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Strength of materials
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FF
Faculty of Engineering and Materials Science
Prof. Dr. E.I. Imam Morgan Head of Mechatronics Department
Vice Dean of EMS for Academic Affairs
C7.110
Strength of Materials I
(1)
mailto:[email protected]
Office hours: Monday, 2nd slot (10:00 12:00)
Teaching Assistant: Eng. Bola George, Bishoy Emil, Micheal Wahba [C7:116]
Eng. Mona Nader [C7:101], Eng. Sara Elkhamisy [C3:123]
Eng. Andrew Nagy [C3:127]
Text book:
Beer, F.P., and Johnston, E.R. Mechanics of Materials,
fourth Edition (2006), Fifth edition is now available.
McGraw Hill Publishing Co., New York, NY, ISBN 007-124999-0.
and:
J.M. Gere Mechanics of Materials,
sixth Edition (2004), eights edition is available.
Brooks/Cole, Belmont, CA, ISBN 0-534-41793-0.
Course Assessment:
10% 4 Assignments
25% 2 Quizzes (best 2 out of 3)
25% Mid term exam
40% Final exam.
Attendance: 75% of the course must be attended
How to drop:
Dropping of the course not later than 2 weeks after the start by
notice to the admission office. (2)
Chapter 1
Introduction
3
This chapter is devoted to the study of stresses occurring in many of the elements contained in the shown excavator, such as two-force members, axles, bolts, and pins.
(4) Prof. Dr. Imam Morgan
Head of MCTR Department
Strength of Materials
or, (Mechanics of Deformable Bodies)
or, (Mechanics of Materials) ??
The Strength of Materials is the branch of applied mechanics that deals
with the behavior of elastic bodies subjected to various types of loading.
Bars (axial loading)
Shafts (torsion)
Beams (bending)
Columns (compression)-buckling
Objective Is to provide the future engineer with the means of
analyzing and designing the various components of
any structure or any machine such that they must
operate safely.
The bodies under investigations represent the components of a machine or
a structure.
These components may be
1 Introduction
Prof. Dr. Imam Morgan
Head of MCTR Department
(5)
Beams (B.M load)
Shafts
(torsion load)
The most important
concepts in Strength of
Materials are:
Stress and Strain
These concepts will be
illustrated (in this chapter)
by considering a prismatic
bar subjected to axial
forces.
Later, these loadings
produce other types of
stresses.
But, there are other
types of loadings such
as torsion (in shafts)
and bending moment
loads (in beams).
Bars (axial load)
(6) Prof. Dr. Imam Morgan
Head of MCTR Department
The first step in studying an element is to determine the
loads applied on it in order to decide whether it can support these loads or not.
Therefore, methods of statics are used to determine the static
forces in the supports of the structure as well as the forces applied on the component itself. In machines, the applied dynamic forces are calculated using a similar technique
based on the method of inertia forces (Later in theory of machines).
The steps of study (analysis and design) of a certain structure as well as the concept of stress are investigated using the following example.
2 Static Review
Prof. Dr. Imam Morgan
Head of MCTR Department
(7)
The shown structure is designed
to support a 30 kN load
The structure consists of a boom AB and rod CB joined by pins at the junction B and supports A and C.
Prof. Dr. Imam Morgan
Head of MCTR Department
(8)
Static Analysis
Perform a static analysis to
determine the internal force in
each structural member and the
reaction forces at the supports.
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 an axis between the
force application points, equal in
magnitude, and in opposite directions.
Therefore, in this case we can use the
method of equilibrium of pin at B (or resolve
the 30kN force into 2 directions)that had been
given in the previous semester [Mechanics I].
Prof. Dr. Imam Morgan
Head of MCTR Department
(9)
Joints must satisfy the conditions for static equilibrium of concurrent forces
which may be expressed in the form of a force polygon (triangle):
kNF
.FF
kNF
.FF
AB
ABx
BC
BCy
40
080500
50
030600
Note : According the obtained positive values we conclude that:
member AB is in compression. member CB is in tension.
kN50
kN40
3
30
54
BC
AB
BCAB
F
F
FF
Or, from force
polygon
(10) Prof. Dr. Imam Morgan
Head of MCTR Department
Can the structure safely support
the 30 kN load?
A question is waiting
for an answer!!
