It is expressed as vehicles per kilometer.  Traffic capacity is the ability of a lane to accommodate traffic volume. It is a measure of maximum number of vehicles in a lane can pass a given point in unit time.  Theoretical maximum capacity of simple lane given by C= Where, V = Speed in kmph S = Average centre to centre spacing of vehicles. C = Capacity is vehicle per hour.  It is also given by C= Where Ht = minimum time headway (in secs). THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 268


Transportation Engg

 The fundamental relationship, between traffic volume, density and speed is given by q = KVs Where, q = average volume of vehicles passing a spot during a specified period of time. VS = space – mean speed (in kmph) K = average density (in vehicles/km).  The maximum speed is called free mean speed and is denoted by VSf. The maximum density occurs at zero speed and is called j am density which is denoted by kj. The maximum capacity occurs when the speed is qmax =

x

and density in .

=



Transportation Engg

6.2 Highway Materials

 Group Index of soil is used to classify the fine grained soils and for judging their suitability as subgrade material. GI = 0.2 a + 0.005 ac + 0.01 bd. Where, q = Portion of material passing 0.074 mm sierve, greater than 35 and less than 75%. It is expressed as whole number from 0 to 40. b = Portion of material passing 0.074 mm sierve, greater than 5 and less than 35%. It is expressed as whole number from 0 to 20. c = Value of liquid limit in excess of 40 and less than 60. It is expressed as a whole number ranging from 0 to 20. d = Value of plasticity index exceeding 10 and less than 30. It is expressed as a whole number ranging from 0 to 20.

California Bearing Ratio test (CBR)  It is a penetration test used for evaluating the stability of soil sub grade and other flexible pavement materials. .

CBR (in %) =

.

 Pavement thickness (t), can be determined using CBR value from the given formula. t=√ *

.

+

⁄

Where, t = Pavement thickness (in cm) P = Wheel load (in kg) CBR = California Bearing Ratio (in %) δ = Tyre pressure (in kg/cm2)

California resistance value method The pavement thickness can be determined using this method as per following formula T=



Transportation Engg

Where, T = Total thickness of pavement (in cm). K = 0.166 TI = Traffic index = 1.35 (EWL)0.11 R = Stabilometer resistance value. C = Cohesion meter value. Rigid Pavements  Radius of relative stiffness l=*

+

Where, E = Modulus of elasticity of cement concrete (in kg/cm2) μ = oisson ratio of concrete = 0.15 h = Slab thickness (in cm). K = Sub grade modulus (in kg/cm3).  Equivalent radius of resisting section, b = √1. a

h

0.

b=a

5h for a < 1.724h for a > 1.724h.

Where, a = radius of wheel load distribution (in cm). h = slab thickness (in cm). Westergaard’s stress equation for wheel loads  Interior loading Si =

.

* log

( )

1.0

( )

0. 5 +

+

 Edge loading Se =

.

* log

 Corner loading Sc =

[1

(

√

)

.

]



Transportation Engg

Where, h = Slab thickness (in cm). P = Wheel load (in kg). a = Radius of wheel load distribution (in cm). l = Radius of relative stiffness (in cm). b = Radius of resisting section (in cm). Si = Maximum stress at interior loading (in kg/cm2) Se = Maximum stress at edge loading (in kg/cm2) Sc = Maximum stress at corner loading (in kg/cm2). Warping Stresses St(i) =

*

+

St(e) =

or

St(c) =

√

whichever is higher

Where, St(i) = warping stress at interior (in kg/cm2) St(e) = warping stress at exterior (in kg/cm2) St(c) = warping stress at corner (in kg/cm2) e = thermal coefficient of concrete per 0C t = temperature difference between top and bottom of slab (in 0C) Cx = coefficient based on

in desired direction

Cy = coefficient based on

in right angle to desired direction

μ = Poisson’s ratio of concrete taken as 0.15 a = radius of contact l = radius of relative stiffness



Transportation Engg

Frictional stresses Sf = Where, Sf = unit stress developed in cement concrete pavement (in Kg/cm2). W = unit weight of concrete (in Kg/cm2). f = coefficient of subgrade restraint (taken as 1.5). L = slab length (in meters).

