introdcution to managment 101 for engineers PDF

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introdcution to managment 101 for engineers
introdcution to managment 101 for engineersintrodcution to managment 101 for engineers
introdcution to managment 101 for engineers...

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Civil Engineering Mechanics CVG2149 Elena Dragomirescu

What is Civil Eng. Mechanics? Civil Eng. Mechanics is the science which describes and predicts the conditions of rest or motion of bodies under the action of forces.

• Categories of Mechanics: - Rigid bodies - Statics - Dynamics - Deformable bodies - Fluids • Mechanics is an applied science - it is not an abstract or pure science but does not have the empiricism found in other engineering sciences. • Mechanics is the foundation of most engineering sciences and is an indispensable prerequisite to their study.

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Examples of mechanics applied in civil engineering Static Structural Analysis

Columns remained intact Bridge deck collapsed!!

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Quebec Bridge

–

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Due to a design flaw the actual weight of the bridge was heavier than its carrying capacity, which caused it to collapse twice, first in 1907, then in 1916. DEATHS: 95, from both tragedies

http://www.time.com/time/photogallery/0,29307,1649646_1421688,00.html

https://simple.wikipedia.org/w iki/Millennium_Bridge

https://www.youtube.com/watch?v=eAXVa __XWZ8

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A collapse which led to development

1940 - Failure of Tacoma Bridge only 4 months after opening

Science of Equilibrium

2D Equilibrium

Equilibrium of a Rigid Body

3D Equilibrium

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Newtonian Mechanics

Mass Absolute Concepts

Space Time

(independent from each other) Force depends on the mass of the body, variation of velocity

Force

In Newtonian Mechanics, space, time, and mass are absolute concepts, independent of each other. The force acting on a body, however, is related to the mass of the body and the variation of its velocity with time.

1. Fundamental concepts:

z

A. Particle: The very small amount of matter which may be assumed to occupy a single point in space.

3D

P y

z x

B. Rigid body: combination of large number of particles occupying fixed positions with respect to each other .

y

x

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• C. Free-Body Diagrams

Space Diagram: A sketch showing the physical conditions of the problem.

Free-Body Diagram: A sketch showing only the forces on the selected particle.

3. Fundamental Principles 3.1. Newton’s First Law 3.2. Newton’s Second Law 3.3. Newton’s Third Law

3.5 Addition of vectors: Parallelogram Law

3.4. Newton’s Fourth Law of Gravitation

3.6 Principle of Transmissibility

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Review: Forces in Plane Forces in Space

Forces and Vectors Replacing multiple forces acting on a particle with a single equivalent or resultant force Relations between forces acting on a particle that is in a state of equilibrium. • Force: action characterized by its point of application, magnitude, line of action, and sense. • Force is a vector quantity.

• Vector: parameters possessing magnitude and direction which add according to the parallelogram law. Examples: displacements, velocities, accelerations. • Scalar: parameters possessing magnitude but not direction. Examples: mass, volume, temperature

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Review of Trigonometric functions Definitions: c a cosA  b b sin A tan A  cosA

C

sin A 

Law of sines:

Law of cosines

a A c

B

a b c   sin A sin B sin C a  b cosC  c cos B

a 2  b 2  c 2  2bc cosA

b  a cosC  c cos A c  a cos B  b cos A

b2  a 2  c2  2ac cos B c 2  a 2  b 2  2ab cosC

• Trigonometric solution - Apply the triangle rule. From the Law of Cosines,

R 2  P 2  Q 2  2PQ cos B  40 N  2  60 N  2  240 N 60 N cos 155

R  97.73N

From the Law of Sines,

sin A sin B  Q R sin A  sin B

Q R

 sin 155

60N 97.73N

A  15.04

  20  A

  35.04

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Rectangular Components of a Force: Unit Vectors • May resolve a force vector into perpendicular componentsso thatthe resulting parallelogram is a rectangle. Fx and F y are referred to as rectangular vector components and

   F  Fx  Fy • Define perpendicular unit vectors parallel to the x and y axes.

  i and j which are

• Vector components may be expressed as products of the unit vectors with the scalar magnitudes of the vector components.

   F  Fx i  F y j



• Fx and Fy are referred to as the scalar components of F

Example Method SOLUTION: • Resolve each force into rectangular components. • Determine the components of the resultant by adding the corresponding force components.

Four forces act on bolt A as shown. Determine the resultant of the force on the bolt.

• Calculate the magnitude and direction of the resultant.

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SOLUTION: • Resolve each force into rectangular components.

force mag  150 F1  80 F2  110 F3  100 F4

x  comp  129.9  27.4 0  96.6 199.1

y  comp  75.0  75.2  110.0  25.9 14.3

• Determine the components of the resultant by adding the corresponding force components.

R  199.12  14.32

R  199.6 N

• Calculate the magnitude and direction.

tan  

14.3 N 199.1 N

  4.1

Rectangular Components in Space •

If the force is in space and does not belong to neither plans: xoy, zox, yoz then decompose the force successively until rectangular components are reached

 • The vector F is contained in the plane OBAC.



• Resolve F into horizontal and vertical components.

Fy  F cos y Fh  F sin  y

• Resolve Fh into rectangular components

F x  Fh cos  F sin  y cos  F y  Fh sin   F sin  y sin 

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Rectangular Components in Space

 • With the angles between F and the axes, Fx  F cos x Fy  F cos y Fz  F cos z     F  Fxi  Fy j  Fzk      F cos xi cos y j  coszk F 













  cos x i  cos y j  cos z k





•  is a unit vector along the line of action of F , and cos z are the direction and cos x , cos  y cosines for F

Expression of F based on unit vectors i, j, k

F  Fx i  Fy j  Fz k scalar components

unit vectors

F  vector

F  F (cos x i  cos y j  cos z k ) scalar

unit vector, 

  cos x i  cos  y j  cos z k

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Force F defined by two points (Position Vectors) Method of Relative Position

How can we determine the vector of the resultant if the application point is not in the origin or if position is known (not the magnitude of the force)?

Tutorial Vectors http://www.screencast.com/users/Elndrag/folders/Mechanics/media/baf11b3e-d6ab4414-8a83-6fc21743806e

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(Position Vectors) Method of Relative Position Is MN vector not F vector! Direction of the force is defined by the location of two points,

M x1 , y1 , z1  and N x2 , y2 , z 2 

d x  x2 - x1

d y  y2 - y1

d z  z2 - z1

d  d x2  d y2  d z2

 M N  d xi  d y j  d z k  MN 1   (d i d y j d zk )  MN d x

  F    F  F  (d xi  d y j  d zk ) d Fd Fd Fdx Fz  z Fx  Fy  y d d d

cos x 

Determine the position vector directed from A to B and its  vector. Scalar components Fx, Fy, Fz

cos y 

dx d dy

d dz cos z  d

Example SOLUTION: • Based on the relative locations of the points A and B, determine the unit vector pointing from A towards B. • Apply the unit vector to determine the components of the force acting on A.

The tension in the guy wire is 2500 N. Determine:

• Noting that the components of the unit vector are the direction cosines for the vector, calculate the corresponding angles.

a) components Fx, Fy, Fz of the force acting on the bolt at A, b) the angles  x, y, z defining the direction of the force

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Example SOLUTION: • Determine the unit vector pointing from A towards B.

• Determine the components of the force.

Example • Noting that the components of the unit vector are the direction cosines for the vector, calculate the corresponding angles.

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