From a statics analysis
FAB = 40 kN (compression)
FBC = 50 kN (tension) dBC = 20 mm
Consider, for example the rod BC. Its ability to withstand the internal
tensile force FBC depends on: the internal force in the rod, FBC the cross section area of the rod, A the material of which the rod is made
Compare
and
decide
3 Concept of Stress
Combined in
one parameter, called Stress
Each material has its
own mechanical
properties (parameters) Prof. Dr. Imam Morgan
Head of MCTR Department
(11)
In general, the force per unit area, or intensity
of the force distributed over a given section is
called the stress (sigma).
As shown in figure:
A
P Pa2 ,orm/N
we have:
1 kPa = 103 Pa
1 MPa = 106 Pa commonly used
1 GPa = 109 Pa Animation
(12) Prof. Dr. Imam Morgan
Head of MCTR Department
../Animation/Normal_and_shear_stresses.swf
Back to the example:
Analysis From a statics analysis FAB = 40 kN (compression)
FBC = 50 kN (tension)
MPa159m10314
N105026-
3
A
PBC
At any section through member BC, the internal force is 50 kN with a force intensity
or stress of
Conclusion: the strength of
member BC is adequate (safe).
MPa 165all
From the material properties for steel, the allowable stress is
allBC
dBC = 20 mm
Now assume that the rod BC is
made of steel.
Prof. Dr. Imam Morgan
Head of MCTR Department
(13)
Design of new structures requires selection
of appropriate materials and component
dimensions to meet performance
requirements
For reasons based on cost, weight, availability,
etc., the choice is made to construct the rod from
aluminum all= 100 MPa). What is an
appropriate choice for the rod diameter?
mm2.25m1052.2
m1050044
4
m10500Pa10100
N1050
226
2
26
6
3
Ad
dA
PA
A
P
allall
An aluminum rod 26 mm or
more in diameter is adequate
Design
Choose
material
d = ?
for safety
(14) Prof. Dr. Imam Morgan
Head of MCTR Department
Normal Stress
Shear Stress
Bearing Stress
Types of Stresses
Prof. Dr. Imam Morgan
Head of MCTR Department
(15)
The resultant of the internal forces for an axially
loaded member is normal to a section cut perpendicular to the member axis.
A
P
A
Fave
A
0lim
The force intensity (force per unit area) on that
section is defined as the normal stress.
The normal stress at a particular point may not be
equal to the average stress but the resultant of the
stress distribution must satisfy
A
ave dAdFAP
The detailed distribution of stress is statically
indeterminate, i.e., can not be found from statics
alone.
Animation
4 Normal Stress
(16) Prof. Dr. Imam Morgan
Head of MCTR Department
../Animation/Saint_Venants_principle.swf
If a two-force member is eccentrically loaded, then the resultant of the stress distribution in a section must yield an axial
force and a moment.
The stress distributions in eccentrically loaded
members cannot be uniform or symmetric.
This case will be studied in Ch.4.
A uniform distribution of stress in a section implies that the line of action for the resultant of the internal forces passes through the centroid of the section.
A uniform distribution of stress is only
possible if the concentrated loads on the end
sections of two-force members are applied at
the section centroids. This is referred to as
centric loading.
Normal Stress in case of Centric and Eccentric Loadings
(17)
Example (1) P = 115 kN
L
A short post constructed from a hollow
circular tube of aluminum supports a
compressive load of 115 kN. The outer and
inner diameters of the tube are do= 115 mm
and di= 100 mm, respectively. Determine the compressive stress in the post
(average stress).
Solution
MPa
mmddA
NP
A
P
io
4.4591.2532
10115
91.253210011544
10115
3
22222
3
Note that: Substituting P in N and the area A in
mm2
will give, directly, the
stress in MPa.
Prof. Dr. Imam Morgan
Head of MCTR Department
(18)
Example (2)
Two solid cylindrical rods AB and BC are welded together at B and loaded as shown. Determine the
average stress at the midsection of:
(a) rod AB, (b) rod BC.
-80 kN
+
+160 kN
Solution
At first determine the axial force in each
rod. So, draw the Normal Force Diagram.
a- rod AB (rod 1)
N. F. D.
MPa
mmA
NP
A
P
1.1886.4417
1080
86.4417754
1080
3
2
22
2
3
2
2
22
b- rod BC (rod 2)
MPa
mmA
NP
A
P
49.815.1963
10160
5.1963504
10160
3
1
22
1
3
1
1
11
+ tensile stress
Compressive
stress
Prof. Dr. Imam Morgan
Head of MCTR Department
(19)
Forces P and P are applied transversely to the member AB.
