Engg Mathematics and Sciences Formulas Series
Engg Mathematics and Sciences Formulas Series
Engg Mathematics and Sciences Formulas Series
Definitions
Theorems and Properties of Triangles
Area of Triangle
Rectangle
Square
General Quadrilateral
Parallelogram
Rhombus
Trapezoid
Cyclic Quadrilateral
Ptolemys Theorem
Polygons
Theorems in Polygons
Regular Polygons
Circle
Theorems on Circles
Area of Circle
Sector
Segment
Parabolic Segment
Spandrel of Parabola
Ellipse
Radius of Circles
Circumscribed about a Triangle
Inscribed in a Triangle
Escribed in a Quadrilateral
Circumscribed about a Quadrilateral
Inscribed in a Quadrilateral
Area by Approximation
Trapezoidal Rule
Simpsons One-Third Rule
Area by Coordinates
Polyhedrons
Regular Polyhedron (Platonic Solids)
Eulers Polyhedron Theorem
Prism
Rectangular Parallelepiped
Cube
Truncated Prism
Pyramids
Frustum of a Pyramid
Cylinders
Right Circular Cylinder
Cone
Right Circular Cone
Frustum of a Cone
Frustum of Right Circular Cone
Sphere
Spherical Segments
Spherical Cone
Spherical Lune and Wedge
Spherical Polygons
Spherical Pyramid
Solids of Revolution
Pappus Theorems
Ellipsoid
Prolate and Oblate Spheroids
Paraboloid of Revolution
Prismatoid
Prismoidal Formula
Volume of Some Prismatoid
Similar Solids
Limits
L Hospitals Rule
Short Technique on Limits
Differentiation Formulas
Algebraic Functions
Logarithmic and Exponential Functions
Trigonometric Functions
Inverse Trigonometric Functions
Hyperbolic Functions
Inverse Hyperbolic Functions
Slope of the Curve
Rate of Change
Curvature and Radius of Curvature
Circle of Curvatrure
Graph of a Function
Relative Maximum and Relative Minimum
Points of Inflection
Applications of Maxima and Minima
Steps in Solving Maxima & Minima Problems
Common Variable Relationship for Maximum
and Minimum Values
Time Rates
Integration Formulas
Algebraic, Exponential, & Logarithmic
Functions
Trigonometric Functions
Inverse Trigonometric Functions
Hyperbolic Functions
Integration by Parts
Trigonometric Substitution
Wallis Formula
Examples
Wallis Formulas
Double Integration
Triple Integration
Integration by Parts
Algebraic Substitution
Trigonometric Substitution
Integration by Partial Fraction
Plane Areas
Using Horizontal Strip
Using Vertical Strip
By Polar Coordinates
Area of Some Polar Curves
Length of Arc
Centroid of Plane Areas
Moment of Inertia
Polar Moment of Inertia
Product of Inertia
Mass Moment of Inertia
Properties of Common Shapes
Solids of Revolution
Using Circular Disk
Using Hollow Cylindrical Shell
Surface Area
Volume of Other Solids of Known Cross Section
Centroid of Volume
Work
Work to Stretch a Spring
Work in Winding Up Load
Chapter Eight: Differential Equation
Variable Separable
Homogeneous First Order Differential Equation
Linear First Order Differential Equation
Differential Equation
Bernoullis Equation Type
Finding the Differential Equation from a General
Solution
Applications of Differential Equation
Population Growth
Exponential Growth and Decay
Cooling and Heating
Flow Problems
Continuous Compound Interest
Motion Problems
Newtons Second Law of Motion
Statics
Force Systems
Resultant of Forces
Resultant of Forces in Space
Resultant of Parallel Forces
Resultant of Non-coplanar forces
Equilibrium of Forces
Cables
Parabolic Cables
Catenary
Cables Under Concentrated Loads
Friction
Belt Friction
Properties of Sections
Centroid
Center of Gravity of Flat Plates
Centroids of Composite Figures
Moment of Inertia
Polar Moment of Inertia
Moment of Inertia With Respect to Inclined
Axis
Mohrs Circle
Dynamics
Kinematics
Translation
Rectilinear Translation
Uniform Motion
Variable Acceleration
Constant Acceleration
Free-falling Body
Curvilinear Motion
Motion Curves
Rotation
Kinetics
Newtons Laws of Motion
D Alemberts Principle
Centrifugal Force
Conical Pendulum
Banking of Curves
Ideal Angle of Banking
Horizontal Rotating Platform
Work and Energy
Work and Energy Equation
Impulse and Momentum
Impulse and Momentum Equation
Law of Conservation of Momentum
Coefficient of Restitution
Simple Stress
Normal Stress
Shearing Stress
Bearing Stress
Thin Walled Pressure Vessels
Cylindrical Vessel
Spherical Vessel
Thick Walled Cylinders
Simple Strain
Stress Strain Diagram
Axial Deformation
Shearing Deformation
Poissons Ratio
Biaxial Deformation
Triaxial Deformation
Thermal Stress
Torsion
Helical Spring
Spring in Series
Spring in Parallel
Shear and Moment in Beams
Shear and Moment Diagrams
Moving Loads
Stresses in Beams
Radius of curvature
Flexure Formula
Shearing Stress
Superimposed Beams
Spacing of Rivets or Bolts in Built-up Beams
Economic Sections
Combined Stresses
Combined Axial and Flexure
Kern of a Section
Combined Axial and Shearing Stress
Mohrs Circle
Combined Torsional and Flexural Stresses
Beam Formulas (Moment, Deflection, and
Rotation)
Simple and Cantilever Beam
Propped Beamasasas
Fully Restrained Beam
Dynamic (Impact) Loading
Properties of Fluids
Unit Weight and Density
Specific Volume
Visocosity
Surface Tension
Capillarity
Bulk Modulus of Elasticity
Compression of Gases
Pressure Disturbances
Unit Pressure
Total Hydrostatic Pressure
On Planes Surfaces
On Curved Surfaces
Buoyancy
Statical Stability of Floating Bodies
Relative Equilibrium of Fluids
Rotating Vessels
Fluid Flow in Pipes
Reynolds Number
Energy Equation
Bernoullis Energy Theorem
Head Lost in Pipe Flow
Pipes in Series
Pipes in Parallel
Equivalent Pipe
Orifice and Tubes
Unsteady Flow
Weir
Cipolletti Weir
Triangular Weir
Suttro Weir
Unsteady Flow
Hydrodynamics
Drag Force
7. Distributive Law
a ( b + c ) = ab + ac
Basic Laws of Equality
1. Reflexive Property
a = a
2. Symmetric Property
If a = b, then b = a
3. Transitive Property
If a = b and b = c, then a = c. That is, things
equal to the same thing are equal to each
other.
4. If a = b and c = d, then a + c = b + d.
That is, if equals are added to the equals, the
results are equal.
Inequality
Theorems of Inequalities
1. a 0=0
2. If a b = 0, then either a = 0 or b = 0 or both
a and b are zero.
0
3. = 0 if a = 0
a
a
4. = undefined
0
a
5. =0
Laws of Exponents (Index Law)
1. an = a a a (n factors)
2. am an = am + n
am
3. = am - n
an
n
4. ( am ) = a m n
5. ( abc )n = an bn cn
a n an
6. ( ) =
b bn
m
n
7. an = am
1 1
8. a-m = and am =
am a-m
9. a0 = 1
10. If am = an , then m = n provided a 0
Properties of Radicals
1
n
1. an = a
m m
= am = a
n n
2. a n
n
a = a
n
3.
n n
a b = ab
n
4.
n
a n a
5. n = , provided that b 0
b b
Properties of Logarithms
Expanding Brackets
( a + b + c )( d + e ) = ad + ae + bd + be + cd + ce
Factorization
2x2 - 6x + 4 = 2 x2 + 3x + 2
= 2(x-2)( x -1)
1. ( x + y )( x - y ) = x2 - y2
2. ( x + y )2 = x2 + 2xy + y2
3. ( x - y )2 = x2 - 2xy + y2
4. ( x + y + z )2 = x2 + y2 + z2 + 2xy + 2xz + 2yz
5. x3 + y3 = ( x + y ) x2 - xy + y2
6. x3 - y3 = ( x - y ) x2 + xy + y2
3 3 2
7. x6 - y6 = x2 - y2 = ( x2 - y2 ) [ x2 +
2
x2 ) ( y2 + y2 ] = ( x + y ) ( x - y ) ( x4 +
x2 y2 + y4 )
Division of Polynomials
Example 1-1
Divide x4 - 10x2 - 9x - 20 by x 4
x3 + 4x2 + 6x + 15 remainder 40
x-4 x4 - 10x2 - 9x - 20 1. x4 x = x3
x4 - 4x3
6x2 - 9x 3. 6x2 x = 6x
6x2 - 24x
15x - 20 4. 15x x = 15
-15x - 60
Remainder 40
1 0 -10 -9 -20 4
4 16 24 60
1 4 6 15 40
Factor Theorem
Example:
Factorize 2x3 + 5x2 - x - 6
Solution:
Remainder Theorem
Example 1-2
Find the remainder when x4 - 10x2 - 9x - 20 is divided
by (x - 4).
Solution:
f(x) = x4 - 10x2 - 9x - 20
x-r = x-4,r = 4
Remainder = f(4) = (4)4 - 10(4)2 - 9(4) - 20 = 40
Example 1-3
Find k such that x-3 is a factor of kx3 - 6x2 + 2kx - 12.
Solution:
Remainder = f(3) =k(3)3 - 6(3)2 + 2k(3) - 12 = 0
k=2
Properties:
1. The number of terms in the expansion n + 1.
2. The first term is an and the last term is bn ,
3. The exponent of a descends linearly from n
to 0,
4. The exponent from b ascends linearly from
0 to n,
5. The sum of the exponents of a and b in any
of the terms is equal to n,
6. The coefficient of the second term and
second from the last term is n,
Pascals Triangle
rth term of ( a + b )n
n!
rth term= an - r+1 br+1
( n - r + 1 ) ! ( r - 1)!
n
r = + 1
2
Example 1-4
5
Find the 3rd term in the expansion of x2 + y .
Solution 1:
(Using the properties and Pascals Triangle)
5 5 4 3
x2 + y = ( x2 ) + 5 x2 y + 10 x2 y2
= x10 + 5x8 y + 10x6 y2
n!
rth term = an - r + 1 b r - 1
(n-r+1)!(r-1)!
r = 3, a = x2, n = 5, b = y
5! 5-3+1 3-1
3rd term = (x2 ) y
(5-3+1)!(3-1)!
6 2
=10x y
Example 1-5
Expand (x + y)8
Solution:
By principle, the first term is x8, the second term is
8x y. The variable part of the third term is x6y2. To get
7
the coefficient:
Example 1-6
Find the sum of the coefficient of the variables in the
expansion of (2x + 3y - z)8.
Solution:
Sum = [ (2)(1) + (3)(1) - 1 ]8 = 48 = 65536
Proportion
a c
a : b = c : d or =
b d
where:
b and c are called the means
a and d are called the extremes
d is the fourth proportional to a, b and c
Mean Proportional = ab
Properties of Proportion:
1. Proportion by Inversion
a c b d
If = then =
b d a c
2. Proportion by Alteration
a c a b
If = then =
b d c d
3. Proportion by Composition
a c a+b c+d
If = then =
b d b d
4. Proportion by Division
a c a-b c-d
If = then =
b d b d
Quadratic Formula
- B B2 - 4AC
x=
2A
B
Sum of the roots, x1 + x2 = -
A
C
Product of the roots, x1 x2 =
A
Partial Fraction
2x2 + 4x - 5
Proper Fraction:
5X3 + 6x2 - 2x - 1
3x2 - 2x + 1 4x2 - 2x + 3
Improper Fraction: ;
2x2 + 6 3x + 2
Improper fractions may be expressed as the sum of a
polynomial and a proper fraction.
12x2 - 13x - 9 5
For example, = 3x + 2 +
4x - 7 4x - 7
x-4
Proper fractions such as can be expressed as
2x2 - 4x
the sum of partial fraction, provided that the
denominator will factorize.
2 3 2 (2x - 4) - 3x x-8
- = =
x 2x - 4 x (2x - 4) x (2x - 4)
4x2 + 7x + 8 A B C D
3
= + + 2
+ 3
x (x + 2) x x + 2 (x + 2) (x + 2)
3 2
4x + 7x + 8 = A (x + 2) + Bx (x + 2) +
2
Cx (x + 2) + Dx
x4 - x3 + 14x2 - 2x + 22 A Bx + C Dx + E
= + 2 + 2
(x + 1)(x2 + 4)(x2 - 2x + 5) x+1 x +4 x - 2x + 5
x4 - x3 + 14x2 - 2x + 22 =
A(x2 + 4)(x2 - 2x + 5) + (Bx + C)(x + 1)(x2 - 2x + 5)
+ (Dx + E)(x + 1)(x2 + 4)
4x + 1 1 2
dx = dx + dx
2x2 + 5x - 3 2x - 1 x+3
Variation
Direct Variation
x is directly proportional to y:
x y or x = ky
k = constant of proportionality
Inverse Variation
x is inversely proportional to y:
1 k
x or x =
y y
Joint Variation
x is directly proportional to y and inversely
proportional to the square of z:
y ky
x or x =
z2 z2
Elements:
a1 = first term
an = nth term
am = any term before an
d = common difference
d = a2 - a1 = a3 - a2
s = sum of all terms
an = a1 + (n - 1) d or an = am + (n - m) d
n n
S= (a + an ) or S = [ 2a1 + (n - 1)d ]
2 1 2
an = a1 rn-1 or an = am rn-m
a2 a5
Common ratio, r = =
a1 a4
Sum of n terms of a G.P.
a1 (rn - 1)
S= , when r >1
r-1
a1 (1 - rn )
S= , when r < 1
1-r
Harmonic Progression
Example 1-8
Find the 12th term of the series 6, 3, 2.
