# Engineering Applications

Analytical and Numerical Calculation with MATLAB

Wiley (Verlag)
• 1. Auflage
• |
• erschienen am 24. März 2021
• |
• 368 Seiten

 E-Book | ePUB mit Adobe-DRM | Systemvoraussetzungen
978-1-119-09364-0 (ISBN)

A comprehensive text on the fundamental principles of mechanical engineering

Engineering Applications presents a comprehensive text to the fundamental principles and applications of the statics and mechanics of materials in the design of complex mechanical systems. The book uses the modern tool of MATLAB to help solve problems with numerical and analytical calculations. The authors noted experts on the topic offer an understanding of the static behaviour of engineering structures and components considering the mechanics of materials knowledge as an essential part (most important) for their design.

The authors explore the concepts, derivations and interpretations of the general principles and discuss the creation of mathematical models and the formulation of the mathematical equations. The practical text highlights the solutions of the problems that are solved analytically and numerically using MATLAB. The figures generated with MATLAB reinforce visual learning for students (and professionals) as they study the programs. This important text:

Shows how mechanical principles are applied to engineering design

Covers basic material with both mathematical and physical insight

Provides an understanding of classical mechanical principles

Offers the modern tool of MATLAB to solve problems

Helps to reinforce learning using visual and computational techniques

Written for students and professional mechanical engineers, Engineering Applications helps hone reasoning skills in order to interpret data, generate mathematical equations and learn different methods of solving them for evaluating and designing engineering systems.
 Auflage: 1. Auflage Sprache: Englisch Dateigröße: 18,36 MB Schlagworte: ISBN-13: 978-1-119-09364-0 (9781119093640)
weitere Ausgaben werden ermittelt
1 Forces 1

