STRENGTH OF MATERIALS
TRAN MINH TU - University of Civil Engineering,
Giai Phong Str. 55, Hai Ba Trung Dist. Hanoi, Vietnam
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2• Ass. Prof. Tran Minh Tu, PhD. Eng.
• Mon., Fri.: 12.15 -14.45 A.M. at Room 202 H1
• Office: 1. Floor – Lab. Building
• E-mail: tpnt2002@yahoo.com
• Tel: 04.8691 462 (off.). Handphone: 0912101173
• Course notes:
•
• Ref. : Ferdinand P. Beer, Jr. J. T. DeWolf, D. F.
Mazurek. Mechanics of Mate

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rials. Mc Graw Hill . 2009
• Office hours: Tuesday 8:00-11:00 A.M. or by
appointment
GENERAL INFORMATION
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3
Sample reading list:
Russell C. Hibbeler, Mechanics of Materials, 6/E
(required text)
Roy R. Craig, Jr (1996), Mechanics and Materials
Bedford, Fowler & Liecht (2003), Statics and
Mechanics and Materials
Beer & Johnson, Mechanics and Materials
James H. Gere, Mechanics and Materials, 6th
edition
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Strength of Materials
This course discusses stress calculation due to loads.
Description includes stress-strain concept, tensile test, stress
and strain due to axial loading, statically indeterminate case
related to axial loading, introduction to plasticity and residual
stress, stress and strain due to other loads, such as torsion,
bending moment, and shear force, Mohr’s circle of stress,
failure theory, deflection, statically indeterminate structures, and
energy method.
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Princeton ： Grading:
Mid Term Exam - 20%
Design Project - 30%
Take Home Final Exam -
25%
Problem set(s) - 15%
Other (See Instructor) - 10%
MIT： Grading
Homework 25%
Lab Assignments 30%
Quiz 15%
Case Study and Presentation 10%
Final Exam 20%
Class Attendance*
Stanford： Grading
Homework Assignments
25%
Lab Reports 10%
Midterm 25%
Final Exam 40%
Berkeley: Grading
Weekly homework
assignments(25%) Two Midterm
Examinations(20%+20%)
Final examination (35%).
GENERAL INFORMATION
1
CHAPTER
Introduction –
Concept of Stress
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1.1. Introduction
1.2. Review of Static
1.3. Equilibrium of deformable body
1.4. Concept of Stress
1.5. Stress Under General Loadings
1.6. Strain
1.7. Types of loading
1.8. Assumptions
1.9. Principle Superposition
Contents
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1.1. Introduction
- Consider a diving board as an example of a deformable body
- From rigid-body equilibrium, given the weight of the diver, and the
lengths L1, L2, we can determine the diving board support reactions at
A and B. Questions of the following type can only be answered by
employing the principles and procedures of strength of materials
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1.1. Introduction
1.1.1. Strength of materials
• A branch of mechanics
• It studies the relationship of
– External loads applied to a deformable body,
and
– The intensity of internal forces acting within the
body
• Are used to compute the deformations of a body
• Strength of Materials is a field of study that
determines strength, stiffness, & stability
1.1. Introduction
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1.1. Introduction
Mechanics
Rigid Body
Mechanics
Deformable Body
Mechanics
Strength of MaterialsStatic
(Dynamic)
Kinetic
Kinematic
Fluid Mechanics
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(2) Kinetic:
∑F = ma
(1) Equilibrium
∑Fx = 0; ∑Fy = 0; ∑Fz = 0;
∑Mx= 0; ∑My= 0; ∑Mz= 0
Rigid Body
• Static
Deformable Body
• Strength of Materials
(1) Equilibrium
∑Fx = 0; ∑Fy = 0; ∑Fz = 0;
∑Mx= 0; ∑My= 0; ∑Mz= 0
(2) Stress – Strain relationship:
s = Ee
1.1. Introduction
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1.1.2. Classification of Structural element
1.1. Introduction
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1.1. Introduction
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1.1. Introduction
Classification of Structural elements
Structural elements compose a structure and can be classified as by
their forms (shapes and dimensions).
• Three dimensional body
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• Plates and Shells
1.1. Introduction
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1.1. Introduction
• Rods, Bars
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1.1. Introduction
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1.1. Introduction
1.1. Introduction
1.1.3. External loads
• Surface forces
– Area of contact
– Concentrated force
– Linear distributed force
– Centroid C (or geometric
center)
• Body force (e.g., weight)
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1.1. Introduction
1.1.4. Support reactions
• for 2D problems
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1.2. Review of Static
1.2.1. Structure Free-Body Diagram
The first step towards solving an engineering problems is drawing the
free body diagram of the element/structure considered
• The structure is designed to
support a 30 kN load
• Perform a static analysis to
determine the internal force
in each structural member
and the reaction forces at the
supports
• The structure consists of a
boom and a rod joined by
pins (zero moment
connections) at the junctions
and supports
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1.2. Review of Static
• The structure is detached from the
supports and the loads and reaction
forces are indicated
• Ay and Cy can not be determined from
these equations
• Conditions for static equilibrium:
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0 0.6 30 0.8m kN mC xM A
40kNxA
0x x xF A C
40kNx xC A
0 30 0kNy y yF A C
30kNy yA C
1.2. Review of Static
• In addition to complete the structure,
each component must satisfy the
conditions for static equilibrium
• Results:
40 40 30kN kN kNx x yA C C
0 0.8 0mB y yM A A
• Consider a free-body diagram for the
boom:
kN30yC
substitute into the structure
equilibrium equation
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0 0.8mB yM A 0yA
1.2.2. Component Free-Body Diagram
1.2. Review of Static
x, y, z coordinates x, y plane
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1.2.3. Equilibrium of rigid body
Internal forces: The forces set up within a body to balance the effect of
the externally applied forces.