Solution
The reciprocals are 1/6, 1/3, which forms an A.P.
with a common difference d of 1/6.
Work Problem
1
Rate=
Time to finish the work
Example 1-9
Solution:
Rate of A =
Rate of B = 1/8
Work done in 2 hours =
( 14)(2) + ( 18)(2) = 34
This means or 75% of the work was done.
Principle:
If 8 persons can do a job in 6 days, the
number of man-days to finish the job is (8)(6)
= 48 man-days
Thus, if 12 persons will do the job, it will take
them 48/12 = 4 days to finish it.
Example 1-10
A job could be done by eleven workers in 15 days. Five
workers started the job. They were reinforced with four
more workers at the beginning of the 6th day. Find the
total number of days to finish the job.
Solution
Clock Problem
Example 1-11
How many minutes after 2 oclock will the hands of the
clock be perpendicular for the first time?
Solution:
x = 10 + x12 + 15
x = 27.273 min.
Mixture Problem
Example 1-12
How many grams of gold must be added with 500
grams of an alloy containing 30% gold and 70% silver
in order to produce another alloy analyzing 40% gold
and 60% silver?
Solution
Represent the different mixtures by boxes
500 x 500 + x
30% G 40% G
70% S 100% G 60% S
Motion Problem
n!
P(n , r) =
(n - r) !
and P(n , n) = n !
Note: 0! = 1
Example 1-13
How many permutations can be made out of the letters
on the word DIEGO taken 3 at a time?
Solution:
n=5,r=3
5!
P(5 , 3) = (5 - 3)!
= 60 ways
n!
P=
q! r!
Example 1-14
How many permutations can be made out of the letters
in the word ENGINEERING?
Solution:
P = (n - 1) !
Combination
P(n , r) n!
C(n , r) = =
r! (n - r)! r !
and C(n , n) = 1
Example 1-15
How many ways can you draw 3 Queens and 2 Kings
from a deck of 52 cards?
Solution:
A deck of 52 cards has 4 Queens and 4 Kings, thus
C = C (4,3) C (4,2) = 24 ways
C = 2n - 1
Example 1-16
How many ways can you invite any one or more of your
five friends to your birthday party?
Probability
Single Event
f
q=
h+f
And
p+q=1
Example 1-17
For a single question in the board exams, there are
four choices and only one of which is correct. By
guessing, what is the probability that you will get the
correct answer?
Solution:
The event here is to get the correct answer and there
are four trials. Out of four trials, the event (correct
answer) can happen only once, and can fail 3 times.
Thus, the probability that the event will happen is
1 1
p= =
1+3 4
Multiple Events
P = P1 P2 P3
P = P1 + P2 + P3 +
Example 1-18
A box contains 4 blue chips and 5 red chips.
a. If one chip is drawn at random, what is the
probability that it is blue?
b. If two chips are drawn at random, what is the
probability that both are red?
c. If two chips are drawn at random, what is the
probability that one is blue and the other is
red?
Solution:
a. Single Event. There are four blue chips out
of nine chips. P = 4/9
b. Multiple Events. The events (getting red) are
to occur twice.
First draw red: There are five red chips out
of nine chips. P1 = 5/9
Second draw red: There are now only four
red chips out of eight chips. P2 = 4/8 or
Thus P = P1 P2 = (59)(12) = 15/18.
c. Mutually Exclusive. The event here is to get
a red and a blue ball in two draws. This can
happen in two ways (first draw red and
second draw blue) and another is (first draw
blue and second draw red), but these two
cannot happen in the same time, hence they
are mutually exclusive events.
First draw Red, Second draw Blue
5 4 5
P1 = =
9 8 18
First draw Blue, Second draw Red
4 5 5
P2 = =
9 8 18
5 5 10
Thus, P = P1 + P2 = + =
18 18 18
Repeated Trials
Example 1-29
There are ten questions in an examination. The
probability that an examinee will get the correct
answers is 0.25. What is the probability that he will get
(a) exactly 7, and (b) at least 7 correct answers?
Solution
a.) There are 10 questions, n = 10 with p = 0.25 and q
= 0.75.
The probability of getting exactly 7 is
P(10,7) = C (10, 7) 0.257 0.7510 - 7
= 0.00309 or 405/131072
b.) At least seven means can be exactly
7, 8, 9, or 10.
Pat least 7 = P(10,7) + P(10,8) + P(10,9) + P(10,10)
Pat least 7 = C (10, 7) 0.257 0.7510 - 7 + C
(10, 8) 0.258 0.7510 - 8 +C (10, 9) 0.259
0.7510 - 9 +C (10, 10) 0.2510 0.7510 - 10 =0.00351
P=1-Q
Example 1-20
The probability of getting a credit in each of the three
examinations is 0.65. What is the probability of getting
at least one credit?
Solution:
There are three trials (n=3) with p = 0.65 and q = 0.35.
Thus, the probability of getting no credit at all is Q =
(0.35)(0.35)(0.35) = 0.042875. Thus,
P = 1 - Q = 0.957125
Matrices and Determinants
Matrix
Classification of Matrices
1. Equality of Matrices
Two matrices are equal if they have the same
number of rows and columns and their
corresponding entries are also equal.
Example 1-21
Add
1 4 1 3 0 2
7 1 6 + 2 5 6
-3 0 4 9 1 1
3. Multiplication of Matrices
Multiplication of matrix can be done only if the
number of columns of the left-hand matrix is
equal to the number of rows of the right-hand
matrix. Multiplication is accomplished by
multiplying the elements in each right-hand
matrix column, adding the products, and then
placing the sum at the intersection point of the
involved row and column.
Example 1-22
2 (2)(2)+(1)(4)+(5)(1)
2 1 5 13
4 = =
1 4 7 (1)(2)+(4)(4)+(7)(1) 25
1
4. Division of Matrices
Division of matrices can be accomplished only
by multiplying the inverse of the denominator
matrix.
Example 1-23:
Determine the transpose of A
1 6 9 1 2 7
'
A = 2 3 4 ; Hence A = 6 3 1
7 1 5 9 4 5
Properties of Determinants
1 4 0
4 5 0 =0
1 7 0
1 2 5
4 6 1 = 0
1 2 5
1 4 6 1 2 1
2 5 2 = 4 5 7
1 7 9 6 2 9
1 4 5 1 22 5 1 2 5
4 6 1 = 4 32 1 = 2 4 3 1
2 8 4 2 42 4 2 4 4
1 4 2
5 1 10 =0
3 6 6
2 1 6 2 6 1
5 4 7 = - 5 7 4
1 3 9 1 9 3
1 4 5 1 4 5 + (1)(3) 1 4 8
4 6 1 = 4 6 1 + (4)(3) = 4 6 13
2 8 4 2 8 4 + (2)(3) 2 8 10
Example 1-24
Solve for x: (2nd order)
4 5
x= = (4)(3) - (2)(5) = 2
2 3
Example 1-25
Find the cofactor of -2 in the following matrix.
2 7 3
-2 5 6
3 4 7
Solution:
-2 is at column 1 row 2. The resulting matrix
is
(-1)1 + 2 7 3
4 7
The cofactor is:
7 3
-1 = - [ (7)(7) - (4)(3) ] = - 37
4 7
Example 1-26
Determine the classical adjoint of
2 3 -4
0 -4 2
1 1 5
Solution:
After Solving the cofactors of each entry, the
matrix of the cofactors is
-18 2 4
-11 14 5
-10 -4 -8
Example 1-27
4 5
Determine the inverse of .
2 3
Solution:
The determinant is
4 5 (4)(3) (5)(2)
D= = - =2
2 3
The inverse is
1 3 -5 32 -52
=
2 -2 4 -1 2
Solution to System of Linear Equations using
Determinants (Cramers Rule)
Nx Ny
x= ;y=
D D
Where:
D = determinant of the coefficient of the variables
Nx = determinant taken from D replacing the
coefficients of x by their corresponding constant terms
leaving all other terms unchanged.
Ny = determinant taken from D replacing the
coefficients of y by their corresponding constant terms
leaving all other terms unchanged.
Example 1-28
Solve for x, y, and w in the following equations:
3x - 2y + w = 11
x + 5y - 2w = - 9
2x + y - 3w = - 6
Solution:
3 -2 1 3 -2
D = 1 5 -2 1 5 = - 46
2 1 -3 2 1
11 -2 1 11 -2
Nx = -9 5 -2 -9 5 = - 92
-6 1 -3 -6 1
3 11 1 3 11
Ny = 1 -9 -2 1 -9 = 46
2 -6 -3 2 -6
3 -2 11 3 -2
Nw = 1 5 -9 1 5 = - 138
2 1 -6 2 1
Nx - 92
x= = =2
D - 46
Ny 46
y= = =-1
D - 46
Nx - 138
x= = =3
D - 46
Complex Numbers
a + bi
Addition or Subtraction
Example 1-30
Simplify i30 - 2i25 + 3i17
Solution:
15 12 8
=i2 - 2 i2 i + 3 i2 i
=(-1)15 - (2) (-1)12 i + (3) (-1)8 i
=-1 - 2i + 3i = -1 + i
Example 1-31
(3 + 2i) (4 - 3i) = 12 - 9i + 8i - 6 = 18 i
Conjugate of a Complex Number
Number Conjugate
2 + 3i 2 - 3i
3 - 5i 3 + 5i
- 5 + 2i - 5 - 2i
Example 1-32
(2 + 3i)(2 - 3i) = 4 - 9i2 = 4 - 9 ( - 1) = 13
Example 1-33
3 + 4i 2 + i 6 + 3i + 8i + 4i2
=
2-i 2+i 2- i2
6 + 11i - 4 2 + 11i 2 11
= = = + i
4 - ( - 1) 5 5 5
Complex Equation
Example 1-34
Solve for x and y if 3x - 2yi = 6 + 8i
Solution:
3x = 6 , - 2y = 8 ; Hence x = 2 and y = - 4
a
a + bi
Imaginary axis, b
r b
Real Axis, a
In the Argand chart shown:
r = a2 + b2
b
tan =
a
a + bi = r cos + r sin i
a + bi = r
Multiplication
r1 1 r2 2 = r1 r2 (1 + 2 )
Example 1-35:
5 30 6 45 = 30 75
Division
r1 1 r1
= (1 2
r2 2 r2
Example 1-36:
45 67
=3 50
15 17
De Moivres Theorem
[r ]n = rn n
Example 1-37:
[5 15]3 = 53 (3)(15) = 125 45
Example 1-38
Find 5 + 12i
Solution
There are two roots, each are 360/2 = 180 apart.
Exponential Form
a + bi = r ( cos + i sin )
ei = r ( cos + i sin )
e-i = cos - i sin
a + bi = rei
Example 1-39
Find ln (3 + 4i)
Solution:
Convert 3 + 4i to polar form (a = 3, b = 4)
r= 32 + 42 = 5, tan = 4/3, = 53.13 = .9273 rad
3 + 4i = 5 53.13 = 5e0.9273i
ln (3 + 4i) = ln 5e0.9273i = ln 5 + ln e0.9273i
ln(3 + 4i) = 1.609 + 0.9273i
Venn Diagram
Example 1-40
An engineering professor conducted a survey
regarding the favorite subjects of the students. The
following data were gathered: 60 students like the
subject algebra, 50 like the subject calculus, and 45
likes the subject physics. Thirty students like both
algebra and calculus subjects, 25 students like both
calculus and physics subjects, and 20 students like
both algebra and physics subjects. Only 15
students like all the three subjects. How many
students were surveyed?
