# Aggregates

Topics: Force, Shear stress, Shear strength Pages: 24 (2312 words) Published: April 15, 2013
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

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

Maximum Stresses

Shearing Stress

State of Stress

Shearing Stress Examples

Factor of Safety

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
• 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.

1- 3

Fourth
Edition

MECHANICS OF MATERIALS

Beer • Johnston • DeWolf

Review of Statics
• The structure is designed to
• 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

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

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

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

FBC = 50 kN
1- 7

Fourth
Edition

MECHANICS OF MATERIALS

Beer • Johnston • DeWolf

Stress Analysis
Can the structure safely support the 30 kN
• 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...