# MAE314 Solid Mechanics Fall, 2008

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MAE314 Solid Mechanics Fall, 2008

Dr. Yuan

1-1
Chapter 1. Introduction to Mechanics of Materials
1.1 What is Mechanics of Materials?
Objectives : To study mechanical behavior of solid bodies subjected to various types
Approaches : Develop analysis techniques (models) from basic principles
Structures : bar, beam, column, shaft, pressure vessel.
Issues : Strength, stiffness, stability.
Ultimate Goal : Design Safe Structures
Modeling is a process of deliberately neglecting the insignificant phenomena which
consciously capturing the important ones.
Mechanics

(A) Mechanics of Rigid Bodies

Static and dynamic behavior of undeformable bodies under external forces and/or
moments
(B) Mechanics of Deformable Solids

Internal forces, moments (stresses) and associated change in geometry of the
bodies (strains, displacements) under external forces and/or moments
Strength : determine whether the bodies will fail in service
Stiffness : determine whether the amount of deformation will be acceptable
Stability : determine whether the structure will collapse
These equations of statics are directly applicable to deformable bodies. The deformations
tolerated in engineering structures are usually negligible in comparison with the overall
dimensions of structures (except stability). Therefore, for the purposes of obtaining the forces in
members, the initial undeformed dimensions of members are used in the analysis.
1.2 The Fundamental Equations of Deformable-Body Mechanics
In general, three types of equations are used to solve deformable body problems
A. Equilibrium equations need to be satisfied;
B. Geometry of deformation must be prescribed; (boundary conditions, geometric
compatibility, etc.)
MAE314 Solid Mechanics Fall, 2008

Dr. Yuan

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C. Type of materials needs to be known.
1.3 Problem-solving Procedures
For a given problem, three steps are used to solve the problems correctly:
A. Plan the solution
: plan strategy and steps in solving the problems
B. Carry out the solution
: find the proper tools to solve the problems using equilibrium
equations, geometry of deformation, and material type.
C. Review the solution
: Does the dimension make sense? Are the quantities in a reasonable
manner including sign and magnitude? Does the solution violate the assumption you assumed
before solving the problem?
1.4 Review of Static Equilibrium; Equilibrium of Deformable Bodies
External and Internal Forces and Moments

Forces or moments that are applied to a structure are described as external (e.g., a weight
attached to the end of a rope). Forces and moments that are developed within a structure in
response to the external force systems present in the structure are described as internal (e.g., the
tension in a rope resulting from the pull of an attached weight).

(a) Free-body diagram of block

(b) Free-body diagram of beam

(c) Free-body diagram of beam

Fig. 1 Reactions and free-body (or equilibrium) diagrams. The weight of the beam has been
neglected in Fig. 1c.
The external forces systems
can be categorized into applied and reactive forces. Applied
forces are those that act directly on a structure (e.g., snow). Therefore, the applied forces move
MAE314 Solid Mechanics Fall, 2008

Dr. Yuan

1-3
as the structures deform. Reactive forces are those generated by the action of one body on
another and hence typically occur at connections and supports. Consequently, the reactive forces
support the structures and do not move. The existence of reactive forces follows from Newton's
third law, which generally states that whenever one body exerts a force on another, the second
always exerts on the first a force which is equal in magnitude, opposite in direction, and has the
same line of action shown in Fig. 1.a. In the figure 1.b, the force on the beam causes downward
forces on the foundation and upward reactive forces are consequently developed. A pair of action
and reaction forces thus exist at each interface between the beam and its foundation. In some
cases like Fig. 1.c, moments form part of the reactive system as well. The diagrams in the Fig. 1
which show the complete system of applied and reactive forces acting on a body, are called free-
body diagrams. The reactive forces and moments are presented as arrows with a slashed symbol.
In later chapters, the reactive forces and moments are drawn in single arrows for simplicity.
In broad sense, a free body diagram is a pictorial representation often used by physicists and
engineers to analyze the forces acting on a free body
. It shows all contact
and non-contact
forces

acting on the body.
After establishing the nature of the complete force system consisting of both applied and
reactive forces acting on the structure, the next step is to determine the nature of the internal
forces and moments developed in the structure as a consequence of the action of the external
forces.
Internal forces and moments
are developed within a structure due to the action of the
external force system acting on the structure and serve to hold together, or maintain the
equilibrium of, the constituent elements of the structure.
If the solid body as a whole is in equilibrium, any part of it
must be in equilibrium. For
such parts of a body, however, some of the forces necessary to maintain equilibrium must act at
the cut section. These considerations lead to the following fundamental conclusion: the
externally applied forces to one side of an arbitrary cut must be balanced by the internal forces
developed at the cut.

Fig. 2 The six internal forces and moments on an arbitrary cross section of a slender member.
MAE314 Solid Mechanics Fall, 2008

Dr. Yuan

1-4
Equilibrium Equations: (Using free body diagrams, FBD)

0,0,0
x y z
F F F
=
= =
∑ ∑ ∑

0,0,0
x y z
M M M
=
= =
∑ ∑ ∑

Support Reactions

1.5

A Short Review of the Methods of Statics
Two-bar structure: Points A, B, and C are hinged (no moment)
MAE314 Solid Mechanics Fall, 2008

Dr. Yuan

1-5
What is the two-force member
?
Force is along the direction of the member. Otherwise, it cannot satisfy the equilibrium.
Example 1.2.1

Free body diagram (FBD)

BY looking the original beam configuration, F
BC
is in tension, F
AB
is in compression.

Two equations are enough to determine all the
internal forces by using summing the forces
0=

x
F
,
030
5
3
=−
BC
F
,
kNF
BC
50=

0=

y
F
,
0
5
4
=−
ABBC
FF
,
kNF
AB
40=

The internal forces can be determined from the
moment equilibrium:

= 0
A
M
,
0
5
4
60030800 =×−×
BC
F
,
kNF
BC
50=

= 0
C
M
,
060030800 =
×

×
AB
F
,
kNF
AB
40=

The question: Can bars AB and BC sustain the load? Are they safe?
Problem 1.4-11 (a) determine the reactions at A and D, and (b) determine the internal resultants
(axial force, shear force, and bending moment) on the cross section at B.

MAE314 Solid Mechanics Fall, 2008

Dr. Yuan

1-6
Ans: A
x
= 0, A
y
= 5w
0
L/27, D
x
= 0, D
y
= 4w
0
L/27
(b) F
B
= 0, V
B
= 11w
0
L/108, M
B
= 17w
0
L
3
/324.