notes
DESCRIPTION
solid mech lecture notesTRANSCRIPT
![Page 1: Notes](https://reader031.vdocument.in/reader031/viewer/2022020812/563db919550346aa9a99fb48/html5/thumbnails/1.jpg)
7/17/2019 Notes
http://slidepdf.com/reader/full/notes-568ea80fc5783 1/19
CHAPTER ONE
Introduction—Concept of Stress
1.1 Introduction
1.2 Forces And Stresses
1.3 Axial Loading: Normal Stress
1.4 Shearing Stress
1.5 Bearing Stress In Connections
1.6 Application To The Analysis Of Simple Structures
1.7 Stress On An Oblique Plane Under Axial Loading1.8 Stress Under General Loading Conditions: Components Of Stress
1.9 Ultimate And Allowable Stress: Factor Of Safety
Review And Summary
1.1 INTRODUCTION
The main objective of the study of the mechanics of materials is to
provide the engineer with the means of analyzing and designing various
machines and load-bearing structures. Both the analysis and the design
of a given structure involve the determination of stresses and
deformations. This first chapter is devoted to the concept of stress.
1
![Page 2: Notes](https://reader031.vdocument.in/reader031/viewer/2022020812/563db919550346aa9a99fb48/html5/thumbnails/2.jpg)
7/17/2019 Notes
http://slidepdf.com/reader/full/notes-568ea80fc5783 2/19
1.2 FORCES AND STRESSES
Forces ( external and internal ): Consider the structure in
Fig. 1.1, from the knowledge of statics we know that theboom AB and rod BC are two-force members. The forces
in the two members can be determined by “Equilibrium
Conditions” using the “Free-body Diagram” as the
following.
Fig. 1.1
2
![Page 3: Notes](https://reader031.vdocument.in/reader031/viewer/2022020812/563db919550346aa9a99fb48/html5/thumbnails/3.jpg)
7/17/2019 Notes
http://slidepdf.com/reader/full/notes-568ea80fc5783 3/19
Fig. 1.2 Fig. 1.3
9
Fig. 1.4
From the free-body diagram of pin B and the force triangle (Fig. 1.3),
we obtain:
3
![Page 4: Notes](https://reader031.vdocument.in/reader031/viewer/2022020812/563db919550346aa9a99fb48/html5/thumbnails/4.jpg)
7/17/2019 Notes
http://slidepdf.com/reader/full/notes-568ea80fc5783 4/19
Stresses: The force per unit area, or the intensity of the forces
distributed over a given section, is called the stress on that section and
is denoted by Greek letter σ . The stress formula:
=
(1.1)
Fig. 1.5 Fig. 1.6
The sign of stress: + tensile stress (member in tension)
− compressive stress (member in compression)
The unit of stress: =
, a Pascal. SI metric unites are used.
1 kPa = 103 Pa = 103 /2
1 MPa = 106 Pa = 106 /2
1 GPa = 109 Pa = 109 /2
4
![Page 5: Notes](https://reader031.vdocument.in/reader031/viewer/2022020812/563db919550346aa9a99fb48/html5/thumbnails/5.jpg)
7/17/2019 Notes
http://slidepdf.com/reader/full/notes-568ea80fc5783 5/19
1.3 AXIAL LOADING: NORMAL STRESS
• Axial loading: Forces directed along the axis of
the rod
• Normal stress: Average of the force per unit area
in a member under axial loading
=
(1.1)
• Stress at point Q:
• The relation between the axial load P
And stress (by equilibrium)
• The actual distribution of stresses in any
given section is statically indeterminate.
Fig. 1.7
Fig. 1.8
5
![Page 6: Notes](https://reader031.vdocument.in/reader031/viewer/2022020812/563db919550346aa9a99fb48/html5/thumbnails/6.jpg)
7/17/2019 Notes
http://slidepdf.com/reader/full/notes-568ea80fc5783 6/19
• A uniform distribution of stress is possible only if the line of action
of the concentrated load P pass through the centroid of the section
considered.
Fig. 1.9
• If a two-force member is loaded axially , but eccentrically as
shown in Fig. 1.11 , then the internal forces in a given section
must be equivalent to a force P applied at the centroid of the
section and a couple M of moment M = Pd , the distribution of
stresses cannot be uniform.