1.1 Terminology and Notation 1

1.2 Resolution of Forces 2

1.3 Angle Between Two Forces 4

1.4 Force Vector 5

1.5 Scalar (Dot) Product of two Forces 5

1.6 Cross Product of two Forces 6

1.7 Examples 6

2 Moments and Couples 15

2.1 Types of Moments 16

2.2 Moment of a Force About a Point 16

2.3 Moment of a Force about a Line 19

2.4 Couples 21

2.5 Examples 23

3 Equilibrium of Structures 59

3.1 Equilibrium Equations 60

3.2 Supports 61

3.3 Free-Body Diagrams 63

3.4 Two-Force and Three-Force Members 65

3.5 Plane Trusses 65

3.6 Analysis of Simple Trusses 67

3.6.1 Method of Joints 68

3.6.2 Method of Sections 70

3.7 Examples 73

4 Centroids and Moments of Inertia 137

4.1 Center of the Mass and Centroid 137

4.2 Centroid and center of the mass of a solid region, surface or curve 138

4.3 Method of decomposition 141

4.4 First moment of an area 142

4.5 The Center of Gravity 143

4.6 Examples 144

5 Stress, Strain and deflection 195

5.1 Stress 195

5.2 Elastic Strain 195

5.3 Shear and Moment 196

5.4 Deflections of Beams 199

5.5 Examples 203

6 Friction 223

6.1 Coefficient of Static Friction 224

6.2 Coefficient of Kinetic Friction 225

6.3 Friction Models 226

6.3.1 Coulomb friction model 226

6.3.2 Coulomb model with viscous friction 228

6.3.3 Coulomb model with stiction 229

6.4 Angle of Friction 230

6.5 Examples 232

7 Work, Energy and Power 269

7.1 Work 270

7.2 Kinetic Energy 271

7.3 Work and Power 272

7.4 Conservative Forces 274

7.5 Work Done by the Gravitational Force 274

7.6 Work Done by the Friction Force 275

7.7 Potential Energy and Conservation of Energy 275

7.8 Work Done and Potential Energy of an Elastic Force 276

7.9 Potential energy due to the gravitational force 277

7.9.1 Potential energy due to the gravitational force for a particle 277

7.9.2 Potential energy due to the gravitational force for a rigid body 278

7.10 Examples 279

8 Simple Machines 315

8.1 Load and Effort, Mechanical Advantage, Velocity Ratio and Efficiency of a

Simple Machine 315

8.2 Effort and Load of an Ideal Machine 317

8.3 The Lever 318

8.4 Inclined Plane (Wedge) 319

8.5 Screws 320

8.6 Simple Screwjack 321

8.7 Differential Screwjack 324

8.8 Pulleys 324

8.9 Differential Pulley 329

8.10 Wheel and Axle 329

8.11 Wheel and Differential Axle 331

8.12 Examples 334

# 1Forces

## 1.1 Terminology and Notation

A force exerted on a body tends to change the state of the body, that is, if the body is rigid the force tends to move the body, but when the body is elasto-plastic the force tends to deform the body.

A force can be defined as a vector quantity that is defined by magnitude and direction. The direction of a force is specified by its orientation (also known as the line of action) and sense. The magnitude of a force is a positive scalar. A scalar is a number expressed in specific units of measure.

Vectors (forces) are usually denoted by boldface letters. If the starting point and the end point of a vector (force) are given, the vector (force) could be denoted by or more simply . It is also usual to denote the magnitude of the vector (force) by or by . Some other notations for vectorial quantities could be , , or .

Graphically a force is represented by a straight arrow as shown in Figure 1.1. The point is named the application point or the origin of the force and the line passing through and is named the action line of .

There are some possible operations regarding vectors.

Equality of forces

Two forces and are equal to each other when they have the same magnitude and direction, that is

(1.1)

If the forces and are equal but are acting at different locations on the same body it will not cause identical motion.

Multiplication of a Force by a Scalar

The product between a force and a scalar written as , is a force having the same orientation as , the same sense as if if and opposite sense if , and the magnitude .

Figure 1.1 Vector representation.

Zero Force

A zero force, usually denoted by , has a zero magnitude and an undefined direction.

Unit Vector (Force)

A unit vector has its magnitude equal to unity, that is, . Any force can be written as a product of a unit vector having the same orientation and sense as the force and its magnitude or equivalent

(1.2)

The sum of a two forces and is a new force named resultant. The sum of the forces and is the force represented graphically by the diagonal of the parallelogram shown in Figure 1.2 with its tail connecting the tail of the force and head connecting the head of the force .

The sum is named the difference of the two forces as shown in Figure 1.3.

Figure 1.2 Parallelogram law of vector addition.

Figure 1.3 Parallelogram law of vector subtraction.

## 1.2 Resolution of Forces

If the unit vectors , , have the same application point (origin) and are perpendicular to each other, as shown in Figure 1.4, they form a Cartesian reference frame.

Any force can be expressed with respect to the unit vectors , , by where , , and are the , , components of the force.

The magnitude of can be written as

Addition and subtraction of forces could be easily manipulated using the resolution of forces into components. Considering the forces and , one can calculate

and

## 1.3 Angle Between Two Forces

The angles between the forces and , and respectively and - in the range between and - are usually denoted by Greek letters such as  and , as shown in Figure 1.5.

The direction of a force in a Cartesian frame is given by the direction cosines (Figure 1.6) of the angles between by the force and the associated unit vectors , , , written as

A unit force (of magnitude 1) having the same direction as can be written as

Figure 1.4 Resolution of a force.

Figure 1.5 The angles and between the forces and , and respectively and .

Figure 1.6 Direction cosines.

## 1.4 Force Vector

The position force (vector) shown in Figure 1.7 of a point relative to a point can be written as

(1.3)

The position force (vector) shown in Figure 1.7 of the point relative to a point is calculated with

(1.4)

Figure 1.7 Position forces (vectors).

## 1.5 Scalar (Dot) Product of Two Forces

Definition. The dot product of two forces and is

(1.5)

where is the angle between the forces and .