Spacing of contraction joints LC =

x 104

Where, LC = spacing between contraction joints (in cm). f = coefficient of friction (taken as 1.5). W = unit weight of slab in kg/m3. SC = allowable stress in tension in cement concrete (in Kg/cm2).  Spacing of contraction joints when reinforcements is provided is given by LC =

.

Where, As = total area of steel (in Kg/cm2 per meter width). b = slab width (in meters). h = slab thickness (in cm). SS = allowable tensile stress in steel (in Kg/cm2). Design of Tie bars  As = Where, As = area of steel per meter length (in cm2). b = distance between the joint and nearest free edge (in meter). h = thickness of pavement (in cm). SS = allowable working stress in tension for steel (in kg/cm2). W = unit weight of cement concrete (in kg/cm3).



Transportation Engg

 Total length of tiebar, Lt = Where, SS = allowable stress in tension (in kg/cm2). Sb = allowable bond stress in concrete. This is equal to 24.6 kg/cm2 in case of deformed bears and 17.5 kg/cm2 in plain bars. d = diameter of tie bar (in cm).



Surveying

Part – 7: Surveying Determination Of Relation Position And Plotting By Means Of Different Type of Measurement Plane: Curvature ignored

Surveying Geodetic:Curvature of earth taken into account Plane surveying Types 1. 2. 3. 4. 5. 6. 7. 8.

Chain surveying Compass surveying Plane table surveying Theodolite surveying Tacheometric surveying Triangulation surveying Aerial surveying Photogrammetric surveying

Map R.F: Representative factor Vernier scale: Least count-Difference between smallest division on main scale and the smallest division on vernier scale. Corrections (

)

(

)

Levelling: Determination of relative attitudes of the point or elevation of a point w.r.t benchmark Instrument used: Level, levelling staff, dumpy level Part of dumpy level 1. 2. 3. 4. 5. 6. 7. 8. 9.

Levelling head Telescope Eye piece Diaphragm screws Focusing screws Ray shade Longitudinal/Attitude bubble tube Bubble tube adjusting screws Cross bubble tube



Surveying

Technical term used a) b) c) d) e) f) g)

Level surface: Surface parallel to the mean spheroidal surface of the earth Horizontal surface: Surface tangential to the level surface Vertical line: Line perpendicular to the level line and the direction of plumb line Datum: Reference to which vertical measurement are done Mean Sea level Reduced level: Height above or below the assumed datum Line of sight: Line through the optical centre of the line and inter section of cross hair

Back sight: Dot observation made at the point of known elevation (used to obtain of instrument) Fore sight: Sight taken at a point of unknown elevation Classification of levelling 1. Simple levelling 2. Differential levelling (fly levelling or continuous levelling) instrument is set at different points and successive elevation difference is calculated Difference of elevation between points A and B =B.S. - F.S R.L. of point B= R.L. of point A  ( B.S. -  F.S) Booking and Reduction of level (A) Rise and Fall method:- B.S. - F.S= Rise-Fall=Last R.L- First R.L Check on intermediate sight is done as each is included in rise and fall calculation (B) Height of collimation method:- B.S. - F.S= Last R.L- First R.L No such check Gradient of line: Can be calculated by diving the elevation by the total length Classification of spirit levelling 1. 2. 3. 4. 5. 6.

Differential levelling Profile levelling  like road Check levelling Cross sectional levelling Precision Reciprocal levelling : Elevation determined by two sets of reciprocal observations



Surveying

Horizontal line a1

Line A collimators

Level Lines

a

b1

True difference of level between A&B ( ) ( )

B

( a2

)

(

)

Error is equal to the half of the difference of the apparent differences of level

b2

 Curvature and Refraction corrections 1. Curvature correction (Approx) =0.0785 d2metre Curvature correction is negative and always subtracted from the staff reading 2. Correction for refraction

0.0785D2 (curvature)

Line of collimation

0.0112d2 (Refraction)

A B

3. Combined correction for reparation curvature 0673 d2m(-ve) (d is in k.m)  Compass surveying Direction measured by compass Length measured by taping or chaining



