MECH 260: Mechanics of Materials
Topic 1: Forces, Deformation, and
stress
Assistant Prof. Mauricio Ponga
Department of Mechanical Engineering, UBC
August 2021
1
Force, Deformation, Stress and Strain
Suggested Reading & Practice:
1-Beer and Johnston, Mechanics of Materials, 8th edition:
Chapter 1: pages 1-35.
2-Problems:
Chapter 1, problems 3, 7, 17, 25, 29, 37, 48, 52
All pictures are reproduced with permission from
correspond to the book MoM by Beer et al., 8th edition
McGrawl-Hill Education. They
Force, Deformation, Stress and Strain
Outline:
1- Concept of Force.
2- Deformation/Displacements.
3- Review of Statics
4- Free-body diagrams (FBD).
5- Concept of Stress
6- Concept of Strain
All pictures are reproduced with permission from
correspond to the book MoM by Beer et al., 8th edition
McGrawl-Hill Education. They
Why Mechanics of Materials?
• 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 forces, stresses and displacements and strains. This topic is devoted to
these concepts.
Team work!
What is a force? (Widely used term) !
What is a displacement? (Widely used term) !
What is a stress? (Engineering terms) "
What is a strain? (Engineering terms) "
Instructor’s job is to facilitate
the tools and instruments to
the students
Instructor
Students
•
•
•
•
Future engineers should be able to
understand and apply these
concepts!
Force concept
• Two or more objects, they interact with
each other. This interaction is manifested as
a FORCE between two objects. When the
interaction ceases, the two object do not
experience any forces.
•
•
•
There are two type of forces, surface and body forces.
Surface forces appear at the surfaces of bodies. Examples are
friction force, pressure forces, etc. Most of the forces in Mechanics of
Materials are of this type.
Body forces are related to the mass of a body, with the mots famous
body force being the gravity. Unless otherwise specified we will
neglect body forces.
Force concept and laws
First and second law
•
Forces have to be described by their
intensity (magnitude) and direction.
Forces are denoted using vectors. A
vector is denoted by a bold greek letter.
•
•
Forces obey Newton’s laws.
First law: A particle will conserve its state
of rest or motion if the force is zero.
•
Second law: F = ma (notice the bold
symbols showing the vectors).
•
Third law: A force comes with a reaction
force in a way they have the same
magnitude, and opposite direction.
Third law and its consequences
Deformation and Displacements
Force
Structure
•
Materials deform upon the application of
forces, in order to generate a system of
internal forces to compensate the external
force.
•
Structures are designing to work under
safe conditions for the system of forces
during the service of it.
Displacement
Reference configuration
Deformed configuration
Deformation: the process of changing in shape and/or
distorting, especially through the application of forces.
Deformation and Displacements
•
Notice that the deformation is manifested by changing the length and shape .
These are the effects of the force.
Reference configuration
Deformation of the body
Deformed configuration
Displacements
Reference configuration
Reference system
Displacement
Deformed configuration
Let the point in the reference configuration be denoted by
Let the point in the deformed configuration be denoted by
Then, the displacement is defined as
.
Using a Cartesian triad, we can compute the displacement as
,
, and
.
, and
,
Forces in Statics
•The structure is designed to support a 30
kN load
•The structure consists of a boom AB and
rod BC joined by pins (zero moment
connections) at the junctions and supports
•Perform a static analysis to determine the
reaction forces at the supports and the
internal force in each structural member
Forces in Statics
Free-body diagram: The structure is removed from its constrains and we show their forces.
Forces in Statics
Free body diagram for rod AB
Forces in Statics
Forces in Statics
What is the nature of the forces acting on
the boom structure?
Will the boom resist safely this
forces? Need to compute stresses in
the rods.
Force in Statics
Example application: Pratt bridge truss
For the bridge in the figure, obtain the
forces acting on the bars and the
reactions forces. Compute the stresses if
the area is A=5.87 in2
Forces in Statics
FyAB F AB
FxAB
αABC
F AC
Individual rods have to satisfy equilibrium
∑
Fx = 0,
∑
Fy = 0
∑
Fx = 0, F AC + FxAB = 0
∑
Fy = 0, R A + FyAB = 0
RA
Results: F AB has the opposite direction.
Forces in Statics
FyAB = RA
F AB
FxAB = F AC
αABC
RA
F AC
Individual rods have to satisfy equilibrium
∑
Fx = 0,
∑
Fy = 0
∑
Fx = 0, F AC + FxAB = 0
∑
Fy = 0, R A + FyAB = 0
Forces in Statics
Keep cutting the structure and perform equilibrium
F BC = 80
F AC
F CE = F CA
F BD
F AB
F BE
F BC = 80
Forces in Statics
Keep cutting the structure and perform equilibrium
F BD
F BE
F AB
F BC = 80
Forces in Statics
Keep cutting the structure and perform equilibrium
Normal stresses
The resultant of the internal forces for an
axially loaded member is normal to a
section cut perpendicular to the member
axis.