## 1.6 Cross Product of Two Forces

The cross product of two forces and is another force defined by (Figure 1.8)

(1.6)

where is a unit force normal to and having its direction given by the right-hand rule.

The magnitude of the cross product is given by

When and the cross product can be calculated using

(1.7)

Figure 1.8 Cross product of two forces and .

## 1.7 Examples

Example 1.1

Figure 1.9 shows three forces , , and , and the angles of the forces with the horizontal , , and . The forces have the magnitudes  kN,  kN, and  kN. Find the resultant of the planar forces and the angle of the resultant with the horizontal.

Solution

The input data are introduced in MATLAB with:

clear; clc; close all F1 = 1; % kN F2 = 3; % kN F3 = 2; % kN % angle of force F1_ with x-axis theta1 = pi/6; % angle of force F2_ with x-axis theta2 = pi/3; % angle of force F3_ with x-axis theta3 = pi;

Figure 1.9 Graphical representation the forces , and .

The components of the forces on and axes are

(1.8)

or in MATLAB:

% components of forces F1_, F2_, and F3_ F1x = F1*cos(theta1); F1y = F1*sin(theta1); F1_ = [F1x F1y]; F2x = F2*cos(theta2); F2y = F2*sin(theta2); F2_ = [F2x F2y]; F3x = F3*cos(theta3); F3y = F3*sin(theta3); F3_ = [F3x F3y];

The numerical values are:

F1_ = [ 0.866 0.500] (kN) F2_ = [ 1.500 2.598] (kN) F3_ = [-2.000 0.000] (kN)

The resultant is calculated with

(1.9)

and the angle of the horizontal with the horizontal axis is

(1.10)

With MATLAB the resultant and the angle are calculated with:

R_ = F1_+F2_+F3_; phi = atand(R_(2)/R_(1));

and the results are

% R_ = F1_+F2_+F3_ = [ 0.366 3.098] (kN) % phi = atan(Ry, Rx) = 83.262 (deg)

The MATLAB representation of the forces is shown in Figure 1.10 and it is obtained with:

sa = 4; hold on axis([-sa sa -sa sa]) axis square quiver(0,0,F1_(1),F1_(2),0,'Color','k','LineWidth',1.2) quiver(0,0,F2_(1),F2_(2),0,'Color','k','LineWidth',1.2) quiver(0,0,F3_(1),F3_(2),0,'Color','k','LineWidth',1.2) quiver(0,0,R_(1),R_(2),0,'Color','r','LineWidth',2) text(F1_(1),F1_(2),' F_1',... 'fontsize',14,'fontweight','b') text(F2_(1),F2_(2),' F_2',... 'fontsize',14,'fontweight','b') text(F3_(1),F3_(2),' F_3',... 'fontsize',14,'fontweight','b') text(R_(1),R_(2),' R',... 'fontsize',14,'fontweight','b') grid on xlabel('x'), ylabel('y'),

Example 1.2

Figure 1.11 shows a system of spatial forces with the magnitudes  N,  N,  N, and  N. The parallelepiped has the dimensions  m,  m, and  m. Find:

1. (a) the resultant of the system of forces;
2. (b) the angle between the forces and ;
3. (c) the projection of the force on the force ; and
4. (d) calculate .

Figure 1.10 MATLAB representation of the forces , and .

Solution

1. 1)
2. (a) The input data in MATLAB are: clear; clc; close all a = 2; % m b = 3; % m c = 5; % m F1 = 15; % N F2 = 30; % N F3 = 10; % N F4 = 15; % N

A Cartesian reference frame is selected as shown in Figure 1.11. The position force (vector) of the application point of the force is

(1.11)

The position vector of the application point of the force is

(1.12)

The position (vector) of the application point of the force is

(1.13)

The position vector of the application point of the force is

(1.14)

Figure 1.11 System of four spatial forces , , and...

Schweitzer Klassifikation
DNB DDC Sachgruppen
BISAC Classifikation
Warengruppensystematik 2.0

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