PDF Introduction to Mechanics of Materials: Part I

Iz, Iy the second moment, or moment of inertia, of area A about the z or y axis Not the polar moment of inertia of area A. At the end of each chapter the reader can find unsolved problems to practice their understanding of the subject discussed.

1 INTRODUCTION – CONCEPT OF STRESS

INTRODUCTION
A SHORT REVIEW OF THE METHODS OF STATICS
DEFINITION OF THE STRESSES IN THE MEMBER OF A STRUCTURE The results of the proceeding section represent the first necessary step in the design or analysing
BASIC STRESSES (AXIAL, NORMAL, SHEARING AND BEARING STRESS)
APPLICATION TO THE ANALYSIS AND DESIGN OF SIMPLE STRUCTURES
METHOD OF PROBLEM SOLUTION AND NUMERICAL ACCURACY Every formula previously mentioned and derived has its own validity. This validity predicts the
COMPONENTS OF STRESS UNDER GENERAL LOADING CONDITIONS

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MASTER IN MANAGEMENT

DESIGN CONSIDERATIONS AND FACTOR OF SAFETY

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2 STRESS AND STRAIN – AXIAL LOADING

INTRODUCTION

NORMAL STRESS AND STRAIN UNDER AXIAL LOADING

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STRESS-STRAIN DIAGRAM, HOOKE’S LAW, AND MODULUS OF ELASTICITY

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POISSON’S RATIO

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GENERALISED HOOKE’S LAW FOR MULTIAXIAL LOADING

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SAINT VENANT’S PRINCIPLE

In other words, except for the vicinity of load application point, the stress distribution function can be assumed independent of the load application mode. Download free ebooks at bookboon.com Click on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to to read moreClick on the ad to read more on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read more to readClick on the ad to read moreClick on the ad to read more.

DEFORMATIONS OF AXIALLY LOADED MEMBERS

2.4, the deformation depends on the position of any point Q, so we must use equation (2.2) to calculate the deformation. Download free eBooks at bookboon.com. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. per ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad. ad to read more.

PROBLEMS INVOLVING TEMPERATURE CHANGES

If the temperature of the rod grows by ∆T, we will observe the elongation of the rod by ∆LT, see Fig. Download free eBooks at bookboon.com Click on the ad to read more. Click on the ad to read more. read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read more readClick on the ad to read more on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read more.

Fig. 2.27 Vertical and horizontal displacements

EXAMPLES, SOLVED AND UNSOLVED PROBLEMS Problem 2.1

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EXPERIENCE THE POWER OF FULL ENGAGEMENT…

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RUN FASTER

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SETASIGNThis e-book

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3 TORSION

INTRODUCTION

Torsion members are encountered in many engineering applications and are primarily used to transfer power from one point to another. Both the force vectors and the applied torque act in the direction of the axis of the elements, see the figure.

Fig. 3.3 Transmitting shafts, [http://www.directindustry.com]

DEFORMATION IN A CIRCULAR SHAFT

Let's imagine a circular part consisting of two wooden plates which represent the faces on the front of the part. Due to the determination of the inner radius ρ, the shear stress reaches its maximum on the outer surface of the shaft, where ρ = R.

STRESS IN THE ELASTIC REGION

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ANGLE OF TWIST IN THE ELASTIC REGION

In the case of variable cross-sectional area along the shaft, as in Fig.3.12, the strain depends on the position of the arbitrary point Q, therefore we need to apply a similar equation to (2.2) for the calculation of the shear strain.

STATICALLY INDETERMINATE SHAFTS

The main problem in this problem is that the number of unknown reactions is greater than the number of equilibrium equations. In our case, the angle of rotation at point C is equal to zero, and corresponds to the total angle of rotation.

DESIGN OF TRANSMISSION SHAFTS

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TORSION OF NON-CIRCULAR MEMBERS

Both are on a free surface and there is no load in the x-axis direction. Download free eBooks at bookboon.com. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. per ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad to read more. Click on the ad. ad to read more Click on ad to read more Click on ad to read more Click on ad to read more Click on ad to read more Click on ad to read more Click on ad to read more Click on ad to read more Click on ad to read more Click on ad for more Click on ad for more Click on ad for more Click on ad for more Click on ad for more Click on ad for more Click on ad for more Click on ad for more Click on the ad for more Click on the ad to read more Click on the ad to read more

THIN-WALLED HOLLOW MEMBERS

If we focus on the balance of part DE in the longitudinal x direction, the shear law states that the shear forces 국어, 국어 act on the surfaces D and E, see the figure. Longitudinal shear forces 국어, 국어 act on small areas of areas ȟ�� and ��.

EXAMPLES, SOLVED AND UNSOLVED PROBLEMS Problem 3.1

Determine the torque exerted on the shaft by each of the struts and determine the maximum shear stress. Two shafts of the same length and made of the same materials are connected by a welded rigid beam. A moment coupling given by force F is applied to the ends of the rigid beam.

APPENDIX

CENTROID AND FIRST MOMENT OF AREAS

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SECOND MOMENT, MOMENT OF AREAS

A.7, we note that the center of gravity C must be located on the y-axis, since this axis is the axis of symmetry than z 0=. Download free ebooks at bookboon.com Click on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read moreClick on the ad to read more.

PARALLEL AXIS THEOREM

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PRODUCT OF INERTIA, PRINCIPAL AXES Definition of product of inertia is

If the cross-sectional area has an axis of symmetry, which is taken to be the y or z axis (Figure A.19), the product of inertia is zero. In the general case, for each point of an arbitrary cross-sectional area, two perpendicular axes can always be found, so that the product of inertia vanishes behind them.

STRAIN ENERGY FOR SIMPLE LOADS

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