Surveying

Traverse: series of connected lines. Types 1. Closed: Originates and terminates on same station 2. Open  Classification of traverses based on instrument used 1. Chain traversing or chain angles method 2. Compass traversing: Angular measurements by surveying compass 3. Plane table traversing: Plotted graphically on plane table using alidade 4. Theodolite traversing: Theodolite used. 5. Tacheometric traversing  Types of surveying compass 1. S y ’ m : Needle is of edge bar type and also acts as an index. Qradrantal bearing system used. Graduated ring is attached to the box and not to need lesightin and rading can be done simultaneously 2. Poismative compass: Needle in broad needle type but does not act as index. Graduated ring attached to the needle graduation in W.C.B system. Sighting and reading can be done simultaneously  Meridian: Reference from which angle is measured Types 1. True meridian: Line passing through true north and south 2. Grid meridian 3. Magnetic meridian: Line passing through that point and magnetic north and south  Bearing: Angle between reference meridian and the line, measured clockwise.  Azimuth: Smaller angle of the line from the true meridian  Magnetic bearing: Angle made with the magnetic meridian  Designation of Bearings 1. Whole circle bearings: Azimuthal system clockwise direction numerical value vary from 00 to 3600 N 1 E 2

3 S

2. Quadrilateral bearing system: (Reduced Bearings) Numerical value between 00 to 900. S Measured from north or south, which is nearest N

W

1

E 3

𝛉S

𝛉2

2



Surveying

 Fore and Back Bearings: Both expresses in W.C.B System and differ each other by 1800. The measurement of bearing in the direction of survey is called fore bearing Back bearing = fore bearing  1800

1 1+

N

Local attraction: Needle deflects due to local attraction like magnets, cable carrying current, iron, ores, etc. Error= observed back bearing-calculated back bearing Magnetic declaration: Horizontal angle between true north and magnetic north  Theodolite: Classification of theodolites 1. Transit theodolite : Telescope can be rotated through a complete revolution about its horizontal axis in a vertical plane 2. Non-transit theodolite 3. Vernier theodolite 4. Glass arc theodolites  Parts of transit theodolite 1. Levelling head 2. Lower plate 3. Upper plate 4. Standards 5. Telescope 6. Vertical circle 7. Plumb bob 8. Tripod

 Grades 1. Up grade or positive grade 2. Down grade or negative grade

Curves: Curve followed by highway or railway alignment is called a curve Types of curves

P

1. Simple curve: Curve of single are of a circle A C

B P

R

D O



Surveying

2. Compound curve: Two or more or is of difference circles with different rad Q P A

B R1

O1

R2

O

3. Reverse curves: Centre is on the opposite sides of the curve at turning point then in only one common tangent R1

O2

A

B

C R2 O1 B1 Vertex

I

T1 Backward tangent

-Angle of deflection

B

T2 /2 /2

A

Forward tangent R

R

Tangent length=R tan/2 Length of the long chord = T1T2=2T1E=2Rsin/2 Length of the curve, l=R(in radius) ⁄ (

) (



)

 Degree of curve: Defined w.r.t a fixed length of an arc of the curve or with respect to a fixed length of a normal chord of the curve 1. Based on fixed length of an arc: Degree of curve in the central angle of the curve, that is subtended by an arc of 30 in length 30m

R

00



Surveying

2. Based on fixed length of a chord  ⁄

Q

P

⁄

15 m 15 m 1 1

⁄ R

O

 Methods of curve ranging 1. Linear or chain and tape method 2. Angular or instrumental methods 

B

 Setting out curves 1. Chain and tape i. Offset from to long chord O  = +( )  √ ( ) √

(

√

√

O

⁄ )

B

Offsets from the tangents (a) perpendicular method  

T2

E

T1

a

ii.

A Q1

Q

(

)

Ox E T2

T1 N

√ √

iii.

Successive bisections of chords ( ) Or = =R(

O

2. Angular measurement methods i. Offset from chord produced: Adopted when a theodolite is not available ii. Ranking method of tangential deflection angles: Curve set out by the tangential angle. iii. Deflection angles from the point of curve and point of tangency using two theodolites



Surveying

B j j1

 A

O2

C1

R

 

2

O

(

)

(

)

(

)

 Tachometric surveying Horizontal & vertical distance are determined by using instruments. Instruments used is tachometer and it theodolite having diaphragm fitted with two additional wires fitted called stadia hairs

Methods of tachometric measurement 1. Stadia hair method i. Fixed hair method ii. Movable hair method 2. Tangential method 3. Substance bar (movable hair) 1. Stadia hair method i. Distance and devotion formulae for horizontal sights by fixed hair method (a) Horizontal distance of staff position i O b

A

B 1

i S

F a d

A 1

B D

( )

(

)

⁄



Surveying

(b) Elevation of the staff station: Elevation of the staff station= Elevation of the instrument axis - Control hair reading ii.

