CHAPTER

MECHANICS OF

MATERIALS

Ferdinand P. Beer

E. Russell Johnston, Jr.

John T. DeWolf

Introduction –

Concept of Stress

Lecture Notes:

J. Walt Oler

Texas Tech University

© 2006 The McGraw-Hill Companies, Inc. All rights reserved.

Fourth

Edition

MECHANICS OF MATERIALS

Beer • Johnston • DeWolf

Contents

Concept of Stress

Bearing Stress in Connections

Review of Statics

Stress Analysis & Design Example

Structure Free-Body Diagram

Rod & Boom Normal Stresses

Component Free-Body Diagram

Pin Shearing Stresses

Method of Joints

Pin Bearing Stresses

Stress Analysis

Stress in Two Force Members

Design

Stress on an Oblique Plane

Axial Loading: Normal Stress

Maximum Stresses

Centric & Eccentric Loading

Stress Under General Loadings

Shearing Stress

State of Stress

Shearing Stress Examples

Factor of Safety

© 2006 The McGraw-Hill Companies, Inc. All rights reserved.

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MECHANICS OF MATERIALS

Beer • Johnston • DeWolf

Concept of Stress

• The main objective of the study of the mechanics

of materials is to provide the future engineer with

the means of analyzing and designing various

machines and load bearing structures.

• Both the analysis and design of a given structure

involve the determination of stresses and

deformations. This chapter is devoted to the

concept of stress.

© 2006 The McGraw-Hill Companies, Inc. All rights reserved.

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MECHANICS OF MATERIALS

Beer • Johnston • DeWolf

Review of Statics

• The structure is designed to

support a 30 kN load

• The structure consists of a

boom and rod joined by pins

(zero moment connections) at

the junctions and supports

• Perform a static analysis to

determine the internal force in

each structural member and the

reaction forces at the supports

© 2006 The McGraw-Hill Companies, Inc. All rights reserved.

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MECHANICS OF MATERIALS

Beer • Johnston • DeWolf

Structure Free-Body Diagram

• Structure is detached from supports and

the loads and reaction forces are indicated

• Conditions for static equilibrium:

∑ M C = 0 = Ax (0.6 m ) − (30 kN )(0.8 m )

Ax = 40 kN

∑ Fx = 0 =Ax + C x

C x = − Ax = −40 kN

∑ Fy = 0 = Ay + C y − 30 kN = 0

Ay + C y = 30 kN

• Ay and Cy can not be determined from

these equations

© 2006 The McGraw-Hill Companies, Inc. All rights reserved.

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MECHANICS OF MATERIALS

Beer • Johnston • DeWolf

Component Free-Body Diagram

• In addition to the complete structure, each

component must satisfy the conditions for

static equilibrium

• Consider a free-body diagram for the boom:

∑ M B = 0 = − Ay (0.8 m )

Ay = 0

substitute into the structure equilibrium

equation

C y = 30 kN

• Results:

A = 40 kN → C x = 40 kN ← C y = 30 kN ↑

Reaction forces are directed along boom

and rod

© 2006 The McGraw-Hill Companies, Inc. All rights reserved.

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MECHANICS OF MATERIALS

Beer • Johnston • DeWolf

Method of Joints

• The boom and rod are 2-force members, i.e.,

the members are subjected to only two forces

which are applied at member ends

• For equilibrium, the forces must be parallel to

to an axis between the force application points,

equal in magnitude, and in opposite directions

• Joints must satisfy the conditions for static

equilibrium which may be expressed in the

form of a force triangle:

r

FB = 0

∑

FAB FBC 30 kN

=

=

4

5

3

FAB = 40 kN

© 2006 The McGraw-Hill Companies, Inc. All rights reserved.

FBC = 50 kN

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MECHANICS OF MATERIALS

Beer • Johnston • DeWolf

Stress Analysis

Can the structure safely support the 30 kN

load?

• From a statics analysis

FAB = 40 kN (compression)

FBC = 50 kN (tension)

• At any section through member BC, the

internal force is 50 kN with a force intensity

or stress of

dBC = 20...