A
Pave
The corresponding average shear stress is,
The resultant of the internal shear force
distribution is defined as the shear of the section
and is equal to the load P.
Corresponding internal forces act in the plane of section C and are called shearing forces.
Shear stress distribution varies from zero at the member surfaces to maximum values that may be much larger than the average value (Ch.6).
The shear stress distribution cannot be assumed uniform.
Our interest in this chapter
5 Shearing Stress
Prof. Dr. Imam Morgan
Head of MCTR Department
(20)
P
y
z
x
2
1 1
2
Shearing stress distribution due
to transverse load (Ch. 6).
Note that P acts in plane of the section.
Prof. Dr. Imam Morgan
Head of MCTR Department
(21)
A
F
A
Pave
Single Shear
Examples of Shearing stress
A
F
A
P
2ave
Double Shear
F/2
F/2
2/FP
Prof. Dr. Imam Morgan
Head of MCTR Department
(22)
Bolts, rivets, and pins create
stresses on the points of contact
or bearing surfaces of the members they connect.
dt
P
A
Pb
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.
The bearing area is defined as the projected area of the
curved bearing surface
6 Bearing Stress
Prof. Dr. Imam Morgan
Head of MCTR Department
(23)
We Would like to determine
the stresses in the members
and connections of the
structure shown.
Must consider :
maximum normal
stresses in AB and BC, shearing stress at pins,
bearing stress, on
members at each pinned
connection
From a statics analysis:
FAB = 40 kN (compression)
FBC = 50 kN (tension)
Example (3) Application to Analysis of Simple Structure
Prof. Dr. Imam Morgan
Head of MCTR Department
(24)
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.
MPaA
P
mmA
end,BC 167300
1050
300254020
3
2
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 = d2/4 = 314.15 mm2) is
BC = +159 MPa.
A- Normal Stresses: (rod BC and boom AB)
Rod BC
Boom AB
FBC
Prof. Dr. Imam Morgan
Head of MCTR Department
(25)
The cross-sectional area for pins at
A, B, and C,
222
4912544
mmd
A
The pin at A is in double shear with a total force equal to the force exerted by
the boom AB,
MPa.10
A
P 3
ave,A 740491
20
B- Shear Stresses: (pin supports)
MPaA
Pave,C 102
491
1050 3
The force on the pin at C is equal to the force exerted by the rod BC. The pin has single shear.
Pin C
Pin A
Prof. Dr. Imam Morgan
Head of MCTR Department
(26)
Divide the pin at B into sections to determine the section with the largest shear force,
(largest) kN25
kN15
G
E
P
P
MPa.10
A
P 3Gave,B 950
491
25
Evaluate the corresponding average shearing
stress,
kN50BCF
Pin B
Note that we can consider that the pin is
subjected to double shear with P=50/2 kN
Prof. Dr. Imam Morgan
Head of MCTR Department
(27)
To determine the bearing stress at A in the boom AB, we have t = 30 mm and d = 25 mm,
MPa.
10
td
P 3
b 3532530
40
To determine the bearing stress at A in the bracket, we have t = 2(25) = 50 mm and d = 25 mm,
MPa.
10
td
P 3
b 0322550
40
C- Bearing Stresses: (at supports)
Support at A
Bracket Boom
The bearing stresses at B in member AB, at B and C
in member BC, and the bracket at C are found in a similar way.
Prof. Dr. Imam Morgan
Head of MCTR Department
(28)
2000 N
250 mm 125 mm
Example (4)
The shown system is used to support 2000N. The upper
portion of link ABC is 10 mm thick and the lower portions
are each 6 mm thick. Epoxy resin is used to bond the two
portions at B. The diameters of pins at A and C are 10 mm
and 6 mm, respectively. Determine:
a- the shearing stress in pin A,
b- the shearing stress in pin C,
c- the largest normal stress in link ABC,
d- the average shearing stress on the bonded surface at B,
e- the bearing stress in the link at C
Solution:
Draw the F. B. D. for the entire
system. Link ABC is two force member.
tensionNF
FM
AC
ACD
3000
02503752000:0
2000 N 250 mm
125 mm
175 mm
45 mm
150 mm
30 mm
Prof. Dr. Imam Morgan
Head of MCTR Department
(29)
(a) Shear stress at pin A: (A)
pin A is single shear:
MPa.
A
F
A
ACA
238
104
3000
2
(b) Shear stress at pin C: (C)
double shear.
MPa.