Distance and elevation formula for inclined sight by fixed hair method a. For inclined sight with staff vertical A1B1 = ABcos = Scos Inclined distance OC (

)

A1

A S B1 B h v

L d

E

O D

 Elevation formula (

)

Elevation of the staff station R.L. of staff station= R.L. of line of horizontal sight V-h  Distance and elevation formulae for inclined sight with staff normal Case 1: Line of sight at an angle of elevation (a) Horizontal distance formula L= KS+c  D= KScos +cos +hcos (b) Elevation of the staff station V= Lsin = (KS+c)



Surveying

A

L E1



O

h

B g

V

E

Case 2: Line of sight at an angle of depression (a) Horizontal distance formula D= (KS+c) cos -hsin (b) Elevation of the staff station V= (KS+C) sin

V L V D

The telescope used in stadia theodolitis are of the following three types 1. Internal focusing telescope 2. External focusing telescope [Theodolite with this is known as stadia theodolite] 3. Internal focusing telescope fitted with an analytic lens (c=0) [tacheometer]  Substance theodolite (Movable hair method) Staff intercept  constant Distance between stadia hairs  variable (

) (

)



Surveying

2. Tangential method Distance are computed by observing vertical angles of two point at fixed known distance S on the staff

S B

2

-S

1 D

(1)

S S S

(

V

)

S

(2)

Similarly for other cases results can be obtained 3. Substance Bar method

⁄ D S ⁄

Used where direct chaining becomes difficult S⁄ S⁄

S

(

⁄ S

(

) )



Surveying

 heory of Errors and Adjustment 1. Mistakes 2. Systematic errors 3. Accidental or random errors  True value: Free from all types of error  Observed value  True error= True value-observed value  Triangulation : Process of measuring the angles of a network of triangles Principles of triangulation If 3 angles and one side in known, remaining parameters can be obtained.

Classification of triangulation 1. Primary triangulation 2. Secondary triangulation 3. Tertiary or 3rd order triangulation or topotriangulation Correction 1. Correction for standard length L= measured length, l=Designed length

2. Correction for alignment (

)

(

B

(

)

)

(

A

2

1

)

l2

l2

Always subtractive If A&B are not indivisible, then √ B

3. Correction for scope (

Z

h

) A

D

Always subtractive



4. Correction for tension (

Surveying

)

Applied pull is more, tension correction is positive if less tension correction is negative 5. Sag correction (

)

6. Reduction to M.S.L



Reference Books

Reference Books Mathematics



Higher Engineering Mathematics By Dr. BS Grewal



Advance Engineering Mathematics By Erwin Kreyszig



Advance Engineering Mathematics By Dr. HK Dass

Geotechnical Engineering  Soil Mechanics and Foundation Engineering By K.R.Arora 

Foundation Engineering By P.C.Varghese



Principles of Foundation Engineering By Braja M.Das



Geotechnical Engineering By Venkatamaiah. C

Transportation Engineering  Highway Engineering By S.K.Khanna, C.E.G. Justo 

Highway Engineering By Rangwala



Principles and Practice of highway engineering By L.R. Kadiyali

Surveying  Surveying By Dr. B.C.Punmia, Ashok Kumar Jain, Kr. Jain 

Textbook of Surveying By Venkatramaiah.C



Surveying By A.R.Abhyankar, S.S.Rajput THE GATE ACADEMY PVT.LTD. H.O.: #74, KeshavaKrupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 288


Reference Books

Environmental Engineering 

Water Supply Engineering Environmental Engineering (Volume – 1) By SK Garg



Noise Pollution By Agarwal



Environmental Engineering By Ruth F.Weiner, Robin A.Matthews, P.Aarne Vesilind


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