1.3. Equilibrium of deformable body
1.3.1. Method of Sections
To obtain the internal loading acting on specific region – “cut” by
imaginarily section through the region where internal loadings are to be
determined.
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1.3. Equilibrium of deformable body
Two parts of the body are separated => A free-body diagram of one of
the parts is drawn.
Internal loadings
The Internal force is distributed on the area of section and represent
effects of the material of the top part of the body acting on the adjacent
material of the bottom part.
The Resultant force FR and moment MR0 at any specific point O is
shown in the fig. c.
If a member is long and slender (rod, beam), the section considered is
generally taken perpendicular to the longitudinal axis – is called a cross
section 27
Define resultant force (FR) and moment (MRo) in 3D:
• Normal force, N
• Shear force, Q
• Torsional moment or torque, T
• Bending moment, M
1.3. Equilibrium of deformable body
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1.3. Equilibrium of deformable body
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- Coplanar Loadings. If the body is subjected to a coplanar system of
forces (fig. a), then only normal-force, shear-force, and bending-moment
components will exist at the section (fig. b).
- If we use the x, y, z coordinate axes, as shown on the left segment.
0 ...zF N 0 ...yF Q
0 0 ...M M
1.3.2. Internal resultant loadings
• For coplanar loadings:
– Apply ∑ Fx = 0 to solve for N
– Apply ∑ Fy = 0 to solve for Q
– Apply ∑ MO = 0 to solve for M
1.3. Equilibrium of deformable body
Procedure for Analysis
• Method of sections
1. Choose a segment to analyze
2. Determine the Support Reactions
3. Draw a free-body diagram for whole body
4. Apply the equations of equilibrium
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The resultant internal loadings at a point located on the section of a
body can be obtained by using the method of section. This requires the
following steps.
1.3. Equilibrium of deformable body
Procedure for analysis
• Free-body diagram
1. Keep all external loadings in exact locations before “sectioning”
2. Indicate the unknown resultants, N, Q, M, and T at the section,
normally at the centroid C of the sectioned area
3. Coplanar system of forces only include N, Q, and M
4. Establish x, y, z coordinate axes with origin at the centroid
• Equations of equilibrium
1. Sum moments at section, about each coordinate axes where
resultants act
2. This will eliminate unknown forces N and Q, with direct solution
for M (and T)
3. Resultant force with negative value implies that the assumed
direction in reality is opposite to that shown on the free-body
diagram
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Example 1.1
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Example 1.1
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1.4. Concept of Stress
• The main objective of the study of
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.
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1.4. Concept of Stress
Stress: The distribution of the internal force over the area on which it
acts is expressed as the force intensity, that is force per unit area.
There are two types of stress: normal stress and shear stress
Consider the sectioned area subdivided in
to small area such as DA
Very small force DF acting on DA, will be
replaced by three components
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1.4.1. Stress Analysis
1.4. Concept of Stress
The sign convention for the normal stress is as follows:
- A positive value for s indicates tensile stress, that is, the stress due
to a force DF that pulls on the area on which it acts.
- A negative value for s indicates compressive stress – push on
1.4.3. Shear stress
• Intensity of force, or force per unit area, acting tangent to ΔA
• Symbol used for the shear stress is t (tau)
1.4.2. Normal stress
• Intensity of force, or force per unit area, acting normal to ΔA
• Symbol used for the normal stress, is σ (sigma)
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1.4. Concept of Stress
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Units (SI system)
• Newtons per square meter (N/m2)
or a pascal (1 Pa = 1 N/m2)
• kPa = 103 N/m2 (kilo-pascal)
• MPa = 106 N/m2 (mega-pascal)
• GPa = 109 N/m2 (giga-pascal)
1.4.3. Axial Loading: Normal Stress
1.4. Concept of Stress
1.4.4. Average shear stress:
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1.4. Concept of Stress
A
F
A
P
avet
Single Shear
A
F
A
P
2
ave t
Double Shear
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1.4.5. Shearing Stress Examples
1.4. Concept of Stress
• Bolts, rivets, and pins create
stresses on the points of
contact or bearing surfaces
of the members they
connect.
dt
P
A
P
bs
• Corresponding average
force intensity is called the
bearing stress,
• The resultant of the force
distribution on the surface is
equal and opposite to the
force exerted on the pin.