A
F
C
ACC
153
64
2
3000
2 2
(c) Largest Normal stress in link ABC: The largest stress occurs at cross section at A
(smallest area)
The area at A is: Anet= 10 x (30-10) = 200 mm2
MPa
A
F
A
net
ACA
15
200
3000
3000 N
Prof. Dr. Imam Morgan
Head of MCTR Department
(30)
2000 N 250 mm
125 mm
175 mm
45 mm
150 mm
(d) Average shear stress at B: (B)
MPa.
mmA
N/F:sideoneFor
A
F
B
B
BB
111350
1500
13504530
150023000
2
1
1
(e) Bearing stress in link ABC at C : For each portion:
F1 = 1500 N bearing area = 6x6 = 36 mm2
MPa.
A
F
b
b
741
36
15001
Prof. Dr. Imam Morgan
Head of MCTR Department
(31)
2000 N 250 mm
125 mm
175 mm
45 mm
150 mm
3000 N
Example (5)
The steel bar is to be designed to support a
tension force of P = 120 kN when bolted
between double brackets at A and B. The bar will
be fabricated from 20 mm thick steel plate. The
maximum allowable stresses for this steel are
= 175 MPa, = 100 MPa, and b = 350 MPa.
a- Determine the diameter d of the bolt.
b- Determine the dimension b at each end of the
bar.
c- Determine the dimension h of the bar.
Solution:
Given: Required:
P = 120 kN d = ? (for bolt) t = 20 mm b = ? (at end of the bar) allowable stresses: h = ? (for the bar)
= 175 MPa = 100 MPa b = 350 MPa
20 mm
h
Prof. Dr. Imam Morgan
Head of MCTR Department
(32)
(a) Diameter of the bolt:
at A or B we have double shear
mmd:usemm.d
d
N/PF:whereA
Fall
28627
4
1060100
10602
2
3
3
Check (for bearing stress on bolt or bar):
all,bb
b MPaMPatd
P
350214
2820
10120 3
OK
Given: Required:
P = 120 kN d = ? (for bolt)
t = 20 mm b = ? (at end of the bar)
allowable stresses: h = ? (for the bar)
= 175 MPa
= 100 MPa
b = 350 MPa
Prof. Dr. Imam Morgan
Head of MCTR Department
(33)
(b) Dimension b:
mm.b
.adb
mm.a
adbt
A
P
netall
362
14172282
1417
220
1012010120175
33
(c) Dimension h:
mmh:usemm.h
h
th
P
35334
20
10120175
3
Given: Required:
P = 120 kN d = ? (for bolt)
t = 20 mm b = ? (at end of the bar)
allowable stresses: h = ? (for the bar)
= 175 MPa
= 100 MPa
b = 350 MPa
Prof. Dr. Imam Morgan
Head of MCTR Department
(34)
However, we shall show that either axial or transverse forces may
produce both normal and shear stresses with respect to a plane other than one cut perpendicular to the member axis.
Axial forces on a two force
member result in only normal
stresses on a plane cut
perpendicular to the member axis.
Transverse forces on bolts and
pins result in only shear stresses
on the plane perpendicular to bolt or pin axis.
7 Stresses on an Oblique Plane Under Axial Loading
Prof. Dr. Imam Morgan
Head of MCTR Department
(35)
Pass a section through the member forming
an angle with the normal plane. Consider the left part. Note that +ve is in CCW direction from the vertical.
cossin
cos
sin
cos
cos
cos
00
2
00
A
P
A
P
A
V
A
P
A
P
A
F
The average normal and shear stresses on
the oblique plane are:
sincos PVPF
Resolve P into components normal and tangential to the oblique section,
From equilibrium conditions, the distributed
forces (stresses) on the plane must be
equivalent to the force P.
Animation Prof. Dr. Imam Morgan
Head of MCTR Department
(36)
Animation/Normal_and_shear_stresses.swf
cossinA
Pcos
A
P
0
2
0
Normal and shearing stresses on
an oblique plane:
The maximum normal
stress occurs when the
reference plane is
perpendicular to the
member axis,
00
m A
P
The maximum shear
stress occurs for a plane
at + 45o with respect to
the axis, ,
00 245cos45sin
A
P
A
Pm
Maximum Stresses
Prof. Dr. Imam Morgan
Head of MCTR Department
(37)
= ? for which
A member subjected to a general
combination of loads is cut into two
segments by a plane passing through
an internal point Q. Assume that the
plane is parallel to yz plane.