Fig. 1.10 Fig. 1.11
6
![Page 7: Notes](https://reader031.vdocument.in/reader031/viewer/2022020812/563db919550346aa9a99fb48/html5/thumbnails/7.jpg)
7/17/2019 Notes
http://slidepdf.com/reader/full/notes-568ea80fc5783 7/19
1.4 SHEARING STRESS
• Shearing Force or Shear
• Average Shearing Stress
• Bolt, Pins and Rivets
• Single shear and Double shear
Fig. 1.12 Fig. 1.13
• Average shearing stress:
• The distribution of shearing stresses across the section cannot be
assumed uniform
• Rivet – single shear (free body diagram analysis)
Fig. 1.14 Fig. 1.15
A
P=aveτ
7
![Page 8: Notes](https://reader031.vdocument.in/reader031/viewer/2022020812/563db919550346aa9a99fb48/html5/thumbnails/8.jpg)
7/17/2019 Notes
http://slidepdf.com/reader/full/notes-568ea80fc5783 8/19
“Equilibrium Condition” “Free-body diagram”
Fig. 1.16
Fig. 1.17
• Double shear (free-body diagram)
AF
AP
2ave
==τ
8
![Page 9: Notes](https://reader031.vdocument.in/reader031/viewer/2022020812/563db919550346aa9a99fb48/html5/thumbnails/9.jpg)
7/17/2019 Notes
http://slidepdf.com/reader/full/notes-568ea80fc5783 9/19
1.5 BEARING STRESS IN CONNECTIONS
Fig. 1.18
Fig. 1.19
• Bearing stress:
d t P
AP == b
σ
9
![Page 10: Notes](https://reader031.vdocument.in/reader031/viewer/2022020812/563db919550346aa9a99fb48/html5/thumbnails/10.jpg)
7/17/2019 Notes
http://slidepdf.com/reader/full/notes-568ea80fc5783 10/19
1.6 APPLICATION TO THE ANALYSIS OF
SIMPLESTRUCTURES
Fig. 1.20
• Determination of the normal stress in two-force members
First to calculate the force in each of the members by free-body
diagram analysis and equilibrium equation for a rigid body:
Σ = 0, Σ = 0, Σ M = 0
• Calculate the tensile stress in each cross-section of rod BC
10
![Page 11: Notes](https://reader031.vdocument.in/reader031/viewer/2022020812/563db919550346aa9a99fb48/html5/thumbnails/11.jpg)
7/17/2019 Notes
http://slidepdf.com/reader/full/notes-568ea80fc5783 11/19
• Determination of the shearing stress
in various connections
Single shear (Pin at C)
Double shear (Pin at A)
For example:
MPa102m10491
N105026
3
=×
×==
− A
PPinC aveτ
Fig. 1.22
For example: Pin at B, the largest shearing stress
occur in section G and H, where
MPa9.50m10491
kN2526, =
×==
− A
PGave Bτ
• Determination of bearing stresses in members, pins and
brackets.
Fig. 1.21
Fig. 1.23
11
![Page 12: Notes](https://reader031.vdocument.in/reader031/viewer/2022020812/563db919550346aa9a99fb48/html5/thumbnails/12.jpg)
7/17/2019 Notes
http://slidepdf.com/reader/full/notes-568ea80fc5783 12/19
1.7 STRESS ON AN OBLIQUE PLANE UNDER AXIAL
LOADING
• Axial force cause both normal and
shearing stresses on planes not
perpendicular to the axis of the member.