Algebra 30
Calculus
60 50
1
25 10
15
5 10
20 25
15
Physics
45
Solution:
The diagram shows the following information:
25 like the subject algebra only
15 like the subject calculus only
10 like the subject physics only
15 like the subject algebra and calculus
subjects
10 like the subject calculus and physics
subjects
5 like the subject algebra and physics
subjects
15 likes all three subjects
Plane Trigonometry
c
a
opposite side o
sin = = (soh)
hypotenuse h
adjacent side a
cos = = (cah)
hypotenuse h
opposite side o
tan = = (toa)
adjacent side a
adjacent side a
cot = = (tao)
opposite side o
hypotenuse h
sec = = (cha)
adjacent side a
hypotenuse h
csc = = (sho)
opposite side o
Pythagorean Theorem
c 2 = a2 + b2
Trigonometric Identities
x2 - 4 = 0 Conditional
equation
True only for
x = 2
Basic Identities
c
a
a ac sin
tan = = =
b bc cos
b bc cos
cot = = =
a ac sin
c c c 1
sec = = =
b bc cos
c c c 1
csc = = =
a ac sin
Pythagorean Relations
a2 + b2 = c2
a2 b2 c2 a 2 b 2
2
+ 2 = 2 or + = 1
c c c c c
Then;
sin2 + cos2 = 1
tan2 + 1 = sec2
1 + cot2 = csc2
Sum and Difference of Two Angles
tan x + tan y
tan ( x + y )=
1 - tan x tan y
tan x - tan y
tan ( x - y )=
1 + tan x tan y
Half-Angle Formulas
x 1 - cos x
sin =
2 2
x 1 + cos x
cos =
2 2
x 1 - cos x sin x
tan = =
2 sin x 1 + cos x
1 - cos x
=
1 + cos x
Powers of Functions
1 - cos 2x 1 + cos 2x
sin2 x = ; cos2 x =
2 2
1 - cos 2x
tan2 x=
1 + cos 2x
Product of Functions
1
sin x cos y = [ sin(x + y) + sin (x - y)]
2
1
sin x sin y = [ cos(x - y) - cos (x + y)]
2
1
cos x cos y = [ cos(x + y) - cos (x - y)]
2
Sum and Difference of Functions (Factoring
Formulas)
x+y x-y
sin x + sin y = 2 sin cos
2 2
x+y x-y
sin x - sin y = 2 cos sin
2 2
x+y x-y
cos x + cos y = 2 cos cos
2 2
x+y x-y
cos x - cos y = -2 sin cos
2 2
sin ( x + y )
tan x + tan y =
cos x cos y
sin ( x - y )
tan x - tan y =
cos x cos y
Oblique Triangles
a
b
A
B c
Sine Law
a b c
= =
sin A sin B sin C
Cosine Law
Law of Tangents
A-B B-C
a-b tan b-c tan
= 2 ; = 2
a+b A+B b+c B+C
tan tan
2 2
C-A
c-a tan
= 2
c+a C+A
tan
2
Mollweides Equations
A-B A-B
a-b sin a+b cos
= 2 ; = 2
c C c C
cos sin
2 2
How to get the other trigonometric functions with
one function known.
Example 2-1
k
1
k2 - 1
1 k
tan = ; sec =
k 2 - 1 k 2 - 1
Angles
Angle,
Units of Angle
r
r
1 radian
Area of a Triangle
a c
C A
b
Given base b and altitude h,
1
Area = bh
2
1
Area = ab sin
2
Area = s ( s - a ) ( s - b ) ( s - c )
a + b + c
s =
2
a2 sin B sin C
Area =
2 sin A
The area under this condition can also be solved by
finding one side using sine law and apply the
formula for two sides and an included angle.
Median of a Triangle
Altitude
b
to side c Median
c to side
aa a
ac ab
C B
a
Incente
A Angle
Angle bisector
bisector to to side a
side c
b
bc c Angle
bisector to
side b
bb
ba
C B
a
Spherical Trigonometry
Spherical Triangle
a
C
a
B b b
c
c A
R2 E
A =
180
E = A + B + C - 180
Or
E s s-a s-b s-c
tan = tan tan tan tan
4 2 2 2 2
Where
a + b + c
s = (semi-perimeter)
2
c
B
c a
A b
C a
A b
NAPIERS CIRCLE
Napiers Rules
= tan c tan b
sin A
or cos A = cot c tan b
Oblique Spherical Triangles
Law of Sines
Napiers Analogies
1 1
sin (A-B) tan (a-b)
2 = 2
1 1
sin (A+B) tan c
2 2
1 1
sin (a-b) tan (A-B)
2 = 2
1 1
sin (a+b) cot C
2 2
1 1
cos (A-B) tan (a+b)
2 = 2
1 1
cos (A+B) tan c
2 2
1 1
cos (a-b) tan (A+B)
2 = 2
1 1
cos (a+b) cot C
2 2
The Terrestrial Sphere
Definitions:
Triangle
Right Triangles
A
C
Similar Triangles
1. Two triangles are similar if the
angles of one are respectively equal
to the angles of the other, or if two
angles of one are respectively equal
to two angles of the other.
2. Two triangles are similar if an angle
of one equals an angle of the other
and the sides including these
angles are proportional.
3. Two triangles are similar if their
sides are in the same ratio.
4. If two triangles have their sides
respectively parallel, or respectively
perpendicular, each to each, they
are similar.
Area of a Triangle
B
c
h
C A
b
1
Area = bh
2
1
Area = ab sin
2
Given three sides a, b, and c (Heros Formula)
Area = s ( s - a ) ( s - b ) ( s - c )
a + b + c
Semi-perimeter, s =
2
a2 sin B sin C
Area =
2 sin A
B B
Y
Y
A A
X C C
X
Area = ab
d a
Perimeter, P = 2(a + b)
Diagonal, d = a2 + b2
b
Square
Area = a2
d
a
Perimeter, A = 4a
Diagonal, d = 2
a
General Quadrilateral
C
b
B d
a
d2 d1 D
A c
1
Area = d d sin
2 1 2
Given four sides a, b, c, d and sum of two opposite
angles
1 1
Area = ab sin B + cd sin D
2 2
Parallelogram
Area of a Parallelogram
1
Area = d d sin
2 1 2
Area = ab sin A
Rhombus
C D
d1
A rhombus is a parallelogram with
a
four equal sides. The diagonals of a d2
rhombus bisect each other at an 90
angle of 90. B A
a
Area of a Rhombus
1
A= d d
2 1 2
A = a2 sin A
Trapezoid
a
are parallel.
b
a+b
Area = h
2
a
Length of the dividing
line parallel to the A1
c
h
two parallel sides.
A2
b
ma2 + nb2 m A1
c = ; =
m+n n A2
Cyclic Quadrilateral
B c
d1
D
d2
a
d
A + C = 180 ; B + D = 180
d1 d2 = (a)(c) + (b)(d)
Polygons
otherwise, it is called a
a
1
1 6
Theorems in Polygons:
Sum, = (n - 2) 180
= 360
Number of Diagonals, D
n
D= (n - 3)
2
Regular Polygons
Circumscribing Circle
x
Inscribed Circle
x R x
R
Apothem
x
x
x
x=side
=angle subtended by the side from the
center
R=radius of inscribed circle, also called
the apothem
n=number of sides
360
=
n
n
Area = R2 sin
2
Perimeter = P = (n)(x)
n-2
Interior angle = = 180
n
360
Exterior angle = = =
n
Tangent
Circle
Secant
Chord
Radius
Diameter
Arc
Theorems:
tangent
r
L
r
r
L
Intercepted Arc
C O
B
1
= or = 2
2
90
90 90
Chord Arc
=
Tangent
O C
D
B
(OA)(OB) = (OC)(OD)
1
= (ArcAC - ArcBD )
2
ADC = ABC; BAD = BCD
O
C
()2 = (OA)(OB)
1
= (ArcBC - ArcAc )
2
1
= ArcAC
2
11. A perpendicular from a point on the
Circumference to a diameter of a circle is
a mean proportional between the
segments of the diameter.
Diameter
h
a b h2 =ab
Area of Circle
Circumference = 2r = d
r
D
Area = r2 = D2
4
Sector of a Circle
r
Length of arc, C = rradians =
180
r r 1 2 r2 1
O Area = r = = Cr
2 radians 360 2
Segment of a Circle
C
Area = Asector - Atriangle
1 1
A = r2 - r2 sin
r
r
2 2
1
O
A = r2 ( - sin )
2
= 360 -
Area = Asector + Atriangle
r r 1
A = r2 ( + sin )
2
B 2
Area = bh
3
h b2
Length of ABC = [ me + ln (m + e)]
A C
8h
b 4h
m= , e = 1 + m2
b
Spandrel of a Parabola
B b
h
1
A= bh
3
h
F C
E
A
D
L
Areashaded = AADB - AADE + ABFC + ACEF
2
Areashaded = (AC) (h)
3
a a
Ellipse
a
b
c
V2 V1
F2 c F1
b
Area = ab
2 + 2
Perimeter = 2
2
a2 = b 2 + c 2
d3 c
Eccentricity (first eccentricity), e = = > 1.0
d4 a
c
Second eccentricity, e' =
b
c
Angular eccentricity, =
a
a-b
Ellipse flatness, f =
a
a-b
Second flatness, f' =
b
Radius of Circles
Circumcenter
of the triangle
abc a
r
r= c
4AT b
c r r a
r
A C
b
At a+b+c
r= ; s=
s 2
Circles Escribed about a Triangle [Excircles]
ra
c
rc rb
a
At At At
ra = ;r = ;r =
s-a b s-b c s-c
b
r
c
a
d
(ab + cd)(ac + bd)(ad + bc)
r=
4Aquad
Aquad = (s - a)(s - b)(s - c)(s - d)
a+b+c+d
Semi-perimeter, s =
2
Aquad a+b+c+d
ra = ; Aquad = abcd ; s =
s 2
Area by Approximation
d
Area = [ h + 2(h2 + h3 + ) + hn ]
2 1
d
Area = [ h1 + 2hodd +4heven + hn ]
3
Area by Coordinates
(x1,y1)
(x3,y3)
(x8,y8)
(x4,y4) (x2,y2)
(x6,y6)
(x5,y5) (x7,y7)
1 x1 x2 x3 xn - 1 xn x1
A= y y2 y3 yn - 1 yn y1
2 1
1
A= [x1 y2 + x2 y3 + - x2 y1 + x3 y2 + ]
2
Polyhedrons
Regular Polyhedrons
Regular Regular
Dodecahedron Icosahedron
Let m = number of polygons meeting at a vertex
n = number of vertices of each polygon
f = number of faces of the polyhedrons
e = number of edges of the polyhedron, and
v = number of vertices of the polyhedron
2
3
Tetrahedron 4 4 4 3 3
62
Hexahedron 6 12 6 3 62 3
3 2
Octahedron 8 12 8 4 22 3
3
3 + 5 3 15 + 75
Dodecahedron 12 30 20 3 152
5 5 4
53 3 + 5
Icosahedron 20 30 12 5 52 3
12
f = 2f=2v
+e-v
Prism
Volume, V = Ab h = AR L
Lateral area, AL = PR
Rectangular Parallelepiped
c
d2
d1
b
a
Volume, V = Ab h = abc
Lateral area, AL = AR h=2(ac + bc)
Total surface area, As = At + 2Ab = 2(ab + bc + ac)
Face diagonal, d1 = a2 + c2
Space diagonal, d2 = a2 + b2 + c2
d2 c
d1
b
a
Volume, V = Ab h = a3
Truncated prism
Lateral area, AL = 4a2
Total surface area, A s = 6a2
Face diagonal, d1 = a2
Space diagonal, d2 = a3
Truncated Prism
Pyramids
Where ; 1
Volume = Ab h
Ab = area of the base 3
Frustum or Pyramid
h
Volume = (A +A +A1 A2 )
3 1 2
where:
A1 =lower base area;
A2 =upper base area
h=altitude
Cylinders
L 90
h=L h
Volume, V = Ab h = AR L
Lateral Area, AL = PR L
r
h
Volume, V = Ab h = r2 h
Lateral area = AL
AL = Base perimeter*h = 2rh
Cone
1
Volume, V= Ab h
3
where:
Ab =base area
h=altitude
Slant height, L = r2 + h2
1 1
Volume, V = Ab h = rL
3 3
Lateral area = AL = rL
where:
r = base radius
h=altitude
Frustum of a Cone
h
where: Volume, V = (A1 +A2 +A1 A2 )
3
A1 =lower base area
A2 =upper base area
h=altitude
r r
L
where:
Surface Area = = 4r
4
Volume, V =
3
Azone = 2rh
h2
Volume, V = (3r - h)
3
1
Volume = A r
3 zone
2 2
Volume = r h
3
Spherical Polygons
n = no. of sides
r = radius of sphere
E= spherical excess
r2 E
Area =
180
E = sum of angles - (n-2)180
Spherical Pyramid
B
A
D
C
r
r
r2 E
Volume =
540
As = L C
C = R (Angle of rotation in radians, )
If = 360, As = L (2R)
Second Proposition of Pappus
Volume = AC = A (R )
If = 360, Volume = A(2R)
Ellipsoid
b a
4
Volume = abc
3
Prolate Spheroid
4
Volume = ab2
3
arcsin 2 2
As = 2b2 + 2ab ( ); =
Spheroid
4
Volume = a2 b
3
2
b 1+e
As = 2a2 + ( ln )
e 1-e
Paraboloid of Revolution
1
Volume = r2 h
2
3
4r r2 2 r 2
A= + h2 -
3h2 4 2
Prismatoid
General Prismatoid
Prismoidal Formula
L
Volume = [ A1 + 4Am + A2 ]
6
Prismatoid Theorem
16 3
Volume = r
3
16 3
Volume = r
3
Solid with circular base of radius r and every
cutting plane perpendicular to a certain diameter is
an EQUILATERAL TRIANGLE.