For equilibrium, an equal and
opposite internal force and stress
distribution must be exerted on
the other segment of the member.
A
V
A
V
A
F
xz
Axz
xy
Axy
x
Ax
limlim
lim
00
0
The distribution of internal stress
components may be defined as,
Stress Components
Q
8 Stress Under General Loading Conditions
Prof. Dr. Imam Morgan
Head of MCTR Department
(38)
Stress components are defined for the planes
cut perpendicular 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
0
zyyzzyyz andsimilarly,
Consider the moments about the z axis:
Prof. Dr. Imam Morgan
Head of MCTR Department
(39)
Finally: The General State of Stress at a point Q is completely
defined by 6 independent stress components, namely:
x , y , and z Normal stresses x .. Stress along x-axis Similar definitions for y , and z.
xy ,yz , and zx Shear stresses xy . Stress perpendicular to x-axis and directed along y-axis. Similar definitions for yz and zx.
Prof. Dr. Imam Morgan
Head of MCTR Department
(40)
Structural members or
machines must be designed
such that the working stresses
(allowable 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
stressallowable
stress ultimateFS
safetyof FactorFS
all
u
load allowable
load ultimate
P
PFS
def.ealternativan or,
all
u
Ultimate Stress: is the stress at which the specimen will break or begins to
carry less load. (see next chapter).
9 Factor of Safety
Prof. Dr. Imam Morgan
Head of MCTR Department
(41)
Example (6)
150 mm
200 mm
dB = dD = 10 mm
dC = 12 mm
Diameter of rod AB is
dA = 11 mm
The rigid beam BCD is attached by bolts to a control rod at B, to a hydraulic cylinder at C, and to a fixed support at D. Each bolt acts in double shear and is made from steel with U = 280 MPa. The control rod
is made of steel with U = 420 MPa. If the minimum
factor of safety is to be 3 for the entire unit, determine the largest upward force which may be
applied by the hydraulic cylinder at C.
Solution:
200 mm 150 mm
F.S. must be 3 in each of the three bolts and in
the control rod.
Statics: Find the force C in terms of the force B
and the force D.
233.20350150:0
175.10200350:0
DCDCM
BCCBM
B
D
Prof. Dr. Imam Morgan
Head of MCTR Department
(42)
150 mm
200 mm
dB = dD = 10 mm
dC = 12 mm
Diameter of rod AB is
dA = 11 mm
Calculation of C on the bases of:
Control Rod: For F.S = 3 we have,
MPaallall
U 14034203
kNCSoCFrom
kNNd
B Ball
28.23,3.1375.1:1
3.136.133044
11140
4
22
Bolt at B : For F.S = 3 we have,
kNCCFrom
kNN
d
areashearedBsheardoubleFor
MPa
Ball
all
Uall
66.2566.1475.11
66.144.14663
4
10233.93
42
33.9332803
22
233.2
175.1
DC
BC
Prof. Dr. Imam Morgan
Head of MCTR Department
(43)
150 mm
200 mm
dB = dD = 10 mm
dC = 12 mm
Diameter of rod AB is
dA = 11 mm
Bolt at D : For F.S. = 3 we have,
kNCCFrom
kNN
areashearedBDsheardoubleFor
MPaAgain
all
Uall
16.3466.1433.22
66.144.14663
33.9332803,
Bolt at C : For F.S. = 3 we have,
kNCNC
d
areashearedCsheardoubleFor
MPa
Call
all
all
11.217.21110
4
12233.93
42
33.93
22
Conclusion: We have calculated, separately, four maximum allowable
values of C. Therefore we choose the smallest value:
C = 21.11 kN
233.2
175.1
DC
BC
Prof. Dr. Imam Morgan
Head of MCTR Department
(44)
Revision Problem
Given:
8x36 mm for the four links joining
BD and CE.
16-mm diameter pins at B, D, C, E thickness of lever ABC is 10 mm
Required:
a) maximum normal stress in links
connecting: points B and D points C and E b) normal stress at mid points of the four links.
c) Average shear stress at each pin.
d) At B, find the bearing stress on links and on lever.
Answer:
a) +101.56 and -21.7 MPa
b) +56.42 (BD) & -21.7 (CE) MPa
c) 80.82 (B,D) & 31.08 (C,E) MPa
d) 126.95 MPa on links BD
and 203.13 MPa on lever
Prof. Dr. Imam Morgan
Head of MCTR Department
(45)