The force intensity on that section is
defined as the normal stress.
Normal stresses
Analyze the nature of the stresses in the Pratt bridge.
Shearing stresses
Forces P and P’ are applied transversely to the member AB.
Corresponding internal forces act in the plane
of section C and are called shearing forces.
The resultant of the internal shear force distribution is defined
as the shear of the section and is equal to the load P.
The corresponding average shear stress is,
Shearing stresses
Single shear
Bolt subjected
to single shear
Force is distributed in the
cross-section.
Double shear
Bolt subjected
to double shear
Force is distributed in two
cross-sections.
Bearing stresses in Connections
Bolts, rivets, and pins create stresses on
the points of contact or bearing
surfaces of the members they connect.
The resultant of the force distribution
on the surface is equal and opposite
to the force exerted on the pin.
Stress analysis and design example
Rod and Boom normal stress
Would like to determine the stresses in the
members and connections of the structure shown.
Stress analysis and design example
Rod and Boom normal stress
Stress analysis and design example
Pin shearing stress
Stress analysis and design example
Pin shearing stress
Stress analysis and design example
Pin shearing stress
Stresses components
Normal stresses
Shearing stresses
Stresses in arbitrary planes
Stress state and components of the stress
Stresses in arbitrary planes
Stress state and components of the stress
From the previous analysis it is evident
that the component of the stress will
depend on the specific plane that we
“virtually cut” the material.
Thus, the character of the stress components
—normal or shearing stresses— give only
partial information.
However, the stress state — imposed due to
the external loads— remains the same!
The physical interpretation of this statement is that values of interest, such as maximum
(minimum) normal stresses, and maximum (minimum) shearing stresses at a point remain
unchanged. We just need to find the location (point and plane) where these happen.
Stresses in arbitrary planes
Stress state and components of the stress
Normal and shearing stresses on an oblique plane
Stress tensor under general conditions
Stress tensor and its components
Stress tensor under general conditions
Stress tensor and its components
P3
y
ΔF y
ΔVzy
ΔVxy
P1
z
P2
x
Exercise: Virtually cut the body with a plane parallel
to z and sketch the forces and area for that cut.
What are the resultant stresses?
Stress tensor under general conditions
Stress tensor and its components
σxx τxy τxz
σij = τyx σyy τyz
τzx τzy σzz
Cauchy stress tensor (Force per unit Area). i, j=1,2,3.
The tensor is represented with an array of nine scalars, in a matrix form.
The first line denote the force intensity when we cut along the x-direction,
The second when we cut along the y-direction and,
The third one when we cut along the z-direction.
Normal stresses are placed in the diagonal of the matrix, while shearing stresses are
placed in the off-diagonal terms.
Units: [σij
= force⋅length-2 =N/m2 =Pa]
Stress tensor under general conditions
Stress tensor and its components
σxx τxy τxz
σij = τyx σyy τyz
τzx τzy σzz
Analyze how many components are independent: (Analyze equilibrium of moments)
Stress tensor under general conditions
Stress tensor and its components
σxx τxy τxz
σij = τSym
yx σyy τyz
τzx τzy σzz
Six independent components
σxx τxy τxz
σij = τxy σyy τyz
τxz τyz σzz
Stress tensor under general conditions
Stress tensor and its components in 2D
σxx τxy
σij = τ σ
[ xy yy]
Three independent
components
Some questions remain to be answered:
1- How one can determine the maximum (minimum) normal stresses?
2- How one can determine the max (min) shearing stresses?
3- On which planes and directions these max (min) stresses act?
We will postpone the answers for later (Topic #6 and #7). We will focus on
understanding single stress components.
Factor and Margin of Safety
Structural members or machines must be designed such that the working stresses are less than
the ultimate strength of the material.
Factor of safety considerations:
• uncertainty in material properties
• uncertainty of loadings
• uncertainty of analyses
• number of loading cycles
• types of failure
• maintenance requirements and deterioration effects
• importance of member to integrity of whole structure
• risk to life and property
• influence on machine function
FS = Factor of Safety
FS =
σY
Yield stress
=
> 1 to be safe
σall allowable stress
Yield stress = σY (Material property)
allowable stress = max (σ) in structure
MS = Margin of Safety
σY
− 1 = FS − 1
σall
MS > 0 to be safe
MS =
Factor and Margin of Safety
Application: Obtain the Factor and Margin of Safety (FS and MS, respectively) for the boom
structure analyzed in the previous slides. Assume the structure is made of Aluminum 2024 T4
with σY = 250 MPa of yield strength.