4r3
Volume =
3
8 3
Volume = r
3
Solid with circular base of radius r and every
cutting plane perpendicular to a certain diameter is
an ISOSCELES RIGHT TRIANGLE with
hypotenuse on the plane of the base.
4 3
Volume = r
3
CHAPTER 5: PLANE ANALYTIC GEOMETRY
Ordinate
2
d = (x2 - x1 )2 + y2 - y1
Straight Line
x2 x1
O
(a, 0)
Ax + By + C = 0
rise y2 - y1
Slope, m = =
run x2 - x2
Where:
m is positive if the line is inclined upwards to
the right.
1. Point-slope form
Given a point P1(x1, y1) and its slope m:
y - y1 = m(x - x1 )
2. Slope-intercept form
Given a slope m and y-intercept:
y = mx + b
3. Intercept form
Given x-intercept a and y-intercept b:
x y
+ =1
a b
4. Two-point form
Given two points P1(x1,y1) and P2(x2,y2):
y - y1 y -y
= 2 1 =m
x - x1 x2 - x2
= 2 - 1
m2 - m1
tan =
1 + m1 m2
Y Line = Ax + By +C = 0
d
(x1, y1)
O
X
Ax1 + By1 + C
d=
A2 + B2
Use of Sign:
That is,
If B is positive and the point is above the line, then
use (+) (+) = (+)
If B is positive and the point is below the line, then
use (+) (-) = (-)
If B is negative and the point is above the line, then
use (-) (+) = (-)
If B is negative and the point is below the line, then
use (-) (-) = (+)
If only the distance is required, use the absolute
value:
Ax1 + By1 + C
d=
A 2 + B 2
Y d Line = Ax + By + C1 = 0
O
X
Line = Ax + By +C2 = 0
C2 - C1
d=
A 2 + B 2
Division of Line Segment
P2 (x2, y2)
r2
L2
r1 r1 = L1 / L
P (xp, yp) r2 = L2 / L
L1
P1 (x1, y1)
x1 r2 + x2 r1
xp =
r1 + r2
y1 r2 + y2 r1
yp =
r1 + r2
x1 + x2
xm =
2
y1 + y2
ym =
2
Area of Polygon by Coordinates
Let (x1, y1), (x2, y2), (x3, y3) (xn-1, yn-1), (xn, yn) be
the consecutive vertices of a polygon arranged in
counterclockwise sense. The area is:
1 x1 x2 x3 xn-1 xn x1
A= y y2 y3 yn-1 yn y1
2 1
1
A= [x1 y2 + x2 y3 + - x2 y1 + x3 y2 + ]
2
Conic Sections
Ax2 + Cy2 + Dx + Ey + F = 0
From the foregoing equations:
If B2 < 4AC, the conic is an ellipse
If B2 = 4AC, the conic is an parabola
If B2 > 4AC, the conic is an hyperbola
Y Hyperbola, e > 1
Parabola, e = 1
Vertex of the cone
Circle, e = 0
F
X
Ellipse, e < 1
The parabola antenna
used in communication is
our example.
Circle
(h, k)
O X
r = radius
(h , k) = center
Ax2 + Ay2 + Dx + Ey + F = 0 or
x2 + y2 + Dx + Ey + F = 0
Center at (h, k)
(x - h)2 + (y - k)2 = r2
Center at (0, 0)
x2 + y2 = r2
For circle Ax2 + Ay2 + Dx + Ey + F = 0
-D -E D2 + E2 - 4AF
h= ;k= ;r=
2A 2A 4A2
Parabola
d1
Focus
d2
Latus Rectum =4A
a 2A
Directrix
2A
2A
Ax2 + Dx + Ey + F = 0 or
x2 + Dx + Ey + F = 0
A=0
Cy2 + Dx + Ey + F = 0 or
y2 + Dx + Ey + F = 0
Eccentricity
e=1
Latus Rectum, LR
LR = 4A
Vertex at (0, 0)
y2 = 4ax x2 = 4ay
y2 = - 4ax x2 = -4ay
Vertex at (h, k)
-D D2 - 4AF -E
h= ;k= ;a=
2A 4AE 4A
E2 - 4CF -E -D
h= ;k= ;a=
4CD 2C 4C
Ellipse
The locus of the point that moves such that the sum
of its distances from two fixed points called the foci
is constant. The constant sum is the length of the
major axis, 2a. It can also be defined as the locus
of the point that moves such that the ratio of its
distance from a fixed point, called the focus and a
fixed line, called the directrix, is constant and is less
than one (1).
y
d d
Directrix
d4
b
Minor Axis, 2b
d3 a F1
V2 F2
Directrix
c V1
x
c LR
d1
d2
P(x, y)
a a
Major axis, 2a
Elements of an Ellipse
a2 = b2 + c 2
d3 c
Eccentricity (first eccentricity), e = = < 1.0
d4 a
a
Distance from the center to the directrix, d =
e
2b2
Length of Latus rectum, LR =
a
c
Second eccentricity, e' =
b
c
Angular eccentricity, =
a
a-b
Ellipse flatness, f =
a
a-b
Second flatness, f' =
b
Ax2 + Cy2 + Dx + Ey + F = 0 or
x2 + Cy2 + Dx + Ey + F = 0
Center at (0, 0)
x2 y2 x2 y2
+ =1 + =1
a2 b2 2 a2
b
Center at (h, k)
-D -E
h= ;k=
2A 2C
Hyperbola
Directrix
d4
d2
Conjugate axis, 2b
d3 d1
b
F2 V2 F1 LR
c V1 x
b b
a
d d
Transverse Axis, 2a
a a Asymptote
d2 - d1 = 2a c c
Elements of a Hyperbola
c 2 = a2 + b2
d3 c
Eccentricity (first eccentricity), e = = > 1.0
d4 a
a
Distance from the center to the directrix, d =
e
2b2
Length of Latus rectum, LR =
a
Equation of Asymptote
y-k =m(x-h)
Ax2 - Cy2 + Dx + Ey + F = 0 or
x2 - Cy2 + Dx + Ey + F = 0
x2 y2 x2 y2
- =1 - =1
a2 b2 b2 a2
-D -E
h= ;k=
2A 2C
Point of Tangency
Tangent
Y
(x2, y2)
Normal
Conic
O X
Pole Axis
Sign Convention
is positive (+) if measured
counterclockwise.
is negative ( - ) if measure clockwise.
r is positive ( + ) if laid off at the terminal
side of .
r is negative ( - ) if laid off prolongation
through O from the terminal side of .
(r1, 1)
d
r1
(r2, 2)
= 2 - 1
r2
O X
Pole
d = r1 2 + r2 2 - 2r1 r2 cos ( 2 - 1 )
Relationship between Polar and Cartesian
Coordinate Systems
x (r, )
y r
Pole Axis
r2 = x2 + y2
y
x = r cos ; y = r sin ; tan =
x
Polar Curves
Cardioid
r=a (1 - sin )
y=1/x y =| 1 / x| y = ax3 y3 = ax2
SPACE ANALYTIC GEOMETRY
P (r, , z)
X
r
Y
Relations between Rectangular and Cylindrical
Coordinates System
Limits
Theorems of Limits
n
lim f(x)= lim f(x) = A , provided
n n
6.
xa xa
n
A is a real number
L Hospitals Rule (Intermediate type 0/0)
That is;
x4 - 81
Example 6-1: lim ; Substitute x = 2.9999
x3 x - 3
(2.9999)4 - 81
(2.9999) - 3
= 107.99 108
1 1
Example 6-2: lim - ;
x0 x ex - 1
Substitute x = 0.0001
1 1
- = 0.49999 0.5
0.0001 e0.0001 - 1
cotx
Example 6-3: lim ; Substitute x = 0.0001
x0 cot2x
cot ( 0.0001 )
=2
cot [2 (0.0001)]
ex + e-x - x2 - 2
Example 6-4: lim ;
x0 sin2 x- x2
Substitute x = 0.0001 radian
Note: set your calculator to radian mode
e0.0001 + e-0.0001 - ( 0.0001 )2 - 2
2 = 0.25
sin2 ( 0.0001 ) ( 0.0001 )
Differential Formulas
In the following formulas u, v, and w are
differentiable functions of x and a and n are
constants.
Algebraic Functions
d
1. (c)=0
dx
d d(u)
2. ( cu ) = c
dx dx
d d(u) d(v)
3. (u+v)= +
dx dx dx
d d(v) d(u)
4. ( uv ) = u +v
dx dx dx
d d(u)
5. (u )=n
nun - 1
dx dx
d(u)
d
6. u = dx
;u0
dx u
d(u) d(v)
u v -u
7. d = dx
2
dx
;v0
v v
d(v)
c -c dx
8. d =
v v2
Trigonometric Functions
d d(u)
15. ( sin u) = cos u
dx dx
d d(u)
16. ( cos u) = -sin u
dx dx
d 2 d(u)
17. ( tan u) = sec u
dx dx
d 2 d(u)
18. ( cot u) = -csc u
dx dx
d d(u)
19. ( sec u) = sec u tan u
dx dx
d d(u)
20. ( csc u) = -csc u cot u
dx dx
d(u)
d dx
22. ( arccos u) = -
dx 1 - u2
d(u)
d dx
23. ( arctan u) =
dx 1 + u2
d(u)
d dx
24. ( arccot u) = -
dx 1 + u2
d(u)
d dx
25. ( arcsec u) =
dx
uu2 - 1
d(u)
d dx
26. ( arccsc u) = -
dx
uu2 - 1
Hyperbolic Functions
d d(u)
27. ( sinh u) = cosh u
dx dx
d d(u)
28. ( cosh u) = sinh u
dx dx
d d(u)
29. ( tanh u) = sech2 u
dx dx
d 2 d(u)
30. ( coth u) = -csch u
dx dx
d d(u)
31. ( sech u) = - sech u tanh u
dx dx
d d(u)
32. ( csch u) = - csch u coth u
dx dx
Where:
ex - e-x ex + e-x
sinh x = cosh x=
2 2
sinh x 1
tanh x= coth x=
cosh x tanh x
1 1
sech x= csch x=
cosh x sinh x
d(u)
d dx
35. ( arctanh u) =
dx 1 - u2
d(u) d(u)
d -
dx dx
36. ( arccoth u) = = 2
dx u2 - 1 1- u
d(u)
d -
dx
37. ( arcsech u) =
dx 2
u1 - u
d(u)
d -
dx
38. ( arccsc u) =
dx u1 + u2
Where:
arcsinh x = lnx+ x2 + 1
arccosh x = ln x+ x2 - 1
1 1+x
arctanh x = ln
2 1-x
1 x+1
arccoth x = ln
2 x-1
1 + 1 - x2
arcsech x = ln
x
1 + 1 + x2
arcsech x = ln x >0
x
dy
Slope at any point = y =
dx
Rate of Change
Curvature
Curvature refers to the rate of change of the
direction of the curve. Thus, a circle with a smaller
radius has great curvature, or is sharply curved.
y"
K= 3
[1 + (y')2 ] 2
Radius of Curvature
The radius of curvature, , is the reciprocal of the
curvature, k, or:
3
1 1 + (y')2 2
= =
k | y" |
Circle of Curvature
At any point on a curve y = f(x), where y and y
exist and y 0, there is associated with the curve
a circle, which is called the circle of curvature with
the following equation:
( x h )2 + ( y k )2 =
y' 1 + (y')2
h=x-
y"
1 + (y')2
k=y+
y"
3
1 + (y')2 2
=
| y" |
B y = f(x)
C E
A
D
x
dy
= y' = 0
dx
Points of Inflection
A point of inflection is a point at which the curve
changes from concave upward to concave
downward or vice verse (see point E from the
figure). At these points, the tangent changes its
rotation from clockwise to counterclockwise or vice
versa.