Considering the equilibrium of
Fig. 1.26 (b) and resolving P into normal
component F and shearing component V ,
we have:
θ θ sincos PV PF ==
The corresponding average normal and
shearing stresses are
θ θ
θ
θ
τ
θ
θ
θ σ
θ
θ
cossin
cos
sin
cos
cos
cos
00
2
00
A
P
A
P
A
V
A
P
A
P
A
F
===
===
Fig. 1.24
Fig. 1.26
12
![Page 13: Notes](https://reader031.vdocument.in/reader031/viewer/2022020812/563db919550346aa9a99fb48/html5/thumbnails/13.jpg)
7/17/2019 Notes
http://slidepdf.com/reader/full/notes-568ea80fc5783 13/19
Or:
σ =
02 = 0
2
=
0 = 0
We can see that:
At = , σ = =
= 0
and 0 when 90°
At = and = °, = 0
and at = 45°,
= =
2=
1
20
At = °, = σ =
2
Fig. 1.27
13
![Page 14: Notes](https://reader031.vdocument.in/reader031/viewer/2022020812/563db919550346aa9a99fb48/html5/thumbnails/14.jpg)
7/17/2019 Notes
http://slidepdf.com/reader/full/notes-568ea80fc5783 14/19
1.8 STRESS UNDER GENERAL LOADING
CONDITIONS: COMPONENTS OF STRESS
Fig. 1.28
• Consider a body subjected to several
loads 1, 2, etc. Now let’s define the stress at a
point Q inside the body. Define the stress on a
small area Δ normal to the x axis:
A
F x
∆
∆,
A
V x
y
∆
∆,
A
V x
z
∆
∆
Letting Δ 0, we define the stress
components at point Q as
A
V
A
V
A
F
x
z
A
xz
x
y
A
xy
x
A x
∆
∆=
∆
∆=
∆
∆=
→∆→∆
→∆
limlim
lim
00
0
τ τ
σ
• The sign of stresses
Fig. 1.30
Fig. 1.31
14
![Page 15: Notes](https://reader031.vdocument.in/reader031/viewer/2022020812/563db919550346aa9a99fb48/html5/thumbnails/15.jpg)
7/17/2019 Notes
http://slidepdf.com/reader/full/notes-568ea80fc5783 15/19
Similarly we can define the stress
components on the small area (or
point Q) normal to the y and z axes:
, ,
, ,
• Important relations betweenshearing stress components:
Considering the six equilibrium
conditions of small cube of side a
0
0
===
===
∑∑∑
∑∑∑
z y x
z y x
M M M
F F F
From
( ) ( ) 0=∆−∆=∑ a Aa A M yx xy z τ τ
We have yx xy τ τ =
Similarly, zy yz zy yz τ τ τ τ ==
Fig. 1.32
Fig. 1.33
Fig. 1.34
15
![Page 16: Notes](https://reader031.vdocument.in/reader031/viewer/2022020812/563db919550346aa9a99fb48/html5/thumbnails/16.jpg)
7/17/2019 Notes
http://slidepdf.com/reader/full/notes-568ea80fc5783 16/19
• Only 6 stress components are required to define the
condition of stress at a given point Q , instead of 9 as originally
assumed
3 Normal Stresses , ,
3 Shear Stresses , ,
• At a given point, shear cannot take place in one plane only; an equal
shear must be exerted on another plane perpendicular to the first
one
Fig. 1.35
• Same loading condition may
lead to different
interpretations of stress
situation at a given point, depending upon the orientation of the
element considered. More detail will be
discussed in Chapter 6.
Fig. 1.36
16
![Page 17: Notes](https://reader031.vdocument.in/reader031/viewer/2022020812/563db919550346aa9a99fb48/html5/thumbnails/17.jpg)
7/17/2019 Notes
http://slidepdf.com/reader/full/notes-568ea80fc5783 17/19
1.9 ULTIMATE AND ALLOWABLE STRESS: FACTOR
OF SAFETY
• Concept of ultimate load and its determination
Ultimate normal stress and ultimate shear stress
• Concept of allowable load —working load or design load
• Factor of safety = F.S.= ultimate loadallowable load
Factor of safety =ultimae stress
allowable stress
• The determination of the factor of safety should be based upon
many considerations.
Fig. 1.37 Fig. 1.38
17
![Page 18: Notes](https://reader031.vdocument.in/reader031/viewer/2022020812/563db919550346aa9a99fb48/html5/thumbnails/18.jpg)
7/17/2019 Notes
http://slidepdf.com/reader/full/notes-568ea80fc5783 18/19
REVIEW AND SUMMARY
(On PP.44-46 of the text book)
xial loading. Normal stress
Transverse forces. Shearing stress
Single and double shear
18
![Page 19: Notes](https://reader031.vdocument.in/reader031/viewer/2022020812/563db919550346aa9a99fb48/html5/thumbnails/19.jpg)
7/17/2019 Notes
http://slidepdf.com/reader/full/notes-568ea80fc5783 19/19
Bearing stress
Stresses on an oblique section
Stress under general loading
Ultimate strength
Ultimate load and allowable load
Factor of safety
19