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Fourth Edition

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.

1- 2

Fourth
Edition

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.

1- 3

Fourth
Edition

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.

1- 4

Fourth
Edition

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.

1- 5

Fourth
Edition

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.

1- 6

Fourth
Edition

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
1- 7

Fourth
Edition

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...
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