2 2 2 L
L 3 =h 3 +x 3
h
x
For maximum area (to admit the
most light) of a Norman window r
of given perimeter
x x y
y= ;r=
2 2
h=x
x
h=y h
x = ab b
x
Minimum length L of a line
tangent to an ellipse
L
b
L=a+b a a
b
x = 2y
x = y (a square)
Diameter = 2 height D
D = 2h
Maximum volume of an OPEN (one end)
cylindrical tank of given surface area, or minimum
surface area for given volume
Diameter = 2 height
D = 2h
For maximum volume of a
closed rectangular box with
given sum of all edges or given z
total area y
x
x = y = z (a cube)
Maximum volume of open (one end)
rectangular box of square base and
h
given surface area, or minimum
surface area for given volume. y
x
x = 2y
For a rectangle of maximum
area or perimeter that can be cut
r
from a circle of radius r
y
x = y (a square)
2r = b 3
b
For maximum capacity of a
trapezoidal gutter or canal of x
x = 2y (a half-rectangular hexagon)
y = b
2
22a
y=
3
Wcyl = W
3
Vcyl = V
3
Largest rectangle that
can be cut from a h
given triangle x
x = b2
b
y = h2
2 2 2
L 3 =x 3 +y 3 x
h
sin = 13 r
h
y=
3
For a minimum cost of closed
cylindrical tank of known volume or h
x d
= b
a a+b
d
K
x= n
m+n
K
y= m
m+n
Time Rates
dS
Velocity, v =
dt
dV
Discharge, Q =
dt
d
Angular Speed, =
dt
Integration Formulas
Trigonometric Functions
7. sin u du = - cos u + C
8. cos u du = sin u + C
9. tan u du = ln|sec u| + C
= - ln|csc u+ cot u| + C
13. sec u tan u du = sec u + C
du u
17. = arcsin + C
a2 -u2 a
du 1 u
18. 2 2 = arctan + C
a +u a a
du 1 u
19. = arcsec + C
2
uu -a 2 a a
20. arcsin u du = u arcsin u + 1-u2 + C
du
34. - = ln u+ u2 a2 + C
u2 a2
du 1 a+u
35. 2 2 = ln + C, u2 < a 2
a -u 2a a-u
du 1 ua
36. 2 2 = ln + C, u2 < a2
u -a 2a u + a
u a2 u
37. a2 -u2 du = a2 -u2 + arcsin + C
a 2 a
2 u+
u a
38. u2 a2 du = u2 a2 + ln 2 2 + C
2 2 u +a
39. udv = uv- vdu
Trigonometric Substitution
Use To obtain
a 2
a-bx2 x2 = sin a1-sin = a cos
2
b
a
a+bx2 x2 = tan2 11+tan2 = a sec
b
a
bx2 -a x = sec2
2
asec2 -1 = a tan
b
Wallis Formula
/2
[(m-1)(m-3)(2 or 1)][(n-1)(n-3)(2 or 1)]
sinm cosn d=
(m+n)(m+n-2)(2 or 1)
0
Where:
= /2 when both m and n are even
= 1 if otherwise
m & n = positive integer, not equal to 1
Examples
Wallis Formula
/2 (4)(2) 8
Ex.7-1: sin5 xdx = 1 =
0 (5)(3)(1) 15
Double Integration
1 3x 1 4
y 3x
Ex.7-2: y3 dydx = dx
0 0 0 4 0
1 (3x)4 1
81 4
= dx = x dx
0 4 0 4
81x5 1 81
= =
4(5) 0 20
Triple Integration
3 2 y 3 2
y
Ex. 7-3: dxdydz = [x] dydx
0 0 0 0 0 0
3 2 3 2
y 2
= ydydx = dx
0 0 0 2 0
3
3
= 2dx = 2x| = 6
0 0
Integration By Parts
Ex.7-4: xsin xdx
Let u = x, du = dx
dv = sinxdx,v = -cos x
Algebraic Substitution
1
xdx
Ex. 7-5:
0 (x+1)
8
2 (z-1)dz 2
= z-7 -z-8 dz
1 z8 1
1 -6 1 -7 2
= z - z =0.0223
-6 -7 1
Trigonometric Substitution
dx
Ex.7-6 : Find -
x2 4+x2
let x2 = 4tan2
x = 2tan
dx=2sec2 d
dx 2sec2 d
=
x2 4+x2 4tan2 4+4tan2
1 sec
= d
4 tan2
1 1
= sin-2 cosd= - +C
4 4sin
dx 1 1
= - +C = -
x2 4+x2 4sin 4
x
4+x2
dx 4+x2
=- +C
x2 4+x2 4x
Plane Areas
y = g(x)
Yu
yL
x2
A = yu - yL dx
x1
yU = f(x) ; yL = g(x)
Using Horizontal Strip
y = g(x)
XL
XU
y = f(x)
y2
A = (xR - xL )dy
y1
xR = g(y); xL = f(y)
By Polar Coordinates
1
2
1
A= r2 d
2
1
Length of arc
ds
1 dy
dx
x2
dx 2 y2
dx 2
s = 1 + dx or s = 1 + dy
x1 dy y dy
1
Centroid of Plane Areas
x
dA
xG cg
xC yc
yG y
x2 y2
A XG = xc dA A YG = ydA
x1 y1
x2 y2
A XG = xdA A YG = yc dA
x1 y1
Centroid of Parabolic Segment and Spandrel
x1
y1
Parabolic
Segment
h
hbx
Spandrel
Spandrel
3 2
x= b y = h
8 5
1 3
x1 = b y1 = h
4 10
2 1
Aparabola = bh Aspandrel = bh
3 3
dA
dA
y
y2 x2
Ix = y2 dA Iy = x2 dA
y1 x1
J = Ix +Iy
Product of Inertia
Ixy = xydx
1
I= Mr2
2
M = mass; r = radius
1
I= M(R2 - r2 )
2
R = outer radius
r = inner radius
1
Iy = M L2 y L/2 y L/2
3
1 2
Iy' = ML
12
Solid Sphere
2
I = M r2
5
Spherical Shell
2
I= M r2
3
r = mean radius
Right Circular Cone
3
I= M r2
10
r = base radius
1
I= Ma2 + b2
4
b
a
Triangle
a+b h
y a xc = yc =
3 3
1
Area= b h
xc
h 2
cg
bh3 bh3
yc Ix = Igx =
12 36
b/2 b/2
Rectangle
y Area = bd
bd3 db3
Ix = Iy =
d 3 3
cg
bd3 db3
Igx = Igy =
x 12 12
b
Circle
y
1 2
r Area = r2 = D
4
cg
r D4
x 4
Igx = Igy = =
4 64
D
Quarter Circle 1 2
Area = r
4
y 4r
xc = yc =
3
r
xc
r4
cg Ix = Iy =
yc 16
r x
Igx = Igy = 0.055r4
Semi - Circle
1 2
Area = r
2
4r
yc =
y 3
r4
Ix = Igy =
8
cg
yc x
Igx = 0.11r4
r r
Ellipse
Area = ab
y
ab3
Igx =
a
b 4
cg
a
x
ba3
b Igy =
4
Half Ellipse
1
Area = ab
2
4b
y yc =
3
ab3
b
cg
yc Ix = Igx = 0.11ab3
a a x 8
ba3
Igy =
8
Quarter Ellipse
1
Area = ab
y
4
4a 4b
xc = yc =
3 3
b xc cg ab3 ba3
yc Ix = Iy =
a x 16 16
Igx = 0.055ab3
Igy = 0.055ba3
Sector of a Circle
1 2
Area = r (2) = r2
y 2
r
2 rsin
xc =
x
3
r4 1
r Ix = - sin2
4 2
xc r4 1
Iy = + sin2
4 2
Parabolic Segment
2
y x = ky2
Area = bh
3
2 3
xc = h yc =
b
5 8
2 2
cg
xc
Ix = hb3 Iy = bh3
yc 15 7
x
h
Spandrel
1
y Area = bh
n+1
1
y = kxn xc = b
h
n+2
xc
n+1
yc = h
yc 4n + 2
x
b
Segment of an Arc
length of arc = 2r
y
r rsin
xc =
cg x
When
= 90(semicircle)
xc 2r
xc =
Solids of Revolution
Volume
xR
xL
yL
yR
V = xR 2 -xL 2 dy
y1
x2
or V = yR 2 -yL 2 dx
x1
dy
y2
V = 2y(xR - xL ) dy
y1
x2
or V = 2xyR - yL dx
x1
Surface Area
r
ds
d2
As = 2rds
d1
ds = 1 + (dx/dy)2 or ds = 1 + (dy/dx)2
y2
VYG = ydV
y1
Constant Force
x2 x2
Work = (dF)(x) or Work= F(x)dx
x1 x1
Work required to stretch a spring
1
Work = kx2 2 - x1 2
2
Where, k = spring constant or stiffness in N/m
Variable Separable
dy
A. Differential equations of type dx = f(x)
dy
Differential equations of type dx = f(x) can
be solved direct integration by writing it in the
form
dy = f(x) dx
Example 8-1
Solve the differential equation dydx = 2x + sin 3x
Solution
dy = (2x + 3 sin x) dx
y = x2 (1/3) cos 3x + C (general solution)
dy
B. Differential equations of type dx = f(y)
dy
Differential equations of type dx = f(y) can
be solved direct integration by writing it in the
form
dy
dx =
f(y)
Example 8-2
Solve the equation (y2 1) dy/dx = 3y,
given that y = 1 when x = 13/6.
Solution
y2 - 1 y 1
Rearrange dx = dy = - dy
3y 3 3y
1 1
x= y2 - ln y + C (general solution)
6 3
13
When x = , y = 1 then:
16
13 (1)2 1
= - ln(1) + C ; C=2
16 6 3
1 1
x= y2 - ln y + 2 (particular solution)
6 3
C. Differential equation of type
dy
dx = f(x) g(y)
dy
Differential equations of type dx = f(x) g(y)
can be solved direct integration by writing it in
the form
dy
= f(x)dx
g(y)
Q = C ekt
Where C is constant
Example 8-3
Solve the equation dydx = 3y
Solution
Q=y dQ = dy,
t=x k=3
then y = Ce3x
Homogeneous First Order Differential Equation
Example 8-4
Solve the equation; x2 xy + y2 dx - xy dy = 0
Solution
The coefficients are of the second degree.
Let y = vx
dy = v dx + x dv
then x2 - x2 v + v2 x2 dx - x2 v (v dx + x dv) = 0
Removing x2
(1 v + v2) dx v (v dx + x dv) = 0
dx v dx + v2dx v2dx vx dv = 0
(1 v) dx vx dv = 0
Separating variables
dx vdv
+ =0
x v-1
dx 1
Or + 1 + dv = 0
x v-1
Integrating:
ln x + v + ln(v - 1) = ln C
C
ln =v
x(v - 1)
c
= ev
x(v - 1)
Finally, substituting v = y / x
y y
x - 1 ey/x = C or (y - x)ex = C
x
Linear First Order Differential Equation
y e Pdx = e Pdx Q dx
Example 8-5
Solve the equation dy + 4xy dx = 2x dx
Solution
Rearranging:
dy / dx + 4xy = 2x
P = 4x and Q = 2x
y = A ex + B ex
y = (Ax+B)ex
y = ex [ C cos x + Dsinx ]
Example 8-6
d2 y dy
Solve the equation 2 + 5 - 3y = 0
dx2 dx
Solution
Writing in D-operator form:
2D2 + 5D - 3y = 0
M N
=
y x
Example 8-7
Solve the equation (2x + 4y + 6)dx +
4x - 2y - 5dy = 0
Solution
Check for exactness
M = 2x + 4y +6 N = 4x -2y 5
M N
=4 =4
y x
M N
Since = , the equation is exact
y x
Integrate:
x2 - y2 + 6 5 + 4 xy = C
dy
Equation of type + P(x) y = Q(x) y" has a
dx
general solution of
Where v = y1 n
ln y = (Q - P) dx + C
Solution
x2 + y2 y2
Isolate c: c = = x+
x x
1. Reflexive property
a=a
2. Symmetric property
If a = b, then b = a
3. Transitive property
If a = b and b = c, then a = c. that is, things
equal to the same thing are equal to each
other.
4. If a = b and c = d, then a + c = b + d. That is, if
equals are added to equals, the results are
equal.
Population Growth
dP
=kP
dt
P = Po ekt
dQ
=kQ
dt
Q = Qo ekt
dT
= k (T - ts )
dt
T = ts + (To - ts )e-kt
Flow Problems
Qin
h1
h2
Qout
dQ
= Rate of inflow - Rate of outflow
dt
t2 h2
dQ
dt =
t1 h1 Qin - Qout
dP
=rP
dt
P = Po ert
Motion Problems
ds
Velocity, v =
dt
dv
Acceleration, a =
dt
v dv = a ds
Newtons Second Law of Motion
dv
F=M
dt
CHAPTER 9 ENGINEERING MECHANICS
(PHYSICS)
ENGINEERING MECHANICS
STATICS DYNAMICS
Force
Applications Kinematics Kinetics
System
Force System
A force system is any arrangement where two or
more forces act on a body or on a group of related
bodies.