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1.4.6. Bearing Stress in Connections
• The rod is in tension with an axial force of 50 kN.
• The boom is in compression with an axial force of
40 kN and average normal stress of –26.7 MPa.
• The minimum area sections at the boom ends are
unstressed since the boom is in compression.
MPa167
m10300
1050
m10300mm25mm40mm20
26
3
,
26
N
A
P
A
endBCs
• At the flattened rod ends, the smallest cross-
sectional area occurs at the pin centerline,
• At the rod center, the average normal stress in
the circular cross-section (A = 314x10-6m2) is sBC
= +159 MPa.
1.4. Concept of Stress
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1.4.7. Rod & Boom Normal Stresses
1.4. Concept of Stress
• The cross-sectional area for pins at A,
B, and C,
26
2
2 m10491
2
mm25
rA
MPa102
m10491
N1050
26
3
,
A
P
aveCt
• The force on the pin at C is equal to
the force exerted by the rod BC,
• The pin at A is in double shear with
a total force equal to the force
exerted by the boom AB,
MPa7.40
m10491
kN20
26,
A
P
aveAt
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1.4.8. Pin Shearing Stresses
1.4. Concept of Stress
• Divide the pin at B into sections to
determine the section with the largest
shear force,
(largest) kN25
kN15
G
E
P
P
MPa9.50
m10491
kN25
26,
A
PG
aveBt
• Evaluate the corresponding average
shearing stress,
FBC = 50kN
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• To determine the bearing stress at A in the boom
AB, we have t = 30 mm and d = 25 mm,
MPa3.53
mm25mm30
kN40
td
P
bs
• To determine the bearing stress at A in the
bracket, we have t = 2(25 mm) = 50 mm and d
= 25 mm,
MPa0.32
mm25mm50
kN40
td
P
bs
1.4. Concept of Stress
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1.4.9. Pin Bearing Stresses
1.5. Stress Under General Loadings
• A member subjected to a
general combination of loads is
cut into two segments by a
plane passing through Q
A
V
A
V
A
F
x
z
A
xz
x
y
A
xy
x
A
x
D
D
D
D
D
D
DD
D
limlim
lim
00
0
tt
s
• The distribution of internal
stress components may be
defined as,
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1.5. Stress Under General Loadings
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• For equilibrium, an equal and opposite internal force and stress
distribution must be exerted on the other segment of the member.
• Stress components are defined for the planes
cut parallel to the x, y and z axes. For
equilibrium, equal and opposite stresses are
exerted on the hidden planes.
• It follows that only 6 components of stress are
required to define the complete state of stress
• The combination of forces generated by the
stresses must satisfy the conditions for
equilibrium:
0
0
zyx
zyx
MMM
FFF
yxxy
yxxyz aAaAM
tt
tt
DD 0
zyyzzyyz tttt andsimilarly,
• Consider the moments about the z axis:
1.5. Stress Under General Loadings
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• Displacement is the relative movement of a point on a rod with
respect to a point in space
• Deformation is the relative movement of two points on a rod.
L
P
d
A
B
Point B displaces a distance d down.
The deformation of the rod between
points A and B is d.
1.6. Strain – Displacement and deformation
1.6. Strain
1.6.1. Deformation:
When A force is applied to a body, it will tend to change the body’s
shape and size - deformation
1.6.2. Strain:
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The strain is the geometrical expression of deformation caused by
the action of stress on a physical body. The strain is calculated by first
assuming a change between two body states: the beginning state and
the final state
Basically Strain are of three types
1. Longitudinal Strain: -It is defined as the increase in length (d) per
unit original length (L) when deformed by the external force.
Longitudinal Strain = d / L
2.Volumetric Strain:-It is defined as the change in volume (v) per unit
original volume (V), when deformed by external force.
Volume Strain = v / V
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- Normal Strain (longitudinal strain)
3. Shear Strain:-When the change takes place in the
shape of the body, the strain is called a shear strain.
1.6. Strain
- Shear Strain
1.7. Types of loading
Mechanics of materials is a branch of applied mechanics that deals with
the behaviour of solid bodies subjected to various types of loading
Compression Tension (stretched) Bending Torsion (twisted) Shearing
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• Continuous
• Homogeneous
• Isotropic
• Perfectly elastic
• Small deformation
1.8. Basic Assumptions
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• Continuous
1.8. Basic Assumptions
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• Homogeneous
1.8. Basic Assumptions
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• Isotropic
Isotropic Anisotropic
1.8. Basic Assumptions
1/10/2013 56
1.8. Assumptions – Hooke’s Law
• "Hooke's Law" is about stretching
springs and wires.
• Hooke's Law states:- the extension is
proportional to the force
• the spring will go back to its original
length when the force is removed
• so long as we don't exceed the
elastic limit.
Hooke’s Law
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s = E e t = G g
1.8. Assumptions – Hooke’s Law
Young’s modulus Shear modulus
1.9. Principle Superposition
Conditions must be satisfied:
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+1.9. Principle Superposition
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THANK YOU FOR
YOUR ATTENTION !
1/10/2013 63

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