Parallelogram Law
Resultant, R F2
F2
F1
R2 = F1 2 + F2 2 - 2F1 F2 cos
F2 R
=
sin sin
Resultant of Two or More Concurrent Coplanar
Forces
Resultant, R
F2
F1
Analytical Solution
Rx = Fx = F1 x + F2 x +
Ry = Fy = F1 y + F2 y +
Ry
R = Rx 2 + Ry 2 ; tan x =
Rx
Resultant of Concurrent Forces in Space
F2
F1
F2z
F1z
z
F2y
F1x
F2x
F1y
y
F3y x
F3x
F3z R
F3 x
y
Rx = Fx = F1 x + F2 x + F3 x +
Ry = Fy = F1 y + F2 y + F3 y +
Rz = Fz = F1 z + F2 z + F3 z +
R = Rx 2 + Ry 2 + Rz 2
Rx Ry Rz
cos x = cos y = cos z =
R R R
Example 9-1
Given the three concurrent forces which pass
through (1, -3, 4) and the indicated points:
Solution
2 2 2
F1 = 150 N, L1 = (5 - 1) + (-6 + 3) + (2 - 4) = 5.385
2 2
F2 = 340 N, L2 = (4 - 1) + (0 + 3)2 + (-3 - 4) = 8.185
2 2
F3 = 280 N, L3 = (-1 - 1) + (2 + 3)2 + (6 - 4) = 5.745
Rx = Fx = F1 x + F2 x + F3 x
5-1 4-1 -1 - 1
Rx = 150 + 340 + 280 = 138.56 N
5.835 8.185 5.745
Ry = Fy = F1 y + F2 y + F3 y
-6 + 3 0+3 2+3
Ry = 150 + 340 + 280 = 284.76 N
5.835 8.185 5.745
Rz = Fz = F1 z + F2 z + F3 z
2-4 -3 - 4 6-4
Rz = 150 + 340 + 280 = -248.99 N
5.835 8.185 5.745
R = Rx 2 + Ry 2 + Rz 2 = 402.84 N
x4
x3
x2
F2
x1 F1
F3 F4
R = F = F1 + F2 + F3 +
R x = F x = F1 x1 + F2 x2 + F3 x3 +
Resultant of Non-Concurrent Forces
P1
P2
P5
O
d
P4
R P3
Rx = Px Ry = Py
R = Rx 2 + Ry 2
Equilibrium of Forces
Fx = 0 Fy = 0
Mo = 0
Equilibrium of Two Forces
Two forces are in equilibrium
if they are equal, collinear,
and oppositely directed.
P Q Q
R
Force Polygon
R
P Q
S
Q
R
T
P T
S
Equilibrium of Three Coplanar Forces
x1 x2
T1 A B T2
y1 y2
C
w (N/m)
x1 2 x2 2
=
y1 y2
The solution is to cut a segment from the support to
the lowest point of the cable.
x/2 x/2
T
A
T
W W
y
H
H
C
w (N/m)
W=wx
MA = 0
x
Hy=W
2
From the force polygon;
T2 = W2 + H2
W y
tan = =
H x2
T sin = W
Symmetrical Cable
L
L/2 L/2
T A B T
d
C
w (N/m)
w L2
Tension at C = H =
8d
wL 2
Tension at A = T = H2 +
2
Length of Cable S:
L2
Exact S = [ m k + ln( m + k ) ]
8d
4d
m= ; k = 1 + m 2
L
8d2 32d4
Approximate S = L + -
3L 5L3
Catenary
T2
T1
S2
S1
y2
y1 w (N/m)
c y
x
x1 O x2
T1 = H2 + (w S1 )2 T2 = H2 + (w S2 )2
y1 2 = S1 2 + c2 y2 2 = S2 2 + c2
S1 + y1 S2 + y2
x1 = c ln x2 = c ln
c c
Friction
Elements
f
tan = =
N
The maximum angle that a plane may be inclined
without causing the body to slide down is:
= = arctan
Belt Friction
T1
T2
T1
= ef
T2
T1
ln =f
T2
Where:
f = coefficient of friction
= angle of contact in radius
e = 2.71828
T1 = tension in the tight side
T2 = tension in the slack side
Properties of Sections
x3
XG
x1
1
G 3
y1
YG y3
2
y2
O
x2
A XG = a x = a1 x1 + a2 x2 +
A YG = a y = a1 y1 + a2 y2 +
Center of Gravity of Flat Plates
z y W
w1
w2
XG w3
cg
O YG
W XG = w x = w1 x1 + w2 x2 +
W YG = w y = w1 y1 + wy2 +
z y W
w1
XG w2
cg
O YG
x
W XG = w x = w1 x1 - w2 x2
W YG = w y = w1 y1 - wy2
dA
x
r
y
Ix = y2 dA Iy = x2 dA
J = r2 dA = Ix + Iy
Radius of Gyration
I I
x rx = x ry = y
A A
XO
cg
Ix = Ig + Ad2
V U
O x
Ix
Imin
R Ixy
C 2
Ix - C
Iyx R
Iy
Imax
Iyx = Ixy
Ix + Iy Imax + Imin
C= =
2 2
R = Imax C
2
R = Ixy + (Ix - C)2
Imax = C + R
Imin = C - R
Dynamics
BRANCHES OF DYNAMICS
Kinematics
The geometry of motion. This term is used to
define the motion of a particle or body without
consideration of the forces causing the motion.
Kinetics
The branch of mechanics that relates the force
acting on the body to its mass and
acceleration
KINEMATICS
Motion of Particles
I. Translation
The motion of a rigid body in which a straight line
passing through any two of its particle always
remain to be parallel to its initial position
II. Rotation
The motion of a rigid body in which the particles
move in circular paths with their centers on a fixed
straight line called the axis of rotation.
III. Plane Motion
The motion of a rigid body in which all particles in
the body remain at a constant distance from a fied
reference plane
Notations
s = distance
x = horizontal displacement
y = vertical displacement
v = velocity at any time (final velocity)
vo = initial velocity (velocity at time = 0)
a = acceleration
g = acceleration due to gravity (9.81 m/s2, 32.2
ft/s2)
t = time
TRANSLATION
Rectilinear Translation
Motion along a straight line.
s=vt
B. Variable acceleration
dv ds
a= v=
dt dt
v dv = a ds
C. Constant acceleration
v = vo + at
1 2
s = vo t + at
2
v2 = vo 2 + 2as
v = gt
1 2
s= gt
2
v2 = 2gh
Curvilinear Translation (Projectile Motion)
Resolve the motion into x and y-components and
use the formulas in rectilinear translation.
vy = 0, v = vx
A
B vx
vo vy
H v
y
voy
vy = -voy
O vox x C
R D
At any point B:
x-component of motion (ax = 0)
vx = vox x = voxt
vy = voy - gt
1
y = voy t - gt2
2
vy 2 = voy 2 - 2gy
gx2
y = x tan -
2vo 2 cos2
At the summit A: (vy = 0)
voy 2
H=
2g
voy
t=
g
At point C:
y=0 vy = -voy v = vo
2
vo sin 2
R=
g
2voy 2vo sin
t= =
g g
Motion Curves
zero acceleration
a-t
diagram
A1
zero slope
v-t
diagram zero velocity
A1
A2
t
s
zero slope
s-t
diagram
A2
t
The relationships between these curves are:
ds
v=
dt
Pnet
Pnet
t
Net force diagram
xG
a = Pnet/M Time where the v
and s are required
Area
cg
t
Acceleration diagram
v = vo + area
S = vo t1 + area xG
ROTATION
Notation
= angular displacement, radians
= angular speed, rad/sec
= angular acceleration, rad/sec2
t = time
Uniform Motion
=t
Uniform Acceleration
= o + t
1
= o t + t2
2
2 = o 2 + 2
r
a
r
v
r
s=r
v=r
a=r
KINETICS
F F
a or a = k (where k = 1)
M M
F=ma
DAlemberts Principle
W
The resultant of the external a
Friction pulley
w1 + w2
a = w2 - w1 a
g a
W1
W2
w1 < w2
y
y
W 1x x x
a
W1 a
W2
f
w1 + w2
- Forces
a = Forces = W2 f - W1x
g
Centrifugal Force (Reverse Normal Effective
Force)
r W
an
CF
T
Normal acceleration, an = 2 r = v2 / r
CF = M an = M 2 r
W 2 W v2
CF = r=
g gr
W
T = M at = r
g
Conical Pendulum
T W
L
h
CF
r T
W CF
v
CF 2 r v2
tan = = =
W g gr
g g
cos = 2 for >
L L
W
T=
cos
h
Time to complete one revolution, t = 2
g
L
t max = 2
g
Rod uniform mass of length L rotated about one
end:
2g
cos = for > 2g/L
2 L
L
Banking of Curves
v2
tan ( + ) =
gR
v2
The ratio is also known as the impact factor or
gR
centrifugal ration.
Horizontal Rotating Platform
2 R v2
tan = =
g gR
Impulse-momentum Equation
When a body of weight W moving with an initial
velocity vo changes its velocity to vf over a period of
t along a straight line,
W
(+) Impulse - (-) Impulse = (vf - vo )
g
V1 V2 V1 V2
M1 M2 M1 M2
Coefficient of Restitution
h'
e=
h
tan
e=
tan '
1 M1 M2
Loss in KE = 1 - e2 (v1 - v2 )2
2 M1 + M2
CHAPTER 10 STRENGTH OF MATERIALS
SIMPLE STRESS
Normal Stress
Stress is defined as the strength of a material per
unit area or unit strength. It is the force on a
member divided by the area, which carries the
force, formerly expressed as psi, now in N/mm2 or
MPa.
P
=
A
Where P is the applied normal load in Newton and
A is the area in mm2. The maximum stress in
tension or compression occurs over a section
normal to the load.
P
Bar in Compression
Bar in Tension
P
P
Shearing Stress
V
=
A
Single Shear
P
P
Double Shear
P
P
P
Bearing Stress
Bearing stress is the contact pressure between
separate bodies. It differs from compressive
stress, as it is internal stress caused by
compressive forces.
Pb Pb
Ab
Pb
b =
Ab
ST
pressure SL SL
p
ST
pD
Tangential stress, ST =
2t
pD
Longitudinal stress. SL =
4t
p = internal pressure - external pressure
Spherical Shell
If a spherical tank of diameter D and thickness t
contains a gas under a pressure of p, the stress at
the wall can be expressed as:
pD
Wall stress, S =
4t
THICK-WALLED CYLINDER
In thin-walled cylinders,
the thickness of the
wall is very small po
compared to the tank b
diameter. If in the case pi
of thick-walled a r
cylinders, the
tangential stress ST
and radial stress SR at
any distance r from the
center is given by the following equations:
a2 pi - b2 po a2 b2 ( pi - po )
SR = 2
- 2
b - a2 (b - a2 ) r2
a2 pi - b2 po a2 b2 ( pi - po )
ST = 2
+ 2
b - a2 (b - a2 ) r2
Simple Strain
Also known as a unit deformation, strain is the
ratio of the change in length caused by an applied
force, to the original length.
L P
=
L
R
Y
E
P
Yield point
Elastic limit
Proportional limit
O Strain,
or = K
=E
Elastic Limit
The elastic limit is the limit beyond which the
material will no longer go back to its original shape
when the load is removed, or it is the maximum
stress that may be developed such that there is no
permanent or residual deformation when the load
is entirely removed.
Elastic and Plastic Ranges
The region in the stress-strain diagram from O to
P is called the elastic range. The region from P to
R is called the plastic range.
Yield Point
The point at which the material will have an
appreciable elongation or yielding without any
increase of load.
Ultimate Strength
The maximum ordinate in the stress-strain
diagram is the ultimate strength or tensile strength.
Rapture Strength
The strength of the material at rapture. This is also
known as the breaking strength.
Modulus of Resilience
Modulus of resilience is the work done on a unit
volume of material as the force is gradually
increased from O to P, in N-m/m3. This may be
calculated as the area under stress-strain curve
from the origin O up to the elastic limit E. The
resilience of a material is its ability to absorb
energy without creating a permanent distortion.
Modulus of Toughness
Modulus of toughness is the work done on a unit
volume of material as the force is gradually
increased from O to R, in N-m/m3. This may be
calculated as the area under the entire stress-
strain curve (from O to R). The toughness of a
material is its ability to absorb energy without
causing it to break.
Working Stress, Allowable Stress, and Factor
of Safety
Working stress is defined as the actual stress of a
material under a given loading. The maximum safe
stress that a material can carry is termed as the
allowable stress. The allowable stress should be
limited to values not exceeding the proportional
limit. However, since proportional limit is difficult to
determine accurately, the allowable stress is taken
as either the yield point or ultimate strength
divided by a factor of safety. The ratio of this
strength (ultimate or yield strength) to the
allowable strength is called the factor of safety.
Axial Deformation
In the linear portion of the stress-strain diagram,
the stress is proportional to strain and is given by
= E . Since = P / A and = / L, then P / A = E
/ L, or
PL L
= =
AE E
P
y y
t
dx
x dx
P L dx
=
E O A
g L2 M g L
= =
2E 2AE
P
k=
Shearing Deformation
s
L
G=
VL L
s = =
As G G
Poissons Ratio
When a bar is subjected to a tensile loading there
is an increase in length of the bar in the direction
of the applied load, but there is also a decrease in
the lateral dimension perpendicular to the load.
The ratio of the sidewise deformation (or strain) to
the longitudinal deformation (or strain) is called the
Poissons and is denoted by . For most steel, it
lays in the range 0.25 to 0.3, and 0.20 for
concrete.
x
y
P
P
y z
=- =-
x x
x y x + y E
x = - or x =
E E 1 - 2
And
y x y + x E
y = - or y =
E E 1 - 2
Triaxial Deformation
If an element is subjected simultaneously by three
mutually perpendicular normal stresses x, y, and
z, which are accompanied by strains x, y, and z,
respectively.
1
x = - y + z
E x
1
y = - (x + z )
E y
1
z = - x + y
E z
E
G=
2 ( 1 + )
E
K= =
3 ( 1 2 ) V / V
Where V is the volume and V is change in
volume. The ratio V/V is called volumetric strain
and can be expressed as:
V 3 ( 1 2 )
= =
V K E
Thermal Stress
Temperature changes cause the body to expand
or contract. The amount of linear deformation, T,
is given by:
T = L ( Tf - Ti )= L T
a
L
L T = a +
E
Torsion
Consider a bar to be rigidly attached at one end
and twisted at the other end by a torque or twisting
moment T equivalent to F x d, which is applied
perpendicular to the axis of the bar, as shown in
the figure. Such a bar is said to be in torsion.
F
d
r
T L
T
Torsional Shearing Stress
For a solid or hallow circular shaft subject to a
twisting moment T the torsional shearing stress Ss
at a distance from the center of the shaft is:
T Tr
Ss = and Ss max =
J J
TL
= in radians
JG
Centerline of wall
The torque applied in thin-walled tubes is
expressed as:
T=2Aq
q T
Ss = =
t 2At
Helical Springs
When close-coiled helical spring, composed of a
wire of round rod diameter d wound into a helix of
mean radius R with n number of turns, is
subjected to an axial load P gives the following
stresses and elongation:
16 PR d
Ss = 1 +
d3 4R
16 PR 4m - 1 0.615
Ss = +
d3 4m - 4 m
64 P R3 n
=
G d4
P G d4
k= = in N/mm
64 R3 n
Springs in Series
1 1 1
= + +
k k1 k2
Where k1, k2 are the spring constants for
different springs.
Springs in Parallel
k = k1 + k2 +
Definition of Beam
A beam is a bar subject to forces or couples that
lie in a plane containing the longitudinal of the bar.
According to determinacy, a beam may be
determinate or indeterminate.
Statically Determinate Beam
Statically determinate beams are those beams in
which the reactions of the supports may be
determined by the use of the equations of static
equilibrium. The beams shown below are
examples of statically determinate beams.
Cantilever Beam
Simple Beam
Overhang Beam
R3 Propped Beam
M
R2
R1
Restrained Beam
Constinuous Beam
Types of Loading
Loads applied to a beam may consist of a
concentrated load (load applied at a point),
uniform load, uniformly varying load, or an applied
couple or moment. These loads are shown in the
following figures:
Concentrated Loads Uniformly Distributed
Load
x x
M
w (N/m) w (N/m)
A B A
C CV
R1 L R2
R1
R1 = R2 = wL/2
of the beam to the right of C be removed. The
portion removed must then be replaced by a
vertical shearing force V together with a couple m,
to hold the left portion of the bar in equilibrium
under the action of the force R1 and wx.
The couple M is called the resisting moment or
moment and the force V is called the resisting
shear or shear. The sign of V and M are taken to
be positive if they have the senses indicated
above.
dM wL
= - w x = V (shear)
dx 2
Therefore, the rate of change of the moment with
respect to x is the shearing force, or the shear is
the slope of the moment diagram.
dV
= - w (load)
dx
Sign Conventions
The customary sign conventions for shearing force
and bending moment are represented by the
figures below. A force that tends to bend the beam
downward is said to produce a positive bending
moment. A force that tends to shear the left
portion of the beam upward with respect to the
right portion is said to produce a positive shearing
force.
An easier way for determining the sign of the
bending moment at any section is that upward
forces always cause positive bending moments
regardless of whether they act to the left or to the
right of the exploratory section.
Load
Load
Shear Diagram
Moment Diagram
(PL - Ps d)2
Mmax =
4PL
P = Ps + Pb
E I Ec
= =
M fb
Flexure Formula
The stresses caused by bending moment are
known as bending or flexural stresses, and the
relation between these stresses and the bending
moment is expressed by the flexure formula.
At any fiber at distance y from the neutral axis, the
bending (flexural) stress is:
My
fb =
I
Mc M
fb max = =
I S
I
Section modulus, S =
c
Shearing Stress Formula
VQ
fv =
It
Q = A y
Shear Flow
If the shearing stress fv is multiplied by the width t,
we obtain a quantity q, known as the shear flow,
which represents the longitudinal force per unit
length transmitted across a section at a level y1
from the neutral axis.
VQ
q = fv t =
I
Application of Flexural and Shearing Stresses
to Rectangular Section
For rectangular beam of width b and depth d:
6M
fb max =
bd2
3V 3V
fv max = =
2bd 2A
Superimposed Beams
If a beam is composed of two or more thin layers
placed on each other without any attachment, the
separate layers would slide past each other and
the total strength of the beam would be the sum of
the strengths of the various layers. The strength of
this beam is lesser than a solid beam having the
same cross-sectional area.
Neglecting friction between any two adjacent
layers, the following relationships may be used:
M = M1 + M2 + M3
I = I1 + I2 + I3
M
Curvature, = constant
EI
M M1 M2 M3
= = =
E I E I1 E I2 E I3
Spacing of Rivets or Bolt in built-Up Beam
When two or more thin layers of beams are
fastened together with a bolt or a rivet so that they
act as a unit to gain more strength, it is necessary
to design the two sizes or spacing of these bolts or
rivets so that it can carry the shearing force acting
between each adjacent layers.
Consider the beams shown in the figure
RI
s=
VQ
Where R is the total shearing force to be resisted
by the bolts and is equal to the allowable shearing
stress x area x number of bolts in the group. R
may also be taken as the bearing capacity at the
section.
Economic Sections
From the flexure formula fb = M y / I, it can be seen
that the bending stress at the neutral axis, where y
= 0, is zero and increases linearly outwards. This
means that for a rectangular or circular section a
large portion of the cross section near the middle
section is under stressed.
For steel beams or composite beams, instead of
adopting the rectangular shape, the area may be
arranged to give more area on the outer fiber and
maintaining the same overall depth, and saving a
lot of weight.
When using a wide flange or I-beam section for
long beams, the compression flanges tend to
buckle horizontally sidewise. This buckling is a
column effect, which may be prevented by
providing lateral support such as a floor system so
that the full allowable stresses may be used,
otherwise the stress should be reduced. The
reduction of stresses for these beams will be
discussed in steel design.
Combined Stresses
In the previous sections, we studied the three
basic types of stresses; the axial or normal stress,
torsional shearing stress, and flexural stress, in
which we assume that only one of these stresses
acts on a member. However, these stresses may
act simultaneously and there are four possible
combinations of these stresses: (1) axial and
flexure; (2) axial and torsional; (3) torsional and
flexural; and (4) axial, torsional, and flexural.
Combined Axial and Flexure
The simplest of the four combinations is the axial
and flexural because it combines only normal
stresses, which can be added arithmetically.
A member subject to bending moment M and an
axial load P causes a flexural stress of fb = M y / I
at any point from its neutral axis, and a normal
stress of P / A which is uniformly distributed over
its entire area, respectively. The combination of
these stresses results if the member is
eccentrically loaded as in a column or a
prestressed beam. The combined stress
developed is:
1 = +
P My
f=
A I
Where m = Pe, e is the eccentricity of P from the
neutral axis
Kern of a Section
I
a=
Ae
fx + fy fx - fy
f= - cos 2 + sxy sin 2
2 2
sx - sy
s= sin 2 + sxy cos 2
2
Principal Stresses
2
fx + fy fx - fy 2
fmax = + + sxy
2 2
2
fx + fy fx - fy 2
fmin = - + sxy
2 2
Principal Planes
- 2sxy
tan 2p =
fx - fy
There are always two values of p that will satisfy
this equation. The maximum stress occurs on one
of these planes, and the minimum stress occurs
on the other. The planes defined by the angle p
are known as principal planes.
2
fx - fy 2
smax = + sxy
min 2
The angle between the x-axis and the planes on
which the maximum and minimum shearing
stresses occur are given by the equation
fx - fy
tan 2s =
2sxy
Mohrs Circle
A visual interpretation of the formulas in the
preceding sections, devised by the German
engineer Otto Mohr in 1882, eliminates the
necessity of remembering them. In this
interpretation, a circle is used; accordingly, the
construction is called Mohrs Circle.
The following are the rules for applying Mohrs
circle to combined stresses, given fx, fy, and sxy.
1. On the rectangular f-s axes (f along x-axis
and s along y-axis), plot points having the
coordinates (fx, sxy) and (fy,syx). In plotting
these points, assume tension as plus,
compression as minus, and shearing stress
as plus when its moment about the center of
the element is clockwise.
Te = M2 + T2
1
Me = (M + Te )
2
Where M is the actual bending moment and T is
the actual torque. Once Me and Te is computed,
the formulas for flexural and torsional stresses can
be used
Me c 4Me
Max fb = =
I r3
T 16 Te
Max s = =
J d3
w (N/m)
A
B
C
wL2
Mmax = Mmid =
8
5wL4
max = mid =
384 EI
wL3
A = B =
24 EI
P
L/2 L/2
B
A
L
PL
Mmax = Mmid =
4
PL3
max = mid =
48 EI
PL2
A = B =
16 EI
a b
B
A C L
Pab
Mmax = Mp =
L
L 2 - b2
x=
3
3
Pb L2 - b2 2
max = C =
93 E I L
Pb
mid = 3L2 - 4b2 when a > b
48 EI
w (N/m)
y
B
A C L
2.5wL4
mid =
384 EI
8wL3 7wL3
B = ; =
360 EI A 360 EI
wx
EI y = 7L4 - 10L2 x2 + 3x4
360L
y
w (N/m)
A B
L
wL2
Mmax =
12
wL4
max = mid =
120 EI
5wL3
A = B =
192 EI
wx
EI y = 25L4 - 40L2 x2 + 16x4
960 L
for 0 < x < L/2
0.577L
M
x
A B
L
Mmax = M
ML3
max = at x = 0.577L
93 EI
ML ML
A = ; =
6 EI B 3 EI
Mx
EI y = (L - x)(2L - x)
6L
y
A L
Mmax = MA = - PI
PL3
max = B =
3 EI
PL2
B =
2 EI
Px2
EI y = (3L - x)
6
y
a b
x
A L
Mmax = MA = - Pa
Pa2
max = B = (3L - a)
6 EI
Pa2
B =
2 EI
Px2
EI y = (3a - x) for 0 < x < a
6
Pa2
EI y = (3x - a) for a < x < 1
6
y
w (N/m)
x
L
A
wL2
Mmax = MA = -
2
wL4
max = B =
8 EI
wL3
B =
6 EI
wx2
EI y = 256 - 4Lx + x2
24
y
w (N/m)
x
L
A
wL2
Mmax = MA = -
6
wL4
max = B =
30 EI
wL3
B =
24 EI
Mx2
EI y =
2
PROPPED BEAM FORMULAS
M
x
L
A
Pa2 3L - a
R=
2L3
P a2 b
MA = - 2
b2 a +
L 2
a b
A B
L
R
5P
R=
16
3PL
MA = -
16
w (N/m)
A B
L R
7wL
R=
16
wL2
MA = -
8
w (N/m)
A B
L R
7wL
R=
128
9wL2
MA = -
128
w (N/m)
A B
a b
L R
3
wb
R= (4L - b)
8L3
wa2
MA = RL -
2
w (N/m)
A B
L
R
wL
R=
10
wL2
MA = -
15
w (N/m)
A B
L
R
11wL
R=
40
7wL2
MA = -
120
w (N/m)
L
R
11wL
R=
64
5wL2
MA = -
64
3 EI
MA =
L2
a b
A B
L
Pab2
MA = -
L
Pba2
MB = -
L
Pb2
mid = (3L - 4b)
48 EI
P
L/2 L/2
A B
L
PL
MA = M B = -
8
PL3
max =
192 EI
w (N/m)
A B
L
wL2
MA = M B = -
12
4
wL
max =
384 EI
w (N/m)
A B
L/2 L/2
5wL2
MA = -
192
11wL2
MB = -
192
w (N/m)
A B
L
wL4 wL2
max = MA = -
768 EI 30
2
wL
MB = -
30
wL4
mid =
768 EI
w (N/m)
A B
L
5wL2
MA = M B = -
96
4
7wL
max =
3840 EI
M
a b
A B
L
Mb 3a
MA = - 1
L L
Ma 3b
MB = - - 1
L L
6 EI
MA = -
L2
6 EI
MB =
L2
P = y dx
Any loading
a b
A B
x2
Pab2
MA = -
x1 L2
x2
Pba2
MB = -
x1 L2
a=x;b=L-x
P = y dx
For varying load, y = f(x)
P
k= (N/mm or kN/mm)
PL3 3 EI
Static deformation, st = ,k= 3
3 EI L
If a load P is dropped from a height of h, the
resulting deformation can be computed from:
2h
=1+ 1+
st st
Impact Stress
2h
=1+ 1+
st st
Properties of Fluid
Weight of Fluid
=
Volume
Mass of Fluid
=
Volume
Density of Gases
p
=
RT
where:
p = absolute pressure of gas in Kpa
R = gas constant in joule/ kg-K
For air, R = 287 joule/ kg-K
T = absolute temperature in degree
Kelvin K = C +273
Specific Volume, Vs
1
VS =
Specific Gravity, s
fluid fluid
s = =
water water
Viscosity
The property of a fluid which determines the
amount of its resistance to shearing force. A
perfect would have no viscosity.
= (Pascal-second or poise)
dV/dy
Note: 1 poise = 0.01 Pa/s
= (m2 /s or stokes)
4
p=
d
where:
= surface tension in N/ m
d = diameter of the droplet in m
p = gage pressure in Pascal
Capillarity
4 cos
h=
d
4
h=
d
where:
h = capillary rise or depression
d = diameter of the tube
= unit weight
= surface tension
dp'
E= ( lb /in2 or Pa)
-dv/v
where:
dp = change in pressure
dv = change in volume
v = volume
Compression of Gases
pvn = p1 v1 n = constant
pv = p1 v1
p1 v 1 k = p2 v 2 k
Pressure Disturbances
EB
c= (m/s or ft /s)
where:
c = celerity or velocity of pressure wave
in m/s or ft/s
EB = bulk modulus of elasticity if the fluid
in Pa or lb/ft2
Unit Pressure
Variations in Pressure
p2 - p1 = h
h
p = h
Air Pressure = p
h1 Liquid 1
Liquid 2
h2
h3 Liquid 3
pbottom = h +p
pbottom = 1 h + 2 h2 + 3 h3
1
Total Hydrostatic Pressure
cg
cp
F = pcg A or F = hA
Ig h
e= y=
AY sin
where :
pcg = pressure at the centroid of the
plane
Ig = centroidal moment of inertia of the
plane
A = area of the plane surface
= angle that the plane makes with the
horizontal
D C
FV
cg
FH
FH = pcg A
Fv = VABCD
F = FH 2 + Fv 2
FV
tan =
FH
where:
FH = total force acting on the vertical
projection of the curved surface.
Fv = weight of imaginary or real fluid
directly above the curved surfaces.
Bouyancy
VD
VD
BF
BF
For a homogenous body floating on a
homogeneous liquid, the volume displaced is:
body Sbody
VD = Vbody = V
liquid Sliquid body
S
MBo =
VD sin
RM or OM = W(x) = W ( Mg sin )
Where:
= volume of the wedge of immersion
s = horizontal distance between the
centroid of the wedges
VD = volume displaced
= angle of tilting
B2 tan2
MBo = 1 +
12D 2
where: B = width D = draft
Metacentric Height
I
MBo =
VD
Horizontal Acceleration
a
a
tan =
g
Inclined Acceleration
ah
tan =
g + av
aH = a cos
aV = a sin
Vertical Acceleration
a
p = h 1
g
Rotation
2 x2
y=
2g
dy
tan =
dx
2 x
tan =
g
Volume of Paraboloid
V = 1/2r2 h
Flow rate
Q
1
3
2 Q
Incompressible Fluid
Q1 = Q2 = Q3
A1 V1 = A2 V2 = A3 V3..
Compressible Fluid
1 Q1 = 2 Q2
Reynolds Number
Energy Equation
ZB
A
ZA
Datum
E1 + HA HE HL = E2
Darcy-Weisbach Formula
fL v2
hf = in ft or meter
D 2g
64
For Laminar Flow, f=
R
For non-circular pipe, use D=4R
v2 8Q2 fL 8Q2
= hf =
2g 2 gD4 D 2 gD4
0.0826fLQ2
For S.I. units, hf = a
D5
1 2/3 1/2
v= R S
n
10.29n2 LQ2
hf =
D16/3
10.67LQ1.85
hf = 1.85
C1 D4.87
1 vn 2
hn = 2 - 1
Cv 2g
p2 -p1
HL =
H
HL = H
Pipes in Series
A 1 B 2 C 3 D
Q1 Q2 Q3
Q1 = Q2 = Q3
HL = hL1 + hL2 + hL3
Pipes in parallel
1 Q1
2 Q2
3 Q3
Q = Q1 + Q2 + Q3
HL = hL1 = hL2 = hL3
Equivalent pipe
For a pipe or system of pipes (O), the equivalent
single pipe (E) is must satisfy the following
conditions.
QE = Qo
and HLE = HLo
h2
As dh
t=
Qi -Qo
h1
If Qi = 0
h1
As dh
t=
Qo
h2
If the flow is through an orifice under a variable
head H:
Qo = CAo 2gH
2As
t= H1 - H2
CAo 2g
Weir
General Formula
2
Q = C2gL(H + hv )3/2 - hv 3/2
3
or Q = Cw L(H + hv )3/2 - hv 3/2
Va 2
where hv = velocity head of approach
2g
C = coefficient of discharge
Cw = weir factor
Neglecting va:
2
Q= C2gLH3/2 or Q = Cw LH3/2
3
Considering va:
Neglecting va:
Q=1.84LH3/2
Cipolleti weir
Q = 1.859 L H3/2
= 75.9637 = 7557' 50"
= 14.0363 =142'10"
Triangular V-notch Weir
8
Q= C2gtan H5/2
15 2
X
H
Y
q = CK2gH
K = x y
Unsteady flow weir (Variable Head)
H1
As dH
t=
Qo
H2
2As 1 1
t= -
Cw L H2 H1
2As 1
t=
Cw L H1
Hydrodynamics
Force against a fixed flat plate held normal to
the jet
Q
F= v = Qv
g
RX
V1
Q
Fx = (v - v )
g 1x 2x
Q
Fy = v - v
g 1y 2y
F = Fx 2 + Fy 2
where:
v1 = velocity of the jet before hitting the
vane
v2 = velocity of the jet as it leaves the
vane
u v
RY
v
RX v
'
u
u v
y
v1
v'
v
2x
v'
Q' Q'
Fx = (v1x -v2x ) Fy = v1y -v2y
g g
Q' = Au
u = v1 -v'
Dynamic Force
Q' Q'
Fx = (v1x - v2x ) Fy = v1y - v2y
g g
Total Force
Q'
Fx = (v1x - v2x ) = F1x + Rx - F2x
g
Q'
Fy = v1y - v2y = F1y + Ry - F2y
g
Drag Force
2 2
DF = =
2 2
Where:
= Drag Coefficient
= Density of the Fluid
A = Area normal to the direction of motion
Chapter Twelve
Engineering Economics
Interest
0 1 2 3 4 5
0 1 2 3 4 5
P30T
Simple Interest
F=P+I
I = Prt
F = P +I = P+ Prt
F = P(1 + rt)
Value of t
Example:
4 years; t=4
3 months; t=3/12 or
90 days
Ordinary simple interest, t= 90/360
Exact simple interest, t=90/365 or
90/360 for leap years
2 years and 4 months; t=2+4/12
=2.3333
Compound Interest
P1000
i = r/m
I = interest earned
I=F-P
F = P(1 + i)n
F
P=
(1 + i)n
1
The term , also denoted as (P/F, i, n) is called
(1 + i)n
the single payment present-worth factor.
Values of i and n
Compounded annually ( m = 1)
i = 0.12/1= 0.12
n = 5(1) = 5
Compounded semi-annually (m = 2)
i = 0.12/2 = 0.06
n = 5(2) = 10
Compound quarterly (m = 4)
I =0.12/4 = 0.03
n = 5(4) = 20
Compounded monthly (m = 12)
i = 0.12/12 = 0.01
n = 5(12) = 60
Compounded bi-monthly
i = 0.12/6 = 0.02
n = 5(6) =30
Continuous Compounding (m )
F = Pert
Annuity
Elements of Annuity
A=periodic payment
P= present worth of all periodic payments
F or S= future worth or sum of all periodic
payments after the last payment is made
i= interest rate per payment
n=number of payments
Types of Annuity
A. Ordinary Annuity
A A A A A
A[(1 + i)n - 1]
F=
i
(1 + i)n - 1
The factor is called equal-payment-series-
i
compound-amount factor and is denoted as (F/A, i,
n)
The value of A if F is known is:
Fi
A=
(1 + i)n - 1
i
The factor (1 + i)n - 1
is known as equal- payment
sinking- fund factor and is denoted as (A/F, i, n)
F A[(1 + i)n - 1]
P= =
(1 + i)n (1 + i) n i
(1 + i)n - 1
The factor (1 + i)n i
, is known as equal- payment-
series- present- worth factor and is designated as
(P/A, i, n)
P(1 + i)n i
A=
(1 + i)n - 1
(1 + i)n i
The factor (1 + i)n - 1
, is known as the equal-
payment- series- capital- recovery factor and is
designated as (A/P, i, n)
B. Deferred annuity
0 1 2 3 4 5
A A A A
P n=5
C. Annuity due
A A A A A A
S
F
n=6
P
n=5
A[(1 + i)n - 1]
P=
(1 + i)n i
Perpetuity
A
P=
i
Uniform gradient
Arithmetic Gradient
0 1 2 3 4 5 n
A
A+G
A+2G
A+3G
A+4G
A+nG
A+G
A+G(1+r)
A+G(1+r)2
A+G(1+r)3
A+G(1+r)n
1+
Let =
1+
Present worth
If w 1
A[(1 + i)n - 1] G 1 - wn
P= +
(1 + i)n i 1+i 1-w
If w = 1(for r = i)
A[(1 + i)n - 1] Gn
P= +
(1 + i)n i 1+r
F = P(1 + i)n
0 1 2 3 4 5 n
OM OM OM OM OM OM
FC
RC
Capitalized cost, K
OM RC - SV
K = FC + +
i (1 + i)n - 1
Annual Cost, AC = K i
(RC -SV)i
AC = (FC)i + OM + n
(1 + i) - 1
Example 12-1
Solution:
OM RC-SV
K = FC + + n
i (1-i) -1
5,000 300,000 - 5,000
K = 300,000 + + 15
0.05 (1+0.05) -1
K = P 631,711.44
Cost Comparison of different alternatives
Example 12-2
Elements of Depreciation
FC = first cost
SV = salvage value or trade-in value
d = depreciation charge
n = economic life of the property in years
m = any time before n
BVm = book value after m years
Dm = total depreciation for m years
Dm
Cost D
Curved
FC
BVm
SV
m
time
n
The book value of the property at any time m is:
BVm = FC - Dm
n SV
Constant Percentage, K = 1-
FC
BVm = FC(1 + K)m
dm = FC(1 - K)m - 1 K
2
Depreciation charge to date = BV at the
n
beginning of the year
2 m
BVm = FC 1 - SV
n
2
dm = BVm - 1 and BVn = SV
n
Example 12-3
A mine costs P21M, and will last for 20 years. Its
plant has a salvage value of P1M, at the end of
the time. The mine will yield an equal dividend at
the end of each year. What is the annual dividend,
if it is sufficient to pay interest annually at the rate
of 6% on the original investment and to
accumulate a replacement fund, invested at 4%?
(RC - SV)i
D = (FC)r + n
(1 + i) - 1
(21,000,000-1,000,000)(0.04)
D = (21,000,000)(0.06) +
(1 + 0.04)20 - 1
D = P1,931,635.00
Bond
Elements
F = face value or par value of the bond
C = redemption value on a specified redemption
date
r = bond rate or dividend rate
D = periodic dividend
D=Fr
C D[(1 + i)n - 1]
P= n = n
(1 + i) (1 + i) i
Example 12-4
A P100, 000.00, 6% bond, pays dividends semi-
annually and will be redeemed at 110 % on July
19, 1999. Find its price if bought on July 1, 1996,
to yield an investor 4%, compounded semi-
annually.
Solution:
Face value of the bond, F = P100, 000.00
Redemption Value, C = 10 %( 100,000) = P110,
000.00
Bond Rate, r = 0.06/2 = 0.03
Periodic Dividend, D=F x r=100,000(0.03)
= P3, 000.00
Investors rate of return (per semi- annual),
I = 0.04/2 = 0.02
Number of dividends, n = 3(2) = 6
C D[(1 + i)n - 1]
P= +
(1 + i)n (1 + i)n i
110,000 3000(1 + 0.02)6 - 1
P= +
(1 + 0.02)6 (1 + 0.02)6 (0.02)
P=P114,481.14
Elements
R = pN
C=total cost
C = f + aN
Cost = f + aN
Break-even Point
Loss
Break-even
Quantity
Example 12-5
Solution:
Given f = fixed cost = P69, 000 per month
a = marginal cost = P7.00 +P2.00 + P1.50
a = P10.50
p = marginal revenue = P45.00
f
N=
p-a
69, 000
N=
45-10.5
Prefixes
Yocto (y) .10-24
Zepto (z) .10-21
Atto (a) ....10-18
Femto (f) 10-15
Pico (p)....10-12
Nano (n)..10-9
Mirco (u)..10-6
Centimilli (cm).10-5
Decimille (dm).10-4
Milli (m).10-3
Centi (c)10-2
Deci (d).10-1
Deca (D)...101
Hecto (h)...102
Kilo (K)..103
Mega (M)..106
Giga (G)109
Tera (T).1012
Peta (P).1015
Exa (E)..1018
Zetta (Z)1021
Yotta (Y)...1024